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by: Dr. Simeon Wiza


Dr. Simeon Wiza
GPA 3.96

Stephen Ellis

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Stephen Ellis
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This 42 page Class Notes was uploaded by Dr. Simeon Wiza on Wednesday September 9, 2015. The Class Notes belongs to PHYS 227 at University of Washington taught by Stephen Ellis in Fall. Since its upload, it has received 29 views. For similar materials see /class/192436/phys-227-university-of-washington in Physics 2 at University of Washington.

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Date Created: 09/09/15
Lecture 7 Appendix B Some sample problems from Boas Here are some solutions to the sample problems assigned for Chapter 3 Sections 3 6 7 and 9 33 5 Solution We want to find the determinant of a 5x5 matrix by first simplifying the matrix using the general invariances of determinants e g adding and subtracting rows and then use the Laplace development We have 21 22 21 22 7 0 1 3 5 7 0 1 3 5 2 1 0 1 4 2 1 0 1 4 7 3 2 1 4amp5 3 0 8 8M 8 6 2 7 4 8 6 2 7 4 1 3 5 7 5 1 3 5 7 5 7 0 1 3 5 7 0 1 3 5 2 1 0 1 4 2 1 0 1 4 15 3 0 8 815 3 0 8 8 22 6 0 13 14 22 6 0 13 14 1 3 5 7 5 36 3 0 8 30 2 1 1 4 2 1 1 4 115 3 8 8 xiii 31 5 0 40 22 6 13 14 R4R48R1 48 7 0 66 36 3 8 30 52 5 0 62 31 5 40 31 5 40 148 7 66 gt48 7 665 48 61 31 40 52 5 62 21 0 22 51056 1386 7682 840 16501106 544 33 17 Solution Here we apply Cramer s rule to the equations described the Lorentz transformation of Special Relativity to find the inverse transformation We have Physics 227 Lecture 7 Appendix B 1 Autumn 2007 x yx vt y t 7t vxc2 G W02 Z x v t 7 7x39vt39 t x 7 W 721VzCz 7xv W02 2 j 7 7 x vc2 t39 7tVxCZ 2 t 7 7V 721 v2cz 7tvxC W62 2 It should no surprise that the form of the transformation is what we would obtain by boosting in the opposite direction v gt v 36 2 Solution Here we practice finding various combinations of 2 matrices We find 2 5 1 4 2 0 8 10 2 2 A 3 AB 1 3 0 2 10 46 1 2 2 4 512 6 17 BA 2 0 2 06 2 6 2 1 54 1 1 AB 10 32 1 5 21 5 4 3 9 A B 1 0 3 2 1 1 Ali 2W 5 Physics 227 Lecture 7 Appendix B 14 45 10 15 9 25 2 3 59 5 2 Autumn2007 32 1 4 1 4 10 48 1 4 X 0 2 0 2 00 04 0 4 y 2 5 10 25 1 4 3 12 1 3 5 15 0 2 0 6 We see immediately that AB 7 BA and that 3 9 1 1 12 48 A BAB 1 1 1 5 2 6 1 1 3 9 4 10 ABA B 1 5 1 1 8 14 2 2 9 25 1 4 8 29 A B 5 14 0 4 5 10 Finally we look at the various determinants We find 1 a 2 5 1 4 detA 3 6 51detB 0 2 2 2 2 1 2 6 detAdetB 1 2 and also that 2 2 6 17 detAB 2 42 2detBA 3634 1 1 detAB 1 5 5 14 detAdetB1 2 1 Taking the determinant is not a linear operation Similarly we have Physics 227 Lecture 7 Appendix B 3 Autumn 2007 10 25 det5A 5 15 150 12525 5detA5 det5A 52 detA 251 25 3 12 det3B O 6 18 0 18 3det13 6 det3B32detB9 2 18 Thus n 2 as expected for 2X2 matrices The Mathematica version of most of these steps can be found on the web page 36 22 Solution Now we use matrices to solve a set of simultaneous equations We have in standard matrix notation 4 1 1 1 x 4 1 2 1 1 y 1 3x 2y 22 5 3 2 2 z 5 x y z 2x y z So now we work out the inverse of the coefficient matriX using our new technology The pieces are 1 1 1 detM2 1 1122 14314 347112 3 2 2 7 1 4 4 0 4 4 0 C 4 1 5 CT 7 1 3 M1 7 1 3 0 3 3 1 5 3 1 5 3 It is informative to check this result Via Physics 227 Lecture 7 Appendix B 4 Autumn 2007 1 11440 MM lz 2 l l 7 l 3 3 2 2 l 5 3 47l 4l 5 0 33 l 0 0 2i 8 7 1 8 l5 03 3 0 l 0 12 l42 12 2 10 066 0 0 1 So finally we have x 4 4 0 4 l6 40 l y 2 7 1 3 1 2i 28l15 1 z 1 5 3 5 4515 2 It is easy to confirm the validity of this result by plugging back into the original equations 37 35 Solution We want to think about the following matrix as a combination of rotations and re ections First we check the determinant to nd 0 1 0 1 0 0 1 1 0 1 0 0 1 So we conclude that this transformation contains a re ection We also note that the 2X2 sub matriX in the upper vector left corner corresponds to an active 90 degree rotation about the z aXis x gt y y gt x see Eq 712 while the lower right corner means a re ection through the xy plane 2 gt z 0 l 0 x y l 0 0 y x 0 0 1 Z Z Physics 227 Lecture 7 Appendix B 5 Autumn 2007 39 4 Solution Now we think about more properties of matrices We find the transpose inverse complex conjugate and the transpose Hermitian conjugate of the specified matrix We have 0 2i 1 0 i 1 0 2i 1 A i 2 03AT2i 2 0Az 2 0 3 0 0 1 0 0 3 0 0 0 i 1 ATATAT 2i 2 0 1 0 0 Finally we construct the inverse and demonstrate its correctness with 0 2i 1 detA i 2 0 2i0 1 66 3 0 0 0 0 6 0 0 2 C 0 3 6iCT0 3 i 2 i 2 6 6i 2 0 0 2 A lz 0 3 z 6 6i 2 Checking we have Physics 227 Lecture 7 Appendix B Autumn 2007 0 2i 1 0 0 2 6 6i 6i 22 AA lzl i 2 0 0 3 i210 6 2i2i 6300 66i 2600 6 100 010 001 002 02i l 6 0 0 AM1 0 3 i i 2 0 21 3i3i 6 0 6 6 6i 2 3 0 0 6 6 12i12i6 100 010 001 398 Solution We want to verify some properties of the Hennitian conjugate a From Eq 910 in Boas we have ABT BTAT Here we go through the same argument including the conjugation We have AB AB AB kj 211213 XIX1 3 T J z Ely1 Blll jk 2 ABT BlAl b Now we apply the transpose or the Hermitian conjugate to a product of more that 2 matrices Eq 911 in Boas by using the associative property of matriX multiplication see Eq 98 in Boas We have ABCDT ABCDT BCDTAT CDT BTAT DTCTBTAT Physics 227 Lecture 7 Appendix B 7 Autumn 2007 Lecture 7 Appendix A Dirac Bra Ket Notation In Lectures 6 and 7 we have developed some facility with vector and matrix notation including an effort to be able to think in terms of general vectors and linear operators In concrete notation ie with an explicit choice of basis vectors these objects are represented by Ntuples of numbers 1 D arrays and matrices 2D arrays In this Appendix we will attempt to connect the previous discussion to a notation due to Paul Dirac that is standard in discussions of Quantum Mechanics and in some pure mathematics In QM we are typically interested in describing the state of a quantum mechanical system and the function that provides this description is called the state vector The fairly standard Dirac notation for such a vector is in terms of the ket f2 elw Al In similar notation the Hermitian conjugate vector is represented by a bra r gt x A2 and the familiar scalar product become the braket bracket expression i gtWgt A3 If we operate on the state llgt with a linear operator 0 we have a in principle new state vector 0w gt A4 and a matrix element of the operator given by MWlt gtltx0w A5 If the state vector describes an electron and we want to find the wave function of the elector as a function of its location we can do so by projecting onto a state vector corresponding to a localized position x In the Dirac notation this looks like XIV I x A6 Physics 227 Lecture 7 AppendixA 1 Autumn 2007 where w x is the desired wave function Physics 227 Lecture 7 AppendiXA 2 Autumn 2007 Lecture 4 Complex Variables I See Chapter 2 in Boas Although it is not immediately obvious an extremely important and use Jl extension of our usual study of real functions of real variables all physical quantities are real after all 7 is to consider the corresponding complex functions of a complex variable So where do complex numbers come from The underlying idea is that we want to be able to make sense of fractional powers of negative numbers Typically we rst see this issue raised in the context of quadratic equations you should commit this result to memory if you have not already done so ax2bxc0 x bibz 4ac 41 2a Thus we want to understand what happens if the discriminate the argument of the square root is less than zero b2 4610 lt 0 In particular we need a de nition of l or by extension 16 6 lt l The rst step is to give this quantity a symbol z39e a label 139 E J l the symbol j is also sometimes used typically in contexts where z39 is the electric current By de nition we have the following properties ia si2 1i3 ii41 42 Starting with two real numbers x and y we construct a complex number in the form 2 x z39y where x is called the real part and y is called the imaginary part Thus a complex number is associated with two real numbers The properties of complex numbers are very similar to the familiar properties of twodimensional vectors The algebra of complex numbers addition y subtraction and multiplication by real constants is identical to that of twodimensional vectors x y but the multiplication of two complex numbers is 39 not identical to the multiplication of two vectors as we will see As with the usual two dimensional vectors we can represent complex numbers as points in a twodimensional plane were the real and imaginary parts play the roles of components as in the gure This realization of complex numbers is called the rectangular Physics 227 Lecture 4 1 Autumn 2007 form The corresponding twodimensional plane is called the complex plane Such plots of complex numbers are often called Argand diagrams As with usual twodimensional vectors complex numbers can also be represented in cylindrical coordinates or polar form The length of the radius in this form is called the modulus or absolute value of the complex number r 2 modz The corresponding polar angle is called the phase of the complex number rzvx2y22izl xrcos6 6 tan 1zj yrsm9 43 x zrcos6isin6 This structure is illustrated in the gure to the right Using the Euler formula from the last lecture we can write the compact expression Reei6 cos6 e e cos 9is1n6 Ime 6 sin6 44 3 z rem The choice of whether to use the rectangular form or the polar form depends on the context z39e we use the representation that simpli es the discussion recall that we are lazy and smart One of the goals in this course is to learn to let the mathematics do the work for us Note that in order for the above expressions to make sense we require that Physics 227 Lecture 4 2 Autumn 2007 A 9 is dimensionless so that power series expansion for the exponential makes sense so we use radians and not degrees B 9 is not in general singlevalued Since em cos 27m 139 sin 27m l the angles 0 00 i 27m n 01 23 are all equivalent in this expression Typically we use just the smallest positive value the principal angle 0 g 9 lt 271 or that angle minus 21 but other usages can be found in literature These different choices for the angle 9 will not change the value of the cosine or sine function but will be important when we take into account fractional powers C The complex number 2 is closely related to the complex number obtained by re ection in the real x axis This re ection changes the sign of the imaginary part or equivalently changes 9 to 9 or z39 to 7139 This new complex number is denoted 2 or Z and is called the complex conjugate of z z xiy rcos6 isin6 rem 45 2x zyrcos6 isin6rcos 6isin 6r6716 Again we can represent these relations pictorially as to the right Next consider how we can manipulate complex numbers 1 We can multiply by a real constant which just multiplies the real and imaginary parts separately by the constant just like multiplication by a constant for a two dimensional vector which simply changes the absolute value but not the phase of Physics 227 Lecture 4 3 Autumn 2007 the resulting complex number 02 cx icy 2 crew 46 2 We can add or subtract complex numbers just like we add or subtract two dimensional vectors zlizzzxl ixziy1iyz 47 3 We can multiply two complex numbers needing only to be careful about the factors of z39 This process is typically simplest to consider using the polar form 2122 x1ly1x2 iy2 xixz ix1y2 y1x2 izy1y2 xixz ylyZ ix1y2 y1x2 116101 rzeiez yirzei6162 48 3 xixz y1y2 7irzcos9192 xiyz y1x2 yirz Sin91quotquot92 The real part of this expression is similar except for the minus sign to the usual twodimensional scalar product of 2 vectors while the imaginary part is analogous to the usual twodimensional vector product of 2 vectors again except for the minus sign For example consider 21 1 139 Hem4 22 2 2J i 46 3 We have the product 2122 2 26 i2 2J 45le 5464i1464 49 Note that multiplication of a complex number by it s complex conjugate yields the modulus squared z xiyx iyx2y2 r2 23922 410 which really is like the usual scalar product Likewise the product of one complex number by the complex conjugate of another complex number is again like a scalar Physics 227 Lecture 4 4 Autumn 2007 product plus 139 times a vector product but now with the usual signs 2152 x1iylx2 iy2 xlx2 i xly2 y1x2 i2y1y2 xixz y1y2 ix1y2 y1x2 411 6162 71726 4 Finally we can also divide complex numbers which again is typically simplest using the polar form r K 1 I 412 29 Z zghrzcoswl 62i3r2s1n61 62 Z Consider the speci c example 1 1H 1i22J i 2 2 i22J 2 2 2 2J i 42 42 lll i1 l 413 ew4 ei77r12 While the rst version in rectangular form requires several steps the polar version essentially takes only a single step An equation involving compleX numbers like a twodimensional vector equation means that there are really 2 equations one for the real part and one for the imaginary Part ReLHS ReRHS 414 ImLHS ImRHS Physics 227 Lecture 4 5 Autumn 2007 Complex equations or inequalities serve to de ne points lines or regions in the complex plane For example the equation 2 3 de nes a circle about the origin of radius 3 Rez gtl de nes the open region to the right of x 1 all y z 3 4139 speci es the single point at x 3 y 4 The modulus of the difference between two complex numbers 21 22 x1 x2 2 y1 y2 2 just speci es l Z Z l the d1stance between the 2 pomts 1n the complex plane 1 2 Next let us consider functions of a complex variable Take your favorite function of the real variable x and replace x by z In general the function f 2 will be complex valued fz E Xz iYz Rel 21 Xz1mfz Yz To make use of our knowledge of power series de nitions of functions from the previous lectures we need to understand powers of complex numbers This is easiest to evaluate using the polar form 415 zquot x z39yquot Vela quot We 2 rquot coan isinn9 416 The interested reader is encouraged to check this feature of the exponential explicitly eg ewe 1i0 022li0 022 12i0 4022 e Life gets more interesting for noninteger powers eg n 1 m m integer where we nd multiValued expressions Due to the fact that 9 is itself multiValued recall point B above there will always be m distinct possibly complex roots to zW As a speci c example consider 3 14 and z 1 We have Physics 227 Lecture 4 6 Autumn 2007 gin2 1 21 3214211 417 147 e17 1 39 e427 eii Z Z Y This set exhausts the 4 distinct roots note that em 14 e 3 2 z39 although only 2 of the roots are real numbers The roots are always arrayed symmetrically around a circle of radius r lzl as in the gure X A more interesting example is 1 3 Sew Zel 3 i 7 13 in 3 8 8e 3 2e Sea7 13 ZeiizrS 2ei57r3 418 These roots are illustrated in the gure at the right Y Note that in this case only one root is a real number ln 16 1 In general for the quantlty z re a there are always n roots b the roots lie evenly spaced on a circle of radius lzln c the phases of the roots are given by 9n9n 2 n9n 4 n9n2 39n ln If our original function is just a polynomial in x then the compleX form is easy to determine For an initially real in nite power series the b are real we can write Physics 227 Lecture 4 Autumn 2007 ReSz Xz ibnmosne 419 110 ImSz Yz ib rquot sian 110 For this expression to be useful both of the series must converge Again we consider rst the question of absolute convergence which treats both the real and imaginary parts simultaneously r 420 If z converges then S 2 X z and Y z all converge absolutely Note that lrcoanl S r and lrsianl S r When the complex series does not converge absolutely it is still possible for X andor Y to converge conditionally due to the alternating signs Consider the series 5 1 Z J pquot WWW 2n 421 1 1 For 2 139 12 2 2 l the series 339 diverges On the other hand the real and imaginary parts are cos l E amp A ReSiZ n 4 1mszSinm 7 422 E 1 m gM8 N Physics 227 Lecture 4 8 Autumn 2007 where we have used the facts that cos2n l7r2 0 cosmr 1 sin 2n l7r2 1quot and sinmr 0 Both ofthese series satisfy the alternating sign test test 5 and converge conditionally In then follows that the series S 2 converges absolutely everywhere inside the unit circle in the complex plane r lt l and converges conditionally at certain discrete points on the unit circle eg z ii In fact the behavior on the unit circle is even richer At the point 2 1 r 19 0 the real part diverges while the imaginary part vanishes At 2 l r 19 71 the imaginary part again vanishes but the real part converges conditionally Clearly the concept of interval of convergence that was so useful for real series of real variables is to be replaced with the concept of a circle of convergence for compleX series of compleX variables If a compleX series of a compleX variable 2 20 converges absolutely inside of a circle about the point 20 in the compleX plane 2 20 l lt R we say the radius of convergence of the series is R The behavior on the circle 2 20 R is typically complicated as in the above example analogous to what happened at the ends of the interval of convergence and generally we study the behavior on the circle pointbypoint although this is not typically required In the region inside the circle of convergence where the series converges absolutely we can apply the 4 properties of convergent power series listed in Lecture 3 for real power series to manipulate and make use of the compleX power series In particular let us consider the issue of range of convergence for the ratio of two series Consider two power series in 2 20 0 S1z and S 2 where S1z converges for 2 lt171 and S 2 converges for 2 ltR y but S 0 0 We want to consider the ratio S1 2 S 2 and so we are especially interested in the points where S 2 0 Let the point z 2 be the point closest to the origin for which S 2 0 and de ne A R3 lzl The circle of convergence z39e the radius of convergence of the ratio S1 2 S 2 is given by the minimum of the three radii R1RR3 see the gure which in general could be any one of the three Physics 227 Lecture 4 9 Autumn 2007 As a simple example consider the ratio sin zl 2 Both the numerator and denominator converge for 2 lt oo R1 R2 the denominator is just a polynomial while the denominator vanishes at z z2 1 Hence the ratio converges z39e the ratio is well de ned for 2 lt R3 1 Now we will remind ourselves of various useful compleX functionspower series eXpansions Probably the most useful is the compleX eXponential which is just like the real case n 2 3 co Z Z Z ezlz E 423 2 3 quot0n As in the real case this converges everywhere z39e the radius of convergence is in nite zl lt R 00 Thus it follows from this convergent series representation that just as for the real case d d o0 Zn 00 anl 00 Zm z 9 dz quot0 n quot1 quot10 m 2 2 Z Z 621622 1zljmj 1zzjmj l lz1 zz3zl2 2le2 zzz 424 z1z2 8 ASIDE This last result that the product of 2 eXponentials is an eXponential with eXponent equal to the sum of the two original eXponents is true as long as the two 4 77 eXponents commute 2122 E 2122 2221 0 z39e the eXponents are c numbers and not matrices As we noted in the previous chapter for an imaginary argument we have Physics 227 Lecture 4 10 Autumn 2007 e y cosyisiny 3 eriy eiyf 2 0037 isin y cosy isiny 425 iy iy iy e iy 300sy27 51H 2 2239 These last exponential expressions for the sinusoidal functions are often useful Note that they lead directly to the usual double angle formulae 1 6 7 6 7 6 511161 62Ee2612 z e2122 l e261262 ez6e7z6z e7261e26Z e7261e26Z e261e7262 e72617202 2239 4239 4239 2239 Le261 e7261e26Z e726zez6l e7261ez6Z e726Z 42 sin01cos02 cos01 sin 02 1 6 6 7 7 6 cos616225e212 z e26122 ez6z6z I e26 e726Z e726e26z e726 6262 e261e7262 e72617262 2 4 I 4 2 426 i el l e 191e262 e 192ii2ez l e726 Xe Z 162 cos01cos02 sin01sin02 For a complex eXponent a little arithmetic with the series yields 2 xi w xiyn w xn w 13 39 2 y e e Z n n m 427 210 2 e eiy e cosy z39siny Physics 227 Lecture 4 11 Autumn 2007 The real part of the exponent x tells us about the modulus of the exponential xz e y ex while the imaginary part of the eXponent de nes the phase of the exponential tan 1Ime y Reemy y A related and important function is the compleX logarithm which calls out to be studied in terms of the polar form lnz i6 Z 78 Relnz lnr 428 3 Imlnz 6 3 lnz lnr 16 ii27m n 012 Note that the logarithm is multivalued due to the y multivalued phase We will eventually represent this by de ning the logarithm to have in nitely many branches in the compleX plane see Chap 14 in Boas The principal value or principal X branch is de ned by 0 S 9 lt 271 with the socalled branch cut along the positive real aXis as indicated by the wavy line in the gure although sometimes it is useful to choose instead 7r lt 9 S 71 with the branch cut along the negative real aXis Thus the logarithms of several familiar compleX numbers are 0 mian39 429 We interpret the i27m to be describing the points on the various sheets each sheet is labeled by a value of n which are reached by passing through the branch cut We have already used this structure without mentioning the branch cuts in our discussion of fractional powers and will return to it later Now let us return to our discussion of compleX eXponentials We saw in Eq 425 that purely imaginary eXponents led to the sine and cosine functions with real Physics 227 Lecture 4 12 Autumn 2007 arguments What about sines and cosines of imaginary arguments which are related to real eXponentials We have the following de nitions 1W 40y y 7y 2 4 COSW1W Ecoshy iiy iiy y 7y 3 5 e e e e Sln1yTl 2 1 ym zisinhy 430 eiy coshy isinhy ey e y 1 2 coshzy sinh2 y 3 cosh2 y 1 sinh2 y icoshy sinhy isinhy coshy dy dy These are the hyperbolic functions which like the sinusoidal functions are de ned by power series that converge for all nite complex arguments On the other hand the behaviors of the hyperbolic functions themselves with real arguments are different from the sinusoidal functions Instead of exhibiting periodic bounded behavior the hyperbolic functions are monotonic with ranges given by Is coshy lt 00 0S M lt 00 coshy cosh y 431 00 lt smhy lt OOsinhy sinh y oo lty lt 00 We also have the following relations Physics 227 Lecture 4 13 Autumn 2007 cosz cosx iy cosxcosiy sin xsiniy cosxcoshy isinxsinhy sinz sinx iy sin xcosiy cosxsiniy s1nxcoshyzcosxs1nhy 432 em cosz iisinz cosxcoshy 1 sinhy isinxsinhy 1 coshy e cosx iisinx We can now combine our new knowledge of complex exponentials and logarithms to explore the behavior of complex powers and roots being careful to include the issue of multiValued phases Consider two complex numbers 21 x1 z39y1 rle g and 22 x2 z39y2 ride2 and evaluate zzlnzxiyx zlz eZ eKeK ezcosy3zs1ny3 x3 Rezzlnzl 2 x2an1 y2 61 i27m1 433 y3 Imzzlnzl y21nr1 x2c91i27ml Thus such an expression will in general be multiValued Let s consider some speci c examples First consider x1 l y1 lx2 14 y2 0 1 J x3 yan 1 eAanH eln 21A27rm 3 8 y3 76 i y gen7 434 2 8 X e1716 e25i6 Physics 227 Lecture4 14 Autumn 2007 z39e just the 4 roots we expected for a complex number to the real power 1A For a more interesting result consider xl 1y1 1 x2 0 y2 2 7Z39 1i21 821ln izmn 3 x3 A 4 y3 ln2 435 i1112 4 87 47 47 787 e e X e e le e So in this case there are an in nite number of distinct values of this expression An in nite number of values is the typical result whenever Imzz y2 0 or if Rezz x2 is an irrational number Finally consider an example of the full case x1 19y1 19x2 13ay2 1 1 1 27 i2 1iii eiiln1i eiiln ir2mn 3 x3 II A 7m y372lni27r 1 436 2 eeifh gtlt 2 gtlt e4 62 1872 64 e27r Note that the multiple values of y3 yield only 3 distinct phases while there are an in nite number of overall magnitudes To nish this particular discussion we raise some nal warnings about the care needed when dealing with complex numbers Consider the product involving 3 complex numbers 21 sz which you might expect could be rewritten as 22 Z1 Z 7 22 hizzz 123 ZIZ 32612 3 437 If we just use our de nitions and keep all multivalued variables we nd the logarithms of the two sides of the equation have the forms Physics 227 Lecture 4 15 Autumn 2007 lnRHS 22 23lnr1 191 i27rin lnLHS 22 1an 191 i27z39in1 23 1an 191 i27z39in2 438 Thus the lefthandside has more values than the righthandside z39e the cases when 2 n1 n2 i n The same wamlng app11es to express1ons 11ke 21232 7 2122 3 and z 23 ZZZ 39 39 211 21 As a spec1 c example cons1der ii 27 01 ei izil in if eilni eiW m e m 439 14quot eilni eiili2 nll2 m2 e yiz rmie rIZI The rst expression and the last expression agree only for n2 0 You must use care for such expressions Typically in physics applications the situation will help de ne how to proceed z39e which of the possible values contribute Actually this issue is not unfamiliar Consider what happens when we invert an exponential by taking a logarithm In the complex variable world we must use In ez z i Zm39n which leads to the questions above z39e new values appear that were not present when we lived just on the real axis Similarly for other inversions we have 7221 6 21 e 2 2 440 321 ilnz2 i lzzz 1i27m 1 21 cos z2 lt2 and 1 e zls1n 22gt z2 21 321 ilniz2 i lzzz li27m Physics 227 Lecture 4 16 Autumn 2007 441 where the principal value is the logarithm alone familiar from real analysis Physics 227 Lecture 4 17 Autumn 2007 Lecture 1 Introduction The Big Picture Let s begin with a brief discussion of what this course is about As you strengthen your understanding of physics it is imperative that you master the mathematical tools necessary to make full and quantitative use of that knowledge At the core of this goal is the desire to be able to solve partial differential equations subject to a variety of boundary conditions Such equations describe systems of interest in classical mechanics quantum mechanics and electromagnetism and in other subject areas that we will not discuss A particularly important example which we will address multiple times in various contexts is the harmonic oscillator familiar from freshman physics For the case of a mass m on a spring described by spring constant k Newton s law for the ldimensional motion of the mass with coordinate xt is maFgtm kx 11 The minus sign on the righthandside tells us that the spring is supplying a restoring force and the system oscillates near the origin x 0 with natural frequency mo 2 km This suggests that sinusoidal time dependence plays an important role and that we can write the time dependence of the location of the mass in the form xt x0 cosa0t p0 Note in particular the appearance of 2 arbitrary constants x0 and p0 These allow us to fit arbitrary initial conditions for x0 and x0 dxdtt0 Note that 550 dzxdtz by the differential equation We will eventually spend some time reviewing Fourier expansion techniques To obtain this simple harmonic solution with a single frequency it is essential that the spring is in the linear regime force proportional to displacement which will be a good approximation for small enough magnitude of the displacement x For larger displacement of the mass the nonlinear response of the spring terms proportional to x2 x3 and higher powers may become relevant and eventually the spring breaks Approximate and exact power series expansions are very useful tools and we will review the subject of series first this quarter H and all higher derivatives are specified If our mass also experiences velocity dependent or viscous damping Newton s equation now looks like m56bxkx0 12 The corresponding frequency is complex Physics 227 Lecture 1 1 Autumn 2007 i 13 where the imaginary part b2m leads to the damping of the motion xt 0C 6717mm To see where this result comes from consider trying the Ansatz a sophisticated guess xt Recem with 3 a complex constant Substituting this expression in Eq 12 yields the result in compleX notation 1a2 m ia2b 161 2 0 gt co2m ia2b k 0 i at 0 14 Note that our Ansatz has served to convert a di rential equation into an algebraic one Note also the essential role of the eXponential form with trivial derivative deafd1 616 quot The solution of the algebraic equation is just the result in Eq 13 Substituting back into our Ansatz and taking the real part define 2 xoeiq 0 yields Rexoeieiylwgb4mz ib2mt eribtzm COS b4m2t p0 This result clearly suggests that compleX numbers are a very useful concept Do not be concerned if the manipulations we just performed do not look familiar We will review them shortly after looking at series Finally we also want to be able to analyze a driven harmonic oscillator m56bxkxFOt 16 which we will be able to accomplish in complete generality using Fourier and Laplace techniques to describe the time dependence of the driving force along with the concept of linear superposition Since nearly any system near equilibrium ie for small oscillations around equilibrium we are assuming a stable equilibrium can be described by a system of coupled harmonic oscillators characterized by uncoupled normal modes our understanding of the above equations will be of very general applicability For this Physics 227 Lecture 1 2 Autumn 2007 reason we will also review the solution of multiple coupled equations Via algebraic matrix techniques ie the normal modeeigenfunction analysis Note that the above equations are linear a single power of the variable x and either homogeneous a single power of x or its derivatives in every term as in Eqs 11 and 12 or inhomogeneous as in 16 the nonzero righthandside We will want to generalize such equationssystems to more than one spatial dimension which will require that we review vector calculus and partial differential equations We will also consider nonlinear systems like our friend the UV potential of gravity and EampM with a 1 r2 force We will advance from Newton s description of mechanical systems to the more general formalism of Lagrange including ideas from the calculus of variations The concept of symmetries transfomiations that leave the physics unchanged will play a central role and lead us to consider the ideas of group theory Note also the important idea that once we have worked out the solution of the l r problem in mechanics we can carry the structure of that solution over to the analysis of EampM only the names of the functions will change In the words of Richard P Feynman The same equations have the same solutions Throughout the quarters we will focus on practicing with these concepts on nontrivial problems in the homework and on the exams It is essential that we become proficient solving quantitative exercises that involve multiple steps This situation is to be contrasted to what you experienced in your introductory physics course where many exercises required only the single step of matching the problem description to an equation in the book Since we all make mistakes conceptual arithmetic and transcription it is essential that we develop good selfchecking habits As we perform each step in an analysis we should ask ourselves if the result makes sense39 are the units correct is the sign correct etc Note the simple but remarkable constraint that equations must have the same units on the two sides of the sign We will attempt to master mathematical tools that are by construction both simple and selfchecking so that the math tools themselves do all the hard work We want to be both lazy and smart This will be our central motto We will start working towards the above goals by considering two basic building blocks of mathematical analysis the properties of series in the next Lecture and those of complex variables in Lecture 3 We will use the tools inherent in these two topics for the rest of the year Physics 227 Lecture 1 3 Autumn 2007 Lecture 10 Appendix B Some sample problems from Boas Here are some solutions to the sample problems assigned for Chapter 313 313 6 We consider the group formed by the integers 01 23 with multiplication defined by addition mod 4 eg12212332 5mod4l Solution Using this rule we obtain the following multiplication table We recognize this table as being of order 4 only 4 elements and essentially the same as isomorphic to the tables in Eqs 131 and 132 in Boas Thus this group is isomorphic to the cyclic group of order 4 3 13 11 Consider the symmetry group of the rectangle like the square but with unequal sides Solution Unlike the case of a square the rotations through 90 degrees 270 degrees and the re ections through the diagonals are no longer symmetries We are left with just the following transformations the identity the rotation through 180 degrees the re ection through the XaXis change the sign of y and the re ection through the y aXis change the sign of X These can be represented by the following 2X2 matrices 1 0 1 0 1 0 1 0 1 12 1px 13 0 1 0 1 0 1 y 0 1 with multiplication table Physics 227 Lecture 10 Appendix B 1 Autumn 2007 We recognize this multiplication table to be that of the 4 s group as in exercise 3134 The symmetry group of the rectangle is isomorphic to the 4 s group Physics 227 Lecture 10 Appendix B 2 Autumn 2007 Lecture 9 7 Appendix A To ensure that we are all on the same page let s do the single pendulum problem in some detail with various coordinate choices and compare the Newtonian and Lagrangian approaches A mass m is suspended in a gravitational eld on a rigid massless bar of length l A de ection I angle 9 is de ned with respect to the downward direction De ne fc as 7 the horizontal direction and j as the vertical up direction The e acceleration of gravity is then in the 7 direction The tricky part of J m analyzing this problem in rectangular coordinates is the a priori 2 unknown constraint force due to the bar that ensures that the motion of the mass is along a circle of radius 1 The simplest way to think about the problem is in terms of cylindrical or 2D polar coordinates de ned by the angle 9 The rectangular coordinates of the mass with respect to the relaxed down position 9 0 are then de ned as yll cos9 xlsin9 A39l We obtain the equation of motion most easily by considering the angular acceleration moment of inertia and torques with respect to the point where the pendulum is suspended Since the constraint force due to the suspending bar acts through the suspension point it provides no torque The moment of inertia of the mass with to the suspension point is ml2 while the torque due to gravity is r mgl sin0 Thus the angular version of Newton s equation is moment of inertia times angular acceleration torque mIZQz mglsine sine 0 A2 In the small angle limit 9 ltlt 1 this becomes the familiar harmonic oscillator equation 6Q 1 A3 with natural frequency coo gl Physics 227 Lecture 9 Appendix A 1 Autumn 2007 To obtain Eq A3 directly in rectangular coordinates we must go through a free body analysis for the mass In There is a vertical force on the mass due to gravity Fg mgf The component of this gravity force along the direction of the bar denoted by 2 5c sine j cos6 directed away from the point of suspension is F5 13 mg 0089 Since the mass cannot accelerate in the direction of the bar because the bar is rigid with length l the bar must supply a constraint force such that the radial component of acceleration vanishes Until we have solved the problem the only thing we know about this constraint force is its direction ie along the bar We can write the force applied to the mass by the bar as Fm 6 FW 6 where we allow the magnitude of the force to vary with the orientation 6 and FM 6 gt 0 corresponds to the bar being under tension With this notation the rectangular version of Newton s equations become 8 2 mi mlsin6 m19 cos6 92sm6 g F bar lpbar 2 93myml lcos6ml6lsin66lzcos6 A4 Fg F barjgt 2 mgEar 60036 The complexity of this result should serve as an indication of how the correct choice of coordinates above simplifies the problem To simplify this system of rectangular equations we multiply the first equation by cos 6 and the second by sin 6 and sum This procedure eliminates the contribution of the constraint force and we have cos6mllt6lcos6 6lz sin6sin6ml6lsin6 92cos6 m167 2 cosQ mgcosQ s1n6s1n6 mg mg cos 6 mgs1n6 A5 3 67 sin 6 0 This component of the vector Newton s equation which is just the component orthogonal to the direction of the support bar reproduces Eq A2 If we take the small angle limit we get back to Eq A3 Physics 227 Lecture 9 AppendiXA 2 Autumn 2007 Of course the lazy but smart physicist will simply choose coordinates to line up with the bar and automatically achieve the above separation In particular we can use cylindrical coordinates Here we will use the nonstandard notation 39 the standard choice is p ie usually represents an azimuthal angle 0 S lt 271 while 6 usually represents a polar angle 0 S 6 S 7 and define p x2y2Qtan ll x y A6 A 3 fcsinQ jcosQ Q 320056 jsin6 To eliminate any role for the constraint force along the suspension bar we consider only the component of Newton s equations this is just the projection we made above in the rectangular case With F representing the location of the mass with respect to the origin we find f1319gt713 1319 sizz ze 2si ga m l A A A mg stQ cost Far6p mlt lt gt b gt l gsin6 sin60 3192w gcosegtFar6mgcosem192 m The equation is the familiar one from above while the equation allows us to solve for the constraint force once we know the angle as a function of time ie it also involves the centripetal force which we must solve for It is actually easier to solve the last equation using the conservation of energy If we write the total energy in terms of its fixed energy is conserved total value and as a sum of kinetic and potential energy we have Physics 227 Lecture 9 AppendiXA 3 Autumn 2007 Em TV z mlze 2 mgll cos6 2 2 AS 3 6 WETOT mgl mglcos6 Thus the constraint force is given by note that it depends on the total energy 2 Em 6 mg 0056 7ETOT mg mglcosQ A9 mg3cos 2 This discussion is meant to suggest that as smart but lazy physicists we would really prefer to analyze mechanics in terms of scalar quantities rather vector quantities This is precisely what Mr Lagrange has done for us The analysis due to Lagrange which was introduced in the Lecture and will be developed more completely in Lecture 20 is presented entirely in terms of scalar quantities and is based on the calculus of variations see Chapter 9 in Boas The underlying physics principle is the Principle of Least Action or Hamilton s Principle The Action S in physics is de ned by the time integral of the Lagrangian L SzjdszjdzU V AIO Where as above T is the kinetic energy describing a system and Vis the potential energy Typically T and Vare functions of coordinates e g of point particles ik and the corresponding velocities 55k Application of the calculus of variations ie the EulerLagrange equation to nd the extreme value of the action leads to Lagrange s Equation for the 1D case d aL aL0 a g All as a simpler replacement for Newton s vector equation We have replaced having to worry about vector components with having to worry about partial derivatives We Physics 227 Lecture 9 AppendiXA 4 Autumn 2007 can apply this formalism to the pendulum by treating 9 9 as the relevant coordinates and writing T 2ZQZVmgll cos9gtLlQ2 mgll cos9 d 2 EEOll 9 mg1sm90 A12 9sin9 0 As advertised Lagrange does lead to the same equation of motion as Newton Note that we have explicitly used the constraint to make this a ldimensional problem no motion in the radial direction just as we did in the Newton analysis at the beginning of this Appendix We could also keep the problem 2D and use the method of Lagrange multipliers to eXplicitly solve for the constraint force Physics 227 Lecture 9 AppendiXA 5 Autumn 2007 Lecture 6 Review of Vectors Physics in more than one dimension See Chapter 3 in Boas but we try to take a more general approach and in a slightly different order Recall that in the previous two lectures we used the twodimensional vector analog to study complex numbers The concept of a vector is much more general Vectors give us a notation for handling ordered sets of numbers generally real numbers but we can also consider complex valued vectors The rules for handling vectors are rules for manipulating these sets of numbers Making use of our experience with vectors in the real world we can often make use of explicit geometric interpretations of these manipulations to guide our intuition Note that the numbers which represent a specific vector can have different forms depending on our choice of basis vectors this is very similar to the use of the rectilinear representation versus polar representation of complex numbers e g for 3D there standard choices x yzrectilinear ltgt r6 spherical ltgt p Zcylindrical 61 Also note that the quantities preserved during manipulations of vectors are a property of the geometry For example while the individual components of a vector are changed by a rotation the length of the vector is not changed in spherical coordinates r is not changed by rotations but the other coordinates are changed Such invariant quantities specify the symmetry properties of the system ie the geometry Symmetries play an essential role in the understanding of physics An N dimensional vector or Nvector corresponds to an ordered set of N numbers arrayed linearly Thus we can use a single index to label the individual elements or components in this linear array with the index running from 1 to N A ltgt Agpm 62 The symbol a is being used here to signify the fact that a vector with the over arrow label 1 is associated with an Ntuple of ordinary numbers and vice versa The two expressions are not strictly equal The lefthand side of this expression is meant to be abstract while the righthand side is concrete Once we have defined how to add and subtract these N dimensional arrays and multiply by a constant we have defined a Ndimensional ND linear vector space see 310 in Edition 3 and 38 in Edition 2 of Boas Physics 227 Lecture 6 1 Autumn 2007 Let s consider a 3D example 7 Cgt r1rzr3 xyz Consider first multiplication by the constant c C ltgt cxcycz 63 Next consider two vectors which we can add or subtract component by component r1 ir2 Q 11 irzy ig irzgarm ir23x1ix2y1iy2zli22 64 Just as in the case of 1D vectors the familiar scalars objects unchanged by rotations addition and subtraction of vectors of higher dimensionality eXhibit the properties of being associative and commutative for addition they can be performed using any grouping and in any order air i173 I71i172if3 I71i172 i173associative 65 I71I72I commutativebutI I I I To further define our vector space we need to be able to say something about the lengths of vectors and they relative directions ie we want to say something about the geometry of the space For that purpose we define products of vectors There are two types of multiplication of vectors ie two types of products which find many uses in physics The first is the scalar or inner or dot product which no surprise results in a scalar quantity ie a single number that is unchanged by rotations It is written as 3 f1 r2 2 xle ylyz 2122 Zwm 66 k and is commutative and distributive but not associative il r1 2 r2 r1 commutatlve r 2 13 2171 171 distributive 67 F1 The scalar product provides a simple way to calculate the length or norm of a vector Hi A U l i 2 171172173 not associative Physics 227 Lecture 6 2 Autumn 2007 f fJff 68 The geometric interpretation of the scalar product is a2 a a r rr W2 lflll lc0s912 69 where 912 is the polar angle between the directions of the two vectors Thus for vectors with nonzero length the scalar product vanishes 171 172 0 if and only if 912 7r2 ie if the vectors are orthogonal Note that Eq 69 which is valid in any number of Euclidean dimensions is guaranteed to satisfy our expectation that c03612 S1 by the familiar Schwartz Inequality N 2 N 2 a a SyZ qZ k l lllrzl 610 kl kl Also note that if we consider complex valued vectors ie the components rug are l l39r2lz N Z ikr2k k1 allowed to take complex values we want to still define the scalar product in such a way that vectors have positive real lengths Hence in the complex case we define 3 3 gt gt a r1 39r2 221151 or 24151 2 r k1 k1 ASIDE 1 If we want a picture of what is happening with the complex scalar product we can think of the second vector as being concretely represented by the column vector of its coordinates for some set of basis vectors while the first vector is represented by the complex transpose Hermitian conjugate of its coordinates ie a row vector T r11 11 A a a T T 13 gt r12 all gt13 r1 2 r1 ruarrbrry 612 Hg r23 r13 Then the scalar product is in the form of usual matrix multiplication of a row times Physics 227 Lecture 6 3 Autumn 2007 a column element by element A 21 gt 3 i39rzzliializali r22 Z ikr2k 613 H k r23 ASIDE 2 Strictly speaking when we define the scalar product we should consider the general bilinear form see Chapter 10 in Boas defined in N dimensions in terms of an NXN matriX or metric g we will use a little matriX notation here Fir EZZ lkgklr2l 614 N N k1 11 We should think of the metric as defining the geometry of the vector space For the familiar rectilinear coordinates defined by fixed orthogonal unit basis vectors more on this later the metric is just the unit matriX and we obtain the earlier expressions The geometry defined by the unit matriX is called a Euclidean geometry 0 0 0 0 l 0 0 gkl 615 0 0 l 0 0 0 0 kl As you may know from your studies of special relativity the metric describing 4D spacetime has minus signs and the geometry is called a Minkowski geometry In general relativity the metric has even more structure and gravity arises as a result of the response of the metric to the presence of a nonzero energymomentum density ie the metric is treated as a dynamically varying quantity The second type of product is the cross or vector product and results in a new pseudo vector ie the resulting quantity is another Ntuple that changes under rotations but does not change under re ection as we will discuss later and is special to 3D also called an aXial vector Physics 227 Lecture 6 4 Autumn 2007 fix 2 lt3 0122 y2219z1x2 sz1 x1y2 x2y1 63916 The vector product is not commutative x z szflifzxfl 617 ie it is antisymmetry but is associative and distributive in the sense that i x ix xia a a a a a a a 618 r1gtltr2r3r1gtltr2 r1gtltr3 Note the essential feature that the 123 ordering is maintained in each expression ASIDE A particularly handy notation for the vector product of 3vectors employs the completely antisymmetric 3tensor the LeviCevita symbol in 3D skim 8lkm 8ka 8mlk 8123 1 3 WWW Z Skimnkrzlrsm 619 klml 3 r2 X 3 k 2 Z gklmr2lr3m E gklmr2lr3m lml The last step introduces the common notation that repeated indices are understood as being summed over The geometric interpretation of the cross product in Eq 617 is that it defines a vector with magnitude given by compare Eq 69 M X g I71HI72sin612 620 where again 912 is the polar angle between the directions of the two vectors The vector direction of I71 X F is orthogonal to both if and F ie orthogonal to the plane defined by if and E Recall that any 2 nonparallel vectors define a plane in which they both lie The sense of the vector along this direction is given by the infamous righthandrule ie the direction a right handed screw would advance if we turned it Physics 227 Lecture 6 5 Autumn 2007


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