Wilcox Chapter3R

3.1 Chapter 3: Electrostatics: Cylindrical and Spherical Coordinates 3.1 Reduced Green function technique in cylindrical coordinates and Bessel functions Basic problem in extending reduced G.f. method to cylindrical & spherically symmetric geometries: have to beable to separate the variables in order to reduce it to a 1-D problem. Techniques in this chapter largely based on methodsfrom Schwinger. Let's go back to the wall problem. Reminder: + zGeometry is indep. of x&y => nin zEq Set GD(x≥,x≥') = 4π ∫ d2k (2π)2 eik≥.(x≥ - x≥')⊥ g(z,z',k) (3.1) based upon 4π δ(x≥ - x≥') = 4π ∫ d2k (2π)2 eik≥.(x≥ - x≥')⊥ δ(z-z'). (3.2) Since -∇2GD(x≥,x≥') = 4π ∫ d2k (2π)2 eik≥.(x≥ - x≥')⊥ (k2 - ∂2 ∂z2) g(z,z',k), (3.3) => (k - ∂2 ∂z2) g = δ(z-z'). (3.4) 3.2 Now try it for cylindrical coordinates: (same situation w/cyl. sym.) φφ' α⊥⊥ ⊥≥ ≥x x' k≥ x (now give k a z- component also.)≥ Geometry is indep. of φ & z => eqn in ρ. k≥.(x≥ - x≥') = k z(z-z') + k ⊥ρcos(φ - α) - k ⊥ρ'cos(φ'- α), (3.5) (ρ = |x≥ ⊥|, ρ' = |x≥'⊥|, k⊥ = |k≥ ⊥|) ∂(x≥ - x≥') = ∫ dkz (2π) eikz(z-z') ∫ k⊥dk⊥dα (2π)2 eik⊥ρcos(φ-α) e-ik⊥ρ'cos(φ'-α). (3.6) It is obvious that ek⊥ρcos(φ-α) ≠ f(kρ)g(φ). (3.7) The reduced G.f. method depends upon a separation of variables. Brute force it: k⊥ρ ≡ t, (3.8) eiφ ≡ 1 i u (a generic φ). (3.9) Then eik⊥ρcosφ = et/2(u-1/u). (3.10) Imagine expanding in powers of u. It must be of the form 3.3 et/2(u-1/u) = ∑ m=-∞∞ um Jm(t), (3.11) where the J m(t) are simply the coefficients of the expressions. These coefficients, which are functions of thet variable, turn out to be Bessel functions of integer order.LHS above is called a generating function. We have now separated; the price paid is an infinite sum. From theinvariance of the gen. funct. under u .- 1 u , (3.12) we learn that ∑ m=-∞∞ um Jm(t) = ∑ m=-∞∞ (-1)m u-m Jm(t) = ∑ m=-∞∞ (-1)m um J-m(t), (3.13) => J-m(t) = (-1)m Jm(t). (3.14) Therefore (cosφ = (u - 1 u)/2i) eitcosφ = ∑-∞ ∞ im eimφ Jm(t) = J 0(t) + 2 ∑ m=1∞ im cosmφ Jm(t). (3.15) Can solve for m(t) by usingJ : ∫ 02π dφ 2π e-im'φ eimφ = ∂mm'. (3.16) (Can do the integral directly to see this.) This gives imJm(t) = ∫ 02π dφ 2π ei(tcosφ-mφ). (3.17) Called an integral representation. From either the generating or integral eqns, can now show that (m -> 0) 3.4 Jm(t) = ∑ n=0∞ (-1)n (t/2)m+2n n!(m+n)!. (3.18) We can develop the diff. eqn that J m(t) satisfies as follows. Consider d dt Jm(t) = 1 im∫ 02π dφ 2π (icosφ)ei(tcosφ -imφ). (3.19) But Jm-1(t) = 1 im-1 ∫ 02π dφ 2π ei(tcosφ-(m-1)φ), " = 1 im ∫ 02π dφ 2π (ieiφ)ei(tcosφ-mφ), (3.20) and Jm+1(t) = 1 im ∫ 02π dφ 2π (-ie-iφ)ei(tcosφ-mφ). (3.21) Therefore Jm-1(t)-Jm+1(t) = 1 im ∫ 02π dφ 2π i(eiφ + e-iφ)ei(tcosφ-mφ) = 2d dt Jm(t). (3.22) Likewise Jm-1(t)+Jm+1(t) = 1 im ∫ 02π dφ 2π (-2sin φ)ei(tcosφ-mφ) = 2 im 1 it∫ 02π dφ 2π (d dφ eitcosφ)e-imφ) by parts = 0 - 2 im {1 it∫ 02π dφ 2π eitcosφ d dφe-imφ} -ime-imφ 3.5 = 2 m t Jm(t). (3.23) These are called recurrance relations. So, we have 2 d dt Jm(t) = J m-1(t) - J m+1(t), (3.24) 2 m t Jm(t) = J m-1(t) + J m+1(t), (3.25) => J m+1(t) = (m t - d dt)Jm(t), (3.26) J m-1(t) = (m t + d dt)Jm(t). (3.27) Therefore (m-1 t - d dt)(d dt + m t)Jm(t) = J m(t). (3.28) Jm-1(t) In more standard form, this gives [d2 dt2 + 1 t d dt - m2 t2 + 1]Jm(t) = 0. (3.29) [This is the dimensionless form of the Bessel function. Re- instating t=k ⊥ρ gives us the more physical dimensionful form (what we get when we separate variables in Laplace's eqn in cylindrical coordinates): (k ⊥ . k here) [d2 dρ2 + 1 ρ d dρ - m2 ρ2 + k2]Jm(kρ) = 0. ] 3.2 Completeness of Bessel functions What have we shown? We know 3.6 δ(x≥ - x≥') = ∫ dkz 2π eikz(z-z') ∫ dk⊥k⊥dα (2π)2 eik⊥ρcos(φ-α) e-ik⊥ρ'cos(φ'-α). (3.30) Expand eik⊥ρcos(φ-α) = ∑ -∞∞ imeim(φ-α)Jm(k⊥ρ), (3.31) e-ik⊥ρ'cos(φ'-α) = ∑ -∞∞ (-i)m'e-im'(φ'-α)Jm'(k⊥ρ'). (3.32) Perform the α-integration (just showing the relevant part) ∫ 02π dα 2π e-iα(m-m') = δmm' (3.33) so δ(x≥ - x≥') = ∫ dkz 2π eikz(z-z') ∫ dk⊥k⊥ 2π ∑ m=-∞∞ eim(φ-φ')Jm(k⊥ρ')Jm(k⊥ρ) (3.34) δ(x≥ - x≥') = δ(z-z') 1 ρ δ(ρ-ρ')δ(φ-φ'). (3.35) (The form of the last line uses the UVW δ(u-u')δ(v-v')δ(w-w') form I talked about in Chapter 1.) Since we can easily identify the ∂(z-z') part above, we must have 1 ρ δ(ρ-ρ')δ(φ-φ') = ∫ dk⊥k⊥ 2π ∑ m=-∞∞ eim(φ-φ')Jm(k⊥ρ)Jm(k⊥ρ'). (3.36) (In an equation like this where k ⊥ is just an integration variable that has lost its geometric meaning, I probably should just put k ⊥ . k.) Now mult. by e-im'(φ-φ') and integrate over φ: (m' . m) ∫ 0∞ dkkJm(kρ)Jm(kρ') = 1 ρ δ(ρ-ρ') ("completeness"). (3.37) 3.7 (Notice rhs is indep. of m; true for each m.) It now follows that ∑ m=-∞∞ 1 2π eim(φ-φ') = δ(φ-φ') (more completeness). (3.38) Completeness in ρ implies (t=k ρ) F(t) = ∫ 0∞ du uJ m(ut) f(u), (3.39) => ∫ 0∞ dt tF(t)J m(tv) = f(v). (3.40) Similarly for φ. These results are useful only if the full range of φ or ρ values is allowed in some problem. 3.3 Zeros and orthogonality properties of Bessel functions The dimensionful eqn satisfied by J m(kρ): [d2 dρ2 + 1 ρ d dρ - m2 ρ2 + k2] Jm(kρ) = 0, (3.41) => 1 ρ d dρ(ρ dJm dρ) + (k2 - m2 ρ2)Jm = 0. (3.42) We have that -∫ 0a dρρJm(k'ρ)1 ρ d dρ (ρ dJm(kρ) dρ) = ∫ 0a dρρ(k2 - m2 ρ2)Jm(k'ρ)Jm(kρ). (3.43) On the other hand (k<=>k') -∫ 0a dρρJm(kρ)1 ρ d dρ(ρ dJm(k'ρ) dρ) = ∫ 0a dρρ(k'2 - m2 ρ2)Jm(kρ)Jm(k'ρ). (3.44) Integrate by parts on lhs of (3.44): ∫ 0a dρJm(kρ)d dρ(ρ dJm(k'ρ) dρ) 3.8 = ρJm(kρ) dJm(k'ρ) dρ|a 0 - ∫ 0a dρρ dJm(kρ) dρ dJm(k'ρ) dρ. (3.45) The combination ρJm(kρ) dJm(k'ρ) dρ vanishes at ρ=0. Can see this from our earlier result: Jm(kρ) ____. kρ<<1 (kρ)m 2mm!, (3.46) => ρJm(kρ) dJm(k'ρ) dρ ____. k'ρ<<1 m(kk')m (2mm!)2 ρ2m. (3.47) (Actually vanishes like ρ2 for m=0 also) As for the other limit, we simply impose Jm(ka) = 0. (3.48) Call k mna ≠ xmn => J m(xmn) = 0, n = 1,2,3... . (3.49) There are an infinite number of such roots for each m value. Such a condition, we will see, is appropriate if we arebuilding a function w/Dirichlet B.C. However, we can alsoimpose d dρ Jm(kρ) |ρ=a = 0. (3.50) Define k mna = ymn => d dx Jm(x)|x=ymn = 0. (3.51) Therefore, (3.43) & (3.44) become (for either set of roots) (3.43 3.44)=> ∫ 0a dρρ dJm(kmnρ) dρ dJm(kmn'ρ) dρ 3.9 = ∫ 0a dρρ ((k2mn k2mn')- m2 ρ2) Jm(kmnρ)Jm(kmn'ρ). (3.52) Subtracting yields (k2mn - k2mn')∫ 0a dρρJm(kmnρ)Jm(kmn'ρ) = 0, (3.53) => ∫ 0a dρρ Jm(kmnρ)Jm(kmn'ρ) = 0 , n ≠n'. (3.54) Orthogonality for n ≠n' (for a given m). One can also show that ∫ 0a dρρ [Jm(kmnρ)]2 = a2 2 [Jm+1(xmn)]2 (3.55) when we are dealing with zeros of Bessel functions, and ∫ 0a dρρ (Jm(kmnρ))2 = a2 2 (1 - m2 y2mn)(Jm(ymn))2(3.56) for zeros of derivatives of Bessel functions. Material on how one obtains these integrals now follows. We are dealing here with the Dirichlet case, (3.55), only.Eq.(3.56) is derived in an analagous fashion. [d 2 dρ2 + 1 ρ d dρ - m2 ρ2 + k2] Jm(kρ) = 0, (3.57) (kmn = xmn a) => ∂J"m ∂a + 1 ρ ∂J'm ∂a + [- m2 ρ2 + k2] ∂Jm ∂a + ∂k2 ∂a Jm = 0, (3.58) => ρ ∂J"m ∂a Jm + ∂J'm ∂a Jm + [- m2 ρ2 + k2] ρ∂Jm ∂a Jm + ρ∂k2 ∂aJm2 = 0. (3.59) (J'm(x) ≡ dJm(x) dx, etc.) Multiply (3.57) by ρ ∂Jm ∂a : 3.10 ρ J"m ∂Jm ∂a + J'm ∂Jm ∂a + [- m2 ρ2 + k2] ρJm∂Jm ∂a = 0. (3.60) Subtract (3.59) - (3.60): ρ ∂J"m ∂a Jm- ρ J"m ∂Jm ∂a + ∂J'm ∂a Jm - J'm ∂Jm ∂a = - ∂k2 ∂a ρ Jm2. (3.61) However d dρ [ρ∂J'm ∂a Jm - ρJ'm ∂Jm ∂a] = LHS of (3.61) = - ∂k2 ∂a ρ Jm2. (3.62) Can now integrate: ρ[∂J'm ∂a Jm - J'm ∂Jm ∂a]|a 0 = - ∂k2 ∂a ∫ 0a dρρ Jm2. (3.63) As expressed before: (m=0 separate) ρ Jm(kρ) dJm(kρ) dρ ____. ρ .0 m(kk')m (2mm!)2 ρ2m. (3.64) (Derivatives wrt "a" don't change the power of ρ & therefore don't change ρ .0 limit.) Also (Dirichlet case) Jm(kmma) = 0, (3.65) xmn => ρ ∂Jm ∂ρ ∂Jm ∂a |ρ=a = ∂k2 ∂a ∫ 0a dρρ Jm2, (3.66) ∂Jm ∂ρ = dJm(t) dt ∂t ∂ρ = dJm dt (x a), (3.67) (t = kρ = xρ a) ∂Jm ∂a|ρ=a = dJm(t) dt ∂t ∂a|ρ=a = - xρ a2 dJm dt |ρ=a = - x a dJm dt |t=xmn. (3.68) 3.11 ∂k2 ∂a = ∂ ∂a ⎝⎜⎛ ⎠⎟⎞x2 a2 = - 2x2 a3, (3.69) => a dJm dt x a ⎝⎜⎛ ⎠⎟⎞-x a dJm dt = - 2x2 a3 ∫ oa dρρ Jm2, (3.70) => ∫ 0a dρρ Jm2(kmnρ) = a2 2 (d dt Jm)2|t=xmn. (3.71) However 2 d dt Jm = Jm-1 - Jm+1 (2 m t Jm = Jm-1 + Jm+1), (3.72) => d dt Jm = - J m+1 (at J m = 0), (3.73) => ∫ 0a dρρ Jm2(kmnρ) = a2 2(Jm+1(xmn))2. (3.74) Assuming completeness, we may expand f(ρ) = ∑ n=1∞ Amn Jm(kmnρ), (3.75) for any m. Using orthonormality and the above integral, we may now invert this for A mn: Amn = 2 a2 Jm2(xmn) ∫ 0a dρ'ρ' Jm(xmn aρ')f(ρ'). (3.76) Plugging A mn in above, what this says is f(ρ) = ∫ 0a dρ'ρ' f(ρ') ∑ n=1∞ 2 a2 Jm+12(xmn) Jm(kmnρ)Jm(kmnρ), (3.77) => ∑ n=1∞ 2 a2 Jm+12(xmn) Jm(xmnρ' a) Jm(xmnρ a) = 1 ρ ∂(ρ-ρ'). (3.78) Similarly, we find 3.12 ∑ n=1∞ 2 a2 (1- m2 ymn2) Jm(ymnρ' a) Jm(ymnρ a) = 1 ρ ∂(ρ-ρ'). (3.79) To save some wear and tear, define J1m(kmnρ) ≡ √⎯2 a Jm(kmnρ) Jm+1(xmn), (3.80) J2m(kmnρ) ≡ √⎯2 a√⎯⎯⎯1 - m2 ymn2 Jm(kmnρ) Jm(ymn). (3.81) Then the above reads (for each m) ∫ 0a dρρ J1,2m(kmnρ)J1,2m(kmn'ρ) = ∂nn', (3.82) or ∑ n=1∞ J1m(kmnρ')J1m(kmnρ) = 1 ρ ∂(ρ-ρ'), (3.83) ∑ n=1∞ J2m(kmnρ')J2m(kmnρ) = 1 ρ ∂(ρ-ρ'). (3.84) 3.4 Reduced Green function for the conducting cylinder First problem we will consier is the conducting cylinder: L y xzV(ρ,φ) 3.13 Will do this interior problem as a Green function problem first, then as a B.V. problem. We have a Dirichlet problem with restricted values of ρ,z. Therefore, choose 4π δ(x-x≥') = 4π ρ δ(ρ-ρ') δ(φ-φ') δ(z-z') = 4π∑ n=1∞ ∑ m=-∞∞ J1m(kmnρ')J1m(kmnρ) eim(φ-φ') 2π δ(z-z'). (3.85) Similarly, assume GD(x≥,x'≥) = 4π∑ n=1∞ ∑ m=-∞∞ J1m(kmnρ')J1m(kmnρ) eim(φ-φ') 2π gmn(z,z'), (3.86) Using (from Eq.(3.42)), 1 ρd dρ (ρdJm dρ) = (m2 ρ2-k2)Jm , (3.87) we have => -∇2GD(x≥,x≥') = 4π∑ n=1∞ ∑ m=-∞∞ J1mJ1m eim(φ-φ') 2π[ ⎝⎜⎛ ⎠⎟⎞-m2 ρ2+k2mn+m2 ρ2 -∂2 ∂z2]gmn(z,z). (3.88) Therefore, we must solve [xmn2 a2 - ∂2 ∂z2]gmn(z,z') = ∂(z-z'), (3.89) subject to gmn|z=0,L = 0. (3.90) We have done exactly this problem before for the parallel plates (k .kmn, a .L): gmn(z,z') = sinh(k mnz<)sinh(k mn(L-z>)) kmnsinh(k mnL). (3.91) 3.14 Full solution: GD(x≥,x≥') = 2∑ n=1∞ ∑ m=-∞∞ eim(φ-φ')J1m(kmnρ')J1m(kmnρ) x sinh(k mnz<)sinh(k mn(L-z>)) kmnsinh(k mnL). (3.92) Now get the solution of our BV problem: ∂ ∂z' gmn(z,z')|z'=L = - sinh(k mnz) sinh(k mnL). (3.93) Potential is Φ(x≥) = ∑ n,m Cmneimφ J1m(kmnρ)sinh(k mnz), (3.94) where Cmn = 1 2π 1 sinh(k mnL)∫ 0a dρ'ρ'∫ 02π dφ'e-imφ'J1m(kmnρ')V(ρ',φ'). (3.95) 3.5 The cylinder as a boundary value problem Can also get the above result directly from BV point of view using separation of variables. (Did this before in 2-D.)Laplace eq n: ∂2Φ ∂ρ2 + 1 ρ ∂Φ ∂ρ + 1 ρ2 ∂2Φ ∂φ2 + ∂2Φ ∂z2 = 0. (3.96) Assume Φ(ρ,φ,z) = P( ρ)Q(φ)Z(z). (3.97) Leads to d2Z dz2 - k2Z = 0, (3.98) 3.15 d2Q dφ2 + ν2Q = 0, (3.99) d2P dρ2 + 1 ρ dP dρ + (k2 - ν2 ρ2) P = 0. (3.100) Solutions to the first two are Z(z) ~ e¶kz, (3.101) Q(φ) ~ e¶iνφ. (3.102) If the full azimuthal range is allowed for φ => ν = integer. ν ≡ m. Other eqn is just dimensionful form of Bessel's eqn. Solns: P(ρ) ~ Jm(kρ), Nm(kρ) where (m integer;"Neumann function") Nm(x) ≡ lim ν .m [Jν(x)cosνπ - J-ν(x) sinνπ]. (3.103) (Jν and J -ν are linearly indep. if ν ≠ m. When ν = m we know J -m = (-1)mJm; however, the above defn of Nm does give a second linearly indep. soln. Form is indeterminate if we just plug in. Must use l'Hospital's rule in order ν. Explicit form of the Nm(x) is somewhat sickening.) Remind ourselves of the BC's: Φ|ρ=a = 0 => J m(ka) = 0 => k mn = xmn a, (3.104) Φ|z=0 = 0 => Z(0) = 0 => Z(z) ~ sinh(k mnz). (3.105) Most general form: Φ(ρ,φ,z) = ∑ n,m Cmneimφ J1m(kmnρ)sinh(k mnz). (3.106) 3.16 Last BC: V(ρ,φ) = ∑ n,m Cmneimφ J1m(kmnρ)sinh(k mnL). (3.107) Use our orthogonality formulas ∫ 02π dφ 2π eiφ (m-m') = ∂mm', (3.108) ∫ 0a dρρ J1m(kmnρ)J1m(kmn'ρ) = ∂nn'. (3.109) (Notice the last statement requires a single m value.) => Cmn = 1 2πsinh(k mnL) ∫ 0a dρ'ρ'∫ 02π dφe-imφJ1m(kmnρ')V(ρ',φ'), (3.110) as before. This way is quicker, but less general. 3.6 Bessel functions of imaginary argument Bessel function eqn again: ( ν arbitrary; dimensionful form) [1 ρ d dρ (ρd dρ) - ν2 ρ2 + k2] (Jν(kρ) Nν(kρ)) = 0. (3.111) ÷ oscillating at large ρ This is if the separation constant, k, is real. If it is imaginary, we get [1 ρ d dρ (ρd dρ) - ν2 ρ2 - k2] (Iν(kρ) Kν(kρ)) = 0. (3.112) ÷ Exponential at large ρ "Bessel functions of imaginary argument": 3.17 ⎩⎪⎪⎨⎪⎪⎧Iν(x) ≡ i-ν Jν (ix), Kν(x) ≡ π 2iν+1 H(1) ν(ix). (3.113) (H(1) ν(x) ≡ Jν(x) + iN ν(x)) Situation is similar to d2x dt2 + ω02x = 0, (3.114) ω02 > 0 , x ~ sin t, cos t, ω02 < 0 , x ~ sinh t, cosh t. Unlike the J m, Nm, the I ν, Kν are not complete for ρ in the range [0+,∞] (just like sin t, cos t being complete for t 1- [-∞,∞], but sinh t, cosh t are not.) Also, can not build functions in a finite interval from them. In general, can use either form in a given problem, but in a different sense. Usually (solution in φ, e+-imφ, is oscillatory for 0< _φ 0<_2π): oscill. non-oscill. (reduced G.f.) Jm(kρ), Nm(kρ) e+-kz e+-ikz Km(kρ), Im(kρ) The Km(kρ) go to zero for ρ -> ∞, while I m(kρ) is regular at the origin (except for m=0). 3.7 Free charge solution using Bessel functions Let's do a simple problem to illustrate the use of these functions. We will also see the Wronskian here for the firsttime, which is a great labor saving technical device.Isolated charge in free space (use cyl. coord.'s). One way: ∇ 2G(x≥,x≥') = -4 πδ(x≥-x≥'), (3.115) 3.18 4πδ(x≥ - x≥') = 4π ρ δ(ρ - ρ')δ(φ - φ')δ(z - z'), = 2 ∫ 0∞ dkk ∑-∞∞ eim(φ-φ') Jm(kρ)Jm(kρ')δ(z - z'). (3.116) Assume G(x≥,x≥') = 2 ∫ 0∞ dkk ∑-∞∞ eim(φ-φ') Jm(kρ)Jm(k'ρ)gm(z,z'), (3.117) => -∇2G(x≥,x≥') = -2 ∫ 0∞ dkk ∑-∞∞ [1 ρ d dρ (ρd dρ) + 1 ρ2 ∂2 ∂φ2 + ∂2 ∂z2] x eim(φ-φ')Jm(kρ)Jm(k'ρ)gm(z,z'), (3.118) => [k2 - ∂2 ∂z2]gm(z,z') = ∂(z - z'). (3.119) Easy way to apply continuity: gm(z,z') = ⎩⎪⎨⎪⎧f(z')ψ1(z), z -< z' g(z')ψ2(z), z -> z'. (3.120) However, gm(z,z') = g m(z',z), (3.121) => g m(z,z') = ⎩⎨⎧ψ1(z)ψ2(z'), z -< z', ψ2(z)ψ1(z'), z -> z'. (3.122) or g m(z,z') = ψ1(z<)ψ2(z>). (3.123) We must now have that - dgm dz |z=z'+ + dgm dz |z=z'- = 1, (3.124) => -ψ'2(z')ψ1(z') + ψ'1(z')ψ2(z') = 1. (3.125) Define: W[ ψ1,ψ2] ≡ ψ1ψ'2 - ψ2ψ'1 "Wronskian". (3.126) 3.19 Explicitly, here ψ1(z') = C1ekz (finite as z .-∞), (3.127) ψ2(z') = C2e-kz (finite as z .+∞), (3.128) => 2kC1C2 = 1, C1C2 = 1 2k, (3.129) => gm(z,z') = 1 2k e-k(z> - z<). (3.130) Put the pieces together (prob. 3.14(b) ): «non-oscill . 1 |x≥-x≥'| = ∫ 0∞ dk ∑-∞∞ eim(φ-φ') Jm(kρ)Jm(kρ') e-k(z> - z<). (3.131) ÷oscillatory Another way: (sect. 3.11) δ(z-z') = 1 2π∫-∞∞ dkeik(z-z') = 1 π ∫ 0∞ dk cos[k(z-z')], (3.132) δ(φ-φ') = 1 2π ∑-∞∞ eim(φ-φ'), (3.133) 4πδ(x≥-x≥') = 4π ρ δ(ρ-ρ')δ(z-z')δ(φ-φ'), (3.134) 4πδ(x≥-x≥') = 1 2π2 ∫ 0∞ dk∑-∞∞ cos[k(z-z')]eim(φ-φ') 4π ρ δ(ρ-ρ'). (3.135) Assume G(x≥,x≥') = 1 2π2 ∫ 0∞ dk ∑-∞∞ cos[k(z-z')]eim(φ-φ') gm(ρ,ρ'),(3.136) => 1 ρ d dρ (ρ dgm dρ) - (k2 + m2 ρ2)gm = - 4π ρ δ(ρ-ρ'). (3.137) Solutions: I m(kρ), Km(kρ) gm(ρ,ρ') = ψ1(kρ<)ψ2(kρ>). (3.138) 3.20 BC's: gm(ρ,ρ')|ρ=0 = finite => ψ1(kρ) ~ Im(kρ), (3.139) gm(ρ,ρ')|ρ .∞ = 0 => ψ2(kρ) ~ Km(kρ), (3.140) => g m(ρ,ρ') = A I m(kρ<)Km(kρ>). (3.141) ÷unknown Build in discontinuity: - dgm dρ|+ + dgm dρ|- = 4π ρ' . (3.142) => A[-K' m(kρ')Im(kρ') + K m(kρ')I'm(kρ')] = 4π ρ'. (3.143) W[Im(kρ'),Km(kρ')] -k Can show (from known forms for I m,Km for x<<1 or x>>1): W[I m(x), K m(x)] = - 1 x, (3.144) => A = 4 π. (3.145) => 1 |x≥-x≥'| = 2 π∫ 0∞ dk ∑-∞∞ eim(φ-φ') Im(kρ<)Km(kρ>)cos[k(z-z')]. (3.146) With the use of the Wronskian, continuity is automatic, and one can compute the necessary scale factor as in (3.141)by comparison of the discontinuity condition with theexplicitly computed Wronskian. In summary, if one has g m(z,z') = C ψ1(z<)ψ2(z>), for ψ1(z) and ψ2(z) known solutions, one rewrites the boundary condition as 3.21 C(ψ1(z')ψ'2(z') - ψ'1(z') ψ2(z')) = D(z'), => C = D(z') W[ψ1,ψ2]. Explicitly evaluating the Wronskian then gives the constant C. 3.8 Reduced Green function/image method: conducting wedge Not all cyl. sym. problems need involve Bessel functions. Wedge problem (for a .∞ becomes Green function for wedge problem in Ch.2): a A 2D problem. Need to solve: ∇2G(x≥,x≥') = -4 π 1 ρ δ(ρ-ρ') δ(φ-φ'). (3.147) Can not assume δ(φ-φ') = 1 2π ∑-∞∞ eim(φ-φ'), (3.148) because φ,φ' have a restricted range. Solution is to imagine: Φ=0 automaticallyβ β+1 -1 3.22 A combined image/reduced G.f. method again (like box problem), except here we are working in φ rather than x. So solve ∇2G(x≥,x≥') = -4 π 1 ρ δ(ρ-ρ')[δ(φ-φ') - δ(φ+φ')], (3.149) in the larger region. What happens, however, when β > π ? We are forcing a periodicity on the problem which can not be realized physically. This is OK, as long as we realize 0 < φ < β for physics. Think of the image extension as a sort of Riemann sheet, the "cut" for which is at φ = 0. The sheets may lay on top of one another (if β > π), that is, we are allowing the potential to become a mult. valued function (in non-physical regions) in order to force a symmetry on theproblem. What all this means is that we take (a .β in earlier expression) 1 2 [δ(φ-φ') - δ(φ+φ')] = 1 β ∑ n=1∞ sin(nπφ β)sin(nπφ' β), (3.150) for -β < φ <β instead of the original range for φ,φ'. Also assume G(x≥,x≥') = 8π β ∑ n=1∞ sin(nπφ β)sin(nπφ' β)gn(ρ,ρ'), (3.151) => -∇2G(x≥,x≥') = 8π β ∑ n=1∞ sin(nπφ β)sin(nπφ' β) x [-1 ρ d dρ (ρ d dρ) + n2π2 ρ2β2]gn(ρ,ρ'), (3.152) => [1 ρ d dρ (ρ d dρ) - n2π2 ρ2β2]gn = -1 ρ δ(ρ-ρ'). (3.153) I guess this can still be considered Bessel's eqn but for k=0. We know that (see for example Jackson, Section 3.7) lim k .0 Jγ(kρ) ~ (kρ)γ , (3.154) 3.23 lim k .0 Nγ(kρ) ~ (kρ)-γ (γ ≠ 0), (3.155) which suggests: ( γ ≡ nπ β) gn ~ ργ,ρ-γ . (3.156) We let ψ1(ρ') = ρ'γ, (3.157) ψ2(ρ') =(ρ'γ - a2γ ρ'γ). (3.158) Then gn(ρ,ρ') = C ψ1(ρ<)ψ2(ρ>), (3.159) One has that - dgn dρ|+ + dgn dρ|- = 1 ρ', (3.160) gives -C W[ ψ1(ρ'),ψ2(ρ')] = 1 ρ' . (3.161) Explicitly W[ψ1(ρ'),ψ2(ρ')] = γρ'γ(ρ'γ-1 + a2γ ρ'γ+1 )-γ(ρ'γ - a2γ ρ'γ )ρ'γ-1 = 2γa2γ ρ', (3.162) which means that C = - 1 2γa2γ , (3.163) => g n(ρ,ρ') = 1 2γ ργ<(-ργ> a2γ + 1 ργ>), (3.164) 3.24 => G(x≥,x≥') = ∑ n=1∞ 4 n ργ<(1 ργ> - ργ> a2γ )sin(nπφ β)sin(nπφ' β). (3.165) Of course, we have that G(x≥,x≥') = ∑ n=1∞ 4 n (ρ< ρ>)γ sin(nπφ β)sin(nπφ' β), (3.166) in the a .∞ limit, which gives the Green function for the wedge problem of Ch.2. Using this form near the origin, we have G(x≥,x≥') ~ 4(ρ ρ')π/β sin(πφ β) sin(πφ' β). (3.167) This is of the same form as in Ch.2: Φ(ρ,φ) = V + a 1ρπ/β sin(πφ β). (3.168) ÷can evaluate by comparison 3.9 Construction of spherical harmonics: Schwinger technique We've beat cylindrical symmetry to death. Next: spherical symmetry. Start off like cylindrical case. Goteverything from e ik≥.x≥ = eikzzeik⊥ρcos(φ-α). (3.169) Deceptively simple structure. Try the same thing here. Write eik≥.r≥ = ei(a≥.r≥ + b≥.r≥), (3.170) (notice the r≥ instead of x≥ here) where k≥ = a≥ + b≥, (3.171) =>k≥2 = a≥2 + b≥2 + 2a≥.b≥. (3.172) 3.25 However, require a≥2 = b≥2 = 0. (3.173) a≥, b≥ now necessarily complex. Also require b≥ = a≥*, (3.174) => 2a≥.b≥ = 2|a≥|2 = k≥2. (3.175) Can we really do this? Count parameters: a≥ + b≥ = 2Re(a≥) = k≥, (3.176) ⎭⎪⎬⎪⎫ a1 = 1 2 kx + iC1 a2 = 1 2 ky + iC2 a3 = 1 2 kz + iC3 C1,2,3 real (3.177) There are 3 real parameters, but a≥2 = 0 represents 2 conditions. => 1 arbitrary parameter allowed. Still leaves onefree to choose. Let C 3 = 0. (3.178) Gives an explicit realization of a≥: (verify yourself) a≥ = (1 2(kx - ik k⊥ ky), 1 2(ky + ik k⊥ kx),kz 2), (3.179) where k ⊥2 = kx2 + ky2, k2 = kz2 + k⊥2. Now notice that (∂2 ∂x2 + ∂2 ∂y2 + ∂2 ∂z2)eia≥.r≥ = -a≥2eia≥.r≥ = 0, (3.180) => eia≥.r≥ is a soln of Laplace's eqn. Since 3.26 eia≥.r≥ = ∑ l=0∞ (ia≥.r≥)l l!(3.181) each monomial (a≥.r≥)l is also a soln to this eqn. Let's let a 1 + ia2 ≡ 2ξ-2, (3.182) a 1 - ia2 ≡ -2ξ+2, (3.183) a 3 = 2ξ+ξ-. (3.184) ξ+,- two arbitrary complex nos. 6 equations but there are 2 conditions on a≥ from a≥2 = 0. Indeed a≥2 = (a1 + ia2)(a1 - ia2) + a32 = -4 ξ-2ξ+2 + 4ξ+2ξ-2 = 0, (3.185) is automatic. Concentrate on (a≥.r≥)l : a≥.r≥ r = a1x r + a2y r + a3z r = a 1sinθ(eiφ+e-iφ 2) + a2sinθ(eiφ-e-iφ 2i) + a3cosθ = (a1 - ia2) 2 sinθeiφ + (a1 + ia2) 2 sinθe-iφ + a3cosθ = - ξ+2sinθeiφ + ξ-2sinθe-iφ + 2ξ+ξ-cosθ. (3.186) Therefore (a≥.r≥)l = rl(a≥.r≥ r)l = rl(ξ+2eiφ sinθ)l [(ξ- ξ+ sinθe-iφ)2 + 2ξ- ξ+ sinθe-iφcosθ - sin2θ]l = rl(ξ+2eiφ sinθ)l [(ξ- ξ+ sinθe-iφ + cosθ)2 - 1]l. (3.187) Do a Taylor series expansion in ξ ≡ ξ- ξ+ . Reminder: 3.27 f(ξ) = ∑ n=0∞ ξn n! (d dξ)nf(ξ)|ξ=0. (3.188) In this case, the highest power = 2 l. So [(ξsinθe-iφ + cosθ)2 - 1]l = ∑ n=02 l ξn n! (d dξ)n [(ξsinθe-iφ + cosθ)2 - 1]l|ξ=0. (3.189) Define cosθ~ = ξ sinθe-iφ + cosθ, (3.190) => ξ = eiφ sinθ (cosθ~ - cosθ), (3.191) => dξ = eiφ sinθ dcosθ~. (3.192) Therefore (d dξ )n [(ξsinθe-iφ + cosθ)2 - 1]l|ξ=0 = (sinθ eiφ d dcosθ~)n [cos2θ~ - 1]l|ξ=0 = (sinθe-iφ)n ⎝⎜⎛ ⎠⎟⎞d dcosθn (cos2θ - 1)l. (3.193) So, we have [(ξsinθe-iφ + cosθ)2 - 1]l = ∑ n=02 l (ξsinθe-iφ)n n! (d dcosθ)n (cos2θ - 1)l. (3.194) Now define m ≡ l-n, - l≤ m ≤ l (2 l+ 1 values) and put it all together: 3.28 (a≥.r≥)l = rl(ξ+2eiφ sinθ)l ∑ m=- ll ((ξ-/ξ+)sinθe-iφ)l-m ( l-m)! (d dcosθ)l-m (cos2θ - 1)l (3.195) or (a≥.r≥)l = rl2l l! ∑ m=- ll ξ+l+m ξ-l-m √⎯⎯⎯⎯⎯⎯⎯⎯⎯( l+m)!( l-m)! √⎯⎯( l+m)! ( l-m)! (sinθ)-meimφ x (d dcosθ)l-m (cos2θ - 1)l 2l l!. (3.196) Define spherical harmonics: Y lm(θ,φ) ≡ √⎯⎯⎯⎯(2 l+1)( l+m)! 4π( l-m)! eimφ (sinθ)-m (d dcosθ)l-m (cos2θ - 1)l 2l l!. (3.197) so (a≥.r≥)l = rl2l l! ∑ m=- ll ξ+l+m ξ-l-m √⎯⎯⎯⎯⎯⎯⎯⎯( l+m)!( l-m)! √⎯⎯4π 2 l+1 Y lm(θ,φ), => eia≥.r≥ = ∑ l=0∞ ∑ m=- ll ilrl2l ξ+l+m ξ-l-m √⎯⎯⎯⎯⎯⎯⎯⎯( l+m)!( l-m)! √⎯⎯4π 2 l+1 Y lm. (3.198) Since eia≥.r≥ satisfies Laplace eqn: (coeff's of ξ+l+mξ-l-m vanish) => ∇2 [rlY lm] = 0, (3.199) ∇2 = 1 r2 ∂ ∂r (r2 ∂ ∂r )+ 1 r2sinθ ∂ ∂θ (sinθ∂ ∂θ) + 1 r2sin2θ ∂2 ∂φ2. (3.200) Now 3.29 1 r2 ∂ ∂r (r2 ∂rl ∂r ) = l r2 ∂ ∂r rl+1 = l( l+1)rl-2, (3.201) =>[ l( l+1)rl-2 Y lm + rl-2(1 sinθ ∂ ∂θ (sinθ ∂Y lm ∂θ) + 1 sin2θ ∂2 ∂φ2 Y lm)]=0. (3.202) => [(1 sinθ ∂ ∂θ (sinθ ∂Y lm ∂θ) + 1 sin2θ ∂2Y lm ∂φ2]= - l( l+1)Y lm. (3.203) Notice also ∂2 ∂φ2 Y lm = -m2 Y lm, => [1 sinθ ∂ ∂θ (sinθ ∂ ∂θY lm) - m2 sin2θ Y lm]= - l( l+1)Y lm. (3.204) Y lm is the solution of the separated Laplace eqn in spherical coordinates. There is an alternate form for Y lm. Remember (a≥.r≥)l = rl [-ξ+2sinθ eiφ + ξ-2sinθe-iφ + 2ξ+ξ-cosθ]l. (3.205) It is unchanged under: ξ+ ↔ ξ-, θ . -θ , φ . -φ. Implies that ∑ m=- ll ξ+l+mξ-l-m √⎯⎯⎯⎯⎯⎯⎯⎯( l+m)!( l-m)! √⎯⎯4π 2 l+1 Y lm(θ,φ) = ∑ m'=- ll ξ-l+m'ξ+l-m' '√⎯⎯⎯⎯⎯⎯⎯⎯⎯( l+m')!( l-m')! √⎯⎯4π 2 l+1 Y lm'(-θ,-φ). (3.206) Coeff's of like powers must be the same: => Y lm(θ,φ) = Y l-m(-θ,-φ). (3.207) So 3.30 Y lm(θ,φ) = √⎯⎯2 l+1 4π √⎯⎯( l-m)! ( l+m)! eimφ (-sinθ)m (d dcosθ)l+m (cos2θ-1)l 2l l!. (3.208) Usual defn for "Legendre polyns": P l(x) = 1 2l l! (d dx)l (x2-1)l, => Y lm(θ,φ) = √⎯⎯2 l+1 4π √⎯⎯( l-m)! ( l+m)! eimφ(-sinθ)m (d dcosθ)m P l(cosθ) (m ≥ 0) (3.209) => Y lm(θ,φ) = √⎯⎯2 l+1 4π √⎯⎯( l+m)! ( l-m)! eimφ(sinθ)-m (d dcosθ)-m P l(cosθ) (m ≤ 0) (3.210) Can only use the top form when m ≥ 0 and bottom when m ≤ 0. Combine them: Y lm(θ,φ) = (-1)(m+|m|)/2 √⎯⎯(2 l+1) 4π √⎯⎯⎯( l-|m|)! ( l+|m|)! eimφ x (sin θ)|m| (d dcosθ)|m| P l(cosθ). (3.211) From the last form we can read off that Y lm(θ,φ) = (-1)m Y l-m(θ,-φ). (3.212) But since Y lm*(θ,φ) = Y lm(θ,-φ), (3.213) this gives Y lm*(θ,φ) = (-1)m Y l-m(θ,φ). (3.214) 3.31 We also have that (get diff. eq n for P l(cosθ) by setting m=0 in Y lm eqn) Y l0(θ,φ) = √⎯⎯(2 l+1) 4π P l(cosθ). (3.215) 3.10 Orthogonality properties of spherical harmonics All this started with an exponential. Go back to it: ∫dΩ eik≥.r≥ = ∫dΩ eia≥*.r≥+a≥.r≥ = ∑ l1,l2=0∞ il1il2 ∫dΩ (a≥*.r≥)l1 l1! (a≥.r≥ )l2 l2! . (3.216) Use our expansion: ∫dΩ eik≥.r≥ = ∑ l1,l2=0 il1il2 ∑ m1=- l1l1 ∑ m2=- l2l2 rl1rl22l12l2 ξ+* l1+m1 ξ-* l1-m1 √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯( l1+m1)!( l1-m1)! x ξ+l2+m2 ξ-l2-m2 √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯( l2+m1)!( l2-m2)! √⎯⎯4π 2 l1+1 √⎯⎯4π 2 l2+1 ∫dΩ Y l1m1*(θ,φ)Y l2m2(θ,φ). (3.217) On the other hand ∫dΩeik≥.r≥ = 2π∫ 0π dθsinθeikrcosθ = 4π kr sin(kr). (3.218) We can write 4πsin(kr) kr = 4π kr ∑ l=0∞ (-1)l (kr)2 l+1 (2 l+1)! = 4π ∑ l=0∞ (-1)l (r2k2)l (2 l+1)!. (3.219) But k2≥ = 2a*≥.a≥, 3.32 => ∫dΩeik≥.r≥ = 4π ∑ l=0∞ (-1)l 2l(r2)l (2 l+1)! (a*≥.a≥)l. (3.232) But again(a*≥.a≥) l = 2l(ξ+*ξ+ + ξ-*ξ-)2 l = 2l ∑ m'=02 l 2 l! (2 l-m')!m'! (ξ+*ξ+)2 l-m'(ξ-*ξ-)m'. (3.221) Let m= l-m' (a*≥.a≥)l = 2l ∑ m=- ll 2 l! ( l+m)!( l-m)! (ξ+*ξ+)l+m(ξ-*ξ-)l-m, (3.222) so ∫dΩeik≥.r≥ = ∑ l=0∞ ∑ m=- ll 4π 2 l+1 (-1)l(2lrl)2 (ξ+*ξ+)l+m(ξ-*ξ-)l-m ( l+m)!( l-m)!. (3.223) Comparing (3.217) & (3.223), we see that in (3.223) the coeff. of the various ξ's gives us information about the integral in (3.217). Zero unless: ⎭⎪⎬⎪⎫ ξ+* : l1+m1 = l+m ξ-* : l1-m1 = l-m > l1= l,m1=m (3.224) ⎭⎪⎬⎪⎫ ξ+ : l2+m2 = l+m ξ- : l2-m2 = l-m > l2= l,m2=m (3.225) => only non-zero terms are when l1= l2= l, m1=m2=m. Therefore ∫dΩ Y l1m1*(θ,φ)Y l2m2(θ,φ) = ∂ l1l2∂m1m2. (3.226) 3.33 Since we know that Y l0 = √⎯⎯2 l+1 4π P l(cosθ), we also have (m1=m2=0) ∫ -1 1 d(cosθ) P l(cosθ)P l'(cosθ) = 2 2 l+1 ∂ ll'. (3.227) Now consider (cos γ=cosθcosθ'+ sinθsinθ'cos(φ-φ')) 1 |r≥-r≥'| = 1 √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯r2 + r'2 - 2rr'cos γ = 1 r √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯1 + (r' r)2 - 2(r' r)cosγ = 1 r' √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯1 + (r r')2 - 2(r r')cosγ. (3.228) Do a Taylor series expansion in x = r< r> : 1 |r≥-r≥'| = ∑ l=0∞ r<l r>l+1 1 l (d dx)l 1 √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯1 + x2 - 2xcos γ |x=0. (3.229) Define Q l(cosγ) ≡ 1 l (d dx)l 1 √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯1 + x2 - 2xcos γ |x=0. (3.230) Clear that Q l(cosγ) is a polynomial in cos γ of degree l. Can show Q l(cosγ) = P l(cosγ)! 3.11 The Coulomb expansion, completeness of spherical harmonics and the "Addition Theorem" Cushing argument ("Applied Analytical Mathematics for Physical Scientists", p.172): (| μ|,|ν| < 1) I(μ,ν) =∫ -1 1 dx √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯1 - 2μx + μ2 √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯1 - 2νx + ν2. (3.231) 3.34 Direct, but tedious (see for example Gradshteyn & Ryzhik, (2.261)): I(μ,ν) = 1 √⎯⎯μν ln (1 + √⎯⎯μν 1 − √⎯⎯μν) " = 2 ∑ l=0∞ (μν )l (2 l+1). (3.232) On the other hand 1 √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯1 - 2μx + μ2 = ∑ l'=0∞ μl'Q l'(x), (3.233) 1 √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯1 - 2νx + ν2 = ∑ l=0∞ νlQ l(x). (3.234) Comparison shows ∫ -1 1 dxQ l(x)Q l'(x) = 2 2 l+1 ∂ ll'. (3.235) Now Pn(x) = ∑ m=0n αnmQm(x), (3.236) since n,m are the order of the polynomials P n(x) and Q m(x). => ∫ -1 1 dxPn(x) Qr(x) = 2 2r+1 αnr , r=0,1,2,...,n . (3.237) However ∫ -1 1 dxPn(x)xm = 1 2nn! ∫ -1 1 dx dn[(-1+x2)n] dxn xm by parts = 0 , m<n . (3.238) 3.35 => αnr = 0 ¢ r < n, (3.239) => αnnQn(x) = P n(x). (3.240) Set x=1 on both sides: P l(1) = 1 . (3.241) On the other hand 1 |r≥-r≥'| = ∑ l=0∞ r<l r>l+1 Q l(cosγ). (3.242) If cosγ = 1 => 1 |r≥-r≥'| = 1 |r-r'|. (3.243) But 1 |r-r'| = ∑ l=0∞ r<l r>l+1 >Q l(1) = 1, (3.244) => P l(x) = Q l(x). (3.245) So "Coulomb expansion" 1 |r≥-r≥'| = ∑ l=0∞ r<l r>l+1 P l(cosγ). (3.246) Now we know that ∇2 1 |r≥-r≥'| = -4π ∂(r≥-r≥'). (3.247) Substituting (3.246) in (3.247) and integrating across r=r'gives 3.36 [-r2 ∂ ∂r ∑ l=0∞ r<l r>l+1 P l(cosγ)]|r'+ε r'-ε = 4πδ(cosθ-cosθ')δ(φ-φ'), (3.248) which results in ∑ l=0∞ 2 l+1 4π P l(cosγ) = δ(cosθ-cosθ')δ(φ-φ'). (3.249) This is actually a statement of completeness of the Y lm(θ,φ). To see why, consider the following. For r≥ ≠ r≥', 1 |r≥-r≥'| is a soln to Laplace's eqn. The only linearly indep. solns to Laplace's eqn of the form rlx something( θ,φ) are: rl Y lm(θ,φ) (2 l+1 of them). Likewise, the only linearly indep. solns of the form r- l-1x something( θ,φ) are: r- l-1 Y lm(θ,φ) (2 l+1 of them). Therefore, considered as functions of r, θ, φ we must have r<r': 1 |r≥-r≥'| = ∑ l,m rl r'- l-1Y lm(θ,φ)A lm , (3.250) r>r': 1 |r≥-r≥'| = ∑ l,m r'l r- l-1Y lm(θ,φ)B lm . (3.251) Continuity across r=r' demands A lm = B lm, giving the unified form 1 |r≥-r≥'| = ∑ l,m r<l r>l+1 A lm Y lm(θ,φ). (3.252) 3.37 Comparison with (3.246) just shows the P l(cos γ), considered as a function of the r, θ, φ, must be a linear combination of the form, P l(cos γ) = ∑ m A lm Y lm(θ,φ). (3.253) Substitution above in (3.249) gives ∑ l,m (2 l+1) 4π A lm Y lm(θ,φ) = δ(cosθ-cosθ')δ(φ-φ'). (3.254) Using orthonormality ∫dΩ Y lm(θ,φ)Y*l'm'(θ,φ) = δ ll'δmm', => A lm = 4π 2 l+1 Y*lm(θ',φ'). (3.255) Putting this in (3.254) gives us completeness for sph. harmonics: ∑ l,m Y lm(θ,φ)Y*lm(θ',φ') = δ(cosθ-cosθ')δ(φ-φ'), (3.256) => 1 |r≥-r≥'| = ∑ l,m r<l r>l+1 4π 2 l+1 Y lm(θ,φ)Y*lmθ',φ'). (3.257) In addition, (3.253) gives us the "addition theorem": P l(cos γ) = 4π 2 l+1 ∑ m Y lm(θ,φ)Y*lm(θ',φ'). (3.258) 3.12 Green function for concentric spheres Now do this problem: 3.38 ab concentric spheres Will get special results for a .0, b .∞ limits. We have 4πδ(r≥-r≥') = 4π∑ l,m Y lm*(θ',φ')Y lm(θ,φ) 1 r2 δ(r-r'), (3.259) G(r≥,r≥') = 4π∑ l,m Y lm*(θ',φ')Y lm(θ,φ) gl(r,r'), (3.260) => - §≥2 G(r≥,r≥') = 4π∑ l,m Y lm*(θ',φ') Y lm(θ,φ) ⎝⎜⎜⎛ ⎠⎟⎟⎞ - 1 r2 ∂ ∂r(r2∂gl ∂r) + l( l+1) r2 gl. (3.261) Eq≥n will be satisfied if -∂ ∂r (r2∂gl ∂r) + l( l+1)gl = δ(r-r'). (3.262) When r≠r' solutions are gl ~ rl, r- l-1 , r≠r'. Therefore gl = C(r<l- a2 l+1 r<l+1 ) (r l > - b2 l+1 r>l+1 ). (3.263) Since 3.39 -r2∂gl ∂r |+ _ = 1, (3.264) we have -Cr'2 [ ⎝⎜⎜⎛ ⎠⎟⎟⎞ lr'l-1+( l+1) b2 l+1 r'l+2 ⎝⎜⎜⎛ ⎠⎟⎟⎞ r'l -a2 l+1 r'l+1 - ⎝⎜⎜⎛ ⎠⎟⎟⎞ lr'l-1+( l+1)a2 l+1 r'l+2 ⎝⎜⎜⎛ ⎠⎟⎟⎞ r'l- b2 l+1 r'l+1] = 1, => -Cr'2 [a2 l+1 r'2 ()- l-( l+1) + b2 l+1 r'2 ()( l+1)+ l] = 1, (3.265) => C = 1 (2 l+1)[]a2 l+1 - b2 l+1. (3.266) => gl(r,r') = 1 (2 l+1) ⎣⎢⎢⎡ ⎦⎥⎥⎤1-(a b)2 l+1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ r<l- a2 l+1 r<l+1 ⎝⎜⎜⎜⎛ ⎠⎟⎟⎟⎞1 r>l+1 - r>l b2 l+1. (3.267) Examine some special cases: 1. a .0, b .∞. (Free Green’s function) - 1 |r≥-r≥'| = ∑ l,m r<l r>l+1 4π 2 l+1 Y lm*(θ',φ')Y lm(θ,φ). (3.268) Compare this with 1 |r≥-r≥'| = ∑ l r<l r>l+1 Pl(cos γ), (3.269) get (“addition theorem” again) 3.40 Pl(cos γ) = 4π 2 l+1 ∑ m Y lm*(θ',φ')Y lm(θ,φ). (3.270) 2.a .0, b finite (Interior sphere) G(r≥,r≥') - . ∑ l,m ⎝⎜⎜⎜⎛ ⎠⎟⎟⎟⎞r<l r>l+1 - rlr'l b2 l+1 4π 2 l+1Y lm*(θ',φ')Y lm(θ,φ). (3.271) Notice that rlr'l b2 l+1 = b r' rl ⎝⎜⎛ ⎠⎟⎞b2 r'l+1 = b r' rl r- bl+1 (r- b>r), (3.272) so we can perform the sums by the above result G(r≥,r≥') - . 1 |r≥-r≥'| - b r' 1 |r≥-r-≥ b|, (3.273) where r-≥ b = ⎝⎜⎛ ⎠⎟⎞ b2 r' ,θ',φ'. (3.274) locates the image charge. (outside sphere) 3. b .∞, a finite (outside sphere) G(r≥,r≥') - . ∑ l,m ⎝⎜⎜⎜⎛ ⎠⎟⎟⎟⎞r<l r>l+1 - a2 l+1 rl+1r'l+1 4π 2 l+1Y lm*(θ',φ')Y lm(θ,φ). (3.275) As before a2 l+1 rl+1r'l+1 = a r' ⎝⎜⎛ ⎠⎟⎞a2 r'l rl+1 = a r' r- al rl+1 (r- a< r), (3.276) 3.41 G(r≥,r≥') - . 1 |r≥-r≥'| - a r' 1 |r≥-r-≥ a|, (3.277) where r-≥ a = ⎝⎜⎛ ⎠⎟⎞ a2 r',θ',φ'. (3.278) is the image charge (inside sphere). We can see the surface delta function in the expansion, ∂G ∂n'|r'=a = - ∂G ∂r'|r'=a = -∑ l,m ⎝⎜⎜⎜⎛ ⎠⎟⎟⎟⎞ lal-1 rl+1 + ( l+1)a2 l+1 rl+1al+2 x 4π 2 l+1Y lm*(θ',φ')Y lm(θ,φ), (3.279) = -4 π∑ l,m al-1 rl+1Y lm*(θ',φ')Y lm(θ,φ), (3.280) => limr .a+ - 1 4π ∂G ∂n'|r'=a = a-2 δ(cosθ-cosθ')δ(φ-φ'). (3.281) We now see it explicitly (compare with Eq.(1.89)). 3.13 Conducting sphere in a uniform field Of course, can do B.V. problems in spherical coordinates as well. Do the same problem as in Ch.2: sphere in a uniformfield. Start: 1 r 2 ∂ ∂r ⎝⎜⎛ ⎠⎟⎞r2∂Φ ∂r + 1 r2sinθ ∂θ ⎝⎜⎛ ⎠⎟⎞sinθ∂Φ ∂θ + 1 r2sin2θ ∂2Φ ∂φ2 = 0. (3.282) Assume Φ = R(r) P(θ) Q(φ). (3.283) Get 3.42 PQ d dr ⎝⎜⎛ ⎠⎟⎞r2dR dr + RQ sinθ d dθ ⎝⎜⎛ ⎠⎟⎞sinθ dP dθ + RP sin2θ d2Q dφ2 = 0. (3.284) Multiply by sin2θ/Φ: sin2θ R ⎣⎢⎢⎡ ⎦⎥⎥⎤d dr ⎝⎜⎛ ⎠⎟⎞r2dR dr + sinθ P d dθ ⎝⎜⎛ ⎠⎟⎞sinθ dP dθ + 1 Q d2Q dφ2 = 0, (3.285) => 1 Q d2Q dφ2 = -m2, (3.286) => Q = e+-imφ. (3.287) If full range 0 ≤ φ ≤ 2π is allowed, must have m=0, +-1,+-2,.... Now have: 1 R d dr ⎝⎜⎛ ⎠⎟⎞r2dR dr + 1 Psinθ d dθ ⎝⎜⎛ ⎠⎟⎞sinθ dP dθ - m2 sin2θ = 0 (3.288) r only θonly => 1 Psinθ d dθ ⎝⎜⎛ ⎠⎟⎞sinθ dP dθ - m2 sin2θ = - l( l+1) or 1 sinθ d dθ ⎝⎜⎛ ⎠⎟⎞sinθ dP dθ + ⎣⎢⎢⎡ ⎦⎥⎥⎤l( l+1) - m2 sin2θ P = 0, (3.289) and 1 R d dr ⎝⎜⎛ ⎠⎟⎞r2dR dr = l( l+1), => d dr ⎝⎜⎛ ⎠⎟⎞r2dR dr - l( l+1)R = 0. (3.290) Solutions: 3.43 "Legendre funct. of the 2ndkind"; infinite at end points, θπ.=0, P ~ Pm l(cosθ), Qm l(cosθ) R(r) ~ rl,r- l-1. Both with l = 0,1,2,... Simple problem: a sphere+z General solution: Φ(r,θ) = ∑ l=0∞ AlrlPl(cosθ) + ∑ l=0∞ Blr- l-1Pl(cosθ). (3.291) B.C.: uniform electric field in +z direction of magnitude E0. No φ dep.=> m=0 only B.C.: 1. Φ(r,θ)|r .∞ = -E0z, (3.292) 2. Φ(r,θ)|r=a = V0, (V0=0 if grounded) (3.293) P0(cosθ) = 1, P1(cosθ) = cos θ, 1. => A1 = -E0 , Al = 0, l >1. Charge on sphere: 3.44 Q = ∫ dΩσ =- 1 4π o ∫ ∂Φ ∂r|r=adΩ = ( l +1) 4π ∑ l=0∞ Bla- l-2∫ dΩ Pl 4πδ l0 => B0 = Q. 2. => A0 - E0aP1(cosθ) + ∑ l=0∞ Bla- l-1Pl(cosθ) = V0, so A0 = V0 - Q a, l=0, B1 = E0a3, l=1, Bl = 0, l >1. Determines Φ(r,θ) = V 0 - Q a + Q r - E0 ⎝⎜⎛ ⎠⎟⎞r -a3 r2 cosθ. (3.294) Notice that V 0 and Q are independent parameters here. A grounded sphere (V 0 = 0) in this case does not necessarily have zero charge because the potential is not required tovanish at infinity. 3.14 Method of last resort: eigenfunction expansions Method of last resort: Eigenfunction expansion. Consider a more general eqn than we usually solve. §≥2ψ(x) + k2ψ(x) = 0. (3.295) (# of dims. is arbitrary.) Given a certain confined geometry, allows only discrete values of k2: §≥2ψn(x) + k2 nψn(x) = 0. (3.296) 3.45 Normalized as ∫ V ψ*n ψm d3x = δnm. (3.297) Assume these are complete. Thus G(x≥,x≥') = ∑ n an ψ*n(x')ψn(x), (3.298) where we have enforced the symmetry G(x≥,x≥') = G(x≥',x≥) as usual. G(x≥,x≥') solves §≥2G(x≥,x≥') = -4 π δ(x≥-x≥'), (3.299) so -∑ n an ψ*n(x≥')ψn(x≥)k2 n = -4π δ(x≥-x≥'). (3.300) Mult. by ψ*m and integrate over V: an ψ*n(x≥')kn2 = 4πψ*n(x≥'), => an = 4π kn2, (3.301) => G(x≥,x≥') = 4π∑ n ψ*n(x≥')ψn(x≥) kn2. (3.302) Trivial example. Solve §≥2ψ(x) + k2ψ(x) = 0, (3.303) in a 2D box with ψ|s = 0. (Will give us the Dirichlet G.F.) Geometry: 3.46 aby x Solve ⎣⎢⎢⎡ ⎦⎥⎥⎤ ∂2 ∂x2 + ∂2 ∂y2 ψ + k2ψ = 0. (3.304) The B.C. are ψ()0a,y = 0, (3.305) ψ()x,0b = 0. (3.306) Obviously separable. Get ( ψ(x,y)=ψx(x)ψy(y)) 1 ψx d2ψx dx2 + 1 ψy d2ψy dy2 = -k2, (3.307) => d2ψx dx2 = -kx2ψx, (3.308) d2ψy dy2 = -ky2ψy. (3.309) where kx2 + ky2 = k2. (3.310) Look at the x-direction, ψ(x) = C1sin kxx + C2cos kxx, B.C.: 0 = C1sin kxa, 3.47 => kx a = lπ , l=1,2,3,... Similarly ky b = mπ , m=1,2,3,... Correctly normalized eigenfunction is ψnm(x≥) = √⎯4 ab sin ⎝⎜⎜⎛ ⎠⎟⎟⎞ lπx a sin ⎝⎜⎛ ⎠⎟⎞mπy b. (3.311) Plug in: G(x≥,x≥') = 16 πab ∑ l,m=1∞ sin ⎝⎜⎜⎛ ⎠⎟⎟⎞ lπx a sin ⎝⎜⎜⎛ ⎠⎟⎟⎞ lπx' a sin ⎝⎜⎛ ⎠⎟⎞mπy b sin ⎝⎜⎛ ⎠⎟⎞mπy' b ⎩⎪⎨⎪⎧ ⎭⎪⎬⎪⎫ l2 a2 + m2 b2 (3.312) A double infinite series! (i.e., almost useless, except for theorists!) One more problem: a cylinder of radius = a Want to solve 1 ρ d dρ ⎝⎜⎛ ⎠⎟⎞ρ∂ψ ∂ρ + 1 ρ2 ∂2ψ ∂φ2 + k2ψ = 0. (3.313) The B.C. is ( ψ=ψ(ρ,φ)) 3.48 ψ(a,φ) = 0. (3.314) (Also that ψ(ρ,φ) is single-valued in φ.) Assume separable, get (ψ=ψρψφ ; mult. by ρ2 ψ) ρ2 ⎣⎢⎢⎡ ⎦⎥⎥⎤ 1 ψρρ d dρ ⎝⎜⎛ ⎠⎟⎞ρdψρ dρ + k2 + 1 ψφ d2ψφ dφ2 = 0, (3.315) -m2 => d2ψφ dφ2 = -m2ψφ , ψφ = 1 √⎯⎯2π eimφ , Also => 1 ρ d dρ ⎝⎜⎛ ⎠⎟⎞ρ dψρ dρ + ⎝⎜⎜⎛ ⎠⎟⎟⎞ k2 - m2 ρ2ψρ= 0. (3.316) ψρ ~ J1m(kmnρ), Nm(kmnρ), ⎝⎜⎛ ⎠⎟⎞J1m(kmnρ) = √⎯2 a Jm(kmnρ) Jm+1(xmn) where kmn = xmn a , Jm(xmn) = 0. (3.317) Correctly normalized eigenfunction with our B.C.’s:: ψmn(x≥) = J1m(kmnρ) 1 √⎯⎯2π eimφ . (3.318) (n=1,2,3,... and m=0, +-1,+-2,+-3,...) Plug in: 3.49 GD(x≥,x≥') = a2 π ∑ n=1∞ ∑ m =-∞∞ J1m(kmnρ)J1m(kmnρ') eim(φ - φ') xmn2. (3.319) Comparison of such eigenfunction solutions with other known solutions allows interesting summation formulas to bederived. For example, comparison of (3.312) with a reducedGreen function solution gives a formula for one of the sumsthere, whereas comparison of (3.319) with an image solutionfor the same problem (prob. 2.17) gives a formula for the sumthere. 3.50 Problems 3.1.1 By evaluating the expression (the “ ⊥” notation indicates the x and y components of a vector; “ α” is the cylindrical angle associated with k≥; see below), ⌡⎮⎮⌠ 02π dα (2π)2 eik≥ ⊥.(x≥-x≥')⊥ two different ways (or any other way you can think of), show the Bessel function “addition theorem”, J0(kD) = ∑ m Jm(kρ)Jm(kρ')eim(φ-φ'), where D = |(x≥-x≥')⊥|, ρ = |x≥ ⊥|, ρ' = |x≥'⊥|. The picture for this is: φ' α⊥⊥ ⊥≥ ≥x x' k≥ xφy 3.1.2 Using the generating function for Bessel functions, eit cosφ = ∑ m =-∞∞ im eimφ Jm(t), prove the sum rules: (a) ∑ m =-∞∞ J2 m(t) = 1, (b) J 0(2t) = ∑ m =-∞∞ (-1)m J2 m(t). 3.51 3.1.3 Show that (a) ⌡⎮⎮⌠ 0a dρ ρm+1Jm(kρ)= 1 k am+1 Jm+1(ka), (b) ⌡⎮⎮⌠ 0∞ dkk (2π) J0(kD) = δ(x-x')δ(y-y'), (D2 ≡ (x-x')2 + (y- y')2). 3.3.1 Show that for zeros of derivatives of Bessel functions (k-≠kmn, y-≠ ymn = k-a) ⌡⎮⎮⎮⌠ 0a dρρ Jm2(k-ρ) = a2 2 (1-m2 y-2) Jm2(y-). Use the method described in the notes after (3.56). 3.4.1 Using the reduced Green function technique, show that the Dirichlet Green function for the infinite conducting wallproblem, +zz=0 plane may be solved in cylindrical coordinates to give 3.52 G(x≥,x≥')=4∫ 0∞ dkk g(z,z') [1 2J0(kρ)J0(kρ')+∑ m =1∞ Jm(kρ)Jm(kρ')cos(m( φ-φ'))]. Find g(z,z'). 3.4.2 Find the Dirchlet Green's function for an infinite, flat conducting plate in infinte space. Take the surface of the conductor as the z-x plane . Show that the Green function may be written in cylindrical coordinates ( ρ,z,φ) as GD(x≥,x≥') = 4∫ 0∞ dk ∑ m =1∞ Jm(kρ)Jm(kρ')sin(m φ)sin(mφ')ek(z<-z>), where z< (z>) is the lesser (greater) of z and z'. 3.7.1 Show directly (using, say, the differential equations and large x expansions) the Wronskian relation W[Im(x),Km(x)] = - 1 x . 3.7.2 Evaluate the Wronskian ( ν is a positive non-integer): W[Jν(t),J-ν(t)] = ? [Hints: The differential equation satisfied by both J ν(t) and J-ν(t) is, [d2 dt2 + 1 t d dt - m2 t2 + 1]Jν,−ν(t) = 0. The large t expansion of J ν(t) is Jν(t) = √⎯2 πt cos(t + νπ 2 - π 4).] 3.53 3.7.3 Consider an infinitely long conducting cylinder of radius a, with the z-axis of coordinates along the axis ofsymmetry of the cylinder. +1a+z xy Assuming the form for the Dirichlet Green function, GD(x≥,x≥') = 4π∑ m =-∞∞ eim(φ-φ') 2π 1 π ∫ 0∞ dk cos [k(z-z')]g m(ρ,ρ'), and with 4πδ(x≥-x≥') = 4π∑ m =-∞∞ eim(φ-φ') 2π 1 π ∫ 0∞ dk cos [k(z-z')] 1 ρ δ(ρ-ρ'), find the reduced Green function, g m(ρ,ρ'), for the volume exterior to the cylinder. 3.7.4 Consider again the infinite conducting cylinder in three dimensions, with the z-axis of coordinates along theaxis of symmetry of the cylinder. 3.54 a+1 Assuming the same form for the Dirichlet Green function and delta function as above, find the reduced Green function, gm(ρ,ρ'), for the interior solution. This may be done either using the Wronskian technique or as an eigenfunction expansion. 3.7.5 Consider a infinitely long vacuum-filled wedge-shaped hole (wedge radius "a") inside a conductor in three dimensions. (We are working with a point charge, not a linecharge.) a+1 circular cross-section Use the reduced Green function method to derive a form for the Dirichlet Green function, G D(ρ,φ,z;ρ',φ',z'), in cylindrical coordinates. It is possible to derive a differential equation in either the ρ or z variables. [Hint: Starting out with a correct form of the three-dimensional delta function is the key to doing this problem.] 3.7.6 Consider the region between flat parallel conducting plates and outside of a conducting cylindrical post of radius 3.55 "a" connecting the plates. The distance between the plates is "D". D+1z' ρ' cylindrical post of radius "a" Find the Green function for the region between the plates and outside of the cylindrical post. [Hint: Choose the non- oscillatory direction for the Green function to be ρ.] 3.7.7 Consider the interior region of a cylindrical toroid of rectangular cross section. The height is L, and the inner and outerradii are A and B, respectively: LAB Construct the Dirichlet Green function in cylindrical coordinates (ρ,z,φ) for the interior region of the cylinder. [Hint: The key as always is an appropriate representation of the Dirac delta function in the interior. There are actually two ways of proceeding here, 3.56 depending on whether one chooses the z or ρ variable as oscillatory.] 3.8.1 Consider the closed conducting wedge shown. It opens with angle β and is bounded by circular conducting radii b and c (b < c). β b c+1 β Find the Dirichlet Green function for a line charge of unit charge density in this geometry by the reduced Green functionmethod. 3.8.2 In Eq.(3.166) of the script we found that the Green's function for a line charge in the volume bounded by conducting walls intersecting at arbitrary angle β, β could be written as the infinte sum, G(x≥,x≥') = ∑ n=1∞ 4 n (ρ< ρ>)γ sin(nπφ β)sin(nπφ' β), 3.57 where ρ< (ρ>) is the lesser (greater) of ρ and ρ' and γ = nπ β. Using the summation method of Ch.2, show that this expression can be summed exactly into the closed form G(x≥,x≥') = ln ⎣⎢⎢⎢⎡ ⎦⎥⎥⎥⎤1 + ⎝⎜⎛ ⎠⎟⎞ρ< ρ>2π/β - 2 ⎝⎜⎛ ⎠⎟⎞ρ< ρ>π/β cos(π β(φ+φ')) 1 + ⎝⎜⎛ ⎠⎟⎞ρ< ρ>2π/β - 2 ⎝⎜⎛ ⎠⎟⎞ρ< ρ>π/β cos(π β(φ-φ')) . [Hints: cos(x+y) = cosx cosy - sinx siny, ln(1-x) = - ∑ n=1∞ xn n .] 3.8.3 By using a reduced Green function technique or other means, show that the two-dimensional Green function for Dirichlet boundary conditions in the cylindrical region 0 < _ ρ <_ a has the expansion, G = - ln(ρ>2/a2) + 2 ∑ m=1∞ cos[m(φ-φ')] m (ρ<m)(1 ρ>m - ρ>m a2m), where ρ> (ρ<) is the greater (lesser) of ρ and ρ'. [Comparing with (3.319) of the text gives some interesting sums over Bessel functions.] 3.9.1 With the help of the generating function G(x,t) ≠ 1 √⎯⎯⎯⎯⎯⎯⎯⎯1-2xt + t2 = ∑ n =0∞ tn Pn(x), for the Legendre polynomials, P n(x), obtain the relation x d dx Pn+1 - d dx Pn = (n + 1)Pn+1. 3.58 3.9.2 Show by direct substitution that Pn(x) ≠ 1 2nn! (d dx)n(x2-1)n satisties dP l+1(x) dx = (2 l+1)P l(x) + dP l-1(x) dx . (b) Give an argument which shows that Pn(1) = 1. 3.12.1 A capacitor consists of two thin hemispherical conducting caps at potentials V and -V separated by a thingap, as shown. the radius of the sphere is "a". +V -Vgap (a) Define the capacitance as C ≠ Q V , where Q is the charge on the upper cap. Treating this either as a boundary value problem or using the Green function forthis geometry, show that 3.59 C = a 2 ∑ j=1∞ [P2j-2(0) - P 2j(0)]2, where P j(0) is the value of the Legendre polynomial at x=0. (b) Given that 1 √⎯⎯⎯⎯1+μ2 = 1 + ∑ n=1∞ (-1)n1.3.5...(2n-1) 2.4.6...(2n) μ2n, and comparing with the Coulomb expansion, show that C = a 2π ∑ j=1∞ 4(j-1 4)2 (Γ(j-1 2) j!)2 where Γ(j-1 2) is the gamma function with argument j-1 2. 3.12.2 (a) Find the electrostatic Neumann Green function, G(x≥,x≥'), for the outside region of a sphere of radius “a”. For simplicity, take the +1 charge on the +z axis: +z axis+1 a The “Coulomb expansion” for 1 |x≥-x≥'| should be helpful here. (b) What is the dipole charge density, D(x≥), on the sphere’s surface? 3.12.3 (a) Solve the Neumann interior problem for the region inside two concentric spheres of interior radius "a" and 3.60 exterior radius "b". Show that the most general form of the solution is (the P l are the Lengendre polynomials) ∇ 2GN(x≥',x≥) = -4πδ (x≥-x≥'), GN(x≥',x≥) = ∑ l g l(r,r') P l(cos γ) + ∑ l P l(cos γ)(α lr'l + β lr'-( l + 1)), where ( l > 0 only) g l(r,r')= l( l+1) b2 l+1 - a2 l+1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ r<l l + a2 l+1 ( l+1)r<l+1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ r>l l + b2 l+1 ( l+1)r>l+1, and where r < (r>) is the lesser (greater) of r and r'. Notice g0(r,r') is not specified: can you determine it? Is it unique? What are the values of α l and β l ? Be sure to refresh yourself on the unusual properties of the interior Neumann Green function from Ch.1. (When b-> ∞ we have an outside problem. Will your results necessarily agree with prob. 3.12.2(a) above in this limit?)(b) Verify that the surface delta functions discussed inSection 1.7, Eq.(1.78), are present in the final answer ofprob. 3.12.3. 3.12.4 (a) Using the Green function for a sphere of radius "a" or other means, show that the outside potential, V( θ,φ), from a fiven potential, Φ(x≥), on the surface of a sphere can be written as Φ(x≥) = ∑ l,m ⎝⎜⎛ ⎠⎟⎞a rl+1 Y lm(θ,φ) B lm, where B lm = ∫sodΩ ′ V(θ',φ')Y lm*(θ',φ'). 3.61 What is the inside for for Φ(x≥)? (b) Show that this is also equivalent to (+ for inside, - for outside) Φ(x≥) = +- a(a2-r2) 4π ∫sodΩ ′ V(θ',φ') (a2+r2-2ar cos γ)3/2, where cosγ ≅ r≥.r≥' rr'. 3.13.1 A conducting sphere of radius "a" is split into two unequal pieces which are insulated from one another. The "top" piece subtends a uniform polar angle, θ0, as measured from from the center of the sphere and is raised to potential V. The rest of the sphere has potential -V. V -V+z Oθ0 Find the electric field, E≥, at the center of the sphere, O. 3.14.1 Using the method of eigenfunction expansions or other means, show that another expression for the Dirichlet Greenfunction for the infinite parallel plate capacitor problem, 3.62 +1 (x',y',z')z=a z=0 is GD(x≥,x≥’) = 4 a ∫ 0∞ dk k J0(kD)∑ n=1∞ sin(nπz a)sin(nπz’ a) k2 + (nπ a)2, where D ≠ |(x≥-x≥’)| and J 0 is the Bessel function of order zero.