411-exam.final.cover.pages

411-exam.final.cover.pages - Physics 411 Final...

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Unformatted text preview: Physics 411 Final Name__________________ cover p_1 1.________(80 points) course ave ______ 2.________(35 points) course grade ______ 3.________(35 points) total .________(/150 points) __________________________________________________________________________________ 4 6 DÝr, t Þ = 4^k 1 O o _ 0 Ýr, t Þ DÝr Þ= PEÝr Þ ßD 1 Ýr, t Þ ? D 2 Ýr, t Þà 6 n = aÝr, t Þ 4 6 BÝr, t Þ = 0 /BÝr, t Þ /t J /DÝr, t Þ + 4^k J J Ýr, t Þ 4 × H Ýr , t Þ = k 2 W o O o 2W 0 o k1 /t 4 × EÝr, t Þ = ?k 3 ßB 1 Ýr, t Þ ? B 2 Ýr, t Þà 6 n = 0 n ×ÝE 1 Ýr, t Þ ? E 2 Ýr, t ÞÞ = 0 n × ÝH 1 Ýr, t Þ ? H 2 Ýr, t ÞÞ = FÝr, t Þ ®Ý r Þ = 1 4^O o Ýv XXX |_?rrÞ| d 3 x + ® o v r V k?k v ?Ýg?g v Þtà Cover pages 4 6 JÝr, t Þ+ /_Ýr, t Þ = 0; /t x NÝgÝxÞÞ= > i NÝdg ? x i Þ i F = qßE +k 3 v × B à; NÝ k ? k v Þ NÝ g ? g v Þ = 1 | dx | x=x i Ý 2^ Þ 4 XXXX all space, time e ?ißr6 d 3 rdt df = dr 6 4fÝr Þ system esu (Electrostatic) emu (Electromag.) Gaussian Heaviside-Lorentz SI (MKSA) SI (MKSA) quantity Ampere Oo = Wo = Wo Oo dq Coul.s ?1 dt 10 7 4^c 2 ?7 DÝr Þ= PEÝr Þ. J k3 1 1 c c 1 1 1 1 c 1 c XXX contains r fÝr v ÞN Ý3Þ Ýr ?r v Þdx v dy v dz v = fÝr Þ B k1 1 c2 1 1 4^ 1 4^O o k2 1 c2 q statcoul. statcoul. statcoul. statcoul 1 1 c2 1 4^c 2 Wo 4^ gauss (G) 1 Coulomb (C) 1 ” V/ohm = V/I tesla ” 10 ?7 c 2 10 ?7 1 8.854 187 10 ?12 Farad/m 1.256 637 10 ?6 Henry/m 376.730 610 5 /c 2 6.582 10 ?16 eVs e2 ci (SI) 4^10 376.730 ohms (IÞ 1.0546 10 ?34 Js ÝSIÞ = 1 137.04 i e e2 4^O o ci 1.602 10 ?19 Coulomb 4.80 10 16 statcoul Ýesu, GaussianÞ Ýesu, GaussianÞ Ýesu, GaussianÞ 0.529177 Å 0.529177 10 ?10 m e2 2r H Bohr radius, r H e2 4^O o Ý2r H Þ ÝSIÞ = 13.605 eV ........................................................................................................................................................cover p_2 _ 0 Ýr v Þ Ý r ? r v Þ 3 v 4 6 D 0 Ýr Þ = 0 DÝr Þ = 1 XXX d r + D 0 Ýr Þ, SI units 4^ |r ? r v | 3 dÝ r Þ = 1 4^ P XX surface aÝr v Þ dS v + d 0 . SI units |r ? r v | _Ýr Þ= aÝr ÞNÝFÝr ÞÞ|4FÝr Þ | ! EÝzÞ = n a 0 2P d + Ýr Þ ? d ? Ýr Þ = ? 1 ^ d Ýr Þ P å nÝr Þ6ÝP 1 E 1 Ýr Þ ? P 2 E 2 Ýr ÞÞ= aÝr Þ; å nÝr Þ6ÝP 1 E 1 Ýr Þ ? P 2 E 2 Ýr ÞÞ= 0; dÝ r Þ = ? 1 P N¸0 ?a 0 z = d x , y , z || Ý Þ 2P lim aÝr, N ÞN = ^ d Ýr Þ; XX ^ d Ýr Þ Ýr ? rÞ 6 dS v = ? 1 P |r ? r v | 3 v XX ^ d Ýr Þ dI 4 2 dÝ r Þ = ? 1 _Ý r Þ P 4 2 gÝr, r v Þ = N Ý3 Þ Ýr ? r v Þ; dÝ r o Þ = dÝ r Þ = 1 P XXXß?_Ýr v Þà gÝr, r v Þ d 3 r v ?1 + F Ýr , r 0 Þ 4^|r ? r 0 | gÝr, r 0 Þ = XXX gÝr, r o Þß?_Ýr Þ/Pàd 3 r+ XX volume bounding surface ! ßdÝr Þ4gÝr, r o Þ ? gÝr, r o Þ4dÝr Þà6nÝr ÞdS d D Ýr 0 Þ = 1 P XXX G D Ýr, r 0 Þß?_Ýr Þàd 3 r volume + XX bounding surface dÝr s Þ4G D Ýr s , r 0 Þ 6 dS 4 2 G N Ýr, r 0 Þ = N Ý3 Þ Ýr ? r 0 Þ; XX bounding surface ! 4G N Ýr, r 0 Þ 6 nÝr ÞdS=1 dÝ r 0 Þ = 1 P XXX G N Ýr, r 0 Þß?_Ýr Þàd 3 r volume + d avg + XX bounding surface ! G N Ýr, r 0 Þ4dÝr Þ 6 nÝr ÞdS ! ß4G N Ýr, r 0 Þà 6 nÝr Þ = d L?D Ýr o Þ = 4F N Ýr, r 0 Þ + 4 ?1 4^|r ? r 0 | ! 6 nÝr Þ = 1 S tot XX bounding surface dÝr s Þ L?D 4G D Ýr s , r 0 Þ 6 dS; d L?N Ýr Þ = ? XX bounding surface G N Ýr s , r 0 Þ4d L?N Ýr s Þ 6 dS SÝr, t Þ = EÝr, t Þ × HÝr, t Þ; W ij = P W =1 2 W electron = e 2 /2a Q i = > C ij V j j=1 N Wi = P 2 C ii = XXX |4d i Ýr Þ |2 d 3 r volume XXX 4d i Ýr Þ 6 4d j Ýr Þd 3 r, volume Qi Vi , Vj= 0 > QiVi= 1 2 i=1 3 N > C ij V i V j i,j=1 N Ißf, 4f à = XXX FÝr, fÝr Þ, 4fÝr ÞÞd 3 r / NF NF ? ? V u Ýr Þ = 0 Nf > /x i NÝ/f//x i Þ > J J i=1 J ....................................................................................................................................................cover p_3 4 2 ®Ý_, j, z Þ = 1/ r 2 /r _ //_ ®Ý_, j, z Þ + _12 //j 2 ®Ý_, j, z Þ + / r 2 /r ®Ýr, S, d Þ ? 1 L 6 L®Ýr, S, d Þ = 0 r2 1/ _ /_ 2 /2 /z 2 ®Ý_, j, z Þ = 0 W m (d)=A m e L 6 L Y(S,d)=§(§+1)Y(S,d) /2 WÝdÞ = ?m 2 WÝdÞ /d 2 imd d dx Ý1 ? x 2 Þ +B m e ?imd ; F §,|m| Ýx Þ=A §,|m| P |§m| (x)+B §,|m| Q |§m| (x) d dx F |lm| Ýx Þ + ߧݧ + 1 Þ ? m2 1?x 2 àF |lm| ÝxÞ = 0.; x = cos S m/2 P m Ýx Þ = Ý1 ? x 2 Þ § P 0 Ý1 Þ = 1 : § d m P § Ýx Þ dx m ,m³0 , P 0 Ý ?1 Þ = Ý ?1 Þ § , § P m Ýx Þ = Ý?1 Þ m P |§m | Ýx Þ, m < 0 § if m ® 0, P m ݱ1 Þ = 0. P m Ý?x Þ = Ý?1 Þ l+m P m Ýx Þ § § § P 0 Ýx Þ = 1 P 1 Ýx Þ = x P 2 Ýx Þ = P 3 Ýx Þ = P 4 Ýx Þ = 1 2 x 2 1 8 Ý3x 2 ? 1 Þ Ý5x 2 ? 3 Þ Ý35x 4 ? 30x 2 + 3 Þ 2l+1 Ýl?m Þ! 4^ Ýl+m Þ! x + 1 m/2 FÝ?§, § + 1, 1 ? m; 1 ? x Þ x?1 2 x + 1 m/2 1 ? x m FÝm ? §, § + m + 1, m + 1; 1 ? x Þ m P § Ýx Þ =constant x?1 2 2 @Ýa + nÞ@Ýb + nÞ n @Ý 1 ? c Þ FÝa, b, c; zÞ = Fn z n!@Ýc + nÞ @ÝaÞ@ÝbÞ @ÝaÞ = Ýa ? 1Þ! Q m Ýx Þ =constant § Y m Ý S , d Þ = Ý ?1 Þ m l P m ÝcosÝS ÞÞe imd l 2^ ^ Y ?m ÝS, d Þ = Ý?1 Þ m Y m ÝS, d Þ D l l Y m Ý^ ? S, 2^ ? d Þ = Ý?1 Þ l Y ?m ÝS, d Þ l l Y 0 Ý S, d Þ = 0 1 4^ 3 Y 0 Ý S, d Þ = 1 4^ Y 1 Ý S, d Þ = ? 3 1 4^ 1/2 1/2 X 0 X 0 sin S Y m v ÝS, d Þ D Y m ÝS, d ÞdSdd=N ll v N mm v l lv 5 4^ Y 1 ÝS, d Þ = ? 15 2 8^ Y 2 ÝS, d Þ = ? 15 2 4^ Y 0 Ý S, d Þ = 2 1/2 1/2 1/2 l > lK 0 > m=?l Y m ÝS v , d v Þ D Y m ÝS, d Þ = NÝd ? d v ÞNÝcos S ? cos S v Þ l l = 3 cos 2 S ? 1 2 2 sin S cos Se id cos S sin Se id 1/2 sin 2 Se i2d 7 1/2 5 cos 3 S ? 3 cos S 4^ 2 2 1/2 Y 1 ÝS, d Þ = ? 1 21 sin Sß5 cos 2 S ? 1 àe id 3 4 4^ 1/2 Y 2 ÝS, d Þ = 1 105 sin 2 S cos Se i2d 3 4 2^ 1/2 Y 3 ÝS, d Þ = ? 1 35 sin 3 Se i3d 3 4 4^ 4 2 ®Ýr, S, d Þ = ?4^_Ýr, S, d Þ; 4 2 ®Ýr, S, d Þ = ?_Ýr, S, d Þ/O units K l rl 1 < m ®Ýr, S, d Þ = 4^q > l=0 > m=?l 2l+1 r l+1 Y l ÝS, d ÞY m ÝS v , d v Þ D l > ————————————————————————————————————————————– ^ ^ X 0 sinÝmzÞ sinÝm v zÞdz = ^ N mm v X 0 cosÝmzÞ cosÝm v zÞdz = ^ N mm v 2 2 Y 0 Ý S, d Þ = 3 .....................................................................................................................................................cover p_4 l K l ^ < GÝr, r v Þ = ?1 > l=0 > m=?l 24+1 rrl+1 Y m ÝS, d ÞY m ÝS v , d v Þ D l l l > 4^ G D Ýr , r v Þ = ?1 4^ 1 |r?r v | + ?R/ r v r?ÝR 2 /r v2 Þr v , K 1 = > l=0 v! |r ?r z| 1 => K > l l=0 m=?l |r ? r v | dÝ r Þ = 1 P l 4^ r < 2l+1 r l+1 > Y m Ý S, d Þ Y m Ý S v , d v Þ D l l rl < r l+1 > P 0 Ýcos S Þ l XXX V _Ýr v ÞGÝr, r v Þd 3 r v + XX S ßdÝr vs Þ4 v GÝr, r vs Þ ? GÝr, r vs Þ4 v dÝr vs Þà 6 dS v l 4^ 2l+1 P l Ýcos L Þ = > m=?l Y m Ý S, d Þ Y m Ý S v , d v Þ D l l ——————————————————————————————————————————/ /2 1/ 1 /2 _ /_ _ /_ ®Ý_, j, z Þ + _ 2 /j 2 ®Ý_, j, z Þ + /z 2 ®Ý_, j, z Þ = 0 d 2 W k Ýz Þ = k 2 W k Ýz Þ; W k Ýz Þ = A k e kz + B k e ?kz Note that k can dz 2 2 d V m Ýj Þ = ?m 2 V m Ýj Þ;V m Ýj Þ = C m e imj + D m e ?imj dj 2 2 _ 2 dd_ 2 F m Ýk_ Þ + _ dd_ F m Ýk_ Þ + Ýßk_à 2 + m 2 ÞF m Ýk_ Þ = 0 be complex, pure imaginary or real. F m Ýk _ Þ = a m J m Ýk _ Þ + b m N m Ýk _ Þ H ± Ýk_Þ ¯ J m Ýk_Þ ± iN m Ýk_Þ m Ýk_/2 Þ m J m k_ ¸0 i @Ý m + 1 Þ N m k_ ¸0 i @Ýp ÞÝk_/2 Þ ?m /^. J m Ýk_ ¸K Þ i N m Ýk_ ¸K Þ i 1 2.405 3.832 7.016 2 ^k_ 2 ^k_ cos k_ ? sin k_ ? m^ 2 m^ 2 ? ? ^ 4 ^ 4 n m=0 m=1 m=2 m=3 m=4 m=5 5.135 8.417 6.379 9.76 7.586 8.780 15.700 Table of x mn where J m Ýx mn Þ = 0 : 2 5.520 3 8.654 11.064 12.339 10.173 11.620 13.017 14373 4 11.792 13.323 14.796 16.224 17.616 18.982 5 14.931 16.470 17.960 19.410 20.827 22.220 J m Ýx ÞN m Ýax Þ ? J m Ýax ÞN m Ýx Þ = 0 a Ýa > 1 Þ J vm Ýx ÞN vm Ýax Þ ? J vm Ýax ÞN vm Ýx Þ = 0 Ýa > 1 Þ X 0 _J m Ýk n v _ ÞJ m Ýk n _ Þd_ = 0.5 N n v n ßaJ m+1 Ýk n a Þà 2 where J m Ýka Þ = 0. X 0 _J m Ýk_ ÞJ m Ýk v _ Þd_ = k ?1 NÝk ? k v Þ I m Ý|k|_Þ = i ?m J m Ýi|k|_Þ I m Ý|k|z ¸0 Þ i 1 @Ým+1 Þ |k|z 2 ?1/2 m K and ; K m Ý|k|_Þ = ^ i m+1 H + Ýi|k|_Þ m 2 K m Ý|k|z ¸0 Þ i @Ým Þ 2 ^ 2|k|z 2 |k|z 1/2 m , m ® 0 ( i ? ln |k|z 2 , m = 0Þ I m Ýkz ¸K Þ i Ý2^|k|z Þ expÝ|k|z Þ K m Ýkz ¸K Þ i expÝ?|k|z Þ. ————————————————————————————————————————————– ....................................................................................................................................................cover p_5 1 GÝr; r v Þ = 4^|?1 r v | 4 2 GÝr; r v Þ = NÝr ? r v Þ = _ NÝ_ ? _ v ÞNÝz ? z v ÞNÝj ? j v Þ r? 1 NÝ_ ? _ v Þ = _ v X kJ p Ýk_ ÞJ p Ýk_ v Þdk NÝz ? z v Þ = 21^ X e ikÝz?z Þ dk = ^ X cosßkÝz ? z v Þàdk, ?K 0 0 v K . NÝj ? j v Þ = 21^ > m=?K e imÝj?j Þ = 21^ á1 + 2 cosßmÝj ? j v Þàâ ———————————————————————————————————————————— v K K K K l GÝr, r v Þ = ?1 > l=0 > m=?l 4^ l 4^ r < 2l+1 r l+1 > Y m Ý S, d Þ Y m Ý S v , d v Þ D l l + > ml= ?l Y m ÝS, d ÞY m ÝS v , d v Þ D l l GÝr; r v Þ = 4^ > l=0 GÝr; r v Þ = GÝr; r v Þ GÝr; r v Þ GÝr; r v Þ GÝr; r v Þ = = = = 1 2^ K b 2l+1 ?R 2l+1 > Ý2l+1 Þ R 2l+1 ?a 2l+1 < Ýrr v Þ l+1 b 2l+1 ?a 2l+1 K > m=?K e imÝj?j Þ X 0 k g m Ýz; z v ÞJ m Ýk_ ÞJ m Ýk_ v Þdk v K 1 2^ 1 2^ 1 2^ 1 2^ K > m=?K e imÝj?j v Þ X 0 k ?1 expÝkz < Þ expÝ?kz > ÞJ m Ýk_ ÞJ m Ýk_ v Þdk 2k K K mn mn mn > m=?K e imÝj?j v Þ > n=1 xa C mn expÝ xa z < Þ expÝ? xa z > ÞJ m v K K x mn D sinhÝ x mn z Þ sinhÝ? x mn z ÞJ > m=?K e imÝj?j Þ > n=1 a mn a< a>m K K n^ n^ zÞ sinÝ n^ z v ÞÞI m n^ _ < imÝj?j v Þ > m=?K e > n L E mn sinÝ L L L K x mn _ J x mn _ v m a x mn _ J a mn _ v x m a a K m m^ _ > L GÝr, r v Þ = > lm X k 2 g l Ýk Þj l Ýkr Þj l Ýkr v ÞY m ÝS, d ÞY m ÝS v , d v Þ D dk l l 0 ®Ýr Þ = 4^ > l=0 > m=?l K l 1 2l+1 K Y m ÝS, d Þ XXX _Ýr v Þ l r l< Y m ÝS v , d v Þ D r v2 sin S v dS v dj v l r l>+1 —————————————————————————————————————————— 4 2 fÝr Þ + k 2 fÝr Þ = 0 r2 K d2 dr 2 fÝr Þ = > l=0 > m=?l M l Ýr ÞY m ÝS, d Þ l M l Ýr Þ = a l j l ÝkrÞ + b l n l ÝkrÞ j l Ýkr Þ = ^ 2kr K l d M l Ýr Þ + 2r dr M l Ýr Þ ? lÝl + 1 ÞM l Ýr Þ + k 2 r 2 M l Ýr Þ = 0. ^ X 0 r 2 j l Ýkr Þj l v Ýk v r Þdr XX Y m ÝS, d ÞY m v ÝS, d Þ D dI = 2kk v NÝk ? k v ÞN ll v N mm v l lv K l > lK 0 X 0 k 2 j l Ýkr Þj l Ýkr v Þdk > m=?l Y m ÝS, d Þ D Y m ÝS v , d v Þ = ^ NÝr ? r v Þ l l 2 = J l+ 1 Ýkr Þ 2 ....................................................................................................................................................cover p_6 ®Ýr Þ = > k > m=?K C km expß±kzà > m=?K e imj ßC km J m Ýk_Þ + D km N m Ýk_Þà ®Ýr Þ = > k > m=?K C km expß±ikzà > m=?K e imj ßC km J m Ýik_Þ + D km N m Ýik_Þà K K K K cover p_7 Solutions to Laplace’s equation: (x,y,z), (r,S, jÞ, Ý_, j, zÞ; Helmholtz equation in (r,S, jÞ Equation separation. const. k=k 1 General solution: sum over all sep. constants conditions 4 2 ®Ýx, y, zÞ = 0 4 ®Ýr, S, jÞ = 0 2 ! ! x +k 2 ŷ +k 3 z F k6k =0 ßcÝkÞe ## k6r + dÝ k Þ e ## k6r ÝÝax +bÞN k1,0 + Ýcy +dÞN k2,0 k 6 k = 0, k can be co + Ýez + fÞN k3,0 Þ + Ýax +bÞÝcy +dÞÝez +fÞN |k|,0 à |m| ² § = 0, 1, 2... ² § = 0, 1, 2... F §,m ßa § r § + b § r ?§?1 àßc §,m Y §,m ÝS, jÞ + d §,m Q §,m ÝS, jÞà Q §,m ¸ K, cos S N m ¸ K, _ ¸ 4 2 ®Ý_, j, zÞ = 0 Ý4 + k Þ®Ýr, S, jÞ = 0 2 2 J, m= 0, ±1, ±2, ... F J,m ßa Jm J m ÝJ_Þ + b Jm N m ÝJ_Þàe imj e ±Jz |m| F §,m ßa m j m Ýk_Þ + b m n m ÝJ_Þàßc §,m Y §,m ÝS, jÞ + d §,m Q §,m ÝS, jÞà n m ¸ K, r ¸ 0 Function separation consts k 1,2 = k3 = e ## k6r ± i|k 1,2 |; k3 real e iÝ|k 1 |x+|k 2 |yÞ e ±k 3 z ===>sin, cos in two variables ,exponential in third e Ýk 1 x+k 2 yÞ e ± i |k 3 |z ===>sin, cos in one variable, exponential in other two m 2§ + 1 ݧ ? mÞ! 1/2 |m| Y §,m ÝS, jÞ |m|² § = 0, 1, 2... (-1) ß à P § Ýcos SÞ e imj ; XX Y §,m D Y § v ,m v dI = N §,§ v N m,m v 4^ ݧ + mÞ! J_ 2j+m K Ý?1Þ j J m ÝJ_Þ J, m= 0, ±1, ±2, ... F j ß à ; X a_ J m Ýa_ÞJ m Ýb_Þd_ = NÝa ? bÞ where J m ÝaÞ = J m ÝbÞ =0 0 j !Ý m + j Þ ! 2 Ý k _Þ § j m Ýk_Þ; n m Ýk_Þ |m|² § = 0, 1, 2... j m Ýk_Þ = ß ^ à 1/2 J m+1/2 Ýk_Þ; j 0 ÝxÞ = sin x ; j § Ýk_Þ k_¸K ¸ x 2k_ Ý2§ + 1Þ!! ± i|k 3 |; k 1,2 real The boundary conditions:the boundary conditions on the solution restrict the separation constants to specific values. Uniqueness: solutions to 4 2 ® = 0 in a finite volume, satisfying the same boundary conditions, are unique. ...
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