INTRO THEORET PHYSICS
INTRO THEORET PHYSICS PHZ 3113
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6 Div grad curl and all that 61 Fundamental theorems for gradient divergence and curl fX fb fltagt iiiiiiiiiiiiiii adxbx Figure l Fundamental theorem of calculus relates dfdz overab and fa fb You will recall the fundamental theorem of calculus says b a dgfhx my rm lt1 in other words it s a connection between the rate of change of the function over the interval a b and the values of the function at the endpoints boundaries of that interval There are equivalent fundamental theorems 7 for line integrals area integrals and volume integrals ln vector calculus we deal with different types of changes of scalar and vector elds eg W5 6 17 and v x 17 amp each has its own theorem We ve discussed line integrals before mostly in the context of the work done along a path but let s remind ourselves of the de nition B n 1302 dF711LIampZF dF 2 A i1 in other words we add up the area of all the little rectangles F d consisting of the vector F at the point F dotted into the path element d see Figure Remember the point is that although we are doing an integral in a 2D space we are constrained to move along a path so there is only one real independent variable Figure 2 Line integral Example from test Consider the triangle in the xy plane with vertices at 10 10 and 01 Evaluate the closed line integral 1 ya x3 m 3 around the boundary of the triangle in the anticlockwise direction Since the vector F has components xy the measure is df idx ydy so y 5 953 df ydx xdy On leg 10 gt 10 we have 3 0 so integral is fj1 ydx 0 On the leg 10 gt 01 we have 3 x 1 so integral is f10 x 1dx f011 ydy 1 On the path 01 gt 10 we have 3 x 1 so integral is f071x 1dx f10y 1dy 1 So total line integral is 2 Let s go back and look at the leg from 10 gt 01 again We could have pararneterized 7 this leg as x 1 t y t 0 lt t lt 1 Then F 1 t c 153 and df 02 3dt We can therefore write this integral as 01 A 1 1 Fdf ta 1 tgj gjdt t1 t 1 4 170 0 0 same answer as above 611 Conservative eld Recall we said that the integral of an exact differential was independent of the path This result can be expressed in our current particular context line integrals of vector elds Suppose you have a vector eld F which can be expressed as the 2 gradient of a scalar eld F Then we have the fundamental theorem for gradients lbwdrlwa wa 5 in other words the integral 13 d doesn t depend on the path between a and b If 13 W5 for some q 13 is conservative integral independent of path Since as you proved on the practice test v x W 0 for any q showing that v x 13 0 is a su icient condition for a force to be conservative as well 612 Surface integrals Figure 3 Arbitrary surface with measure d6 We de ne the surface integral of the vector eld 17 as Udd JEEOZWE m2 6 A i1 where the in nitesimal directed area element dd is oriented perpendicular to the surface locally Sometimes we call the integrand the flux 7 of 17 through A dCID 17 dd This should remind you of the example of the flux of the electric eld through a surface from your Eampll class but it can be de ned for any vector eld Another example suppose 17 is a current density ie represents the number of particles crossing a J plane per unit area per unit time Then the flux dCID 17 dd is the number of particles passing though do per time In general the surface A may not be a plane but may have some curvature but the total flux will just be CIDA27dd 7 Suppose A is a closed surface now and there is a source of particles shooting out in all directions penetrating the surface Clearly the number of particles passing through the surface per time per area is nonzero On the other hand if the source is outside A as shown the flux through the surface looks like it might cancel on the left and right surfaces and be zero 1 In fact we must somehow be able to express the conservation law that the particles which enter A must eventually leave it what goes in must come out The answer is a very general relation between the surface integral of a vector eld and the volume integral of its divergence 6 17 613 Fundamental theorem for divergences Gauss theorem 1 C r 39 z l Figure 4 Left particle source inside closed surface A Flux is nonzero Right source outside closed surface Flux through A is zero Mathematically the divergence of 17 is just 821 39 2 7 Consider the volumes inside A and A A 8V and A BV the symbol 8 here means the boundary of Remembering that we re thinking of 17 as a current density for now let s ask ourselves how much how much comes out passing through the surface We can divide up the whole volume into in nitesimal cubes dxdydz and consider each cube independently First imagine that 17 is constant Then the surface integral over the cube will give zero because the flux in one face will be exactly cancelled by the flux out the other face signs of the dot product 17 dd will be opposite If 17 is not a constant the vectors on opposite faces will not be quite equal eg at x 1You might suspect that this result is related to Gauss7 s law in electromagnetism and if the source were and electric charge and 17 the electric eld you would be right Be careful however Gauss s law is a statement about physics and the relationship between the electric flux and its source the electric charge We reminded ourselves the rst week that this depends sensitively on the 17quot2 nature of the Coulomb force 4 we ll have vx but at x dx we ll have vx dvz So the flow rate out of the cube in this direction will be equal to 23 dx lntegrating over the whole cube will give the whole flow rate out 61 v27dxdydz dd 8 cube 6 cube But if we consider any arbitrary volume T to be a sum of such cubes the surface integrals of the interior in nitesimal cubes will cancel because their normals C are oppositely directed Therefore the total volume integral over T is similarly related to the total surface integral of 8T v5dTJdd 9 739 8739 This is called the divergence theorem or Gauss s theorem not Gauss s lawl Z X Figure 5 Faces of cube in example Let s do an example where the total volume T is an actual cube of length 1 take 17 Ty j2xy 22 k2yz The divergence is V 17 205 3 Volume integralis a 1 1 1 V172dxdydzxy2 10 r 0 0 0 and the integrals through the faces are 1 1 A 1 1dd dydzz 051 17dcTy2dydzgt ygdydz 0 0 1 1 A 1 2dd dydzz 950 27dc7 y2dydzgt ygdydz 0 0 3 1 1 A 4 3dd dxdzj y1 17dd2x22dxdzgt 2x22dxdz 0 0 1 1 A 1 4dd dxdzj y0 d6 22dxdzgt 22dxdz 0 0 1 1 5dd dxdyk 21 17dd2yzdxdygt2 ydxdy1 0 0 6 dd dxdyz 20 17dd 2yzdydz0gt0 and the sum over the 6 faces is g 1 2 as for the left hand sidel 614 Fundamental theorem for curls Stokes theorem Figure 6 Directed area measure is perpendicular to loop according to right hand rule The proof here is very similar to the divergence theorem so I won t belabor it here I recommend you read Ban or Shankar but simply state it In this case we consider a directed closed path in space the boundary of a 2D area A gtlt27dd Udfi 11 A 6A Remarks Stokes and divergence thm 1 De ne the normal to the area according to the right hand rule using the sense of the loop as in Fig 6 Figure 7 The same loop can be the boundary for different open surfaces 2 Many surfaces can have the same boundary see Fig 10 3 H27 y x R then x 0gt ledd0 12 since VVVgtlt dT6VVgtlt dd 13 Some tricks H If you need ff 13 df along a path between A and B check to see if 13 is conservative Do you know or can you check that F or 6 x 13 0 If yes choose the simplest pathl 6x5dd5df Wanda 14 A1 A2 if A1 and A2 are bounded by same loopichoose simplest A 00 lfv1707theanUdcTfAIUd5fA217dcT07soifyoucando fA117dcT you can nd L12 17 da 615 Intuition for vector elds wr k k44 kll l f AA14111 LA439rI39 VV AVquot 1 44VA 4117V IquotV if ffiii r ril Jl Figure 8 Example 1 Diverging77 radial eld 17 F v 17 3 gt 0 but v gtlt 17 0 Vector eld spreads out77 or diverges Flux dd through Closed surface surrounding F 0 obviously nonzero Remarks on divergenceless elds If 6 17 0 everywhere the following condi tions are equivalent 139 fclosed surface 77 39 dd 0 2 17 is the curl of some vector 17 V x 15 Why Note it is obviously not unique we can Vq to 16 for any scalar eld q and still maintain 17 V x 15 This might remind you of gauge symmetry 7 in electromagnetism 3 fopen surface 17 dd is independent of the surface for a given boundary line Remarks on cvrlless conservative elds f6 x 17 0 everywhere 1 fUdr 0 by Stokes 2 17 is the gradient of some scalar 17 W 3 17 dr independent of path between a and b Figure 9 Example 2 Consta nt vector eld 17 11 6 17 any Closed surface is zero 0 N0 spreading Net ux through k r r l x l I V V A V quot Lllll v39v39y AAAlli 4114lf v444fff grvt rr1 1114 4gt4739 Figure 10 EX 3 17 717170 6 17 07 but 6 gtlt 17 002 31 0 Line integral 5517d77 j r r d 27TT2 Compare with integral of curl dotted into surface7 take cylinder for simplicity 6 gtlt 17 d6 2 wrz Stokes works 616 Divergence of Coulomb or gravitational force of point chargemass Consider the vector eld 17 with the form F ie it points radially outward and falls off as 173 This should remind you of the electric eld of a point charge or the gravitational eld of a point mass Here s a little paradox claim 6 17 0 a xyz i 1 3x2 16 x2 y2 ZQ32 7 x2 y2 2232 x2 y2 2252 015 lt gt y f lt gt Z 17 2x2 32 22 x2 32 Z252 95 H 3 95 H Z 18 0 19 OK so this eld is divergenceless It looks like something is owing out but on the other hand the magnitude of 17 is getting smaller as you go out so maybe it s ok BUT what about the divergence theorem If I calculate 27 dd 12 ma sin0d0d 4w 20 sphere If So is the divergence theorem wrong or did we miss something The answer is that we were a little careless in calculating the divergence and declaring it to be zero It is zero almost everywhere However if you look at our equations exactly at the point x y z 0 you will see that the derivatives diverge ie are in nite you may need to do L Hospital s rule to convince yourself but it s right So we have encountered a very peculiar situation a function of space which is in nite at one point but zero everywhere else This type of function2 was explored in the physics context by Paul Dirac so we refer to it as the Dirac delta function and give it the symbol 6 de ned as 00 695 000 8 such that 1 6xdx 1 21 00 2mathematically it is not a function it s called a distribution 22 and WW 55y5z So since a f f a 7V da47T we have determined that 24 a f 34 V 247T6 Alternate de nitions of 605 you will see de ne a sequence of normalized functions eg 7 0 i n 72I2 605 i n lt 21 or 605 i We 25 for which one can show that 6nxdx 1 and lim 6nxfxdx f0 26 for any function f Then de ne 695 as 27 6xfxdx E lim 6nxfxdx Note limnnoo does not exist but integral does That s What makes 695 a TL distribution not a function according to mathematicians 617 Properties of 6function o 6x0ifx7 0 ff 55 am 1 o 6ax Where an are the zeros of 995 0 EX 1 5 2 053605 3dx 33 but 053605 3dx 0 0 0 EX 2 951 65xdx g 051 695 3 EX 3 0 1W fr TE1 dm 1Awg x 1dx oox36x1 28 29 30 31 32 33 9 Linear algebra Read Boas Ch 3 91 Properties of and operations with matrices M x N matrix with elements AZj AM1 AMQ AMj AMN De nitions 0 Matrices are equal if their elements are equal A B ltgt AZj Bij o A BM Aij Bij o kAZj kAZj for k const 0 ABM 2 Aingj Note for multiplication of rectangular matrices need MxNNgtltP o Matrices need not commute AB not nec equal to BA A B E AB BA is called commutator of A and B If A B 0 A and B commute o For square mats N x N det A A ZN sgmr A1W1A2W2 ANMN Where sum is taken over all permutiations 7T of the elements 1 N Each term in the sum is a product of N elements each taken from a different row of A and from a different column of A and sgn 7T Examples A11 A12 A A A A 2 A21 A22 11 22 12 217 A11 A12 A13 A21 A22 A23 A11A22A33A12A23A31A13A21A32 A31 A32 A33 141214211433 141314221431 141114221432 o det AB detA det B but detA B y detA det B For practice with determinants see Boas 0 Identity matrix I A A VA Iij 627 o Inverse of a matrix A A 1 A 1A I o Transpose of a matrix ATM A 0 Formula for nding inverse 1 71 T 4 det A 7 Where C is cofactor matrix An element OH is the determinant of the N 1 x N 1 matrix you get when you cross out the row and column i7j and multiply by 1i 1j See Boas ie it s Adjoint of a matrix Al is adjoint of A has elements All A conjugate transpose Don t worry if you don t know or have forgotten What conjugate means ABT BTAT AB 1 B lA 1 0 row vector is 1 x N matrix a b c n 0 column vector is M x 1 matrix 6 f g 5 m 0 Matrix is diagonal if AZj Ail617 Trace of matrix is sum of diagonal elements Tr A Aii Trace of produce is invariant under cyclic permutations checkl Tr ABC Tr BOA Tr CAB 6 92 Solving linear equations Ex x yz 4 2xy z 1 3x2y22 5 2 may be written 1 1 1 x 4 2 1 1 y 1 7 3 2 2 z 5 Symbolically A F What we want is F A lg So we nd the determinant det A l2 and the cofactor matrix 4 71 04 1 5 8 035 1 4 4 0 413 7 1 3 9 1 5 5 so a 1 4 4 0 4 1 y E 7 13 1 1 10 z 1 5 5 5 2 So 05 l y 1 z 2 Alternate method is to use Cramer s rule see Boas p 93 1 4 1 1 953 1 1 1 1 11 5 2 2 and similarly 7 for y and 2 Here the 12 is det A and the determinant shown is that of the matrix of coef cients A with the x coef cients in this case replaced by k Q what happens if g 0 Eqs are homogeneous gt detA 0 93 Rotation matrices Here s our old example of rotating coordinate axes F 0573 is a vector Let s call F ac7y the vector in the new coordinate system The two are related by 3 Figure l Rotation of coordinate axes by 0 the equations of coordinate transformation we discussed in week 4 of the course These may be written in matrix form in a very convenient waycheck 05 cos6 sin6 x 3 sinO cos0l y 12 V W 7 R9 F where R9 is a rotation matrix by 6 Note the transformation preserves lengths of vectors F as we mentioned before This means the rotation matrix is orthogonal 13129 I 13 These matrices have a special property group property which we can show by doing a second rotation by 6 R R i cos 6 sin 6 cos6 sin6 9 6 T sin 6 cos 6 sin0 cos6 i cos 6 cos6 sin 6 sin6 cos 6 sin0 sin 6 cos6 14 T sin 0cos6 cos 6 sin0 cos 0cos6 sin 6 sin0 i cos0 0 sin6 0 i T sin6 0 cos0 0 T ROTH 15 Thus the transformation is linear More general def A01B02D 01AB 02AD 94 Matrices acting on vectors U1 A11 A12 AU A1N U1 112 A21 A22 A2j A2N U2 Ail 1422 Aij Al39N DEV AN1 AN2 ANj ANN UN or more compactly N U ZAZ39J39UJ39 jl Another notation you will often see comes from the early days of quantum mechan ics Write the same equation w Am 18 So this is a ket 7 7 or column vector A bra 7 7 or row vector is written as the adjoint of the column vectorlz ltvl ltva 2 v v up 19gt v1 v2 UN if 112 are real 20 NB In this notation the scalar product of vgt and is vwgt and the length of a vector is given by v2 ltvvgt 95 Similarity transformations Suppose a matrix B rotates rgt to r1gt7 r1gt Brgt Now we rotate the coordinate system by some angle as well so that the vectors in the new system are rgt and M eg Rr1gt What is the matrix which relates rgt to M ie the transformed matrix B in the new coordinate system Id RIM RBI RBRJRIM RBR WRW RBRJWt 21 so the matrix B in the new basis is B RBI2 1 22 1The name comes from putting bra and ket together to make the word bracket no one said physicists were any good at language This is called a similarity transformation of the matrix B To retain o similarity transformations preserve traces and determinants Tr M Tr M det M det M 0 matrices R which preserve lengths of real vectors are called orthogonal RRT l as we saw explicitly 0 matrices which preserve lengths of complex vectors are called unitary Suppose vgt Uvgt require UUl 1 then WM UleUlv lelUvgt ltvvgt 23 A similarity transformation with unitary matrices is called a unitary transfor mation o If a matrix is equal to its adjoint it s called self adjoint or Hermitian Examples 1 123 A245 24 356 is symmetric ie A AT also Hermitian because it is real 2 0 2 3 A 2 0 5 25 3 5 0 is antisymmetric and anti self adjoint since A AT Al 3 AU 26 is Hermitian A Al Al h 27 is antiHermitian A Al Checkl en 11 is unitary A 1 Al 1 1 1 E l 1 1 l 29 is orthogonal A 1 AT Checkl 951 Functions of matrices Just as you can de ne a function of a vector like Fl you can de ne a function of a matrix M eg CLM2 MW5 Where a and b are constants The interpretation here is easy since powers of matrices can be understood as repeated matrix multiplication On the other hand What is expM It must be understood in terms of its Taylor expansion expM Zn Mnnl Note that this makes no sense unless every matrix element sum converges Remark note eAeB y eAB unless A7 B 01 Why Hint expand both sides 3 More applications of derivatives 31 Exact 85 inexact differentials in thermodynamics So far we have been discussing total or exact 7 differentials Bu Bu du dx dy l 805 y 83 I but we could imagine a more general situation du MalW Maggt612 2 If the differential is exact M y and N 2 7 By the identity of mixed partial derivatives we have I 8M i 82 i 8N 3 8y I T 80583 T 895 y Ex ldeal gas pV RT for 1 mole take V VTp so 9V 9V R RT Now the work done in changing the volume of a gas is dW pdV RdT dp 5 Let s calculate the total change in volume and work done in changing the system between two points A and C in p T space along paths AC or ABC 1 Path AC ZE sodT idp 6 255 2 T Tl Zltp p1gt lt7 so 8 dV gi idp gm P P1ldp Tl i Zplwp 9 WV Tl A ppmlp 10 1 P Figure 1 Path in pT plane for thermodynamic process Now we can calculate the change in volume and the work done in the process P2 T 1 T T R2P1 1P2 A V2 V1 dV RT1 A P1 11 AC P P m P1P2 AT 102 T T W1 W2 pdV RT1 p1lnp Ruin2 12 AC AP 101 P2 P1 P1 2 Path ABC Note along AB dT 0 while along BC dp 0 av av P2 av T2 av V V lt gt lt gt CH gt W gt W 2 1 ABC 3T p 8P T p1 81 T T1 8T p P2 RT T2 R T T 2 1dp dT 12 2191 1 13 01 p T1 p2 P1192 P2 av T2 RT av p2 W W d dT RTl RT T 2 1 p1PltapgtT PT1 Vlt8Tgtp 1np1 2 1 Note the change in V is independent of the path 7 the volume is characteristic of a point T in equilibrium 7 but the work done in the process is notl What s the difference In the system with p T as independent variables dV is an exact di erential while dW is not How can you see the difference Go back and examine the forms we had for dV and dW in 4 and In the case of dV we had dVMdpNdT withM andN R P P 8M7 R 8N7 R 27 7192 W7 19 lt14 lt15 2 which is indeed exact whereas dW Mdp NdT M R N 16 BM 8N R a W 7 7 is not This is a demonstration we won t use the word proof that for a for a thermodynamic process involving changes in the p T plane the volume of the system is a state variable 7 ie for 1 mole of gas it simply depends on what p and T are if you have speci ed p T you know the volume of the system The change in volume between two points will therefore always be V2 V1 independent of which path is chosen The work done in the same process is however not independent of the path of integration 32 Maximaminima problems with constraints Lagrange multipliers I m going to make a mathematical detour before coming back to thermodynamics in order to give you some tools you need to solve homework problems In physics we often need to nd the extrema of a function of several variables subject to a constraint of some kind In math a simple example would be the distance function in 3 space d V952 32 22 Of course the minimum of this function over all 3 space is just 0 achieved at x y z 0 But suppose we were to look for the minimum of d over the constrained 7 set of points de ned by the plane 05 2y 22 3 The usual way to proceed is often the simplest if it works express 3 yx then set dydx 0 and solve But sometimes one can t solve for yx explicitly so one can try the method of implicit differentials see Boas ch 4 or use an elegant technique called the method of Lagrange multipliers Joseph Lagrange 1736 1813 was a Hench mathematical physicist The idea is if you can express the problem in terms of the minimization of a function fxy together with a constraint gxy 0 to imagine you are solving the unconstrained problem of nding the minimum of F x y A x 3 Agx y Then you can simply treat A as an additional variable and minimize with respect to it as well In the process of solving the problem you eliminate A from the solution Ex Coming back to our problem with the plane let s rst make our life a bit easier by recalling that if we minimize d x2 y2 22 the square root will also be a minimum So de ne Fx2y222x 2y 22 3 18 and set all the derivatives equal to zero 2 1 2x0 19 539 1 2y 2A0 20 2 1 22 2A0 21 2 1 x Zy 22 30 22 Note the equation BFBA 0 is always just the constraint equation itself Now we have a problem with 4 equations and 4 unknowns and it can be solved The usual idea is to eliminate the constraint as quickly as possible The rst eqn tells us that A 2x so 23 49 0 23 224a 0 24 x 2y 22 3 0 25 which we can solve to nd 0573 z 13 23 23 33 Differentiation of integrals LeibnitZ7 rule for differentiating integrals d W dv du 711 8 f t dt dt 26 dag I m gt mock much I 895 lt gt EX 21 It I 67m 27 d 6121 611 2x It 2 2 2 2 z z re e gt lt gt 34 Laws of thermodynamics Let s come back to the idea of exact and inexact differentials in thermodynamics Here s another example of an inexact differential d c dT A d 30 Q 10 p P heat absorbed heat capacity const p latent heat 31 We showed already that the work done is also not an exact differential however the combination of the two is dU E dQ dW 32 op wz wn lAP pc zudp This is called the internal energy and this equation is sometimes referred to as the 1st law of thermodynamics expressing energy conservation ie the change in internal energy of a gas in a thermodynamic cycle goes either into heating the system 6162 is the in nitesimal heat absorbed by the system is the or into doing work done by the system Another exact differential is 13 dQ T Remark I sometimes you will see the notation dQ and JW for in nitesimal heat absorbed and work done This is just a more careful notation to remind you that they are inexact differentials Remark 2 since the work done in a thermodynamic process depends on the path it is really nonsense what I wrote above W2 W1 You are to think of this as the work performed by the system in the process of going 1 gt 2 but W2 and W1 have no independent meaning since they are not characteristic of a macrostate 35 Legendre transformation If for a function fxy we have the differential df pdx qdy where p and q are equal to BfBx and BfBy respectively we might ask the question how do we make a simple change of variables to a new function 905 q with q one of the the independent variables We simply make use of the fact that df qy df dqy qdypdxqdy dqy qdypdfr ydq 34 This function is by de nition associated with an exact differential dg You explored this on the HW for various thermodynamic quantities 5 2 Differential calculus 21 Asymptotics of functions blackbody radiation One of the hardest things to teach students is how to have a qualitative feel for the important aspects of different functions The rst thing to always do when you are studying a function is consider its asymptotics ie its behavior at 0 and 00 and possibly near other as where the function is singular Let s take the example of the blackbody radiation function MA This is the density of radiation per unit wavelength emitted by a blackbody ie a perfectly absorbent body in thermal equilibrium at temperature T i 87Thc l ud i d A5 ehcAkT 1 7 1 where h is Planck s constant c the speed of light k Boltzmann s constant and A the wavelength of light omitted The easiest thing to think about is an object with a cavity inside with radiation of all frequencies emitted by the blackbody lling the cavity Actually the rst thing to do is to analyze the equation to make sure it s dimensionally correct n has the dimensions of angular momentum ie h mLZt The quantity he therefore has dimensions ngtg and the quantity hcA dimensions mLZtg which has dimensions of an energy Therefore the exponent in the denominator of Eq 1 is indeed dimensionless as it should be before we can raise it to a power The factor which has dimensions hcd 5 then has dimensions ngt2L4 or 771 L152 which is the same dimensions as energy volume so that checks The equation is therefore ok Our analysis of dimensions gives us hints about how to approach understanding the function It contains a lot of physical quantities but most of these are con stants We are interested in how u depends on wavelength A Let s de ne a new dimensionless variable x E AkThc which is proportional to A In terms of as the formula reads kT 5 1 5 1 ux 87Thc e1I 1 2 Let s ignore the x independent prefactor and just look at the function of 05 Large wavelengths clearly correspond to large as and small wavelengths to small 05 But large and small relative to what The claim is now that the dimensionless function Was eaI l 1Qc5 has no obvious scale for x in it therefore the most important values of x ie where the function is big must be of order 1 This may seem like an outlandish claim so let s check it We can expand using a Taylor expansion in either 05 or 1 05 to nd 11 5 a e 05 05 ltlt 1 3 11 4 05gtgt1 ux expl1xx5 025 05 075 l 125 15 175 2 X Figure 1 Energy density E of blackbody as function of reduced wavelength x with asymptotic short and long wavelength approximations Note a couple of tricks we used to obtain these results In the 05 ltlt 1 limit the exponential is huge and dominates the 1 in the denominator which can be neglected But the exp 105 left over is divided by 055 The numerator is going to zero denominator alsoiwhich wins Of course formally we should use l Hospital s rule what you will nd is that the derivatives of the power on the bottom get bigger but the derivatives on the top remain the same hence the ratio indeed approaches zero as 05 gt 0 This is a special case of a general rule that when exponentials and power laws compete exponentials always winl 7 just as powers always win 7 over logs In the 05 gtgt 1 limit the exponential exp 105 gets very close to 1 so we can expand in powers of 105 exp 105 2 1 105 10522 Hence the exp105 1 2 105 and we get the result in Eq 3 A full plot of the function as seen in the gure reveals in fact that there is a peak around 05 02 which is indeed of order 1 ie not of order 10 1 or 101 As you see this criterion was pretty vague because 02 is indeed closer to 01 than to 1 but a physicist would say ok well it might be 127T so if we take 27F equal 2 1 it works This may sound ludicrous but it is simply the statement that there is no other scale in the function What do I mean by scale Well suppose I had a function 1 x 10 2342 1 4 995 which cropped up in some problem Then I would say there is a scale or dimension less number 10234 which is pretty important for the physics here for some reason In our blackbody problem however it doesn t exist The function is peaked near 05 1 Does this mean a blackbody always radiates with a maximum intensity at a particular wavelength Of course not Depending on the temperature T the A for which the maximum intensity is achieved can be quite different To keep 05 N 1 if the temperature goes up the wavelength has to go down A 3 Visible o 3 Infrared m 3 a f Fealt Wavelengm ID u Irrlenslty curve I E A for each 3 temperature a 5 l 5 Wavelan qth l urn Figure 2 Energy density curve vs wavelength in pm for different temperatures in Kelvin 22 Some tricks for doing integrals 221 Differentiation With respect to a parameter Suppose a function of x depends on a parameter a EX sin am If an integral converges we can differentiate under the integral sign to obtain 1 mm lt5 Z I 2de 6 If you know F a you can derive results for the family of integrals f 8 x a 8a dx Sometimes you know an integral and would like to use it to do a related harder integral you can do so by introducing a parameter and differentiating with respect to it EX suppose you know or are given fowmg l 1dx 7T2 but you want to know fowmg l 2dx which is not so obvious Introduce 0 dx 1 7 a 2 H2 lt gt then express in terms of u xa 1 0 du 7T 1 8 a aO U2 1 2a Now differentiate the original expression 81a 0 dx 81a i 87T2a 7T 0 22a butalso i 9 8a 952 a2 8a 8a 2a2 00 dx 7T 2 10 0 x2 a22 a 2a2 0 dx 7T gt 11 0 x2 a22 4a3 Now note that the result holds for any value of a we like so we could put a l to get 00 dx 7T 12 0 2 1 4 lt gt so in a pinch eg if we were cast away on a desert island without tables of integrals or the internet we could in principle derive this result or corresponding ones for higher powers of the denominator by taking further derivatives 4 3 21 an quot I l 4 quot x 39 o I x I 3 I I l II I I I I I quotX 2 x I I 00 Figure 3 Paths from 00 to 21 222 Integrals along a path We consider the integral of a function of x and 3 along a path in 2 space see Ch 6 Sec 8 EX 1 271 I xydx ygdy 13 00 from 00 to 2 1 along different paths as shown The paths are 1 1 straight line 2 parabola y 952 3 2 straight lines along axes 4 x 2153 152 Q Will the value of the line integral be the same along all paths A No Path 1 y 052 gt dy dx 11fdxx x x22gt ferEgg 1 14 5 Path 3 3a 00 to 01 x 0 d0 011 fol yZdy 13 3b 01 to 21 3 1 dy 0 so 13 f02 xdx 2 SO 3 Igajgb Paths 2 amp 4 7 try yourselves before looking in book EX 2 01 1 0 00 10 Figure 4 Paths frorn l0 to 10 Find the integral Jxdy ydx 15 x2y2 along the two paths shown 1 Along semicircle pararneterize x oos0 y sin 6 So d d 0 Md6gtJd6 7r 10 x2 y2 W 2 Along straight lines Check yourself using f 151 tan lu Should nd J1 J2 23 Fun with partial differentiation see Boas ch 4 17 6 Notation for f fry is the partial derivative of f with respect to x with 3 held xed constant Ex1f052 y2 957 cos6 yrsin6 af 2 18 lt8xgty x l A little trickier what is Bf8009 or Bf86 Method 1 express f in terms of x and 6 rst f xg y2x21 y2x2 x21 tan26 19 9f 7 2 9f gt a 7 2051 tan 6 y y 20 Method 2 works in principle even if method 1 fails because you can t solve for fx 6 in closed form use chain rule 8 8 f rm df doc dy 21 in this case df 2x dx 23 dy 22 but if we consider f fx6 then i 3f 3f df 7 ltagt6d0 de 23 Now we need to compare 22 and 23 Now 3 xtan6 so dy can be expressed in terms of x and 6 as follows i 8y 8y dy i ltagt6da lt gtxd6 24 tan6dx xsec26d6 25 So putting these together we have df 29 dx 23 dy 26 2x dx 2ytan6 dx 05 sec2 6d6 27 2051 tan2 6 dx 2Qc2 tan6 sec2 6 d6 28 v 3f 9f 29 71 Questlon and frequent source of confusion is g 7 In a situation where there is only one dependent variable this works eg take 05 tan 6 0 tan 1 x it is easy to show that dOdx dxdO 1 But in general Answer no only true if the same variable is held constant in both cases For example 895 895 xrcos0gt rs1n67 y so this is 30 but also 71y 30 ll3L 2 3 96 Oitan Eiaimi sothlsls ay 71 Since the same variable is not held constant we do not nd 231 Chain rule implicit differentiation EX 2 Suppose x e t What is 327 If we could nd 95t explicitly no problem but we can t Use differentials 1 dx d Id dtgt1 Id dtgt 32 xe x ex dt 1ez To get second derivative diVide differential by dt and differentiate wrt t dx dx dgx dgx dx 2 I 1 gt 1 I 0 33 dtle dt on dtgle ltdtgt d2 I solve to nd x e 34 dt2 1 e 3 EX 3 If w ax by show that new Let u ax by We have w u so 8w 7 deu 7 8f natam lt36 8w 7 deu 7 8f i Budy i bdul 37 so Eq 35 follows 232 Thermodynamics Equation of state fp V T 0 eg pV RT 0 ideal gas laW or p 0 van der Waals gas Differential f 8f 8f 8f i dfa etwemo Now do 3 different experiments 1 hold p xed dp 0 so Si 5 gflkpd 39gt 8V ltg fgtpV 1 lt5 3 1 T a 40gt 2 hold V xed dV 0 so i lt 9PgtTv i 1 41 8p 7 7 V 6 lt6 3 hold T xed dT 0 so lt gtT 42 8V lt3 gtTV ltgtT Now examine product of two such derivatives using above results E H 5 Ty W M P Ty So independent of any particular equation of state f we always have relation among partial derivatives 8p 8V 8T Wt W 87 1 lt45 6 r r a lt43 ltgtpT ltgtpv la lily av T 8T p i f 14 Fourier analysis Read Boas Ch 7 141 Function spaces A function can be thought of as an element of a kind of vector space After all a function f is merely a set of numbers one for each point x of the underlying space We can add functions in this way componentwise like vectors Mac f 905 and we will show below we can de ne a metric or distance function on the set of all functions as well It s simplest to think about 1D rst a nite interval 0 S x 3 2L and imagine discretizing 7 this space so the N points in it are separated like the gradations on a ruler by an amount A E 2L N A vector 7 in function space fgt is therefore de ned to be the set of components f1 f2 fN representing the values of the function f at the points x1 x2 Now if we choose a basis of this space called a position basis we de ne a vector 0 0 0 0 where the 1 is in the ith position in other words the vector represents the position 052 This is clearly a basis for the vector space since each vector is linearly inde pendent and the whole space is spanned The function f may now be represented as lfgtf10gtf21gtf34gtquot39leNgt7 2 ie the function has the value f1 E ag at 951 and so on Note this space is nite dimensional but we can make L as large as we like or choose N as large as we like Now suppose I wanted to de ne the product of two functions on this space Well a function has been represented as a vector so the idea is obvious de ne a scalar product as you would between two vectors N ltfggt Z fZgz lt3 1 Now however the idea is to take the limit N gt 00 so that we have an in nite dimensional vector space functions still live on a nite interval howeverl We can do this in such a way such that the limit is well behaved if we de ne N 2L Umzmma fwmmm w Once we have an inner product we can de ne the lengths of vectors functions ie 2L MEVM A mowm A function is said to be normalized if its length is one ie mAlmm4 A space of functions where all elements are normalized is called a Hilbert space after mathematician David Hilbert We can also de ne the notion of orthogonality ie two functions are orthogonal over the interval 0212 if 2L maa fwwmo aimmmmml a There can be many different sets of orthogonal functions Here are some exam ples l emmxL over interval 0 2L De ne l mgt mermL m0l 1i2 8 You can check that i 2L eiirwnzLeirrnzde 6mm 9 2L 0 2 Legendre polynomials on interval 1 l 1 2 11 denxPmx 2n 6m 10 2 Note as de ned these functions are orthogonal set but not normalized since their square integral isn t I These are solutions to a particular differen tial equation which arises in electromagnetic theory and quantum mechanics Without further discussion about Where the Pn come from I can give you the rst few so that you can test Eq 10 13095 1 13195 x 13295 3052 12 13395 5x3 3x211 142 Fourier series Figure 1 Example of function satisfying Dirichlet conditions Let s focus on the mgt set a little more closely This set of functions form an orthonormal basis for functions obeying Dirichlet conditions Fig l o periodic period 2L 0 single valued on interval 0 nite number of max min 0 nite number of discontinuities o 02L fx2dx nite In other words any function f obeying these conditions can be expanded in this basis m Z cmeWIL 12 721700 compare N If 20mm 13 3 How do we gure out what the cm are For ordinary vectors we can just use the orthonorrnality of the basis take the inner product of 13 with to nd N U Zcmltnmgt0n 14 m7 6m 15 We can use the same trick with our complex exponential basis the coe icient cm in 12 is 1 2L cm ltmfgt E WWIW 16gt These are called FOUM39GT coe icwnts Note if L 7T we have an expansion 00 1 27f Z cnezmcm e mx xwx 17 27T0 TLOO period 27F This is called an exponential Fourier series or just Fourier series 143 Sine and cos Series Recall elm cos 1195 1 sin 1101 so 00 71 00 Z cncos 1195 1 sin 11x co Z Zcncos 1195 1 sin 11x Tl00 71700 711 co 0171 cos 1195 1cn c171 sin 1105 711 E Egan cos 1195 bn sin 11x 19 where 1 27f 1 27f an on 071 cos nxdx bn 1cn c171 sin nxdx 7T 7T 0 Ex 1 Piecewise continuous function Suppose you want to Fourier analyze the function 0 7T lt x lt 0 0ltxlt7r 20 4 Figure 2 First extend it periodically as in the gure Then the Dirichlet conditions are ful lled and we can immediately write 1 27r 1 7r 1 7r a0 fxdx dx 1 an cosmcdx0 nl22l 0 7T 0 7T 0 bn sinmcdx 20 n Odd 22 0 0 71 even So Fourier series for f is 1 2 sinx sin3x 23 flxlilglt 1 l 3 Now it should seem slightly crazy to you that we can add a bunch of sine functions and get something at and piecewise continuous But it works How it works ie how the series converges to the right answer 7 in this case is shown in the next gure Figure 3 How one builds up a square wave function as a sum of sine waves fxx2 7Tltxlt7r 24 Note that f is even in x We therefore know in advance that the series for f is a cosine series only bn 0 1 7r 22 a0 acgdaci 7T 3 1 7r 2 2 an x2 cosmc dx l 7T 1 2 7T 7T7r n2 71 So we get a Fourier series expansion for 952 Figure 4 x 0ltxlt7r f1mi 05 7Tltxlt0 fgw ac 0ltxlt7r x 7Tltxlt0 25 26 27 28 29 The function fl is even in 05 so we expect a priori to nd it represented by a cosine series ie all bn0 On the other hand f2 is odd so it will be represented by a sine series Note that over the interval 0 lt x lt 7T they represent the same functionl f2X q F O F X Figure 5 EX 3 f1x and f2 144 Fourier integral Now here we have just analyzed a function over a symmetric interval 7T 7T and we can clearly do the same over a symmetric interval L L 11 Complex numbers Read Boas Ch 2 Represent an arb complex number 2 E C in one of two ways x iy x y E R rectangular or Cartesian form 2 2 rel r0 E R polar form 1 Here 139 is l engineers call it j ychhl The height of bad taste If 21 22 both real and imaginary parts are equal 951 952 and y1 yg This implies of course that 0 E C is the complex number with both real and imaginary parts 0 The complex conjugate of 2 x iy is de ned to be 2 x iy Boas sometimes calls the same thing 2 Note 22 952 y a real number 2 0 The modulus or magnitude of 2 is then de ned to be W x 952 32 has obvious analogies to the distance function in Euclidean space Some useful relations you can easily work out are 11112 Z 2122 2 z Re2 You should practice simplifying complex numbers which are not given explicitly in the form x iy For example ilt1igtw 4 1z39 l z39 l z39 2 2 2 3 Polar form Just as all real numbers can be represented as points on a line complex numbers can be represented and manipulated as points in a 2D space spanned by the real and imaginary parts 05 and y Euler theorem Recall the expansion of the exponential function e Zn xnnl which converges for arbitrary size of 05 Let s de ne a complex exponential e2 in the same way and choose in particular 2 2396 Then 29 7 0 19 7 0 1 02 0 1 92 1 e i Z i n H 2n1 even terms cos6 odd terms sin0 cos6 139 sin 6 4 zxiyrei9 zxiyre39iU Figure 1 Representation of a complex number 2 and its conjugate 2 where 1 used 1392 1 1393 z39 1394 1 etc and identi ed the series for sin and cos of a real variablel Then going back and forth between rectangular and polar form is as easy as Z 6 My 715 My ricosemmei re 5 r 7 Note 2 6 2 6 2 6 2 6 COSOReei6 sin61mei6 2396 6 2 Now we can see an interesting relationship between these functions and their hy perbolic analogs 6 6 6 6 e T e sinh6 g 7 cosh 0 2 2 111 Roots and powers of complex variable Powers 2 a ib r629 z rnemg 8 1 03 ew4V 46m 4 Roots require a bit more care since they are fractional powers 7 rlneilt92mgtn m 0 i1 i2 9 This seems a little paradoxical Adding 2m7T to the argument of the exponential doesn t change 2 Nevertheless it has different nth roots according to how you choose n Any of em yawn7 are perfectly acceptable nth roots of 2 These are called the branches of the complex root function 2 5 Wk Figure 2 Roots of 372 2239 Note there are precisely 3 distinct roots in the complex plane Ex 2 2139 ei13 816eilt2 Tmgt m 0 1 2 10 m View4 11 m I 6211 12 m eig g 13 m same as m 1 14 7714 same asm2 15 16 Q How about 71 112 Complex power series functions of complex variable General Zn anzn Ex 1 z222 233 17 Ratio test for absolute convergence remember before we require that ratio of successive terms be less than 1 if limit exists an1z an 39 p lim TLHOO converges if p lt 1 18 so series in our example above converges if lt 1 This determines a disk of radius p in complex plane the disk of convergence The exponential series 2 00 Zn 6 E H 19 has an in nite radius of convergence since nlz n 1 n 1 for any 2 Similarly we de ne other complex functions by their power series eg gt p gt 0 20 sinhz E i COShZE i 21 2 2 sinz E l 00822 i 22 21 2 Note cos2 zsin2 z 1 remains valid for complex 2 as do all trig relations you know for real 2 Now we see relations between hyperbolic and ordinary trig functions sinz39z isinhz sinh 2392 isinz 23 cosiz coshz coshz39z cosz 24 Log function 00 1nl l l E n 25 M z n z lt gt and we have 61 2 as for real nos However consider 2 rel ln 2 lnrei9 Lnr 10 27Tn 26 ie in nitely many values The Ln written with large L is the natural log for real argument n 0 gives what is called the principal value of the log function lnz Lnri0 27 but don t forget that the other values or branches are also allowed Ex ln l ln 6i Ln 1 2397T 2mT lz397T l32397r 28 but the principal value is W Note Zln eilnz 6Lnrit92m7r Tuna3 29 113 Complex exponents and roots Ex 20PM e1iln2i e1iln2lni Now 139 itself may be written 139 eie m so lnz39 l 27Tn E 13901 So eliln2lni eln27ailn2a 2eiaeiln2a 114 Applications of complex numbers in physics Here I just give a few random examples where these concepts are useful 0 Adding harmonic waves with xed phase Suppose you want to know what the sum of a bunch of waves is I ll leave out the physical dimensions but you can imagine electric eld height of a water wave it doesn t matter much S sin x0 sinx x0 sin2x x0 sin3x x0 sinNx x0 32 This is hard to get a closed form for because there are many terms and one can t combine terms without using trig identities which are messy How about this let s write S lmSl where 81 eizo eizzo ei21zo eiNzzo e 1z2223zN7 33 where z e The terms in square brackets are just a geometric series with sum 1 zN11 So we could obtain a closed form expression iNz S lm 62101 6 34 1 6 Circular motion in complex plane and its uses If we take a function of time zt rem it is clear that z is following a circle of radius r in the complex plane starting at r70 at 150 and going counterclockwise The speed of rotation 7 is dzdt an etc So we can use 6 to represent an object going in a circle at angular frequency w or Reem to represent an object oscillating back and forth with position x rcos wt etc This has important technical uses in physics If something is oscillating with time with sinusoidal or cosinusoidal form we represent the signal 7 as the real or imaginary part of a complex exponential Often the resulting equations are much simpler to solve due to the nice mathematical properties of exponentials and we can always take the real part of the solution at the end of the calculation to obtain the physical observable desired Here s an example 5 0 Consider a series LRC circuit as shown The voltage drop across the resistor is R where I is the current through the resistor across the capacitor is 620 where Q is the charge on the capacitor and across the inductor is CH dtL due to Faraday s law Since the current leading to the capacitor has nowhere else to go it must be that I dQdt We can therefore express Kirchoff s law for the sum of the voltage drops around the circuit including the voltage source Vt equaling zero as L R Vocos wt 35 We assumed a cosinusoidal form for the driving voltage for simplicity but this is not essential Figure 3 Series circuit with inductor L resistor R and capacitor You probably solved a circuit like this in elementary Eampll but let s review First consider a simple case with no capacitance or inductance so only a loop with the single element R Then from Ohrn s law we know that the current and the voltage are proportional I VR 36 gooswt 37 Note both the current and the voltage have the same cosine forrn Second consider the case B 0 ie only L and C present Guess1 Q Q0 cos wt and substitute into 35 We get L w2Q0 cos wt cos wt Vocos wt 38 1note this way the differential equation is converted to an algebraic equation with parameters Q0 here to be determined by substitutionivery useful strategy which has the solution V0 Q0 2 w L C Note in this case the charge and voltage have the cosine form 39 General case Now if all 3 circuit elements are present Q Q0 cos wt is not a solution and neither is Q0 sin wt We can make a linear combination of the two and get the right answer but it s tedious Another way is to free our minds a bit and imagine that the voltage is complex oscillatory ie V Voem We will simply remember that the real voltage applied is the Re part of this expression and use the fact that the mathematics becomes much simpler rememberlng always to take the Be part of all quantltles at the end of the calculation So then let s make the ansatZ Q Qoem and substitute into the differential equation We obtain V0 1 2 39 zwt zwt L R V 40 l w w ClQoe 0e gtQo WLwR lt gt so we can write Weiwt t 41 w2Lle dQ Voeiwt Voeiwt I t E 42 0 dt mLR Z Z l gt so the current the complex current not the physical one looks just like our Ohm s law expression for the pure resistive case B only but with an effective resistance Z which we call the total impedance of the circuit We notice that it consists of three terms Z ZR ZL Z0 one associated with each of the circuit elements 1 ZR39 Z 39 ZtwL 43 R 7 C 7 L Now don t forget that the physical current is the real part of the result we obtainedl So the physical current is I ll be sloppy with notation and keep the same symbol Weiwt I t R 44 lt gt e Z lt gt It s easier to take the real part if we put the impedance in polar form Z Zei45 Imam tan1W 45gt 7 So V0 V0 15 Re elm 15 coswt Q5 46 Z Z So the current lags the driving voltage by a phase 5 which depends on the impedance of the circuit R L1 C1 R2 L2 02 ammo Figure 4 Parallel circuit with inductors L1 and L2 resistors R1 and R2 and capacitors 01 and 02 Now we could have guessed a form I olt coswt q plugged it into 35 and solved for 5 without all the complex number nonsense But it gets progres sively harder to make such guesses as the circuits become more complicated By contrast the complex voltage method is easily systematized Consider a circuit element consisting of two impedence branches in parallel as in Fig 4 In the complex voltage approach we note that the voltage across both legs is the same and the current is the sum of the current in both legs Therefore for the complex currents l l 1VZ1 2VZ2 IIl12v 7 Z1 Z2 so parallel impedances add just like resistances in parallel l l l arallel circuits 48 2m 21 22 p l l The impedance of each leg is Z L Z R Z0 in the example shown since the 3 elements are in series with each other as before Looking at general series circuits it s also easy to show Zwt Z1 Z2 series circuits 49 8 Electrodynamics Read Boas Ch 6 particularly sec 10 and 11 81 Maxwell equations Some of you may have seen Maxwell s equations on t shirts or encountered them brie y in electromagnetism courses These equations were written down for the rst time by Scottish physicist James Clerk Maxwell in his Treatise on Electricity and Magnetism 1873 and they caused a stir because his new equations proved that light was an electromagnetic phenomenon Imagine that you had no clue that light and all the phenomena of electricity and magnetism you knew from the laboratory were related and someone showed you that you could calculate the speed of some weird electromagnetic wave solutions to these differential equations and show that the speed was exactly that of light which had been measured astronomically Spectacular With the new tools you possess you can understand the equations at a deeper level Suppose charge is increasing at some rate within a given volume 739 Assume that we have no sources or sinks of charge in the system This means that it has to come from outside the region The amount created inside per time has to show up as a ux of the charge through the boundary of the volume coming from outside make sure you understand the sign 3 pFtdT idd vjd7 1 at vol 739 surf 8739 739 where the last equality follows from the divergence theorem Now since we did this for an arbitrary volume 739 it must hold locally 8p a a 390 2 atVJ the so called equation of continuity This is not thought of as one of Maxwell s equations because it doesn t contain the electromagnetic elds E and B but merely expresses the conservation of charge Here are the standard 4 l Gauss s law For pt charge A a A dc E7 Edd4i 2a1 716071 Closed 7T60 r 60 d9 10 00 Divergence theorem then says v m l pFdT E 3 60 60 q Maxwell 1 There are two mathematical subtleties l swept under the rug in this proof First I did it for a point charge but expressed things in terms of a general charge density at the end You can go back and convince yourself that if you say the E eld is a sum of many small charge elements dq each producing a eld falling off like 17 2 from itself you get the same answer Secondly in the last step we jumped from a statement about equality of integrated quantities to a statement about the equality of the integrands Normally this is a no no but here it s ok since we are talking about a relation which is valid for any p distribution We ll see the underlying reason for this kind of argument when we talk about function spaces No magnetic charge V B 0 Maxwell ll 4 You can make an analogous mathematical argument for the magnetic eld but since we don t know about any particles which carry magnetic charge yet we set the right hand side equal to zero Faraday s law A d A Edf Bdd 5 op surface A8 loop Now use Stokes7 theorem o a 4 8E vXEddfEdF dd 6 where we assumed that the loop over which the 39 was taken was xed in time only eld was changing so we could bring the time derivative inside the integral and apply it to the g only But now again we have integrals left and right over the same surface this time and the surface A is arbitrary so the only way the equation can hold is if 7 Maxwell Ill 2 4 Ampere s Law g 39 UOIencl IU O loop A enclosed by loop Using our previous reasoning and applying Stokes7 theorem we might come to the conclusion that vx ddMO5dd 7 x p0j 7 9 A A You might worry about this conclusion because if we take the divergence of both sides we see that 6 j 0 always whereas the continuity equation says there has to be another term dpdt when the charge density is changing locally in time What happened to it Well turns out the version of Ampere s law we started with wasn t the most general one it s valid only for stationary currents and elds We need to add something else if things are changing with time To guess what to add consider the AC circuit shown Figure 1 Geometry for calculation of displacement current 0 is a loop containing a wire leading to a capacitor S is an open surface enclosing C as is S Only 5 encloses the capacitor however The current owing through the wire is the time derivative of the charge Q on the capacitor I dQdt If we apply Stokes7 theorem to S and use the stationary form of Ampere s law we get dF xgd5p05ddp01 10 C S S no problem but if we were to do the same thing with S we would get zero since there is no current actually owing through 3 Remember we can t argue with Stokes7 lawithat s mathematics The paradox 7 suggests that there is a missing term in Ampere s law which turns on 7 when there is a time changing electric eld such as that which exists on the capacitor plate when charge is building up there Maxwell guessed a generalization of Arnpere s law v x g M06JD 5d E 60 11 Now applying Stokes7 law we get that the right hand side should be just like 10 except we replace jwith 5 id The second term id is called the displace rnent current it has the dimensions of a current but does not correspond to transport of charge And we nd that it doesn t depend any more which surface we choose since Sltiidgt la Sltiidgt dc7 12 For the rst term 96 0 over the surface S which may be taken far from the capacitor the displacement current is zero but the physical current is nonzero For 3 which passes rnostly near the capacitors surface there s no physical current but the charge is building up so E is changing with time Thus the current is exclusively displacement in nature Have we xed the continuity problern Take the divergence of both sides of 11 and using V V x 17 0 V17 we nd a a a 8 a 0 LUV 39 39 V 106an a T 8p 7 M0ltV 3 14gt where in the last step I used Coulomb s law 6 E peo So continuity is satis ed charge conserved so we can all rest easy in our beds For complete ness let me then record our answer from Maxwell for the modi ed Arnpere s law a A A 7 V X B 0 0605 15 Maxwell lV Some examples of the math of Eamp M Gauss law Ball of radius R constant chg density p could sum up the electric eld contributions dE from all in nitesimal charge elements dq or use divergence theorem and Poisson equation V E peo For a Gaussian sphere 7 A radius r gt R E dd Erda 47T7 2Er pd739 TFRSE 16 A A T 3 60 ngp Q 47mm 7 Em 17 47mm Where recall the symmetry argument that the eld must be radial due to the spherical symmetry of the charge distribution was crucial For 7 lt R L7 18 gt E 360 4 47T7 2E gm g p 60 Reminder how do we nd the potential q given E B B A Ed 19 A Choose reference point Q50 gt oo 0 i T Q Q T gt R i 00 47T r 260 47mm 20 T P P 2 2 R R d R 21 w lt gt lt gt R 36076 r 66006 gt lt gt Q 72 R2 22 660 7TR3 ooooo HNUJgtJgtU1 05 1 15 2 25 3r Figure 2 Potential of solid sphere of charge in units of QTFEOR 0 EM waves Maxwell s great achievementl Consider free space with p 0 j 0 such that V E 0 V B 0 Consider Ampere s law and Poisson eqn for B eld no magnetic monopoles 7 law 610 vx m4 ea 9 a A 9 BE 8 a a a a V X X Ll060V X W LlOEOE X U0 0 B o o a 9 8E vltv B v23 E 24 where I used a common vector identity setting V B 0 and the last step follows from Faraday s law This is now a differential equation for the components of B only 2 H 92 H V B B 25 06082 Compare with the wave equation we discussed earlier for propagating waves in 1D with speed c a 1282 J nim w 8052 c 81 and you will see that each component of B obeys its own wave equation meaning that propagating magnetic waves are a property of Maxwell s equations By comparing with 26 we see they have a speed 1 8 7 W 7 2997925 x 10 ms 27 The quantities uo and 60 are measured in the laboratory and 2997925 x 108 m s is very close to the known speed of light from astronomical ob servations in Maxwell s time It s vital to understand that the conclusion that light waves are electromagnetic waves so mundane sounding to us today was a dramatic discovery in Maxwell s day It uni ed the under standing of disparate phenomena light electricity amp magnetism which had previously been thought independent Note also that the speed in the equations is not given relative to some medium a fact which had profound consequences for Einstein s thinking Finally note that you will derive an identical wave equation for the electric eld B on the homework A prop agating EM wave in free space has the same magnitude of E and B with their polarizations perpendicular 6 Gauge transformations a a Maxwell H V B 0 Maxwell HI 6 x E lt25 scalar potential 30 Note 41 are not unique If A is any function of space and time we can make the changes 8A W 31 fi n i 7 gt Without changing the elds E g Check 5 Vector and scalar elds 51 scalar elds A scalar eld is a fancy name for a function of space ie it associates a real number with every position in some space eg in 3D p xyz We ve already encountered examples without calling them scalar elds eg the temper ature Txy in a metal plate or the electrostatic potential p xyz The gravitational potential is another and it s frequently convenient to think about potential landscapes imagining that a set of hills is a kind of paradigm for a varying potential since the height in this case scales with the potential mghx y itself Formally scalar is a word used to distinguish the eld from a vector eld We can do this because a scalar eld is invariant under the rotation of the coordinate system W05 1 2 015217 2 1 In other words I may label the point on top of one of the hills by a different set of coordinates but this doesn t change the height I assign to it This is in contrast to a vector eld where the values of the components do change in the new coordinate system as we have discussed 511 gradients of scalar elds If you re standing on the hill somewhere say not on the top there s one direction in my space which gives you the direction of the fastest way down This vector is Vq where p is the gravitational potential Consider the differential dq in 2D d g dx is j dy lg j 32 ids jdygt E W d W W d v drcoss 2 where 0 is the angle between the gradient W and the change in position dr so we see that the general change of p is the projection of the gradient onto the direction of whatever change one is making this is sometimes called a directional derivative One important way to remember about gradients of scalar elds is that they are always perpendicular to lines of constant scalar eld You know this if you ve 1 Figure l Contours of constant b and gradient a ever used a topographical map to navigate in the woods Figure 1 shows such a set of contours lines of equal height or gravitational potential which are sort of concentric The arrows shown are the gradients of the height or potential at the points shown Note a couple of things 0 the arrows point out so the map 7 must be of a valley in the center since the gradient points in the direction of steepest ascent o Vq points perpendicular to the lines of constant q o The arrows are longer Vq is bigger where the rise is steeper ie the contours are closer together Let s do some examples EX 1 rx2y222so v 39 39 k 1 zgt rz w z 8m 383 82 7 y so 7 increases fastest along f 7 no surprise EX 2 Fermi velocity In a metal the electrons make up a kind of a gas almost free Even at T 0 they are moving because the Pauli principle prevents them from being in the lowest zero momentum state The gure shows the allowed quantum mechanical energy states of an electron in a metal There can only be one spin up and one spin down 2 in each momentum state somewhat like an atom For a given hunk of material which has a given number of electrons in it the highest such occupied level has an energy referred to as the Fermi energy 6F In a typical light metal the Fermi energy is quite large of order a few eV In general this de nes some surface in momentum space because the energy momentum relation in the metal is For concreteness let s assume pi 193 p2 2mi Zm 6153 4 where 77mi and 77m7 don t really represent the components of a vector mass mass is a scalar right but some effective coe icients coming from solving the Schrodinger equation of quantum mechanics properly which happen to have di mensions of mass 2 VF Figure 2 Left states allowed by Pauli principle in a metal and de nition of Fermi energy 6F as highest occupied state Right ellipsoid representing surface in momentum space where 65 6F The Fermi velocity is now de ned as FEQWFEF If m Lmz it would point radially in the 13 direction but for the ellipsoidal case as shown we can calculate it to be a g A Pr V PeeF 1 ml fpy Apz k 5 ymi mzl F where the p95 py and p2 are the values of these quantities on the ellipsoid pz py p2 6F The Fermi velocity is always perpendicular to the Fermi surface In a simple metal this typical velocity of a conduction electron has a magnitude of about 1 100 to 110 of the speed of light 512 Transformation of scalar elds under rotations How does a scalar eld transform when the coordinate system is rotated Unlike the components of a vector eld see last week s notes a scalar eld transforms as x7y7Z39 m7y7z ie it is invariant Consider What this really means suppose you have a map and are looking at the height of a particular hill which on your map has coordinates x and 3 If someone else has a map with a rotated coordinate system coordinates 05 and 3 the height of that particular hill doesn t change just the coordinates 12 Analytic functions Read Boas Ch 14 121 Analytic functions of a complex variable Def A function f is analytic at 2 if it has a derivative there AZHO AZ which exists and is independent of the path by which one lets AZ gt 0 W Figure 1 Left function of 1 real variable Derivative does not exist at no because limit fz Ax 7 fzA is different from left or from right Right possible ways to approach 2 in the complex plane To clarify the importance of the path independence consider the complex func tion This looks smooth enough since we can write it as fxy x2 32 But considered as a complex function it is not analytic as we can see by applying the de nition fz AZ z Azz Az 22 zAz Azz lim E AZHO AZ AZ AZ 95 tyAx tAy a tyAx tAy i Ax tAy ZxAx ZyAy Ax tAy 2 Consider now path 1 approaching 2 Ag 0 A05 gt 0 Then derivative gt 205 On the other hand on path 2 A05 0 Ag gt 0 derivative gt 2iy So this is not an analytic function at any Z In general simple functions of z itself not z have regions where they are analytic If a function is analytic and single valued within a given region we call it regu lar If it is multivalued there are places where the function is not analytic called branch cuts 7 O zxiy out X Figure 2 Branch out of w 212 consider the function w ew2 w is a complex number let s call it w pew So we see that p TV and 2 6 We can see that the mapping is not 1 1 since both q and q 7T two different points in the w plane correspond to 0 and 0 27F ie the same 2 We can de ne a new function w which is single valued by restricting the value of 0 to lie between 0 and 27T This is a new complex function which is identical to the rst in this range of theta If we specify a branch cut 7 in the 2 plane as in Figure 2 the restriction of 0 amounts to a statement that we never cross 7 this when taking the square root M Cauchy Riemann conditions Theorem say f uz A neces sary and suf cient condition for f to be analytic is that a u and a u 3 8m 8y 8m 83 are satis ed 0 Check ex 1 22 so u x2 y2 and v 2953 i 81 81 8 11 895 T 83 895 T T 0 ex 2 ux andv y 8 8 8 8 1 gt not analytic 5 Sometimes we need the C R conditions in polar coords Again f u iv 2 6 2 re 8f dig ew i 9v 5 Br i 82 f 8r 15 6 8f 7 deziBf 2678 01 n E 6 7 df Bu Bu 1 Bu 01 29 gt 6 dz 8r Z9r ir lt80 180 8 8u71 8117 181 9 i an 7 8r r 196 8r r 86 Remarks o If f is regular in a region R derivatives of all orders exist there 0 A Taylor series is possible about any point 20 E R a0 a1z 20 a2z 202 1 with no fzo an f zo 10 n Region of convergence of series about 20 is a circle of radius equal to the distance 7 in the complex z plane between 20 and the nearest singularity o If is regular in R then u and v satisfy Laplace s equation eg Vgu Ol They are called harmonic functions 122 Line integrals and Cauchy s theorem lntegrals of analytic functions along paths have some spectacular properties which we will try to prove here in a somewhat nonrigorous fashion Let s rst make a statement Claim If f is analytic in a region R the line integral f 2612 11 C along any contour C connecting 21 and 22 which does not leave this region is the same In particular we can deform the contour C shown to new contour C 3 Figure 3 If fz is analytic in a region the line integral of f along any path between two points 21 and 22 is the same without changing the integral Let s leave this as a claim for now and prove a related proposition which clearly follows from it If we note that fmZhZZ dz fowzwl dz it is easy to see that if we integrate from 21 to 22 along C and come back to 21 along C we have made a closed loop If the integrals along the two paths from 21 to 22 were originally the same the integral around the closed loop must be zero Therefore another property about a line integral of a function analytic in a given region is fzdz 0 12 Proof Now call any closed curve in a region where uxy ivxy is analytic C see Fig 4 fzdz u ivdx idy udm vdy ivdm udy 13 C C C 1 Q 14 Now consider 1 and write the integrand as 13 df where F uf v 0 and apply Stoke s theorem 39 F df fA V x F dd where A is the area enclosed by C But you can easily calculate that v x F a v 9yu 0 15 9 U gt mgt oSDPw H where the last equality follows from the Cauchy Riemann conditions You can show by similar arguments that integral 2 is also zero 4
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