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# Electromagnetism PHY 712

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This 110 page Class Notes was uploaded by Miss Lucienne Hamill on Wednesday October 28, 2015. The Class Notes belongs to PHY 712 at Wake Forest University taught by Natalie Holzwarth in Fall. Since its upload, it has received 59 views. For similar materials see /class/230730/phy-712-wake-forest-university in Physics 2 at Wake Forest University.

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February 29 2000 Notes for Lecture 20 Magnetic eld due to electrons in the vicinity of a nucleus According to the Biot Savart law the magnetic eld produced by a current density Jr is given by I I up J r x r r Br 7 d 7quot 3 1 4W r r In this case we assume that the current density is due to an electron in a bound atomic state With quantum numbers nlmlgt described by a wavelunction 10mm 139 Where the azimuthal quantum number ml is associated With a factor of the form emW For such a wavelunction the quantum mechanical current density operator can be evaluated ah I I I a J r l 27m wrahmv WWW Wnlmzv filan 2 Since the only com lex art of this wavefunction is associated With the azimuthal uantum l l 1 number this can be written bah 392 i J ca W 8 8 W winmy 139 f W W l W l W 9 2mmquot sin 9 In 899 n m n m 899 77 m7quot sin 9 We need to use this current density in the Biot Savart law and evaluate the eld at the nucleus r O The vector cross product in the numerator can be evaluated in spherical polar coordinates g x r 7quot 32 cos 9 cos 9 37 cos 9 sin 9 2 sin 9 4 Thus the magnetic field evaluated at the nucleus is given by the integral p chm 30 Zwm l d3rl 32 cos 9 cos 9 37 cos 9 sin 9 2 sin 9 7quot sin 9 7J3 5 2 7 lm I In evaluating the integration over the azimuthal variable 9 the 5c and 7 coi nponents vanish leaving the simple result ifquot A 130 Ill I 1 a ca 1 2 0 I Wnlmz I 47777 L 6 3 quotPMquot 01011 xpx1LXxLX W7 1 WM LEWNM W7HWWWW7 Hz 39 1 z w Sanpm asap 0 Suonnqynum R A X3IXIX 30 sadzxquot mo aw awa lip lip 0 Wm mwmwmmwm mdgAmamm m1 39 T v u 1 I In u mun x x um 31 T r u Ml h33 d 1gt h p 9 5 WV asgqum 0 11452 S 2 311427103 will amqm A L v mmwamw aq o3 summing asaq 31123 am 31 39suopautg aduqs am 30 mm 319 uodn spuadap 5pm WM 30 mm aql nsuap 3an am 30 mm 319 moug o3 paw am asm mum am U1 4319 103 pm 7374 m mafiang mp awnxma amp aq o3 paw am tgasn aq 03 mme Sm m mpm uI 39qsgum mum Smpunoq am 31219 paumsszz wzq am mxsm Sm txgugmqo 1 p 39fi xyup fi xV Mipxp E 7319 pm fi qu JA fi x73q Aipxp E Wm amqm a 9 MD Layaway 3 mm am 30 Iqumd qua w xmug p Sumos Sq a 107 va fi mowv ZA uog31gtnba uossgod am 30 uogmxos uaAg p m paugmwmp aq um mum apm d1mgt aql 39fi 29Lq uogmun aduqs mp mm pawgaosmz apnmdum am s3tmsaxdax Q01 amqm L 1 5 2lt Llt Jlt a Z 3 kava mmj am sag Ixogsmzdxa Sgtp Suogsuamgp MA 1 fi ka M1qu paxg 30 Summing umoug 30 8mm Hg Hammad mesmmap umougun am 30 Ixogsmzdxa In no pastzq Sg JROJddR mamap anng aql poqqam 1119111919 91mm 8 answer 105 saqu 000239 12 Sumqu 1 39qazzoxddp mumng 03mg 0113 Sugsn panqu 030113 03 mgmgs aw 3mm 0113 312113 008 01 216111919960 069912260010 lt1 102299116190 088119991100 lt1 10A 9210 mm 104 9821171198180 661 8188860228 6221711986610 811 9816669099 1922921611789 quotlt1 8199860699 8617912201090 0q 0 punt aw snnsax 33mm 0113 pm suopmxba 0301 03 uogmxos m l 0 lt1 88 86 21 88 0 0 0 lt1 81 88 21 21 0 0 0 69 81 86 88 88 21 221 0 89 81 21 21 88 21 21 1 90 0 0 21 88 88 221 1 0 0 21 21 21 88 axdpm Sq 8185112th 103 mm xmmn 01 03ng 3nd aq um an Inqqmd sgq 103 Ipmuddp ummap 031ng aql 39OA 71 371 171 3121 summ sgq g 0305 mmaaq ug Sugxapug 011st mp Sum1 39OA ka o3 uomppp Hg 0 11 1 xqgt axgnbax pm 11 I S ii S 0 101801 01 03 8185112th mgwnnu mo pumxa um am 1 S ii S 0 101801 011 ug pa3saxa3txg Kym aw am mugs wmmoq 39b3 30 uommgwp mp 113111 ummsum mu 81 0A ka 30 0mm Sx1gtpunoq aql 39uonmggpom mm 0mm 0 01011 011 g 0305 mtmaq u passmsgp uqumd mum 5x12punoq 0113 01108 03 anbgqum 1113 asn o3 mpm 1 39qu sapnmdtmz mgwmg at 0 12 u 0 msn M um um 39b 01 san 12A Kw IHKK at th m In sum md xo 3 H y El I I 1 GSEMJGIQO 0 a v 1 11 1101th 12 1 31103 x 1 1 pm 1 31 103 731 g 8 2x11312sz 01 103 satqm pupsgp umtmoj 01 03 p201 Sumgpm ug Jgstzq asaql 11 0 11 x1 x1 1141 472r 4 2 U 1 11 WWW 95511 1419 11 1 4 11 58p 1 4 3 T 4 72 4 2 GI a l l I l 1141123 pm January 29 2000 Notes for Lecture 6 Complete function expansion methods for solving Poisson and Laplace equations As an example consider the 2 dimensional Laplace equation in the square lattice discussed in Lecture notes5 It can be shown that a series solutions takes the form sin2n l7r1a sinh2n l7rya 271 l l7r sinh2n l7r 1 103 2 2 4V0 n0 Refer to the beginning of the maple le lecture5mws to see plots of ltlgt1 y for 2 different summations of the series February 147 2008 Notes for Lecture 12 Dipole elds The dipole moment is de ned by p czarw 1 with the corresponding potential l p ltlgt if 2 r We 2 lt gt and electrostatic eld 1 3fp f39 7 p 47139 3 E 7 7 6 3 r We T3 3 p r lt gt The last term of the eld expression follows from the following derivation We note that Eq 3 is poorly de ned as r a 07 and consider the value of a small integral of EI39 about zero For this purpose7 we are supposing that the dipole p is located at I39 0 In this case we will approximate E1 0 m We Erd3rgt 631 4 First we note that LR Erd3r iRZ lt1gtrido 5 rR This result follows from the Divergence theorrn v Vd3r VdA 6 vol surface In our case7 this theorem can be used to prove Eq 5 for each cartesian coordinate if we choose V E X I r for the xi component for example Vltlgtrd3r 2 v EltIgtd3r y v ylt1gtd3r 2 v2lt1gtd3r 7 TSR SR SR SR which is equal to LR ltIgtrR2d 2 2 as y my 2 m LR ltIgtrR2in 8 Thus7 ASRErd3r 7 LR VltIgtrd3r iRZ lt1gtrido 9 rR Now7 we notice that the electrostatic potential can be determined from the charge density pr according to 1 pr 1 47139 Tl lt1gt 7d3 7 7 d 7ltY Ym 10 a 4 m r rirq 4 023ml rpo rglimagtz a lt gt lm We also note that the unit vector can be written in terms of spherical harrnonic functions 11 f sin0 cos sinz5 jL cos62 V4agmmjyumm Therefore7 when we evaluate the integral over solid angle 9 in Eq 57 only the l 1 term contributes and the effect of the integration reduced to the expression 1 47TH2 47reo 3 7132 lt1gt d9 d3 37 12 Mltm wm r ltgt The choice of rlt and rgt is a choice between the integration variable r and the sphere radius R If the sphere encloses the charge distribution pIquot7 then rlt r and rgt B so that Eq 12 becomes 1 47TH2 1 3 R2 L 38039 13 7R2 ltIgtrid9 d3r pr r E 7 47reo If the charge distribution pIquot lies outside of the sphere7 then rgt r and rgt B so that Eq 12 becomes 1 47TH2 pr 47TR3 7 2 7 3 7 7 R TR lt1gtI39I39dQ 47m 3 Rd 139 139 7 3 14 January 257 2008 Notes for Lecture 4 Examples of solutions of the onedimensional Poisson equation Consider the following one dimensional charge distribution 0 forxlt7a ipo for 7a lt z lt 0 1 p0 for0ltlta 0 forgta M96 We want to nd the electrostatic potential such that d2ltIgtz 7 2 dz 80 7 with the boundary condition 700 0 The solution to the differential equation is given by 0 forlt7a 7 a2 for7altxlt0 3 xi 72xia2p2 for0ltlta gaz forgta The electrostatic eld is given by 0 forxlt7a 77a for ialtxlt0 1513x7a for0ltlta 0 forgta This particular example is one that is used to model semiconductor junctions where the charge density is controlled by introducing charged impurities near the junction A plot of the results is given below Electric charge density Electric potential Electric field The solution of the above differential equation can be determined by piecewise solution within each of the four regions or by use of the Greens function Gxx 47rzlt7 where7 i 1 i 4713980 W96 L Gz spamy 5 In the expression for Gxx zlt should be taken as the smaller of z and x It can be shown that Eq 5 gives the identical result for ltIgt as given in Eq 3 Notes on the onedimensional Green7s functions The Greens function for the Poisson equation can be de ned as a solution to the equation V2Gzx 747T6 7 x 6 Here the factor of 47139 is not really necessary but ensures consistency with your text7s treat ment of the 3 dimensional case The meaning of this expression is that x is held xed while taking the derivative with respect to x It is easily shown that with this de nition of the Greens function 6 Eq 5 nds the electrostatic potential ltIgt for an arbitrary charge density In order to nd the Greens function which satis es Eq 6 we notice that we can use two independent solutions to the homogeneous equation v2 z 0 7 where 239 1 or 2 to form 4 Ga w WW 1zlt 2zgt s This notation means that zlt should be taken as the smaller of z and x and zgt should be taken as the larger In this expression W is the Wronskian d 195 45295 WE d 2z7 1 d 9 We can check that this recipe77 works by noting that for z 31 z Eq 8 satis es the de ning equation 6 by virtue of the fact that it is equal to a product of solutions to the homogeneous equation 7 The de ning equation is singular at z z but integrating 6 over z in the neighborhood of x z 7 E lt z lt x 6 gives the result dG Ls dG Lz In our present case we can choose 1z z and 2x 1 so that W 1 and the Greens function is as given above For this piecewise continuous form of the Greens function the integration 5 can be evaluated 1 Hz 4713980 1 Gltzz gtpltz gtdz emu3mm lt11 which becomes 1 m 00 ltIgtz 87 zpxdz z pxdx 12 0 700 m Evaluating this expression we nd that we obtain the same result as given in Eq In general the Greens function Gx solution 5 depends upon the boundary conditions of the problem as well as on the charge density In this example the solution is valid for all neutral charge densities that is 30 pzd 0 April 5 2009 Notes for Lecture 27 Derivation of the Li nardWiechert potentials and elds When we previously considered solutions to the inhomogeneous electromagnetic wave equa tions in the Lorentz gauge7 chapter 6 in Jackson we were using MKS units We keep these units in the following derivations Consider a point charge q moving on a trajectory Rqt We can write its charge density as and the current density as where peweRm HM qRqt53r RAW Evaluating the scalar and vector potentials in the Lorentz gauge7 i 1 i 47reo 131quot7 t and laweaehe w w Art dgr dt 17 W 7 t7 lr rlc 5 We performing the integrations over rst dgr and then dt 7 and make use of the fact that for any function of t L dt ft 6 t e t e lr e Rqt lcgt where the retarded time77 is de ned to be We nd and f t Rqti39rRqti 7 6 c g tT lrR l where we have used the shorthand notation R E I39 7 Rqt7 and V E Rqt7 In order to nd the electric and magnetic elds7 we need to evaluate 3A t Em 7Vlt1gtrt7 A 10 3t and Brt V gtlt Art 11 The trick of evaluating these derivatives is that the retarded time 7 depends on position r and on itself We can show the following results using the shorthand notation de ned above R Vt 7m 12 lt J and at R a ltR i 13 Evaluating the gradient of the scalar potential7 we nd 2 7Vlt1gtrti3 R1712 3 Riv R m 14 47TEO Rig 0 c c c and 6Art i q 1 VR 12 V R V R VR V R 15 3t 747m Ri f c 02 RC 02 02 c These results can be combined to determine the electric eld 1 R 2 R 13rti 3 R7Lgt 1712 ltRXltR7LgtXXZgt 16 47TEO R 7 0 c c c We can also evaluate the curl of A to nd the magnetic eld 7R 2 R R Bm mi ltgtgfltgh gtltgtwg39 0 0 R 7 L R 7 L One can show that the electric and magnetic elds are related according to R gtlt Er t B t 18 n w ltgt Two formulations of electromagnetic elds produced by a charged particle moving at constant velocity In Chapter 11 of Jackson page 559 7 Eqs 11151 2 and Fig 1187 we derived the electric and magnetic eld of a particle having charge q moving at velocity 1 along the 21 axis The results are for the elds at the point r big are U Yt l kaz E z z z 7t E 0b0t q 1 2 3 b21 yt232 19 and V5559 b2 W2 for the electric and magnetic elds respectively The denominators of these expressions are easily interpreted as the distance of the particle from the eld point as measured in the particles own reference frame On the other hand we can consider the same physical problem from the point of view of Lienard Wiechert potentials B172737t 3075707t q 20 Consider the electric eld produced by a point charge q moving on a trajectory described by r0t with prt E Q63r7r0t Assume that V0t E 6r0t6t and 62r0t6t2 0 Using the previously derived results for the Lienard Wiechert potentials changed into Gaussian units the electric eld can be written in the form 13070 q 170302R7V0Rc 7 q1 711302R7V0Rc7 21 47TEO 7 V0 RC3 Gaussian units 7 V0 RC3 where R E lRtl Rt E I39 7 r0t and where all quantities which depend on time on the right hand side of the equation are evaluated at the retarded time t E t 7 Rtc ln Gaussian units the corresponding magnetic eld is given by R gtlt E B 7 22 R lt gt If we evaluate this result for the same case as above Fig 118 of Jackson V0 E wig and Rt 711ter big In order to relate this result to Eqs 19 and 20 above we need to express t in terms of the known quantities Noting that Rt Ct 7 t m 2 b2 23 we nd that It must be a solution to the quadratic equation t3 7 272th yztz 7 VZbZCZ 0 24 t v wt 7 lt25 Now we can express the length parameter which appears in Eq 21 as with the physical solution B y lt7 v yt WW b2 26 We also can show that the numerator of Eq 21 can be evaluated R 7 VORc 77mg big 27 and the denominator can be evaluated 1 t 2 b2 R 7 v0 Rc L 28 Y Substituting these results into Eqs 21 and 22 we obtain the same electric and magnetic elds as given in Eqs 19 and 20 from the eld transformation approach April 9 2008 Notes for Lecture 31 Two formulations of electromagnetic elds produced by a charged particle moving at constant velocity In Chapter 11 of Jackson page 559 7 Eqs 11151 2 and Fig 1187 we derived the electric and magnetic eld of a particle having charge q moving at velocity 1 along the x1 axis The results are for the elds at the point r bxz are U Yt l Yb z E z z z t E 0b0t q 1 2 3 b21 yt232 1 and 2 b2 WW2 for the electric and magnetic elds respectively The denominators of these expressions are easily interpreted as the distance of the particle from the eld point7 as measured in the particles own reference frame On the other hand7 we can consider the same physical problem from the point of view of Li nard Wiechert potentials B172737t 30757070 9 Consider the electric eld produced by a point charge q moving on a trajectory described by r0t with prt E 163r 7 r0t Assume that V0t E 6r0t6t and 62r0t6t2 0 Show that the electric eld can be written in the form 1722R7R 1722R7R EM q lt voc gtlt v03 cgt a qlt voc gtlt v03 cgt7 3 4713960 R 7 V0 Rc Gaussian units R 7 V0 RC where R E lRt7l R02 E r 7 r002 and where all quantities which depend on time on the right hand side of the equation are evaluated at the retarded time t E t 7 RtTc In the same notation7 the magnetic eld is given by RgtltE B7 4 R ltgt If we evaluate this result for the same case as above Fig 118 of Jackson7 V0 E 0x1 and R03 7111 1 big In order to relate this result to Eqs 1 and 2 above7 we need to express t in terms of the known quantities Noting that Rt Ct 7 t m 2 1927 5 we nd that It must be a solution to the quadratic equation t3 7 27 yztz 7 yzbzcz 0 6 with the physical solution MW b2 mvltw lt7 0 Now we can express the length parameter which appears in Eq 3 as R y i v yt 1702 b2 8 We also can show that the numerator of Eq 3 can be evaluated R 7 VoRC 711321 bkz and the denominator can be evaluated m b2 Riv Rc 0 V 10 Substituting these results into Eqs 3 and 4 we obtain the same electric and magnetic elds as given in Eqs 1 and 2 from the eld transformation approach January 15 2009 Notes for Lectures 2 amp 3 Electrostatic energy Section 111 in Jackson7s text The total electrostatic potential energy of interaction between point charges at the positions r1 is given by igj 1 1 1 1117131 8713980 7 1117131 1 W 7 47Teo Ki In this expression the rst form explicitly counts all pairs while the second form counts all interactions and divides by 2 to compensate for double counting For a nite system of charges this expression can be evaluated directly however for a large or in nite system the expression 1 does not converge and numerical tricks must be used to evaluate the energy In fact for the in nite system one has an in nite amount of charge and the energy of interaction is unde ned If the system is neutral it is possible to de ne a meaningful interaction energy by use of an Ewald transformation The basic idea of the Ewald approach is as follows The error function erfx and its com plement erfc are de ned as 2 m 2 2 0 2 erfx O 6 dt and erfcx 1 7 erfz 6 dt 2 Ewald noted that f 1 f 1 1 er 7 7 er c e r 7 2W 2 t 3 r r T In this expression the rst term goes to a constant MnW as r a 0 but has a long range tail as r a 00 The second term has a singular behavior as r a 0 but vanishes exponentially as r a 00 This is illustrated in the plot below where 77 was chosen to be 77 4 elrfxx 9 1 erfxx 39 1x 39 8 7 6 Thus7 Ewald7s idea is to replace a single divergent summation with two convergent summa tions The rst summation has a convergent summation in the form of its Fourier transform and the second has a convergent direct summation Thus the calculation of the electrostatic energy would be evaluated using 1 1 W87 9 1 EagammarjgtZqiqjerfcltmra13gt 0 7 in 7 rjl 7 871780 7 in 7 rjl 7 in 7 rjl 4 For an appropriate choice of the parameter 77 the second summation in Eq 4 converges quickly and can be evaluated directly The rst term in the summation of Eq 4 must be transformed into Fourier space In order to described these summations explicitly7 we assume that we have a periodic lattice so that every ion can be located by 391 Ta T a location Ta within a unit cell and a periodic translation vector T7 In this way7 the summation becomes ZNE 5 ij 049T where N denotes the number of unit cells in the system Since we have a periodic system7 N is in nite7 but the energy per unit cell WN is well de ned The other identity that we 1Note that for any two lattice translations T and Tj T 7 Tj Tk7 where Tk is also a lattice translationi must use is that a sum over lattice translations T may be transformed into an equivalent sum over reciprocal lattice translations G according to the identity 1 26 e T 7 2e lt6 T Q G where 9 denotes the unit cell volume The proof of this relation is given in the appendix The rst term of Eq 4 thus becomes 2 qiqjerfo lri rjl N gqaq Z af x lm 7 n Tl 7 EggZ 7 7 W lriirjl T lTa T l Tl where the last term in Eq 7 comes from subtracting out the self interaction 239 j term from the complete lattice sum Using the short hand notation7 73904 E Ta 7 71 the lattice sum can be evaluated f l a T fl a Zer ZWlT JV dsrz saiT ZWlT JVI39l 8 T lTa Tl T lTa rl This becomes7 dareiGrerf l7a rl i 41 Z g iGmn 570202 l diu 9 9G lTa rl 79 0 G2 2 0 u3 7 where the last term7 which is in nite7 comes from the G 0 contribution If the last term of Eq 9 cannot be eliminated7 it is clear that the electrostatic energy is in nite The term can be eliminated if and only if the system is neutral Thus7 it is only meaningful to calculate the electrostatic energy of a neutral periodic system If the actual system has a net charge of Q E Ea qm we could calculate a meanful energy if we add to the actual charge density7 a compensating uniform density charge density of iQQ Taking all of the terms into account7 we nd the nal Ewald expression to be W qaq 47139 e iG39M 6 027 77 erfcl l7a Tl 417er i 7 L5a 2 7 N g 9 2 Ga 71 19 2 87m 0 T lug Tl 87TEOQ77 10 where the in the summation over lattice translations T indicates that all self interaction terms should be omitted Appendix I 7 comments on lattice vectors and reciprocal lattice vectors In this discussion7 will assume we have a 3 dimensional periodic system It can be easily generalized to 1 or 2 dimensional systems In general7 a translation vector can be described a linear combination of the three primitive translation vectors T17 T27 and T3 T 711T1 71sz 713T37 11 where 711712713 are integers Note that the unit cell volume 9 can be expressed in terms of the primitive translation vectors according to Q gtltT3gtl The reciprocal lattice vectors G can generally be written as a linear combination of the three primitive reciprocal lattice vectors G1 G2 and G3 G m1G1 szz m3G37 13 where 77117771277713 are integers The primitive reciprocal lattice vectors are determined from the primitive translation vectors according to the identities G1 Tj 2W61j Note that the volume of the primitive reciprocal lattice is given by lG1 G2 gtlt G3l 15 Some examples of this are given below Proof of Eq 6 Consider the geometric series if Mam Sin M G1 391quot 16 k7M T sinG1 r2 39 The behavior of the right hand side of Eq 16 is that it is small in magnitude very except when the denominator vanishes This occurs whenever G1 r2 n17r7 where 711 represents any integer If we take the limit M a 007 we nd that the function represents the behavior of a sum of delta functions sin G1 rgt 2 6 G 7 T 17 Mirnoo SinltG1r2 7T 1 139 I11 1 The summation over all lattice translations n1T1 is due to the fact that sinG1 r2 0 whenever I39 anl Carrying out the geometric summations in the right hand side of Eq 6 for all three reciprocal lattice vectors and taking the limit as in Eq 177 2611GT 271393 Z 39 I39 7 n1T16G2 39 I39 7 n2T26G3 39 I39 7 1 13T3 G 711712713 Finally7 the right hand side of Eq 18 can be simpli ed by eliminating the reciprocal lattice vectors from the 6 functions 6iGr 31397 317 lG1G2xG3l6 T 9N T 19 which is consistent with Eq 6 Appendix II 7 examples In these examples denote the length of the unit cell by a CsCl structure There are two kinds of sites 7 73905 0 and 73901 3 T1 a T2 357 T3 32 2 2 2 G1 12 G2 1y G3 12 21 a a a It is convenient to de ne 77 E uzaz Also we will denote by q the unit charge In these terms Eq 10 becomes for this case W q2 N gr 72 22 where 2 47f 2 2 2 1 7 6i7rm1mgmg6 7m1m2m3 713 Z r m2m2m2 7 23 m1m2mg70 1 2 3 and T i 2erfc n 71 Z 2erfcg 7112 7122 713 2 7 m2 i 7122 i ms2 mmm eo Mn 71 71 mmm 24 In this expression the sum over 04 and B has a total of 4 contributions 7 2 pairs of identical contributions For Cs Cs or 01 01 interactions 73904 0 and qa q resulting in repulsive contributions For Cs Cl or Cl Cs interations 73904 igb 3 2 and qa q resulting in attractive contributions This expression can be evaluated using Maple which gives the result W 2 q N 7 8W80a 4071 25 NaCl structure There are two kinds of sites 7 TNa 0 and 73901 T1 7xy T27y2 T37x2 26 G17x 72 G277x 2 G37x7S2 27 4 A A algal 942 gm i WQQE I w quot4 r4 j 7 A 4 v ll A I I I February 117 2009 Notes for Lecture 12 Dipole elds The dipole moment is de ned by p dsrpltrgtn 1 with the corresponding potential l p 10 R77 2 and electrostatic eld i l 3fpf 7p 47139 3 ml 7 We 7 3p 6 r lt3 The last term of the eld expression follows from the following derivation We note that Eq 3 is poorly de ned as r a 07 and consider the value of a small integral of EI39 about zero For this purpose7 we are supposing that the dipole p is located at I39 0 In this case we will approximate E1 0 m We Erd3rgt 631 4 First we note that E d3 7R2 ltIgt d9 5 SR 1 r ltrgtr lt gt This result follows from the Divergence theorm v Vdgr VdA 6 vol surface In our case7 this theorem can be used to prove Eq 5 for each cartesian coordinate if we choose V E xltlgtr for the x7 component for example LR Vltlgtrd3r xTSR v xltlgtd3r yTSR v yltlgtd3r 2792 v 2qgtd3r 7 which is equal to LR ltIgtrR2d 2 2 as y my 2 m LR ltIgtrR2doi 8 Thus7 LR Erd3r 7 LR Vltlgtrd3r iRZM ltIgtrido 9 Now7 we notice that the electrostatic potential can be determined from the charge density pr according to 71 3pr71 47quot 3 Tllt 3 Mr 7 47TEO d T 1quot 7 r l 7 47TEO gm 2l1d Tprrlgt15lmrnmr39 10 We also note that the unit vector can be written in terms of spherical harrnonic functions sin0 cos gt sin0 sin cost92 i 4w 2 up 06710 511 Y10i2gt 11 Therefore7 when we evaluate the integral over solid angle 9 in Eq 57 only the l 1 term contributes and the e ect of the integration reduced to the expression 1 47TR2 47TEO 3 7R2 lt1gtrid9 d3r pr LjE 12 7 Tgt The choice of rlt and rgt is a choice between the integration variable r and the sphere radius R If the sphere encloses the charge distribution pIquot7 then rlt r and rgt B so that Eq 12 becomes 1 47TR2 1 p 747m 3 dgrMr f E 7g 13 7R2 ltIgtrid9 39rR If the charge distribution pI39 lies outside of the sphere7 then rgt r and rlt B so that Eq 12 becomes 1 4 R2 4 R3 732 ltIgtrid9 77 W Rdgr r E LEW 14 712 47TEO 3 r 3 January 24 2000 Notes for Lecture 3 The mean value theorem for solutions to the Laplace equation Consider an electrostatic eld ltlgtr in a charge free region so that it satis es the Laplace equation V2ltlgtr 0 1 The mean value theorem value theorem states that the value of ltlgtr at the arbitrary charge free point r is equal to the average of ltlgtr over the surface of any sphere centered on the point r see Jackson problem 110 One way to prove this theorem is the following Consider a point r r u where u will describe a sphere of radius R about the xed point r We can make a Taylor series expansion of the electrostatic potential ltlgtr about the xed point r 1 2 1 3 1 4 ltlgtr u ltlgtr u Vltlgtr 5u V ltlgtr u V ltlgtr 1u V ltlgtr 2 According to the premise of the theorem we want to integrate both sides of the equation 2 over a sphere of radius R in the variable u 2W 1 dSu R2 dgbu dcos u 3 sphere 0 1 We note that 2W H R2 czqu dcos u1 47TR2 4 0 1 2W 1 192 czqu dcos9uu v 0 5 0 1 2W 1 4 4 122 czqu dcos9uuV2 7T3 v2 6 0 1 2W 1 R2 czqu dcos9uu V3 0 7 0 1 and 2 1 4 R4 7 122 czqu dcos9uuV4 2 7 v4 8 0 1 Since V2ltlgtr 0 the only non zero term of the average it thus the rst term 2W 1 RZO dgbu 1 dcos ultlgtru 47TR2ltIgtr 9 0r 1 2 1 ltlgtr 47TR2 1220 dgbu 1 dcos ultlgtru 10 Since this result is independent of the radius R we see that we have proven the theorem February 207 2009 Notes for Lecture 15 Vector potentials in magnetostatics The vector potential which vanishes at in nity and corresponds to a con ned current density distribution JI39 is given by M0 3 J 1quot A I39 7 d r 7 1 o M kg m This expression is useful if the current density JI39 is con ned within a nite region of space Consider the following example corresponding to a rotating charged sphere of radius 17 with p0 denoting the uniform charge density within the sphere and w denoting the angular rotation of the sphere 7 powxr forr ga Jltrgt T 0 otherwise 2 In order to evaluate the vector potential 1 for this problem7 we can make use of the expansion 1 47139 Tl Z 77ltmmltrmtnltw 3 gt r7rquot lm 2l1r Notingthat 4w xiy ixiy rr 7 Y I39 7Y r 7 31mgt um we see that the angular integral result takes the simple form mma a m2mmm wa Therefore the vector potential for this system is MOPOW gtlt 1quot 37 lt A 7 d 7 6 m Orr ltgt which can be evaluated as 2 2 7MP 7 570 for r S 1 A0quot 7 7031 for r 2 a As another example7 consider the current associated with an electron in a spherical atom In this case7 we assume that the current density is due to an electron in a bound atomic state with quantum numbers nlm as described by a wavefunction zbnlmxr where the azimuthal quantum number ml is associated with a factor of the form em For such a wavefunction the quantum mechanical current density operator can be evaluated 76h JO wilmlvwnlmz 7 Ibnzmlvzbilm 8 277m Since the only complex part of this wavefunction is associated with the azimuthal quantum number this can be written 7 h a a 7 h JI39 6 wilml wnlmg 7 wnlml wzlmlgt b 6 ml 2mmquot sint9 lwnlmg 2 39 mgr sin 6 where m5 denotes the electron mass and 6 denotes the magnitude of the electron charge For example consider the lnlm 211gt state of a H atom 1 r 1 2110quot 74 647mg Ee TZ Z s1n 0e 15 10 7 76h 7 64m57ra5 and e Tla 2 gtlt r 11 10 where 1 here denotes the Bohr radius Using arguments similar to those above we nd that 7 h A 00 Am M dr 3 eir a 7 12 192m57ra5r 0 T This expression can be integrated to give iehuoi gtlt I39 7 r r2 r3 A 1 7 W 1 7 7 7 13 r Smewrs e lt a 2a 813 Much m 2005 Notes for Lecture 22 y for anisotropic media 7 Extension oi Section 73 in t Re ectivit n s Jackso V tax WV 4i k WV hi I k above We Win amine that the penneahihty is in and that the pennittmty s is a berm whmh can be en in arm hi a iinitia dieiettnc tam K ee Fm simplicity we Win assume that the dielamc berm y the nieaiiim is diagonal and is yven by o o 0 Kw 0 1 o 0 K Win be claim by hzgmimy 2 h A Le so 4 electric eld inside the medium is given by E Em Eyy Ezzeiltwwgtm 3 In terms of this electric eld and the magnetic eld H BMO7 where H is assumed to have the same complex spatial and temporal form as 37 the four Maxwell7s equations are given by VH0 VHE0 4 VgtltE7zwu0H0 VgtltH weOHE0 Using these equations7 we obtain the following equations for electric eld amplitudes within the medium Hm 7 n nmny 0 Ex nmny Hyy 7 n 0 E 0 5 0 0 Is 7 E1 Once the electric eld amplitudes are determined7 the magnetic eld can be determined according to i 1 H 7 M00 Ezmyg 7 My Eynm i Emny eilt wwwygtm 6 The incident and re ected electro magnetic elds are given in your textbook In the notation of the gure the wavevector for the incident wave is given by kl 8sin 239 cos 239377 7 c and the wavevector for re ected wave is given by W i kR 7s1n 2x 7 cos 2y 8 0 In this notation7 Snell7s law requires that 711 sinz The continuity conditions at the y 0 plane involve continuity requirements on the following elds H0zt7 Ewz0zt7 EZ0zt7 and Dyx0zt7 9 at all times t Below we consider two different polarizations for the electric eld Solution for s polarization In this case7 Em Ey 07 and n Is 7 The elds in the medium are given by t 1 n E Ezieznxznyy7wt H W 7 817nxznyy7wt39 0 The amplitude E1 can be determined from the matching conditions E0 E E1 11 E0 7 E3 cosz39 Ezny E0 E3 sinz E1711 In this case7 the last equation is redundant The other two equations can be solved for the re ected amplitude E cost 7 n 0 y 12 E0 cos 2 my This is very similar to the result given in Eq 739 of Jackson for the isotropic media Solution for p polarization In this case7 E1 0 and 41212 7132 70 7 13 R212 In terms of the unknown amplitude Em7 the electric eld in the medium is given by E Ex 2 7 Kmmnm y einwmnyy7iwt39 14 Kyyny The corresponding magnetic eld is given by H 7 7281nxmnyy7iwt39 05 my The amplitude Em can be determined from the matching conditions E0 7 E3 cosz39 E1 16 E0 E3 LLJEm E0 E3 sinz Mn Er y Again7 the last equation is redundant7 and the solution for the re ected amplitude is given by E0 Hm cosz 7 my 17 E0 Hm cosz39 my This result reduces to Eq 741 in Jackson for the isotropic case February 21 2000 Notes for Lecture 5 Methods for solving Poisson equation There are are large nui nber of tools for solving the Poisson and Laplace equations 1 Green s function methods 1 3 1 84gt 8Gr r lt1gtr d Iquotpr39Grr E gnaw mow da 1 2 Coi nplete function expansions Fourier series etc 3 Variational methods 4 Grid based methods Introduction to gridbased methods The basis for most grid based methods is the Taylor s expansion 1 2 1 3 1 4 a lt1gtr u lt1gtr uVlt1gtr V lt1gtr Vlt1gtr V lt1gtr 2 We Will work out some explicit formulae for a 2 dii nensional regular grid With h denoting the step length For the 2 dii nensional Poisson equation we have 02 02 pm 1 lt1gt 39 3 W W e y lt gt We note that a sum of 4 surrounding edge values gives SA E 11 3 I y lt1gtx 3 1 lt1gtxy It 4 82 82 114 84 84 27 for if for Ly 1 8362 832 Ly 12 8364 834 Ly h Sir nilarly a sum of 4 surrounding corner values gives 5 E 11 3 1 I 3 1 11 3 1 I y It 5 2 84 84 82 82 8 82 114 v 22 7 7 39 i 7 7 bu Ly h 8362 832 Ly 6 8364 824 83628212 Ly h We note that we can combine these two results into the relation 1 3 14 SA a5 5lt1gtxy V2lt1gtxy gvzvqum 119 6 This result can be written in the form 3112 1 1 10331 55A i513 710plxry l M 114 2 20 MV Plx y a In general the right hand side of this equation is known and most of the left hand side of the equation except for the boundary values are unknown It can be used to develop a set of linear equations for the values of lt1gtxy on the grid points For example consider a solution to the Laplace equation in the square region 0 3 x 3 a 0 g y g a which lt1gtx0 lt1gt0y lt1gtay 0 and lt1gtxq H We will first analyze this systei n with a mesh of 9 points In this case 95 E is unknown while 91 92 93 1 and 94 96 97 98 99 0 13xaV0 c c amp L a 0 9 6 h v s 9 e Ix00 For this example Eq 7 states 1 1 3 95 F 992 94 96 98 091 93 97 99 jlo 8 3 This results is within 20 of the exact answer of 025 If analyze this same systei n with the next more accurate grid using the syi ni netry of the systei n lt1gtxy lt1gtq xy we have now 6 unknown values 5Q Q 8Q g9 119 12 and boundary values 91 92 93 1 and 94 97 9510 9513 9514 9515 ltIXaV0 0 ltIgtay Igtx00 This results in the following relations between the grid points 1 1 9 5 9 2 39139 9 4 39139 9 6 39139 9 8 209 1 39139 9 3 9 7 9 9 09 3 1 1 99c 50193 o 9 3 09 092 92 9 8 998 0 10 1 1 9 8 9 5 39139 9 79 9 39139 9 11 39139 9 6 39139 9 10 9 12 09 11 3 1 1 9 9 50 s 39139 9 8 39139 9 8 39139 9 12 9 5 39139 9 5 39139 9 11 39139 9 11 09 12 1 1 9 11 9 8 39139 9 10 39139 9 12 39139 9 14 WW7 39139 9 9 9 13391 9 15 09 3 1 1 9 12 9 9 39139 9 11 39139 9 11 39139 9 15 9 8 39139 9 8 39139 9 14 991401 I 0 These equations can be cast into the form of a matrix problem WlllCll can be easily solved using Maple 1 15 15 120 0 0 95 310 425 1 110 15 0 0 96 310 15 120 1 15 15 120 98 i 0 F 110 15 25 1 110 15 99 0 0 1 0 0 15 120 1 15 an 0 0 0 110 15 425 1 1912 0 The solution to these equations and the exact results are found to be 95 96 9911 9912 04628135839 05566467694 01920222635 012615955473 007150923611 010012503012 Vb exact 4320283318 5405292183 1820283318 025 06797166807 09541411792 We see that the accuracy has ii npi oved considerably with the new mesh m 15 January 16 2002 Notes for Lecture 1 1 Introduction 1 Textbook and course structure 2 Motivation 3 Chapters I and 1 and Appendix of Jackson Units SI vs Gaussian b Laplace and Poisson Equations c Green s Theorm 2 Units SI vs Gaussian Coulomb s law has the form F aw 1 712 Ampere s law has the form F 131 X 132 X flag 12 where the current and charge are related by 7C1 dqldt for all unit systems The two constants Kc and K14 are related so that their ratio KcKA has the units of 7725 and it is experimentally known that in both the SI and CGS Gaussian unit systems it the value CKA 02 where c is the speed of light The choices for these constants in the SI and Gaussian units are given below GGS Gaussian SI 7 L A C 1 4m 0 I i L A A 69 47 Here E 10 7NA2 and 1 02 10 7NA2 898755 gtlt 109Nm2C2 4mg Below is a table comparing SI and Gaussian unit systems The fundamental units for each system are so labeled and are used to de ne the derived units Variable SI Gaussian SI Gaussian Unit Relation Unit Relation length m fundamental cm fundamental 100 kg fundamental gm fundamental 1000 time 3 fundamental 3 fundamental 1 force N kg 77228 dyne gm C77L28 105 current A fundamental statampere statcoulomb 5 1 charge C A 5 statcoulomb U1 One advantage of the Gaussian system is that all of the eld vectors E D B Hg P M have the same dimensions and in vacuum B H and E D and the dielectric and permittivity constants 6 and u are unitless VxII4CJJ FqE gtltB u8EDBH S ExII u i CGS Gaussian SI VD47rp VDp VB0 VB0 VxE VxE VXHJ F qEvgtltB 1EDBII SEgtltH Proof of the identity Eq 131 V2 47r63r 1quot 3r rE Noting that fsmall sphere 137quot 63 101quot NH about Iquot we see that we must show that 39 3 2 small sphere d T V ET 1quot about Iquot We introduce a small radius a such that lim 6 I II r r tau EIIE2a2 For a xed value of a 1 3a2 2 H V I r 2a2 if T 2a25239 I If the function is continuous we can make a Tayor expansion about the point r Iquot J ackson s text shows that it is necessary to keep only the leading term The integral over the small sphere about Iquot can be carried out analytically by changing to a coordinate system centered at Iquot u r Iquot 8 so that 1 3a2 3 2 3 fsrtrliall sphere d T V i1 15 N r Algal R d a25239 9 a out 1 We note that 3a2 3 3a2u2 R3 3 r ultR d a2 4 0 du U2 a2 47 93239 10 If the in nitesimal value a is a lt R then R2 332 R3 and the right hand side of Eq 10 is 47r Therefore Eq 9 becomes fsmall sphere 137quot V2 1 x 47F 11 1quot T i about Iquot which is consistent with Eq 5 January 18 2008 Notes for Lectures 2 amp 3 Electrostatic energy Section 111 in Jackson7s text The total electrostatic potential energy of interaction between point charges at the positions r1 is given by igj 1 1 1 1117131 8713980 7 1117131 1 W 7 47Teo Ki In this expression the rst form explicitly counts all pairs while the second form counts all interactions and divides by 2 to compensate for double counting For a nite system of charges this expression can be evaluated directly however for a large or in nite system the expression 1 does not converge and numerical tricks must be used to evaluate the energy In fact for the in nite system one has an in nite amount of charge and the energy of interaction is unde ned If the system is neutral it is possible to de ne a meaningful interaction energy by use of an Ewald transformation The basic idea of the Ewald approach is as follows The error function erfx and its com plement erfc are de ned as 2 m 2 2 0 2 erfx O 6 dt and erfcx 1 7 erfz 6 dt 2 Ewald noted that f 1 f 1 1 er 7 7 er c e r 7 2W 2 t 3 r r T In this expression the rst term goes to a constant MnW as r a 0 but has a long range tail as r a 00 The second term has a singular behavior as r a 0 but vanishes exponentially as r a 00 This is illustrated in the plot below where 77 was chosen to be 77 4 elrfxx 9 1 erfxx 39 1x 39 8 7 6 Thus7 Ewald7s idea is to replace a single divergent summation with two convergent summa tions The rst summation has a convergent summation in the form of its Fourier transform and the second has a convergent direct summation Thus the calculation of the electrostatic energy would be evaluated using 1 W 7 8713980 M 1 Zqiqjer i rirjZqiqjerfC rirjgt39 j lri 7 I39jl T 87750 13974 lri 7 I39jl i 39 lri 7 I39jl 4 For an appropriate choice of the parameter 77 the second summation in Eq 4 converges quickly and can be evaluated directly The rst term in the summation of Eq 4 must be transformed into Fourier space In order to described these summations explicitly7 we assume that we have a periodic lattice so that every ion can be located by 391 Ta T a location Ta within a unit cell and a periodic translation vector T7 In this way7 the summation becomes ZNE 5 ij 049T where N denotes the number of unit cells in the system Since we have a periodic system7 N is in nite7 but the energy per unit cell WN is well de ned The other identity that we 1Note that for any two lattice translations T and Tj T 7 Tj Tk7 where Tk is also a lattice translationi must use is that a sum over lattice translations T may be transformed into an equivalent sum over reciprocal lattice translations G according to the identity 1 Z530quot T Ze G39 6 T Q G where 9 denotes the unit cell volume The proof of this relation is given in the appendix The rst term of Eq 4 thus becomes 2 Qiner Eeri 13D N anq Z er lh 7 Tl 7 Zqz 77gt 7 7 049 04 7 lriirjl T lTaiT Tl 0 7139 where the last term in Eq 7 comes from subtracting out the self interaction 239 j term from the complete lattice sum Using the short hand notation7 73904 E Ta 7 71 the lattice sum can be evaluated Z MEWWH Tl dgr63r 7 T erf l7a rl 8 T lTa Tl lTa rl 39 This becomes7 1 dSTEiGrerf l7a rl 7 47139 Z e iG39M 5 02712 1 w7 du 9 9G lTa rl 79 0 G2 2 0 U3 7 where the last term7 which is in nite7 comes from the G 0 contribution If the last term of Eq 9 cannot be eliminated7 it is clear that the electrostatic energy is in nite The term can be eliminated if and only if the system is neutral Thus7 it is only meaningful to calculate the electrostatic energy of a neutral periodic system If the actual system has a net charge of Q E Ea gm we could calculate a meanful energy if we add to the actual charge density7 a compensating uniform density charge density of iQQ Taking all of the terms into account7 we nd the nal Ewald expression to be W M119 47f e iG39M 6 027 77 8rf0l l7a Tl 47TQ2 7 7 Lga 2 N g 9 2 Ga 71 19 Z 8713980 0 T lug Tl 8711609777 10 where the in the summation over lattice translations T indicates that all self interaction terms should be omitted Appendix I 7 comments on lattice vectors and reciprocal lattice vectors In this discussion7 will assume we have a 3 dimensional periodic system It can be easily generalized to 1 or 2 dimensional systems In general7 a translation vector can be described a linear combination of the three primitive translation vectors T17 T27 and T3 T 711T1 71sz 713T37 11 where 711712713 are integers Note that the unit cell volume 9 can be expressed in terms of the primitive translation vectors according to Q gtltT3gtl The reciprocal lattice vectors G can generally be written as a linear combination of the three primitive reciprocal lattice vectors G1 G2 and G3 G m1G1 szz m3G37 13 where 77117771277713 are integers The primitive reciprocal lattice vectors are determined from the primitive translation vectors according to the identities G1 Tj 2W61j Note that the volume of the primitive reciprocal lattice is given by lG1 G2 gtlt G3l 15 Some examples of this are given below Proof of Eq 6 Consider the geometric series if 6MltG1ri sin G1rgt 16 k7M T sinG1 r2 39 The behavior of the right hand side of Eq 16 is that it is small in magnitude very except when the denominator vanishes This occurs whenever G1 r2 n17r7 where 711 represents any integer If we take the limit M a 007 we nd that the function represents the behavior of a sum of delta functions sin G1 rgt 2 6 G 7 T 17 MILD SinltG1r2 7T 1 139 I11 1 The summation over all lattice translations n1T1 is due to the fact that sinG1 r2 0 whenever I39 anl Carrying out the geometric summations in the right hand side of Eq 6 for all three reciprocal lattice vectors and taking the limit as in Eq 177 2611GT 271393 Z 39 I39 7 n1T16G2 39 I39 7 n2T26G3 39 I39 7 1 13T3 G 711712713 Finally7 the right hand side of Eq 18 can be simpli ed by eliminating the reciprocal lattice vectors from the 6 functions 6iGr 31397 317 lG1G2xG3l6 T 9N T 19 which is consistent with Eq 6 Appendix II 7 examples In these examples denote the length of the unit cell by a CsCl structure There are two kinds of sites 7 73905 0 and 73901 3 T1 a T2 357 T3 32 2 2 2 G1 12 G2 1y G3 12 21 a a a It is convenient to de ne 77 E uzaz Also we will denote by q the unit charge In these terms Eq 10 becomes for this case W q2 N gr 72 22 where 2 47f 2 2 2 1 7 6i7rm1mgmg6 7m1m2m3 713 Z r m2m2m2 7 23 m1m2mg70 1 2 3 and T i 2erfc n 71 Z 2erfcg 7112 7122 713 2 7 m2 i 7122 i ms2 mmm eo Mn 71 71 mmm 24 In this expression the sum over 04 and B has a total of 4 contributions 7 2 pairs of identical contributions For Cs Cs or 01 01 interactions 73904 0 and qa q resulting in repulsive contributions For Cs Cl or Cl Cs interations 73904 igb 3 2 and qa q resulting in attractive contributions This expression can be evaluated using Maple which gives the result W 2 q N 7 8W80a 4071 25 NaCl structure There are two kinds of sites 7 TNa 0 and 73901 T1 7xy T27y2 T37x2 26 G17x 72 G277x 2 G37x7S2 27 llll Illl 4 r 4 E 7 Y I II x gill LL 4 43 F LI I A j I I l February 19 2002 Notes for Lecture 15 Vector potentials in magnetostatics The vector potential corresponding to a current density distribution J r is given by Mo 3 I J0 A r 7 f d 7 7 1 47139 r r This expression is useful if the current density J r is con ned within a nite region of space Consider the following example corresponding to a rotating charged sphere of radius a with pa denoting the uniform charge density within the sphere and w denoting the angular rotation of the sphere 2 paw X r for 7quot g a Mr 0 otherwise In order to evaluate the vector potential 1 for this problem we can make use of the expansion 1 477 7 A m A I Emlehmvrzw lt3gt Sr r lm Noting that a g 3493 y31quot y moi 4 we see that the angular integral result takes the simple form A m A 7quot fdQ39ZYlmrlmr r39 r in a m Therefore the vector potential for this system is WOW XI39 I 37 lt A 2 139 39 7 6 r 37 u 7 7 g2 which can be evaluated 07mg for 7 g 1 Am lt7 5 for 739 2 a As another example consider the current associated with an electron in a spherical atom In this case we assume that the current density is due to an electron in a bound atomic state with quantum numbers Enlml described by a wavefunction vnlmlr where the azimuthal quantum number M is associated with a factor of the form em For such a wavefunction the quantum mechanical current density operator can be evaluated 67z Jltrgt m i mam lt8 6 Since the only complex part of this wavefunction is associated with the azimuthal quantum number this can be written eh A chmlc y 2 Jr am am 17 2 27nez7 smg nlm10 nmz nm10 nlm iersmgwnmz 9 Where me denotes the electron mass and 6 denotes the magnitude of the electron charge For example consider the inlm 211 state of a H atom I Eh ir a A I 1 Jr 64 muse zgtltr 0 Where 1 here denotes the Bohr radius Using arguments similar to those above we nd that Eh 0i X 139 00 I 3 7 l 739lt A 139 39 T m 7 11 r 1927mm57 0 7 7 O 7 This expression can be integrated to give ehpui X r 7 7 7 2 7 3 A 1 7 1 7 7 7 12 r 8mm 3 O a 2112 8113 March 297 2008 Notes for Lecture 28 Electromagnetic wave guides In order to understand the operation of a wave guide7 we must rst learn how electromagnetic waves behave in a dissipative medium A plane wave solution to Maxwell7s equations of the form k E Eoeik riiwt and B 712 X EoeikRriiwt w for the electric and magnetic elds7 with the wave vector k satisfying the relation k2 w2M82Rz I 2 We can determine the complex wavevector k 2k according to k 7 xR2IZRgt12 7 7 f xRZ 12 772 12 and ki 3 The form of the frequency dependent constants R and I depend on the materials For the Drude model at low frequency Eq 7567 R wzpeb and I Mm for example The value of ki determines the rate of decay of the eld amplitudes in the vicinity of the surface7 with the skin depth given by 6 E 1ki In the limit that I gt R as in the case of a good conductor at low frequency7 6 2wpa12 For an 77ideal conductor I a 007 so that the elds are con ned to the surface Because of the eld continuity conditions at the surface ofthe conductor7 this means that7 Btangential 31 0 because there can be a surface current7 Enomal 31 0 because there can be a surface charge7 but Bnomal 0 and Etangential 0 Suppose we construct a wave guide from an 77ideal conductor7 designating 2 as the propa gation direction We will assume that the elds take the form E Exyeikz m and B Bxyeikz m 4 inside the pipe7 where now k and e are assumed to be real Assuming that there are no sources inside the pipe7 the elds there must satisfy Maxwell7s equations 816 which expand to the following aBm aBy i 696 E szz 7 0 5 aEm aEy i 696 E zkEz 7 0 6 6E 6y 7 ME mew 7 zkEm 7 66L z wBy 8 7 66E szz 9 661 7 ME 7m8wEm 10 z kBm 7 6611 7mewa 11 7 665 7m8sz 12 Combining Faraday7s Law and Arnpere7s Law7 we nd that each eld component must satisfy a two dimensional Helrnholz equation 52 52 2 2 w a k 8 lm 0 13 with similar expressions for each of the other eld components For the rectangular wave guide discussed in Section 84 of your text a solution for a TE mode can have Ezxy E 0 and Bzxy B0 cos WWW cos 7 14 a 2 7 2 2 Wm 2 TM 2 7 With k km new 7 T From this result and Maxwell s equations7 we can determine the other eld components For example Eqs 7 8 simplify to k k Bm 77 d B 7Em 15 w 2 an 2 w These results can be used in Eqs 10 11 to solve for the elds Em and E and B1 and By in 7 QBZ i 7 mr 7mm mry E1 7 7By 7 k2 i Mng ay 7 b B0 cos lt a gts1n 16 a b w m QBZ m m7r 7mm mry E 7 Ex B 7 17 y k2 7 new as a Osmlt a gtCOSlt b gt One can check this result to show that these results satisfy the boundary conditions For exarnple7 Etangential 0 is satis ed since Emz7 0 Emz7 b 0 and Ey0y Eyay 0 This was made possible choosing Vszsu ace 07 where f1 denotes a unit normal vector pointing out of the wave guide surface January 19 2000 Notes for Lecture 2 Proof of the identity Eq 131 v2 1 47r63r r39 1 lr r l Noting that small sphere 370 63139 rlfr Z frla 2 about r we see that we must show that 1 fsmall sphere d3 V2 1 fr 47Tfrl 3 about r We introduce a small radius a such that 1 1 lim 4 r r a gt0 rr1 2la2 For a xed value of a 1 3a2 V2 5 W m2 a2 r rr cow If the function f r is continuous we can make a Tayor expansion about the point r r Jackson s text shows that it is necessary to keep only the leading term The integral over the small sphere about r can be carried out analytically by changing to a coordinate system centered at r u r r39 6 so that 2 f H h d3 V2 1 fr39 pal 7 S ts ere lr r l ultR u2 a252 a ou r We note that 3a2 3a2u2 R3 R 3 KR d uu2 a252 47T0 du U2 a252 47TR2 a23239 8 If the in nitesimal value a is a lt R then R2 0232 R3 and the right hand side of Eq 8 is 47r Therefore Eq 7 becomes small sphere d3 V2 lt about r 1 lr r l fr s fem 470 9 which is consistent with Eq 3 Examples of solutions of the onedimensional Poisson equation Consider the following one dimensional charge distribution 0 for 1 lt a p0 for a lt 1 lt 0 pct p0 for 0 lt 1 lt a 10 0 for513gta We want to nd the electrostatic potential such that d2lt1gtltccgt pm 332 8 0 11 with the boundary condition ltlgt oo 0 In class we showed that the solution is given by 0 for 1 lt a pg250a a2 for a lt 1 lt 0 W 00250I a2 p0a250 for 0 lt 1 lt a 39 12 p0a250 for 1 gt a The electrostatic eld is given by 0 for 1 lt a 00501 a for a lt 1 lt O 13 pg501 a for0lt513lta 0 for1gta The electrostatic potential can be determined by piecewise solution within each of the four regions or by use of the Green s function 0513 13 1lt where ltlgt1 i foo GI139p139d139 14 80 00 In the expression for G11 1lt should be taken as the smaller of 1 and 13 It can be shown that Eq 14 gives the identical result for ltlgt1 as given in Eq 12 January 287 2008 Notes for Lecture 6 The Greens function allows us to determine the electrostatic potential from volume and surface integrals lt1gt1 80 V dgrprGrr i S Grr v ltigtr 7 r v Grr f dzr 1 This general form can be used in 17 27 or 3 dimensions In general7 the Greens function must be constructed to satisfy the appropriate Dirichlet or Neumann boundary conditions For some special cases7 we can use the results of the method of images to construct Dirichlet Green7s functions as described in Section 26 of your text Orthogonal function expansions and Green7s functions Suppose we have a complete set of orthogonal functions de ned in the interval 1 3 x 3 2 such that 2 unum dab 6W 2 1 We can show that the completeness of this functions implies that 6Q 7 x 3 This relation allows us to use these functions to represent a Green7s function for our system For the 1 dimensional Poisson equation7 the Greens function satis es 62 7 G7 i 747T6z7 4 Therefore7 if d2 where also satisfy the appropriate boundary conditions7 then we can write the Greens functions as com 2W 6 For example7 if 4 2a sinn7Tza7 then 87139 Z sinn7rxa sinn7rx a Gxx 7 2 7 a n e These ideas can easily be extended to two and three dimensions For example if vnx and denote the complete functions in the x y and 2 directions respectively then the three dimensional Green7s function can be written 7 ulul9 vmyvmywn2wn2 waqwz iwg MWW o where d2 d2 d2 11K CWML mymyv and Vnwrdzl 9 See Eq 3167 in Jackson for an example An alternative method of nding Green7s functions for second order ordinary differential equations is based on a product of two independent solutions of the homogeneous equation u1z and u2 which satisfy the boundary conditions at 1 and 1 respectively 47139 du 2 dul Gxx Ku1zltu2xgt where K E 1H 112 10 with zlt meaning the smaller of z and x and zgt meaning the larger of z and x For example we have previously discussed the example of the one dimensional Poisson equation with the boundary condition ltIgt0 0 and dig 0 to have the form Gz z 47Tzlt 11 For the two and three dimensional cases we can use this technique in one of the dimensions in order to reduce the number of summation terms These ideas are discussed in Section 311 of Jackson For the two dimensional case for example we can assume that the Greens function can be written in the form G7z yy ZUnWWnWMmw l 12 If the functions satisfy Eq 5 then we must require that G satisfy the equation 2 VZG Zun96Un 7a gnyy 747695 i z 6y 7 y 13 The yidependence of this equation will have the required behavior if we choose 62 7Jmwimmiw m which in turn can be expressed in terms of the two independent solutions Um and vn2y of the homogeneous equation d2 dig 7 an m 07 15 and a constant related to the Wronskian 47139 Kn E d dual 16 UnZ ma WW2 If these functions also satisfy the appropriate boundary conditions7 we can then construct the 2 dimensional Green7s function from GWJCyw ZUnUn Knvm 98 11gt 17 For example7 a Green7s function for a two dimensional rectangular system with 0 S x S a and 0 S y 3 b7 which vanishes on each of the boundaries can be expanded await ismem1153sinhewn 18 As an example7 we can use this result to solve the 2 dimensional Laplace equation in the square region 0 S x S 1 and 0 S y S 1 with the boundary condition lt1gt0 lt1gt0y lt1gt1y 0 and lt1gt71 Vb In this case7 in determining x7y using Eq 1 there is no volume contribution since the charge is zero and the surface77 integral becomes a line integral 0 S x S 1 for y 1 Using the form from Eq 18 with a b 17 it can be shown that the result takes the form sin2n 17rxsinh2n 17ry 2n 17Tsinh2n 17T 19 WW Z 4 n0 February 11 2008 Notes for Lectures 10 85 11 Introduction to gridbased methods for solving Poisson and Laplace Equations Finite difference methods The basis for grid based nite difference methods is a Taylor s series expansion ltIgtr u ltIgtr u VltIgtr V2ltIgtr V3ltIgtr V4ltIgtr 1 We will work out some explicit formulae for a 2 dimensional regular grid with h denoting the step length For the 2 dimensional Poisson equation we have 82 wrwpw W 872 60 We note that a sum of 4 surrounding edge values gives 5AE Maw17y ltmih7y 7yh 7yih 82 82 I14 84 84 7 2 6 4ltIgtzy h 78302 7812 lt1gtampy E 78354 7814 m h Similarly a sum of 4 surrounding corner values gives SB xhyhltIgtx7hyhltlgtxhy7hltlgtxihyih 82 82 h4 84 84 82 82 7 2 4ltIgtzy2h lt78278y2gtlt1gtzy 6 lt 8y4 8x4 We note that we can combine these two results into the relation 1 312 2 774 2 2 6 SA 153 5 I957y TV 19079 t gv V 19079 t 71 This result can be written in the form 1 1 312 PWW BSA SB 7 4 2 20 1080p V pm 24 7ym if 6 6828y2gtlt1gtmyh 2 6 In general the right hand side of this equation is known and most of the left hand side of the equation except for the boundary values are unknown It can be used to develop a set of linear equations for the values of Mm g on the grid points lt1gtx2h V0 0 0 D2hy D0y 9 I i Figure 1 3 x 3 grid for solution of the Poisson equation within a 2 dimensional square Ix0 0 For example7 consider a solution to the Laplace equation in the square region 0 g x g 017 0 g y g a which ltIgtm0 ltIgtOy ltIgtay 0 and ltIgtza V0 We will rst analyze this system with a mesh of 9 points generated with a grid spacing of h In this case7 101 h E ltIgt9 9 is unknown while lt1gt02h ltIgth2h ltIgt2h2h V0 and ltIgt00 Wm lt1gt2h0 ltIgt07 h 101 2h O For this example7 Eq 6 states lt1gt5 ltIgth0lt1gt0hlt1gt2hhlt1gth2hlt1gt00ltIgt2h0lt1gt02hlt1gt2h21 v0 lt7 This results is within 20 of the exact answer of Mg 025 If analyze this same system with the next more accurate grid7 h using the symmetry of the system Mm y PC17m7 y7 we have now 6 unknown values 1017 hltIgt2hhltIgth7 2hltIgt2h2hlt1gth3hltIgt2h3h and boundary values lt1gt04h ltIgth 4h ltIgt2h4h V0 and lt1gt03h lt1gt02h lt1gt0h ltIgt00 ltIgth 0 ltIgt2h 0 This results in the following relations between the grid points ltIgth 3hlt1gth 4hlt1gt03hlt1gt2h 3hlt1gth 2h7lt1gt04hlt1gt2h 4hlt1gt2h 2hlt1gt0 21 07 ltIgt2h 3hlt1gt2h4hlt1gt3h7 3hlt1gth3hlt1gt2h 2hlt1gth4hlt1gt3h 4hlt1gt3h 2hlt1gt8h 21 07 ltIgth2h7lt1gth3hlt1gt02hlt1gt2h2hlt1gth7h7073hgtm2h73hgt lt07h 2h7h9 07 ltIgt2h 2hlt1gt2h3hlt1gth 2hlt1gt3h 2hlt1gt2h hlt1gt3h73hlt1gth 3hlt1gt3h Max112 07 Ml migmwhllmi hlt mhv hgt h700 2hlt1gt2h 2hlt1gt00lt1gt2h0 12 ltIgt2h07lt1gt2h 2hlt1gt3hhlt1gth hlt1gt2h07lt1gth 2hlt1gt3h 2hlt1gth0lt1gt3h0 0 13 x2h V0 CIX0 0 Figure 2 5 x 5 grid for solution of the Poisson equation within a 2 dimensional square These equations can be cast into the form of a matrix problem which can be easily solved using Maple 1 715 715 7120 0 0 ltIgth3h 310 725 1 7110 715 0 0 ltIgt2h3h 310 715 7120 1 715 715 7120 ltIgth2h 0 V0 14 7110 715 725 1 7110 715 ltIgt2h2h 0 0 0 715 7120 1 715 ltIgthh 0 0 0 7110 715 725 1 ltIgt2hh 0 The solution to these equations and the exact results are found to be ltIgth7 3h 04628135839 4320283318 ltIgt2h3h 05566467694 5405292183 ltIgth7 2h 01920222635 1820283318 V0 exact V0 15 ltIgt2h7 2h 02615955473 025 101 h 007150923611 06797166807 ltIgt2h7 h 01001250302 09541411792 We see that the accuracy has improved considerably with the new mesh Introduction to Finite element method The nite element approach is based on an expansion of the unknown electrostatic potential in terms of known grid based functions of xed shape In two dimensions using the indices 23 to reference the grid we can denote the shape functions as gti7 The nite element expansion of the potential in two dimensions can take the form 47r 0lt1gt907y Z ij ijlt7y7 16 H where 11 represents the amplitude associated with the shape function jugz The amplitude values can be determined for a given solution of the Poisson equation 7V2 47r60lt1gtm 47rpm y 17 by solving a linear algebra problem of the form ZMW Jij GM 18 27 where Mk1 Edzdyv klzy V gtijxy and GM Edzdy gtklxy 47rpmy 19 In obtaining this result we have assumed that the boundary values vanish This will be ensured by our choice of the functional form of the shape functions gtijz In order for this result to be useful we need to be able evaluate the integrals for MM and for GM In the latter case we need to know the form of the charge density The form of Mk1 only depends upon the form of the shape functions If we take these functions to be gtij7y E Xiyjy7 20 where 7 m7 7 lt lt Mag 1 h forzl hizizHJi 21 0 otherw1se and my has a similar expression in the variable y Then 7 mm pgm I I d311ltygtdyjltygt MW dz dy dz dz aimmay maxim dy dy lt22 There are four types of non trivial contributions to these values zih 1 alarm h lt17 lubzdu 3 lt23 miih 71 3 mih 1 h lt2 lt2w1ltzgtgtd2 h lt17 ugtudu g lt24 ziih 0 2M pam 2 1 1 2 Ark lt 192 gt Mfg71 MT 25 2M d2zd2c1z 1 1 71 AH lt7 7d gtdm77EO duff 26 These basic ingredients lead to the following distinct values for the matrix 2 fork1andlj MM 7 for 17111 andorliji1 27 0 otherwise nah t For problems in which the boundary values are 0 Eq 18 then can be used to nd all of the interior amplitudes 11172 In our case7 we have the boundary conditions ltIgtm0 ltIgt0y ltIgtay 0 and ltIgtma V0 Using the same indexing as in Fig 27 this means that 10471 1017471 II2I17 41 V0 The nite element approach for this problem thus can be put into the matrix form for analysis by Maple 83 713 713 713 0 0 01311 1 723 83 723 713 0 0 ltIgt2I13I1 1 713 713 83 713 713 713 01211 0 W 28 723 713 723 83 723 713 ltIgt2I12I1 0 0 0 713 713 83 713 ltIgtI1I1 0 0 0 723 713 723 83 ltIgt2I1I1 0 The solution to these equations and the exact results are found to be 1017 311 05070276498 4320283318 ltIgt2I13I1 5847926267 5405292183 1017 211 01928571429 1820283318 V0 exact V0 29 ltIgt2I12I1 02785714286 025 ltIgtI1I1 007154377880 06797166807 ltIgt2I17 I1 01009216590 09541411792 We see that the results are similar to those obtained using the nite difference approach January 257 2009 PJotes ar Lecture 4 Examples of solutions of the onedimensional Poisson equation Consider the following one dimensional charge distribution 0 forxlt7a ipo for 7a lt z lt 0 1 p0 for0ltlta 0 forgta M96 We want to nd the electrostatic potential such that mi 9 m dz 80 7 with the boundary condition 700 0 The solution to the differential equation is given by 0 forlt7a 7 a2 for7altxlt0 3 xi 72xia2p2 for0ltlta gaz forgta The electrostatic eld is given by 0 forxlt7a 77a for ialtxlt0 1513x7a for0ltlta 0 forgta This particular example is one that is used to model semiconductor junctions where the charge density is controlled by introducing charged impurities near the junction A plot of the results is given below Electric charge density Electric potential Electric field The solution of the above differential equation can be determined by piecewise solution within each of the four regions or by use of the Greens function Gxx 47rzlt7 where7 i 1 i 4713980 W96 L Gz spamy 5 In the expression for Gxx zlt should be taken as the smaller of z and x It can be shown that Eq 5 gives the identical result for ltIgt as given in Eq 3 Notes on the onedimensional Green7s functions The Greens function for the Poisson equation can be de ned as a solution to the equation V2Gzx i 747T6 7 x 6 Here the factor of 47139 is not really necessary but ensures consistency with your text7s treat ment of the 3 dimensional case The meaning of this expression is that x is held xed while taking the derivative with respect to x It is easily shown that with this de nition of the Greens function 6 Eq 5 nds the electrostatic potential ltIgt for an arbitrary charge density In order to nd the Greens function which satis es Eq 6 we notice that we can use two independent solutions to the homogeneous equation v2 z 0 7 where 239 1 or 2 to form 4 Ga w WW 1zlt 2zgt s This notation means that zlt should be taken as the smaller of z and x and zgt should be taken as the larger In this expression W is the Wronskian d 19c d 2 d WE 2z7 1 d 9 We can check that this recipe77 works by noting that for z 31 x Eq 8 satis es the de ning equation 6 by virtue of the fact that it is equal to a product of solutions to the homogeneous equation 7 The de ning equation is singular at z z but integrating Eq 6 over z in the neighborhood of x x 7 E lt z lt x 6 gives the result dGx z dG z mm 57 mz 7574 dgc l dgc 1 7T In our present case we can choose 1z z and 2x 1 so that W 1 and the Greens function is as given above For this piecewise continuous form of the Greens function the integration 5 can be evaluated lt1gt 80 L Ga x px dx Ga zpzd 11 which becomes 1 m 00 sOioozpzdx m pxdx 12 Evaluating this expression we nd that we obtain the same result as given in Eq In general the Greens function Gx solution 5 depends upon the boundary conditions of the problem as well as on the charge density In this example the solution is valid for all neutral charge densities that is 30 pzd 0 April 16 2008 Notes for Lecture 33 Synchrotron Radiation For this analysis we will use the geometry shown in Fig 149 of Jackson A particle with charge q is moving in a circular trajectory with radius p and speed 1 Its trajectory as a function of time t is given by Rm p smmmx p lt1 e max13gt y lt1 lts velocity as a function time is given by Vqt 1 cosvtpx 1 sinvtpy 2 The spectral intensity that we must evaluate is given by the expression Eq 1467 in Jackson d2 7 12612 dwdQ T 47TZC 2 3 12 X f X memt f qucldt foo After some algebra7 this expression can be put into the form d2 2 2 2 Mg 101ltwgt12101ltwgt12 lt4 where the amplitude for the light polarized along the y axis is given by dtsinvtpeiwt7 CostinHtp and the amplitude for the light polarized perpendicular to y and Iquot is given by CL w 00 dt sin 6 cosvtpei t C05951 t 6 We will analyze this expression for two different cases The rst case7 is appropriate for man made synchrotrons used as light sources In this case7 the light is produced by short bursts of electrons rnoving close to the speed of light 1 Cl 71272 passing a beam line port In addition 6 w 0 and the relevant integration times t are close to t m 0 This results in the form shown in Eq 1479 of your text It is convenient to rewrite this form in terms of a critical frequency 3 Boy we W 7 The resultant intensity is then given by d2 3 1272 W2 222 W 223 2 V292 6 223 2 7 1 6 K 1 6 7 K 1 6 dwd 4620 7 l 2615 7 1v262l 26 7 8 By plotting this expression as a function ofw we see that the intensity is largest near w m we The second example of synchroton radiation comes from a distant charged particle moving in a circular trajectory such that the spectrum represents a superposition of light generated over many complete circles In this case there is an interference effect which results in the spectrum consisting of discrete multiples of vp For this case we need to reconsider Eqs 5 and 6 There is a very convenient Bessel function identity of the form 00 e7iasinuz Z Jmae7imuz39 m7oo Here Jma is a Bessel function of integer order m In our case a Efcost and Oz 7 Analyzing the parallel77 component we have c 3 00 7g 2 c 3 wp 1 O lwt Ccos9sin39utp Jm lt7 2 6 7 7 39 H 7mm 3cos0 700 e 7 wp6cos t9 7 c COS 7T w mp 10 In determining this result we have used the identity 00 iw7m3t i 1 dte 7 2W6w 7 m7 11 700 p Eq 10 can be simpli ed to show that CH 2m 2 n cos 19gt 6a 7 my 12 700 C P where Ja E djgsa The perpendicular component can be analyzed in a similar way using integration by parts to eliminate the extra cos1tp term in the argument The result is tant9 2W7 C 2 Jm cos 19gt 6a 7 13 In both of these expressions the sum over m includes both negative and positive values of m However only the positive values of w and therefore positive values of m are of interest and if we needed to use the negative m values we could use the identity Jma 71mea 14 CL 11 Combining these results we nd that the intensity spectrum for this case consists of a series of discrete frequencies which are multiples of vp 2 Jm cos 19gt 15 c 2 2 wp tan 6 J b 01 m C cos vzCz These results were derived by Julian Schwinger Phys Rev 75 1912 1925 1949 The discrete case is similar to the result quoted in Problem 1415 in Jackson7s text It should have some implications for Astronomical observations but I have not yet found any references for that d2 q2w2 2 00 U 5 7 dwdQ 0 W20 w mp January 23 2002 Notes for Lecture 3 Form of Green s function solutions to the Poisson equation According to Eq 135 of your text for any two threedimensional functions 0r and meun uu wMVDJW f waywa wmvaawwn w Surf where i39 denotes a unit vector normal to the integration surface We can choose to evaluate this expression with 0r ltIgtr the electrostatic potential and Grr and also make use of the identities vM 9 m 50 and V2Gr r 47r6r r Then the Green s identity 1 becomes I I pg I I I I I 1 4f 27 47rV01ltIgtr6r r Grr Lyra d3quot urflt1gtrVGrr Grr V39ZM d 4 This expression can be further evaluated If the arbitrary position r is included in the integration volume then the equation becomes 1 39 ltIgtr orr39 rd3r 7 f Grr VltIgtr ltIgtrVGrr 4f39d2r 5 Vol 47r o 47F Surf This expression is the same Eq 142 of your text if we switch the variables r ltgt r and also use the fact that Green s function is symmetric in its arguments Gr r E Gr r Mean value theorem for solutions to the Laplace equation Consider an electrostatic eld ltIgtr in a chargefree region so that it satis es the Laplace equation V2ltIgtr 0 6 The mean value theorem value theorem problem 110 of your textbook states that the value of ltIgtr at the arbitrary chargefree point r is equal to the average of ltIgtr over the surface of any sphere centered on the point r see Jackson problem 110 One to prove this theorem is the following Consider a point r r u where u will describe a sphere of radius R about the xed point r We can make a Taylor series expansion of the electrostatic potential ltIgtr about the xed point r 1 5 According to the premise of the theorem we want to integrate both sides of the equation 7 over a sphere of radius R in the variable 11 ltIgtr u ltIgtr 11 4 VltIgtr 114V2ltIgtr gm 4 V3ltIgtr in A V4ltIgtr A A A 7 23 1 2 u s u 8 Lphorodsu R do 7 dcos9 We note that 2T 1 R2 dau dcos9u1 4M2 9 t D t 1 23 1 R2 dau dcos9uu V 0 10 0 r 1 271 1 47rR 1 2 v A I 2 7 2 R don7 dcos9uu V 3 v 11 23 1 A R2 dau dcos9uu4V3 0 12 0 r 1 and 6 v 13 23 1 R2 d 39u d Qu A V 4 0 a 1 m u gt Since V2ltIgtr 0 the only nonzero term of the average it thus the rst term 23 1 R2 D dau dcos9ultlgtr u 47rR2ltIgtr 14 t t 1 01 1 2 23 1 v 141 R don7 dcos9ultlgtru 19 Since this result is independent of the radius R we see that we have proven the theorem April 8 2002 Notes for Lecture 26 Re ectivity for anisotropic media 7 Extension of Section 73 in Jackson s text 0 El 0 50 Consider the problem of determining the re ectance from an isotropic medium as shown above For simplicity we Will assume that the dielectric tensor for the medium is diagonal and is given by Km 0 0 H E 0 Hwy 0 t 1 0 0 If We Will assume also that the wave vector in the medium is con ned to the L39 9 plane and will be denoted by w kt E C 9 We will assume that the electric eld inside the medium is given by E Ex Eyy Ezieiltnwnyygt 3 In terms of this electric eld and the magnetic eld H B um Where H is assumed to have the same complex spatial and temporal form as 3 the four Maxwell s equations are given by V4H0 V H E30 VXE iwugH0 VXHiweng0 Using these equations we obtain the following equations for electric eld amplitudes within the medium mm n 911 0 Efr nxny 11W 0 Ey OI 0 Eu Ez Once the electric eld amplitudes are determined the magnetic eld can be determined according to 1 i IILTIEdq ngaamx Enq d w wwt m 0 The incident and re ected electromagnetic elds are given in your textbook In the notation of the gure the wavevector for the incident wave is given by w A A k FSanX cos 2y 7 and the wavevector for re ected wave is given by W A A k3 Fsin 2x cos 2y 8 In this notation Snell39s law requires that 911 sin The continuity conditions at the y 0 plane involve continuity requirements on the following elds Hw02 Exc027 Ezlc027 and Dyw0z 9 Below we consider two different polarizations for the electric eld Solution for spolarization In this case Ex Ey 0 and n mu The elds in the medium are given by A m 1 A A m E Ezze1quot quotyyl H H mm um WWW 10gt Q The amplitude Ez can be determined from the matching conditions mma OD Ea E Q cosi Ezny Eg Eg sini 53an In this case the last equation is redundant The other two equations can be solved for the re ected amplitude Equot cosi n i44 m Ea cos 2 y This is very similar to the result given in Eq 739 of Jackson for the isotropic media Solution for p polarization In this case Ez 0 and H 91 04 13 H w In terms of the unknown amplitude E1 the electric eld in the medium is given by E Efr 5 Cinxnyy 14 yyny The corresponding magnetic eld is given by E m H ii01n fnyy 15 we y The amplitude Efr can be determined from the matching conditions E0 E3 cosi Efr 16 E0 E3 E0 E Q sini my Ext n Again the last equation is redundant and the solution for the re ected amplitude is given by Equot cos in n 70 m yr 17 E0 cos mm y This result reduces to Ed 741 in Jackson for the isotropic case April 15 2000 Notes for Lecture 34 Two formulations of electromagnetic elds produced by a charged particle moving at constant velocity In Chapter 11 of Jackson page Eqs 11151 2 and Fig 118 we derived the electric and magnetic field of a particle having charge 1 moving at velocity v along the 51 axis The results are for the fields at the point r big are 39U f51 7 322 ExxixtE00tq l 2 l 32 9702 1 and 2 A 7 sz 32 9702 for the electric and magnetic fields respectively The denoi ninators of these expressions are easily interpreted the distance of the particle from the field point i39neasured in the particle s own reference frame On the other hand in Chapter 6 and in HVV16 we considered the same physical problem from the point of view of Lienard VViechert potentials Bx1x2x3t 130 MM 11 2 Consider the electric field produced by a point charge 1 moving on a trajectory described by rot with pr t E 1631quot rot Assume that Vot E 8rot8t and 82rot8t2 0 Show that the electric field can be written in the form 1 1 vgCZMR VoRc gt 1 vgCZMR VoRc E 7 4 7 rd 47760 R V0 Rj3 Gaussian units 1 R V0 Rj3 where R E Rtr RM E r row and where all quantities which depend on time on the right hand side of the equation are evaluated at the retarded time tr E t Rtrc If we evaluate this result for the same case above Fig 118 of Jackson v0 E toil and RM ionic big In order to relate this result to Eqs 1 and 2 above we need to express tr in terms of the known quantities Noting that HUT Ct tr ctr2 I 4 we find that tr must be a solution to the quadratic equation if 272 72 725262 0 5 with the physical solution C 39Ipzzoxddp IXOQRIILXOJSURH may am mm 39b In muff a 3mm a 11st at 3 mm 0 am 391 mm s nsm amt 3 3IH3IQQS n 3 I H H E I i I E El I i I b I 39 oH0A H 2 ZWMV pamnmm aq um xomugmouap am pm 6 8 ZXQ 1w oHGA H IGRHRAG Gq U23 339 b 30 103121811le Gq 312g MOQS U123 081 6AA A 39 2 AM mg L H g bg ug smaddp ngm mmmmzd 98qu mp ssmdxa um am MON April 8 2002 Notes for Lecture 27 Electromagnetic wave guides In order to understand the operation of a wave guide we must rst learn how electromagnetic waves behave in a dissipative medium A plane wave solution to Maxwell s equations of the form E Eoemf PW and B gtlt Eoeik 39FW 1 for the electric and magnetic elds with the wave vector k satisfying the relation k2 bogus a 73 71 2 We can determine the complex wavevector k 71 according to 1 7 2IQ7 1 12 Ra 73 1 2 k and hi 3 2 2 The form of the frequency dependent constants R and I depend on the materials For the Drude model at low frequency Eq 756 R 021155 and I who for example The value of hi determines the rate of decay of the eld amplitudes in the vicinity of the surface with the skin depth given by 6 E 119 In the limit that I gtgt 73 as in the case of a good conductor at low frequency 6 z 2 wua1 j2 For an ideal conductor I gt 00 so that the elds are con ned to the surface Because of the eld continuity conditions at the surface of the conductor this means that Biangemm a 0 because there can be a surface current Emma a 0 because there can be a surface charge bun Bnormal 0 and Eiangeniial Suppose we construct a wave guide from an ideal conductor designating i as the propa gation direction We will assume that the elds take the form E 1390v Mews W and B Bcv ye quot39W 4 inside the pipe where now k and 5 are assumed to be real Assuming that there are no sources inside the pipe the elds there must satisfy Maxwell s equations 816 which expand to the following as 833 7 103 aw 69 7 0 o 813 65 7 E aw 69 7k 0 6 8E ikEy m5 7 39 ikEm 8E2 iny 8 870 if 8 inz 9 8615 ikBy 7J wE 10 7ka mawa 11 if 21 4mm 12 Combining Faraday s Law and Ampere s Law we nd that each eld component must satisfy a twodimensional Helmholz equation 82 a2 2 2 with similar expressions for each of the other eld components For the rectangular wave guide discussed in Section 84 of your text a solution for a TB mode can have 1347039 E 0 and 84 y Bo cos Trix cos i 14 2 2 with k2 E kfm nan From this result and Maxwell s equations we can determine the other eld components For example Eqs 7 8 simplify to k k Bx Ey and By 10 These results can be used in Eqs 10 11 to solve for the elds Ex and l5y and Bx and By w 7Cw 88 7Cw n71quot max mry ERR B 7 16 y k2 Mawg ay 72 b o COSlt a sin b V l w 85 mr max mry r T B 7 7 1 Ey 8 k2 1502 890 mwf a Osm a COS b 7 One can check this result to show that these results satisfy the boundary conditions For example Etangential 0 is satis ed since 1349030 Ewe b 0 and Eym y 13301 9 0 This was made possible choosing Vszsumm 0 where f1 denotes a unit normal vector pointing out of the wave guide surface January 257 2008 Notes for Lecture 5 Interesting properties of the Poisson and Laplace Equations Mean value theorem for solutions to the Laplace equation Consider an electrostatic eld ltlgtr in a charge free region so that it satis es the Laplace equation V2lt1gtr 0 1 The mean value theorem77 value theorem problem 110 of your textbook states that the value of ltlgtr at the arbitrary charge free point r is equal to the average of ltlgtr over the surface of any sphere centered on the point r see Jackson problem 110 One way to prove this theorem is the following Consider a point r r u7 where u will describe a sphere of radius R about the xed point r We can make a Taylor series expansion of the electrostatic potential ltlgtr about the xed point r 1 1 1 ltlgtr u ltlgtr u Vltlgtr Eu V2ltlgtr u V3ltlgtr 1u V4ltlgtr 2 According to the premise of the theorem7 we want to integrate both sides of the equation 2 over a sphere of radius R in the variable u 27r 1 7 2 Aphere dSu 7R 0 don71 dcost9u 3 We note that M 1 2 7 2 R 0 ddhi1 dcos6u1i 4m 4 27r 1 2 7 R 0 ddhi1 dcost9uu V 0 5 27r 1 47TR4 2 2 2 R 0 day71 dcos0uu V 3 V 6 27r 1 2 3 7 R 0 day71 dcost9uu V 0 7 and 2 1 4 R6 2 7r 4 7 7T 4 RA day71 dcos0uu V 7 5 V 8 Since V2ltlgtr 07 the only non zero term of the average it thus the rst term R2 A2 dab 1 dcos6ultlgtr u 47TR2ltlgtI39 9 01 1 27r 1 1 77122 du d 97m 27 dSultIgt 10 47TR2 0 71 cos r u 47TR2 sphere r u Since this result is independent of the radius R7 we see that we have proven the theorem ltIgtI39 Form of Green7s function solutions to the Poisson equation According to Eq 135 of your text for any two three dimensional functions 151 and 71 7 V 1 lt rV2wr 7 ZI39V2 r d3r 7 u rwr 7 rv r f39d2r 11 where f39 denotes a unit vector normal to the integration surface We can choose to evaluate this expression with 151 ltIgtI39 the electrostatic potential and 7139 G031quot and also make use of the identities V2ltIgtr 7 12 and V2Grr 747T6r 7 I39 13 Then7 the Greens identity 11 becomes 47mg 74wwltlt1gtr5r 7 r 7 Gr ramp d3r 7 mflt1gtrVGrr 7 Grr VltIgtr f39d27quot 14 This expression can be further evaluated If the arbitrary position7 Iquot is included in the integration volume7 then the equation 14 becomes lt1gtr 7 Vol Grr d3r 7 i u Gr r vlt1gtr 7 lt1gtrvcr r f39dzr 15 This expression is the same as Eq 142 of your text if we switch the variables Iquot gt I39 and also use the fact that Green7s function is symmetric in its arguments GI39Iquot E Gr r January 24 2002 Notes for Lecture 4 Electrostatic energy Section 111 in Jackson s text The total electrostatic potential energy of interaction between point charges 1 at the positions ri is given by 1 4760 Z Iin 1 Z qqj 1 iltj Eri I39ji 875017 il i I39ji IV In this expression the rst form explicitly counts all pairs while the second form counts all interactions and divides by 2 to compensate for double counting For a nite system of charges this expression can be evaluated directly however for a large or in nite system the expression 1 does not converge and numerical tricks must be used to evaluate the energy In fact for the infinite system one has an in nite amount of charge and the energy of interaction is unde ned If the system is neutral it is possible to de ne a meaningful interaction energy by use of an Ewald transformation The basic idea of the Ewald approach is follows The error function erf and its com plement erfcc are defined 2 I if i x if HM fe 6 1t and erfcx 1 erfx 7r 6 it Ewald noted that 1 erfg r erfcf r 7 3 r r r In this expression the rst term goes to a constant T gt 0 but has a long range tail T gt 00 The second term has a singular behavior T gt 0 but vanishes exponentially T gt 00 Thus Ewald s idea is to replace a single divergent summation with two convergent summations The rst summation has a convergent summation in the form of its Fourier transform and the second has a convergent direct summation Thus the calculation of the electrostatic energy would be evaluated using 1 W 1 2ij Iin p 1 Z qiqjorff zzri rjz Zqiqj0rfcb zri rjs 87 ij ri r3 87 ij ri r3 ij Eri T3 4 For an appropriate choice of the parameter 7 the second summation in Eq 4 converges quickly and can be evaluated directly The rst term in the summation of Eq 4 must be transformed into Fourier space In order to described these summations explicitly we assume that we have a periodic lattice so that every ion can be located by ri Ta T a location Ta within a unit cell and a periodic translation vector T In this way the summation becomes1 2 N Z 5 ij am where N denotes the number of unit cells in the system Since we have a periodic system N is in nite but the energy per unit cell I39VN is well de ned The other identity that we must use is that a sum over lattice translations T may be transformed into an equivalent sum over reciprocal lattice translations G according to the identity 1 i Z 530 T i Z GIGquot 5 T Q G where 9 denotes the unit cell volume The proof of this relation is given in the appendix The rst term of Eq 4 thus becomes qquerfl 7 ri rv Y erfl 7M T3T 7 2 Jllx Ema 2f a m 7 m 1753 l J a T Ta 7 T a 7r where the last term in Eq 7 comes from subtracting out the selfinteraction i term from the complete lattice sum Using the short hand notation 7a E Ta 73 the lattice sum can be evaluated erfg irmgFTE d3rz 3frTerf ira3r 8 T Tax T i Tax i i This becomes d3reiGTerfg Tag r Z 871137 eimm g Q G Tagr Q 0 G2 2 u U3 l where the last term which is in nite comes from the G 0 contribution If the last term of Eq 9 cannot be eliminated it is clear that the electrostatic energy is in nite The term can be eliminated if and only if the system is neutral Thus it is only meaningful to calculate the electrostatic energy of a neutral periodic system If the actual system has a net charge of Q E 2a qa we could calculate a meanful energy if we add to the actual charge density a compensating uniform density charge density of QQ Taking all of the terms into account we nd the nal Ewald expression to be W Ia IB 47r e iG W e4 7 5 Orfev7l Ta3 Ti 47rQ2 39 9 Z G2 7r a T EnvyFT 87r5097 10 where the I in the summation over lattice translations T indicates that all selfinteraction terms should be omitted quot a1 87r 0 Gym 1Note that for any two lattice translations T and Tj T Tj Tk where Tk is also a lattice translation Appendix I comments on lattice vectors and reciprocal lattice vectors In this discussion will assume we have a 3dimensional periodic system It can be easily generalized to 1 or 2 dimensional systems In general a translation vector can be described a linear combination of the three primitive translation vectors T1 T2 and T3 T mTl 7ng 713T3 11 where 711 712 713 are integers Note that the unit cell volume 9 can be expressed in terms of the primitive translation vectors according to Q T1T2 XT3 12 The reciprocal lattice vectors G can generally be written a linear combination of the three primitive reciprocal lattice vectors G1 G2 and G3 G 7R1G1 7R2G2 7R3G3 where m17n27n3 are integers The primitive reciprocal lattice vectors are determined from the primitive translation vectors according to the identities Gi 39 Tj 2 27min 14 Note that the volume of the primitive reciprocal lattice is given by 2W 3 G1 39 X G3 Some examples of this are given below Proof of Eq 6 Consider the geometric series 3 l Z eiMGyr 5m 2G1 1 sinG1 r2 I927 The behavior of the right hand side of Eq 16 is that it is small in magnitude very except when the denominator vanishes This occurs whenever G1 r2 n17r where 711 represents any integer If we take the limit M gt 00 we nd that the function represents the behavior of a sum of delta functions sin A1 G1r 2 6 T 17 Ml IPQC sinG1r2 7r G1 r 111 1 gt The summation over all lattice translations 711T1 is due to the fact that sinG1 r2 0 whenever r anl Carrying out the geometric summations in the right hand side of Eq 6 for all three reciprocal lattice vectors and taking the limit in Eq 17 26quot 2703 Z 6G1r n1T16G2 r n2T26G3 r 113 18 G 711712713 Finally the right hand side of Eq 18 can be simpli ed by eliminating the reciprocal lattice vectors from the 6 functions iGT amp 3 3 28 G1G2gtltG36r Tl95 1quot T 19 which is consistent with Eq 6 Appendix II examples In these examples denote the length of the unit cell by a CsCl structure There are two kinds of sites ms U and Tc y T1 ax T2 ay T3 a2 20 2 2 2 G1 12 G2 y G3 12 21 a a a It is convenient to de ne 7 E p2a2 Also we will denote by q the unit charge In these terms Eq 10 becomes for this case W 12 7 7 22 N ES7rquotgr1quotT1 772 gt where 2 1 1 eiwm1m2m3e ir2 mim m 2 71 E 7 2 7 23 7r whmmg m mg mg and 72 E 2erfcdnfn n Z 2erfcdnl2 71227132 uhngmgy o 71 71 71 111112113 7112 7122 7132 24 In this expression the sum over a and 8 has a total of 4 contributions 2 pairs of identical contributions For Cs Cs or C1C1 interactions 7a 0 and 1a 1 resulting in repulsive contributions For Cs Cl or Cl Cs interations 7a i y i and 1a 1 resulting in attractive contributions This expression can be evaluated using Maple which gives the result W q2 4 71 2 N 87r ga 0 gt NaCl structure There are two kinds of sites ma 0 and TC T1g y T27y2 T3gt igt 26 G12 ampy 2 G2 5y2 G37amp y2 27 January 287 2009 Additional Notes for Lecture 6 7 Mean value theorem for solutions to the Laplace equation Consider an electrostatic eld ltlgtr in a charge free region so that it satis es the Laplace equation V2ltlgtr 0 1 The mean value theorem77 value theorem problem 110 of your textbook states that the value of ltlgtr at the arbitrary charge free point I39 is equal to the average of ltlgtr over the surface of any sphere centered on the point I39 see Jackson problem 110 One way to prove this theorem is the following Consider a point Iquot r u7 where u will describe a sphere of radius R about the xed point I39 We can make a Taylor series expansion of the electrostatic potential ltlgtr about the xed point I39 ltlgtr u ltlgtr u Vltlgtr u V2ltlgtr 01 V3ltlgtr 01 V4ltrgtr 2 According to the premise of the theorem7 we want to integrate both sides of the equation 2 over a sphere of radius R in the variable u 27r 1 7 2 Aphere dSu 7R 0 ddhi1L dcost9u 3 We note that M 1 2 i 2 R 0 d u A dcos6u1 74m 4 27r 1 2 7 R 0 day71 dcost9uu v 0 5 27r 1 47TR4 2 2 7 2 R A day71 dcos0uu V 7 3 v 6 133wa 1dcos6 u V3 0 7 0 U 71 U 7 and 2 1 4 R6 2 7r 4 7 7T 4 R A day71 dcos0uu V 7 5 v 8 Since V2ltlgtr 07 the only non zero term of the average it thus the rst term 173302W dab 1 dcos6ultlgtr u 47TR2ltlgtI39 9 or ltlgtr L 17332W d u f1 dcos6ultlgtr u E dSultIgtr u 10 47TR2 0 71 47TR2 sphere Since this result is independent of the radius R7 we see that we have proven the theorem January 31 2009 Notes for Lectures 8 amp 9 7 Introduction to gridbased methods for solving Poisson and Laplace Equations Finite difference methods The basis for grid based nite difference methods is a Taylor s series expansion ltIgtr u ltIgtr u mm u V2qgtr u V3ltIgtr u V4ltIgtr 1 We will work out some explicit formulae for a 2 dimensional regular grid with h denoting the step length For the 2 dimensional Poisson equation we have 82 82 p1 y I 7 7 2 83 82 m 80 lt gt We note that a sum of 4 surrounding edge values gives 5AE Maw17y Wk1720 7yh 7yih 3 82 82 h4 84 84 4ltIgt Wife 777 716 720 8352 egg 17y12lt8m4 Wt x7y Similarly a sum of 4 surrounding corner values gives 53 E 1gth7yh ltmih7yh h7yihmihw h 4 a a h4 a4 a4 82 82 7 2 6 79 M 8352 an MM 6 8354 W 68128112 MW l h quot3939 We note that we can combine these two results into the relation 1 312 114 SA 153 5 I957y TVZ NQEW gVZVWm 716 5 This result can be written in the form 1 1 312 4 lt1gt 7 is 7 is 7 7 2 6 9w 5 A 20 B 1060096721 40 0V We 24 In general the right hand side of this equation is known and most of the left hand side of the equation except for the boundary values are unknown It can be used to develop a set of linear equations for the values of Mm g on the grid points For example consider a solution to the Laplace equation in the square region 0 g x g a 0 g y g a which ltIgtm0 ltIgtOy ltIgtay 0 and ltIgtza V0 We will rst analyze this system with a mesh of 9 points generated with a grid spacing of h In this case ltIgthh E My is unknown while ltIgt02h ltIgth2h ltIgt2h2h V0 and ltIgt00 Wm ltIgt2h0 ltIgt0 h ltIgth 2h O For this example Eq 6 states lt1gtx2h V0 0 0 D0y 9 D2hy I i Figure 1 3 x 3 grid for solution of the Poisson equation within a 2 dimensional square IX0 0 lt1gt5 ltIgth0lt1gt0hlt1gt2hhlt1gth2hlt1gt00ltIgt2h0lt1gt02hlt1gt2h21 v0 lt7 This results is within 20 of the exact answer of Mg 025 If analyze this same system with the next more accurate grid7 h using the symmetry of the system Mm y EhLem7 y7 we have now 6 unknown values 1017 hltIgt2hhltIgth7 2hltIgt2h2hlt1gth3hltIgt2h3h and boundary values lt1gt04h ltIgth 4h ltIgt2h4h V0 and lt1gt03h lt1gt02h lt1gt0h ltIgt00 ltIgth 0 ltIgt2h 0 This results in the following relations between the grid points ltIgth 3h7lt1gth 4hlt1gt03hlt1gt2h 3hlt1gth 2h7lt1gt04hlt1gt2h 4hlt1gt2h 2hlt1gt0 21 0 ltIgt2h 3hlt1gt2h4hlt1gt3h7 3hlt1gth3hlt1gt2h 2hlt1gth4hlt1gt3h 4hlt1gt3h 2hlt1gt8h 21 07 PM wi wh wwloi 271 2 17 WWW hlt1gt0 3hlt1gt2h 3hlt1gt0 hlt1gt2h his 0 will 2h72h 3h h 2h m3hv 2h2h7 h 3h3hltlgth 3hlt1gt3h hlt1gthl0h 07 Ml migmwhllmi WWW hgt h70 210lt1gt0 2hlt1gt2h 2hlt1gt00lt1gt2h0 min mi mh 2hll l3h7 WWW hgt 2h70lt1gth 2hlt1gt3h 2hlt1gth0lt1gt3h01 0 lt13 These equations can be cast into the form of a matrix problem which can be easily solved using CIX0 0 Figure 2 5 x 5 grid for solution of the Poisson equation within a 2 dimensional square Maple 1 715 715 7120 0 0 ltIgth3h 310 725 1 7110 715 0 0 ltIgt2h3h 310 715 7120 1 715 715 7120 ltIgth2h 0 V0 14 7110 715 725 1 7110 715 ltIgt2h2h 0 0 0 715 7120 1 715 ltIgthh 0 0 0 7110 715 725 1 ltIgt2hh 0 The solution to these equations and the exact results are found to be ltIgth7 3h 04628135839 4320283318 ltIgt2h3h 05566467694 5405292183 ltIgth7 2h 01920222635 1820283318 V0 exact V0 15 ltIgt2h7 2h 02615955473 025 ltIgth7 h 007150923611 06797166807 ltIgt2h7 h 01001250302 09541411792 We see that the results obtained with a smaller mesh has is much closer to the exact results than those for the larger mesh Introduction to Finite element method The nite element approach is based on an expansion of the unknown electrostatic potential in terms of known grid based functions of xed shape In two dimensions using the indices 23 to reference the grid we can denote the shape functions as gti7 The nite element expansion of the potential in two dimensions can take the form 47r 0lt1gt907y ZMWMWQL 16 H where 11 represents the amplitude associated with the shape function jugz The amplitude values can be determined for a given solution of the Poisson equation 7V2 47r60lt1gtm 47rpm y 17 by solving a linear algebra problem of the form ZMW Jij Gkh 18 27 where Mk1 Edzdyv klzy V gtijxy and GM Edzdy gtklxy 47rpmy 19 In obtaining this result we have assumed that the boundary values vanish This will be ensured by our choice of the functional form of the shape functions gtijz In order for this result to be useful we need to be able evaluate the integrals for MM and for GM In the latter case we need to know the form of the charge density The form of Mk1 only depends upon the form of the shape functions If we take these functions to be gtijlt907y E Xiyjy7 20 where 7 m7 7 lt lt Mag 1 h forzl hizizHJi 21 0 otherw1se and my has a similar expression in the variable y Then 7 mm pgm I I d311ltygtdyjltygt MW dz dy dz dz WynM maxim dy dy lt22 There are four types of non trivial contributions to these values zih 1 alarm h lt17 lubzdu 3 lt23 miih 71 3 mih 1 h lt2 ltzgt2a1ltzgtgtdz h lt17 ugtudu g lt24 ziih 0 2M pam 2 1 1 2 Ark lt 192 gt Mfg71 MT 25 2M d2zd2c1z 1 1 71 AH lt7 7d gtdm77EO duff 26 These basic ingredients lead to the following distinct values for the matrix 2 fork1andlj MM 7 for 117111 andorliji1 27 0 otherwise nah t For problems in which the boundary values are 0 Eq 18 then can be used to nd all of the interior amplitudes 111172 In our case we have the boundary conditions ltIgtm0 ltIgt0y ltIgtay 0 and ltIgtma V0 Using the same indexing as in Fig 2 this means that II0 411 01411 II211 411 V0 The nite element approach for this problem thus can be put into the matrix form for analysis by Maple 83 713 713 713 0 0 ltIgt01311 1 723 83 723 713 0 0 ltIgt211311 1 713 713 83 713 713 713 ltIgt01211 0 W 28 723 713 723 83 723 713 ltIgt211211 0 0 0 713 713 83 713 ltIgt0111 0 0 0 723 713 723 83 ltIgt21111 0 The solution to these equations and the exact results are found to be 101 311 05070276498 4320283318 ltIgt211311 5847926267 5405292183 101 211 01928571429 1820283318 V0 exact V0 29 ltIgt211211 02785714286 025 ltIgt0111 007154377880 06797166807 ltIgt211 11 01009216590 09541411792 We see that the results are similar to those obtained using the nite difference approach January 23 2007 Notes for Lectures 4 amp 5 Electrostatic energy Section 111 in Jackson7s text The total electrostatic potential energy of interaction between point charges at the positions r1 is given by 1 1in 1 1 1 T 47Teo K7 friirjf T 8713980 7 friirjf39 In this expression the rst form explicitly counts all pairs while the second form counts all interactions and divides by 2 to compensate for double counting For a nite system of charges this expression can be evaluated directly however for a large or in nite system the expression 1 does not converge and numerical tricks must be used to evaluate the energy In fact for the in nite system one has an in nite amount of charge and the energy of interaction is unde ned If the system is neutral it is possible to de ne a meaningful interaction energy by use of an Ewald transformation The basic idea of the Ewald approach is as follows The error function erfx and its com plement erfc are de ned as erfx Om e t2dt and erfcx 1 7 erfz 0 e t2dt 2 Ewald noted that 1 erf r erfc r 7 3 r r T In this expression the rst term goes to a constant 777T as r a 0 but has a long range tail as r a 00 The second term has a singular behavior as r a 0 but vanishes exponentially as r a 00 Thus Ewald7s idea is to replace a single divergent summation with two convergent summations The rst summation has a convergent summation in the form of its Fourier transform and the second has a convergent direct summation Thus the calculation of the electrostatic energy would be evaluated using 1 1 W 1 11 i 1 Zqiqjer y ri rj Zqiqjerf0 riirj 39 8713980 7 ri 13quot T 8 80 1 ri 13quot W ri 13quot 4 For an appropriate choice of the parameter 77 the second summation in Eq 4 converges quickly and can be evaluated directly The rst term in the summation of Eq 4 must be transformed into Fourier space In order to described these summations explicitly7 we assume that we have a periodic lattice so that every ion can be located by 391 Ta T a location Ta within a unit cell and a periodic translation vector T7 In this way7 the summation becomes1 2 N Z 5 ij a T where N denotes the number of unit cells in the system Since we have a periodic system7 N is in nite7 but the energy per unit cell WN is well de ned The other identity that we must use is that a sum over lattice translations T may be transformed into an equivalent sum over reciprocal lattice translations G according to the identity 1 Z530quot T Ze G39 6 T Q G where 9 denotes the unit cell volume The proof of this relation is given in the appendix The rst term of Eq 4 thus becomes qiqjeFfo lri 13D ex lm 7 Tl 2 77 N QaQ i Ia 7 7 lrrrjl E lTwnHTl 7r where the last term in Eq 7 comes from subtracting out the self interaction 239 j term from the complete lattice sum Using the short hand notation7 73904 E Ta 7 71 the lattice sum can be evaluated erf l 739 T erf l 739 I39 zx l 049 dsrz saiT a 8 T lTa Tl T lTa rl This becomes7 1 fl 3 4 iG39M GW 1 lfd G lTa rl Q 0 G2 u3 where the last term7 which is in nite7 comes from the G 0 contribution If the last term of Eq 9 cannot be eliminated7 it is clear that the electrostatic energy is in nite The term can be eliminated if and only if the system is neutral Thus7 it is only meaningful to calculate the electrostatic energy of a neutral periodic system If the actual system has a net charge of Q E Ea gm we could calculate a meanful energy if we add to the actual charge density7 a compensating uniform density charge density of iQQ Taking all of the terms into account7 we nd the nal Ewald expression to be K 7 qaq a 4 e iG39M 5 027 7 6 2 erfc l739a Tgt 7 47TQ2 N T 043 8713980 Q G2 7139 D49 lug Tl 8711609777 10 where the in the summation over lattice translations T indicates that all self interaction terms should be omitted 130 1Note that for any two lattice translations T and Tj T 7 Tj Tk7 where Tk is also a lattice translationi Appendix I 7 comments on lattice vectors and reciprocal lattice vectors In this discussion7 will assume we have a 3 dimensional periodic system It can be easily generalized to 1 or 2 dimensional systems In general7 a translation vector can be described a linear combination of the three primitive translation vectors T17 T27 and T3 T 711T1 71sz 713T37 11 where 711712713 are integers Note that the unit cell volume 9 can be expressed in terms of the primitive translation vectors according to Q XTg The reciprocal lattice vectors G can generally be written as a linear combination of the three primitive reciprocal lattice vectors G1 G2 and G3 G m1G1 szz m3G37 13 where 77117771277713 are integers The primitive reciprocal lattice vectors are determined from the primitive translation vectors according to the identities G1 Tj 2W61j Note that the volume of the primitive reciprocal lattice is given by 1G1G2 gtlt G3 15 Some examples of this are given below Proof77 of Eq 6 Consider the geometric series 2131 Magir Sin M G139rgt 16 k7M sinG1 r2 39 The behavior of the right hand side of Eq 16 is that it is small in magnitude very except when the denominator vanishes This occurs whenever G1 r2 n17r7 where 711 represents any integer If we take the limit M a 007 we nd that the function represents the behavior of a sum of delta functions sin G1 rgt 2 6 G 7 T 17 ernoo SinltG1r2 7T 1 139 I11 1 The summation over all lattice translations n1T1 is due to the fact that sinG1 r2 0 whenever r anl Carrying out the geometric summations in the right hand side of Eq 6 for all three reciprocal lattice vectors and taking the limit as in Eq 177 2611GT 271393 Z 39 I39 7 n1T16G2 39 I39 7 n2T26G3 39 I39 7 1 13T3 G W1 W2 gt713 Finally7 the right hand side of Eq 18 can be simpli ed by eliminating the reciprocal lattice vectors from the 6 functions 6iGr 31quot 317 lG1G2xG3l6 T 9N T 19 which is consistent with Eq 6 Appendix II examples In these examples7 denote the length of the unit cell by a CsCl structure There are two kinds of sites 7 73905 0 and 73901 y T1 a T2 357 T3 32 2 2 2 G1 12 G2 1y G3 12 21 a a a It is convenient to de ne 77 E pzaz Also we will denote by q the unit charge In these terms7 Eq 10 becomes for this case W 12 N MltZ 2 22 where 2 1 7 6i7rm1mgmg6 m mgm 712 Z r m2m2m2 7 23 m1m2mg70 1 2 3 and n1n2n370 M 77a 71 77 n1gtn2gtn3 T i 2erfc n 713 Z 2erfc 711 7122 713 2 7 i m i n2 i ms 24 In this expression7 the sum over 04 and B has a total of 4 contributions 7 2 pairs of identical contributions For Cs Cs or 01 01 interactions7 73904 0 and qa q resulting in repulsive contributions For Cs Cl or Cl Cs interations7 73904 igb y 2 and qa q resulting in attractive contributions This expression can be evaluated using Maple7 which gives the result W 2 1 W 7 TQM 4071 25 NaCl structure There are two kinds of sites 7 TNa 0 and 73901 January 31 2002 Notes for Lecture 7 Methods for solving Poisson equation There are are large number of tools for solving the Poisson and Laplace equations 1 Green s function methods 33 372 ltIgtr39w da39 1 I r 372 3I I I i I drprGrr47TfS Grr 47750 v 2 Complete function expansions Fourier series etc 3 Variational methods 4 Grid based methods Introduction to gridbased methods The basis for most gridbased methods is the Taylor s expansion ltIgtr u ltIgtr 11 4 we u4V2ltIgtr gm 4 V3ltIgtr in 4 Wear A A A 2 We will work out some explicit formulae for a 2dimensi0nal regular grid with h denoting the step length For the 2dimensi0nal Poisson equation we have 82 82 M y ltIgtLy 0 We note that a sum of 4 surrounding edge values gives SA 5 ltIgtIhyltIgtt hyltIgttyhltIgtzty h 32 32 h4 34 34 4ltIgtIJy h2 ltIgtIy E ltIgtIy W Similarly a sum of 4 surrounding corner values gives SB 5 ltIgt1hyhlt1gt1 hyhltIgtIhy hlt1gta hy h 5 r32 32 h4 34 r34 r32 r32 I I 2 I I I I 6 4ltIgtIy 2h 8112 ay2lt1gt1y 6 8114 aw 6811 ay2lt1gt1yh We note that we can combine these two results into the relation 1 3h SA 133 5ltIgtIy 2 4 2 V2ltIgtI h V2V2ltIgtfIJ h6 6 This result can be written in the form 4 V2II y 7 1 1 3h2 rm 33A SB in rm 400 20 1050 In general the right hand side of this equation is known and most of the left hand side of the equation except for the boundary values are unknown It can be used to develop a set of linear equations for the values of ltIgt1 on the grid points For example consider a solution to the Laplace equation in the square region 0 g 1 g a 0 g y g a which ltIgtI0 ltIgt0 ltIgta 0 and ltIgtI V0 We will rst analyze this system with a mesh of 9 points In this case 5 5 Mg is unknown while 5 032 313nd 43 563 73 83 930 xaV0 09 o o e6 0 6 0 x00 0 0 my ay For this example Eq 7 states 1 1 3 5 2 34 36 38 1 3 37 39 VO 8 a 20 10 This results is within 20 of the exact answer of Mg 025 If analyze this same ltIgta 11y we have now 6 unknown values 5 g 9 11 12 and boundary values 51 52 531and43 73 103 133 143 1530 system with the next more accurate grid using the symmetry of the system ltIgtIy xaV0 0 0 my ay This results in the following relations between the grid points 15 1 2 4 1 a 2i01 1 3 7 9 01 9 1 1 11 33559 22ss01 10 8 5 79 11 24 6 10 12 01 11 9 6 s 8 12 25 5 1111 0 12 11 s 10 12 14 27 9 13 15 01 13 12 11 11 15 2i s 8 314 14 0 14 These equations can be cast into the form of a matrix problem which can be easily solved using Maple 1 15 15 120 0 0 155 310 25 1 110 15 0 0 156 310 15 120 1 15 15 120 158 0 V0 15 110 15 25 1 110 15 159 0 0 0 15 120 1 15 11 0 0 0 110 15 25 1 1112 0 The solution to these equations and the exact results are found to be 5 04628135839 4320283318 5 05566467694 5405292183 8 01920222635 1820283318 V0 exact 0amp9 02615955473 025 0 007150923611 06797166807 12 01001250302 09541411792 We see that the accuracy has improved considerably with the new mesh V0 16 March 26 2002 Notes for Lecture 22 Derivation of the LienardWiechert potentials and elds Consider a point charge 1 moving on a trajectory Rqt We can write its charge density as 00quot t 1530quot RM 1 and the current density as V JOquot 0 qRqt53r RM 2 where IR t Rm 2 3 l 3 Evaluating the scalar and vector potentials in the Lorentz gauge 1 3 0r gt i i m d w dt o t t gI r39c 4 and 1 3 1 l Jrltl I Art TKOCZ d 7 dt Er r o t t r r a We performing the integrations over rst 137quot and then dt and make use of the fact that for any function of t grew t Rqt39gtcgtgt lt6 C a where the retarded time is de ned to be tr Et rRqtr We nd 1 L7 Mm m R lt8 and V Amt 9 4776002 R where we have used the shorthand notation R E r Rqt and v E Rqt In order to nd the electric and magnetic elds we need to evaluate Em VltIgtrt Mg 10 and Br t V X Art 11 The trick of evaluating these derivatives is that the retarded time 7 depends on position r and on itself We can show the following results using the shorthand notation de ned above R and at R E RVC 13 Evaluating the gradient of the scalar potential we nd q 1 VltIgtrt 4WEOR 3R1 62 These results can be combined to determine the electric eld EricaQME T1 ltRgtltltR fgtltgtl We can also evaluate the curl of A to nd the magnetic eld 12 V V Br t LI31 RX VIZK 17 7760c RvT c c RvT One can show that the electric and magnetic elds are related according to E t BO 18 CR May 4 2002 Notes for Lecture 32 Two formulations of electromagnetic elds produced by a charged particle moving at constant velocity In Chapter 11 of Jackson page 559 Eqs 111512 and Fig 118 we derived the electric and magnetic eld of a particle having charge 1 moving at velocity v along the 2 axis The results are for the elds at the point r 1252 are Email ybxz Ex 3 3 t E Ub0t 23 qwme 1 and A Ylgbxs 32 2 Wmww for the electric and magnetic elds respectively The denominators of these expressions are easily interpreted the distance of the particle from the eld point measured in the particle s own reference frame On the other hand in Chapter 6 we considered the same physical problem from the point of view of LienardVViechert potentials Bxhx22x39t BUba 09 q Ionsider the electric eld produced by a point charge 1 moving on a trajectory described by r0t with pr t E 163r r0t Assume that Vot E 8r0t9t and 82r0t8t2 0 Show that the electric eld can be written in the form EMF qu memewma qu wampmewma a 47mg R v0 4 Rc3 Gaussian units R v0 4 Rc3 where R E Rt Rt E r roar and where all quantities which depend on time on the right hand side of the equation are evaluated at the retarded time t E t Rtc In the same notation the magnetic eld is given by R X E B 7 4 R m If we evaluate this result for the same case above Fig 118 of Jackson v0 E 911 and Rt 39vt51 big In order to relate this result to Eqs 1 and 2 above we need to express t in terms of the known quantities Noting that Rt ct t vtr 6 C V we nd that t must be a solution to the quadratic equation t3 ngtt Wt 126202 U 6 with the physical solution C 39v thAb trryryt Now we can express the length parameter which appears in Eq 3 R iv v yt We also can show that the numerator of Eq 3 can be evaluated R VoRc 39vtf1 big and the denominator can be evaluated v yt262 R V0 RC 10 Substituting these results into Eqs 3 and 4 we obtain the same electric and magnetic elds given in Eqs 1 and 2 from the eld transformation approach February 257 2008 Notes for Lecture 16 Vector potentials in magnetostatics The vector potential which vanishes at in nity and corresponds to a con ned current density distribution JI39 is given by M0 3 J 1quot A I39 7 d r 7 1 lt gt 4 1 7 r lt gt This expression is useful if the current density JI39 is con ned within a nite region of space Consider the following example corresponding to a rotating charged sphere of radius 17 with p0 denoting the uniform charge density within the sphere and w denoting the angular rotation of the sphere 7 powxr forr ga Jltrgt T 0 otherwise 2 In order to evaluate the vector potential 1 for this problem7 we can make use of the expansion 1 47139 r l Z mgmemw lt3 Noting that 4 xi 7xi A r r 3 mm y Yany mar2 4 we see that the angular integral result takes the simple form r dQ Ym Y 7 6 5 g1 ltrgtlmltrgtr Tr n Therefore the vector potential for this system is MOPOW X T 1 37 lt A 7 d 7 6 r 3f 0 r r lt gt which can be evaluated as LL 39 12 7 2 70 X lt7 7 3170 for r S 1 A0quot 7 W for r 2 a As another example7 consider the current associated with an electron in a spherical atom In this case7 we assume that the current density is due to an electron in a bound atomic state with quantum numbers nlm as described by a wavefunction zbnlmxr where the azimuthal quantum number ml is associated with a factor of the form em For such a wavefunction the quantum mechanical current density operator can be evaluated 76h JI39 202m WW 7 mm mm 8 277m Since the only complex part of this wavefunction is associated with the azimuthal quantum number this can be written 7 76h 7 2mmquot sin 6 767mm 3 3 JI39 wilml wnlm 7 10mm 21m1gt b lwnlml 2 mgr sint9 where m5 denotes the electron mass and 6 denotes the magnitude of the electron charge For example consider the lnlm 211gt state of a H atom 1 74 211039 i 64 3 eirZa Sin 99 10 Ml a 7 76h 7 64m57ra5 and e Tla 2 gtlt r 11 10 where 1 here denotes the Bohr radius Using arguments similar to those above we nd that 7 h A 00 AI39 6 Oz X r d7 r3 e77 7 12 192m57ra5r 0 T This expression can be integrated to give h e 2 3 Am euozxr lieiralt1f 77 13 1 2a 8a3 8m57rr3 March 67 2008 Notes for Lecture 21 Derivation of the Lienard Wiechert potentials and elds Consider a point charge q moving on a trajectory Rqt We can write its charge density as phat 1630quot 7 Rqt7 1 and the current density as 7 J02 1t qRqt53r Rqt7 2 where dR t R t E q 3 go a o Evaluating the scalar and vector potentials in the Lorentz gauge7 7 1 3 pr7t m 4WEO dr dt r w t r rlc 4 and 1 J t r At7d3 dt 6t 7t7 7 5 r gt 4 f062 r rli rm lt lt r rwcgtgt lt gt We performing the integrations over rst dgr and then dzt 7 and make use of the fact that for any function of t f t 00 7 7 7 m fMMtth muwm 17Mmmww ClPRWT where the retarded time77 is de ned to be 7 R tr mate l1quot CA N 7 We nd uw L 1 w 7 47TEO R 7 7 and q V Amt W 7 9 c where we have used the shorthand notation R E I39 7 Rqt7 and V E Rqt7 In order to nd the electric and magnetic elds7 we need to evaluate mm want 7 w 10 and Brt V gtlt Art 11 The trick of evaluating these derivatives is that the retarded time 7 depends on position I39 and on itself We can show the following results using the shorthand notation de ned above R Vt 7 12 c R 7 lt l and at R T 13 at R 7 lt gt Evaluating the gradient of the scalar potential7 we nd 7Vlt1gtrt7q 1 3 R171 3 Riv R RV R 14 47TEO R 7 02 c 2 and 6Art q 1 VR 12 VR vR VB VR 77 3i 7 7 2 7 R 15 3t 47TEO R 7 0 0 RC 0 c c These results can be combined to determine the electric eld B 2 R Bi 1ltRXltRLgt X129 0 c c c We can also evaluate the curl of A to nd the magnetic eld q 1 EI39t KEG R 7 y i 471396062 Brt 7Rgtltv v2 VR RXvc W 1 2 t 72 W 39 One can show that the electric and magnetic elds are related according to mm W 18 Mueh 7A 2003 Notes for Lecture 24 Re ectivity for anisotropic media 7 Extension of Section 73 in Jackson s text above Fm Simplicity we mu assume mm the dielectric berm y the medium is diagonal and is gven by mu be daiw by by In terms of this electric eld and the magnetic eld H BMO7 where H is assumed to have the same complex spatial and temporal form as 37 the four Maxwell7s equations are given by VH0 VHE0 4 VgtltE7zwu0H0 VgtltHZwEOHE0 Using these equations7 we obtain the following equations for electric eld amplitudes within the medium Km 7 n nmny 0 Ex nmny Hyy 7 n 0 E 0 5 0 0 H 7 E2 Once the electric eld amplitudes are determined7 the magnetic eld can be determined according to H i E20121 7 My Eynm 7 Emma eewwwwt 6 The incident and re ected electro magnetic elds are given in your textbook In the notation of the gure the wavevector for the incident wave is given by kl sin 239 cos 2393 7 and the wavevector for re ected wave is given by kR sin 239 7 cos 23937 8 In this notation7 Snell7s law requires that nm sinz The continuity conditions at the y 0 plane involve continuity requirements on the following elds H0z7t7 Ewz707z7t7 EZ707z7t7 and Dyx707z7t7 9 at all times t Below we consider two different polarizations for the electric eld Solution for s polarization In this case7 Em Ey 07 and 7122 Is 7 The elds in the medium are given by u 1 w E Ezielnxmnyy7lwt H 7 7 elnxmnyy7lwt39 M00 The amplitude E1 can be determined from the matching conditions E0 l E2 E0 7 E3 cosz39 Ezny E0 Eff sinz E1711 In this case7 the last equation is redundant The other two equations can be solved for the re ected amplitude Egcosz7ny 12 E0 cosz 712 This is very similar to the result given in Eq 739 of Jackson for the isotropic media Solution for ppolarization In this case7 E2 0 and 41212 2 Kgy 7 13 R212 In terms of the unknown amplitude E7 the electric eld in the medium is given by E Em 7 menw einxmnyyiiwt39 Kyyny The corresponding magnetic eld is given by H EEZGi WHWZFWK 15 05 y The amplitude Em can be determined from the matching conditions EoiEg cosz39 E1 16 E0 Ed HfJEz E0 E3 sinz if Em Again7 the last equation is redundant7 and the solution for the re ected amplitude is given by E0 7 Hm cosz 7 my 17 E0 Hm cos 2 my This result reduces to Eq 741 in Jackson for the isotropic case 50 30 fi A Q 10 51 fi A Q 10 10 1 am 111 9 13 9 390 2311 8111110110 3113 03 13111211313 11311111 915 8110131211133 S AXRI 51813128 3811111 313113 8131311 3113 3d1d 3113 31318111 83311108 011 3112 313113 312113 811111111881 3911231 311 03 133111118812 3112 3 131112 31 MOU 3131111 3d1d 3113 31318111 3m142311a x8 a pup 3m142311a El 2111101 3113 331123 8131311 3113 312113 3111118812 11111 3M 3911013331113 11013128 12d01d 3113 g 811131211818313 103311131103 1123131 1112 111011 3131118 3111211 12 33111381103 3m 3801111115 13111110118 pup 38112113 331211118 12 31 11123 313113 38111233q 0 7 9 1 23901 3 311311113 331211118 12 31 11123 313113 38111233q 0 wu ag mg 312113 811123111 81113 103311131103 311310 331211118 3113 312 8110131131103 5311111131103 131311 3113 0 38111233 393312 1118 31 0 3311 1103 3112 8 3 3 31 121 08 oc lt 10 31131103 123 31 1112 10 3 El 3 i E I E i i I I 1 393 137m a m g K311311b311 MO1 312 103311131103 13008 12 10 38123 3113 111 QL ltlt I 312113 3111111 3113 111 4311 E 1 511 1131118 11311313 111318 3113 1131311 331211118 3113 10 1131111311 3113 111 8313113111111112 131311 3113 10 5123313 10 33121 3113 83111111133313 11 10 31111211 311 3931d11112x3 101 o m I 131112 937131117 2L ggg 39bg 531131113311 MO1 312 13130111 31311111 3113 10 11 8112113312111 3113 110 1311311313 I 131112 2L 831112381103 31131311311313 531131113311 3113 10 111101 311 1 6 gt 11 1m lt Z 11 3 213 3142214 3 2LZIZELN 03 811113103312 311 31 10333A3A12m 31311111103 3113 3111111133313 11123 3M 2 39I1 2L E 371337 3311 21101312131 3113 311151813128 31 1033311 3111211 3113 113111 8131311 313311812111 131112 31133313 3113 101 m 33314111511603 X El pup 33314131511603 El 2111101 3113 10 8110131211133 SJWAAXRI 03 1101311108 3111214 31112111 1 11101133111 3A1312d188113 12 111 3111211311 SGARAA 3133118121110133313 M011 1111231 38111 3811111 311 3131118 3111211 12 10 110131213110 3113 1311123813131111 03 131310 111 89131113 91mm 3139113121110133913 037 mnqoaq 310 saqu 0002 111113 39aazgms apgn 3mm am 30 mo Supugod 103mm mmou mm p smouap g awqm 0 139991 ng A Sugsooqa axqgssod apmn 81m sgql 390 ii My ii 0 g pm 0 q 0 mugs pagsmzs sg 0 wu agl mg axdumxa 101 39suompum Kmpunoq am 339128 smxsm asaq 3mg moqs 03 mme Sm gmqa um auo 21 39 Q Sltgt v IIES 03 T 589 rm 331 r 4 4 A fiuu 2mm Mu Zgg a g 31 x 5 Q IXxS v gt303ng Q z 50 2 231 a 91 Mu 5mm 4 4 a in mg Zgg mg fig pmz 1g pm 43 pm 3 spmg mg m was 03 11 01 39Sbg u past aq um smxsm asaql m 31 03 Kjgdmgs 3 1 39sbg axdumxa 101 39sumuodmm may mtpo am auguxwmp um 91 a W quot39513 3 Qt1Lt am suommba S HGAAXRIA pm mme Sm 111x013 ms 231 E 231 mm 3 z 39 39 39 g n 1 03 030 h x 2 nm 0 E h x 2 1 Mu mm a a I 1 amzq um apom 3 v m Imprst v 3x03 mm 30 3393 Imng ug passmsgp apgn amm XRIK8IXR3JGX mp 101 39s3tmtxltgtmxm may mqm mp 30 qsz m suogssmdxa mgmgs mm 91 0 17 971 1 4 4 uogmnba zwqmpH mmgsuamgp m p 538928 38mm umeodmm may qsz 3mg pug am Awq s awdmv pm M121 S KRpRXR 1 Sugugqmog fig 589 g0 quot3990 950 le x 51 2 av 21 39ngs g 50 230 fig 589 We quot3930 x0 30 01 gm k aw 6 Jam 8 3963 April 14 2009 Notes for Lecture 29 Synchrotron Radiation For this analysis we will use the geometry shown in Fig 149 of Jackson A particle with charge q is moving in a circular trajectory with radius p and speed 1 Its trajectory as a function of time t is given by Rm psinltmpgtlt M1 e WSWp y lt1 lts velocity as a function time is given by Vqt 1 cosvtpx 1 sinvtpy 2 The spectral intensity that we must evaluate is given by the expression Eq 1467 in Jackson d2 q2w2 2 7 7 3 dwdQ 47TZC 00 f X gtlt eiwtif Rqt0dt foo After some algebra7 this expression can be put into the form d2 2 2 2 dwdQ l01wllz WWW 4 where the amplitude for the light polarized along the y axis is given by Sinvtpeiwt7 cos 95in39utp and the amplitude for the light polarized perpendicular to y and f39 is given by CL w L00 dt sin 6 cosvtpei t C05 esin tp 6 We will analyze this expression for two different cases The rst case7 is appropriate for man made synchrotrons used as light sources In this case7 the light is produced by short bursts of electrons rnoving close to the speed of light 1 01 7 1272 passing a beam line port In addition7 because of the design of the radiation ports7 6 m 07 and the relevant integration times t are close to t m 0 This results in the form shown in Eq 1479 of your text It is convenient to rewrite this form in terms of a critical frequency 3 3 we 2 7 2p The resultant intensity is then given by d2 3q2 yz w 2 w 3 2 72192 w 3 2 i 1 2192 2 K lt 1 2192 7 K lt7 1 202 dwdQ 47TZC we ry 23 2wc Y 2 1 7202 13 2w Y 2 8 By plotting this expression as a function ofw we see that the intensity is largest near to m we The second example of synchroton radiation comes from a distant charged particle moving in a circular trajectory such that the spectrum represents a superposition of light generated over many complete circles In this case there is an interference effect which results in the spectrum consisting of discrete multiples of vp For this case we need to reconsider Eqs 5 and 6 There is a very convenient Bessel function identity of the form e7iasinuz Z Jmae7imuz39 m7oo Here Jma is a Bessel function of integer order m In our case a cos0 and Oz Analyzing the parallel component we have c 3 6 OH 7z39wp6cos6 wp 1 0S6 go Jm cos 19gt 2W6w 7 10 00 dteiwltt7 c0595invtp C 00 7 up 3 c In determining this result we have used the identity dteiw mt 2W6w 7 11 Eq 10 can be simpli ed to show that OH 2m 2 n cos 19gt 6a 7 mg 12 where Ja E W The perpendicular component can be analyzed in a similar way using integration by parts to eliminate the extra cosvtp term in the argument The result is tant9 wp 1 C 271397 J 7cost9 6w7m7 13 1 lt p ltgt In both of these expressions the sum over m includes both negative and positive values of m However only the positive values of w and therefore positive values of m are of interest and if we needed to use the negative m values we could use the identity Jma71mea 14 Combining these results we nd that the intensity spectrum for this case consists of a series of discrete frequencies which are multiples of vp w 2 Jm lt7p cos 19gt 15 0 d2 1sz 2 1 wp 2 tan2 6 dwd 7 C 7206M mp Jmlt C cost9gt 1262 These results were derived by Julian Schwinger Phys Rev 75 1912 1925 1949 The discrete case is similar to the result quoted in Problem 1415 in Jackson7s text It should have some implications for Astronomical observations but 1 have not yet found any references for that For information on man made synchrotron sources the following web page is useful http www als lbl govalssynchrotronsources html April 25 2002 Notes for Lecture 35 Synchrotron Radiation For this analysis we will use the geometry shown in Fig 149 of J ackson A particle with charge q is moving in a circular trajectory with radius 0 and speed Its trajectory as a function of time i is given by 1W psinl39vipli M1 COMMpl 1 Its velocity as a function time is given by vq coswip sinvip r The spectral intensity that we must evaluate is given by the expression Eq 1467 in Jackson 121 q2w2 0c 2 g e W 39 viwtifrqtd had 417 r X 1quot X W i l 70c After some algebra this expression can be put into the form 121 q2w262 dwdil 4220 where the amplitude for the light polarized along the y axis is given by ismVa may 4 0c 1 CuW 1iSiilwipkwlt g Wes mmPD a I 70c and the amplitude for the light polarized perpendicular to Sr and f is given by DC CLO LDC di sin 19 cosvitoe quott g mesmlmm 6 We will analyze this expression for two different cases The rst case is appropriate for man made synchrotrons used as light sources In this case the light is produced by short bursts of electrons moving close to the speed of light U z c 1 1 272 passing a beam line port In addition 29 z 0 and the relevant integration times i are close to i z 0 This results in the form shown in Eq 1479 of your text It is convenient to rewrite this form in terms of a critical frequency The resultant intensity is then given by 121 391272 9quot 239 2 2 2 w 39 2 2 2 7292 w 39 2 2 H2 m 7 m 7 7 1 m dwdil 41720 14 9 I m lt2wcl1 9 ll 17292 Al3 2 9 By plotting this expression as a function of car we see that the intensity is largest near w z Lac The second example of synchroton radiation comes from a distant charged particle moving in a circular trajectory such that the spectrum represents a superposition of light generated over many complete circles In this case there is an interference effect which results in the spectrum consisting of discrete multiples of p For this case we need to reconsider Eqs 5 and 6 There is a very convenient Bessel function identity of the form Ciiasina Z Jma07ima Here 1mm is a Bessel function of integer order m In our case a fcos l and a Analyzing the parallel component we have C C 78 OC dieiwlt gmsesmll tm 70 78 i Jm wp cos 9 21760 mg ll iwp6cos9 70c iwp8cos9 7m 0 p 10 In determining this result we have used the identity 9C 39 7 1y U diei w mo 2 6w 77k 11 70c 0 Eq 10 can be simpli ed to show that m w 1 CH 2m J nlt pcos 9 6W pr 12 70C C 0 where J 39a E d39lmla The perpendicular component can be analyzed in a similar way m do using integration by parts to eliminate the extra COS L ip term in the argument The result is taer m w G 2177 1quot lt C fee In both of these expressions the sum over m includes both negative and positive values of m However only the positive values of w and therefore positive values of m are of interest and if we needed to use the negative m values we could use the identity C p 2 quot 2 t9 5 l 7 13 cos Va mp me 1 Jma Combining these results we nd that the intensity spectrum for this case consists of a series of discrete frequencies which are multiples of p 2 2 22 cc 2 1 2 2 aw migtmltwgti Z cfl1mlt cos9lll c mzo p c c These results were derived by Julian Schwinger Phys Rev 75 1912 1925 1949 The discrete case is similar to the result quoted in Problem 1415 in Jackson s text It should have some implications for Astronomical observations but I have not yet found any references for that 39amngtxa UP 103 11033133 ug 191399 ha 635 g 392 m 2 71 pm fimamg maz p x7nx min 3p 3p 3p maqm 4 I ma all we 190 7 1333ng aq um uogmun sgmmg mxogsumngp Mug aq Imq z mpaadsax suonamgp 2 pm fi 2 mp ug summing amxdmm am axump x m pm x a x n 3 axdumxa 101 39suogsuamgp mug pm om o3 paptmxa aq 5mm um smp asaql 2 39 Ty 17003 Imuugs 9 Z mm uaq nxm ugs ng mun 3g axdumxa 101 u 39 u x x t Ixunxun Z I I summing smug am mum um am uaq suompum Smpunoq awudtuddp am 5389128 asp x n amqm gm v x n mi 21 3g 1033qu 9 39 909 x xm 30 sagsmzs uogmun sgmmg mp uogmnba unssgod mxogsuamgpq aq mg 11133858 mo 10 uogmun sgmmg 12 3tmsaxdax o3 suopgutg asaq 3st 0 sn smog gt91an mm 12 39lx 1xun run3 00 3mg sagdm summing Sm 30 ssatmmidmm mp 3mg pamoqs am It I 4ng 17p x1unx uni m 3mg 3th 3x S x S 1x mwmg am ug paung 20 summing mxo oquo 30 ms amidxnoaw p GARq am asoddng suonaun s uaaxf pm suogsuedxa uogqaun 1311030111 4 answer 105 saqu 000239 239 Sumqu 1 An alternative method of finding Green s functions for second order ordinary differential equations is based on a product of two independent solutions of the homogeneous equation um and 11203 which satisfy the boundary conditions at an and x1 respectively I 7K K 477 9 1xx u1xltu2xgt w me W 1 dz dz dz with xlt meaning the smaller of x and x and 33gt meaning the larger of x and 33 For example we have previously discussed the example of the one dimensional Poisson equation with the boundary condition lt1gt0 0 and 0 to have the form Gxx 4W lt 10 For the two and three dii nensional cases we can use this technique in one of the dimensions in order to reduce the number of summation terms These ideas are discussed in Section 311 of Jackson For the two dimensional case for example we can assume that the Green s function can be written in the form Gxx yy ZunwlunW Mdzw l 11 We require that G satisfy the equation 82 A A V20 Z 39unxunx l xn g y y 4mx x y y 12 R lt This form will have the required behavior if we choose 82 I A I 872 mow 4Wgty y 13 If Um and Um are solutions to the homogeneous equation d l xn only 0 14 satisfying the appropriate boundary conditions we can then construct the 2 dii nensional Green s function from Gxxl7 3 3 Z unlxlunlxlKnvm yltvm ygt where the constant Kn is defined in a similar way to the one dimensional case For example a Green s function for a two dimensional with 0 g x g a and 0 g y g I with the potential vanishing on each of the boundaries can be expanded GWWCy yI 8 i sin sin nix sinl1 sinh 0 ygt n1 nsinh 15

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