1.4.3 Elementary solutions to Laplace s equation in the spherical coordinates (Axially symmetric cases) (Griffiths 3.3.2)

Transcription

1 1.4.3 Elementary solutions to Laplace s equation in the spherical coorinates (Axially symmetric cases) (Griffiths 3.3.) In the spherical coorinates (r, θ, φ), the Laplace s equation takes the following form: " r + r r + 1 r sin θ Ã sin θ! + θ θ 1 r sin θ Φ(r, θ, φ) =0 φ We shall consier only the axially symmetric cases in which φ =0. The Laplace s equation is then " r + r r + 1 Ã sin θ! Φ(r, θ) =0 r sin θ θ θ A. General Solution 1/note4

2 Separation of variables Assume Φ(r, θ) =R(r)F (θ) R(r): function of r; F (θ): function of θ. Raial function satisfies: " r + r r k R =0 r an the angular function satisfies: Ã 1 sin θ F! + k F =0 sin θ θ θ Solutions for the raial function R Assume a power function: Raial ifferential equation gives: So R = r l l(l 1)r l +lr l k r l =0 k = l(l 1) + l = l(l +1) Similarly, assume R = 1 r l+1 we obtain k = l(l +1) In this case, we have the same eigenvalue, but two ifferent solutions (inepenent). So R = Ar l + B l =0, 1,, r l+1 /note4

3 Actually l coul be any values at this stage. However, it turns out that only integer numbers will ensure that the potential is single value an perioic in the angular part Φ(θ) in the problems we are iscussing. Angular function à 1 sin θ F! + l(l +1)F =0 sin θ θ θ Variable substitution θ µ =cosθ, ( 1 µ 1) θ = θ = sin θ à sin θ! θ θ = " ³ sin θ θ = sinθ " ³1 µ So the ifferential equation for the raial function is converte to the Legenre ifferential equation " ³1 µ F (µ) + l(l +1)F (µ) =0 For l =0 F 0 (µ) " ³1 µ F 0 (µ) =0 ³ 1 µ F 0 (µ) =A = A 1 µ = A F 0 = A ln à 1+µ 1 µ à 1 1+µ + 1! + B 1 µ! 3/note4

4 Discar the first solution (otherwise F 0 at µ = ±1 orθ = 0, π) F 0 = B Polynomial of orer zero for l =1 an " ³1 µ F 1 (µ) +F 1 (µ) =0 F 1 = µ Polynomial of orer 1 F 1 = µ Ã! 1+µ ln 1 1 µ (iscare) Similarly for l = F (µ) =3µ 1 Polynomial of orer Legenre function (polynomial) Generally, a Legenre function (polynomial) satisfies the Legenre ifferential equation. The n-th orer Legenre polynomial is enote by P l (µ) ansatisfies " ³1 µ P l (µ) + l(l +1)P l (µ) =0 Rorigues formula P l (µ) = 1 l l l! l (µ 1) l Orthogonality of the Legenre polynomial 1 P l (µ)p l 0 (µ) = l +1 δ ll 0 4/note4

5 Legenre Polynomial generating function: 1 = X P 1 µt + t l (µ)t l ( µ 1, t < 1) General solution of the Laplace s equation Φ(r, θ) = µ A l r l + B l P r l+1 l (cos θ) (A) A l an B l are to be etermine by given bounary conitions of particular problems. B. Potential ue to a point charge place on the axis This problem shows the appearance of Legenre polynomials in the potential problem with a more physical picture (It is NOT a bounary value problem). A point charge q is place at (r, θ) =(a, π). What is the inuce potential at an arbitrary point (r, θ)? 5/note4

6 Use Legenre Polynomial generating function, we obtain for r>a q 1 Φ(r, θ) = 4π² 0 r + a +ra cos θ q 1 1 = s 4π² 0 r 1 ( a/r) cos {z } {z θ} +(a/r) t µ q 1 1 = 4π² 0 r 1 µt + t q 1 = P l (µ)t l 4π² 0 r q 1 µ a l = 4π² 0 r ( 1) l Pl (cos θ) r q = ( 1) l a l 4π² 0 r P l(cos θ) l+1 6/note4

7 = q 4π² 0 " 1 r a cos θ + a 3cos θ 1 + r r 3 r>a It contains the terms P l (cos θ)/r l+1 appeare in the solution in eq. (A). The first three terms represent the contribution by a monopole, ipole an quarupole. Similarly for r<a Φ(r, θ) = q " 1 1 r µ r 4π² 0 a a cos θ + 3cos θ 1 + a Each term takes the form r l P l (cos θ) as those terms in eq. (A). r<a C. Potential Due to Oppositely Charge Semi-Spheres (Conucting Shell) 7/note4

8 In this example we have a conucting semi-spherical shell which is oppositely charge, i.e., the potential on the top half is +V anonthebottom half is V. There are no charges anywhere except on the shell. Potential If we choose the axis as shown, the system is axially symmetric. As we iscusse earlier, the solution to the Laplace s equation is P ³ l A r l Pl (cos θ) r<a a Φ(r, θ) = P ³ l+1 B a l Pl (cos θ) r>a r The continuity of the potential requires Φ i (a, θ) =Φ e (a, θ), So A l P l (cos θ) = B l P l (cos θ) 8/note4

9 which inicates A l = B l. Bounary conition: Insie the sphere: ( V 0 < θ < π (0 <µ=cosθ < 1) Φ S = V < θ < π ( 1 <µ=cosθ < 0) π Φ S (θ) = A l P l (cos θ) Multiply sin θp l (cos θ) on both sie an integrate from 0 to π. Z π 0 Φ S P l 0(cos θ)sinθθ = Z π A l P l (cos θ)p l 0(cos θ)sinθθ 0 Variable substitution µ =cosθ, = sin θθ 1 Φ S (µ)p l 0 (µ) = = A l P l (µ)p l 0 (µ) 1 A l l +1 δ ll 0 = A l 0 l 0 +1 So Z l +1 1 A l = Φ S (µ)p l (µ) 1 Since P l ( µ) =( 1) l P l (µ) anφ S (µ) is an o function, A l vanishes for l =k (even l). When l =k +1isanonumber, A l = l +1 1 Φ S (µ)p l (µ) =(l +1)V 0 P l (µ) (1) From mathematical hanbooks, we can fin 0 P l (µ) =( 1) l 1 (l )!! (l +1)!! o l 9/note4

10 So where A l =( 1) l 1 (l )!! (l +1)V o l (l +1)!! n!! = n(n )(n 4) n!! = n(n )(n 4) 1 even n o n Final result Φ i (r, θ) =V Φ e (r, θ) =V " 3 r a cos θ 7 µ r 3 P3 (cos θ)+ 8 a " µ 3 a µ 7 a 4 cos θ P3 (cos θ)+ r 8 r For o l, P l (0) = 0 when θ = π/ onthex-y plane. This inicates that the potential vanishes at the miplane because the up-own antisymmetry. The leaing term for the potential in the exterior region is a ipole term. This is ue to the antisymmetry as well. Raial electric fiel The graient operator in the spherical coorinates is Ã! = er r + 1 e θ r θ + 1 e θ r sin θ φ E r = Φ r Charge on the outer surface Φ e (r, θ) σ e (θ) = D r = ² 0 E r = ² 0 r r=a = ² 0V 3P 1 (cos θ) 7 a P 3(cos θ)+ () Phys 463, E & M III, C. Xiao 10 / note4

11 Similarly on the inner surface Φ (r, θ) σ i (θ) = D r = ² 0 E r = ² 0 r r=a = ² 0V 3 a P 1(cos θ) 1 8 P 3(cos θ)+ (3) D. Determine the coefficients by observation The coefficient A l (an B l ) can be calculate, in principle, through the integral Z l +1 1 A l = Φ S (µ)p l (µ), 1 if the potential is given on the spherical surface as we i in the last example. However, the integral is generally ifficult to calculate. In some cases, simple observation is sufficient to etermine the coefficients. Example: If the potential on a spherical surface is given by Φ(θ) =k sin θ Determine the potential insie the sphere. The general solution is µ Φ(r, θ) = A l r l + B l P r l+1 l (cos θ) To ensure that Φ(0, θ) isfinite at r =0,B l = 0. Using the half-angle formula, we can rewrite Φ(θ) =k sin θ = k (1 cos θ) =k (P 0(cos θ) P 1 (cos θ)) So A 0 = k an A 1 = k. The solution is then a Φ(r, θ) = k µ 1 r a P 0(cos θ) = k µ1 r a cos θ (A) Phys 463, E & M III, C. Xiao 11 / note4