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## Foundations of Wave Phenomena

by: Maurine Stamm

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# Foundations of Wave Phenomena PHYS 3750

Marketplace > Utah State University > Physics 2 > PHYS 3750 > Foundations of Wave Phenomena
Maurine Stamm
Utah State University
GPA 3.92

David Riffe

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Date Created: 10/28/15
Lecture 8 Phys 3750 1D Wave Equation General Solution Gaussian Function Overview and Motivation Last time we derived the partial differential equation known as the one dimensional wave equation Today we look at the general solution to that equation As a speci c example of a localized function that can be useful when studying waves we introduce the Gaussian function Key Mathematics We reacquaint ourselves with the chain rule for taking derivatives and look at the Gaussian function and the integral of the Gaussian function which is known as the error function 1 Solutions to the Wave Equation A GemmlFom of the folm z39m Last time we derived the wave equation 62qxt 2 62qxt atz 6 6x2 from the long wave length limit of the coupled oscillator problem Recall that c2 is a constant parameter that depends upon the underlying physics of whatever system is being described by the wave equation Now it may surprise you but the solution to Eq 1 can quite generally be written in very succinct form as qxt fxctgx ct 2 where f and g are any Hwell behavedH functions We won39t worry about the details of what well behaved means but certainly we certainly want their second derivatives to exist As we previously discussed f xct travels in the x direction at the speed c and gx ct travels in the x direction at the same speed To see that Eq 2 is a solution to Eq 1 let39s calculate the second x and t derivatives of f xct and gx ct To do this we need the chain rule which can be written for the case at hand as 6hkxt Mg ax dk ax ax39 3 Applying this rule to f xct for example we have D M Riffs 1 1232009 Lecture 8 Phys 3750 6fx ct 6x 6x ct f39xct f39xct 4 where f 39xct is the derivative of f xct with respect to its argument Applying the chain rule again to calculate the second derivative of f xct with respect to x give us 6 fx2 ct 6f39x ct fx ct6x ct fx Cr 5 6x 6x 6x Similarly we have for the t derivatives afx6t f39xct ax6t cf39xct 6 6t 6t and 62fxct 6f39xct 6xct 2 6t2 c at cf x ct 6t c f x ct 7 From Eqs 5 and 7 we see that 62fxctc2 62fxct 8 BIZ 6x2 and so fxct indeed solves the wave equation Proof that gx ct also satis es Eq 1 follows from an essentially identical calculation B Mona fpm c folm z39om So the totally unknown functions f xct and gx ct are solutions but only in a very general sense As we shall see any solution has the form of Eq 2 but how do we know what the solution will be in any given situation Well as with earlier problems that we have looked at in this class the situation can be speci ed by initial conditions and the boundary conditions Recall for the coupled oscillator problem the initial conditions were specified by values for qj0 and q Now however the spatial variable is not the discrete index j but the continuous variable x The corresponding initial conditions can thus be written as qx0 ax 9a D M Riffs 2 1232009 Lecture 8 Phys 3750 and 4x50 bx 9b where ax and bx are assumed to be known functions We could have just stuck with qx0 and q39x0 but in the long run we save a bit of notational cumbersomeness by using ax and bx instead For the coupled oscillator system the boundary conditions are q0t 0 and qN1t 0 There will be similar instances for the wave equation where we will be interested in waves on a nite sized system In such cases we will need to specify the condition on qxt at the system boundaries Indeed you have already seen an example of this in Exercise 84 from the last lecture notes II Initial Value Problem IVP for an In nite System Here we write the most general solution to the wave equation given the initial conditions ax and bx To keep things simple at this point we will assume that the ends of the 1D system where these waves exist are at 00 and 00 In this way we do not have to deal with any boundary conditions We will deal with bc39s in the next lecture From Eq 2 qxt fxctgx ct 2 we have for I 0 ax fx gx 10 That is simple enough What about the other initial condition Well taking the t derivative of Eq 2 and setting it equal to bx at I 0 gives us bx le39x g xl 1 1 Now Eq 11 can be formally integrated which gives us 1 1 1 w W fxgx lt12 where x0 can have any constant value If we now take the sum and difference of Eqs 10 and 12 we obtain D M Riffs 3 1232009 Lecture 8 Phys 3750 my axjbx39dx39 133 and goof ax jbx39dx39 13b Using Eq 13 in Eq 2 nally gives us the general solution to the initial value problem 52 qxt1axctax ctl Ibx39dx39 14 2 Cris Notice that the undetermined constant x0 has disappeared Eq 14 is remarkably simple II The Gaussian Function and Two InitialValueProblem Examples A Caz5522174 Fzmcfz39m A very useful function in physics is the Gaussian which is de ned as GT x e XZaz 15 As shown in the picture on the top of the following page the Gaussian is peaked at x 0 and has a width that is proportional to the parameter 039 In fact the full width at half maximum FWHlVI which is the width of the peak at half its maximum height is equal to ZJEO39 H 13860 B IVY folm z39m wit7 Caz5522174 I m39l z39al Poxz39l z39m Let39s see what the solution qxt looks like with the initial conditions ax AGEr x A is just some arbitrary amplitude and bx 0 Physically how would you describe this set of initial conditions Using Eq 14 we rather trivially obtain gm Gxctaaxct 16 or more explicitly D M Riffs 4 1232009 Lecture 8 Phys 3 7 50 GAUSSIAN FUNCTIONS sigma l sigma sigma 5 3 U x A qx I 3 Emaa Emaa L 1 7 So the solution consists of two Gaussian functionsgt one moving in the x direction and one in the x directiongt both at the speed c The amplitude of each function is 12 the amplitude of the initial Gaussian displacement The following picture illustrates this solution as a function of x for several times t For simplicity we have set A 1 a0 1 and c 1 SOLUTION FOR INITIAL GAUSS DISPLACEMENT I 1 qxt U 10 5 0 5 10 D M Riffs 5 1232009 Lecture 8 Phys 3750 B IVY folm z39m wit7 Caz5522174 Im39l z39al Velocigj Emr me z39m Let39s look at another example using the Gaussian function This time let39s have the initial displacement of the system be Zero so that ax0 but let39s have a Gaussian initial velocity function so that bxBGUx where B is some arbitrary velocity amplitude In this case we get from Eq 14 x t qxt EVquot2 dx39 18 1675 So what is the integral of a Gaussian function Well it is known as the error function Speci cally the error function is de ned as erfx fequot2 dx 1 9 The following gure shows a plot of erf x vs x As the graph indicates erf x is de ned such that erfx gt oo 1 erfx gt 00 1 and erf0 0 ERROR FUNCTION erfx 10 5 0 5 10 X All this is fine and well but what do we do about the 039 in Eq 18 which does not appear in Eq 19gt We must do a little math and change variables in Eq 18 Let39s define a new integration variable y x39039 dy dx39039 Then Eq 18 becomes H52 039 qxt 3 0 e y2 dy 20 20 x75 039 D M Riffs 767 1232009 Lecture 8 Phys 3750 We can now use Eq 19 the de nition of the error function to write qx t 4310 erfx ct039 erfx ct039 21 Notice that again we have the sum of two functions each traveling in opposite directions at the speed c The following picture plots the solution vs x for several values oft with B a0 and c all set to 1 SOLUTION FOR INITIAL GAUSSIAN VELOCITY 39 t0 t1 t3 qxt Exercises 81 The chain rule Let qxt x e De ne new independent variables u xt and s x t 3 Find the form of q when written in the new variables Ie nd qxu stu b Using qxt xe calculate figBx and fig t c Starting with qxu stu 3 that you found in a use the chain rule and the derivatives figBu and fig63 to calculate figBx and BqBt 1 Verify that the answers in and c are identical 82 Verify for I 0 that Eq 14 and its time derivative reduce to the initial conditions qx0 ax and x0 bx respectively D M Riffs 7 1232009 Lecture 8 Phys 3750 83 If the initial displacement is Zero then Eq 14 the solution to the initial value 52 problem can be written as qxt 2i bx39dx39 By calculating second derivatives in x c 16752 and t directly demonstrate that this function solves Eq 1 the wave equation M84 The error function a Usin an a ro riate chan e of inte ation variable in the e uation g PP P g gr q x erfx fjeexddx show that erfxo a Ieer a2 dx39 b Using a computer mathematic package plot erfx039 over an appropriate range ofx for 039 1 3 and 5 M85 The initial value problem Consider Eq 14 the general solution to the initial value problem 3 Explain Why Eq 14 in not a function of the variable x39 This is a basic feature of the de nite integral Consult a calculus book if necessary b What are c ax and bx in Eq 14gt c Consider the speci c case where ax 0 and bx 20x1 x4 Using a computer mathematics package and letting c 1 plot bx over an appropriate range of x 1 Using the initial conditions given in c solve Eq 14 The integral can be done either with a change of variable a computer mathematics package or can be looked up in a table of integrals such as found in the CRC Hamdbook of Cvemz39xl g am Ply52a e Show that your solution can be written in the form qxt fxctgx ct Thus identify fx and gx 1 Again using a computer mathematics package and letting c 1 plot your solution qxt as a function of x over an appropriate range for t 0 10 20 and 30 D M Riffs 8 1232009 Lecture 19 Phys 3750 Separation of Variables in Cartesian Coordinates Overview and Motivation Today we begin a more in depth look at the 3D wave equation We introduce a technique for nding solutions to partial differential equations that is known as separation of variables We rst do this for the wave equation written in Cartesian coordinates In subsequent lectures we will do the same using cylindrical and spherical polar coordinates The technique is applicable not only to the wave equation but to a wide variety of partial differential equations that are important in physics Key Mathematics The technique of separation of variables I Separable Solutions Last time we introduced the 3D wave equation which can be written in Cartesian coordinates as 2 2 2 2 iaqzaq q q 1 c2 BIZ 6x2 Byz 622 and we spent some time looking at the plane wave solutions quykyykz Jay at Aexkxxkyykzziwt gt quykyykz xay 251 A elkxxkyykzzl02 gt where a clkl is the familiar dispersion relation Equation 2a describes a wave that travels in the k kx ky 7kzi direction with wavelength A 27rlk while Eq 2b describes a similar wave that travels in the k direction Today we introduce separation of variables a technique that leads to separable also known as product solutions A separable solution is of the form qxy zat X W W 2Tt 3 That is the function qxyzt is a product of the functions Xx Yy 22 and Tt So how do we find these solutions Well let39s just substitute the rhs of Eq 3 into Eq 1 and see what happens With this substitution Eq 1 becomes D M Riffs 1 322009 Lecture 19 Phys 3750 LZXYZTquotXquotYZTXYquotZTXYZquotT 4 c where we have suppressed the independent variables and the double prime indicates the second derivative of the function with respect to its argument Note that because each of these functions is a function of only one variable all derivatives are now ordinary derivatives Equation 4 looks kind of ugly but notice what happens if we divide Eq 4 by qx y zt XxYyZzTt We then get iT quotXquot Yquot Zquot 5 c2 T X Y Z Now here is where the magic happens Notice that each term is only a function of its associated independent variable So for example if we vary I only the term on the lhs of Eq 5 can vary But because none of the other terms depends upon I the rhs cannot vary which means that the lhs cannot vary which means that TquotT is z39 depe dem of t Following the same logic for each of the other terms means that each term is constant So we can write 1 Tquot Xquot Yquot a c2 T X Z 6 6 6d Y y Z a H gt where the constants a y and 6 are known as separation constants However only three of them are independent because Equation 5 tells us that a y 5 7 Notice by demanding a product solution the partial differential wave equation has transformed into four ordinary differential equations We can easily solve all of these equations Let39s start with the equation for Tt Tquotr ac2Tt 0 8 This is essentially the harmonic oscillator equation if ais real and lt0 and so it has the two linearly independent solutions Tt Team2 9a where To is some undetermined constant Similarly the X Y and Z equations have solutions D M Riffs 2 322009 Lecture 19 Phys 3750 xtXoem 9b Yne4 mg ZtZOei z 9d Putting this all together gives us a solution of the form qxyztX0YOZO T0 at ret y 8528543 10 11 Some Physics Added In Notice that the solution in Eq 10 can have all sorts ofbehavior For example if the constant is real and positive then the x dependent part of the solution can either exponentially increase or decrease with increasing x depending upon the sign in the exponent This points out the fact that separation of variables solutions often give you more than you really need in any given situation For example if we are looking for solutions to the wave equation that are physically meaningful as x gt 00 we are probably not going to be interested in a solution that exponentially increases with increasing x With that in mind let39s see what the constraints on the separation constants must be if we are only interested in purely oscillatory solutions all independent variables Let39s consider the x dependent part of the solution If e p is to only oscillate then it must be of the form 6 where kX is real Then if we write a generally complex as l z39 2 we have l l z39 2 z39k which implies l i 2 kf Thus z 0 and l kf That is is real and negative Similarly y and 5 must be real and negative Eq 7 then implies that a is also real and negative So if we rename the constants as 7 6 a w2 gig mi where kX ky kz and a are all real then Eq 10 can be rewritten as qxyzt A etxkxx etzkyy etxkzz 814 12 where AX0Y0 Z0 To is an arbitrary constant Eq 7 can now be written as m2 k k k gt which is the well known dispersion relation Now kx ky and k D M Riffs 3 322009 Lecture 19 Phys 3750 can be positive or negative so we don39t need the i on the spatial function exponents and so we are left with two linearly independent solutions Jay 2 t Ae k W 133 qukykzxyzt A elkWWZM 13b which are the plane wave solutions of Eq 2 So by demanding that the separable solutions to the wave equation oscillate we have ended up with the plane wave solutions that we discussed in the last lecture III Utility of the Separable Solutions Because they are product solutions the separable solutions are pretty specialized That is there are many solutions to the wave equation that cmmol be written as a product solution And so you may ask what is the usefulness of these solutions There are two parts to the answer The first is that sometimes these solutions have intrinsic interest For example in the case of the time dependent Schrodinger equation the separable solutions with some appropriate physics thrown are the energy eigenstates of the system The second and perhaps more important part of the answer is that a basis can be constructed from the set of separable solutions that can be used to represent any solution Let39s think about this statement with regards to the wave equation Let39s first consider the 1D wave equation Back in the Lecture 17 notes we solved the initial value problem on the interval 00 lt x lt 00 The solution to that problem can be written as go ldk WCF elm ampk gram 14 where 51k and are the Fourier transforms of the initial conditions and a clkl is the dispersion relation Notice that in Eq 14 we have written the general solution qxt as a linear combination of the functions e lkw and e lk Although we did not go through the separation of variables procedure for the 1D wave equation to produce these solutions by writing these functions as 6 6 quot and 6 6quot we readily see that they are indeed product solutions Similarly for the 3D wave equation we can write the solution to the initial value problem as a linear combination of the separable plane wave solutions Eq 13 D M Riffs 4 322009 Lecture 19 Phys 3750 gm I d3kampki Llicezkrm 2110 eirm 15 where kr kxxkyykzz Jd3k Jdkxjdkyjdkz a chg k kg and 61k and 5k are the 3D Fourier transforms of the initial conditions In the coming lectures we will look at separable solutions in cylindrical and spherical polar coordinates While we will not do a lot with the general solution written as a linear combination of these solutions we should keep in mind that in principle it can be done Exercises M191 Heat Equation A partial differential equation known as the heat equation which is used to describe heat or temperature ow in an object is given by qu lg where 2 gt0 1 61 a If there is no y or Z dependence to the problem write down a simpli ed version of this equation b Use separation of variables to nd two ordinary differential equations one in x and one in I What are the orders of these two equations Are they linear or nonlinear c Find the general solutions to the two ordinary differential equations Thus write down the general separable solutions to the heat equation Note there should be two linearly independent solutions 1 In solving the heat equation you should have found that X quotxX x C where C is some constant Assume that this constant is real If C gt 0 describe the behavior of the solutions vs x and t In what ways are these solutions like solutions to the wave equation In what ways are they different e If C lt 0 describe the behavior of the solutions vs x and t In what ways are these solutions like solutions to the wave equation In what ways are they different D M Riffs 5 322009 Lecture 19 Phys 3750 M192 Normal Modes Inside a Rectangular Room Here we consider sound waves inside a rectangular room with one comer located at the origin and occupying space as indicated 0 lt x lt L 0 lt y lt Ly 0 lt z lt LZ While we could deal with displacement a vector field it is easier to think about the pressure which is also governed by the wave equation but is a scalar field The boundary conditions on the pressure pxyzt are 2 p0 at x0 and xL 2 p0 at y0 and yLyand x y a p0 at 20 and zLZ 62 a As discussed in the notes the separable l mmlz39 guam solutions to the wave equation can be written q replaced by p as p k k x y zt At e TkXHkkaZM x y z where a chf ky2 kz2 However we could have expressed the separable solutions as the xta dz39 g uaw solutions p xkykz x y zt AX sinkxBx coskx Ay sinkyy By coskyy gtlt AZ sinkzzBZ coskzzei quot Starting with this standing wave form find the speci c solutions that satisfy the above boundary conditions Note these solutions should be labeled with three integers call these integers n ny and 71 Express the allowed frequencies a as a function of n ny and n the wave velocity 0 and the dimensions of the room b The solutions that you found in a describe the 740777741 mode for sound waves in a typical room where L and Ly are the widths and LZ is the height of the room Consider a room of dimensions 10 ft X 11 ft X 8 ft Calculate the frequencies fnxny n2 anx ny nz of the 10 lowest frequency room modes For the speed of sound you may use 330 ms c These lowest frequency modes often wreak havoc with sound reproduction because they serve to amplify through resonance reproduced frequencies near their resonance frequencies This is especially troublesome if two or more modes have frequencies that are close together degenerate For example a 1039 X 1039 X 1039 room would have its three lowest modes at exactly the same frequency Are any of the modes you calculated in degenerate or nearly sogt D M Riffs 767 322009 Lecture 29 Phys 3750 The Uncertainty Principle Overview and Motivation Today we discuss our last topic concerning the Schrodinger equation the uncertainty principle of Heisenberg To study this topic we use the previously introduced general wave function for a freely moving particle As we shall see the uncertainty principle is intimately related to properties of the Fourier transform Key Mathematics The Fourier transform the Dirac delta function Gaussian integrals variance and standard deviation quantum mechanical expectation values and the wave function for a free particle all contribute to the topic of this lecture I A Gaussian Function and its Fourier Transform As we have discussed a number of times a function fx and its Fourier transform hk are related by the two equations fx Ljhoc dk 1a hk I fxe dx 1b f x 6Wquot gt 2 then its Fourier transform hk is also a Gaussian ray3 0k W 3 where the width parameter 039 of this second Gaussian function is equal to 2039 and r so we have the result that the product 00 of the two width parameters is a constant O39O39k2 4 x Thus if we increase the width of one function either fx or hk the width of the other must decrease and vice versa D M Riffs 1 482009 Lecture 29 Phys 3750 Now this result wouldn39t be so interesting except that it is a general relationship between any function and it Fourier transform as the width of one of the functions is increased the width of the other must decrease and vice versa Furthermore as we shall see below this result is intimately related to Heisenberg39s uncertainty principle of quantum mechanics II The Uncertainty Principle The uncertainty principle is often written as Ax Apr 2 5 where Ax is the uncertainty in the x coordinate of the particle Ap is the uncertainty in the x component of momentum of the particle and h h 27239 where h is Planck39s constant Equation 5 is a statement about my state zxt of a particle described by the Schrodinger equation1 While there are plenty of qualitative arguments concerning the uncertainty principle today we will take a rather mathematical approach to understanding Eq The two uncertainties Ax and Ap are technically the standard deviations associated with the quantities x and p respectively Each uncertainty is the square root of the associated variance either may or Apj which are de ned as M Ge 0c 62gt Ami p pa2 6b where the brackets indicate the average of whatever is inside them Experimentally the quantities in Eq 6 are determined as follows We rst measure the position x of a particle that has been prepared in a certain state Lxt We must then prepare an identical particle in exacl the same state zxt time suitable shifted and repeat position measurement exath some number of times We would then have a set of measured position values From this set we then calculate the average position For each measurement x we also calculate the quantity x x2 gt 1 We are implicitly thinking about the 1D Schrodinger equation thus there is only one spatial variable D M Riffs 2 482009 Lecture 29 Phys 3750 and then find the average of this quantity This last calculated quantity is the variance in Eq 6a Finally the square root of the variance is the standard deviation What we want to do here however is use the theory of quantum mechanics to calculate the variances in Eq How do we calculate the average value of a measurable quantity in quantum mechanics Generally we calculate the expectation value of the operator associated with that quantity For example let39s say we are interested in the average value of the quantity 0 for a particle in the state Lxt We then calculate the expectation value of the associated operator 6 which is de ned as2 no 1 x t0 yx t dx 0 40 7 lwltxrgtwltxrgtdx ice The quantity 0 can be any measurable quantity associated with the state the position x for example Notice that the variance involves two expectation values Again consider the position We see that we must first use Eq 7 to calculate x and then use that in the calculation of the second expectation value Finally to get Ax we must take the square root of Eq 6a We can actually rewrite Eq 6 in slightly simpler form as follows Consider Eq 6a We can rewrite it as ext ltx2 2xltxgt ltxgt2gt 8 Now because the expectation value is a linear operation see Eq 7 this simpli es to M xZgtlt2xltxgtltxgtz 9 Furthermore because an expectation value is simply a number ltltxgt2gt ltxgt2 and lt2xltxgtgt 2ltxgtltxgt 2xgt2 Eq 9 thus simpli es to 2 Usually in quantum mechanics one deals with normalized wave functions in which case the denominator of Eq 6 is equal to 1 Rather than explicitly deal with normalized functions we will use Eq 7 as written D M Riffs ese 482009 Lecture 29 Phys 3750 Ax2 x2 x2 10a Similarly for Apx 2 we also have Ami Pgt ltPx2 10b Thus we can write the two uncertainties as Ax x2 x2 11a Apt pZ px2 11b III The Uncertainty Principle for a Free Particle A A Frye Pam z39cle flute Let39s now consider a free particle and calculate these two uncertainties using Eq 11 You should recall that we can write any free particle state as a linear combination of normal mode traveling wave solutions as no 1x t I dk Ck e lkquot k l 12 where the coef cient Ck of the kth state is the Fourier transform of the initial condition Lx0 Ck Idxwx0equotk 13 and the dispersion relation is of course given by akth2m To keep things simple let39s assume that the state we are interested in is a particle moving along the x aXis As discussed in the Lecture 27 notes such a particle can be described with the initial condition wx0 14 As we also saw in those notes this initial condition results in D M Riffs 4 482009 Lecture 29 Phys 3750 WOO x 7kikn2a4 C k 15 0 5 e lt gt which gives us the wave function Wm Idkeiwuvaga ezlkmw 16 B Poxz39l z39m Umm az39 gj Ax With the wave function in Eq 16 we can now principle calculate the expectation values in Eq 11 We start by calculating the uncertainty in x From Eq 11 we see that we need to calculate two expectation values x and x2 Using Eq 6 the de nition of an expectation value we write the expectation value of x as jwxtcLxtdx x W 18 iii1 x tzx t dx foo where we have kept the HhatH on the x inside the integral to emphasize that fc is an operator But when fc operates on zxt it simply multiplies zxt by x Equation 18 then becomes jwxtxzxtdx x 39Z 19 J yxtzx t dx We could now insert Eq 16 for zxt into Eq 19 and calculate away but it will get really ugly really fast But let39s recall the behavior of the free particle state described by Eq 16 As it propagates from negative time it gets narrower up until I 0 and then as it further propagates it becomes broader Given this lets calculate the uncertainty in x when it will be a minimum at t 0 D M Riffs 5 482009 Lecture 29 Phys 3750 Then Eq 19 becomes3 zx0xzx0 dx x w 20 39z8 If we now insert the rhs of Eq 14 the expression for lx0 into Eq 20 we get jxe z z sz dx x 4 21 ferala dx foo You should immediately recognize that the integral in the numerator is Zero why and that the integral in the denominator is not Zero Thus x 0 and so x2 0 This result should not be very surprising at I 0 the probability density is a Gaussian centered at x 0 so the average value of the position is simply Zero We now calculate x2 which is given by Lquotx0x2 zx0 dx x2 0 22 398 As we did in calculating x we substitute the rhs of Eq 14 for Lx0 which gives us 5 As far as all the calculations of the expectation values that we are interested in are concerned I is just a parameter We are free to simply set it to whatever value we might be interested in and calculate all expectation values with it set to that value This would not be true ifwe were interested in a 1 dependent operator such as the energy operator ih 661 D M Riffs 767 482009 Lecture 29 Phys 3750 jixz earla dx x2 23 Ie Z Z E dx foo Now this expectation value is certainly not zero By looking these two integrals up in an integral table or using Mathcad for example we obtain the result x2 4x 24 Given that Ax 1lltx2gt ltxgt2 and x 0 we thus have 0x Ax 2 25 That is the uncertainty in position is simply equal to half of the width parameter at Again this should not be too surprising the more spread out the wave function zxt which is controlled by 0 at t 0 the larger the uncertainty in its position C Mommam Umm az39 gj Apr We now calculate the momentum uncertainty Apx Referring to Eq 11 we see that we need to calculate pi and Notice that both of these are the expectation value of some power of the momentum Now you should have learned in your modern physics course that the momentum operator is given by the differential operator 3 423 26 6x This then implies for any integer n that 2 lt mgtquot 6quot 6x 6x I Before we go ahead and do the calculations of pi and pgt it is worth considering the expectation value p for any integer n D M Riffs 7 482009 Lecture 29 Phys 3750 mquot I zx 0 dx 6x p2 27 To do this we use Eq 12 which is the general form of Lxtfor a free particle Thus this calculation will be applicable to any time t Substituting this into Eq 27 gives us m dx dk39 k39 ecllk39xcaik39 l 6 Jdk ck elerwkt 27239 axn P3 Moo 40 no 7 28 2i I de dk39 k39 ecllk39rew t l Jdk ck elkxizokt 7239 Now we have seen these sorts of integrals before You may remember that things can sometimes get considerably simpler if we do some switching of the order of integration Calculating the derivatives in the numerator and then moving the x integral to the interior both the numerator and denominator produces n Jdk39Ck39e W Jdk kquotCke Wk dee k k39 7139 P3 l 391 391 a 29 Zijdk39ck39ewltk gtt Jdk Cke quotk deeIWV 7139 We now use an expression for the Dirac delta function 6k k39 i I dxe lk k 30 27239 to rewrite Eq 29 as D M Riffs 8 482009 Lecture 29 Phys 3750 hquot dk kquot Cke quotk I dk Ck39e quotltk39 6k k39 0 31 Jdk Cke quotk Jdk39Ck39eWk39 6k k39 foo ltP where we have also switched the order of the k and k39 integrations The k39 integral is now trivially donegt giving hquot dk Ckkquot Ck p S 32 Notice that even though we started with the general time dependentgt free particle wave function zxt in Eq 27 the expectation value of any integer power of the momentum is z39 depe dem of time Perhaps this should not be too surprising For a classical free particle there is no change in momentum of the particle Here we see for a quantum mechanical particle that the expectation value associated with any integer power of the momentum does not change with time In fact the expectation value of my function of the momentum is independent of time for the free particle Let39s now calculate Apx using Eq 32 We rst calculate pxgt which is given by no h dkCkkCk a 33 And using Eq 15 the particular expression for Ck in the case at hand Eq 33 becomes D M Riffs 9 482009 Lecture 29 h Idkke 2k kU2 Px Idk 82071 2 TE4 Phys 3750 34 The integrals can be simpli ed with the change of variable k39 k k0 dk39 dk which produces h Idk39k39k0e 392 34 Px Tdk39e Zk39ZUW Bu inspection it should be clear that this simpli es to prhk0 3 5 3 6 So we see that the average momentum of the particle is just the momentum of the state at the center of the distribution C We lastly need to calculate lt However it is actually much simpler if we directly calculate Apx 2 ltpx pi pi hko 2gt which we can write as no if I dkCkk k0 2 Ck And again using Eq 15 the expression for Ck Eq 37 becomes hi Idkk k02 e39zlk39k lz TXZA w w J dkeiukiknyaga D M Riffs 7107 3 7 3 8 482009 Lecture 29 Phys 3750 Making the same change of integration variable as before k39 k k0 dk39 dk both integrals Eq 38 becomes hz Iidkku eizk39laga w 39gt Idk erzk39za39gA Notice that these integral are essentially the same as in Eq 23 where we calculated x2 Looking them up in the same table as before we nd that W h z 41 0x and so Ap i 42 039 Combining this with Eq 25 for Ax we have for our particular state at t 0 h AxAp 3 43 Notice that this is actually the 7772712th value allowed by the uncertainty principle If you think about our traveling wave packet you will realize that this minimum occurs only at I 0 as discussed above the uncertainty in momentum is time independent but we know that the packet is narrowest in x space at I 0 Thus the minimum uncertainty product only occurs at I 0 Furthermore the mim39imm mcm az39 gj 239 0715 rem wke Ck 239 a Gar5522174 dixl rz39bm z39m For other forms of Ck the minimum uncertainty condition is not possible Lastly we make an observation concerning the coef cients Ck Consider for example the expectation value D M Riffs 711 482009 Lecture 29 Phys 3750 no h dkCkkCk pr 1 44 Notice the striking similarity of this equation and Eq 19 for Because of this similarity and because we interpret Lflj as the probability density in x real space we can interpret CC as the probability density in k space Further because hk is the momentum of the state e lkquot k l the product C C is essentially the pmbabiz39gj demigj z39 momem m 805166 We also emphasize that while Lxtzxt depends upon time the product C kC k is time independent for the free particle That is the momentum probability density is time independent This is the basic reason that functions of the momentum operator have time independent expectation values as discussed above D The Fomz39er Tmmyrom am the Umm az39 gj Primg39ble So what does the Fourier transform have to do with the uncertainty principle Well first recall that the functions zx0 and C k are a Fourier transform pair and that the product of their widths parameters is 0076 2 Now Ax 039x 2 and because Apx hO39X we can also write Apx h 0762 Thus we have at I 0 039 g 45 Axh AP 22 That is the products of the uncertainties associated with the I 0 state is intimately related to the products of the width parameters that govern the Fourier transform pair Lx0 and Ck D M Riffs 712 482009 Lecture 29 Phys 3750 Exercises M291 Uncertainty Principle for a Different Wave Packet Consider the following initial condition for the 1D free particle Schrodinger equation zx0 e k xe 3 Find the corresponding function C b As was done in the notes nd the expectation values x x2 and thus Axthe uncertainty in x at I 0 c Find pi and then the uncertainty Apr 1 Then nd the product AxApx Does the product satisfy the uncertainty principle D M Riffs 713 482009 Lecture 15 Phys 3750 The Dirac Delta Function Overview and Motivation The Dirac delta function is a concept that is useful throughout physics For example the charge density associated with a point charge can be represented using the delta function As we will see when we discuss Fourier transforms next lecture the delta function naturally arises in that setting Key Mathematics The Dirac delta function I Introduction The basic equation associated with the Dirac delta function 6x is I5xfxdxf0 D where f x is any function that is continuous at x0 Equation 1 should seem strange we have an integral that only depends upon the value of the function f x at x 0 Because an integral is Hthe area under the curveH we expect its value to not depend only upon one particular value of x Indeed there is no function 5x that satis es Eq However there is another kind of mathematical object known as a gemmlz39ged ml z39m or dixl rz39bm z39m that can be de ned that satis es Eq A generalized function can be defined as the limit of a sequence of functions Let39s see how this works in the case of 5x Let39s start with the normalized Gaussian functions gltxgt 2 Here n10392 where 039 is the standard Gaussian width parameter These functions are normalized in the sense that their integrals equal 1 jgltxgtdx1 3 for any value of 71 gt0 Let39s now consider the sequence of functions for n1 2 3 D M Riffs 1 2182009 Lecture 15 Phys 3750 1 2 2 7x2 100 7 x2 g1xe gt g2xJez gt gt g100x 78100 gt 4 What does this sequence of functions look likegt We can summarize this sequence as follows As 71 increases a gH 0 becomes larger b gquot x at 0 eventually becomes smaller c the width of the center peak becomes smaller d but Ign xdx 1 remains constant foo The following gure plots some of the functions in this sequence 1 H H H H H o to who o H H H H 900 Let39s now ask ourselves what does the n 00 limit of this sequence look likegt Based on a through d above we would perhaps simplistically say a gw0 b gnx 0 0 c the width of the center peak equals zero D M Riffs 2 2182009 Lecture 15 Phys 3750 d but jgwxdxl Note that a and are not compatible with d if go x is a function in the standard sense because for a function a and would imply that the integral of go x is Zero So how should we think of this sequence of functions thengt Well the sequence is only really useful if it appears as part of an integral as in for example no lira g xfxdx ling gem fxdx 5 foo Let39s calculate the integral and then the limit in Eq The following figure should help with the calculation 1 3 I I g256x 39 39 39 39 39 x 39 I 10 39 A I A 3 En 39 5 39 l l U I I I 2 1 0 l 2 As 71 get large gquot x equot 2 becomes narrower such that it only has weight very close to x 0 Thus as far as the integral is concerned for large enough 71 only f x at x 0 is important We can thus replace fx by f0 in the integral which gives us D M Riffs 3 2182009 Lecture 15 Phys 3750 j fxdx f0 j dx f01iggl f0 lt6 Thus the integral on the lhs of Eq 1 is really shorthand for the integral on the lhs of Eq 6 That is the Dirac delta function is de ned Via the equation j5xfxdx fxdx lt7 foo Now often as physicists we often get lazy and write 5x lim e mz 8 but this is simply shorthand for Eq Eq 8 really has no meaning unless the function e mz appears z39mz39de an integral and the limit lim appears om sz39de the same integral However after you get used to working with the delta function you will rarely need to even think about the limit that is used to define it One other thing to note This particular sequence of functions g x 87an that we have used here is not unique There are infinitely many sequences that can be used to define the delta function For example we could also have defined 5x Via 5x limiM 9 ngtoo n x The sequence of functions sinnx7rx is illustrated in the figure at the top of the next page Notice that the key features of both of these two difference sequences are expressed by a d at the top ofpage 5 11 Delta Function Properties There are a number of properties of the delta function that are worth committing to memory They include the following TEx x39fxdx ma 10 D M Riffs 4 2182009 Lecture 15 Phys 3750 1 i1 HHHHH 4 8 1 3 max sinnx pi x j539xfxdxf390 lt11 6xala6x lt12 The proof of Eq 10 is relatively straightforward Let39s Change the integration variable to y x x39 dy dx which gives I5xx39fxdx I5yfyx39dy lt13 Then using Eq 1 we see that Eq 10 is simply equal to fx39 QED Let39s also prove Eq 12 We do this in two steps for a gt 0 and then for a lt 0 z First we assume that a gt 0 Then D M Riffs 151 2182009 Lecture 15 Phys 3750 IEltxadx TEQlapdx 14 Changing integration variable y xIal dy dx a this last equation becomes 35xadx lalj ltygtdy 15 and Changing variables back to x via x y dx dy gives j5xadx lalj ltxgtdx 16 and so for a gt 0 we have 5xa a5x ii We now assume a lt 0 Then we have 35xadx j xadx 1 7 Changing integration variable y x a dy dx a this last equation becomes I5ltxadx aI5ydy lalj5ydy 18 lali ltxgtdx and so for a lt 0 we also have 5xa a5xQED We leave the proof of Eq 11 as an exercise D M Riffs 767 2182009 Lecture 15 Phys 3750 III Fourier Series and the Delta Function Recall the complex Fourier series representation of a function f x de ned on L S x S L fx 192 quot7w i L rmnxL cquot 2L fxe dx 19b Let39s now substitute cquot from Eq 19b into Eq 19a Before we do this we must change the variable x in either Eq 19a or 19b to something else because the variable x in Eq 19b is just a dummy integration variable Changing x to x39 in Eq 19a and doing the substitution we end up with fx39 imew 20gt Let39s now switch the integration and summation assuming that this is OK to do This produces fx39J ZWW f x dx 21 If we now compare Eq 21 to Eq 10 we see that we can identify another representation of the delta function 6x x ZiLZem xLxL 22 quot7w or setting x39 0 we have 5x 1 26quotquotmL 23 5 quot7w So how is this equation related to the delta function being defined as the limit of a sequence of functions Well we can re eXpress Eq 23 as a sequence of functions via D M Riffs 7 2182009 Lecture 15 Phys 3750 5x m i eWL 24 The following figure plots iZe39Wm for several values of m for L 2 Notice that these functions are quite similar to the function sinmx7r x plotted above 1 m2 40 m4 39 m8 m16 m32 30 2 39E as i as g 0 VJ 39 3a39l3939439 4 10 Iquot 39 quotO39 I u However there is one important difference between these two sequences of functions Because the functions e39mmL that appear in the sum in Eqs 23 and 24 all repeat with on an interval of length 2L Eq 24 is actually a series of delta functions centered at x 0 iZL i4L i6L This is illustrated in the next figure where we have expanded the x axis beyond the limits of L to L Thus the equality expressed by Eq 23 or 24 is only valid on the interval L S x S L 1 We have changed the n to m in the sinnx7r x functions because we are now using m to label the functions in the sequence D M Riffs 8 2182009 Lecture 15 Phys 3750 39E as fa i f 5 10 r u quot l quot I 39 l 39 r39quot I 39 I b39v l39v 0 v u39 l 39 a quotv u39 l Iquot 39nquot f l39 39 quotu 39 39 quot3939 0 I I I I I 396 4 392 0 2 4 6 X Exercises 151 Equation 11 I539xfxdx f390 can be taken as the de nition of the derivative of the delta function Treating the delta function as a normal function show that Eq 11 is true Hint use integration by parts 152 Show that the equation 5x lim mmL can be re eXpressed as 5x lim cos This is perhaps more appealing because the delta function is a real function and the rhs is now explicitly real 153 Find another sequence of functions not based on either the Gaussian or sinxx functions that has as its limit the delta function D M Riffs 9 2182009 Lecture 26 Phys 3750 The 1D Schrodinger Equation for a Free Particle Overview and Motivation Here we look at the 1D Schrodinger equation SE an equation that describes the quantum motion of a nonrelativistic particle The SE can have solutions similar to those of the wave equation In fact as we shall see the normal mode solutions to the SE for a free particle one not subject to any force are very similar to those of the wave equation they are traveling or standing waves that oscillate harmonically in space and time An importance difference though is that the dispersion relation ak is not the same for both equations This leads to key differences in the general solutions that are built up from the normal mode solutions Key Mathematics We use separation of variables to nd solutions to the SE The solution to the initial value problem will be speci ed in terms of the initial conditions using the Fourier transform as we previously did for the WE see Lecture 17 I The 1D Schrodinger Equation The 1D SE E W 2m 6x2 611 V h 1 my 1 a ltgt is a homogeneous and linear partial differential equation for the function 11 Lxt which is often referred to as the wave function While we will not worry too much about the physical signi cance of this equation that will be left to a quantum mechanics class we point out that h hZn39 where h is Planck39s constant m is the mass of the particle being described by the SE and Vxt is the potential to which the particle is subjected But for now simply think of zxt as the solution to Eq II Separation of Variables Let39s look for separable solutions to Eq As before we start with a product function Lxt X xTt and substitute this into Eq 1 which yields hi Xquot I T 7 Vxt 1h 2 Now Vxt is often independent of time in which case Eq 2 becomes T 2 Vxzh 3 D M Riffs 1 3302009 Lecture 26 Phys 3750 The variables x and t are now separated both sides of the equation are thus equal to some constant which is conventionally called E This gives us two ordinary differential equations that are related via the separation constant E rot 00 lt4 hi XquotxVxXxEXx 5 Equation 4 says that the first derivative of T is proportional to T The function T must thus be an exponential and is given by TtT0e h 6 As with the wave equation the time dependence is harmonic but because Eq 4 is a rst order equation there is only one solution not two linearly independent solutions as in the case of the wave equation From Eq 6 we see that the angular frequency ofoscillation 0 equals III Separable Solutions for a Free Particle Equation 5 the differential equation for X x has no solution until we specify the potential Vx Our interest here is in a free particle one with no external force Zero force implies a constant potential Vx we can thus set Vx 0 1 Then Eq 5 becomes Xquotx 2 X0 0 7 which is our old friend the harmonic oscillator equation By this point you should be able to immediately write down the two independent solutions ZmE x 0 7 2 a X Xe l 8 X x 8b 1 As is the case in classical mechanics making the potential some unspeci ed arbitrary constant rather than zero does not change the physics In the case at hand it is simplest to let that constant equal zero D M Riffs 2 3302009 Lecture 26 Phys 3750 As usual we identify the wave vector k as the factor that multiplies x in the argument of the harmonic function excluding the ii Thus k 1l2mEh2 Solving this equation for Eh2k22m and comparing this with the expression for the frequency a Eh gives us the dispersion relation for the free particle SE ak hkz 9 2m Notice that as a function of k this is different than ak ck 10 the dispersion relation for harmonic solutions to the wave equation the wave equation dispersion relation is a i ear function of k while the SE dispersion relation is a qmzdml z39c function of k The significance of this difference will be discussed in a Lectures 27 and 28 Combining the T and X pieces we obtain two linearly independent solutions to the free particle SE 1k 714 10617102 lok 11 x t 1106 6 1106 woe klx39 11a 71k 714 71kxzot lok 11 xt yOe e yOe woequotk 11b These two solutions are harmonic waves that travel with speed vph akhk2m The subscript ph stands for phase more on that later Unlike wave equation solutions which all move with the same speed vph c the propagation speed of these harmonic waves depends upon the wave vector k with vph cx k As de ned above the wave vector is a positive quantity However if we let the wave vector take on negative as well as positive values we see that 11 11 Thus we can write all the separable solutions as Wk Jar Woexerwkt 12 where now 00 lt k lt oo and ak is still defined by Eq 9 the dispersion relation D M Riffs 3 3302009 Lecture 26 Phys 3750 IV The Free Particle Initial Value Problem As it turns out and perhaps shouldn39t be too surprising we can write any solution to the free particle SE as a linear combination of these harmonic solutions Because the index k is continuous the linear combination must be expressed as an integral over k wltxrgtfdkcltkgtwltxrgt lt13 foo where Ck is the coefficient of the kth basis state yk kt Using Eq 12 we can rewrite Eq 13 as Mm Idk Ck e k k 1 14 where we have set 110 1xE2 As in the case of the WE the coefficients Ck are determined by the initial conditions but because there is only one time derivative there is only one initial condition zx0 the value of the wave function at t0 Setting I 0 in Eq 14 we obtain Lx0 Idk Cke k 15 Now this equation tells us that Ck is simply the Fourier transform of the initial condition zx0 We can thus invert Eq 15 to write Ck idxl x 39m 16 Taken in tandem Eqs 14 and 16 are the complete solution to the free particle SE initial value problem V An InitialValueProblem Example Let39s see what happens if we start with the initial condition zx0 1106quot2 2 17 2 The choice We lVZIZ39 normalizes the harmonic basis states See the Lecture 16 notes for details D M Riffs 4 3302009 Lecture 26 Phys 3750 which is a Gaussian with amplitude 10 and width parameter 039 that is centered at the origin As will become apparent this corresponds to a particle whose average position x and average momentum p are both zero Furthermore the particle has an uncertainly in position that is proportional to the width parameter 039 Using Eq 16 we can calculate Ck Ck e k2 TZquot 18 For details of this calculation see the Lecture 12 notes This is pretty cool the Fourier transform of the Gaussian function Lx0 Ck is also a Gaussian function of k Further more the width parameter of Ck is 2039 which is inversely proportional to the width parameter 039 of the original function lx0 Thus the product of the two width parameters is a constant 0392039 2 t 0 t 25 I I I I I I 05 05 RLWX D A mogt u 0 2lwxetwxtle055 I I 05 I 05 I I I 10 0 10 10 0 10 X X t 5 0 t 100 I I I I I I 0 5 0 5 mm mm 0 0 2lwxetwxtle055 05 I A I o5 I 0 D M Riffs 5 3302009 Lecture 26 Phys 3750 118041000 2wxtwxtleO55 05 I I 05 I A l 10 0 10 10 0 10 X X t50 t 100 I I I I I I 05 05 118041000 2wxtwxtleO55 Using Eq 18 in Eq 14 we obtain the solution to this initial value problem Wxt 213 Tdkerkza39zA exkxiwkt 19 foo where ak is given by Eq The previous two gures 3 show graphs of Rel1 Imzj and Lfll the associated probability density for the particle for various times t In the rst gure 039 3 while in the second gure 039 1 As evident from the graphs of fl1 in the second case the particle is initially more localized than in the rst case The following facts are also worth noting 5 These gures show snapshots of the videos 5E Wavepmeel 75112 and 5E Wavepa eel 251122 which are available on the class website D M Riffs 767 3302009 Lecture 26 Phys 3750 2 Although zx0 is real in general zxt has both real and imaginary parts This is due to the presence of 139 in the Eq 1 the SE equation 22 Neither the real or imaginary part of zxt retains a Gaussian shape but the probability density fl1 is Gaussian for all t 222 The wave function Lxt spreads out along x in time This is due to something known as groupvelocity dispersion which depends upon the dispersion relation ak We will discuss this in detail later but for now note that the amuer the initial pulse the fave the pulse spreads in time Let39s think for a moment about how this solution compares to the solution to the WE that has an initial Gaussian displacement and no initial velocity What was the solution in that casegt You should recall that the initial Gaussian pulse split into two Gaussian pulses with the same width as the original displacement and that those pulses traveled in opposite directions away from the origin at the speed c In contrast to the solution here the traveling pulses described by the wave equation do not broaden with time Exercises 261 As illustrated the solution zxt to the SE is in general a complex function Write zxt as a sum of real and imaginary parts zjRxtizIxt substitute into Eq 1 and show that Eq 1 can be written as two 7m coupled equations for lIR and 14 Here you will need to use the fact that the real and imaginary parts of an equation are independent of each other 262 The momentum p of a particle in the state zjk xt yOe lk wlkltl is a well de ned quantity ie it has no uncertainty and is given by p hk Similarly the energy of the particle in this state is well de ned and is simply the parameter E hak Using the dispersion relation ak show that it is equivalent to the energy momentum relationship for a free classical particle 263 The two linearly independent solutions for the X x equation were written as ZVVKE ZVVKE Xx Xoe lh z and X39 x Xoe Jh z Find the appropriate linear combinations of these two solutions that give the other commonly used form of two linearly independent solutions X1x141 cos Z 7x and X2xA2 sing227x D M Riffs 7 3302009 Lecture 26 Phys 3750 264 Show for the Gaussian initial condition zx0 woe XZ z that the probability density Lx0tx0 is also a Gaussian How is the width parameter of Lx0tx0 related to the width parameter 039 of the wave function zx0 M265 A SE Initial Value Problem 3 Starting with the initial condition mo 1 5 1lt x lt1 0 0therw1se nd the formal solution the equivalent of Eq 19 to the initial value problem b Make plots of Lxt for I 0 I 2 and I 4 For purposes of keeping things simple set R 1 and m 1 so that the dispersion relation is simply ak kZZ Hint when numerically evaluating the integral set the limits only large enough so that the integrand is negligible at the endpoints of the integration c Discuss the time dependence of Lxt D M Riffs 8 3302009 Lecture 13 Phys 3750 Vector Spaces Real Space Overview and Motivation We review the properties of a vector space As we shall see in the next lecture the mathematics of normal modes and Fourier series is intimately related to the mathematics of a vector space Key Mathematics The concept and properties of a vector space including addition scalar multiplication linear independence and basis inner product and orthogonality I Basic Properties of a Vector Space You are already familiar with several different vector spaces For example the set of all real numbers forms a vector space as does the set of all complex numbers The set of all position vectors de ned from some origin is also a vector space You may not be familiar with the concept of functions as vectors in a vector space We will talk about that in the next lecture Here we review the concept of a vector space and discuss the properties of a vector space that make it useful A VectorAddz39l z39m A vector space is a set of some kind of quantity that has the operation of addition de ned on it whereby two elements V and u of the set can be added to give another element w of the set1 wuv 1 There is also an additive identity included in the set this additive identity in known as the zero vector 0 such that for any vector V in the space V 0 V 2 The addition rule has both commutative u V V u 3 and associative uvwuvw 4 properties 1 We denote vector quantities by boldface type and scalars in standard italic type This is standard practice in most physics journals D M Riffs 1 2112009 Lecture 13 Phys 3750 B fcalaerl zl39bz39ml z39m The vector spaces that we are interested in also have another operation de ned on them known as scalar multiplication in which a vector u in the space can be multiplied by either a real or complex number a producing another vector in the space Vau If we are interested in multiplying the elements of the space by only real numbers it is known as a real vector space if we wish to multiply the elements of the space by complex numbers then the space is known as a complex vector space Scalar multiplication must satisfy the following properties for scalars a and b and vectors u and V abu aubu 5a abu abu 5b auvauav 5C In u 5d OH 0 5e None of these properties should be much of a surprise I hope C Limar Imlepe deme am Bari A subset of m vectors is linearly independent if none of the subset can be written as a linear combination of the other members of the subset If the subset is linearly dependent then we can write at least one of the members as a linear combination of the others for example um alu1 azu2 amilumil 6 For a given vector space if there is a maximum number of linearly independent vectors possible then that number de nes the dimension N of the vector space2 That means if we have identi ed N linearly independent vectors then any vector in the space can be written as a linear combination of the set of N independent vectors as valu1a2u2aNuN 7 2 If there is not a maximum number of linearly independent vectors then the space is said to have infinite dimension D M Riffs 2 2112009 Lecture 13 Phys 3750 or in more compact notation as v Zanun 8 Furthermore the coefficients3 an in Eq 8 are unique The vectors uquot in Eq 8 are said to form a basis for the space Now Eq 8 should look strangely familiar We have some quantity on the lhs that is written as a linear combination of quantities on the rhs Hum And you might even ask assuming that I know the vectors u in Eq 7 how do I nd the coef cients an 9 D I er Pmd cl This last question is most easily answered after we de ne one more operation on the vector space known as the inner product of two vectors which we denote uv The inner product returns a scalar which is a real number for a real vector space or a complex number for a complex vector space The inner product can be de ned in any manner as long as it satis es the following relationships uv Vll 9a wavbu awvbwu 9b uu20 uu0 u0 9c Note the complex conjugate symbol in Eq 9a If we are dealing with a real vector space then we can just ignore the complex conjugate symbol Also note that Eq 9a implies that the inner product of a vector u with itself is a real number With the properties of the inner product denoted we can de ne the concept of orthogonality Two nonzero vectors u and V are said to be orthogonal if their inner product vanishes ie if uv 0 Note that if two vectors are orthogonal then they are linearly independent This is easy to see as follows Assume the converse that they are linearly dependent Then their assumed linear dependence means that u av where a is some scalar see Eq Then the scalar product vu vav avv cannot be Zero because V is not Zero see Eq 9c Thus they must be linearly independent 5 The coef cient an are also known as the components of V in the basis 11111N D M Riffs 3 2112009 Lecture 13 Phys 3750 The converse is not true two linearly independent vectors need not be orthogonal The proof is given as one of the exercises One last thing regarding the inner product The quantity is generally known as the norm or size of the vector 11 Often we are interested in vectors whose norm is 1 We can quotnormalizequot any vector u with scalar multiplication by calculating 10 uu The HhatH over a vector indicates that the vector39s norm is 1 E Orz kogomlBam Most of the time that we deal with a basis the vectors in that basis are orthogonal That is their inner products with each other vanish In this case it is a simple matter to nd the components an in Eq Let39s say that we want to find the mth component am Then we take the inner product of Eq 8 with 11m and we get uvgtZaltuu 11 So what happensgt Well there will only be one nonzero inner product on the rhs 11 um and so Eq 11 becomes umvamumum 12 and we can now solve for am as a m 13 All of this should now look even more strangely familiar We will get to why that is in the next lecture but right now we will review a vector space with which you should have some familiarity D M Riffs 4 2112009 Lecture 13 Phys 3750 11 1D Displacement Space Let39s look at a simple example to start Assume that we have a line drawn somewhere and on that line we have identi ed an origin 0 as illustrated in the picture at the top of the page The vector space that we are interested in consists of all the arrows that start at O and end someplace on the line The picture also illustrates two of these vectors one denoted u and one denoted V 4 So let39s talk about some of the math introduced above with respect to this vector space We rst have to de ne vector addition which must satisfy Eqs 1 Let39s go with the standard physics de nition of vector addition whereby we add vectors by the tip to tail method where the one of the arrows is translated without any rotation and its tail is placed at the tip of the other arrow as illustrated in the second picture Clearly this produces another arrow whose tails is at the origin and head is on the line and is thus a vector in the space Eq 1 is thus satis ed It should also be clear that we could have translated V rather than u in this example and so this de nition satis es Eq 3 the commutative property of vector addition We will not illustrate it here but you should convince yourself that Eq 4 the associative property is satis ed by the sum of three arrows What about the Zero vectorgt Well if Eq 2 is to be satis ed it must have no length and so it must be the arrow that begins and ends at the origin 4 Note this vector space is not a vector eld A vector eld is the assignment ofa vector to each point in space D M Riffs 5 2112009 Lecture 13 Phys 3750 What about scalar multiplication Again we go with the standard de nition whereby scalar multiplication by a positive number a results in an arrow that points in the same direction and is at times longer than the original arrow Multiplication by a negative scalar b results in an arrow that points in the opposite direction and is b times longer than the original arrow It should be clear that this de nition satis es all parts of Eq What about linear independence and dimension Pick an arrow any arrow Now ask yourself the following question can I nd another arrow that is not a multiple of my rst arrow If the answer is no which it is then the vector space has one dimension and you can use any arrow as the basis for the space For example let39s say you pick the arrow 11 in the above drawing as your basis Then the space is one dimensional because you can write any other arrow v as V au 14 where a is some scalar Although we have not yet de ned what the inner product is notice that if we take the inner product of Eq 14 with V we get V vauaua2uu 15 so that zimvmlziM Jiuaui llul with the sign depending upon the sign of a Now scalar multiplication was de ned as multiplying an arrow39s length by the multiplying scalar Thus a is also the or ratio of the two vector39s lengths Therefore for this space the norm must be proportional to the length of the arrow at 16 So what about the inner product Also notice the following Because this is a one dimensional space this basis 11 is trivially orthogonal and we can use Eq 13 to express the coef cient a in Eq 14 as 1 7 D M Riffs 767 2112009 Lecture 13 Phys 3750 Together Eqs 16 and 17 imply llVi1lllllll llVVli lquot 18 So which sign do we usegt As we now show it depends upon the relative directions of the two arrows Let39s rst consider the case where u and V are in the same direction Then we can write V au where a gt 0 The we have the following uVuauauu 19 Therefore because uugt0 we must use the positive sign if u and V are in the same direction Similarly if u and V are in opposite directions then a lt 0 and we must use the negative sign One last comment notice that nothing we have done here makes us chose the norm to be exactly equal the length of the arrows it must only be proportional to the length of the arrows For this space however the standard de nition of the vector norm is simply the arrow length 111 Real Space R3 What is real spacegt It is the extension of this 1D displacement space that we have been discussing to 3 dimensions That is it is simply the set of all displacement vectors 1 defined with respect to some xed origin Our discussion here will center on the more practical at least from a physics point of view The picture at the top of the next page illustrates the following discussion In dealing with this space we typically de ne a set of three mutually perpendicular axes that pass through the origin which we label x y and z We also denote three special vectors in this space the three unitnorm vectors 5 y and i which are three arrows that point along the three axes respectively The relative orientations of these three unit vectors are de ned by the right hand rule ie cross product through the equation 2 xxy You should convince yourself that these three vectors are linearly independent given our definitions of vector addition and scalar multiplication discussed in the last section It is also true that these three vectors are a basis for our vector space so this vector space is three dimensional Thus we can write any vector in the space as a linear combination of these three vectors as rrxr yri 20 x y z D M Riffs 7 2112009 Lecture 13 Phys 3750 N ltgtquot lt So now we are back to the ever occurring problem of determining the coef cients of some quantity of interest that is expressed as a linear combination of some other quantities To do this we can again use the inner product once it is de ned We use the standard de nition of the inner product of two vectors in this space uv quotvllcost9 21 where 9 is the angle between the directions of the two arrows Notice that this de nition reduces to the de nition that we came up with for two vectors in our 1D space above With Eq 21 it is easy to see that the unit vectors 2 7 and i satisfy the following relationships 9ii1 223 i i 0 22b Eq 22 de nes an orthonormal basis for a three dimensional space That is the basis is made up of unit vectors Eq 22a that are all mutually orthogonal Eq 22b Using Eq 22 we can now eXpress the coef cients in Eq 20 as 7x A rgt ry A IA 7 I gt VI A 1 i rl 23a 23C D M Riffs 8 2112009 Lecture 13 Phys 3750 Note that Eq 23 is the speci c form of Eq 13 for the case at hand Notice also that because the basis vectors have unit norms the coef cients have an especially simple form each coef cient is simply the inner product of the respective basis vector with the particular vector of interest Lastly we remark that the inner product between two vectors 1 and s can be simply written in terms of the components of those vectors in an orthonormal basis Let39s assume that I is given by Eq 20 and s by an analogous equation Then we can write rsrxxry 7risxsysi 24 Now it is probably not obvious that Eq 21 the de nition of the inner product satis es Eq 9 but assuming that it does Eq 24 simpli es to rs rxsx rysy rzsz 25 That is the inner product of two vectors can be simply expressed as the sum of the products of corresponding components of the two vectors Lastly we remark that when working with vectors in real space we often use a more notationally compact form than that in Eq 20 we often simply express the vector 1 as its triplet of components r WM 26 leaving the basis vectors x 7 and i as implied But when using this notation one must keep in mind that lurking in the background is an implied set of basis vectors Exercises 131 The inner product a Show that Eq 9a implies that the inner product of a vector u with itself is a real number b Using Eq 9 show that uav auv and auv auv c Using Eq 9 show that av buw avw buw D M Riffs 9 2112009 Lecture 13 Phys 3750 132 Projection The projection of a vector V onto the direction of another vector uV 1111 basis 111 u2 HS Show that the vector V can be written as uis de ned as pVu 11 Consider an orthogonal but not necessarily normal V pVu1 pVu2 pV113 That is the vector V is simply the sum of its projections onto the orthogonal basis set In physics we often call these projections the vector components of V in the 111 uz 113 basis 133 Consider two linearly independent vectors u and V Show that the vector 11 V w V u 1s orthogonal to u Thls Important result can be used to create an uu orthogonal basis out of any basis 134 Show that two linearly independent vectors need not be orthogonal Hint you may nd the result of Exercise 133 to be helpful here 135 Assuming that Eq 9 and that the results of Exercise apply show that Eq 25 follows from Eq 24 136 Use Eq 25 to nd the norm of the vector 1 rx ft ry 33 r i Does your result look familiargt M137 Real space A vector 1 in real space has components 4 110 in one orthonormal basis In this same basis a set of vectors is given by 1311 f0 fl fig 1 00 1 3 Show that this set of vectors is orthonormal and thus is another orthonormal basis b Find the components of I in this new basis c From the components given in the statement of the problem nd d From the components determined in part b nd Is the same as calculated in part c D M Riffs 710 2112009 Lecture 7 Phys 3750 Long Wavelength Limit Normal Modes Overview and Motivation Today we look at the long wavelength limit of the coupled oscillator system In this limit the equations of motion for the coupled oscillators can be transformed into the partial differential equation known as the wave equation which has wide applicability beyond the coupled oscillator system We also look at the normal modes and the dispersion relation for the coupled oscillator system in this limit Key Mathematics We again utilize the Taylor series expansion 1 Derivation of the Wave Equation A The Lozgg Wamle gfv Limit So why look at the coupled oscillator system at long wave lengthsgt Perhaps the main motivation comes from the fact that we are often interested in waves in systems where the wavelength is much longer than the distance between the coupled objects For example let39s consider audible v 20 to 20000 HZ sound waves in a solid In a typical solid the speed of sound 0 is 2000 ms so audible frequencies correspond to wavelengths A cv between approximately 01 and 100 In These wavelengths are obviously much greater that the typical interatomic spacing d of 2 X 10 10 In As we will see one bene t of the long wavelength limit is that we will no longer need to refer to the displacement of each interacting object the index j will be Htraded inH for the continuous position variable x so that we will be considering displacements as a function of x and t Let39s consider the equation of motion for the j th oscillator which can be any oscillator in the N coupled oscillator system digt dtz qltrgt 2qltrgt 140 1 Let 395 go ahead and trade in the discrete object index j for the continuous position variable x via x jd where d is the equilibrium distance between objects in the chain Then we can rewrite Eq 1 as 62qx t W 52qxdt 2qxt qx dt 2 Notice that the time derivative is now a partial derivative because we are now thinking of the displacement q as a function of two continuous variables x and t D M Riffs 1 1232009 Lecture 7 Phys 3750 Now because d is a small parameter1let39s Taylor series expand the two functions qxdt and qx dt in a Taylor series in d and d respectively about the point x so in doing this expansion we are thinking about x as some fixed point along the chain and qxidt as a function of id The two Taylor series are qx dt qxt aqx tall 62qx t d2 3a 6x 2 6x2 and qx dt qxtm dlm W 3b 6x 2 6x2 Keeping all terms up to order all we consider the validity of this approximation at the end of Sec II and substituting Eq 3 into Eq 2 gives us 2 2 0 196025de 661 92m 4 6t 6x This is the wave equation which as you can see is a homogeneous linear second order partial differential equation There is one more thing we need to do however in order to make Eq 4 more universally applicable The term 52612 d2 5 m on the rhs of Eq 5 is a combination of the fundamental parameters k5 m and d of the coupled oscillator problem Let39s de ne another constant c2 ksmal2 so that we can more generically write the wave equation as 62qxt 2 6q2xt atz 6 6x2 39 1 Technically the small parameter in the long wave length limit is the ratio 61 Generally a parameter can only be large or small if it is unitless Otherwise whether it is small or large or somewhere in between will depend upon the system ofunits being used D M Riffs 2 1232009 Lecture 7 Phys 3750 B GemmlAppz39mbz39lz39gj of the W409 Eqmzl z39m The is the standard form of the wave equation that we will use in this class As we shall see in further study of the wave equation 0 is the propagation speed of waves described by Eq Now if Eq 6 were only useful for studying the long wavelength motion of the coupled oscillator system it really wouldn39t be that interesting Fortunately it is applicable in a wide variety of situations including sound waves in uids and solids transverse waves on a string and electromagnetic waves in vacuum or other nondispersive media In each situation the constant c2 can be related to the underlying physics For example for transverse waves on a string c2 ry where r is the tension in the string and u is the mass per unit length of the string 11 LWL of Coupled Oscillator Solutions A N 0mm Mode Let39s now look at the coupled oscillator normal modes in the LWL As we discussed in Lecture 6 we can write the normal mode solutions indexed by n as a function of x and t as QMXJ SinknxAnelmk Bne m lk 7 where the wave vector is given by 717239 1 k 8 N1d 0 and the dispersion relation is N d akn 2a s1n Ekn J 9 As written Eqs 8 and 9 are expressed in terms of the fundamental or microscopic parameters 615 and Nof the coupled oscillator problem Let39s see if we can re express them in terms of c and other more generic or macroscopic parameters Well the first thing to notice is that the length L of the system a more generic macroscopic parameter can be written in terms of the fundamental parameters N and d as L N1d The wave vector can thus be simply expressed as k 10 D M Riffs 3 1232009 Lecture 7 Phys 3750 Well what about Eq 9 the dispersion relation It is not yet clear what to do with 5 in order to obtain a more generic description This is where the long wavelength limit comes into play again Recall that the wave vector is related to the wavelength via k 1 1 This allows us to write Eq 9 as N d QMquot 2a s1n 7 12 Now the long wavelength limit is exactly the limit dl ltlt 1 Thus in this limit we can replace the sine function by its very small argument so that Eq 12 can be expressed as wl 5d 13 n And now using 5610 and Eq 11 we have the long wavelength limit dispersion relation We ck n 14 Or in a more general form that is applicable to any harmonic wave described by the wave equation not just the normal modes for the coupled oscillator system where k kn is discrete ak ck 15 So we see that because 0 is constant the dispersion relation for waves described by the wave equation is linear vs k Looking back at the dispersion curves for the coupled oscillator system that are plotted in the Lecture 6 notes you should notice that for small kn kN which is equivalent to small dZn that the dispersion curves are indeed linear vs kn The small wave vector limit is thus the same as the long wavelength limit Putting all of this together we can write the normal modes Eq for long wavelengths as D M Riffs 4 1232009 Lecture 7 Phys 3750 qnxt Sin xAnelEH Ll BnermmrLt A remark about the normal modes should be made at this point The function sinn7rxL in Eq 16 came about because of the boundary conditions which we can now express as q0t 0 and qLt0 You can check that these boundary conditions are indeed satis ed by sinmrxL For other boundary conditions we would generally get some linear combination of sinknx and cosknx and perhaps also different allowed wave vectors kn or equivalently allowed wavelengths A 2 B Neglect of H gker Order Term 74 Eqmzl z39oze of Motive 274 the LWL Now that we have an expression for the normal modes in the LWL we can check to see that it is OK to neglect the terms in Eq 3 that are of order higher than all The first thing to notice is that all the odd terms in Eqs 3a and 3b cancel each other when inserted into Eq Thus we need only consider the even terms The second thing to notice is that for even m 6W1mef qnxt 17 This is obtainable using Eq 16 in Eq 3 Eq Therefore for example the ratio of the 4th order term to the 2 order term neglecting numerical factors is given by 61212 which is much smaller than 1 Thus we are justified in neglecting this term and other higher order terms which have an even smaller ratio in Eq 3 the Taylor series expansion of the equations of motion III The Continuum Limit vs the Long Wavelength Limit Recall that for N oscillators that there are N normal modes Well we have been clever and gotten rid of N in all of our expressions So what do we do It depends upon the situation that is being described by the wave equation As a long wavelength limit of a truly discrete system one must simply make sure that the waves being described have wavelengths that are much longer that the appropriate interparticle spacing However there are times when the wave equation is used such as in electromagnetism where there are no underlying oscillators and thus no underlying spacing d to be compared with the wavelength In those cases where the wave equation is believed to describe all waves the long wavelength limit is replaced by a more mathematically technical limit known as the continuum limit In this limit one 2 Think about the normal modes for sound waves in a pipe that is open on both ends vs a pipe with one end open and one end closed This is explored in Exercise 74 D M Riffs 5 1232009 Lecture 7 Phys 3750 actually takes the limit d gt 0 The result is that the terms in Eq 3 of order greater than all can be neglected exactly Furthermore in the limit d gt 0 for xed L we see that because LN1d this limit is equivalent to N gtoo That is in the continuum limit the number of normal modes becomes in nite From here on out when working with normal modes of the wave equation we will assume that there are indeed an in nite number of normal modes iVIore details on the continuum limit can be found in Dr Torre s text Exercises M71 The wave equation for NN and NNN coupled oscillators a Analogous to Eq 2 write down the equation of motion for the coupled oscillator system that has both NN and NNN springs see the Lecture 7 notes b Similar to what was done here for the NN system take the long wavelength limit of this equation of motion and derive the wave equation c For this system what is the constant c2 in terms of the fundamental parameters 5 539 and d P 72 The long wavelength limit a Derive Eq 17 which follows from Eq 16 the expression for the normal mode solutions qnxt b Given your result in a Show that the ratio of the a4 to all terms in the Taylor series expansion of qx dt 2qxt qx dt which appears on the rhs of the equation of motion as expressed in Eq 2 is equal to 272392 d tn 2 c Thus argue that compared to the d2 term the d4 term can be neglected in the equation of motion in the LWL 73 Referring to the Lecture 6 notes show for large N that the ratio kn kN is equal to 2612quot thus proving that the long wavelength and small wave vector limits are equivalent M74 Normal modes for an openclosed pipe Consider the following general form for a 1D standing wave qS x t a sinkx b coskx sina t gt a Show that this is a solution to the wave equation only if a and k are related by the dispersion relation a ck b Let39s assume that the normal modes for sound waves in a pipe are of the form qsxt given above For a pipe that is open on one end at x 0 and closed on the D M Riffs 767 1232009 Lecture 7 Phys 3750 other at x L the appropriate boundary conditions are aq o I 0 and qL I 0 x Starting with the above general form of the standing wave solutions qsxt use the x 0 boundary condition to show that the normal modes of this system have the more speci c form qx t b coskx sinckt gt c Further using the x L boundary condition show that the wave vector k can only take on the discrete values kn rut2L where n 1 3 5 Notice that these wave vectors are not the same as the normal mode wave vectors for the coupled oscillators given by Eq 10 Thus show that the normal modes of the organ pipe can be written as mz39 mz39c xt bcos x sin t q 2L J 2L 4b 1 Thus show that the wavelengths for the normal modes of the organ pipe are in 4Ln D M Riffs 7 1232009 Le cture 25 Phys 3750 Energy Density Energy Flux Total Energy in 5D Overview and Motivation In this lecture we tmd the discussion of the metgy assooated With wave motion to waves descnbed by the 3D wave equation In fact the fustpatt ofthe discussion is actly the same as the 1D case lust tmded to 3D r t Mathematics Some 3D calculus espeoally the divetgmce theomn and the sphencal coordinates version of the gradimt 1 Density Flux and the Continuity Equation As in the 1D case let s assume that we are interested in some quantity Qt that has an assooated dmsity pxt Since we are dealng With a dmsity that hves in a 3D space the units of dmsity Will be the units of Q d1v1ded by m3 That is p Qm3 Let s consider a volume 2 mclosed by a surface S as illustrated In the followxng flgUfE t each pmnt vector sh enclosed volume 2 The amount of Q containechn 2 can be wnttm as gavelawuzr 1 n As in the 1D case1f Q is a conserved quantity then the change in Q inside 2 n Mm 717 3272009 Lecture 25 Phys 3750 dQQt 6p7t 3 ch 47m 9 must be equal to the net ow of Q into Q de th f7t s7dS 3 S The vector quantity is again known as the Q current density or the Q ux The dimensions of are the dimensions of p time a velocity so pms Thus also Q1m2s Note that the rhs of Eq 3 can be interpreted as the total Q current owing out through the surface S Equating the rhs39s of Eqs 2 and 3 gives us Idsr masds 4 Q We can now use the divergence theorem which is one of several 3D extensions of the fundamental theorem of calculus ham Samsws 5 Q S to rewrite Eq 4 as ll ap tviltwgtd3r 6 0 Now because the volume Q is arbitrary the integrand must vanish Thus apg tvjr0 7 Equation 7 is the 3D version of the continuity equation which again is a local statement of the conservation of Q D M Riffs 2 3272009 Lecture 25 Phys 3750 11 Energy Density and Flux for 3D Waves We now apply this discussion to the energy associated with 3D waves In this case Q represents the energy associated with a wave within some volume For waves described by the 3D wave equation the energy density can be written as pltmgt 3Zczltwgtzl s where q7t is the variable that is governed by the wave equation The rst term on the rhs of Eq 8 is the kinetic energy density p7 and the second is the potential energy density pV Now Eq 8 is fairly general as long as u is suitably interpreted If q is a true displacement then u will be a parameter with the units of mass density If q represents something else say an electric eld then it will have some other units From Eq 8 it is fairly easy to see that the units of u are generally given by Lu oule szm3 q2 It is not hard to show that the energy ux which can be written as i a 6 rt cha Vq 9 together with the energy density in Eq 8 satisfy Eq 7 the continuity equation 111 Several Examples Let39s look at some examples that involve spherically symmetric waves A fpverz39ml 31217151ng W409 Let39s look at a standing wave example You may recall that a spherical coordinates separable solution that is nite everywhere is of the form qklm r 6 Cklm ji1P1mcos6Cm em Dmeimw Akelk Bkeilk gt where fl is a spherical Bessel function of the first kind and Pl is an associated Legendre function of the first kind The parameter m is an integer whose absolute value can be no larger than the nonnegative integer l If we want a solution with spherical symmetry then there can be no 9 or dependence This means that both I and m must be zero because the only associated Legendre function independent of 9 is P00cost91 Thus the spherical Bessel function in Eq 10 must be jokr sinkrkr and so Eq 10 simpli es to D M Riffs 3 3272009 Lecture 25 Phys 3750 sinkr kr qk00 ragal at Ck00 Akelk Bkeilk 39 11 If we simplify this further by letting Bk A so that the solution is explicitly real then we have quoyo rt9 t 2Akava0 sniff coskct 12 Because all parts of the system oscillate with the same phase this is a spherically symmetric version of a standing wave Using Eqs 8 and 9 we can calculate the kinetic and potential energy densities and the energy ux associated with the wave in Eq 12 To do this in a fairly simple manner we can use the spherical coordinates version of the gradient imlim 1 1A 13 Vfr 9 6r r 66 rsinlt9gt 12 where f 9 gt and 3 are unit vectors in the r 9 and directions respectively The nice thing about spherically symmetric solutions is that only the first term on the rhs of Eq 13 contributes to the gradient A video of q p7 pV and for the wave in Eq 12 E71er 274 3D 31217151ng Wamam39 is available on the class web site As the video shows the displacement is indeed a standing wave Unfortunately the energy densities and ux fall off with the radial distance r so fast that it is hard to really see their behavior Given this we have made another video E71er 274 3D Aquot 47151ng Wawe 2402 which plots the surface integrated density and ux1 Dr 3 p7dS 14 1 The video separately shows the kinetic and potential contributions to D M Riffs 4 3272009 Lecture 25 Phys 3750 1r fF sds lt15 where the surface S is of radius r centered at the origin Now because a spherically symmetric solution is independent of the two angles 9 and this amounts to multiplying the density p and ux by the factor 47239 r2 which is the surface area of the sphere S The quantity Dr which has units of Joule m can be though of as a linear energy density ie the energy per unit length along the radial direction while the quantity 1r which has units of Joules is the total energy current owing through S Notice that this new video is very similar to the 1D standing wave video that we looked at in the last lecture B Spherical Trapdng W409 Let39s also look at a spherically symmetric traveling wave If we are thinking about sound waves this is the sort of wave that would result from a pulsating sphere centered at the origin We can construct a traveling wave solution from a linear combination of 2 linearly independent standing waves We thus need to use both kinds of spherical Bessel functions The linear combination that produces a spherically symmetric outgoing traveling wave is2 qk00 r 6 a t Ck00 10 kr 005k6t yo kr Sink6tl39 1 6 which can be written in terms of sine and cosine functions as qk 0 0r t9 t Ck 0 0 Finger coskct M sinkct 17 39 39 39 39 kr kr The video E eigy 274 3D Tramng Wamam39 shows Dr and 1r for this wave Indeed away from the origin the wave appears to be an outgoing traveling wave However at the origin something rather different seems to be happening something with some standing wave character perhapsgt Well as it turns out the current density has terms with two types of behavior The first type has a lr2 dependence These terms describe the radiative part of the wave which carries energy off to infinity Because the radiative part of varies as 1 r2 the radiative part of 1r does not vanish as r gt oo However there are also nonradiative terms which vary as 1 r3 These terms act more like a standing wave 2 Note that this solution is only valid in the region of space outside the source In the video you may think of the source as being infinitesimally small so that the solution is valid infinitesimally close to the origin D M Riffs 5 3272009 Lecture 25 Phys 3750 the energy associated with these terms just oscillates back and forth and never really goes anywhere Because of the lr3 behavior to the nonradiative part of j the current 1r associated with these terms vanishes as lr as r gtoo These terms are thus sometimes called the local elds associated with the source In the video E eigj 274 3D Tra elz39 g Wave 24M we separately show the current 1r associated with each type of term Notice that the radiative piece looks essentially like a 1D traveling wave while the nonradiative piece is really only important in the vicinity of the origin Exercises 251 Show that the expressions for the density p and j in Eqs 8 and 9 respectively satisfy Eq 7 the continuity equation 252 Write the wave in Eq 17 qk 0 0r t9 t Ck 0 0 Finger coskct 0080 sinkct 39 39 39 39 kr kr as an explicit function of r ct thus showing that it is a traveling wave moving outward from the origin 253 Plane Wave Energy Density Consider the lane wave solution to the 3D wave equation qx y z t q0 expikx kyy kzz kct a Calculate the kinetic potential and total energy densities p7 xt pV xt and pxt respectively and the energy current density jxt b Show that the 3D continuity equation is satisfied by your expressions for pxt and jxt M254 Current Density for Spherical Traveling Wave a For the s herical travelin wave q rt9 t q Slug cos kct 0080 sin kct P g 1600 0 kr kr show that the radiative and nonradiative components of the current density can be written as um W13 com 4m and jNRrt3q3 coskr kctsinkr kctf Iquot 2 r3 respectively D M Riffs 767 3272009 Lecture 25 Phys 3750 b Calculate the time average of each of these current density components de ned T as Jfrtdt where T is one period of oscillation and show that the average of the 0 radiative part points in the positive rA direction while the time average of the nonradiative part is Zero Note neither answer should have any dependence on T M255 Spherical Standing Wave Energy Density Consider the spherically symmetric standing wave solution to the 3D wave equation sin kr qrt9 qo L a Calculate the kinetic potential and total energy densities p7 rt9 t pV rt9 t and prt9 t respectively b Show for large distances from the origin kr gtgt 1 that the total energy density for coskct this wave is approximately pr t9 t gqgkc21lsinkct krcoskct r D M Riffs 7 3272009 Lecture 3 Phys 3750 Two Coupled Oscillators Normal Modes Overview and Motivation Today we take a small but significant step towards wave motion We will not yet observe waves but this step is important in its own right The step is the coupling together of two oscillators Via a spring that is attached to both oscillating objects Ke Mathematics We ain some experience with coupled linear ordinary differential equations In particular we find special solutions to these equations known as normal modes by solving an eigenvalue problem I Two Coupled Oscillators Let39s consider the 39agrarn shown below which is nothing more than 2 copies of an harmonic oscillator the system that we discussed last time We assume that both oscillators have the same mass m and spring constant k Notice however that because there are two oscillators each has it own displacement either 41 or qz kg m m k 11 t12 ql 0 42 0 Based on the discussion last time you should be able to immediately write down the equations of motion one for each oscillating object as ijl 52 0 and 1a in 5 0 1b where 52 kxm As we saw last time the solution to each of theses equations is harmonic motion at the angular frequency 5 As should be obvious from the DM Riffs 717 192009 Lecture 3 Phys 3750 picture the motion of each oscillator is independent of the other oscillator This is also re ected in the equation of motion for each oscillator which has nothing to do with the other oscillator Let39s now make things a bit more interesting by adding in another spring that connects the two oscillating objects together as illustrated in the following picture To make things even more interesting we assume that this new spring has a different constant k However to keep things simple we assume that the middle spring provides no force if 41 iqz 0 That is this spring is neither stretched or compressed if its length is equal to the its length when two objects are at equilibrium k5 m kr m k5 41 qz gt ql 0 42 0 Thinking about this picture we should realize that the two equations of motion will no longer be independent That is the equation of motion for the first object will depend somehow upon what the second object is doing and vice versa Let39s use Newton39s second law to write down the equation motion for each object Recall that Newton39s second law for either object i12 can be written as 4 m 2 where E is the net force on object 139 The tricky part if there is a tricky part is to determine the sum E on each object The net force on the first object comes from the spring on the left and the spring in the middle With a little thought you should realize that this net force F1 is DM Me 727 192009 Lecture 3 Phys 3750 Fl ksq1 kiq1 q239 33 Make sure that you understand the signs of all the term on the rhs of this equation Notice that the force provided by the middle spring depends not only on the first object39s displacement but also on the second object39s displacement Similarly the net force on the second object is F2 ksq2 kiq2 q1 Substituting these two forces into Eq 2 once for each object we obtain the two equations of motion l52ql 5392q1 q20 4a for the first object and 252q25392q2ql0 4b for the second Here 5392 k39m Given the symmetry of the problem it might not surprise you that you can obtain one equation of motion from the other with the transformation 1 gt 2 in the subscripts that label the objects So now we have a considerably more complicated problem as expected from looking at the drawing above the equation of motion for each object depends upon what the other object is doing Speci cally each equation of motion depends upon the displacement of the other object 11 Normal Modes A Hamom39c Amal g So what are the solutions to these differential equations Well we will eventually write down the general solution next lecture But right now we are going to look at a special class of solutions known as normal mode solutions or simply normal modes A normal mode is a solution in which both masses harmonically oscillate at the same frequency We state why these special solutions are extremely useful at the end of the lecture For now let39s see if we can find them We use the complex form of harmonic motion and write qlt qme g and 53 D M Riffs 3 192009 Lecture 3 Phys 3750 qz I qozelm Notice that the unknown frequency of oscillation Q of both oscillators is the same a key feature of a normal mode Also because we are using the complex form of harmonic motion the amplitudes gel and gel may be complex but they too are unknown at this point Keep in mind that Eq 5 is only the form of a normal mode solution it is only a solution if it satis es the equations of motion In other words we now need to nd values of the frequency Q and amplitudes gel and gel that satisfy Eqs 4a and 4b the equations of motion So let39s substitute Eq 5 into Eq 4 and see what that tells us about 2 gel and gel Carrying out the substitution and calculating the derivatives yields qume Q 52q016 9 a7 qme Q qoze g 0 and 6a quoze g zqoze g a7 qoze g qme g 0 6b Dividing by 6 Is this legal and rearranging some terms gives 525392 Qzq01 5392q02 0 and 7a a7392q0152 5 22q02 0 7b So what do we have heregt We have two algebraic equations and three unknowns Q gel and gel The problem seems a bit underspeci ed and it is as we shall see below we will only be able to solve for Q and the ratio qmqo2 B E ge pal e Pmblem If you have previously studied differential equations and linear algebra you may be inclined to write Eq 7 in matrix notation as NZ N72 N72 a N WN39 gm 92 q01 w2 w2w2 qoz qoz which you would recognize as an eigenvalue problem Generally an eigenvalue problem is one where some linear operator this case a matrix operating on some object this case a 2D column vector produces a constant this case S times the original object Generally an N dimensional linear algebra eigenvalue problem D M Riffs 4 192009 Lecture 3 Phys 3750 has N solutions which consist of special values of the constant or eigenvalue 92 qu and amplitudes qmq0n or eigenvector E qON C E ge pal ex Well that was a lot of terminology but what about the solutiongt Well let39s rewrite Eq 8 as 525392 oz 5392 gm 0 9 5392 anal 22 qoz 39 Now this is interesting Expressed in this way we have the product of two quantities equal to Zero There are two ways that Eq 9 can be true The rst which is the trivial ie uninteresting solution is qm qoz 0 Physically this corresponds to no motion of the system pretty uninteresting The nontrivial way that Eq 9 can be satis ed is if the determinant of the 2 X 2 matrix is zero That is 52 5 422 5392 512 52 512 92 10 AB For a 2gtlt2 matrix the determinant is easily calculated C AD BC so in this case Eq 10 can be expressed as 525392 QZ2 5394o 11 Eq 11 or Eq 10 is known as the characteristic equation for the eigenvalue problem This is great We now have an equation for the eigenvalue 92 and thus the normal mode frequency Q 52 5 QZ ia7392 12 Solving Eq 12 for 92 produces the two eigenvalues 22 2525222a7392 13 which gives us four solutions for the normal mode frequency D M Riffs 5 192009 Lecture 3 Phys 3750 o i5 4352 25 14 D E ge pecforx am NomalModex So now that we have the eigenvalues 52 and 522a7392 we need to nd the eigenvector associated with each eigenvalue To do this we substitute each eigenvalue into either Eq 7a or 7b It doesn39t matter which you get the same equation in either case 7 m Mama mode For the rst eigenvalue 92 522 this substitution produces angel alzqoz 0 which gives us the result for the amplitudes q01 qoz 16 Physically Eq 16 tells us that both oscillators oscillate identically check out Eq 5 with the result of Eq 16 if this normal mode is excited That is the objects oscillate with exactly the same amplitude and the same phase Now because the eigenvalue 92 522 corresponds to two normal mode frequencies Q i5 this rst eigensolution of the linear algebra problem gives us to two linearly independent solutions to the equations of motion qltAle 72 and qz tAle 72 17a17b is the rst solution and q1 t 316quotquot and qz t 1316 2 18a18b is the second where the amplitudes A1 and B1 are arbitrary Equations 17 and 18 can be written in linear algebra inspired notation as 0 1 sinsail and D M Riffs 767 192009 Lecture 3 Phys 3750 10 1 io lfame 20 respectively The 1 and 1 denote the 5 and aN solutions However because oscillations at frequency 5 are really just oscillations at 5 any linear combination of these two solutions really just oscillates at the frequency 5 and so any linear combination of Eq 19 and 20 can be thought of as the first normal mode solution at frequency 5 That is the most general way we can write the first normal mode solution is q1t q1t ql 1 8152 7sz q2tllqztl lqztllllAl Bl 21 where A1 and B1 are unspeci ed constants Note that arbitrariness of A1 and B1 is consistent with our knowledge that for an harmonic oscillator the frequency is independent of the amplitude Note also that if we wish to at this point we can specify the solution to be real in which case we would set B1 AK 2 59mm Mama mode Let39s now look at the second normal mode solution which corresponds to the second eigenvalue 52 25 As before we substitute this eigenvalue into Eq 7a which gives us 39zqm E39Zqoz 0 22 or gel q02 So if this normal mode is excited the two oscillators oscillate with the same amplitude but with opposite phase or a phase difference of That is when the first oscillator is moving to the left the second is moving to the right with the same magnitude in its displacement and vice versa As in the case of the first eigenvalue there are two linearly independent solutions corresponding to the two frequencies i 52 25 The first solution is qz t Item and qz I A16 W 24a 24b D M Riffs 7 192009 Lecture 3 Phys 3750 and the second is q1t Bzequot gum and qz I Bzequot hm 25a25b or in linear algebra notation zixiww and q1t 1 6715144312 mi32L As before the general form of this normal mode solution is q1t 119 q1t 1 81E2w22 WWW qz zlqz h lqztlz 1A1 Bl 28 The graphs on the following page plot the time dependent amplitudes q1t and qz t for the two normal modes for the following values of the arbitrary constants A1 B1 12 for the first normal mode and A2 B2 12 for the second normal mode With these choices the solutions are real For these graphs we have also set m kg and k equal to 1 so that 51 and 45225392 15 Admittedly we have not specified units here but if standard SI units are used for the mass spring constants and time then the unit for displacement is meters As the graphs show in the first normal mode the two objects oscillate identically while in the second normal mode they oscillate exactly oppositely As we will see in the next lecture a great usefulness of the normal mode solutions is that ANY solution of Eqs 4a and 4b the equations of motion for this coupled oscillator system can be written as a linear combination of these two normal mode solutions Indeed this property of normal mode solutions is so important that it will be a theme throughout the course D M Riffs 8 192009 Lecture 3 Phys 3750 FIRST NORMAL MODE 2 I E Z E 39i I I39 Il 1 m 39 L 39 39 0 39 quot 39 39 39 quot 3 E E i 39 2 Q4 0 1 39 39 39 B 39a 39 m I 0 15 30 TIME F E DJ 0 lt1 g m 52 Q Exercises 31 Starting with the Euler formula 6 9 cost9i sint9 and its complex conjugate write cost9 and sint9 in terms of e and e 32 Write the expression Ae Z 136quot2 in the form CcosaNtDsinaNt That is nd C and D in terms of A and B From this result show that if B Aquot then C and D are both real which means that Ae 2 136quot2 is real 33 In the graph of the rst normal mode solution q1t and qz I both look like cosine functions Show for A1 B1 12 that the solution qllttgt 1 3 6 51 canin ee ewri enas qllttgt q2tlI JAle Bl d db tt l 1 N qzltt 1 Jcoswt 1 D M Riffs 192009 Lecture 3 Phys 3750 34 If we take the limit k gt 0 in the coupled oscillator problem what does this correspond to physically What happens to the normal mode frequencies Does this make sense Note if two normal modes have the same frequency then they are said to be degenerate and any linear combination of those two normal modes is also a normal mode M35 Three coupled oscillators In this problem you will nd the normal modes of three coupled oscillators as illustrated below Assume that each object has mass m and each spring constant is k l h h The following steps lead you through the same procedure as is used in the notes to solve the two oscillator problem in order to solve this problem It will be most helpful if you carefully study that procedure before tackling this problem 3 Using Newton39s second law write down the equation of motion for each object in the form of Eq 4 in the notes b Assume a normal mode type solution and find the three algebraic equations equivalent to Eq 7 in the notes that govern 22 and the amplitudes qm gel and qm c Write your equations in b in matrix form equivalent to Eq 9 in the notes 1 Find the characteristic equation equivalent to Eq 11 in the notes that determines the three eigenvalues Hint you will need to calculate the determinant of a 3gtlt3 matrix e Find the three eigenvalues for this problem I For each eigenvalue find the eigenvector associated with that eigenvalue g Write down the 3 normal mode solutions in the form of Eq 21 in the notes h In words describe the motion of the three objects for each of the normal modes D M Riffs 710 192009 Lecture 31 Phys 3750 Maxwell39s Equations Overview and Motivation Maxwell39s M39s equations along with the Lorentz force law constitute essentially all of classical electricity and magnetism E and One phenomenon that arises from M39s equations is electromagnetic radiation that is electromagnetic waves Here we introduce M39s equations discuss how they should be viewed and see how M39s equations imply the wave equation We will look at a plane wave solution to M39s equations We will also see that the conservation of electric charge is a direct result of Maxwell39s equations Key Mathematics We will gain some practice with HdelH V used in calculating the divergence and the curl of the electric and magnetic vector elds I Maxwell39s Equations The basic Maxwell39s equations are typically written in SI unitsl as VErt PS0quot 1 VBrt0 2 VXErt BBquot 3 v x 3a 0 jrt 080 6m 4 61 These are coupled first order linear partial differential equations for the electric and magnetic vector elds Ert and Brt respectively The two constants in the equations 80 and 0 are the fundamental constants of E and M respectively You gt probably rst encountered these two constants in your introductory physics class when you studied the electric force from a point charge and the magnetic force form a long straight wire carrying a constant current The other two quantities in these equations are the electric charge density prt and electric charge current density jr t 1 There have been no fewer than 5 systems of units traditionally used for E and M electrostatic esu electromagnetic emu Gaussian cgs HeavisideiLorentz and Rationalized MKSA now known as SI Beware when reading the literature D M Riffs 1 462009 Lecture 31 Phys 3750 So what is the meaning of Eqs 1 4 The way these equations are written you should think of the quantity on the rhs as giving rise to the quantity on the lhs of each equation So Eq 1 tells us that electric charge density is the source of electric eld Similarly Eq 2 tells us that there is no such corresponding magnetic charge density Equation 3 tells us that a time varying magnetic eld can produce an electric eld and Eq 4 says that both an electric charge current density and a time varying electric eld can produce a magnetic eld Maxwell was actually only responsible for the last term in Eq 4 but the term is key to E and M because without it there would be no electromagnetic radiation Often one considers the charge density prt and current density jrt to be given quantities That is one assumes that there is something external to the problem that controls prt and jrt and so they are simply treated as given source terms for the equations However in some problems the dynamics of prt and jrt are determined by the elds themselves through the Lorentz force equation FqErrvarr lt5 which describes the force on a particle with charge q and velocity V In such cases Eqs 1 5 must be solved self consistently for both the time varying elds and charge and current distributions II The Conservation of Electric Charge One of the consequences of M39s equations is the conservation of electric charge If we start with Eq 1 and takes its time derivative we obtain V 6 after switching the order of the divergence and time derivative on the lhs Equation 4 can be rearranged as 6Ert 6t 08 0 VXBrtgiJrt 7 Using this equation to substitute for 6E6t in Eq 6 then yields v 1 varr ijrt iaplr t 8 1080 so 80 it D M Riffs 2 462009 Lecture 31 Phys 3750 But this simpli es considerably because VV gtlt V 0 for any vector eld V Thus we have apr tVjrt0 9 61 the continuity equation for the charge density and the charge current density As we discussed in an earlier lecture the continuity equation is the local form of the statement of the conservation of the particular quantity this case electric charge that corresponds to the given density and current density 111 Wave Equations for the Electric and Magnetic Fields As we now demonstrate M39s equations imply waves equations for both the electric and magnetic elds To keep things simple we consider M39s equations in a charge freegt current free region of space Then Eq 1 4 become VErt0 10 VBrt0 11 V X Er t 6B2 I 12 V X Brt 1080 6E2 I 13 These are know as the homogeneous M39s equations because either E or B appear linearly in every nonzero term in the equations Let39s derive the wave equation for the magnetic eld To do this we rst take the curl of Eq 13 to produce VgtltVgtltBrt 08on aEg t 14 Using the identity V X Vgtlt Vr VVVr V2Vr on the lhs and switching the order of the curl and time derivative on the rhs then produces VVBrt V2Brt 080 15 Now using Eq 11 to substitute for VBrt on the lhs and Eq 12 to substitute for V X Ert on the rhs yields D M Riffs 3 462009 Lecture 31 Phys 3750 62Br t V2Br t 080 BIZ 16 the wave equation for the magnetic eld Brt where the standard constant 02 in the wave equation is equal to 1 1080 One can similarly derive the corresponding wave equation for the electric eld 62Er t V2Er t 1080 BIZ 1 7 Notice that each of these wave equations is for a vector quantity and so in essence each of these equations is really three wave equations one for each component of the electric or magnetic eld Another thing to note is that while M39s equations imply the wave equations for E and B the two elds are not independent That is all solutions to Eqs 16 and 17 will not necessarily satisfy M39s equations Another way to think about this is that by taking a derivative the curl of Eq 13 to derive the B eld wave equation we lost some information originally contained in that equation IV Plane Wave Solutions to M39s Equations We already know that a plane wave is one possible type of solution to the 3D wave equation So let39s assume that we have an electric eld of the form ErtEocoskr mt 18 Remember the wave vector k points in the direction of propagation and k E k 2721 where A is the wavelength Substituting Eq 18 into Eq 17 gives us as usual the dispersion relation which in this case is a2ck where c1 1080 is known as the speed of light As far as the wave equation is concerned the dispersion relation is the only constraint that needs to be satis ed in order for Eq 18 to be a solution However M39s equations put a further constraint on the electric eld Let39s substitute Eq 18 into Eq 10 Then we obtain VEocoskr at 0 19 To see what this equation implies for our plane wave solution let39s rewrite it using Cartesian coordinates D M Riffs 4 462009 Lecture 31 Phys 3750 6 2 oniE0y IEozicoskxkyykzz wt o 20 Calculating the derivatives produces E k E k E k sinkxkyykZz m o 21 0x I 0y y Oz 2 which can be written in coordinate independent notation as EDksinkr at 0 22 For this to be true for all values of 1 and I this then implies E0 k 0 23 So what does this tell us Because k points in the direction of the waves propagation this tells us that the electric eld must be perpendicular or transverse to the direction of propagation That is there is no longitudinal component to the electric eld for a plane wave solution to M39s equations So what about the magnetic eldgt Somewhere we should have learned that electromagnetic radiation consists of both propagating electric and magnetic elds In order to see what else M39s tell us let39s assume that the magnetic eld associated with the electric eld given by Eq 18 is of the form Brt B0 cosk39r a39t 39 24 Let39s now see what M39s equations tell us about B0 k a and 39 As with the electric eld we can use Eq 11 to tell us that B0 is also perpendicular to the direction of propagation We can learn more by substituting Eqs 18 and 24 into Eq 13 After a bit of algebra and differentiation we end up with the result B0 gtlt k39sink39 r aft 39 zE0 sink r mt 25 c where we have used 1080 1c2 For this to hold for all values of 1 and t we must have the following relationships k39k a a 39 and B0 gtltkac2 E0 The first three relationships tell us that the electric and magnetic elds have the same wavelength frequency and phase and propagate in the same direction The last relationship is a bit more interesting The last relationship tells us that E0 is D M Riffs 5 462009 Lecture 31 Phys 3750 perpendicular to both k which we knew already and B0 Thus because both E0 and B0 are perpendicular to k all three vector are perpendicular to each other Furthermore because B0gtltkHE0 we must also have EOXB0Hk and kXE0HB0 And lastly because B0 and k are perpendicular the relationship B0 gtltkaCZE0 tells us that Bok wc2 E0 or because a ck B0 E0 c Thus the magnetic eld can be expressed as BrtkgtltE0coskr at 26 where k kk Or we can simply write for our plane wave solution to M39s equations 1 A a Br t k X Ert c Exercises 311 As was done in the notes for the magnetic eld derive the wave equation for the electric eld 312 The product 1080 appears in the wave equations where lc2 traditionally appears Look up 0 and so and calculate c 1 J 1080 What do you getgt 313 Derive the dispersion relation a ck by substituting Eq 18 into Eq 17 M314 Conditions on E and E for a plane wave 3 Substitute Eqs 18 and 24 into Eq 13 and derive Eq 25 b Substitute Eqs 18 and 24 instead into Eq 12 and again derive Eq 25 This shows that in this situation the information in Eq 12 is redundant M315 A travelingwave solution to Maxwell39s equations Show that the traveling wave elds Ert oncoslltz an and Brt coskz an c satisfy all four homogeneous Maxwell equations D M Riffs 767 462009 Lecture 5 Phys 3750 Linear Chain Normal Modes Overview and Motivation We extend our discussion of coupled oscillators to a chain of N oscillators where N is some arbitrary number When N is large it will become clear that the normal modes for this system are essentially standing waves Key Mathematics We gain some more experience with matrices and eigenvalue problems I The Linear Chain of Coupled Oscillators Because two oscillators are never enough we now extend the system that we have discussed in the last two lectures to N coupled oscillators as illustrated below For this problem we assume that all objects have the same mass m and all springs have the same spring constant k QN0 Q1 0 Q2 0 Our rst goal is to nd the normal modes of this system At the beginning we approach this problem in the same manner as for two coupled oscillators we nd the net force on each oscillator nd each equation of motion and then assume a normal mode type solution for the system Let39s consider some arbitrary object in this chain say the j th object The force on this object will depend upon the stretch of the two springs on either side of it With a little thought you should be able to write down the net force on this object as F ksqj qu ksq qwl 1 or upon simplifying E 1656114 2 qml 2 D M Riffs 1 1142009 Lecture 5 Phys 3750 You might worry that this equation is not valid for the first j 1 and last j N objects but if we assume that the j 0 and j N1 objects the walls have in nite mass so that go and qm1 are identically Zero then Eq 2 applies to all N objects We shall refer to these two conditions q0 0 and qm1 0 as boundary conditions bc39s on the chain of oscillators With the expression for the net force on each object we can write down the equation of motion Newton39s second law for each object as a 529111 261 qj1 0 3 where 1S jSN and as before 52 kSm Notice that each equation of motion is coupled the equation of motion for the j th object depends upon the displacement ofboth the j 1 and 391 objects 11 Normal Mode Solutions We now look for normal mode solutions where all masses oscillate at the same frequency by assuming that1 q guyea 4 If we substitute Eq 4 into Eq 3 after a bit of algebra the equations of motion become 92 52q0171 2 gown 0 5 Now keep in mind that what we have here are N equations of motion one for each value of j from 1 to N As in the two oscillator problem the set of equations can be expressed in matrix notation 252 52 0 0 9701 go 52 252 52 0 qo2 go 0 52 252 52 q 92 q 6 0 0 52 252 qoy4 qoy4 1 We have slightly changed notation here We now write the amplitudes gov with a comma between the zero and the mass index so that terms such as go 1 are unambiguous D M Riffs 2 1142009 Lecture 5 Phys 3750 Or 252 oz 52 0 0 c1701 a72 252 92 a72 0 902 0 a72 2522 22 a72 qw 0 7 0 0 52 252 oz qoyz So as before nding the normal modes reduces to nding eigenvalues 22 and 1 eigenvectors gm of the N gtltN matrix in Eq Recall that the eigenvalues are found by solving the characteristic equation that arises when we set the determinant of the NgtltN matrix in Eq 7 to Zero LOWEST FREQUENCY EIGENVECTOR I 336 v wan I We VG EIGENVECTOR c c 390 0 10 20 30 40 50 SECOND LOWEST FREQUENCY EIGENVECTOR EIGENVECTOR POSITION mass index D M Riffs 3 1142009 Lecture 5 Phys 3750 A E ge aecforx Now for N any larger than 3 solving the characteristic equation for the eigenvalues and eigenvectors by hand is not advisable So let39s turn to a computer mathematics program such as Mathcad and see what insight we can gain into this problem Given the matrix in Eq 6 with a speci c value for 52 Mathcad can calculate the eigenvalues and eigenvectors of that matrix In the graphs on the previous page we plot the eigenvectors corresponding to the three lowest eigenvalues for the N 50 problem The key thing to notice is that these eigenvectors look like standing waves on a string for example That is as a function of position ie mass index j the components qoyj of the eigenvector appear to be a sine function which must equal zero at the ends of the chain j 0 and j N 1 because of the bc39s This observation inspires the following ansatz for the eigenvectors gov Asin j 8 where A is some arbitrary amplitude for this sine function it could be complex because we are dealing with a complex form of the solutions and is some real number that will be different for each normal mode2 Now Eq 8 obviously satis es the go 0 bc on the lhs of chain but not necessarily the rhs bc qm1 0 To satisfy this bc we must have q0N1 ASinl N 1l 0 which is true only for N 1 717239 where n is an integer That is we must have 717239 N1 10 where the integer n labels the normal mode solution Now because any integer n in Eq 10 produces a value for that satis es Eq 9 it looks like n can take on an in nite number of values this seems to imply an in nite number of normal modes Well this can39t be right because we know that there are only two normal modes for 2 Recall for a standing wave on a string the spatial part of the standing wave can be written as sin x so the parameter is obviously related to the wavelength of the normal modes in some manner 7 more detail on this later D M Riffe 4 1142009 Lecture 5 Phys 3750 the two oscillator problem In fact because the N oscillator problem involves an N dimensional eigenvalue problem there are exactly N normal modes The solution to this conundrum lies in the fact that the sine function Eq is periodic It can be shown that there are N 1 unique solutions but because the n 0 solution is trivial there are only N unique nontrivial solutions In fact we can specify these N unique solutions by choosing n such that ISnSN 11 Combining Eqs 8 10 and 11 we can write the N eigenvectors as mz39 gov Asm N1 j ISnSN 12 B E ge pal ex So we have now speci ed the eigenvectors What about the eigenvalues We can obtain these by inserting Eq 8 into Eq 5 which produces QZ sin j w2sin j 1 2sin j sin j 0 13 Now it looks like this equation depends upon j but it does not Using some trig identities it is not dif cult to show that Eq 13 simpli es and can be solved for 92 as 92 452 sini 14 And remembering that only takes on the discrete values given by Eq 10 we have the N eigenvalues Qz 42 2 7 15 n a s1n 2ltN1 where n 1 2 3N The normal mode frequencies are thus given by on izasin 203 1 16 It is interesting to plot the positive frequencies as a function of mode number 71 Such a graph is shown below for several values of N As with previous graphs we have set m1 and ks 1 D M Riffs 5 1142009 Lecture 5 Phys 3750 As we will discuss in the next lecture these graphs are essentially graphs of frequency vs inverse wavelength and as such are known as dispersion relations or dispersion curves As we will see as we work our way through the course the dispersion relation is extremely useful in understanding the propagation of waves associated with that dispersion relation Also as we will discuss in the next lecture the dispersion relation also contains information on the interactions between the springs connecting l39 r gt l Ar 0 Q I A A Z l 39 A g 1 I II Aquot 2 1quot I 1 av F D l o I 2 39 1 i 0 i A 7 7 2 3 l 1 39G N72 a i ll N10 g A39 tA N20 l U0 5 10 15 20 NORMAL MODE INDEX 11 the oscillating objects Notice that the frequencies plotted for two oscillators N 2 equal our previous results 21 1 and 22 xg for the special case of m 1 and k 1 III The Initial Value Problem Lastly we discuss how the initial value problem can be solved using the normal modes Quite generally using the above results and including both the QM and Q frequencies we can write the n th normal mode solution as qr sinus 1 q2r M13312 39 Mr 19 71042 3 n n q I SlnN13 A 8 3 8 17 gm smug N 1142009 D M Riffe Lecture 5 Phys 3750 where Q Q arbitrary complex numbers3 The general solution can thus be written as a linear combination of the normal modes as and the constants A and B which have replaced A above are ngt c110 sintcill c120 N sintcil lt13 r sinus Ane quot Bnequot w 18 E H I M sintsaN Equation 18 is the extension of Eq 3 of the Lecture 4 notes As before the arbitrary amplitudes A and B depend upon the initial conditions of all the oscillating objects To see exactly how the Aquot and Bquot are determined let39s consider the N 3 case As in the two oscillator case let39s make the normal modes explicitly real by setting Bquot A For three oscillators Eq 18 then becomes 110 3 sin1 ngt 2 sin2 Ane w Aequot9 19 We can now apply the initial conditions which gives us a manner similar to the two oscillator case qzltogt 22 sum W gt 20 qg 0 sin3 and M0 3 Sin 1 420 2 2n sin2 ImAn 21 q 0 quot1 sin T3 5 Notice that the column vector of the rhs of Eq 17 is the n th eigenvector of the associated eigenvalue problem D M Riffs 7 1142009 Lecture 5 Phys 3750 So we see that the real part of the amplitudes A depend upon the initial positions of the three objects while the imaginary part of the amplitudes depend upon their initial velocities So where do we go from heregt You may remember that for the two oscillator problem we applied the normal mode transformation to the equivalent of Eqs 20 and 21 which allowed us to find the amplitudes see p 5 6 of the Lecture 4 notes There is an equivalent transformation here that will allow us to nd the Aquot 395 To most easily see what it is let39s explicitly write out Eq 20 as 6110 SM sin sin37 q20 2 sing ReA1 sin7r ReA2 sin37 ReA3 22 u a v7 sin37 sin7 q3 0 sin a and evaluate the sine functions which gives us M0 5 2 15 q20 2 ReA1 0 ReA2 2 ReA 23 q30 K5 2 FE vv Now notice what now happens if we multiply this equation by the first eigenvector sine sin when it is written as a row vector sin sin Sin 1E 2 We MST obtain 1 3J5q02q205q304ReA lt24 Notice the very nice result that the terms containing the amplitudes A2 and A3 produce zero when multiplied by the rst eigenvector row form We can now solve for the real part of A1 in terms of the initial positions as Ira1335 q102q20 q30 25 This equation is equivalent to the first row of Eq 18 or Eq 20a in the Lecture 4 notes for the two oscillator problem To obtain ReA2 and ReA3 it should be obvious that one needs to multiply Eq 23 by the respective row eigenvectors D M Riffs 8 1142009 Lecture 5 Phys 3750 Further to find the imaginary parts of the amplitudes one similarly multiplies Eq 20 by the row eigenvectors This HtrickH of multiplying by the row eigenvector to obtain the corresponding amplitude is probably the most important part of this lecture We will repeat it many times throughout the course when we discuss Fourier series we will use this trick to find the Fourier coefficients of a function when we talk about vector spaces this trick will be recognized as the inner product and when we talk about Fourier transforms this trick will be known as quotinversionquot As we discuss these topics you should keep in mind this little trick that allowed us to find the amplitudes A So we know how to find the An39s but what about the normal mode transformation mentioned abovegt Well it is lurking about here If we now create an N gtltN matriX by stacking the row eigenvectors then we indeed have that transformation So for the three oscillator problem the transformation matrix would be J5 2 J5 2 0 2l 2 2 J2 2 6 If one multiples Eqs 20 and 21 by Eq 26 then one obtains two equations that are equivalent to Eqs 18 and 19 of the Lecture 4 notes for the two oscillator case 6110 Also if one multiplies the column vector qz t by Eq 26 then one obtains the Mt Q10 normal mode coordinates Q2 t for the three oscillator case Q30 D M Riffs 9 1142009 Lecture 5 Phys 3750 Exercises 51 Using the appropriate trig formulae obtain Eq 14 for 92 from Eq 13 52 Only N unique eigenvalues and eigenvectors Eqs 12 and 15 specify the N eigenvector and eigenvalues for the N oscillator problem The condition 1 S n S N implies that values of 71 outside of this range simply give a solution that is the same as one of the solutions inside the range 1 S n S N 3 Starting with Eq 15 and using the angle addition formula for the sine function show for example that 9 is the same as 9 Hint write N as N1 1 and N2 as N11 b Starting with Eq 12 show also that the eigenvector for n N 2 is the same as the eigenvector for n N 53 Similar to the second gure in the notes graph the three eigenvectors for three coupled oscillators 54 Four coupled oscillators a What are the eigenvalues Qi for four coupled oscillators b Similar to the second gure in the notes find and then graph the four eigenvectors for four coupled oscillators 55 Similar to Eq 25 find the real and imaginary parts of all the Aquot 39s for the N 3 system of coupled oscillators 56 Apply Eq 26 the normal mode transformation to Eqs 20 and 21 to obtain the two equations for three oscillators that are equivalent to Eqs 18 and 19 of the Lecture 4 notes for two oscillators 57 Show that the square of Eq 26 the normal mode transformation is proportional to the identity matriX Thus find the inverse of the normal mode transformation 58 For the three oscillator problem nd the normal mode coordinates Q1 t Q2 t and Q30 in terms of the displacements q1t qz t and q3 D M Riffs 710 1142009 Lecture 20 Phys 3750 The Wave Equation in Cylindrical Coordinates Overview and Motivation While Cartesian coordinates are attractive because of their simplicity there are many problems whose symmetry makes it easier to use a different system of coordinates For example there are times when a problem has cylindrical symmetry the elds produced by an infinitely long straight wire for example In this case it is easier to use cylindrical coordinates So today we begin our discussion of the wave equation in cylindrical coordinates Key Mathematics Cylindrical coordinates and the chain rule for calculating derivatives 1 Transforming the Wave Equation As previously mentioned the spatial coordinate independent wave equation 1 6 c2 BIZ V2q 1 can take on different forms depending upon the coordinate system in use In Cartesian coordinates the Laplacian V2 is expressed as 62 62 62 2 6x2 6y2 622 Our rst goal is to re express V2 in terms of cylindrical coordinates p z which are defined in terms of the Cartesian coordinates x y 2 as p x2 y2 12 23 arctan J 2b Z Z 2c The following picture illustrates the relationships expressed by Eq For the point given by the vector rxxyyzi the coordinate p is the distance of that point from the z axis the coordinate is the angle of the projection of the vector onto the x y plane from the x axis toward the y axis and Z is the signed distance of the point from the x y plane Note that p Z 0 and we can restrict 0 S lt 27239 D M Riffs 1 342009 Lecture 20 Phys 3750 x In order to express V2 in terms of these new coordinates we start with a function fp z and consider it to be function of x y and Z fW Wgv the variablespgt and Z by writing f fpxyz xyz zxyz 3 Then for example using the chain rule we can write iziipi i 4 6x 6 6x 6 6x 62 6x Notice that this equation as well as some later equations have two types of terms The first type is a derivative of the function f while the second type is a derivative of a new coordinate with respect to an old coordinate The goal here is to use the relationship between the two coordinate systems Eq to write the second type of term as a function of the new set of coordinates p and z Then equations such as Eq 4 will be entirely expressed in terms of the new coordinate system For the particular case at hand the transformation is a bit simpler than the general case because see Eq p pxygt xy and z 22 Thus Eq 3 simpli es to f f pxyl xylzz 5 D M Riffs 2 342009 Lecture 20 Phys 3750 so that Eq 4 reduces to iziimi 6 6x 6 6x 6 6x Let39s now express the rst term Ff x2 as a function of cylindrical coordinates We do this by calculating another derivative of Eq 6 with respect to x 31 iim 7 6x 6x 6x 6 6x 6 6x 39 Now we must be a bit careful here We must use the chain rule on the functions if6p and if13 because they are functions of p and as is f see Eq However we do not need to use the chain rule on the terms BpBx and 6 6x because they are both functions of x and y as are p and see Eq With this in mind Eq 7 becomes 62f62f6p 62f Waergasz 62f 6p62f 6 62 8 6x2 apz ax Map ijg 6 6x2 apw ax M ijg 6 6x2 which exp ands to azfazfap1 62f 6 6p I if azpl 62f 6p6 l62ff jz162 9 6x2 62 6 x 39a ap 6x 6x 39 6 6x2 39apa 6x 6x 39 w tax a axz39 You can see that this process is rather tedious Now to eliminate the old variables x and y from Eq 9 we must calculate the quantities 6p a w W 6x 6x 6x2 6x2 103 10d and express them in terms of p and Using Eq 2a we calculate the rst of these terms as 6p6x2yzl2 x l 2 2 12 6x 6x 2x y 2x 1 Now x pcos and p2 x2 y2 so Eq 11 becomes D M Riffs 3 342009 Lecture 20 Phys 3750 2 pcos COS 12 Similarly using 6arctanu 1 Bu 13 6x 1u2 6x and Eq 2b we have 6arctanyx 1 y y 14 6x 6x 1yx2 x2 x2yzi Using y psin and p2 x2 y2 we now re eXpress Eq 14 in terms of the new coordinates as Q psin sin 15 6x p2 p 39 In similar fashion one can express the second derivatives Eq 9c and Eq 9d in terms of p and as 62p 2 sin2 16 6x2 p and 62 200s sin 17 6x2 p2 39 Ifwe now insert Eqs 12 15 16 and 17 into Eq 9 we obtain 6x2 6 6p p 6 p 62 f sin cos l azfsin2 I af 2cos sin 39 6pm p 39 M p2 39 w p2 z cosz 62f sin cos isinz 6 lt18 So we have now expressed the rst term of the Laplacian acting on a function f Ff13x2 in terms of cylindrical coordinates D M Riffs 4 342009 Lecture 20 Phys 3750 In a manner analogous to the procedure that we have just carried out one can also derive the result1 sinszr 662 f sin cos i cos2 6f 62 Mp p 6p p 19 L 62f sin cos L 62f cos2 6f 200s sin 39 6p 13 p 6W p2 13 p2 And of course we also trivially have 20 622 622 Putting Eqs 18 20 together then gives us the fairly simple result 2 2 2 6 f6f6 f 6x2 Byz 622 2 2 2 39 6fiiL0f6f 62 pap p26 2 622 y2f 21 Thus in cylindrical coordinates the wave equation becomes 2 Z Z Z iaqaqia qiaqaq 22 c2 BIZ 6p2 p 6 p2 6W 622 Where now q qp zt II Separation of Variables To look for separable solutions to the wave equation in cylindrical coordinates we posit a product solution qp zat Rp ZZTt 23 Substituting this into Eq 22 produces izRqDZTquotRquotqgtZTiR39qgtZTRqgtquotZTRqgtZquotT 24 c p p 1 Notice that Eq 19 is the same as Eq 18 with sin gt COS and COS gt sin D M Riffs 5 342009 Lecture 20 Phys 3750 If as in the case of Cartesian coordinates we now divide by q which is now written as RQDZT this reduces to 1 RquotiR39 c2 T p R p2 d3 Z Next time we shall look at the four ordinary differential equations that are equivalent to Eq 25 Three of these equations will be rather simple again essentially harmonic oscillator equations but one of them for Rp will be a new equation Its solutions are known as Bessel Functions Exercises 2 39 2 96201 Derive Eqs 15 and 16 That is using the chain rule show that 2 6 Sln x p and 200s sin 6x2 2 202 Suppose that a function fxyz only depends upon the distance p le y2 from the z aXis That is fxyz flx2 y2 Show that if13 0 3 directly in cylindrical coordinates easy and b using the chain rule starting with Cartesian coordinates D M Riffs 767 342009 Lecture 4 Phys 3750 Normal Mode Coordinates Initial Value Problem Overview and Motivation We continue to look at the coupledioscillator problem We extend our analysis of this problem by introducing functions known as normal mode coordinates These coordinates make the coupledioscillator problem simple because they transform the coupled equations of motion into two uncoupled equations of motion Using the normal modes we then solve the general initialivalue problem for this system Key Mathematics We gain experience with linear transformations and initial value problems I Normal Mode Solutions A Summmyfmm Lari Lamar The problem that we studied last time is shown in the following diagram There are two objects each with mass m and three springs The springs on the ends have spring constant k and the one in the center has spring constant k39 q10 420 Last time we found the two normalimode solutions which can be written as t l 41 X14160 Bleexm lt1 121 1 W and DM Riffs 717 1142009 Lecture 4 Phys 3750 q1t 1 81922 871922 ml l llAz 32 2 2 N where 21 a7 and 92 522 2a are the normal mode frequencies of oscillation As we will show below any solution to this problem can be written as a linear combination of these two normal modes Thus we can write the most general solution to this problem as q1 3141691 Blequot91 11Aze 91 Bze z 3 12 I B NomalMode Coordiml ex Let39s now consider the following linear transformation1 of the displacements qlt and q2t Q10 l l 910 QM b lml 4 L 2 Calculating the rhs of Eq 4 produces2 EZ J ZECE3J 5 Written more pedantically we have mmW lt62 and QWM ab For reasons that will soon become apparent the functions Q10 and Q2 tare known 1 Any linear transformation of an N ivector can be represented as lhs multiplication of that vector by an N XN matrix 2 Note that the normalimode coordinate Q2 as de ned here is the negative of Q2 as defined in Dr Torre s text FW P We de ne it here with this change so as to be more consistent with the later treatment of N coupled oscillators To be honest it also makes some of the equations look prettier D M Riffs 2 1142009 Lecture 4 Phys 3750 as normalmode coordinates Why aren39t they called normal mode functions Don39t ask me Let39s now apply the linear transformation in Eq 4 to the rhs of Eq 3 and see what it tells us Applying the transformation and equating the result to the lhs of Eq 4 yields Q1t 1 1012 M 0 w M This equation may look complicated but in fact it is very simple It says that Q1t harmonically oscillates at the rst normal mode frequency 91 and that Q2 t harmonically oscillates at the second normal mode frequency 92 Pretty cool In fact if Eq 3 is the general solution to this problem more on this below Eq 7 say that no matter what the motion the sum qltq2t always oscillates at 21 and the difference q1t qz I always oscillates at 92 C Eqmzl z39om of Motim Let39s now go back to the coupled equations of motion 152q15392q1q20 83 and 2 52 52 q10 8b and see what happens if we write them in terms of the normal mode coordinates Q1t and Q2 To do that we need the inverse of the transformation in Eq Using illi a make sure that you understand this I we apply this inverse transformation to Eq 4 which gives us after switching the rhs and lhs 343ki m h NIH NIH D M Riffs 3 1142009 Lecture 4 Phys 3750 That is qlt Q1tQ2 t and qz t Q1t Q2 Substituting these results into Eq 8 produces Q1QZ 52Q1Q225l2Q2 0 113 and Q1 Q2 52Q1 Q2 zalez 039 Looks pretty ugly ehgt Well it is about to get much simpler If we take the sum and difference of Eqs 11a and 11b we get the following two equations Q152Q1 0 12a and Q2 a72 25392Q2 0 12b First notice that these two equations are uncoupled the equation of motion for Q1 doesn39t depend upon Q2 and vice versa Furthermore you should now able to recognize each of these equations as the equation of motion for a single harmonic oscillator Thus Eq 12a tells us that Q1 harmonically oscillates at 5 91 and Eq 12b tells us that Q2 harmonically oscillates at V52 252 E22 Of course this is exactly what was expressed earlier by Eq We can also infer something very important from this transformation Because the normal coordinates are governed by Eq 12 which is simply an harmonic oscillator equation for each coordinate we know that the general solution for Q1t and Q2 t is given by Eq Thus the general solution for q1tand qz t is given by the inverse transformation of Eq 7 which is simply Eq This proves that the general solution for q1t and q2t is indeed a linear combination of the normal mode coordinates Q1tand Q2 This is a general result that we will use throughout the course 11 Initial Value Problem Let39s now solve the initial value problem for the coupled oscillator system That is we want to write Eq 3 the general solution to the coupled oscillator problem in terms of the initial conditions ql0 910 q20 and 920 which are all real quantities D M Riffs 4 1142009 Lecture 4 Phys 3750 Now Eq 3 1720 uses the complex form of the harmonic oscillator solution Looking back on p 8 of the Lecture 2 notes we see that we have three choices about how to deal with this Let39s use the rst approach and make Eq 3 manifestly real by setting 3114 and B2 Azl Then we have q1t 1 1 1 1 A 1012 A 71012 1022 71922 I 13 11gt ll 16 16 tl JW W l lt gt We can now apply the initial conditions Setting I 0 in Eq 13 gives us 10 1 1 1 0 1le1 MI dlAz M2 14 while taking the time derivative of Eq 13 and then setting I 0 produces 410 l 1 i 1 1 911 419112 A Equations 14 and 15 are really 4 equations for the 4 unknowns ReAlgt ImAlgt ReA2 and ImAZ To see this we use 22 2Rez and 2 2 2139Imz for any complex number 2 to re eXpress Eqs 14 and 15 as 10 1 1 2 0 2 JIM11 2 1JReMZgt 16 and 310 2o1 i1m14 292 jljlmwz 17 92 A O v The easiest way to solve for the four unknowns ReA ImAgt ReC and ImC is gt 1 7 to each side of these two to apply the normal mode transformation 2 h L 2 L 2 D M Riffs 5 1142009 Lecture 4 Phys 3750 equations which produces 1q10 12 0 2 311411 2 311412 18 2 q10 12 0 and 1 10 20 91 3me 3W2 lt19 2 q10 12 0 From these last two equations we immediately see that Wkw 202 walk W lt20bgt makw and em makw 2 Now we could simply substitute Eq 20 into Eq 13 and we would have our solution for the initial value problem However we can make the form of the solution much simpler if we recognize that Ale gl A1equot91 2 ReAle 91 2ReA1cosQlt ImA1sinQlt 21 Using Eq 21 and an equivalent expression for Azelgl Azle l Eq 13 can be re eXpressed as 2 gum11 cosglt 1mm sinQlt 211lReA2costa ImA1sinta 22 If we now substitute Eq 20 into Eq 22 we nally obtain D M Riffs 767 1142009 Lecture 4 Phys 3750 23ii lwowwwc slt91rgtlWlmltwl 23 Notice that the solution is real as it should be The following graphs show the motion that results for two sets of initial conditions For both graphs m1gt k51 and ks391 so that 21 1 and 925 In the top graph ql01gt q20 1 910 0 and q39z 0 0 What special motion is thisgt In the second graph the initial conditions are ql0 1 qz 0 0 410 0 and q39z 0 0 This motion is quite complicated In fact it is not even periodic it never repeats even though the normal mode coordinates are simply harmonically oscillating at their respective frequencies The nonrepetitive nature of the motion occurs because this example the ratio is of the two normal mode frequencies is not a rational numb er 239 q1T 39 39 39 q2T DISPLACEMENT DISPLACEMENT D M Riffs 7 1142009 Lecture 4 Phys 3750 Exercises 1 l 41 Apply the inverse transformation 1 1 to Eq 7 to recover Eq 42 Assuming q10 0 and qz 0 0 nd the general condition on the initial velocities 910 and q39z 0 that results in only the rst normal mode being excited 43 Assuming q10 0 and qz 0 0 nd the general condition on the initial velocities 910 and q39z 0 that results in only the second normal mode being excited M44 The general solution to the two coupled oscillator problem can be expressed in terms of real quantities as W1AcosgltBsingltf 1 MW K K1lecos 22tDsinozt Starting with this form of the general solution nd the real parameters A B C and Din terms of the initial conditions ql0gt 910gt qz 0 and 920 Check to see that your solution agrees with Eq 23 45 Use Euler s relation to derive Eq 21 D M Riffs 8 1142009 Lecture 18 Phys 3750 3D Wave Equation and Plane Waves 3D Differential Operators Overview and Motivation We now extend the wave equation to three dimensional space and look at some basic solutions to the 3D wave equation which are known as plane waves Although we will not discuss it plane waves can be used as a basis for any solutions to the 3D wave equation much as harmonic traveling waves can be used as a basis for solutions to the 1D wave equation We then look at the gradient and Laplacian which are linear differential operators that act on a scalar eld We also touch on the divergence which operates on a vector eld Key Mathematics The 3D wave equation plane waves elds and several 3D differential operators 1 The 3D Wave Equation and Plane Waves Before we introduce the 3D wave equation let39s think a bit about the 1D wave equation 6261 CZ azq 62 6x2 1 Some of the simplest solutions to Eq 1 are the harmonic traveling wave solutions q xt Ae W 2a qxtBe quot 2b where without loss of generality we can assume that a2ckgt 01 Let39s think about these solutions as a function of the wave vector k First we should remember that k is related to the wavelength via k27rZ Let39s now speci cally think about the solution qxt For this solution if kgt0 then the wave propagates in the x direction and if k lt 0 then the wave propagates in the x direction Thus in either case the wave propagates 274 the dimcfim of k Similarly for the solution q xt the wave propagates in the direction opposite to the direction of k We now introduce the 3D wave equation and discuss solutions that are analogous to those in Eq 2 for the 1D equation The 3D extension of Eq 1 can be obtained by adding two more spatial derivative terms yielding 1 lfwe assume a lt 0 then the two a gt 0 solutions just map into each other D M Riffs 1 2252009 Lecture 18 Phys 3750 6 62 6x2 ayz 622 2 2 2 2 661 20q0q6q 3 where now q qxyzt and x y and z are standard Cartesian coordinates This equation can be used to describe for example the propagation of sound waves in a uid In that case q represents the longitudinal displacement of the uid as the wave propagates through it The 3D solutions tO Eq 3 that are analogous to the 1D solutions expressed by Eq 2 Can be written as gawk xay zt Ae lkx WszZ wl 4a xyztBezkxkyykzzzot gawk As you may suspect the wave equation determines a relationship between the set kwkwkz and the frequency a Substituting either Eq 4a or 4b into Eq 3 yields m2 c2k3kk 5 As above we can assume a gt 0 which gives mck kk the dispersion relation for the Eq 4 solutions to the 3D wave equation The solutions in Eq 4 can be also written in a more elegant form If we de ne the 3D wave vector kk k 7ki 6 and use the Cartesian coordinate form of the position vector rx y 7zi 7 then we see that we can rewrite Eq 4 as q rt Ae k39rquotquot 8a D M Riffs 2 2252009 Lecture 18 Phys 3750 q rt B e lk39m 8b where kr kxxkyykzz is the standard dot product of two vectors The dispersion relation can then also be written more compactly as a cl kl 9 It is also the case that the wavelength A is related to k via lk I 2721 Analogous to the discussion about the direction of the 1D solutions the wave in Eq 8a propagates in the k direction while the wave in Eq 8b propagates in the k direction This is why one usually sees the form in Eq 8a the wave simply propagates in the direction that k points in this case These propagating solutions in Eq 8 are known as plane waves Why is that you may askgt It is because at any given time the planes perpendicular to the propagation direction have the same value of the displacement of q Let39s see that this is so Consider the following picture Keep in mind that the wave vector k is a xed quantity for a given plane wave its direction is indicated in the gure The dotted line in the picture represents a plane D M Riffs 3 2252009 Lecture 18 Phys 3750 that is perpendicular to k and passes through the point in space de ned by the vector 1 Now consider the dot product krkrcosltt9gt 10 This is simply equal to IkIIr0 I where 1 0 is the position vector in the plane that is parallel to k Furthermore for am position vector in the plane the dot product with k has this same value That is for any vector 1 in the plane kI is constant Thus the plane wave function Ae lk39r has the same value for all points 1 in the plane A simple example of a plane wave is one that is propagating in the z direction In that case the q plane wave is qgvovkzztAe kzZquot quot Notice that this wave does not depend upon x or y That is for a given value of z the wave has the same displacement for all values of x and y That is it has the same displacement for any point on a plane with the same value of z 11 Some 3D Linear Differential Operators A The Laplacia The combination of spatial derivatives on the rhs of Eq 3 62 62 62 6x2 Byz 622 1 1 is the Cartesian coordinate version of the linear differential operator know as the Laplacian generically designated as either A or V2 del squared The del squared representation is often used because the Laplacian can be though of as two successive although different applications of the differential expression that is simply known as del which is represented by the symbol V 2 ln Cartesian coordinates visiy3a 12 6x 6y x The Laplacian V2 can thus be written in Cartesian coordinates as 2 As we shall see below V can be used in the representation of several operators It is thus probably best not to think of V itself as an operator D M Riffs 4 2252009 Lecture 18 Phys 3750 vVri iyiilri iyiil 13 6x 6y 6x Kax 6y 6x J We now consider the application of V2 to a function f x yz but we do it Hone del at a timeH That is writing V2fxyz as VVfx y 2 we rst consider the piece Vf x yz Afterwards we look at VVf x yz which we usually simply write as VVfxyz However before we do this let39s make sure that we understand the concept of a eld A eld is simply a mathematical quantity that has a value assigned to each point in space The function fxyz is known as a scalar eld because fxyz assigns a scalar to each point in space A vector eld is a function that assigns a vector to each point in space An electric eld is an example of a vector eld B The Gradiem The quantity Vf is know as the gradient of f Let39s take a closer look at Vf Applying Eq 12 the Cartesian coordinate version of V to f x yz produces Vfxyzi iyii 14 6x By 62 Although not explicitly shown each term on the rhs is a function of x y and 2 Thus Vf is a vector eld because it assigns a vector to each point in space Simply put the gradient of a scalar eld is a vector eld An important property of Vfr is that Vfr is perpendicular to the surface of constant f that contains 1 where I is any position vector Let39s use a bit of vector calculus to show this Consider the picture below The surface S f is the surface of constant f that contains the end of r0 The vector rs is also assumed to lie on this surface So we wish to show that Vfr0 is perpendicular to Sf at 1 0 D M Riffs 5 2252009 Lecture 18 Phys 3750 We now assume that 1 s is close enough to 1 0 that we can write the function frs fxsyszs as a Taylor series expanded about the point r0 x0y0 20 fxs ysazs fxoyoazoi 6x r ZS ZO 15 x3 x0 Now this equation can be expressed in coordinate independent form as frsfr0Vfr0rS r0 16 If we now assume that 1 s is close enough to 1 0 so that the curvature of the surface is negligible then the higher order terms can be neglected Then because f 1 s f 1 0 both vectors are on the surface Sf we have Vfr039rsro039 17 And because rs 1 0 lies in the surface3 we have the result that Vf 1 0 is perpendicular to the surface Sf at the point 1 0 QED A concept closely associated with the gradient is the directional derivative If we have some unit position vector f d then the directional derivative of f 1 in the direction of f d is de ned as Vfrfd 18 5 The surface is essentially planar in the vicinity of 1 s and I 0 because of the proximity of TS to 1 0 D M Riffs 767 2252009 Lecture 18 Phys 3750 Physically the directional derivative tells you how fast the function f 1 changes in the direction of f d What kind of field is this quantitygt Notice that you can also think of the directional derivative as the scalar component of Vfr in the f d direction A straightforward application of the gradient is found in classical mechanics If U x yz is a potential energy function associated with a particle then the force on the particle associated with that potential energy is given by Fx yz VUx yz which is a vector field a force eldl As an example let39s consider the gravitational force on a particle near the Earth39s surface If we de ne our coordinate system such that 2 points upwards and x and y lie in a horizontal plane then the gravitational potential energy of a particle with mass m is given by Uxyz mgz z0 where 20 is an arbitrary constant and g m 98 III2 The force on the particle is then given by VUxyz mgi What are surfaces of constant U x yz in this casegt These are simply horizontal planes each one at a constant value of 2 Notice that these planes are indeed perpendicular to the force field mgi which points downward at all points in space C Dz39wgeme Now that we have some feel for the meaning of Vf 1 let39s now apply the second del to Vfr which gives us VVfr V2fr But before we do this maybe we should first say a few words about V operating on any vector field Vr Writing Vr as Vji Vy 7VZi and using the Cartesian coordinate form of V we have VVr 92V3V9V22 6V 6 6V 6x 6y 6x 1 9 The quantity VVr is called the divergence of the vector field Vr So what kind of object is VVr Because it assigns a scalar to each point in space it is a scalar field4 Thus we see that the gradient of a scalar field is a vector field while the divergence of a vector field is a scalar field If we now let Vr equal Vfr we then get using Eq 14 4 We will discuss the divergence in more detail in a later lecture Stay tuned D M Riffs 7 2252009 Lecture 18 Phys 3750 VVfr i 3 zii i i 6x 6y 6x 6x 6y 6x 2 Z Z 6fafaf 6x2 Byz 622 20 So we see that the Laplacian of f V2 f is the divergence of the gradient of f Thus sz is a scalar eld 111 Some Final Remarks Using the generic form of the Laplacian Eq 3 can be written in coordinate independent form as 1 6 6 26 vzq lt21 In the next lecture we will look at some more solutions to Eq 21 using Eq 3 the Cartesian coordinate representation but after that we will look at solutions to Eq 21 using some different coordinate systems cylindrical and spherical polar The representation of Equation 21 in each coordinate system will look vastly different Nonetheless in each case we will be solving a version of Eq 21 the 3D wave equation Exercises 181 Plane Waves Consider the solution q Lt Ae 1k39r to the 3D wave equation Assume that a gt 0 a Calculate Vql rt In what direction does Vql rt pointgt Thus describe constant surfaces of q rt for some fixed value of t b In what direction does q Lt Ae quotquot movegt What is the wavelength 2 of this wavegt What is the dispersion relation ak for this wavegt c Show that the sum of q Lt Ae 1k39r and q rt Be 1k39m is a standing wave if B A lVlake sure that you write down the specific form of this standing wave What property of the wave equation allows you to combine solutions to produce another solutiongt D M Riffs 8 2252009 Lecture 18 Phys 3750 182 Divergence and Gradient Consider the function f x y z xi2 ylz 212 a Is it appropriate to calculate the divergence or gradient of this function Calculate whichever is appropriate b You should now have a new function that you calculated in part a What kind of function is itgt Calculate either the gradient or divergence of this new function whichever is appropriate 183 Divergence Using the Cartesian coordinate form of V compute the divergence of the following vector elds r a Er r 3 where r le y2 z2 gt 0 Coulomb electr1c eld b Br L I x2 y2 gt 0 magnetic eld outside a long wire x2 y2 x2 y2 c Er 1 electric eld inside a uniform ball of charge x 184 Let f and g be two scalar fields Using the Cartesian coordinate form of delgt show that Vng VfVg fV2g What kind offreld is Vng D M Riffs 9 2252009 Lecture 10 Phys 3750 General Solution using Normal Modes Overview and Motivation Last time we solved the initial value problem IVP for the 1D wave equation on a nite domain with Hclosed closedH bc39s using the general form of the solution qxt fxctgx ct Today we solve the same problem using the normal mode solutions for this system Key Mathematics We utilize some integrations involving harmonic functions I The Problem De ned We are looking for the general solution to the wave equation 62qxt 2 62qxt atz 6 6x2 on the nite domain 0 s x s L subject to the initial conditions qx0 ax and cix0 W 22 2b and the boundary conditions q0t 0 and qLt 0 3a 3b This time we write the solution qxt as linear combination1 of the normal mode solutions q x t qltxrgtqltxrgt lt4 where the normal modes can be expressed as qnxt sinknxAne quot Bnequot quot 5 Here kn mrL is the wave vector and aquot ckquot is the angular frequency As before let39s make the normal mode solutions explicitly real by setting 3 A Then Eq 5 can be written as 1 As written Eq 4 looks like a simple sum not a linear combination but as Eq 5 shows we have kept undetermined amplitudes as part of our normal modes so Eq 4 may be justifiably thought of as a linear combination of normal modes D M Riffs 1 1102008 Lecture 10 Phys 3750 qnx t sinkn x Rean cosant Iman sinant 6 where we have de ned a new amplitude an An2 Using Eq 6 we may thus express the general solution via Eq as no qx t Z sinknxRean cosant Iman sinant 7 quot1 As we shall shortly see the amplitudes an are determined by the initial conditions Recall however that normal modes already satisfy the bc39s and so the general solution as expressed by Eq 7 automatically satis es those bc39s We thus need not consider the bc39s any further 11 The Initial Value Problem Yet Again Let39s now apply the initial conditions and see what we get From Eq 7 we obtain for the initial displacement no ax ZsinxRean 8 quot1 and for the initial velocity we obtain after differentiating Eq 7 no bx Z aquot sinxlman 9 quot1 ln Eqs 8 and 9 we have used kn mrL So what do we have heregt Well perhaps not surprisingly we have two equations for the amplitudes an in terms of the initial conditions However unlike the finite N oscillator case the rhs39s of Eqs 8 and 9 have an in nite number of amplitudes because the wave equation has an infinite number of normal modes This looks pretty grim but perhaps a look back at the N oscillator case will give us some insight into the current problem For the N oscillator problem the equation equivalent to Eq 8 is the extension of Eq 19 from the Lecture 6 notes to N oscillators which we can write as D M Riffs 2 1102008 Lecture 10 Phys 3750 c110 sin1 4120 N sin12 c140 2 sin13 Relaxquot 10 mo sinmlN where we have again made the assignment an Aquot 2 Now recall what we did there To nd Ream notice that the subscript is m not n and m just labels some arbitrary eigenvector we took the m th eigenvector expressed as a row vector and multiplied Eq 10 by that vector We can write this multiplication as sinmfll sinmflz sinm3 sinle c140 Now this equation can be succinctly expressed as ism J qj0 an 0110 Jsinfl J 1 1 11 J Now you should recall that the product of two eigenvectors ie the sum on j on the rhs of Eq 11 is nonzero only if they are the same eigenvector That is the sum on j on the rhs is nonzero only if n m Thus Eq 11 simpli es to N N 2mm 1M 0 ReamZsinzm J 12 F1 F1 Notice that we now only have one amplitude Ream on the rhs We can thus solve Eq 12 for this amplitude obtaining N Z sianl 1M 0 Ream F1 13 2mm 1 j1 D M Riffs 3 1102008 Lecture 10 Phys 3750 So how does this apply to the current problem Well the main difference between the two problems is that x is a continuous variable while j is discrete But if we remember a bit of calculus we should recall that an integral is really just a sum over a continuous variable We thus might expect the sums over j in Eq 11 to be replaced by integrals over x Let39s try it and wee what happens Multiplying Eq 8 by sinx and integrating on x from 0 to L gives us J sin xax dx 1 Rean J sinx sinx dx 14 the continuous variable analog to Eq 11 Further analogous to what happened with Eq 11 the integral on the rhs is nonzero only if n m Thus Eq 14 simpli es to jsinxax dx Ream jsinz ML x dx 15 and so as above we can solve for Ream L Isin L xax dx Ream 0L 1 6 Isinz x dx 0 Notice the striking similarity between Eqs 16 and 13 especially when you recall that ax qx0 Now we can simplify Eq 16 even a bit more Using the fact that J sin2xdx 17 we have the result 2 L M Ream st1n7xaxdx 18 0 D M Riffs 4 1102008 Lecture 10 Phys 3750 So we now have Ream expressed in terms of the initial displacement ax Similarly the imaginary part of am can be expressed in terms of the initial velocity bx as L L J sinxbxdx 19 Imam You may think of Eqs 18 and 19 as the inversions of Eqs 8 and 9 respectively Also note there is nothing special about the index m in these last two equations We could used any variable In particular when thought of as the inversion of Eqs 8 and 9 we would normally write these last two equations as Rean Isin xax dx 20 Iman LJ sinxbx dx 21 La Summarizing Eq 7 which can be written as qx t 2 sin x Rean cosct Iman sinct 22 along with Eqs 20 and 21 are the complete solution to the initial value problem on the nite domain 0 S x S L for the bc39s q0t 0 and qLt 0 111 An Example Revisited Let39s consider the problem that we solved in the Lecture 9 Notes Looking at this problem again will give us a good comparison of the two methods of solving the initial value problem The problem in the Lecture 9 notes has the initial conditions2 2 OIQ so we now have another amplitude A but it is not the same as the amplitudes A n l 2 used earlier in Eq D M Riffs 5 1102008 Lecture 10 Phys 3750 ax Aexpjz expjz 23a bx o 23b Recall that 0x is a Gaussian peak that is vertically shifted so that the bc39s are satis ed The following figure plots 0x for the same values of the width parameter 039 that we investigated in Lecture 9 039 005 039 02 and 039 05 The easy part of this particular problem is solving for Iman Using Eq 21 we immediately see that Iman0 Similarly but not as simplyl using Eq 20 we see that Rean is given by L Rean Jsin x exp 0 16L2 a dx 24 2 exp Unfortunately the integral in Eq 24 has no analytic solution3 Fortunately a program such as Mathcad can numerically solve the integral Unfortunately as Eq 22 indicates we need an infinite number of an 395 Fortunately in most cases we only need to use a finite number of the an 39s in order to get a very good approximation to 5 At least as far as I know Actually itis not too dif cult to show that an 0 if n is even But that still leaves the odd values of n to deal with D M Riffs 767 1102008 Lecture 10 Phys 3750 the exact solution That is in practice we typically use a truncated version of Eq 22 which we can write as qx tM i sinxRean cosct Iman sinct 25 quot1 where qxtM is the M term approximation to qxt For the example at hand Iman 0 so we have M qxtM Zsin xRean cosct 26 quot1 So how many terms do we need to usegt The answer of course depends upon the accuracy that we require But we can get a pretty good idea of the number of terms needed by plotting the absolute value of an vs n as shown in the following graph The three curves correspond to the three values of 039 used in the previous graph Note that the y aXis values have been normalized by lall Also because an 0 for even 71 we have only plotted an for odd values of n n135 lanall magnum nun Now this graph is very interesting it shows that the more compact or sharper the wave as indicated in this case by a smaller value of 039 the more normal modes one must use to accurately describe the wave From the above graph we see that for 0005 we need to use M m 29 to well represent the wave for 03902 we need MmS andfor 03905 weneede3 D M Riffs 7 1102008 Lecture 10 Phys 3750 We can also ascertain the number of normal modes needed by comparing the initial condition ax with the approximation qxtM at I 0 At I 0 Eq 26 becomes axM isinxRean 27 where we have de ned axM qx0M That M m 29 does a good job for 039 005 is illustrated in the following figure which plots axxM for several values of M Notice that as M increases axxM more faithfully represents the function ax sigma 005 ax M POSITION mm M3 M9 M19 M29 The next 2 figures show the same sort of thing for 039 02 and 039 05 but in a more direct fashion This figures on the left plots the difference ax ax5 for 03902gt while the figure on the right plots ax ax3 for do 05 In both cases the error is lt006 for all values of x Given the overall size of ax these also seems like reasonable approximations D M Riffs 8 1102008 Lecture 10 Phys 3750 sigma 02 sigma 05 001 0UU ax ax5 ax ax3 39001 0UU 0 05 1 POSITION POSITION The following two graphs are also somewhat illuminating They plot the initial condition ax along with the individual terms on the rhs of Eq 27 sinxRean for n 1M From the graph for 039 05 it is not hard to imagine that only the n 1 and n 3 are needed to accurately describe ax sigma 02 sigma 05 06 m m a a quot8 quot8 E E E E E E Z Z m N 3 3 S3 S3 390 I I 0 0 5 I 0 0 5 1 POSITION POSITION aX aX 111 n1 113 113 11 5 OK so what about the time dependence Now that we know how many normal modes we need we can simply use Eq 26 where Rean is calculated with Eq 24 with the appropriate value of M On the class web site there are videos of the resulting wave motion for all three values of 039 discussed here Both the 039 02 and D M Riffs 9 1102008 Lecture 10 Phys 3750 039 05 videos also show the motion of the individual normal modes that are used to produce the approximation qxtM Exercises 101 Show that Rean cosant Iman sinant can be expressed more succinctly as Reane quotquot Thus Eq 6 can be alternatively expressed as qnxt sinknxReane quot L 102 In the notes it is stated that J sinxsinxdx is nonzero only if m n 0 where m and n are both integers Using the trig identities for cosx y and cosx y do this integral and show that this statement is indeed true 103 An initial value problem Here we solve an initial value problem with the initial conditions 3 Graph ax Does it satisfy Eq 3 the boundary conditions b Find analytic solutions for Rean and Iman for this problem c Thus write down the solution qxt for this problem What is special about the motion M104 Another initial value problem Here we solve an initial value problem with the initial conditions 4 hx 0SxltL4 ape fh bx0 EL x L4SxSL 3 Graph ax Does it satisfy Eq 3 the boundary conditions b Find analytic solutions for Rean and Iman for this problem Either do the integrals by hand consult an integral table or use a program like Mathcad c Thus write down the solution qxt for this problem D M Riffs 710 1102008 Lecture 2 Phys 3750 Harmonic Oscillations Complex Numbers Overview and Motivation Probably the single most important problem in all of physics is the simple harmonic oscillator It can be studied classically or quantum mechanically with or without damping and with or without a driving force As we shall shortly see an array of coupled oscillators is the physical basis of wave phenomena the overarching subject of this course In this lecture we will see how the differential equation that describes the simple harmonic oscillator naturally arises in a classicalimechanics setting We will then look at several equivalent ways to w1ite down the solutions to this differential equation Key Mathematics We will gain some experience with the equation of motion of a classical harmonic oscillator see a physics application of Tayloriseiies expansion and review complex numbers I Harmonic Oscillations The freshmaniphysics concept of an undarnped undiiven harmonic oscillator HO is something like e following picture an object with mass m attache to an immovable wall with a spring with spring constant k There is no gravity here only the spring provides any force on the object Note that the oscillator has only two parameters the mass m and spring constant k Assuming that the mass is constrained to move in the ho zontal direction its displacement q away from equilibrium as a function of time t can be written as qtBsin5t 1 where the angular frequency 5 depends upon the oscillator parameters m and k Via the relation 5 Jigm The amplitude B and the phase do not depend upon m and k but rather depend upon the initial conditions the initial displacement 40 DM Riffs 717 1162009 Lecture 2 Phys 3750 and initial velocity O 40 of the object Note that the term initial conditions is a technical term that generally refers to the minimal speci cations needed to describe the state of the system at time I 0 You should remember that the angular frequency is related to the plain old frequency V via 5 27239V and that the frequency V and period T are related via V 1T Now because the sine and cosine functions are really the same function but with just a shift in their argument by i 2 we can also write Eq 1 as qt Bcos5tz 2 where the amplitude B is the same but the phase z 7r2 for the same initial conditions It probably is not obvious yet but we can also write Eq 1 as qt D cosaNt E sinaNtgt 3 where the amplitudes D and E depend upon the initial conditions of the oscillator Note that the term harmonic function simply means a sine or cosine function Note also that all three forms of the displacement each have two parameters that depend upon the initial conditions 11 Classical Origin of Harmonic Oscillations A The Hamom39c Potem z39al The harmonic motion of the classical oscillator illustrated above comes about because of the nature of the spring force which is the only force and thus the net force on the mass which can be written as Fq ksq 4 Because the spring force is conservative F can be derived from a potential energy function Vq via the general 1D relationship dV Fltqgt M 5 A potential energy function for the spring that gives rise to Eq 4 for the spring force is D M Riffs 2 1162009 Lecture 2 Phys 3750 l Kq 36qu 6 do the mathl Let39s see how the potential energy represented by Eq 6 arises in a rather general way Let39s consider an object constrained to move in one dimension like the oscillator above However in this case all we know is that the potential energy function has at least one local minimum as illustrated in the following graph of Vq vs q Vq Let39s now assume that the mass is located near the potential energy minimum on the right side of the graph and that its energy is such that it does not move very far away from this minimum Just because we can let39s also assume that this minimum defines where q 0 and V 0 as shown in the picture Now here is where some math comes in If the mass does not move very far away from q 0 then we are only interested in motion for small q Let39s see what the potential energy function looks like in this case For small q it makes sense to expand the function Vq in a Taylor series m V0V390qV 0q2 V 0q3 7 where eg V390 is the first derivative of V evaluated at q 0 Now the rhs of Eq 7 has an infinite number of terms and so is generally quite complicated but often only one of these terms is important Let39s look at each term in order The first term V0 is Zero because we defined the potential energy at this minimum to be Zero So far so good The next term is also Zero because the slope of the function Vq is also Zero at D M Riffs 3 1162009 Lecture 2 Phys 3750 the minimum The next term is not Zero and neither in general are any of the others However if q is small enough these other terms are negligible compared to the third quadratic term1 Thus if the object39s motion is suf ciently close to the minimum in potential energy then we have my Voqz 8 So this is pretty cool Even though we have no idea what the potential energy function is like except that it has a minimum somewhere we see that if the object is moving sufficiently close to that minimum then the potential energy is the same as for a mass attached to a spring where the effective spring constant k is simply the curvature Vquot evaluated at the potential energy minimum q 0 Vquot02 Equation 8 is known as the harmonic approximation to the potential Vq near the B Hamom39c Ox lazl orqual z39m of Motim OK so we see that the potential energy near a minimum is equivalent to the potential energy of an harmonic oscillator If we happen to know how an harmonic oscillator behaves then we know how our mass will behave near the minimum But let39s assume for the moment that we know nothing about the specifics of a harmonic oscillator Where do we go from here to determine the motion of the mass near the minimumgt Well as in most classical mechanics problems we use Newton39s second law which is generally written for 1D motion as mil 9 where a is the acceleration of the object and Fquot is the net force ie sum of all the 22 forces on the object In the case at hand in which the object39s acceleration is qud q and the net force comes only from the potential energy near the minimum Eq 9 becomes 1 In physics we are often interested in comparing terms in expressions such as Eq 7 and we often use the comparators gtgt and ltlt to compare these terms These two comparators actually mean lt X and gt 10 X respectively For example a gtgt b indicates that a gt 10b 2 Exceptions can occur If the potential minimum is so at that Vquot0 0 then the third term will be zero and it will be some higheriorder terms that determines the motion near the potentialienergy minimum The object will oscillate but it will not oscillate harmonically D M Riffs 4 1162009 Lecture 2 Phys 3750 q 10 or identifying Vquot0 as the spring constant ks we can write Eq 10 as q k Sq 0 1 1 m Equation 11 is known as the equation of motion for an harmonic oscillator Generally the equation of motion for an object is the speci c application of Newton39s second law to that object Also quite generally the classical equation of motion is a differential equation such as Eq 11 As we shall shortly see Eq 11 along with the initial conditions q0 and completely specify the motion of the object near the potential energy minimum Note that two initial conditions are needed because Eq 11 is a second order equation Let39s take few seconds to classify this differential equation It is second order because the highest derivative is second order It is ordinary because the derivatives are only with respect to one variable It is homogeneous because q or its derivatives appear in every term and it is linear because q and its derivatives appear separately and linearly in each term where they appear An major consequence of the homogeneity and linearity is that linear combinations of solutions to Eq 11 are also solutions This fact will be utilized extensively throughout this course C HO Im39l z39al Vaae Pmblem The solution to Eq 11 can be written most generally as either Eq 1 Eq 2 or Eq 3 where 5 11k m Let39s consider Eq 3 qt Dcosa7tEsinaNt 3 and see that indeed the constants D and E are determined by the initial conditions By applying the initial conditions we are solving the initial value problem IVP for the HO Setting I 0 in Eq 3 we have 610 D 12 Similarly taking the time derivative of Eq 3 41 5D sin51 5E00sa7t 13 D M Riffs 5 1162009 Lecture 2 Phys 3750 and again setting I 0 gives us 40 5E 14 Hence the general solution to the undamped undriven harmonic oscillator problem can be written as 511 qocosatsmat 15 To summarize for a given set of initial conditions q0 and 40 Eq 15 is the solution to Eq 11 the harmonic oscillator equation of motion III Complex Numbers In our discussion so far all quantities are real number possibly some units such as q 3 cm However when dealing with harmonic oscillators and wave phenomena it is often useful to make use of complex numbers so let39s brie y review some facts regarding complex numbers The key definition associated with complex numbers is the square root of 1 known as 139 That is 139 4 1 From this all else follows Any complex number 2 can always be represented in the form 2 x iy 16 where x and y are both 7m numbers Common notations for the real and imaginary parts of z are xRez and yImz It is also often convenient to represent a complex number as a point in the complex plane in which the x coordinate is the real part of Z and the y coordinate is the imaginary part of Z as illustrated by the picture on the following page As this can be inferred from this picture we can also use polar coordinates r and 19 to represent a complex number as Z rcost9irsint9 rcost9isint9 17 Using the infamous Euler relation which you should never forget D M Riffs 767 1162009 Lecture 2 Phys 3750 Imaginary Axis y z x i y y r 9 I I x Real Axis x z x I y e cost9isint9 18 we see that a complex number can also be written as z re 9 1 9 The last important de nition associated with complex numbers is the complex conjugate of 2 de ned as 2 x iy 20 As is apparent in the diagram above this amounts to a re ection about the real x axis Note the following relationships 22 x Rez 21 and D M Riffs 7 1162009 Lecture 2 Phys 3750 22 x2 y2 r2 22 Note also the absolute value of a complex number which is equal to r is given by lzl r J 23 IV Complex Representations of Harmonic Oscillator Solutions Because the HO equation of motion Eq 11 is linear and homogeneous linear combinations of solutions are also solutions These linear combinations can be complex combinations For example because cosaNt and sinaNt are both solutions to Eq 11 we are not worrying about any particular initial conditions at the moment another solution to Eq 11 is the complex linear combination qt acos5tiasina7t ad 24 where a is some complex number So what is the point here Well as we shall see as we go along it is often convenient to work with complex representations of solutions to the harmonic oscillator equation of motion or to the wave equation that we will be dealing with later So what does it mean to have a complex displacement Nothing really a displacement cannot be complex it is indeed a real quantity So if we are dealing with a complex solution what do we do to get a physical real answer There are at least three approaches 1 The first approach is to up front make the solution manifestly real For example let39s say you want to work with the general complex solution 10 ae z ew 25 where a and are complex numbers You can impose the condition aquot which results in qt being real 2 Another approach is to simply work with the complex solution until you need to doing something such as impose the initial conditions Then for example if we are working with the form in Eq 25 the initial conditions might be something like Req0A Imq0 0 Req390 3 and Imq390 0 These four conditions would then determine the four unknowns the real and imaginary parts of a and and again would result in qt being real D M Riffs 8 1162009 Lecture 2 Phys 3750 3 A third approach is to simply use the real part of a solution when something like the initial conditions are needed and then impose the initial conditions Alternatively because the imaginary part of the solution is also a real number you could use the imaginary part of the solution if you were so inclined Note that these different approaches work because the harmonic oscillator equation of motion is real and linear We will not worry about the details of this at the moment but if you are interested you can read more about this on pp 9 10 of the Dr Torre s text Fowdal z39om of W409 Pve omemz Exercises 21 Show that Eqs 1 and 2 are solutions to Eq 11 the equation of motion for the harmonic oscillator 22 Assuming the form of Eq 1 for the solution to Eq 11 nd the values of B and in terms of the initial conditions q0 and 23 Assuming the form of Eq 2 for the solution to Eq 11 nd the values of B and 11 in terms of the initial conditions q0 and 90 This exercise along with Exercise 22 shows that Eqs 1 and 2 are equivalent M24 Consider the potential energy function Vq 1 cosKq where K is some positive constant a Carefully plot using a computer math program such as Mathcad this function for 7239 lt Kq lt 7239 Make sure that your axes are carefully labeled on this and all other graphs b Find the force F q associated with this potential energy function c Taylor series expand the potential energy function about the point q 0 keep all terms up to the q term d What is the effective spring constant k associated with this potential energy function e What condition on q is necessary so that the q4 term the Taylor series expansion is much smaller ltlt that the q2 term See footnote 1 on p 4 f Replot the original function Vq 1 cosKq and the harmonic approximation to this function from 7239 lt Kq lt 7239 Based on this graph for what values of Kq do you expect the harmonic approximation to be valid How does this compare to your answer in e D M Riffs 9 1162009 Lecture 2 Phys 3750 M25 The Euler relation The Euler relation 6 9 cost9isint9 can be shown to be true by comparing term by term the Taylor series of both sides of the relation Calculate the rst ve Taylor series terms about 9 0 so these will be the 60 91 92 93 9 and 95 terms of each side of the equation and show that they are equivalent 26 Write the following complex expressions in the form xiy a 58 quot b 5 413932139 l C 6 7i d 2817 16erx7r2 27 Write the expressions in Exercise 26 in the form Ae where A and 9 are both real 28 Find the real part of qt ae z e Assume that a and are complex 29 Complex solution and initial conditions a Show that qt ae E e is a solution to Eq 11 the harmonic oscillator equation of motion b Assuming that a and are complex nd a and in terms of the initial conditions q0 and 90 Do this using one of the three methods discussed on p 8 D M Riffs 710 1162009 Lecture 21 Phys 3750 Separation of Variables in Cylindrical Coordinates Overview and Motivation Today we look at separable solutions to the wave equation in cylindrical coordinates Three of the resulting ordinary differential equations are again harmonic oscillator equations but the fourth equation is our rst foray into the world of special functions in this case Bessel functions We then graphically look at some of these separable solutions Key Mathematics More separation of variables Bessel functions I CylindricalCoordinates Separable Solutions Last time we assumed a product solution qp ztRp ZzTt to the cylindric al coordinate wave equation 1 6 62g i6q i62q 62q c2 BIZ 6p2 pg p2 6W 622 which allowed us to transform Eq 1 into iT RquotiR39iigZ p We now go through a separation of variable procedure similar to that which we carried out using Cartesian coordinates in Lecture 19 The procedure here is a bit more complicated than with Cartesian coordinates because the variable p appears in the quot term However as we shall see the equation is still separable A Depemleme 074 Time As with Cartesian coordinates we can again make the argument that the rhs of Eq 2 is independent of t and so the lhs of this equation must be constant Cognizant of the fact that we are interested in solutions that oscillate in time we call this constant k2 where we are thinking of k as real and so we have i2 CZT k2 gt 3 which can be rearranged as WaHT0 D M Riffs 1 3162009 Lecture 21 Phys 3750 By now we should recognize this as the harmonic oscillator equation which has two linearly independent solutions T I Toetlm39 5 And so again we have harmonic time dependence to the separable solution1 For the present case it is simplest to assume that k can be positive or negative thus we really only need one of these solutions So we simply go with T t TOW 6 B Depemleme 074 2 Using Eq 3 the lhs of Eq 2 can replaced with k2 which gives after a small bit of rearranging 7 Notice that the rhs of Eq 7 is independent of z Thus the lhs is constant which we call a2 2 We thus have Zquot Z k2 a2 8 gt which can be rearranged as Zquotk2 a2Z0 9 which yet again is the harmonic oscillator equation The solutions are 2 gem 72 10 Although k and a can be any complex numbers let39s consider the case where both k2 and a2 are real and positive Then these solutions oscillate if k2 gt a2 exponentially grow and decay if k2 lt a2 and are constant if k2 a2 Note that here because the square root is always taken as positive we need to keep both the positive sign and negative sign solutions We can however assume that a Z 0 1 It should perhaps be obvious that this will always be the case for separable solutions to the wave equation 2 2 Why not We can call the constant anything we want It just so happens that a is convenient D M Riffs 2 3162009 Lecture 21 Phys 3750 C Depemleme 074 Using Eq 8 the lhs of Eq 7 can be replaced with a2 which gives us mi 19 11 p R p2 To separate the p and dependence this equation can be rearranged as u 2 E RquotiR39 p p2a2 12 d3 p R Because each side only depends on one independent variable both sides of this equation must be constant This gives us our third separation constant which we call n2 The equation for d3 we can then write as d3quotn2 0 13 which is again the harmonic oscillator equation The solutions to Eq 13 are Oeim4 14 Now we need to use a little physics Because we expect any physical solution to have the same value for 0 and 27239 we must have for the q solution 80 emu 1 or 1 em 1 6 Using Euler39s relation it is easy to see that Eq 16 requires that n be an integer Now because 71 can be a positive or negative integer we do not need to explicitly keep up with the q solution and so we write for the dependence on qgt0em n0i1i2 17 D Depemleme 074 p We are now left with one last equation Also setting the rhs of Eq 12 equal to n2 and doing a bit of rearranging yields D M Riffs 3 3162009 Lecture 21 Phys 3750 sz pRla2p2nZR0 This is de nitely ol the harmonic oscillator equation lts solutions are well know functions but to see what the solutions to Eq 18 are we need to put it in HstandardH form which is a form that we can look up in a book such as Ha dboo ofMal kemal z39ml Fzmcl z39om by Abramowitz and Stegun the definitive concise book on special functions To put Eq 18 in standard form we make the substitution 3 ap Now the substitution would be trivial except that because Eq 18 is a differential equation in the independent variable p we need to change the derivatives in Eq 18 to derivatives in s As usual we do this using the chain rule If we now think of Rp as a function of p through the new independent variable 3 ie as Rsp then using dR dR dR a 19a dp ds dp ds and 19b d2Ri d2R 2alRalzsalzRa2 alp2 dp ds dp dsz dp ds 6in2 6132 we can rewrite Eq 18 as S j Rquotsa2 R39sa 32 n2 Rs 0 20 a a where we now emphasize that R Rs Equation 2 obviously simplifies to 32RquotssR39ss2 n2 Rs0 21 This equation is know as Bessel39s equation lt39s two linearly independent solutions are known as Bessel functions Jns and Neumann functions Yns Sometimes the functions Jns and Yns are called Bessel functions of the rst and second kind respectively The subscript n is know as the order of the Bessel function Although one can define Bessel functions of non integer order one outcome of the d3 equation is that n is an integer so we only need deal with integer order Bessel functions for this problem D M Riffs 4 3162009 Lecture 21 Phys 3750 The following figure plots both 13 and Yquot s for several values of positive 71 1 I I 1 I I 0 5 05 Jn0s Jn1s 0 Yn0 5 Yn1S 3905 05 1 1 0 10 20 30 0 10 20 30 s 0 0 Jn2s Jn10 s Yn2s Yn10s 3905 1 I 0 10 20 30 s 5 There are several facts about Bessel functions that are worth noting some of which can be discerned from these graphs 2 J001 Jn00gt 71 0 Yns diverges as s gt0 Also notice the behavior of these functions as s gt 0 for increasing 7 the functions 13 converge more rapidly while the functions Yquot s diverge more rapidly 22 13 and Yquot s oscillate with decreasing amplitude as s gt 00 iii It is customary to define Jquot 1quotJn and Yquot 1quotYn Notice that the Bessel equation only depends upon 712 so the n solution must be essentially the same as the 71 solution D M Riffs 5 3162009 Lecture 21 Phys 3750 iv Bessel functions have the series representation LAW 21 quot0 where no Fx Jt e WI 22 0 is the Gamma function Note that for integer xgt Fx1 x xx 1x 21 p Yquot 23 More entertaining facts about Bessel functions can be found in Ha dboo of Mafvemal z39mleml z39om by Abramowitz and Stegun Now because sap the solutions to Eq 18 are thus simply Jquot ap and Ynap That is we have the two linearly independent results RiapRJJnap and RapRyYnap 242 24b II Separable Solutions Let39s put all of this analysis together and write down our separable solutions Let39s further assume that 1 we want the z dependence of the solution to oscillate or at least not either grow or decay exponentially and 2 we are only interested in solutions that remain nite as p gt0 We then have the following pieces to our separable solution Tk t Toe k 00 lt k lt 00 24a 2 Zoei mz a s k 24b qquot qgt0em n 0 1 2 24c RanpRo Map 24d D M Riffs 767 3162009 Lecture 21 Phys 3750 Puttmg all of Eq 24 together we can wnte our separable soluuons as qiakma z CiakJnaJe BWW 8 25 Let s look at some graphs of some of these mcuons For sn39nphclty we look at functtons that have no 2 dependence From Eq 25 we see that these are soluuons where k a Wlth thls constraint we can wnte Eq 25 as Irmaan amt Cmm Map 2 26 The followmg plctutes Show graphs w1th 111 of Reqnvavap z0 Jnap Re 1xaa v 270 J IPCOS 811d Re 12aa v 270 J211P0052 D MRffe 777 3162009 Lecture 21 Phys 3750 will Q s For the moment thls ends our d1scuss1on of cylmdncal Coordjnates In the next lecture we move on to studymg the wave equanon 1n sphencalrpolar Coordjnates Exercises 211 Dispersion Relation Consider Eq 25 quavkp zt C j Jnape et W 2 From this equahon 1t 1 clear that the frequency a 1s equal to ck n11 looks llke a dlsperslon relanon hut 1t 1s not really clear what k represents m tlus case Here we show that lfposlhve 1t 1s essennally the magmtude ofawave vector k kp Jrkli where the meamngs o t e umt vectors should he ohmous To do tlus we take advantage of the fact that for large s the Bessel funcnons have the asymptonc expanslons and a Usmg ether of these equatlons show that component of the wave vector In the p dJIeCtJOn 1 k a b ldennfy the 2 component kl ofthe wave vector c men show that k k3 k2 DMRiEEe 787 3162009 Lecture 21 Phys 3750 212 Linear Combinations of Bessel Functions The Bessel functions of the third kind also known as Ham 261 functions are de ned as Hi sJnsiYns and H 2 sJnsiYns 3 Using the asymptotic expansions given in Exercise 211 show for large 3 that Hilsm iexhim and Hi2sm leilkimii 7239s 7239s b Usin the results of a describe the behavior of the waves g qglylva p 2 t H lap em and qggya p 2 t H ElmaJ 6 for large p assuming that k gt 0 That is are they standing or traveling waves etc D M Riffs 9 3162009 Lecture 17 Phys 3750 Fourier Transforms and the Wave Equation Overview and Motivation We rst discuss a few features of the Fourier transform FT and then we solve the initial value problem for the wave equation using the Fourier transform Key Mathematics More Fourier transform theory especially as applied to solving the wave equation I Some Properties of the Fourier Transform In the last lecture we introduced the FT hk of a function f x through the two equations fx Imam dk 1a hk I fxequotk dx 1b Here we wish to point out a few useful properties of the Fourier transform A Tramafim The rst property has to do with translation of the function Let39s say we are interested in f xx0 which corresponds to translation of f x by x0 Then using Eq 1a we can write fx x0 fhltkgtezkbcxul dk i 2 hke kx equot dk Thus we see that the FT of fxx0 is hke kx In other words translation of fx by x0 corresponds to multiplying the FT hk by 6 B DZf I P l i l i The second property has to do with the FT of f 39x the derivative of f Again using Eq 1a we have D M Riffs 1 2232009 Lecture 17 Phys 3750 f39x jbk hke k dk 3 So we see that FT of f39x is ik hk That is differentiation of fx corresponds to multiplying hk by z39k C Im egml z39m Let39s consider the de nite integral of f x de fx j dx dk hke k 4 Switching the order of integration on the rhs produces de fx jdk hk dee 12 5 J dk kx2 6 So if we de ne If x to be the inde nite integral of f x we can rewrite Eq 5 as m 1fxl J dke m an 6 This equation tell us that integration of f x essentially corresponds to dividing the Fourier transform hk by ik 1 1 h k 1 You might think that Eq 6 could be simpli ed to Ifx T J dkge k but this cannot be done 27239 i because inde nite integration produces an undetermined integration constant The constant does not appear in Eq 6 because it is an equation for the difference of Ifx2 and Ifx1 D M Riffs 2 2232009 Lecture 17 Phys 3750 D Cowolm z39m The last property concerning the a function and its FT has to do with convolution Because you may not be familiar with convolution let39s rst de ne it Simply put the convolution of two functions f x and gx which we denote f gXx is de ned as fgx jrltx x39gtgltx39gtdx39 lt7 Perhaps the most common place that convolution arises is in spectroscopy where gx is some intrinsic spectrum that is being measured and f x is the resolution function of the spectrometer that is being used to measure the spectrum2 The convolution f gXx is then the measured spectrum Note that fgx is indeed a function of x and so we can calculate its FT which we denote f l gk Using Eq 1b we can write glkt ldxhdx39 x xvgavlew 8 Now this may not look too simple but we can change variables of integration on the x integral x x39 xquot dx dxquot Then Eq 8 becomes A 1 no no I N f gk ELM de39 fxquotgx39 em 2 9 which can be rearranged as ltfggtltkgtm WWW lt10 2 The resolution function is often quite close to a Gaussian of a particular xed width D M Riffs 3 2232009 Lecture 17 Phys 3750 Notice that Eq 10 is simply the product of the FT of fx and the FT of gx along with the factor of Denoting these FT39s as and we thus have3 respectively gt f Zgk W cl 11 So we see that the FT of the convolution is the product of the FT39s of the individual functions One way you may hear this result expressed is that convolution in real space x corresponds to multiplication in k space 11 Solution to the Wave Equation Initial Value Problem Way back in Lecture 8 we discussed the initial value problem for the wave equation 62qxt 2 62qxt atz 6 6x2 on the interval 00 lt x lt oo For the initial conditions qx0 ax 13a 6 a xo bx 13b we found that the solution to Eq 12 can be written as C 16752 qxt axct ax ctlxftbx39dx39 14 With the help of the Fourier transform we are now going to rederive this solution and along the way we will learn something very interesting about the FT of qxt We start by defining the spatial FT of qxt as no k t J qxtequotkx dx 15a 5 Although we have not done it up to this point it is fairly common practice to denote the FT ofa function such as by You will even find the practice of denoting the FT of as At least we won39t be doing that here D M Riffs 4 2232009 Lecture 17 Phys 3750 so that we also have quip Mmeka dk 15b We also de ne the FT of Eq 13 the initial conditions k0 5106 163 k0 13k 16b Now each side of the Eq 12 is a function of x and I so we can calculate the FT of both sides of Eq 12 no 2 2 6 ggf tequothdx c2 I 6 2f tequotkxdx 17 x On the lhs of this equation we can pull the time derivative outside the integral The lhs is then just the second time derivative of kt The rhs can be simpli ed by remembering that the FT of the x derivative of a function is ik times the FT of the original function Thus the FT of fizqxt6x2 is just k2 times kt the FT of qxt Thus we can rewrite Eq 17 as 1 6 fth Helm 18 This equation should look very familiar to you What equation is itgt None other than the harmonic oscillator equation What does this tell us about kt It tells us that kt for a fixed value of k oscillates harmonically at the frequency a kc Thus we can interpret the function kt as the set of normal modes coordinates for this problem This further means that the FT has decoupled the equations of motion for this system as represented by Eq 12 the wave equation Notice also that the dispersion relation a kc has also fallen into our lap by considering the FT of Eq 12 As we should know by this point the solution to Eq 18 can be written as k r Ake k Bkequotk 19 D M Riffs 5 2232009 Lecture 17 Phys 3750 where Ak and Bk are functions of k And as you should suspect these two functions are determined by the initial conditions as follows First setting I 0 in Eq 19 and using Eq 16a produces 20 AkBk 202 and calculating the time derivative of Eq 19 setting I 0 and using Eq 16b gives us aw 2k 2 We can solve Eqs 20a and 20b for Ak and Bk by taking their sum and difference which yields z39kc 2k2k 212 32 l2k W 21b 2 z39kc which gives us the solution for kt in terms of the initial conditions 2322 2k l2k 22 z 2 z39kc We are essentially done We have now expressed the FT of qxt in terms of the FT39s of the initial conditions for the problem The solution qxt is just the inverse FT of Eq 22 see Eq 15b qxt 1 1 w 2kewwgt 21 EJewljdk 23 1272 2 z39kc z39kc This is the initial value problem solution4 We can make it look exactly like Eq 14 with a little bit more manipulation To see this lets rst rewrite Eq 23 as 4 Notice that qxt as expressed in Eq 23 is the sum of two functions fx CI and gx Ct D M Riffs 767 2232009 Lecture 17 Phys 3750 mm laltkgtewwgtaltkgtewltrEmil L 8mm emu Jlalk 24 ik The rst two terms we recognize as 12axctax ct while we can use Eq 5 to recognize the second half of the rhs of Eq 24 as 12cJ itbx39dx39 Thus Eq 24 can be re eXpressed as qxt axctax ctxfbx39dx39 25 175 which is identical to Eq 14 Exercises 171 FT Properties If the FT of fx is hkgt a show that the FT of e wf is hk k0 b show that the FT of xfx is ih39k c show that the FT of x2fx is hquotk 172 Show that fgx g fx by a directly by manipulating Eq 7 the de nition of convolution b by using Eq 11 the result for the FT of f gx 173 Convolution and the Gaussian The function that has the same form as its Fourier transform is the Gaussian Speci cally if fx e39xzCr2 its FT is given by hk e39k2UZ4l Using this fact show that the convolution of two Gaussian functions f1x eixZcr 2 and f2x eixZcr22 is proportional to the Gaussian function e39xzU Zlagl Hint you need not calculate any integrals to do this problem M174 The Rectangular Pulse Consider the function f x which is a rectangular pulse of height H and width 2L centered at x 0 a Graph f x b Find f k the FT of fx D M Riffs 7 2232009 Lecture 17 Phys 3750 c Graph f d The function f f x the convolution of f x with itself is a triangle function of height 2LH2 and base 4L centered at zero Graph this function 6 Write down the functional form of f f x that you graphed in Then dimcf calculate using your functional form for f Do not set H and L to specific values 1 Graph your calculated transform f quotA f g Lastly use fkand the convolution theorem to find f l f Show that this is equal to the result in part e M175 FT Solution to the 1D Wave Equation Eq 23 qxt jakewwgt20 e kl dk 23 is the formal solution to the initial value problem a What kind of wave is described by the function expikxct traveling wave standing wave etc Be as specific as possible What kind of wave is described by the function expikx ct From a vector space point of view what are these functions b From a vector space point of view where functions are viewed as elements of a vector space what do the terms lcAzk and l51k represent 2 W 2 W C Given your answers in a and b how would you describe the solution qxt as written above Hint the term i ear combi al z39m should appear in your answer 1 How are 51k and related to the initial conditions qx0 and q39x0 That is write down expressions for 51k and in terms of qx0 and q39x0 6 Assume that the initial conditions qx0 and q39x0 are real Using your answer to d show that ampk a k and 13k13 k 1 Using the results from e you can now show that qxt is real if the initial conditions qx0 and q39x0 are both real as follows First in Eq 23 replace 51k and by 51 k and 5 k respectively Then make the change of variable k gt k dk gt dk in the integral taking care with the limits of integration Then compare Eq 23 with your new expression for qxt and notice how they are related From your comparison you should be able to conclude that qxt is real D M Riffs 8 2232009 Lecture 22 Phys 3750 Separation of Variables in Spherical Coordinates Overview and Motivation We look at separable solutions to the wave equation in one more coordinate system spherical polar coordinates These coordinates are most useful for solving problems with spherical symmetry Key Mathematics Spherical coordinates the chain rule and associated Legendre functions including Legendre polynomials I Spherical Coordinates and the Wave Equation As in the case of the cylindrical coordinates version of the wave equation our first job will be to express the Laplacian V2 in spherical coordinates rt9 which are de ned in terms of Cartesian coordinates x yz as rlx2y2 z2 1a 9 arccos 1b arctan lj 1c x The diagram at the top of the next page graphically illustrates these coordinates for the vector 1 The coordinate r is the length of 1 the coordinate 9 known as the polar angle is the angle of the vector 1 from the z axis the coordinate known as the azimuthal angle is the angle of the xy plane projection of 1 from the x axis to the y axis Notice that the coordinate is also used in cylindrical coordinates To write sz where f is some function of r 9 and in spherical coordinates we go through the same procedure as we did for cylindrical coordinates We think of f as a function of x y and 2 through the new coordinates r 9 and f f rx y z 9x y 2 x y 2 and then re express 62f62f62f V2 f 6x2 Byz 622 D M Riffs 1 3182009 Lecture 22 Phys 3750 x in terms of the new coordinates using the chain rule For example to re eXpress the x derivative term we rst use the chain rule to write izi i 3 6x 6r 6x 66 6x 6 6x Using Eq 3 we can then express the second derivative as 62f E0fii i j 4 6x2 6x 5 6x arax 666x 6 6x and then using the chain rule again we can write 62f BZfBrL 62f aayazf awakefair 6x2 arz ax 39 agar ax 39 6 6r axjax 39 6r 6x2 2 2 2 2 6f6rL6f66L6f6 at9L6f66 5 area 6x 662 6x 39 we ax ax 39 66 6x2 39 62f arl 62f 66 I azfa aga I if 62 tam ax 39 aaa ax 39 a 2 axjax 39 a 6x2 Pretty ugly ehgt Actually if you scrutinize Eq 5 you will see that there is a bit of symmetry present switching any of the spherical coordinates results in the same equation D M Riffs 2 3182009 Lecture 22 Phys 3750 Although we will not go through the rest of the procedure you should recall that there are two types of terms in Eq There are derivatives of f with respect to the new variable which remain unchanged and there are derivatives of the new variables with respect to the old variable x We must calculate these second type of derivatives and then express them in terms of the new variables using Eq If we go through this procedure for all three terms in the Laplacian and sum everything up we end up with the spherical coordinates expression for the wave equation 2 2 i6 qii Fla q i sin6i 46 q 6 c2 BIZ r2 6r 6r r2 s1nt9 66 66 r2 s1n2t9 6W II Separation of Variable in Spherical Coordinates As before we look for separable solutions to the wave equation by assuming that we can write qr t9 t as a product solution qr6 rRr 6dgt Tr 7 Substituting this into Eq 6 and dividing the result by Rr t9 Tt yields 1T 1 2 1 1r 39 1 qquot R s 6 8 c2 T rzR r r2 sint9 L r2 sin2t9 d3 A Depemleme 074 Time As with Cartesian and cylindrical coordinates we again make the argument that the rhs of Eq 8 is independent of t and so the lhs of this equation must be constant As with cylindrical coordinates we call this constant k2 where we are thinking of k as real and so we again have Tquotc2k2T0 9 which has the two linearly independent solutionsgt T t TOW 10 B Depemleme 074 Equating the rhs of Eq 8 to k2 multiplying by r2 sin2t9 and doing some rearranging of terms gives us D M Riffs 3 3182009 Lecture 22 Phys 3750 gquot 2 2 2 sin219 2 39sin19 1 q k r s1n 19 R r R s1n19 11 which separates out the dependence from r and 19 Equating the lhs of Eq 11 to the constant m2 gives us the equation for d3 1 q3quotm2qgt0 12 which yet again is the harmonic oscillator equation Equation 12 has the solutions qn qgt0em 13 Again because we require continuous solutions as a function of we must restrict m to integer values m 0 i1 i2 Note that the dependence on is exactly the same as in the cylindrical coordinates case C Depmdemg 074 19 The last variable that we will deal with today is the polar angle 19 If we now equate the rhs of Eq 11 to m2 divide by sin219 and do a bit of rearranging we end up with m2 i 2 I39 2 2 sin26 RrR kr 14 mki y 39 which separates the 19 and r variables Each side of this equation is a constant which by convention is taken to be ll1 This results in the differential equation for 99 wig sin19 39 39 11 1 8 66 0 15 This is de nitely not the harmonic oscillator equation It is however close to the standard form of another well known equation To put Eq 15 in this standard form we make the change of variables 36 cos19 We now think of D as a function of 19 through the variable 3 as D S19 and we write the derivatives of D as 1 You may ask why we do we use W2 for spherical coordinates when we used n2 for cylindrical coordinates Ihave no idea D M Riffs 4 3182009 Lecture 22 Phys 3750 6 L 6 6 2 66 6s6t9 6s 51116 6s 1 S 163 and azo a 6G 62 6s2 66 62s 662 96 6s 66 632 66 6s 662 I 16b 2 1 2 6S2 S 6s S Substituting Eq 16 into Eq 15 yields after a bit of algebra 1 32 39Is 2mm 11 11 22 s 0 17 s This equation is known as the associated Legendre equation As with all second order linear ordinary differential equations there are two linearly independent solutions These solutions are know as associated Legendre functions of the rst m 0 m 2 I I I U I I I 1 070x Pay 10 P10x P372x P20x 0 0 P42x P30x P572x P40x 10 1 I I I U I I I 1 0 5 0 5 1 1 0 5 0 0 5 1 x x m1 m 3 4 I I I U I 39 39 P33x 10039 1 473x P53x 0 1 673x 100 I I I I I I 4 uu 1 0 5 0 0 5 1 391 0 5 0 0 5 1 x x D M Riffs 5 3182009 Lecture 22 Phys 3750 and second kind which are denoted B quot s and Q1 3 respectively Usually we are interested in only the 13quot 3 solutions because the Q1 3 solutions diverge as s gt i1 The gure on the previous page plots some of the B quot s functions for various values ofl and m Notice that these functions are plotted for 1SsS1 because this corresponds to 0 S 9 S 7239 the range of the polar angle 9 The following statements summarize some key feature of the associate Legendre functions some of which are evident in the gure 2 For the Pfquot 3 solutions to Eq 17 to remain nite the parameter I must be an integer and m which is already an integer must satisfy S l You are already likely familiar with this result from quantum mechanics where the angular parts of the separable solutions of the Schrodinger in spherical coordinates are identical to the solutions here 22 For m 0 azimuthal symmetry the solutions 1313EPZO s are known as Legendre polynomials These functions are polynomials in s of order I The rst four Legendre polynomials are P0s1 1313 s Pzs3s2 1 P3s533 3s 18a 18d iii A nice simple formula for calculating the Legendre polynomials known as Rodrigues39 formula is 1 d1 sz 11 BS lt l 19 2390 For the associated Legendre functions Rodrigues39 formula generalizes to 1 m d mll Bm3 1 Szl W W32 11 20 v The rst few m 0 Legendre functions of the second kind can be written as l l l 3 2 l l 3 Q0s ln1 Q1s ln1 1 Q2Slenli 7x 21a 21c D M Riffs 767 3182009 Lecture 22 Phys 3750 m For 121 the m 0 Legendre functions of the second kind can be expressed in terms of the Legendre functions of the first kind as 1 1 1 1 Qltsgtyang Z Pltsgtaltsgt lt22 x M m The following gure plots the first few of these functions Notice that they all diverge as Isl gt 1 although because the divergence involves the logarithm function the divergence is very slow as the graph on the rhs of the figure illustrates m 0 m 0 4 I I I 0 I I I Q0X 2 39 Q0X1 5 39 39 QUIX Q17X1 Q2X 0 Q2xi4 Q3IX Q37X1 Q4X Q4xi Z 2 a I I I I I 1 05 0 05 1 098 0985 099 0995 l x x1 As with the Bessel functions more entertaining facts about associated Legendre functions can be found in Hamlboo of Mathematical Faacl z39om by Abramowitz and Stegun Now that we have some idea of the behavior of these functions we can get back to our solution of the wave equation Because 3 cost9 the solutions to Eq 15 are 5mt9 Plecost9 and 9 I9 99 Q1 cost9 23 or some linear combination of the two solutions Because they remain finite we are usually exclusively interested in solutions involving Legendre polynomials of the first kind The figure at the top of the next page plots some of the B quot cost9 functions as a function of I9 Notice that they are similar but not identical to the functions plotted on p 5 D M Riffs 7 3182009 Lecture 22 Phys 3750 e Lloyoyml P22cose 10 p1ocose P 32 6 p2ocose Cm 0 P42cose p3ocose P 52 6 p4ocose Cm 10 P11cose 2 P33cose 100 p21cose p43cose 0 p31cose 0 p53cose P41cos0 P63cos0 2 39100 Exercises 221 Calculate the change of coordinates derivatives fir6x 666x and 6 6x and express them as functions of the new variables 222 Consult the gure on p 5 For the function Pfquot s how many zero crossings of are there for 1 lt s lt 1 9 That is deduce the formula for the number of Zero crossings as a function of l and m 223 The Legendre polynomials Pf x can be used as a set of orthogonal basis functions on the interval 1 S x S 1 Using the standard definition of the inner product show that P0 P1 and P2 are all orthogonal Find normalized versions of each of these functions D M Riffs 8 3182009 Lecture 11 Phys 3750 Introduction to Fourier Series Overview and Motivation Fourier series is based on the idea that many functions of interest can be represented as a linear combination of harmonic functions Is this cool or whatgt Key Mathematics Fourier Series And some facts about integrals of odd and even functions I An Observation We have already been treading in Fourier series territory You should recall in the last lecture that we wrote the general solution qxt to the wave equation as a linear combination of normal mode solutions Well these normal mode solutions are harmonic both space and time To write another function as a linear combination of harmonic functions is the basic idea of Fourier series As a fairly simple example from last time let39s consider Eq 8 from those lecture notes which can be written as ax 1 Rean sinx 1 This is a profound equation It says that we can write the function ax which is fairly arbitrary as a linear combination of the harmonic functions sinx The price we must pay is that we need an in nite number of these functions to describe ax However as we discussed in the last lecture notes we often need only a few of these functions to accurately describe the function ax You should also recall that last time we found an equation for the coef cient Rean of each harmonic function Without such an equation Eq 1 might be theoretically interesting but it would not be of much use That equation is Rean Isinxax dx 2 As we shall see below equations such as Eqs 1 and 2 are the essence of Fourier Series theory D M Riffs 1 242009 Lecture 11 Phys 3750 Our formal discussion of Fourier series will be limited to one independent variable which we call x The variable x does not necessarily represent a spatial position however There are many cases when one is interested in using Fourier series to represent what is happening in time 11 Fourier Series Equations The theory of Fourier series starts by considering a function which we will call fx on the symmetric interval LSxSL If fx is a HgoodH function1 then we can represent f x as a linear combination of harmonic functions f x a0 2 an cOS quot sinl 3 The amplitudes an and quot are known as the Fourier coef cient of the function f There are a several things to point out here The rst is that the harmonic functions in the series have a period of 2Ln Thus each harmonic function has the periodicity 2L of the interval In fact the sum in Eq 3 includes all linearly independent harmonic functions with periodicity 2L Second the average value of an harmonic function over an interval of periodicity is Zero Thus the coef cient a0 is needed to represent functions whose average value is not zero As we shall shortly see we is the average value of the function f As mentioned above the representation of a function by a linear combination of harmonic functions isn39t that useful unless we know how to calculate the coef cients a0 an and quot Fortunately expressions for the coefficients are fairly simple and are given by do fx dx 4a a ifxcosdx 4b jfxsindx 4c 1 In typical physicist fashion we will dodge the question ofwhat exactly makes a function Hgood lfyou are interested there are plenty of text books that discuss this point including Dr Torre s text FWP D M Riffs 2 242009 Lecture 11 Phys 3750 Now you may be wondering where these equations came from but you have seen the derivation of formulae equivalent to Eq 4 several times before The last time was in the last lecture notes when we obtained Eq 2 from Eq The kg 239 to malfzp Eq 3 7 am of the kamom39cfmcl z39om am z39m egml e owr fkepmper z39m erml As an example let39s derive Eq 4b Starting with Eq 3 we multiply it by 004 notice the m and integrate from L to L which gives us no Jfxcos dx do I 004 dx 2 an I coscos dx quot J sincos dx 7L 7L quot1 5 For m 2 1 there is only one nonzero integral on the rhs of this equation L J coscos dx L 6 7139 Equation 5 thus greatly simplifies to L jfltxgtcosedx am L 7 7L which can be solved for am resulting in Eq 4b after replacing m by 71 111 Some Examples A Tria gle me z39m Let39s rst look at the function fxA LSxlt0 xA OSxSL x lt8 which is plotted in the figure on the top of the next page To use the Fourier series representation of this function we must rst calculate the Fourier coefficients using Eq Before we go ahead and try to calculate the D M Riffs 3 242009 Lecture 11 Phys 3750 391 3905 rlx 0 integrals let39s notice a few things that will make the calculations simpler First Eq 4a tells us that do is simply the average of the function f x on the interval L to L From the graph we see that this is A2 so without doing any math we have 9 Second notice that f x is an even function An even function has the property fmn x fmn Now Eq 4c is the integral of the product of this even function with the odd function sin An odd function is de ned via fwd x fm x Now the product of an odd function and an even function is an odd function and the integral of an odd function over a symmetric interval about zero such as L to L is Zero Thus again without explicitly calculating the integral in Eq 4c we have for this example 0 10 We are left with determining the coef cients an Even here things are simpler than at rst glance we can use a simplifying fact about integrals of even functions over a symmetric interval about x0 The simpli cation is that the integral of an even function over a symmetric interval is equal to twice the inte ral of the function over the positive or negative portion of the interval Now cos T is an even function and the product of two even functions is an even function With the simplifying fact and Eqs 4b and 8 we have an J fxAcosdx 11 0 D M Riffs 4 242009 Lecture 11 Phys 3750 Using Mathcad for example the integral is easily evaluated resulting in an quot32 1 cosmr 12 Using Eqs 9 10 and 12 in Eq 3 produces the Fourier representation of f1x A no 1 cosmr W f1x 3 2A Tcos7 13 As discussed in the last lecture in practice we use a truncated version of an infinite series representation such as that in Eq 13 Following that lecture we write the truncated version of Eq 13 as f1xM 2Ai big Woos 14 So where should we cut off the seriesgt To get some idea let39s graphically look at Eq 14 for several values of M As shown in the following gure the series with M 9 19 and 29 all do a reasonable job of representing the original function with the major difference being the sharpness of the peak at x 0 which is clearly visible in the rhs graph2 That the function near this point is hard to represent with an harmonic series Zoomed in on Peak f1xM A f1xM A 02 0 02 xL xL M1 M1 M9 M9 M19 M19 M29 M29 2 Although hard to see in the lhs graph there is also some rounding of the function at the ends of the interval D M Riffs 5 242009 Lecture 11 Phys 3750 isn39t surprising Because has a kink at x 0 its first derivative is unde ned there Conversely the harmonic functions that make up the Fourier series are differentiable at that point thus for any finite value of M is also be differentiable at x 0 B faytooth chl z39m Let39s finish up this introduction to Fourier series with another example This time we look at the function x A L S x lt 0 L x 15 f2 x A OSxSL which is plotted in the next gure Notice that the function is discontinuous at x 0 Because the harmonic functions are all continuous you might expect some difficulty in representing this function with a Fourier series Indeed there is a major problem as we shall shortly see 0 1 L Again we use Eq 4 to calculate the Fourier coefficients As before the function f2x has enough symmetry to make some of the calculations trivial We first note from the graph that the average value of is Zero so do 0 Notice also that f2x is odd which means f2xcos is odd and so this time an 0 Similarly f2x sin is even so we can the equation for quot simplifies to 2 L quot ZJ x Asindx 16 0 Again Mathcad can do the integral and it gives us D M Riffs 767 242009 Lecture 11 Phys 3750 quot 1cosmr 17 mt So the truncated Fourier representation of the function can be written as f2xM 2A1Lsn sin 18 717239 This function is plotted in the next figure for several values of M Notice that using even a large number of terms does not do justice to the original function The problem is due to the discontinuity as was alluded to above f2xM A 1 05 0 05 1 XL M20 M 100 M 1000 In fact something quite pathological happens near the discontinuity as illustrated in the next figure where we Zoom in on a section of the above graph As the gure clearly illustrates there is an overshoot of the truncated Fourier series and the size of the overshoot does not decrease as the number of terms increases This overshoot which is known as the Gibbs phenomenon happens whenever we try to represent a discontinuous function with a truncated Fourier series Notice that the Gibbs phenomenon also occurs for this function at the two ends of the interval This is because f2 L f2 L whereas the harmonic functions in the Fourier series all have the same value at L and L D M Riffs 7 242009 Lecture 11 Phys 3750 f2xM A 0 L 014 012 01 008 006 004 002 0 XL M 20 M 100 M 1000 Summarizing we have seen how to represent a function on a symmetric interval as a linear combination of harmonic functions that have the periodicity of that interval If the function is continuous the representation works well If the function has a discontinuity then the representation it not without its dif culties Exercises 111 Obtain Eq 4a the expression for do from Eq 112 An integral involving harmonic functions In deriving Eq 6 from Eq 5 L we used the fact that J sincosdx 0 for all integers n and m Using the trig 7 identities for sinx y and sinx y do this integral and show that this equation is indeed true 113 Odd and even functions Using the basic definitions of even and odd function fmn x fmn x and fwd x fm x show that the following statements are true a The product of two even functions is even b The product of two odd functions is even c The product of an odd function and an even function is odd D M Riffs 8 242009 Lecture 11 Phys 3750 114 Integrals of odd and even functions a If fx is an odd function show that dx 0 b If fx is an even function show that dx 2Jfltxgt dx 115 Starting with Eq 11 and using integration by parts where appropriate derive Eq 12 M116 A Fourier series example Consider the function 0 LSxlt L2 f3x A L2SxSL2 15 0 L2sst 3 Carefully graph this function b Find the Fourier coef cients of this function c Plot and the truncated Fourier expansions of for M 1 5 and 10 1 Identify all places where the Gibbs phenomenon occurs 3 gt 762 gt 117 Identify whether the following functions are odd even or neither x e erfx coshx sinhx D M Riffs 9 242009 Lecture 24 Phys 3750 Energy Density Energy Flux Total Energy in 1D Overview and Motivation From your study of waves in introductory physics you should be aware that waves can transport energy from one place to another consider the generation and detection of radio waves for example In the next two lectures we consider some of the details of the energy associated with wave phenomena To keep it initially simple we start out with one dimensional waves In the next lecture we generalize the concepts discussed here to three dimensions Key Mathematics density ux and the continuity equation 1 Density Flux and the Continuity Equation Let s start our by considering some quantity Q that has associated with it a density p Because we are interested in only one spatial dimension the density associated with Q will be so much Q per unit length If we then integrate that density between two points in space x1 and x2 then we will get the total amount of Q between the points x1 and x2 Mathematically we write this as 1C2 dwa kjph n We have explicitly included time because Q may be a dynamic quantity Often Q is a conserved quantity That is it cannot be either created or destroyed If that is the case then the change in Q within the region from x1 to X2 X2 6Qxx2r I Mm dx 2 6t 6t 1C1 must be equal to the net ow of Q into the region That is wm wl I where j is the Q current density or ux Convention is that if j gt0 then the ow is to the right and if j lt 0 then the ow is to the left Note that the units of p are Qm and the units of j are pms Qs For example if Q represents charge then Coulombm and Coulombs D M Riffs 1 3252009 Lecture 24 Phys 3750 If we now equate the rhs39s of Eqs 2 and 3 we get 1C2 Now the rhs of Eq 4 can be written as jltxrgt jltxrgt j lt5 6x I1 and so we can rewrite Eq 4 as 1C2 I W all dx 0 6 6t 6x I1 Now because the limits x1 and x2 are arbitrary the integrand must vanish This gives us an important relationship between the density and ux 0 7 Equation 7 is known as the 1D continuity equation Because the density and ux are local quantities which means that they can be de ned at each point in space as opposed to Q which is a global quantity Eq 7 is a local statement about the conservation of Q 11 Energy Density and Flux for 1D Waves Let39s now apply this discussion to the energy associated with 1D waves That is we let Q be the total energy associated with 1D waves between two points x1 and x2 To be speci c let39s think about transverse waves on a string For this particular physical system where the wave speed c is given by c W where r is the tension in the string and u is the mass density mass per unit length the energy density can be written as1 1 We do not prove this result here For its derivation we refer you to an intermediate mechanics text such CamimDjnamz39m by Marion and Thornton D M Riffs 2 3252009 Lecture 24 Phys 3750 1 0g 2 r 6g 2 at 8 pltxgt 2a 26x 0 The rst term on the rhs is the kinetic energy density p7 while the second is the potential energy density pV So we have the energy density but what about the energy ux j 9 Well whatever it is it must satisfy Eq 7 the continuity equation Let39s thus calculate BpBt and see what happens Using Eq 8 we have 2 2 0P alaq iqaq 0 9 6t 6t BIZ 6x 6t6x Comparing this with Eq 7 we see that we would like to be able to write 1 6261 6g 6261 10 y 6t 6 Tax atax as BBx of some quantity which we could then identify as the ux j To do this we can get some help from the wave equation here we use 02 ry for string waves 62g 139 Big gt 11 62 2 6x2 and the equality of mixed partial derivatives to write Eq 9 as Z Z 6 p r 6 16 1 6 16 1 0 13 6t 6t 6x2 6x Bx t which can be compacted as 6 p3 4616 97 0 14 6t x it 6x D M Riffs 3 3252009 Lecture 24 Phys 3750 We can thus identify the energy ux as 6161 xt rat 6x 15 111 Several Examples Let39s look at three examples a traveling wave a standing wave and two colliding wave packets In each example we consider the two energy densities p7 and pV and the energy ux j A Tramlag Wale To follow along at this point you will need to go to the class web site and bring up the Video le E eigy l 7D Trawll g Wamam which shows these time dependent quantities for the traveling wave qxt qo coskx ckt 16 Notice that as perhaps expected that all quantities move to the right at the wave speed 0 Furthermore the quotwavelengthquot of the energy densities and ux is half that of the displacement q For the energy densities this should be obvious from their de nitions Furthermore because for this traveling wave example it is not hard to show that j cp j looks essentially the same as the total energy density p B flamll g Wale The next Video on the web site E71er l 7D flamll g Wamam shows these same time dependent quantities for the standing wave qx t q0 sinkxcosckt 1 7 This example is a bit more interesting Notice that now the energy oscillates back and forth between kinetic and potential which now have their stationary maxima at different spatial points A careful examination of the ux shows that the energy at a given point flows one direction and then the other as it is converted between potential and kinetic C Collldl g wapepac el x The last example can be found in the Video E71er l 7D Collldl g P lremwl The displacement for this wave is given by D M Riffs 4 3252009 Lecture 24 Phys 3750 qxt qo coskx ckte39ka l2 go coskx ckte39kw Z 18 For most of the time the wave looks like two noninteracting wave packets or pulsesgt each moving at the speed c but in different directions Notice that when the pulses are far apart the behavior of the density and ux is similar to that for a traveling wave but as the pulses overlap the density and ux behave in a manner similar to a standing wave which can of course be described as the superposition of two traveling waves IV Total Energy Now that we have seen some examples illustrating the local quantities pxt and jxt let39s consider the total energy associated with wave motion In particular let39s consider transverse waves on a string on the interval OSXSL with the standard boundary conditions q0t qLt0 In that case we can write any wave on the string as a linear superposition of standing waves as2 qx t Z sinx An cosant Bquot sinant 1 9 711 where aquot mz39cL and the coefficients Aquot and Bquot depend upon the initial conditions Using Eq 1 the total kinetic and potential energies can be expressed in terms of their densities as L L m I pxtdx 5 dx 203 Vt WWW dx 20b Let39s now substitute the general form of the displacement on the rhs of in Eq 19 into Eq 20 and calculate the total kinetic and potential energies For the kinetic energy we have 2 See Lecture 10 notes D M Riffs 5 3252009 Lecture 24 Phys 3750 Tt j sinx a A sinant B cosant 21 X sinx mm Am sinamtBm cosamt dx Notice that we use different indices on the two sums so that we know which quantities go with each sum Switching the orders of integration and summation we can rewrite Eq 21 as Tt 2 an mm A sinant B cosant Am sinamt Bm cosamt n1 m1 L gtlt J sinxsinx dx 0 We can now simplify this considerably because of the orthogonality of the sine functions That is using L Isin xsin quotL x dx 5m 23 2 0 Eq 22 becomes no Tt a An sinant Bn cosant2 24 quot1 Now T t mj An sinant Bn cosant2 25 is simply the kinetic energy contained in the n th normal mode Thus Eq 24 can be simply Viewed as the sum of kinetic energies contained in all of the normal modes D M Riffs 767 3252009 Lecture 24 Phys 3750 Twine lt26 Similarly starting with Eq 20b one can show that the total potential energy can be written as mine 27 n V t TLa An cosantBn sinant2 28 is the potential energy contained in each normal mode Further using Eqs 25 and 28 it is not dif cult to show that the total energy contained in each mode En t Tn t Vquot t is equal to EltrgtwltABgt lt29 and is thus constant That the energy in each normal mode is constant is due to the fact that the normal modes do not interact Xhygt Because the equation of motion for each normal mode is independent of the other normal modes Notice also that the energy is each normal mode is proportional the square of the amplitude A2 35 Finally because En t is constant the total energy EltrgtZEltrgt lt3 is also constant Exercises 241 Energy Density and Current for 2 Traveling Wave Consider the traveling wave solution to the 1D wave equation qxt qo coskx ckt D M Riffs 7 3252009 Lecture 24 Phys 3750 a Calculate the kinetic potential and total energy densities p7 xt pV xt and pxt respectively and the energy current density jxt Show that p7 xt pV xt Further show that jxt cpxt b Does the energy current ow in the direction that you expect Explain c Show that the 1D continuity equation is satis ed by your expressions for pxt and jxt M242 Energy Density and Current for a Standing Wave Consider the particular standing wave solution to the wave equation for transverse waves on a string de ned on 0 lt x lt L qxt qo sin xjsin 3f ctj a Calculate the kinetic potential and total energy densities p7 xt pV xt and pxt respectively and the energy current density jxt b Show that the 1D continuity equation is satis ed by your expressions for pxt and jxt c Find the total kinetic energy Tt and potential energy Vt in the string 1 Show that the total energy Et TtVt is independent of time 243 Total Energy in Vibrating String a As was done for the kinetic energy in Sec IV show that the total potential energy contained in the vibrating string is given by no Vt Z a An cosantBn sinant2 quot1 b Using Eqs 25 and 28 show that the total energy En I in a normal mode is constant D M Riffs 8 3252009 Lecture 12 Phys 3750 Complex Fourier Series Overview and Motivation We continue with our discussion of Fourier series which is all about representing a function as a linear combination of harmonic functions The new wrinkle is that we now use complex forms of the harmonic functions Key Mathematics More Fourier Series And a cute trick that often comes in handy when calculating integrals I The Complex Fourier Series Last time we introduced Fourier Series and discussed writing a function f x defined on the interval L S x S L as fx a0 0 an cos quot sin 1 where the Fourier coefficients are given as 1 L 0 0 Zi dx 23 1 L mac an cosT dx 2b 1 L I mac quot I s1nT dx 2c 7L While there is nothing wrong with this description of Fourier Series it is often advantageous to use the complex representations of the sine and cosine functions 004 e L EmuL 3a sin e L mmL 3a If we insert these expressions into Eq 1 we obtain D M Riffs 1 292009 Lecture 12 Phys 3750 fx a0 Z an emuL ermmL l n emuL gamut11 n1 which can be rearranged as no fx a0 Zan i ne L an z39 n eWLl 5 quot1 This doesn39t look any simpler but notice what happens if we de ne a new set of coef cients which are simply linear combinations of the of the current coef cients an and quot c0 a0 63 Cquot gquot i n 6b ofquot an i n 6c Then we can write Eq 5 as fx c0 Z one WL cine mmL 7 n1 or even more simply as fx ZcKeWL 8 Using Eq 2 it is not hard to show that the coef cients on in Eq 6 are given by cquot fx mmL dx 9 Equations 8 and 9 are known as the complex Fourier series representation of the function f x Notice that with the complex representation there is only one expression needed for all of the Fourier coef cients D M Riffs 2 292009 Lecture 12 Phys 3750 There is another way to obtain Eq 9 which is to use the same trick that we have used several times before to find coefficients of the harmonic functions multiply Eq 8 by the proper function and integrate Let39s say we want to find the mth coefficient cm We then multiply Eq 8 by e mmL notice the minus sign in the exponent and integrate on x from L to L which produces L 0 L Jfxe L dx chJe ml L dx 10 7 quot 7 Now as before only one integral on the rhs is nonzero That is the integral with n m and its value is 2L Eq 10 thus simplifies to L Jfxe L dx cm 2L 1 1 7L which is equivalent to Eq That is pretty much it for the setup of the complex Fourier series 11 An Example Revisited Let39s look at an example that we looked at last time the triangle function xA LSxlt0 12 xA OSxSL x which is plotted on the top of the next page Let39s use Mathcad to evaluate the cquot 39s Inserting Eq 12 into Eq 9 0 L cquot i JxAequot L dxJ xAequotWL dx 13 L 0 and asking Mathcad to evaluate this expression results in c Al cosmr n 14 I12 2 D M Riffs 3 292009 Lecture 12 Phys 3750 391 3905 rlx 0 Not this is OK as long as we do not use it for more than it is worth for any nonzero value of 71 Eq 14 is perfectly ne But what about the case of n0 gt Then this expression is unde ned What this means is that we must explicitly set 71 0 in Eq 13 and reevaluate it But for n 0 we see from Eq 9 that CO is just the average value of the function which is A2 Putting this all together we can represent the function as no A 1 ltxgt3AZ Zfi7 em 15gt 0 u Now this is a valid representation of the function flx but you may be wondering about something We know that the function is real but the rhs of Eq 15 appears to have an imaginary part because e WL cosisin So what is the dealgt Well it is not too difficult to see that the imaginary part of each positive 7 term is exactly cancelled by the imaginary part of the corresponding negative n term So Eq 15 is indeed real III The Gaussian Function Let39s take another look at the Gaussian function and think a bit about representing it as a Fourier Series You should recall that the Gaussian function is defined as GEr x eixz T2 16 where 039 is known as the width parameter Let39s assume in this example that L gtgt 039 Then we have something like the following picture where we have set 039 02 and L 2 D M Riffs 4 292009 Lecture 12 Phys 3750 1 I I 3 05 U quot I I I 2 1 0 1 2 X Gaussian Let39s calculate the coefficients on Using Eq 9 we have 1 L iii0392 rmnxL dx 17 c e e quot 2L Now this integral can be expressed in terms of the error function which is the integral of the Gaussian function but the expression is pretty messy However there is an approximate solution to the integral in Eq 17 that is quite simple and very accurate as we now show For the conditions that we assumed namely L gtgt 039 the Gaussian function is nearly Zero for x 2 L Because of this we can extend the limits of integration in Eq 17 to 00 and with very little loss of accuracy we can write 1 co iii0392 rmnxL c 2L e 6 dx 1 8 We can also take advantage of the properties of integrals of odd and even functions if we write e39mmL cosnmcL isinnmcL which turns Eq 18 into c Taxi 2 cosnmcL isinnmcL dx 19 Now the Gaussian function is even so the integral of equot2 r2 isinnmcL is Zero and so we are left with D M Riffs 5 292009 Lecture 12 Phys 3750 1 a cnZi e cosnmcLdx 19 It so happens that this integral has a nice analytic solution Using the fairly well known result which you can find in any table of integrals Jedi52 cos xdx 039e39 z rz4 20 7w we can identify mrL in Eq 19 as in Eq 20 so we have for the coefficients 6 L einzzLmv 2L n 21 Now this is pretty cool as a function of n cquot is also a Gaussian and its width parameter is 2L7239039 Notice that this width parameter is inversely proportional to the width parameter 039 of the original Gaussian function GEr x e39xzcr2 1 We can now use Eq 21 in Eq 8 and represent a Gaussian function on the interval L S x S L as G x 50 w ernzZLIra39yemmL 22 U 2L quot7w Let39s look at the original Gaussian function Gax and its truncated Fourier representation M GaxM einlzLIrayemmcL 23 nrM So how many terms do we need that is how large does M need to be in Eq 23gt We can get some idea by considering what happens to the coefficients on see Eq 21 as lnl gets larger For a Gaussian function if the argument is several times larger than the width parameter then the Gaussian function is very close to zero Thus we need to choose M such that it is a few times larger than the width parameter 2L7239039 This is illustrated in the next figure where we have used M 20 which is 1 We will see later that this observation is essentially the uncertainty principle of quantum mechanics D M Riffs 767 292009 Lecture 12 Phys 3750 approximately 3X2L7r039 On the scale of this graph the truncated Fourier series certainly does a good job of representing the Gaussian function As above we have again set 039 02 and L 2 05 Gx Four Rep Gaussian Fourier Representation However there is an inherent limitation to using Fourier series to represent a nonperiodic function such as a Gaussian That limitation is illustrated in the next figure which plots the Gaussian and its Fourier series over an interval larger than L S x S L Within the interval the match is very good as we saw in the last graph but outside the interval the match is pretty lousy Xhyigt Well that is because the harmonic functions that make up the Fourier series all repeat on any interval with length 2L Thus the Fourier representation of the Gaussian function has periodicity 2L Well you might say that there is no problem here I39ll just pick a value of L that is 1 7 L 8 M a 05 0 LL 3 D U 3910 6 2 2 6 10 x Gaussian Fourier Representation D M Riffs 7 292009 Lecture 12 Phys 3750 larger than any value of x where I might want to evaluate the original function That might work in practice but we might also ask the question is there a Fourier series representation that will work for all x The answer is yes and we will discuss that in a few lectures after this one Right now I just want to point out that getting to such a representation is not at all trivial Consider the following We have represented the Gaussian as a linear combination of harmonic functions emuL If we want to use a Fourier representation for all x then somehow we must take the limit where L gt 00 What does that mean for the harmonic functions It looks like all of the harmonic functions will simply become equal to 1 which seems pretty badl There is a resolution to this dilemma but this illustrates that taking the L gt 00 limit of the Fourier series representation is somewhat nontrivial Exercises 121 Calculate the integral on the rhs of Eq 10 and show that it is nonzero only if quot7quot 122 Consider the result for the coefficient for the triangle function 0 AM which is unde ned for n 0 Use l39Hospital39s rule to show that as 391 I12 2 71 0 c gtA2 the result for co 123 Using Eq 2 in Eq 6b show that c is given by Eq ex xSO 7 M124 Fourier series example Consider the function f x e x gt 0 a Plot this function Is it odd even or neither one b Find an analytic expression for the Fourier coefficients on for this function c Let L 5 Plot the function and its truncated Fourier representation for several values of M What is the minimum reasonable value for M necessary to represent f x on this interval D M Riffs 8 292009 Lecture 30 Phys 3750 Divergence and Curl Overview and Motivation In the upcoming two lectures we will be discussing Maxwell39s equations These equations involve both the divergence and curl of two vector fields the electric eld Ert and the magnetic eld Brt Here we discuss some details of the divergence and curl Key Mathematics The aim here is to gain some insight into the physical meanings of the divergence and curl of a vector field We will also state some useful identities concerning these two quantities I Review ofquotdelquot We have already discussed the differential object V sometimes called Hdelquot In Cartesian coordinates V can be written as vigiy3z 1 We have seen that this object can operate on a scalar function or scalar field f 1 to produce the gradient of fr denoted Vfr If f is expressed as a function of Cartesian coordinates then the gradient can be written as afA ii 2 x 6x Byy 62 Vfxyx Recall that Vf1 points in the direction of the greatest change in f 1 and is perpendicular to surfaces of constant f Recall also that Vfr is a vector function or vector eld which assigns a vector to each point 1 in real space We have also previously used V on a vector function Vr to calculate the divergence of Vr denoted VVr If V is expressed as a function of Cartesian coordinates then the divergence can be written as 014xayaz WNWJ BVzbayaz 6x I 6y I 62 V Vx y z 3 Recall that VVr is a scalar function A special case is the divergence of a vector eld that is itself the gradient of a scalar function VVfr In this case the vector eld is Vf and so Eq 3 becomes D M Riffs 1 462009 Lecture 30 Phys 3750 VVfx yz 32 1 4 which in coordinate system independent notation we also write as V2 f It may not be obvious but one reason that the gradient and divergence are useful is that they are coordinate system independent quantities That is as functions of position vector 17 they produce a quantity that exists independent of the coordinate system used to calculate them However this does not mean for example that the divergence of a vector field will be have the same functional form in each coordinate system For example the consider the function that is equal to the distance from some reference point the origin say This function will have the form lxz y2 z2 when expressed in Cartesian coordinates lp2z2 when expressed in cylindrical coordinates and simply r when expressed in spherical coordinates II Interpretation of the Divergence Although we previously de ned the divergence of a vector function we did not spend much if any time on its meaning Let39s see what we can say about VVr To gain some insight into the divergence let s consider the divergence theorem I VVrdv vr srda 5 Q S In words this equation says that the volume integral over some volume Q of the divergence of a vector field Vr equals the integral over the surface enclosing that volume of the normal component of that same vector field Vr Note that sr is a unit vector field that is only de ned on the surface S and at every point on S it points outward and is normal to the surface Now this is true for any volume but let39s think about Eq 5 when the volume is infinitesimally small In fact let39s think about a cubic in nitesimal volume dv 6113 If the volume is small enough and the vector eld is not too pathological then we can consider VVr to be approximately constant within the volume On each face of the cube with area da dlz we can also consider Vr s Vi r to be constant Then Eq 5 can be rewritten approximately as VVrdl3 2V rd12 6 5 face 1 D M Riffs 2 462009 Lecture 30 Phys 3750 Or rda 7 The quantity Viylrda is usually interpreted as the ux of V or total amount of V passing through the surface with area da So what does Eq 7 tell us It tells us that the divergence of Vr is equal to the net ux of V coming out of the volume per unit volume Thus if the divergence of V is Zero then there is as much V pointing into the in nitesimal volume as is pointing out of the volume or the net ux of V through the surface is zero If the divergence of V is positive then we say that there is a source of V within the volume Conversely if the divergence of V is negative we say that there is a sink of V within the volume As we shall discuss in more detail shortly one of Maxwell39s equations is vEltrgt p 8 where Er is the electric eld and pr is the charge density With our discussion above we see that the charge density is either a source if positive or sink if negative of electric eld If we are in a region of space without any charge density then VEr 0 which tells us that for any volume there is as much E pointing inwards as outwards Let39s look at very simple example Vx y 2 V0 i 9 which is a constant vector eld that points in the x direction Draw a graphical representation of this eld I hope that at each point you drew a vector pointing in the x direction and that each vector had the same length What do you expect for the divergence Hint think about the net amount of V pointing out of a small cubical region of space and apply Eq Using Eq 3 to calculate VVxyz we see that VVxyz 0 Is this what you expected Let39s look at another example Let39s say the vector eld of interest is D M Riffs 3 462009 Lecture 30 Phys 3750 Vxyz Voxx 10 Draw a graphical representation of this vector eld the x y plane for example To what physical eld and situation might this refer Do you expect a net divergence Let calculate VVxyz Using Eq 3 we see that VVxyzT Thus in any region of space there is a net ux of V out of that region Is this what you expected III The Curl and its Interpretation Another useful rst derivative function of a vector eld Vr is known as the curl of Vr usually denoted VXVI It has this notation due to its similarity to the cross product of two vectors Indeed in Cartesian coordinates it is expressed as 6V 6V VXVxyZ 6V2 y 5H an 6V2 9 y M 2 11 By 62 62 6x 6x 6y Notice that VXVI is itself a vector function of 1 As with the gradient and divergence the curl is an object that is independent of the coordinate system that is used to calculate it An easy way to remember VXVI in Cartesian coordinates is to see that Eq 10 can be written as a quotdeterminantquot VXVxyz 12 S mw EVQJlmtltgt NV Jmz1gt So what is the interpretation of VXVI Again we appeal to a theorem of multivariable calculus Stoke39s theorem in this case which can be written as I v x Vr srda Vri crdl 13 S C In words this says that surface integral of the normal component of the curl of a vector eld Vr equals the line integral around the perimeter of that surface of the tangential component of that same vector eld Vr Again sr is a unit vector eld that is only de ned on the surface S but this time the surface is an open surface so there is no outward direction so one must arbitrarily pick one side of the surface for sr to point from Similarly To 19 is a unit vector eld that is only de ned on the curve C It is tangent to the curve C at all points along C Its direction is found D M Riffs 4 462009 Lecture 30 Phys 3750 by the Hright hand ruleH whereby one39s thumb points along the unit vector sr and ones ngers curl hum in the direction of TOG To use Stoke39s theorem to gain some insight into the curl lets consider the surface to be an in nitesimal at square area with sides of length dl Then assuming that VX Vr is approximately constant on this surface we can rewrite Eq 13 as WM dzz ZVHJrdz 14 1 Or vx Vr 61 22 rm 15 I The quantity on the rhs of Eq 15 is know as the circulation of Vr around the curve C So what does Eq 15 tell us This equation tells us that component of VXVI normal to an infinitesimal area is equal to the circulation of Vr around the perimeter of that area Well maybe that last statement is just words to you So let39s do some examples to see if we can get a feel for the curl Let39s consider the vector field that we discussed above in connection with the divergence Vxyz Vox 16 Again draw a picture of this vector field the x y plane and a then draw a small infinitesimal square loop lying in this plane What do you eXpect the curl to be You should be able to determine the 2 component of VXVI by appealing to Eq 15 Do you get zero What about the other components of VXVI How would you orient your loop to enable you to use Eq 15 to calculate these other components Do you get Zero for these components as well You can confirm that indeed VXVxyz 0 by using Eq 11 or Eq 12 That is all components of the curl of Vr are indeed Zero How ab out the vector field VxyzV0 y x 7 17 D M Riffs 5 462009 Lecture 30 Phys 3750 Can you tell whether this vector eld has Zero curl or notgt Sketch the eld again in the x y plane Does it seem to circulate aroundgt Draw a small square area sitting on the x aXis in the same plane and use Eq 14 to see what that curl is like Do the same for a square sitting on the y aXis You must think about orienting the square in three orthogonal direction in order to obtain all three components of VXVI Again we can resort to Eq 11 or 12 to calculate the curl which gives us VXVx yz 2V 2 So for this example there is a net curl in the z direction that is the same everywhere IV Some Useful Identities We will not prove them here but there are three identities that will be useful when we discuss Maxwell39s equations They are v x ltwltrgtgt o vvar0 18 V X V X Vr vv Vr v2vr 19 We have not previously considered the Laplacian V2 operating on a vector eld but looking at Eq 4 we see that in Cartesian coordinates 62V 62V 62V V2Vxyz 6x2 By 622 20 and so V2Vr is also a vector eld Exercises 301 Using Cartesian coordinates show that a V X Vf x y 2 0 for any suf ciently differentiable function f x y 2 and b VV gtlt Vx yz 0 for any suf ciently differentiable vector eld Vx yz M302 Using Cartesian coordinates show that V X V X Vr VV Vr V2Vr for any suf ciently differentiable vector eld Vx y 303 Is Stoke39s theorem applicable to a Mobius stripgt Why or why notgt D M Riffs 767 462009 Lecture 23 Phys 3750 Spherical Coordinates II A Boundary Value Problem Separation of Variables Summary Overview and Motivation We look at the fourth differential equation that arises in the separable solution to the spherical coordinates wave equation We then review the separable solutions in all three coordinate systems Cartesian cylindrical and spherical Finally we use a separable solution to nd the normal modes of a drumhe ad Key Mathematics More separation of variables spherical Bessel functions and normal modes in polar coordinates 1 Separation of Variables in Spherical Coordinates continued Last time we began our search for separable solutions to the wave equation in spherical coordinates 2 2 922262 1 13ng6q1 M 1 c2 BIZ r2 6r 6r r2 s1nt9 66 66 r2 s1n2t9 6W We assumed a solution of the form qrt9 tRr 6 Tt and then solved three ordinary differential equations which gave us the three functions Tt and 6 Because each of these differential equations is second order linear and homogeneous there are two linearly independent solutions that can be linearly combined to produce the most general form of each solution For the functions Tt and 96 the most general forms can be written as Tk I Akexkct Bkerxkct 2 m Cme m Dmequotm m0 1 21 3 9m 9 Em szcost9Fl7mQquot cost9 I 0 1 2 4 Here A C and F mgt lmgt etc are undetermined constants and Pfquot and Q are associated Legendre functions of the first and second kind respectively Let39s now look at the Rr part of qrt9 t Looking back at p 4 of the Lecture 22 notes we see that the ordinary differential equation for Rr is zz12Rr39kzrz 5 D M Riffs 1 3232009 Lecture 23 Phys 3750 which can be rewritten as rzRquot2rR39k2r2 111 0 6 As with some of the other equations that we have looked at this not quite in standard form Equation 6 can be put in standard form a manner similar to the Bessel equation that arose in cylindrical coordinates by de ning the new independent variable skr With this de nition and the chain rule Eq 6 can be transformed 1nto SZRquotS 23R39ss2 ll1Rs0 7 where we emphasize that R is now a function of the new variable S If you look back at Eq 21 of the Lecture 21 notes Bessel39s equation you will see that Eq 7 is quite similar to that equation The solutions to Eq 7 which are indeed similar to Bessel functions are known as spherical Bessel functions The spherical Bessel functions of the rst and second kind are denoted jls and y The following gure plots these functions for l 0123 jls yls 390 391 The following facts about these functions some of which can be discerned from the graphs are worth noting D M Riffs 2 3232009 Lecture 23 Phys 3750 2 53 are nite everywhere y1s diverge ass gt0 Also notice the behavior of these functions as s gt 0 for increasing 1 the functions jls converge more rapidly while the functions y1s diverge more rapidly ii The functions oscillate with decreasing amplitude as s gt oo iii The spherical Bessel functions can be written in terms of standard Bessel functions of mm39m eger order as J39zS S 97143 yzS S 1643 88 8b iv The jl s functions have the convenient integral representation1 Jcoss cost9sin 1t9 d6 9 S1 211rl 1 1 S 0 a The first three jls39s and yls39s can be represented in terms of sine and cosine functions as yewM 102 10b 10c 10d jzltsgt ljsmltsgt cosltsgt y2ltsgt lcosltsgt sinltsgt lt10egtlt10t Similar although increasingly more complicated formulae can be derived for higher order fl 3 39s and y1s 39s Notice that the formulae for y1s are identical to the formula for jl s with the changes sins gt coss and coss gt sins Let39s now get back to our solution to the wave equation Because 3 kr the solution to Eq 6 the equation for Rr can be generally written as 1 At least Eq 9 can be convenient when using a computer mathematic program such as Mathcad In fact you can use Eq 9 and Mathcad to generate the formulae for jl S in Eq 10 D M Riffs 3 3232009 Lecture 23 Phys 3750 R161 r GkJikrHkzy1kr39 11 So putting Eqs 2 3 4 and 11 together we now have the general form of the separable solution in spherical coordinates qklm 3 9 9 I Gigi11 kr Hkal krl Em sz 0056 l39 Flszm 0056l 12 XCmexm Dmerxm Akexkct Bkerxkct gt where the parameters 1 and m are speci ed by 10 1 2 and 01 2 l The parameter k is unspeci ed II Summary of Separable Solutions We previously wrote down separable solutions in the other coordinate systems but never in as general a form as Eq 12 for spherical coordinates Let39s now do this for the previous coordinate systems For cylindrical coordinates we can write the general separable solution as qakp zat GnJnapHnaYnaJl Ene w Fnequot 13 X Cw ezmz Dweamz Akexkct 13kerxkctgt where J and Y are Bessel functions The parameter 7 is speci ed by n 01 2 but the parameters a and k are unspeci ed Similarly for Cartesian coordinates we have2 7k q xyZt G 6 H 6 kx E e kxy F e W k k k k k k k 1 4 XCk exkxzDk erxkyzl4k k k exc lk kk tBk k k eixc lkbk kgz z z x y z x y Z In this equation none of the parameters kX ky or k are speci ed Some general remarks concerning Eqs 12 14 are in order First notice that all three solutions depend upon three parameters which are essentially the three independent separation constants that arise in the separation of Variables process Second the discrete parameters lm in spherical coordinates n in cylindrical coordinates are discrete because of mathematical considerations related to the coordinate system being used The other parameters k in spherical coordinates ka 2 You have seen an almost general form for Cartesian coordinates in Exercise 192 D M Riffs 4 3232009 Lecture 23 Phys 3750 in cylindrical coordinates k k k in Cartesian coordinates can take on any values x y and the solutions will still satisfy the wave equation For a given problem the physics of the situation may dictate that these parameters take on only certain values For example in Exercise 192 where you found the normal mode standing waves in a rectangular room you found that because of the boundary conditions the parameters kX ky and k could only take on the values k r k y k 2 15a 15c where the 711 39s are integers and the L1 395 are the dimensions of the room Physical considerations such as boundary conditions may also place constraints on the multiplicative factors that appear in Eqs 12 14 III A Vibrating Circular Drumhead Let39s look at another example where physics constrains some of these unspeci ed parameters In particular let39s find the normal modes of vibration of a circular drumhead First we must recognize that this is a two dimensional rather than a three dimensional problem If we were working in Cartesian coordinates we would only need x and y but because we are interested in a circular drumhead we should recognize that it might be better to work in polar coordinates with the origin at the center of the drum head Well if we were to write the 2D wave equation in polar coordinates and do separation of variables the solutions would be the same as the k2 all solutions for cylindrical coordinates in Eq 13 That is the solutions would be qltp rgt GJapHYap lt16 Let39s now apply some physics First we know that we want the displacement q to be everywhere finite So because the functions Y s diverge for 3 gt0 we must have H m 0 Also we want the displacement to be a real quantity With that in mind we explicitly write the and t parts of Eq 16 in terms of sine and cosine functions rather then the complex exponential functions3 We then have qm p t GMJK ap En cosn n sinn Za cosact a sinact 17 5 There are of course other ways to make sure that the solution is real See the discussion on p 8 of the Lecture 2 notes D M Riffs 5 3232009 Lecture 23 Phys 3750 where the quantities with tildes in Eq 17 are real multiplicative factors that can be related to the quantities without tildes in Eq 16 if necessary There is now one more bit of physics to consider the boundary condition at the edge of the drum head For simplicity we assume that this bc is qm p0 t 0 where p0 is the radius of the drumhead Applying this bc to Eq 17 then gives us 0 GM ap0 in cosn 7n sinn Za cosact a sinact 18 So how can Eq 18 be satis edgt The only nontrivial way is to require that a be such that Jnap00 Now each Bessel function J has an in nite number of discrete zeros see the gure on p 5 of Lecture 27 notes so that there are an in nite number of discrete values of a which are different for each n that will satisfy Eq 18 Unfortunately in contrast to the Cartesian coordinate example speci ed by Eq 15 there are no nice simple formulae for the zeros of the Bessel functions Nonetheless the zeros can be found numerically A n 0 NomalModex We first look at normal mode solutions speci ed by n 0 For these solutions the Bessel function JO 3 comes into play This Zero of this function are equal to 2405 5520 8654 11791 4 Thus the bc will be satis ed for o x a 02405 a 05520 a 08654 a 11791 J 0a rho 82 rho 4 The zeros are tabulated in Handkoole ofMalMmalz39mFhmlz39om by Abramowitz and Stegun where else D M Riffs 767 3232009 Lecture 23 Phys 3750 up 2 405 5 520 8 65411 791 20 This IS ulusuated m the gtaph on the ptemous page where we have set anew Nouoe that map 15 mdeed equal to zero at a an for each value of a gweh by Eq 20 Let s now look at the complete solution for n e 0 Simplifying Eq 17 we can wnte 0 hetmal modes as awn amt2quot 20 coshmagt m mm lt21 Where amp IS the x th Zero of Jns Nouoe that m Eq 21 we have changed the labelmg of the soluuon we new label the normal modes thh two mtegers the rst D M mm 777 szsznug Lecture 23 Phys 3750 0 corresponds to n and the second labels the value of a that makes the solution vanish at the boundary Notice that the only spatial variable in Eq 21 is p the dependence on is gone Snapshots of rst four normal mode solutions are shown on the previous page The class web site has animated time dependent versions of these solutions As evident in the animations and should be evident from Eq 21 these solutions are radially symmetric versions of standing waves B n 1 NomalModex For any other value of n the normal modes are found in essentially the same manner We must again satisfy the bc at the edge of the drum head which determines the values of a For n1 the Zeros of J1s are 3832 7016 10173 13325 Thus for n 1 the boundary condition is satis ed for ape 3832 7016 10175 13325 22 This is illustrated in the following figure J 1a rho a03832 a07016 a 10173 a 13325 For the complete solution we thus have q p r 6112215 cos i sin 11A71 cosa1 ct 1 sina1ct1 23 D M Riffs 8 3232009 Lecture 23 Phys 3750 Because the a dependence IS no longer mm this soluhon IS a bxt more comphcated r m H soluuons are lmeaxly mdependent For slmphclty lefs lust oonslder the solunon thh i 0wh1chlmves only the cos term Then we 1 e unWJ sham2 ma a comma who lt24 The followmg gure shows snapshots ofthxs soluhon Ammated Vaslons ate agzun Mame on the class web sxte D M mm 797 szsznug Lecture 23 Phys 3750 IV Concluding Remarks That is it for separation of variables until you run into them again in some later course or perhaps while doing some researchl Summarizing we have seen that we can obtain solutions to the wave equation in three different coordinate systems using this technique Sometimes these solutions are simple sometime they are not so simple at least in terms of functions with which you may be familiar Each solution is labeled by three independent separation constants which may or may not be constrained to certain values Because the ordinary differential equations that give us these solutions are homogeneous and linear there are also undetermined multiplicative factors associated with each part of the solution Often at least some of the separation constants and the multiplicative factors are determined by physical considerations such as imposed boundary conditions In the example that we just did we saw that the parameter a must take on discrete values in order for the boundary condition at the edge of the drumhead to be satis ed While separation of variables solutions can be interesting in their own right as in the case of the drumhead modes I39ll again remind you that they are also quite useful because they can be used to construct a basis for any solution to the wave equation much as we previously discussed for the 1D wave equation This point has been previously discussed in Sec III of the Lecture 19 notes Exercises 231 Using the de nition of s and the chain rule derive Eq 7 from Eq Hint you may wish to review the Lecture 21 notes 232 The gure on p 2 indicates that fl 0 0 Using Eq 106 show that this is indeed the case 233 The two linearly independent combinations of the spherical Bessel functions 1711 S fl 31quot lyz S and h12s J391Siy1s are know as spherical Bessel functions of the 7in 2745 Using these de nitions and Eq 10 express 1111s and 1112s in terms of the functions 6 and 6 234 Normal Modes Inside a Sphere Starting with Eq 12 nd all normal mode solutions to the wave equation inside a sphere that satisfy all of the following conditions the solutions are 1 spherically symmetric 2 finite everywhere and 3 vanish on the boundary of the sphere which has a radius of re D M Riffs 710 3232009 Lecture 6 Phys 3750 Traveling Waves Standing Waves and the Dispersion Relation Overview and Motivation We review the relationship between traveling and standing waves We then discuss a general relationship that is important in all of wave physics the relationship between oscillation frequency and wave vector which is known as the dispersion relation Key Mathematics We get some more practice with trig identities and eigenvalue problems 1 Traveling and Standing Waves A Bail De m39l z39om The simplest de nition of a 1D traveling wave is a function of the form q1xat gxct 13 or qz xt fxct 1b where c is some positive constant1 The constant c is the speed of propagation of the wave The wave in Eq 1 a propagates in the positive x direction while the wave in Eq 1b propagates in the negative xdirection Now the functions g and f in Eq 1 can essentially be any well behaved function but often we are interested in harmonic waves In this case the functions g and f in Eq 1 take on the form gm range and 23 and fxctBsin27 xctz 2b where A and B are the amplitudes and 1 are the phases and l is the wavelength of the wave Now the x and t dependent parts of the sine function arguments are often written as kxi wt and so we can identify the wave vector k as 1 As we shall see the functions in Eq 1 are the general solutions to the wave equation which we will study in short order However we shall also see when we study the Schrodinger equation that not all waves have these functional forms D M Riffs 1 1212009 Lecture 6 Phys 3750 27239 k 3 A lt gt and the angular frequency a as2 27239c 4 w A lt gt You should recall from freshman physics that the speed 6 frequency V mZn39 and wavelength A are related by V 01 So what is a standing wavegt Simply put it is the superposition ie sum of two equal amplitude equal wavelength and thus equal frequency harmonic waves that are propagating in opposite directions Using Eqs 2 3 and 4 and for simplicity setting 11 0 we can write such a wave as qs xt Asinkx wtsinkx wt 5 With a bit of trigonometry speci cally the angle addition formula for the sine function Eq 5 can be rewritten as qS xt 2A sinkxc0sat 6 So instead of being a function of kxi wt a standing wave is apmdhcl of a function of x and a function of t Equation 6 also show us how to identify the wave vector k and angular frequency a in the case of an harmonic standing wave whatever multiplies x is the wave vector and whatever multiplies t is the angular frequency In fact for nonha1monic standing waves it is probably safe to define a standing wave as a wave where all parts of the system oscillate in phase as is the case of the harmonic standing wave de ned by Eq B Comecl z39m to the Cazplea Om39llal or Pmblem Let39s now go back to the coupled oscillator problem and reconsider the nth normal mode solutions to that problem which we previously wrote as 2 Do not confuse this definition of a the angular frequency of the wave with our earlier definition of a7 1lkm that arises when discussing single or coupled harmonic oscillators It is easy to confuse the two definitions because for a single harmonic oscillator 5 is also an angular frequency D M Riffs 2 1212009 Lecture 6 Phys 3750 mmo 920 sin2 gt mm Aew3eiw lt7 N1 amstm where Qquot 25 sinmr2N1 As we mentioned last time these modes are essentially standing waves Let39s see that this is the case by writing Eq 7 in the form of Eq After we do this lets also identify the wave vector k and frequency a for the normal modes As written Eq 7 explicitly lists the motion of each individual oscillator But the nth normal mode can also be written as a function of object index j and time t as mm 131 J39Aew 361 8 where j labels the oscillator Although j is a discrete index we have included it in the argument of the normal mode function because it is the variable that labels position along the chain Equation 8 is almost in the form of Eq In fact if we take the specialized case of A 3 A where A is real then Eq 8 can be written 25 mz39 Qnjt 2As1n N100sQnt 9 This is very close except in Eq 6 the position variable is the standard distance variable x while in Eq 9 we are still using the object index j to denote position However if we de ne the equilibrium distance between nearest neighbor objects as d then we can connect x and j via x jd and so we can rewrite Eq 9 as mz39 x qnxt 2A s1n N 1 cosQnt 10 Now remembering that whatever multiplies x is the wave vector k and that whatever multiplies t is the angular frequency a we have k n i 11 N1d D M Riffs 3 1212009 Lecture 6 Phys 3750 and a Q 25sin 12 mt 2N 1 for the coupled oscillator standing waves III Dispersion Relations A De m39l z39m am fame fimple Example Simply stated a dispersion relation is the function ak for an harmonic wave For the simplest of waves where the speed of propagation c is a constant we see from Eqs 3 and 4 that their dispersion relation is simply ak ck 1 3 That is the frequency a is a linear function of the wave vector k We also see from Eq13 that the ratio mk is simply the propagation speed c As we will discuss in more detail in a later lecture the ratio mk is technically known as the phase velocity3 Now you may be thinking what is the big deal heregt Eq 13 is so simple what could be interesting about itgt Now it is simple if the phase velocity 0 is independent of k But this is usually not the case Recall for example the propagation of light in a dielectric medium such as glass where the index of refraction n 0 c where CO is the speed of light in a vacuum depends upon the wavelength and thus the wave vector4 In this case Eq 13 becomes ak quot1k 14 The dispersion relation now has the possibility of being quite interesting Another example of an interesting nonlinear dispersion relation is found in modern physics In your modern physics class you hopefully studied solutions to the Schrodinger equation The wave function that describes a free particle one with no 3 OK it should probably be called the phase speed but it isn39t Sorry even in physics all quantities are not logically named 4 The dependence of the phase velocity on wave vector leads to dispersion of light by a prism for example Thus the name quotdispersion relation D M Riffs 4 1212009 Lecture 6 Phys 3750 net force acting on it propagating in the x direction with momentum p can be written as ILL Wpx9t oekh Ma where m is the mass of the particle and h is Planck39s constant Comparing Eq 15 with Eq 5 for example we identify the wave vector P k 17 h lt gt and the frequency p2 18 a 2mh The dispersion relation is thus ak W 19 2m Notice that this dispersion relation is quadratic in the wave vector k As we will study later a nonlinear dispersion relation has profound consequences for the propagation of a localized wave often called a pulse or wave packet associated with that dispersion relation First for a nonlinear dispersion relation the propagation speed of the pulse will not be equal to the phase velocity Second a nonlinear dispersion relation typically leads to the spreading of the pulse with time This spreading is also known as dispersion B Comecl z39m to the Congled Om39llal or Pmblem So what is the dispersion relation for our coupled oscillator systemgt By combining Eqs 11 and 12 we see that we can write ak ningc 20 This is another example of a nonlinear dispersion relation D M Riffs 5 1212009 Lecture 6 Phys 3750 Let39s make some graphs of the dispersion relation for the coupled oscillator system Now Eq 20 is correct but not particularly useful for doing this That is because for the coupled oscillator problem k can only have the discrete values kn n 1 N1d 21 where n 1 2 3N are allowed see Eq 115 So we must be a bit clever here Let39s rewrite Eq 12 as akn5sin 1 N 1 25sin l N k 22 2N1N 2NlkN The rhs follows because kn cc 7 Thus nN knkN This makes Eq 12 look like the dispersion relation that we want but in constructing a graph we can simply plot a vs nN keeping in mind that nN is the same as kn kN The following graph plots akn vs knkN for N56 N 5 DISPERSION RELATION l l 0 05 1 knkN For N 5 kn is obviously a discrete variable as the graph shows but there are times when it is useful and appropriate to think of kn as a continuous variable even if it 5 Of course you already knew this because of your familiarity with standing waves on a string although in that case N 00 more on this later 6 When making a graph it is often useful to use normalized unitless quantities for the axes This makes the graph more Widely applicable D M Riffs 767 1212009 Lecture 6 Phys 3750 isn39t To see when this is the case let39s consider Eq 22 for larger N as illustrated in the next figure where the dispersion relation is plotted for N 50 The key observation here is that that the spacing between adjacent values of kn kN becomes smaller as N becomes larger In fact it is not hard to show that the relative spacing between allowed values of k is given by N 50 DISPERSION RELATION 20 lg 2 For very large N say N 1023 as one might be interested if one were modeling atoms in a solid as coupled oscillators the spacing AkkN is indeed truly negligible and one is justified in thinking of k as continuous III Interparticle Interactions and Dispersion Relations Now you may think that our model of coupled oscillators is nothing more than an exercise in classical mechanics This model however contains the essence of vibrational dynamics in solid state materials How can this be Surely the interactions between atoms in a solid are much more complicated than the quadratic potentials of a bunch of springs Yes that is true However let39s think back to the first lecture where we discussed the Taylor series expansion of an arbitrary potential energy function near a minimum We found out that if the object does not stray too far from equilibrium then the potential is effectively quadratic that is the object is pulled back towards equilibrium as if it were attached to a spring Well the same thing is true for atoms in a solid At most temperatures they never stray too far from their D M Riffs 7 1212009 Lecture 6 Phys 3750 equilibrium positions What this means is that a model where the atoms are hooked to other atoms with springs is indeed a pretty good model Thus when thinking of vibrations ie oscillations of atoms really the nuclei because the nucleus contain nearly all the mass of an atom we can thi of the nuclei as if they are attached to other nuclei with springs as the picture above suggests Now because the nuclei are essentially connected together with springs there is a set of normal modes for the system Further as with our case of couple oscillators there is a dispersion relation associated with the normal modes of vibration In fact for every wave vector which is indeed a vector k because of the 3 dimensional nature of the solid there are three normal modes because each nucleus can vibrate in three dimensional space With regards to solidestate physics the most important thing about dispersion relations is that they can be measured and thus compared with theory The figure on the next page shows both experimental the discrete points and calculated the continuous lines dispersion curves for Li For the waveevector directions of propagation shown in the graph there are two normal modes with transverse polarization and one with longitudinal polarization for each value of k although along the 100 and 111 directions the two transverse modes are degenerate have the same frequency Notice that along the 100 and 110 directions that the dispersion curves are similar to the dispersion curve for our simple 1D coupled oscillator system So what is the point Well the measured curves provide insight into the microscopic interactions between the atoms in order to theoretically calculate the dispersion curves one must know which atoms are coupled with springs and what the spring constants are for the different springs D M Riffs 787 1212009 Lecture 6 Phys 3750 5 v A A L o E 5 A T2 5 A 39 Z Ll 8 4 39 L E A T T T1 2 u 900 cm qu 0 0 05 10 05 00 05 WAVE VECTOR qm To get a better sense of how the interactions between atoms affect the dispersion curves let39s go back to our simple 1D system and modify it by adding some more springs to see how the dispersion curve is affected In particular let39s add some next nearestineighbor NNN springs in addition to the nearestineighbor NN springs that We already have This modi cation is illustrated in the following picture Of course so that We have something to play with We let the NN and NNN springs have different spring constants We continue to let the NN springs have spring constant k5 While the NNN springs have spring constant kg of course this is What We would expect Why should NN39s and NNN39s have the same interactions D M lefe 797 1212009 Lecture 6 Phys 3750 The following picture shows the dispersion curves that result when these NNN springs are added These curves are for large N and so are plotted as continuous functions Notice that the dispersion curves are quite sensitive to the value of k the spring constant for the NNN interaction Also notice that the dispersion curve in the graph on the right is similar to the dispersion curve for the longitudinal mode in Li along the 111 direction k1k39025 k1k39095 E E o o E i Y gt quot gt quot U 39G U 39 z z Lu Lu D 0 D 0 g 1 039 C7 039 a DJ 1 a M n M I LL 1 LL M M ltc ltc S S g 139 NN Interactions Only 139 NN Interactions Only ltC W NNN interactions lt 039 W NNN interactions U U 0 05 1 0 05 1 knkN knkN D M Riffs 710 1212009 Lecture 6 Phys 3750 Exercises 61 Using the appropriate trig identities derive Eq 6 from Eq 62 Under the condition Aquot Bquot A show that Eq 9 is equivalent to Eq 63 Show that the superposition sum of the two standing waves Asinkxsinat and Acoskxcosa t is a traveling wave What is the direction of propagation of this standing wavegt M64 Linear chain with NN and NNN interactions Here you will nd the normal mode frequencies for a linear chain of coupled oscillators with both NN springs k and NNN springs k39 a Write down the equation of motion for the j th oscillator for this system Consulting the picture on p 9 may be helpful b Analogous to the steps in Sec II of the Lecture 5 notes find the normal mode frequencies Qquot and show that they can be written as QM i2sinz 114M a 2 a where 5W 539W and n mrN1 c For this problem plot the dispersion curve akn vs kn kN for m 1 kg 1 and k 095 Make the plot for N 10 Check that you get the discrete version of the graph on the right at the top ofp 10 12 65 Show that the spacing between allowed wave vectors for the coupled oscillator problem is given by Eq 20 D M Riffs 711 1212009 Lecture 28 Phys 3750 A Propagating Wave Packet Group Velocity Dispersion Overview and Motivation In the last lecture we looked at a localized solution Lxt to the 1D free particle Schrodinger equation SE that corresponds to a particle moving along the x axis at a constant velocity We found an approximate solution that has two velocities associated with it the phase velocity and the group velocity However the approximate solution did not exhibit an important feature of the full solution that the localization ie width of the wave packet changes with time In this lecture we discuss this property of propagating localized solutions to the SE Key Mathematics The next term in the Taylor series expansion of the dispersion relation ak will be central in understanding how the width of the pulse changes in time We will also gain practice at looking at some complicated mathematical expressions and extracting their essential features We will do this in part by de ning normalized unitless parameters that are applicable to the problem I The FirstOrder Approximate Solution Review In the last lecture we looked at a localized propagating solution that can be described as a linear combination of traveling normal mode solutions of the form e lh lkm wx5t 1100quot no 7k7k ZUZ4 1erwkt dke e 1 25 lf Eq 1 is a solution to the SE then the dispersion relation is akth2m In order to gain some insight into Eq 1 we Taylor series expanded the dispersion relation about the average wave vector k0 associated with zxt mac a2k0a239k0k k0a2quotk0k k02 2 We then approximated ak by the rst two terms the constant and linear k terms in the expansion and obtained the approximate solution war we eilwwlzvz 3 where vph k akk is known as the phase velocity and vgrk a k is known as the group velocity The phase velocity is the speed of the normal mode solution elk quot k k while the group velocity is the speed of the envelope function e quot quotk Z r2 Because it is also the speed of the probability density function D M Riffs 1 462009 Lecture 28 Phys 3750 Wily e zl c a llknltl T2 gt it can be though of as the average speed of the particle that the SE describes However the approximate solution Eq 3 does not exhibit an important property of the exact solution the localization or width of the exact solution varies with time while the localization of the approximate solution is constant and can be described by the width parameter 039 The three Videos 3E Waagbaw el l m 3E Waaepaw el iam and 3E Wampaw el am39 illustrate the time dependent broadening of propagation wave packet solutions Notice that the narrower the initial wave packet the faster it spreads out in time This is one result from the analysis below 11 The SecondOrder Approximate Solution By also including the next term in Eq 2 the Taylor39s series expansion of the dispersion relation we obtain an approximate solution that exhibits the desired feature of a width that changes as the wave packet propagates1 Including the first three terms in the Taylor39s series expansion and substituting this into Eq 1 produces after a bit of algebra the approximate solution m llloo39 71zoknrzo39k klt no 4mm1az4W39knzz zkhkany zxt 258 odke e 5 This is exactly the same as the approximate solution that we obtained in the last lecture except for the term containingaquotk0 But notice where it appears as an additive term to 024 which controls the width of the pulse Thus we might already guess that aquotk0 will affect the width as the pulse propagates Fortunately for purposes of further analysis Eq 5 has an analytic solution M g 4 WONG 2 14 8712arctan2w k 20 2 elxw kntxrw kntza42zd kntzl 039 2a2 k0 t 6 e kulxwknvknzl e39quot 39k zl zl l lquotTB X 1 Actually all of the higheriorder terms can contribute to the broadening of the pulse However if the width parameter 039 is not too small then only the contribution of the quadratic term to the time dependent broadening needs to be considered D M Riffs 2 462009 Lecture 28 Phys 3750 OK so maybe solving the integral wasn39t so fortunate But let39s see what we can do with it First notice that compared to our previous solution we have two new exponential function terms e llZlmtmlm39lk wazl and ez kn quot 39kn lZla 2w kntlzl Notice that they are both equal to 1 when aquotk0 0 However because the exponents in both of these terms are purely imaginary they contribute nothing to the width of the wave packet as we will see below We will thus not worry about them The third exponential term we are already familiar with it is the harmonic traveling wave solution ew xiln hll that propagates at the phase velocity vph k0 Because its exponent is also purely imaginary it too does not contribute to the width of the wave packet It is the fourth exponential function term that has some new interest for us Notice that it is a Gaussian function that again travels with the group velocity a k0 but with a time dependent width 12 3 0 O39l 7 that is a minimum for t0 Notice that if aquotk00 as in the case of the wave equation then the width has no time dependence and is simply 039 2 However in the case of the SE for example aquotk0hm 0 Thus the SE wave packet has a time dependent width For large times we see from Eq 7 that 30 is approximately linear vs time EWW lt8 039 which tells us that 9 That is for long times the rate of broadening is proportional to the second derivative of the dispersion relation and inversely proportional to the width parameter 039 That is the narrower the pulse is at I 0 the faster it broadens with time as the videos above illustrated 2 Because all derivatives of the dispersion relation for the WE higher than rst order are zero Eq 3 is exact for the wave equation D M Riffs 3 462009 Lecture 28 Phys 3750 III Normalized Variables To gain some further insight into this time dependent width let39s make a graph of Et vs I But let39s be smart about the graph let39s construct the graph so that it has amberva applicability To do this we will graph m39l lem mmmlz39ged quantities that are scaled values of 3t and I So how do we normalize 3t and t to make them universal to the problem at hand The answer is in Eq 7 itself First notice that if we divide Et by 039 then we will have a unitless width that is equal to 1 at I 0 So let39s de ne a normalized width O39N as Et 039 What about the variable t Again the answer is in Eq Notice that the quantity 2aquotk0 tO392 is also unitless because its value squared is added to 1 in Eq So let39s de ne a normalized time variable 139 2aquotk0 tO392 With these two universal variables Eq 7 can be re expressed as 0N1r4 2 1 Ah much simpler The gure on the next page plots O39N L39 vs 139 on two different graphs with different scales From the graphs we can visually inspect the behavior of Eq 10 For example we see for r ltlt1 that O39Nrm 1 This means that for r ltlt1 the width is approximately constant vs time The actual time scale seconds over which this is true will of course depend upon the parameters that went into the de nition of r aquotk0 and 039 From the graph we also see that for 139 gtgt 1 O39Nrm Ir gt indicating again that the width changes linearly vs time for large negative or positive times IV Application to the Schrodinger Equation Lastly let39s consider the probability density fl1 assuming that Eq 6 describes a solution to the SE From Eq 6 we calculate Wm zygzxoo Z 2 12 eizlxrwn211a11wkuwall 0 4 2aquotk0t 1 1 Using Eq 7 the definition of 3t Eq 11 can be rewritten more compactly as O 7 my 2 E 2 y ym yOWOT e ll knyl 3 0390 D M Riffs 4 462009 Lecture 28 Phys 3750 1007 1005 q 0999 3901 005 0 005 01 7 011 E 011 Notice that the probability density retains its Gaussian shape as the wave packet propagates but with a time dependent width parameter equal to EtxE Notice that the amplitude 141 aEt that multiplies the Gaussian function is also time dependent as the width Et increases this amplitude decreases The amplitude varies with the width such that that the total probability for nding the particle anywhere along the x axis Pltrgt jwltxrgtwltxrgt lt13 remains constant in time The solution given by Eq 6 and the probability density given by Eq 11 are illustrated for both negative and positive times in the videos 3E Wavgbac e m39 3E Wapepaw el g m39 and 3E Wampaw el am39 As Eq 10 indicates the pulse becomes D M Riffs 5 462009 Lecture 28 Phys 3750 narrower as I 0 is approached and the pulse becomes broader after I 0 Notice especially in 3E wa epaw el g m that there is a time near I 0 during which the pulse width is approximately constant corresponding to 139 ltlt 1 During this time the height of zfzj which is also shown in the videos is also approximately constant Exercises 281 Equation 2 is the Taylor39s series expansion of the dispersion relation about the point k k0 For the dispersion relation appropriate to the SE nd all terms in this expansion Then argue why Eqs 5 and 6 are exact solutions as opposed to approximate solutions to the SE 282 A SE free particle a Rewrite Eq 11 the expression for the probability density with expressions for a k and aquotk that are appropriate for a wave described by the Schrodinger equation b Make several graphs at least 3 of the probability density 11 xttkn xt vs x for several different values of t The graphs should clearly illustrate the change in the width of the wave packet as the wave packet propagates For simplicity you may set R 1 and m 1 283 The Dimensionless Time Variable r a Using dimensional analysis show that the variable 139 2aquotk0 tO392 is unitless b lrl ltlt1 and lrl gtgt1 corresponds to what conditions on t 284 The gure in the notes shows that for lrl lt 01 the width of the wave packet is nearly constant Let39s apply this result to a 10 eV free electron that is being described by the SE To do this nd the value of t seconds that corresponds to r 01 Do this for values of 039 10 nm and 10 pm For these two cases how far does the electron travel in the time corresponding to r 01gt How does each of these distances compare with the respective initial widthgt 285 SE probability density a Substitute Eq 12 into Eq 13 calculate the integral and thus show that the result does not depend upon I b Generally the constant 110 in Eq 12 is chosen so that the total probability to nd the particle anywhere is equal to 1 Using your result in a nd a value for 110 that satis es this condition D M Riffs 767 462009 Lecture 14 Phys 3750 A Vector Space of Functions Overview and Motivation We rst introduce the concept of a linear operators on a vector space We then look at some more vector space examples including a space where the vectors are functions Key Mathematics More vector space math 1 Linear Operators A De m39l z39m am Example The essential nature of a linear operator is contained in its name The operator part of the name means that a linear operator A operates on any vector u the space of interest and produces another vector V in the space That is if u is a vector then V Au 1 is also vector The linear part of linear operator means that AaubVaAubAV 2 is satis ed for all scalars a and b and all vectors u and V in the space Linear operators come in many different forms The ones of interest for any given vector space depend upon the problem being solved When dealing with a finite dimension vector space we can always use standard linear algebra notation to represent the vectors as column matrices of length N and the linear operators as square matrices of size N X N 1 For N 3 for example Eq 1 can be written as C a dl 3 e o p q 0 w c 139 j k a whereV dAl m nandu b e opq c 1 When representing a vector as a column vector the elements of the vector represent the components or coordinates of the vector with respect to some assumed orthonormal basis For example if dealing with vectors in real space the elements in a column vector are the scalar components also known as Cartesian coordinates of that vectorin the K 7 2 basis D M Riffs 1 2102009 Lecture 14 Phys 3750 As we shall see below sometimes we are interested in a vector space where the vectors are functions In that case the linear operators of interest may be linear differential operators An example of a linear differential operator on a vector space of functions of x is ddx In this case Eq 1 looks like gxfx lt4 where fx and gx are vectors in the space B E ge pal e Pmbleim An important vector space problem is the eigenvalue problem We already have some experience with this problem as part of the process of nding the normal modes of the coupled oscillators Simply stated the eigenvalue problem is this for a given linear operator A what are the vectors u and scalars A such that Au All 5 is satis edgt These vectors u and scalars A are obviously special to the operator when operated on by A these vectors only change by the scale factor A These special vectors 11 are known as eigenvectors and the values of A are known as eigenvalues Each eigenvector 11 has associated with it a particular eigenvalue 2 For a vector space of N dimensions where we are using standard linear algebra notation the eigenvalues are solutions of the characteristic equation detA 21 0 8 where I is the identity matrix As we did when solving the N 2 and N 3 homework coupled oscillator problems substituting the eigenvalues one at a time back into Eq 5 then gives us the eigenvectors 11 If the finite dimension vector space is complex then Eq 8 always has solutions2 Now here is the cool thing If the operator is selfadjoint also know as Hermitian which means that its matrix elements satisfy A A then 2 its eigenvalues are real 22 its eigenvectors span the space and iii the eigenvectors with distinct eigenvalues are orthogonal 2 This result is known as the fundamental theorem of algebra D M Riffs 2 2102009 Lecture 14 Phys 3750 Thus if the operator A is self adjoint and all eigenvalues are distinct then those eigenvectors form an orthogonal basis for the space If the eigenvalues are not distinct an orthogonal basis can still be formed from the eigenvectors it just takes a little bit of work Often however the eigenvalue problem of interest is on a real vector space In this case if A is symmetric that is the matrix elements of A satisfy A A then Eq 8 will have N real solutions and again the associated eigenvectors u can be used to form a basis for the vector space A famous eigenvalue problem from quantum mechanics is none other than the time independent Schrodinger equation HlIEl 6 which is an eigenvalue problem on a vector space of functions Here the vectors are the functions zx yz the operator is the differential operator 2 2 2 2 h a 6 6 Vxyz 7 2m 6x2 Byz 622 and the eigenvalues are specific values of E 3 This is perhaps the most important equation in quantum mechanics because the normalized eigenvectors describe the spatial part of the states of the system with a definite value of energy and the eigenvalues E are the energies of those states 11 The Coupled Oscillator Problem Redux Let39s revisit the coupled oscillator problem to see how that problem fits into our discussion of vector spaces We first review the associated eigenvalue problem that we solved when finding the normal modes and then we make some remarks about the initial value problem A The Eg39ge mlm Pmblem Recall in that problem we started with N equations of motion one for each object 52qj12qjqj10 9 5 The function Vx y Z is the classical potential energy a particle ofmass m D M Riffs 3 2102009 Lecture 14 Phys 3750 j1N where qt is the time dependent displacement of the jth oscillator We then looked for normal mode solutions 61 I 6100 10 where all N objects oscillate at the same frequency Q By assuming that the solutions had the form of Eq 10 the N coupled ordinary differential equations became the N coupled algebraic equations 92 5261004 2 qoj1 0 1 1 which we rewrote as 252 a72 0 0 901 901 52 2w2 52 0 go 2 go 2 0 52 252 52 gm 92 q 12 qow qow Notice that this is exactly of the form of Eq 5 the eigenvalue problem where the vectors are N row column matrices the linear operator A is an NgtltN matrix and the eigenvalues l are the squared frequencies 92 As we previously discovered in solving that problem there are N eigenvectorsgt I H qOJ Slnlt1v1 I 71 gm Slnlt1v1 q03 Sinm 1 3 qOJV n 1Ngt and the nth eigenvector has the eigenvalue Q 4072 sin 14 D M Riffs 4 2102009 Lecture 14 Phys 3750 Notice that the eigenvalues are real as they should be for a symmetric operator Also because the eigenvalues are distinct the eigenvectors from an orthogonal basis for the space B The Im39l z39al Vazte Pmblem As part of solving the initial value problem for this system we ended up with the equation4 a0 SMU 0 N may 4130 ZReAn2 sin3 15 n1 2 00 sinfN which we needed to solve for the coef cients ReAn 2 Let39s now place the previous solution of Eq 15 for the coef cients ReAn2 within the context of the current discussion of vector spaces5 As we talked about in the last lecture if we write a vector V as a linear combination of orthogonal vectors 11quot V Zanun 16 a M 17 For the example at hand Eq 15 is equivalent to Eq 16 but in order apply Eq 17 to nd the coef cients ReAm2 in Eq 15 we need the de nition of the inner product of two vectors for this vector space For this space the inner product is de ned as 4 See Eq 19 of the Lecture 6 notes 5 Note that Eq 15 says at its most basic level that the eigenvectors Eq 13 form a basis for the space of initial displacements of the objects D M Riffs 5 2102009 Lecture 14 Phys 3750 uvu u 18 where uquot and vquot are the elements of the vectors u and V6 Notice that the elements of the row matrix in Eq 18 are the complex conjugates of the elements of 11 Of course if we are dealing with real vector then the complex conjugate is simply the element itself Note that Eq 18 can be written in more compact form as N uvZuvj 19 j1 Using the form of the inner product in Eq 19 the application of Eq 17 to the coupled oscillator problem is thus7 imquo ReltA2gtH lt20 0 J j1 III Vectors Spaces and Fourier Series The last vector space example is Fourier Series Recall the complex Fourier series representation of a function f x defined on the interval L to L is fx Zone WL 21a where the coefficients on are given by c fx EmuL dx 21b 6 The basis vectors are indeed orthogonal with this de nition of the inner product which is the standard linearialgebra definition of the inner product on a finiteidimension complex space of column vectors 7 Equation 20 is exactly Eq 13 of the Lecture 10 notes D M Riffs 767 2102009 Lecture 14 Phys 3750 If you have been paying attention to this point ie if you are still awake then you should be thinking HAh ha Equation 21 says that we can write the function fx as a linear combination of the basis functions elmL with coef cients on Looks like a vector space to me And ah ha again It seems that somehow Eq 21b is the equivalent of Eq 17 where the coefficients are expressed in terms of inner products on this spaceH But likely you are now asleep and thinking about other things But if you were awake you would be entirely correct Let39s see that this is the case The vectors in this space are indeed functions on the interval L to L and one set of basis vectors uquot is indeed the set of functions uquot xe L w lt nlt 00 So what is the inner product on this space that makes these basis vectors orthogonal You actually saw the inner product back in Lecture 12 before you knew it was an inner product so let me remind you Denoting for example a function f x as the vector f we de ne the inner product gf in this space as gaf Igxfxdx 22 Again note the complex conjugate in the de nition Also notice the similarity of Eqs 19 and 22 Using Eq 22 Eq 21 can be written in vector space notation as f cnun 23a 6 unaf 23b quot llwlln Lastly let39s revisit the idea of an orthonormal basis within the context of Fourier I l ll l1 and we series Recall a normalized or unit vector u is de ned by C2111 normalize any vector 11 V12 11 lluml Let39s find the normalized version of the basis functions unxe mL Calculating 25 uwun we have D M Riffs 7 2102009 Lecture 14 Phys 3750 L uwun Ie mmL e WL dx 2L 26 7L We can thus turn our artvogwml basis into an 07277074077774 basis by using the normalized vectors maxL A e 11 E If we now write a vector in this space as a linear combination of these normalized basis vectors f ch n 28a c w 2 then the functional expression of Eq 28 results in the Fourier series being written as f x zcnem 29a cquot j f kw m dx 29b To some the Fourier Series written as Eq 29 is more appealing because it has a certain symmetry that Eq 21 lacks Exercises r 139 j 141 Consider the operator A Kk l on a two dimensional vector space Show that for any two scalars a and b and any two vectors 11 full and V V1l that this u 2 2 operator is linear ie that it satis es Eq D M Riffs 8 2102009 Lecture 14 Phys 3750 142 Show that for any two scalars a and b and any two functions f x and gx that the differential operator idi is linear ie that it satis es Eq x 143 In solving the N 3 coupled oscillator problem we found the three eigenvectors to the associated eigenvalue problem which can be written as l l l u xE u 0 and u E Find the normalized versions fl and 1 gt Z gt 1 1 Z l l l 13 of each of these vectors M144 Consider the time independent Schrodinger equation eigenvalue problem 2 2 Hip Ell where H is the operator 5 2 kx2 This is the 1D quantum m x mechanical harmonic oscillator problem The solutions eigenvectors and eigenvalues of this problem can be written as n 012 11 x ax e mz2 where a 1lmk 2 and En h7wnha7 where 5 lkm a For n 0 the ground state show that 110 x is a solution to H 1 E 11 with the appropriate eigenvalue b Repeat part a for n 1 the first excited state c Thus show that the linear combination zx Clzjo x Cit11 x where C1 0 and C2 0 is ol an eigensolution to Hy Ell d For this vector space the inner product of two vectors V and q is de ned as pl f x 11 x dx Show that the 710 and 711 states are ortho onal g c Find the norm of the 710 state Thus construct the normalized eigenvector corresponding to this state 145 Consider two vectors written in terms of some orthonormal basis D M Riffs 9 2102009 Lecture 14 Phys 3750 N N V E vnunw E wmum quot1 m1 a Show that the inner product w V can be expressed in terms of the components N of the two vectors as wv E wZvn quot1 b What is the norm of the vector V expressed in terms of its components D M Riffs 710 2102009 Lecture 32 Phys 3750 Energy Density and the Poynting Vector Overview and Motivation We saw in the last lecture that electromagnetic waves are one consequence of Maxwell39s lVI39s equations With electromagnetic waves as with other waves there is an associated energy density and energy ux Here we introduce these electromagnetic quantities and discuss the conservation of energy in the electromagnetic elds Further we see how the expressions for the energy density and energy ux can be put into a form that is similar to expressions for the same quantities for waves on a string Key Mathematics We will gain some more practice with the HdelH operator V We will also discuss what is meant by a time averaged quantity 1 Energy Density and Energycurrent Density in EM Waves Recall from the last lecture the basic Maxwell39s equations VErt PS0quot 1 VBrt0 2 VXErt BBquot 3 y x 3a 0 jrt 080 am 4 61 As we discussed last time for prt 0 and jrt0 M39s equations imply the wave equation for both Ert and Brt We know that waves transport energy So how is the energy in an electromagnetic wave expressed Well you should have learned in your introductory physics course that the energy density contained in the electric eld is given by1 8 uzl 1 t Er t Ert Er t2 5 Typically this energy density is introduced in a discussion of the energy required to charge up a capacitor which produces an electric eld between the plates Similarly the energy density contained in the magnetic eld is given by 1 In keeping with standard EM notation we use u for the energy density and S for the energy ux D M Riffs 1 3132008 Lecture 32 Phys 3750 umgrt L3rt3rt 1 Brt2 6 2M 2 Typically this relationship is introduced in a discussion of the energy required to establish a current in a toroid which produces a magnetic eld inside the toroid Notice again that the two fundamental constants of E and M 80 and 0 appear in Eq 5 and Eq 6 respectively Thus the total energy urt contained in a region of space with both electric and magnetic elds is rtgoErtz LOBrtZ 7 Because c2 1y080 this can also be written as Brt2 8 Recall for a traveling EM wave in vacuum the electric and magnetic eld amplitudes are related by BEc Equation 8 thus shows that equal amounts of energy are contained in the electric and magnetic elds in such a wave What about the energy current density also known as the energy fluxgt Well another basic fact about electromagnetic radiation that you may or may not have learned in your introductory physics course is that the energy ux in a particular region of space is equal to srtiErtxBrt 9 As we learned in the last lecture the direction of propagation of an electromagnetic plane wave is in the direction of Ertgtlt Brt As expected Eq 9 indicates that the energy ux points in this same direction In E and M the energy ux is known as the Poynting vector convenient because it poim x in the direction of the energy ow D M Riffs 2 3132008 Lecture 32 Phys 3750 II Continuity Equation for u and g If u and g are indeed the energy and energy current densities respectively then we expect that they should be related by the continuity equation 6 r tvsrt0 10 61 Let39s see if M39s equations indeed imply Eq 10 We rst start with Eq 8 and calculate its time derivative which gives us 11 6t 0 c2 6t it In deriving Eq 11 we must remember for example that E2 is really shorthand for EE see Eq Starting with Eq 9 we can also calculate VS which gives us after using the vector identity VA X B B Vgtlt A AV X B vsiBvXE EvXB 12 We can now use M39s Eqs 3 and 4 to replace the curls in Eq 12 which produces after a bit of manipulation and the use of 1080 1 c2 gt VS1E6EB6BEj lt13 0 c2 at at Comparing Eqs 11 and 13 we see that Bu VS E39 14 at J So what happened Why didn39t M39s equations gives us Eq 10gt Well there is a very good reason The energy density u and energy current density S are densities associated with the elds only But energy can also exist in the kinetic energy of the charge density The term Ej is known as Joule heating it expresses the rate of energy transfer to the charge carriers from the elds Notice that this term only contains the electric eld because the magnetic eld can do no work on the charges The term appears with a negative sign in Eq 14 because an increase in energy of the charge carriers contributes to a decrease in energy in the elds Obviously if the D M Riffs 3 3132008 Lecture 32 Phys 3750 homogeneous M39s equations apply prt 0 and jrt 0 then Eq 10 the standard continuity equation is indeed valid III The densities u and S for an EM Plane Wave In the last lecture we looked at the plane wave solution ErtE0 coskr at 15 BrtlXE0coskr at 16 to the homogeneous Maxwell39s equations Let39s calculate u and g for these fields Substituting Eqs 15 and 16 into Eqs 8 and 9 produces 11 coskrm 2 17 and srt coskrm 21 18 respectively Comparing Eqs 17 and 18 we see that Sr t cur tl 19 The agrees with the general expectation for a traveling wave that the energy current flux j rt is related to its associated energy density p rt via j rt p5rtv gt where V is the velocity of p rt The Poynting vector expressed in Eq 18 is a space and time dependent quantity Often however often we are more interested in the time averaged value of this quantity In general the time averaged value of a periodic function with period T is given by ltAltrgtgtjAltrgtdr 20 With this definition the time averaged value of S is D M Riffs 4 3132008 Lecture 32 Phys 3750 4227 so 33 I 005krm 2 dt 21 Because the average value of any harmonic function squared is simply 12 we have SM 9702 22 On last remark about ltSrtt In the optics world ltSrtt is known as the intensity associated with the electromagnetic wave lts dot product with a normal vector to some surface gives the average power per unit area incident on that surface IV An Analogy Between Mechanical and EM Waves We previously studied the energy contained in mechanical waves In particular we looked at transverse waves on a string which have an energy density and energy current density that were essentially expressed as psx t 5l aqlx QT MT 23 201 ix Jxt r mlm 24 6t 6x As they stand these equations do not look particularly like Eqs 8 and 9 for the corresponding electromagnetic quantities The mechanical waves expressions are written in terms of derivatives of the displacement while the electromagnetic quantities are written in terms of the elds However in the theory of electricity and magnetism we can introduce a quantity known as the vector potential Art that in the absence of pt and jrt can be de ned such that it is related to the electric and magnetic elds via 6Ar t E t 25 r gt a lt gt and D M Riffs 5 3132008 Lecture 32 Phys 3750 Br t V X Ar t 26 Substituting these expressions into Eqs 8 and 9 then gives us two equations that now look quite similar to Eqs 23 and 24 rtLlwz VXArt2 27 2M 0 6t grip wxwmm 28 For a more exact analogy let39s go back to Eqs 15 and 16 the plane wave solution to M39s equations Let39s simplify things by choosing the coordinate system so that k points in the x direction E0 points along the y direction which leaves B0 to point along the z direction The electric and magnetic elds for the plane wave can then be written as ExtE0ycoskx at 29 E A Bxt z coskx at 30 c These two elds are consistent with the vector potential E A A Axt ys1nkx at Ayxty 31 a With this vector potential the time derivative and curl of A simplify to me4m zmdeAgw sodmt are VxAYampIA my 3 and D M Riffs 767 3132008 Lecture 32 Phys 3750 6A 6A a AxyxA y y 34 6t 6t ax With these last three expressions we can express u and g for our plane wave solution Eqs 31 and 32 as um 35 sxt i 36 0 6t 6x These expression are now essentially identical to Eqs 23 and 24 the analogous expressions for mechanical waves on a string if the following correspondences are made qlt gtAy and relyo Exercises 321 Show that Eqs 29 and 30 follow from Eq 31 M322 A travelingwave solution to Maxwell39s equations Consider the electric eld Er t on coskz kct a What is the corresponding magnetic eldgt b Calculate the energy density uzt associated with each of these elds c Calculate the Poynting vector Szt associated with these elds 1 Show that uztand Sztsatisfy the appropriate continuity equation D M Riffs 7 3132008 Lecture 9 Phys 3750 General Solution with Boundary Conditions Overview and Motivation Last time we wrote down the general form of the solution to the 1D wave equation We then solved the initial value problem for an in nitely long system Today we use the same form of the solution and solve the initial value problem for a nite system with boundary conditions Key Mathematics We again use chain rule for taking derivatives and utilize the Gaussian function 1 Review of the Initial Value Problem for an In nite System Last time we considered waves on a one dimensional system of in nite extent We wrote down the solution to the wave equation 62 xt 2 62 xt 21a 6 2x2 qxt fxctgx ct 2 where f and g are any well behaved functions In terms of the initial conditions qx0 ax and 6qx06t bx the functions f and g can be written as axlJbx39dx39 3a and ax bx39dx39 3b which results in the solution to the initial value problem qx t ax ct ax ct xrhx39dx39 4 16752 D M Riffs 1 1282009

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