Physics 55 Fll 5 Midtem Solutions This midtem is two hou open ook, open notes exm. Do ll thee polems. [35 pts] 1. A ectngul ox hs sides of lengths, nd c z x c [1] ) Fo the Diichlet polem in the inteio of the ox, the Geen s function my e expnded s G(x, y, z; x, y, z ) = m=1 n=1 g mn (z, z ) mπx y mπx nπy nπy Wite down the ppopite diffeentil eqution tht g mn (z, z ) must stisfy. Note tht kx stisfies the completeness eltion m=1 mπx Hence the Geen s function eqution mπx = δ(x x ) x G( x, x ) = δ 3 ( x x ) hs n expnsion ( x g mn (z, z ) mπx mπx nπy ) nπy = δ(z z ) 4 mπx mπx nπy nπy Woking out the x nd y deivtives on the left-hnd side yields [ d ( mπ ) ( nπ dz ) ] g mn (z, z ) mπx mπx nπy nπy = 16π δ(z z ) mπx mπx nπy nπy
Howeve, ce kx foms n othogonl sis, ech tem in this sum must vnish y itself. This esults in the diffeentil eqution ( d dz γ mn ) g mn (z, z ) = 16π δ(z z ) (1) whee γ mn = π (m/) + (n/) is given in pt c). Note tht the Fouie e expnsion utomticlly stisfies Diichlet oundy conditions fo x nd y. The emining oundy condition is tht g mn (z, z ) vnishes wheneve z o z is equl to o c. [15] ) Solve the Geen s function eqution fo g mn (z, z ) suject to Diichlet oundy conditions nd wite down the esult fo G(x, y, z; x, y, z ). We my solve the Geen s function eqution (1) y fist noting tht the homogeneous eqution is of the fom g (z ) γ mng(z ) = This is second-ode line eqution with constnt coefficients dmitting the fmili solution g(z ) = Ae γ mnz + Be γ mnz Howeve, we wnt g(z ) = when z = o z = c. This motivtes us to wite out the solutions u(z ) = h γ mn z v(z ) = h[γ mn (c z )] < z < z z < z < c As we hve seen, the full Geen s function solution is then given y g mn (z, z ) = 16π A u(z <)v(z > ) whee A is given y the Wonskin, W (u, v) = A/p(z ) whee the Stum-Liouville opeto hs the fom d dz p(z ) d dz + q(z ). In this cse p(z ) = 1, nd it is esy to see tht W = u v u v = γ mn [h γ mn z cosh γ mn (c z ) + cosh γ mn z h γ mn (c z )] Hence = γ mn h γ mn c g mn (z, z ) = 16π γ mn h γ mn c h γ mnz < h γ mn [(c z > )]
nd G( x, x ) = 16π 1 mπx γ mn h γ mn c mπx nπy nπy h γ mn z < h γ mn [(c z > )] () [1] c) Conside the oundy vlue polem whee the potentil on top of the ox is Φ(x, y, c) = V (x, y) while the potentil on the othe five sides vnish. Ug the Geens function otined ove, show tht the potentil my e witten s Φ(x, y, z) = m=1 n=1 whee γ mn = π (m/) + (n/) nd A mn = 4 h γ mn c A mn mπx dx nπy h γ mn z dy V (x, y) mπx nπy Since we only hve to woy out the potentil on the top of the ox (nd ce we ssume thee is no chge inside the ox), the Geen s function solution my e witten Φ( x ) = 1 Φ( x ) G( x, x ) n d = 1 V (x, y ) G( x, x ) n dz dy Note tht the outwd-pointing noml ˆn on the top of the ox is in the +ẑ diection, we compute the noml deivtive of () G( x, x ) n = G( x, x ) z = 16π 1 mπx γ mn h γ mn c mπx nπy nπy = ( h γ mn z γ mn cosh γ mn [(c z )]) 4 mπx h γ mn c mπx nπy h γ mn z nπy Inseting this into (3) then stightfowdly gives the desied esult. Note tht the pimes my e dopped fom the doule integl A mn once it hs een isolted fom the est of the expession. (3)
[3 pts]. The potentil on the sufce of sphee of dius is specified y β {, θ < β V β V (θ, φ) = V, β θ π β, π β < θ π Thee e no othe chges in this polem. [] ) Show tht the potentil outside the sphee my e expessed s Φ(, θ, φ) = ( ) l+1 V [P l+1 (cos β) P l 1 (cos β)] Pl (cos θ) l=,4,6,... whee we tke P 1 (x) =. Note tht Legende polynomils stisfy the eltion (l + 1)P l (x) = P l+1 (x) P l 1 (x). Thee e sevel wys of solving this polem. Pehps the most stightfowd is to elize fom zimuthl symmety tht the potentil necessily dmits Legende expnsion Φ( x ) = ( ) l+1 α l Pl (cos θ) l The oundy conditions t = gives V (θ) = l α l P l (cos θ) This is clely Legende expnsion fo V (θ). The legende othogonlity eltion llows us to wite the expnsion coefficients α l s α l = l + 1 1 1 Fo the specified potentil, this ecomes α l = l + 1 cos β cos β V (cos θ) P l (cos θ) d(cos θ) V P l (cos θ) d(cos θ) = V cos β cos β Ug the Legende eltion given in the polem, we otin α l = V = V cos β cos β [P l+1(x) P l 1(x)] dx (l + 1)P l (x) dx [ P l+1 (cos β) P l 1 (cos β) P l+1 ( cos β) + P l 1 ( cos β) = V (1 + ( )l )[P l+1 (cos β) P l 1 (cos β)] ]
This vnishes unless l is even (which should e ovious fom the z z symmety of the polem). The esult is then Φ(, θ) = ( ) l+1 V [P l+1 (cos β) P l 1 (cos β)] Pl (cos θ) l even Note tht l = is llowed, nd gives the monopole contiution. (Excluding l = fom the sum ws mistke.) This polem could lso hve een solved y ug the Diichlet Geen s function outside sphee G( x, x ) = ) 1 ( l< l+1 1 l + 1 l+1 l,m < > l+1 Ylm(Ω )Y lm (Ω) Afte slight mnipultion, we would hve ended up with simil Legende polynomil integtion. [1] ) Fo fixed V, wht ngle β mximizes the qudupole moment? The qudupole is given y l =. Hence the qudupole moment is elted to the α tem in the expnsion. Since we do not ce out nomliztion (ny only to mximize the moment) it is sufficient to wite q, α P 3 (cos β) P 1 (cos β) We extemize this quntity y tking deivtive with espect to β nd setting the esult to zeo. Noting tht tking deivtive simply undoes the integtion, we end up solving dq, dβ P (cos β) β = 1 (3 cos β 1) β = This is solved y β = nd β = cos 1 (±1/ 3). Clely the qudupole moment vnishes when β = (s the entie sphee is t constnt zeo potentil). Hence the mximum is when β = cos 1 (1/ 3), povided we estict β π/. Note tht β > π/ does not elly mke sense, except in foml mnne whee V V (coesponding to intechnging the integtion limits). Techniclly, the polem should hve sked to mximize the mgnitude of the qudupole moment insted of to mke q, s positive s possile (which would depend on the sign of V ). [35 pts] 3. A line chge on the z xis extends fom z = to z = + nd hs line chge density vying s x λ z α λ z α y λ(z) = { λ z α, < z λ z α, z <
whee α is positive constnt. The totl chge on the < z segment is Q (nd the chge on the z < segment is Q). [] ) Clculte ll of the multipole moments of the chge distiution. Mke sue to indicte which moments e non-vnishing. Noting tht unifomly chged line chge on the +z xis hs chge density we see tht vying line chge yields ρ = λ δ(cosθ 1) π ρ = λ() δ(cos θ 1) π This my e checked y oseving dq = ρ d 3 x = ρ d dφ d(cos θ) = λ()d dφ π cos θ=1 (Note tht thee is no distinction etween nd z fo cos θ = 1.) Hence, fo the positive nd negtive line chge, we hve ρ = λ α [δ(cos θ 1) δ(cos θ + 1)] π We my nomlize λ y integting fom to to otin the totl chge Q Q = λ(z) dz = λ z α dz = λ α+1 α + 1 Hence ρ = Q α + 1 ( ) α [δ(cos θ 1) δ(cos θ + 1)] π The multipole moments e then q lm = l Ylm(Ω)ρ d dω = Q α + 1 ( l ) α d Y lm(ω)[δ(cos θ 1) δ(cos θ + 1)] dω π By zimuthl symmety, only the m = moments e non-vnishing q l, = Q α + 1 l+1 1 α + l + 1 1 = Q l α + 1 l + 1 α + l + 1 [P l(1) P l ( 1)] = Q l α + 1 l + 1 α + l + 1 [1 ( )l ] l + 1 P l(cos θ)[δ(cos θ 1) δ(cos θ + 1)]d(cos θ)
Hence ll moments vnish unless l is odd nd m is zeo. Then q l, = Q l α + 1 l + 1 α + l + 1 π l odd [1] ) Wite down the multipole expnsion fo the potentil in explicit fom up to the fist two non-vnishing tems. The multipole expnsion gives Φ = 1 ɛ l,m = 1 ɛ l odd l + 1 q lm = Q ɛ l odd = Q ( (α + 1) ɛ α + = Q ɛ ( (α + 1) α + l + 1 q l, Y lm (θ, φ) l+1 α + 1 α + l + 1 ( P l (cos θ) l+1 ) l Pl (cos θ) P l (cos θ) (α + 1)3 P 3 (cos θ) + α + 4 4 + cos θ (α + 1)3 5 cos 3 θ 3 cos θ + (α + 4) 4 + [5] c) Wht is the dipole moment p in tems of Q, nd α? Note tht the dipole tem in (4) hs the fom Φ = 1 Q(α + 1) z ɛ α + 3 Comping this with the dipole expession gives φ = 1 p x ɛ 3 p = Q(α + 1) ẑ α + Altentively, one could compute diectly p = xρ d 3 x = xq(α + 1) z α sgn(z) dz α+1 The z component is the only non-vnishing component x=y= ) ) (4) p z = Q(α + 1) z α+1 Q(α + 1) dz = α+1 α +