that any image charges lie on the line joining the center of the sphere to the charge outside. Let the radius of the sphere be a,and choose coordinates with origin at the center of the sphere and the charge outside on the positive y-axis at (0,r,0).A single image charge of g'=-ga/r'and location (0,a2/r',0)does the trick. 「1 or)=imorr-a 0 (63) To see that o vanishes on the sphere note that putting r=ar gives: |r-r=a2+rP-2r'r… (64) r-/r=心+an-2arr=re+2-2rr (65) This potential is proportional to the Dirichlet Green function for the exterior of a sphere.We just have to put the exterior charge g=eo at a general point r'and adjust the normalization: Gn(r,)= 1「1 4mr-r可r-a2r/网] (66) This can be plugged into Eq.(46)to solve the arbitrary Dirichlet problem for a sphere.For this we need the normal derivative of Gp on the surface of the sphere: n.VGD OGD r-a Or a-r2/a 4π(a2+r2-2ar.r'3/2 (67) Then the solution for the boundary condition (r=a,,)=V(0,)is oir,g,p0=-av0,pl+-2nrr0 a(a2-2) (68) Here ds=sin dod,andf.r'=r'(cos cos'+sin sin'cos()).While the method of images is very powerful for the simple geometry of a single sphere,it becomes less so with more complicated geometries.In the homework you will see this for the case of two conducting spheres,which require an infinite number of image charges. Even the case of the region between two concentric spheres requires an infinite number of image charges,in spite of the spherical symmetry.This motivates the development of other approaches to electrostatic problems. 3.2 Method of Separation of Variables Partial differential equations put conditions on functions of several variables.The separa- tion of variables method first tries to find solutions that are products of functions of single variables.For instance o(x,y,2)=X(x)Y(y)Z(z) (69) 15 ©2010 by Charles Thorn
that any image charges lie on the line joining the center of the sphere to the charge outside. Let the radius of the sphere be a, and choose coordinates with origin at the center of the sphere and the charge outside on the positive y-axis at (0, r 0 , 0). A single image charge of q 0 = −qa/r0 and location (0, a 2/r0 , 0) does the trick. φ(r) = q 4π0 1 |r − r 0yˆ| − a r 0 |r − a 2yˆ/r0 | (63) To see that φ vanishes on the sphere note that putting r = arˆ gives: |r − r 0 yˆ| 2 = a 2 + r 02 − 2r 0 r · yˆ (64) |r − a 2 yˆ/r0 | 2 = a 2 + a 4 /r02 − 2a 2 r · yˆ/r0 = a 2 r 02 (r 02 + a 2 − 2r 0 r · yˆ) (65) This potential is proportional to the Dirichlet Green function for the exterior of a sphere. We just have to put the exterior charge q = 0 at a general point r 0 and adjust the normalization: GD(r, r 0 ) = 1 4π 1 |r − r 0 | − a r 0 |r − a 2r 0/r02 | (66) This can be plugged into Eq.(46) to solve the arbitrary Dirichlet problem for a sphere. For this we need the normal derivative of GD on the surface of the sphere: nˆ · ∇GD r=a = − ∂GD ∂r = a − r 02/a 4π(a 2 + r 02 − 2arˆ · r 0 ) 3/2 (67) Then the solution for the boundary condition φ(r = a, θ, ϕ) = V (θ, ϕ) is φ(r 0 , θ 0 , ϕ 0 ) = − Z dΩV (θ, ϕ) a(a 2 − r 02 ) 4π(a 2 + r 02 − 2arˆ · r 0 ) 3/2 (68) Here dΩ = sin θdθdϕ, and rˆ·r 0 = r 0 (cos θ cos θ 0 +sin θ sin θ 0 cos(ϕ−ϕ 0 )). While the method of images is very powerful for the simple geometry of a single sphere, it becomes less so with more complicated geometries. In the homework you will see this for the case of two conducting spheres, which require an infinite number of image charges. Even the case of the region between two concentric spheres requires an infinite number of image charges, in spite of the spherical symmetry. This motivates the development of other approaches to electrostatic problems. 3.2 Method of Separation of Variables Partial differential equations put conditions on functions of several variables. The separation of variables method first tries to find solutions that are products of functions of single variables. For instance φ(x, y, z) = X(x)Y (y)Z(z) (69) 15 c 2010 by Charles Thorn
Plugging such a form into the diff eq reduces the problem to ordinary diff egs in a single variable.After finding these one can find more general solutions by linear superposition of the factorized solutions.For this to work the differential operator of the original equation must be a sum of differential operators on single variables. Cartesian Coordinates In Cartesian coordinates-V2 =-(02/0x2)-(82/0y2)-(62/022),and Laplace's equation leads to 102X102Y102Z =0 X 0x2 Y Oy2 Z0z2 (70) To solve this equation,each term must be a constant,say A,B,C respectively subject to a+b+c=0.Thus we must solve three eigenvalue equations: O2X O2Y 0z2 AX, 0u2=BY. 82Z 022 =CZ=-(A+B)Z, (71) The solutions of these equations are just trigonometric and/or exponential functions.Two of the eigenvalues,say A,B are arbitrary. Eigenvalue problems are familiar from quantum mechanics,in which dynamical variables are represented by linear operators whose eigenvalues are the possible results of measuring those variables.In this context the Laplacian is a sum of three commuting linear opera- tors -02/ox?.If those three operators are hermitian,one can find a basis of simultaneous eigenstates of those three operators. But the Laplace equation requires the three eigenvalues to sum to zero,so we have only a two-fold basis.In Cartesian coordinates we can use these eigenfunctions to construct boundary value solutions for shapes made from planes parallel to the coordinate planes. Consider a rectangular box of dimensions a x b x c,placed with three of the sides in the respective coordinate planes.The general Dirichlet problem would be to specify the potential arbitrarily on each of the 6 rectangular surfaces of the box.But we can build that solution by superposing solutions with the potential vanishing on 5 sides,but arbitrarily given on the sixth.Pick that side to be the one at z=c where the potential is specified to be V(z,y).Then we have Xm(x）= m元 πy n2 sin- Yn(y）=sin Zmn(2)=sinh z m2 6 2 (72) a (x,y,2) Amn sin mna nr m2 n2 -s1n- 2sinhπz1 (73) a + m.n=1 This expresses the solution as a double Fourier series in y,with the Amn determined by the last boundary condition V(x,= Amn sin si sin nry m2 n2 (74) a sinhπcVa+ m,n=1 16 ©2010 by Charles Thorn
Plugging such a form into the diff eq reduces the problem to ordinary diff eqs in a single variable. After finding these one can find more general solutions by linear superposition of the factorized solutions. For this to work the differential operator of the original equation must be a sum of differential operators on single variables. Cartesian Coordinates In Cartesian coordinates −∇2 = −(∂ 2/∂x 2 )−(∂ 2/∂y 2 )−(∂ 2/∂z 2 ), and Laplace’s equation leads to − 1 X ∂ 2X ∂x 2 − 1 Y ∂ 2Y ∂y 2 − 1 Z ∂ 2Z ∂z 2 = 0 (70) To solve this equation, each term must be a constant, say A, B, C respectively subject to a + b + c = 0. Thus we must solve three eigenvalue equations: − ∂ 2X ∂x 2 = AX, − ∂ 2Y ∂y 2 = BY, − ∂ 2Z ∂z 2 = CZ = −(A + B)Z , (71) The solutions of these equations are just trigonometric and/or exponential functions. Two of the eigenvalues, say A, B are arbitrary. Eigenvalue problems are familiar from quantum mechanics, in which dynamical variables are represented by linear operators whose eigenvalues are the possible results of measuring those variables. In this context the Laplacian is a sum of three commuting linear operators −∂ 2/∂x 2 i . If those three operators are hermitian, one can find a basis of simultaneous eigenstates of those three operators. But the Laplace equation requires the three eigenvalues to sum to zero, so we have only a two-fold basis. In Cartesian coordinates we can use these eigenfunctions to construct boundary value solutions for shapes made from planes parallel to the coordinate planes. Consider a rectangular box of dimensions a × b × c, placed with three of the sides in the respective coordinate planes. The general Dirichlet problem would be to specify the potential arbitrarily on each of the 6 rectangular surfaces of the box. But we can build that solution by superposing solutions with the potential vanishing on 5 sides, but arbitrarily given on the sixth. Pick that side to be the one at z = c where the potential is specified to be V (x, y). Then we have Xm(x) = sin mπx a , Yn(y) = sin nπy b , Zmn(z) = sinh πz r m2 a 2 + n 2 b 2 (72) φ(x, y, z) = X∞ m.n=1 Amn sin mπx a sin nπy b sinh πz r m2 a 2 + n 2 b 2 (73) This expresses the solution as a double Fourier series in x, y, with the Amn determined by the last boundary condition V (x, y) = X∞ m,n=1 Amn sin mπx a sin nπy b sinh πc r m2 a 2 + n2 b 2 (74) 16 c 2010 by Charles Thorn
Recall that the functions sin(mT/a)are orthogonal on the interval 0<x<a ra dx sin sin a - a (75) This means we can solve for the Amn m2 n2 4 fa Amn sinh nca da dy sin ab Jo 0 a m( sin (76) For example,if V is a constant m2.n2 Amn sinh ea mm(-() 4V (77) and the potential reads 1 (红，，2)= 16Y I sin maa sin nny sinh zvm2/a2+ (78) mn m,n=odd “b sinh cvm2/a2+n2/ Notice that as long as z<c,i.e.inside the box,the ratio of sinh's gives exponential conver- gence of the double sum.Although the individual terms of this expansion solve Laplace's equation outside the box,this sum will not give the proper outside solution because of bad behavior at large distances. Near the boundaries of the box,the exponential convergence becomes less effective,and it is necessary to include ever more terms to maintain accuracy.Near corners and edges convergence is even worse,because fields and surface charge density can become singular there. The representation of a constant by a truncated Fourier series (keeping the first N terms) of sin functions is always poor near the endpoints (Gibbs phenomenon).The error size doesn't improve for larger N,but the region over which it occurs does decrease with larger N.(See the figure.)Thus as long as x is away from the endpoints,one can approximate the constant arbitrarily accurately by including enough terms. 17 ©2010 by Charles Thorn
Recall that the functions sin(mxπ/a) are orthogonal on the interval 0 < x < a Z a 0 dx sin mxπ a sin nxπ a = a 2 δmn (75) This means we can solve for the Amn Amn sinh πc r m2 a 2 + n 2 b 2 = 4 ab Z a 0 dx Z b 0 dy sin mxπ a sin nyπ b V (x, y) (76) For example, if V is a constant Amn sinh πc r m2 a 2 + n 2 b 2 = 4V mnπ2 (1 − (−) m)(1 − (−) n ) = 16V mnπ2 δm,oddδn,odd (77) and the potential reads φ(x, y, z) = 16V π 2 X∞ m,n=odd 1 mn sin mπx a sin nπy b sinh πz p m2/a2 + n 2/b2 sinh πc p m2/a2 + n 2/b2 (78) Notice that as long as z < c, i.e. inside the box, the ratio of sinh’s gives exponential convergence of the double sum. Although the individual terms of this expansion solve Laplace’s equation outside the box, this sum will not give the proper outside solution because of bad behavior at large distances. Near the boundaries of the box, the exponential convergence becomes less effective, and it is necessary to include ever more terms to maintain accuracy. Near corners and edges convergence is even worse, because fields and surface charge density can become singular there. The representation of a constant by a truncated Fourier series (keeping the first N terms) of sin functions is always poor near the endpoints (Gibbs phenomenon). The error size doesn’t improve for larger N, but the region over which it occurs does decrease with larger N. (See the figure.) Thus as long as x is away from the endpoints, one can approximate the constant arbitrarily accurately by including enough terms. 17 c 2010 by Charles Thorn
Gibbs Phenomenon:N=5,30 0.8 0.6 0.4 0.2 0.2 0.6 0.8 The singular behavior near corners and edges can be captured by replacing them by intersecting planes,which can be analyzed simply by the method of images.The Green function for the quadrant bounded by two perpendicular planes can be found using 3 image charges.That for the octant bounded by three perpendicular planes requires 7 image charges. Spherical Coordinates Spherical coordinates are r,0,,the radius,polar angle from the z-axis,and azimuthal angle about the z-axis measured from the r axis:(r,y,z)=r(sin0 cos sin sin cos). It is helpful to think of the Laplacian in terms of the quantum mechanical angular mo- mentum L=r×V/i: L2=-r.(V×(r×V)=-r7r7+rVr7i =-22-0+9 +9 +20=-22+ 02 Or 83 (79) -2=-1PL2 (80) Then,separation of variables involves factorizing (r,0,)=R(r)Yim(0,)where Yim(0,), the spherical harmonic,is an eigenfunction of L2,L:with eigenvalues I(1+1),m respectively where l=0,1,2...and-l<m <I.Then if o satisfies Laplace's equation,we have 1∂2x7 1(+1) T2 R=0 (81) 18 ©2010 by Charles Thorn
0 0.2 0.4 0.6 0.8 1 1.2 0.2 0.4 0.6 0.8 1 x Gibbs Phenomenon: N=5,30 The singular behavior near corners and edges can be captured by replacing them by intersecting planes, which can be analyzed simply by the method of images. The Green function for the quadrant bounded by two perpendicular planes can be found using 3 image charges. That for the octant bounded by three perpendicular planes requires 7 image charges. Spherical Coordinates Spherical coordinates are r, θ, ϕ, the radius, polar angle from the z-axis, and azimuthal angle about the z-axis measured from the x axis: (x, y, z) = r(sin θ cos ϕ,sin θ sin ϕ, cos θ). It is helpful to think of the Laplacian in terms of the quantum mechanical angular momentum L = r × ∇/i: L 2 = −r · (∇ × (r × ∇)) = −ri∇jri∇j + ri∇jrj∇i = −r 2∇2 − r ∂ ∂r + r ∂ ∂r 2 + 2r ∂ ∂r = −r 2∇2 + r ∂ 2 ∂r 2 r (79) −∇2 = − 1 r ∂ 2 ∂r 2 r + L 2 r 2 (80) Then, separation of variables involves factorizing φ(r, θ, ϕ) = R(r)Ylm(θ, ϕ) where Ylm(θ, ϕ), the spherical harmonic, is an eigenfunction of L 2 , Lz with eigenvalues l(l+1), m respectively where l = 0, 1, 2 · · · and −l ≤ m ≤ l. Then if φ satisfies Laplace’s equation, we have − 1 r ∂ 2 ∂r 2 r + l(l + 1) r 2 R = 0 (81) 18 c 2010 by Charles Thorn
L2Yim =l(l+1)Yim; LzYim mYim (82) The radial equation is easily solved with the most general solution 1 Rr)=Ar+Bi, (83) The angular equations determine the spherical harmonics Yim.We can use our familiarity with quantum mechanics to efficiently analyze these solutions.For example we know that, defining L±≡Lx±iLg, L+Yu=0,L_Yim vi(l+1)-m(m-1)Yim-1 (84) To go further,we need the explicit form of the L's: 1 2- =(-=（品-品 (85 a L士=土之 干(x士刘 (86) We see easily that L+(+iy)=0 and L:(x+iy)=l(+iy).So Yu is proportional to (x +iy)=r!sin!beilo Yu(0,)Csin!beile (87) We normalize the Y's to 1,i.e.fdYim2=1,so 1-1C2 iid(2cPdu(2 =2rC222+1 (02 (88) 2l+1)月 Then Ym(0,p)= (-) (21+1)1 l!v2元 2(21+1) sin!Geil (89) The sign is the standard convention.By applying L-repeatedly,one obtains all the Yim from-1<m <+l.Clearly,Yim has the o dependence eime,and the 0 dependence is a polynomial in cos0,sin0. A typical boundary value problem using spherical coordinates is to find the potential between two concentric spherical shells,say at r =a,b.Then the general solution of Laplace's equation between the shells is -(+) Yim(0,) (90) 19 ©2010 by Charles Thorn
L 2Ylm = l(l + 1)Ylm, LzYlm = mYlm (82) The radial equation is easily solved with the most general solution Rl(r) = Alr l + Bl 1 r l+1 (83) The angular equations determine the spherical harmonics Ylm. We can use our familiarity with quantum mechanics to efficiently analyze these solutions. For example we know that, defining L± ≡ Lx ± iLy, L+Yll = 0, L−Ylm = p l(l + 1) − m(m − 1)Yl,m−1 (84) To go further, we need the explicit form of the L’s: Lz = 1 i x ∂ ∂y − y ∂ ∂x , Lx = 1 i y ∂ ∂z − z ∂ ∂y , Ly = 1 i z ∂ ∂x − x ∂ ∂z (85) L± = ±z ∂ ∂x ± i ∂ ∂y ∓ (x ± iy) ∂ ∂z (86) We see easily that L+(x + iy) l = 0 and Lz(x + iy) l = l(x + iy) l . So Yll is proportional to (x + iy) l = r l sinl θe ilϕ Yll(θ, ϕ) = C sinl θe ilϕ (87) We normalize the Y ’s to 1, i.e. R dΩ|Ylm| 2 = 1, so 1 = |C| 2 Z sin θdθdϕ sin2l θ = 2π|C| 2 Z 1 −1 dz(1 − z 2 ) l = 2π|C| 2 Z 2 0 duul (2 − u) l = 2π|C| 2 2 2l+1 (l!)2 (2l + 1)! (88) Then Yll(θ, ϕ) = (−) l l! √ 2π r (2l + 1)! 2 (2l+1) sinl θe ilϕ (89) The sign is the standard convention. By applying L− repeatedly, one obtains all the Ylm from −l ≤ m ≤ +l. Clearly, Ylm has the ϕ dependence e imϕ , and the θ dependence is a polynomial in cos θ,sin θ. A typical boundary value problem using spherical coordinates is to find the potential between two concentric spherical shells, say at r = a, b. Then the general solution of Laplace’s equation between the shells is φ = X∞ l=0 X l m=−l Almr l + Blm r l+1 Ylm(θ, ϕ) (90) 19 c 2010 by Charles Thorn