Legendre Poynomias - Lecture 8 Introduction In spherica coordinates the separation of variabes for the function of the poar ange resuts in Legendre s equation when the soution is independent of the azimutha ange. ( x 2 ) d2 P dx 2 2x dp dx + ( + )P = 0 This equation has x = cos(θ) with soutions P (x). As previousy demonstrated, a series soution can be obtained using the form; P(x) = a n x n+s Taking the derivatives, substituting into the ode, and coecting the coefficient of the same power in x, one obtains the recursion reation for the coefficients. a n+2 = (n + s)(n + s + ) ( + ) (n + s + 2)(n + s + ) a n The indicia equation must aso be satisfied by seection of the initia coefficient and/or starting power of x. Thus a 0 s(s ) = 0 a ( + s)s = 0 We then have the choice of a = 0 and s = 0 or a 0 = 0 and s = 0. Other choices are incorporated in these two. Thus if we choose a = 0 and s = 0, then; a n+2 = n(n + ) ( + ) (n + 2)(n + ) a n Ony even powers of x are incuded in the series. If a 0 and s = 0 then ony odd powers are incorporated in the series and the recursion reation remains the same. For the series to be convergent for x = it must terminate, again as previousy pointed out. The series then takes the form; P (x) = /2 n=0 ( ) n (2 2n)! 2 n!( n)!( 2n)! x 2k This series resuts in a set of poynomias which may be obtained from the generating function;
g(x, t) = ( 2xt + t 2 ) /2 = P (x) t Some initia poynomias are; P 0 = P = x P 2 = (/2)[3x 2 ] =0 2 Orthogonaity Consider the orthogonaity integra for the Legendre functions. This foows from the genera Sturm-Liouvie probem. Put Legendre s equation in sef adjoint form; d dx [( x2 ) dp (x) dx ] + ( + )P (x) = 0 Then ook at the equation for P n (x) and subtract the equations for P and P n after mutipication of the first by P n and the ater by P. Integrate the resut between ±. This resuts in [( x 2 )P n P ( x2 )P P n ] + [( + ) n(n + )] dxp P n = 0 The first term vanishes and thus the poynomias are orthogona but not normaized to. The normaization integra takes the form; N 2 = dxp n P m Use the Rodgiques formua to obtain; P n (x) = 2 n n! ( d dx )n (x 2 ) n N 2 = ( )m+n 2 m+n m!n! dx [( dm dx m )( x 2 ) m ][( dm dx n )( x 2 ) n ] Integrate by parts m times noting that the surface terms vanish at ±. N 2 = ( )n m 2 n+m m!n! dx ( dn m dx n m )( x 2 ) n 2
This vanishes for n > m (reverse n, m if n < m). If n = m then; N 2 = 2 2n (n!) 2 dx ( x 2 ) n = 2 2n + The orthogonaity integra is for the associated Legendre poynomias is expressed as; dxpr m (j)pk m(x) = 2 (j + m)! 2j + (j m)! The normaization for the Legendre poynomia P m r is found for m = 0. 3 Recurrence Reations The recurrence reations between the Legendre poynomias can be obtained from the generating function. The most important recurrence reation is; (2n + )xp n (x) = (n + )P n+ (x) + np n (x) To generate higher order poynomias, one begins with P 0 (x) = and P (x) = x. The generating function aso gives the recursion reation for the derivative. P n+(x) = (n + )P n (x) + xp n(x) 4 Separation of variabes in spherica coordinates We have previousy ooked at separation of variabes in spherica coordinates, particuary with respect to the radia component. Now consider the component invoving the poar ange with x = cos(θ) when the aximutha ange id incuded in the separation. ( x 2 ) d2 Θ dx 2 2x dθ dx + [n(n + ) m2 ( x 2 ) ]Θ = 0 The above equation is the associated Legendre equation. For the soution to remain finite at x = ± with n integra. The equation may be obtained from the ordinary Legendre equation by differentiation. Thus we find; d m dx m [( x 2 )P n (x) 2xP n (x) + n(n + )P n(x)] 3
P m n (x) = ( x 2 ) m/2 d m P n (x) dx m Then; Pn m (x) = ( ) m (n m)! (n + m)! P n m (x) The generating function is difficut to use; (2m)!( x 2 ) m/2 2 m m!( 2tx + t 2 ) m+/2 = The recursion reation is; Finay; s=0 P m s+m(x) t s (2n + )xp m n (x) = (n + m)p m n (x) + (n m + )P m n+ (x) P m n ( x) = ( )n+m P m n (x) P m n (±) = 0 5 Spherica harmonics In spherica coordinates Lapace s equation has the form; 2 V = (/r 2 ) r (r2 V r ) + r 2 sin(θ) θ Now identify an operator on the anges, L, such that; (sin(θ) V θ ) + 2 V r 2 sin 2 (θ) φ 2 (/r 2 )L 2 V = r 2 (sin(θ) V sin(θ) θ θ ) + 2 V r 2 sin 2 (θ) φ 2 Then when Lapace s equation is soved by separation of variabes; L 2 V = ( + )V Thus identify V = R (r) Y m (θ, φ) where Y m (θ, φ) is an anguar eigenfunction with eigenvaues,, and m. We aready know that this function has the form; Y m (θ, φ) P m (θ) e imφ 4
Now we make this function not ony orthogona, which it must be when integrating over θ and φ, but aso normaized to. Therefore integrating over the soid ange, dω = sin(θ) dθ dφ ; Then; δ nm δ k = π 0 sin(θ) dθ 2π 0 dφ [Y m Y m k ] Y m (θ, φ) = (2 + )( m)! 4π ( + m)! P m (θ, φ) e imφ The first few functions, Y M, which are the spherica harmonics; Y 0 0 = /(4π) Y = 3/(8π)sin(θ) e iφ Y = 3/(8π)sin(θ) e iφ Y 0 = 3/(4π)cos(θ) Note that Y m = Y m. The functions, Y m (θ, φ), are the spherica harmonics, and we wi ater identify the operator, L, as proportiona to the anguar momentum operator in Quantum Mechanics. 6 Second soution We have found a series soutions, P (x) with x = cos(θ), which contain ony odd powers and one containing ony even powers. But whie these are ineary independent, they are not the two ineary independent soutions soutions expected from a 2 nd order ode. In fact, we have demonstrated that a second soution must diverge at the singuar points. Suppose we choose to aow divergence by not etting the series terminate. Thus write; ζ = P + Q In the above, Q wi be the second soution. For the series soution we ook at the form; ζ = P + n= b n x n Now is chosen so that the series in even powers terminates. We then start b n with n = so this series forms even powers of x which cannot terminate because is aready chosen to 5
Figure : A representative exampe of Legendre functions of the first kind make the even series terminate. The recursion reation is; b n+2 = n(n + ) ( + ) (n + 2+)(n + ) b n n odd We can aso generate a series in even powers in the same way. This series represents the second soution to Legendre s equation and is written, Q. Q has singuar points at θ = 0, π. Associated functions, as with the function, P, are defined and written, Q m (θ). The first few functiona forms are; Q 0 = (/2) n[ + x x ] Q = (x/2) n[ + x x ] Q 2 = ( 3x2 4 ) n[ + x x ] 3x/2 Figure shows the behavior of the first few Legendre functions of the first kind, whie Figure 2 shows the behavior of the Legendre functions of the second kind. 6
Figure 2: A representative exampe of Legendre functions of the first kind 7 Green s function in spherica coordinates Now return to the expression of the Green s function in spherica coordinates. This was deveoped severa times previousy but we are now in position to finaize this resut. We assume that; 2 G = 4πδ( r r ) Now expand the δ function using the compete set of spherica harmonics. Thus; δ(θ θ )δ(φ φ ) = a νµ Y ν µ (θ φ) Mutipy by Y m (θ, φ) and integrate over the soid ange dω. Ortho-normaity projects out the coefficients, a m = Y m (θ, φ ). Then the Green s function takes the form; G = m g Y m (θ, φ) Y m (θ, φ ) The deta function is; δ( r r ) = 4πδ(r r ) Y m (θ, φ) Y m (θ, φ ) m Substitution into Lapaces s equation yieds; 7
r d2 dr 2 [rg] ( + )g = 4πδ(r r ) This is soved by requiring the soution to be continuous and the derivative to match the discontinuity at r = r. The two soutions to the homogeneous equation are r and r (+). The resut is; g(r, r ) = 4π 2 + The Green s function is then; G = m 4π 2 + r < r + > Aternativey expand the factor, Choose r > r and r r (+) r < r r r (+) r > r Y m (θ, φ)y m (θ, φ ) r r r r = (/r) (r /r) cos(θ) + (r /r) 2 r r = m r r + P (cos(θ)) When comparing this to the expression for the Green s function above, it automaticay gives the addition theorem. P (cos(α)) = m 4π 2m + Y m (θ, φ)y m (θ, φ ) Return to the expansion of the vector form Appy a power series expansion in powers of r /r when r > r. r r. Figure 3 shows the geometry. R = r[ (r /r)cos(θ) + (r /r) 2 ] /2 R = (/r) + (r /r)cos(θ) + (r /r) 2 [/2(3cos 2 (θ) )] + R = R = m (r /r + ) P (cos(θ)) f m, (r, r ) Y m (ˆr)Y m (ˆr ) 8
z r θ R r x y Figure 3: Geometry of the expansion for r r q a 2q z R y Potentia Point a q x Figure 4: Geometry for the potentia probem 8 Exampe From the figure note that the potentia can be represented by; [ ] V = 4πǫ 2q q R + R a + q R + a Then by expansion in Legendre poynomias where R > a; R a = a R + P (cos(θ)) Substitute to obtain; V = 4πǫ Here x = cos(θ) [ 2q R + q a R + P (x) + q ( ) a R +a ] P (x) 9
V = 2q 4πǫ =2; even a R + P (x) The owest non-zero term in the series represents the mutipoe moment of the charge distribution. 9 Exampe2 In the deveopment of scattering probems, we wi need the expansion of a pane wave in spherica coordinates. Note that the [ane wave is symmetric in azimutha ange. Thus; e ikz = e i k r = n c n Y 0 n Use ortho-normaity to project out the coefficients c n = 2π π 0 sin(θ) dθ Y 0 n eikr cos(θ) Appy the integra representation of the spherica Besse function; This gives; j n (z) = (/2)( i) n π 0 eiz cos(θ) P n (θ) sin(θ) dθ e ikr cos(θ) = i [4π(2 + )] /2 j (kr) Y 0 (θ) By the addition theorem; e ikr cos(θ) = 4π m j (kr) Y m (ˆk)Y m (ˆr) Figure 5 shows the geometry of the addition theorem. 0 Anguar momentum Negecting mass, the momentum operator in QM has the form i. The anguar momentum operator then has the form; L = r i r p The energy operator in spherica coordinates is; 0
z θ k θ α r y x φ φ Figure 5: The geometry of the addition theorem T = 2 2M 2 = 2 2M [ 2 r + 2 ] where the Lapacian operator is divided into operations on the radia variabe, r, and the anguar variabes, (θ, φ). In Cartesian coordinates; L x =, i [y z z y ] L x =, i [x z x z ] L x =, i [x y y z ] The map between Cartesian and spherica coordinates is; x = r sin(θ) cos(φ) y = r sin(θ) sin(φ) z = r cos(θ) Make a change of variabes for the Cartesian anguar momentum operators above. L x = i [sin(φ) θ L y = i [ cos(φ) θ + cot(θ) cos(φ) φ ] + cot(θ) sin(φ) φ ]
L z = i φ Then ook at L 2 = L L L 2 = 2 [ sin(θ) θ [sin(θ) θ ] + sin 2 (θ) This can be obtained by using; L = i r = ˆr r r r r 2 Then we have the eigenvaue equation; L 2 Y m (θ, φ) = 2 ( + )Y m (θ, φ) 2 φ 2 ] The commutation rues for the anguar momentum operator can be worked out as we using the above description of the operators. [L i, L j ] = iǫ ijk L k L ± = L x ± il y [L z, L ± ] = ± L ± [L +, L ] = 2L z L z Y m = m Y m L z L ± Y m = (m ± )L ± Y m L 2 commutes with L ± so from the above, L + (L ) is a raising (owering) operator on the eigenvaue m in the eigenfunction Y m Vector spherica harmonics When we dea with the vector Lapacian, it is usefu to introduce vector spherica harmonics of the form; χ m = ( + ) LY m 2
The operator L wi be the same as the anguar momentum operator without the mutipying constant of. Thus L is defined by L = i[ r ]. Then; L 2 = [ sin(θ) θ (sin(θ) θ ) + sin 2 (θ) L 2 Y m = ( + )Y m 2 φ 2 ] A other operations foow in the same way as was the case of the anguar momentum terms. The χ functions as defined above are ortho-norma. N 2 = Then; LY m ( + ) = i r Y m dω( LY m ) LY m Substitute into the integrand to obtain; N 2 = ( + ) dωr 2 [ Y m m Y ] Integrate by parts to produce a 2 operator operating on Y m. The resut is; dω χ m χ m = δ δ m,m In the case of non-cartesian coordinates the vector Lapacian shoud be used when operating on a vector fied. We wi then use the expansion of a vector fied in a new set of functions caed vector spherica harmonics. Define 3 vector operators to span 3-D space; V m = ˆr[ ( 2 + + )/2 Y m ] +ˆθ[ Y m [( + )(2 + )] /2 θ ] + ˆφ[ im [( + )(2 + )] /2 sin(θ) Y m W m χ m = ˆr[( 2 + )/2 Y m ] + ˆθ[ Y m [()(2 + )] /2 θ ] + ˆφ[ im [()(2 + )] /2 sin(θ) Y m ] = ˆθ[ m Y [()(2 + )] /2 m sin(θ) θ ] + ˆφ[ i Y m [()(2 + )] /2 θ ] These functions satisfy orthogonaity and are normaized; ] dω A m B m = δ ABδ δ mm Where A, B are any of the above functions. 3
2 Expansion of a vector fied Any vector fied can be expanded as; F = Ψ + LΨ 2 + cur L/i Ψ 3 To see this note that any vector can be written; F = Ψ + G Then et Q = i rψ 2 i LΨ 3 Ψ 4. Then take Q. Use the definition L = i[ r ] to show that; Q = LΨ 2 + L i In spherica coordinates; Q r = irψ 2 Ψ 4 r Ψ 3 Q θ = Ψ 3 sin(θ) φ (/r) Ψ 4 θ Q φ = Ψ 3 θ r sin(θ) Ψ 4 φ One can then demonstrate that the above equations may be soved for Ψ 2, Ψ 3, Ψ 4. Then for the vector potentia, A, we can write; A = Ψ + LΨ 2 + L i Ψ 3 However, when considering the magnetic fied using Couomb gauge, A = 0. Given the definition of the operator, L, the ony zero component is represented by, [ LΨ 2 ] = 0 3 Exampe3 Appy the vector spherica harmonics to the cacuation of the magnetic fied of a current oop. From Ampere s aw in steady state; B = A = µ J 4
z V = F( θ ) a V = 0 x Figure 6: The potentia inside a hemisphere over a fat pane In the above, B is the magnetic fied, A is the vector potentia, and J is the current density. We choose the Couomb gauge, A = 0. This reduces the above to; A R(r) χ m Then 2 A separates to obtain; R [r2d2 R dr 2 + (2r) dr dr ] = χ m [L 2 ]χ m = ( + ) The remainder of the soution is obtained by matching boundary conditions for A or by using the Green s function. When m = 0 (azimutha symmetry) and for the soution to remain finite as r the soution wi have the form; F = a (/r + ) [ i dy 0 (( + )) /2 dθ ˆφ] 4 Exampe4 Consider the potentia due to a semi-circuar screw head on a fat conducting pane. The geometry is iustrated in Figure 6. We ook for soutions to Lapace s equation, 2 V = 0, for the potentia in a spherica coordinate system. In genera the soution must have the form; V = m A m r Y m + m B m r (+) Note that whie the genera soution shoud incude the Legendre functions of the 2 nd kind as we as functions of the st kind the boundary condition that the soution be finite aong the z axis excudes the Q functions. The soution is aso symmetric in azimutha ange so m = 0. If we wish soutions inside the a hoow screw head, then the soution must be finite at r = 0. In this case the soution takes the form; V = A r P 5
For the potentia to vanish when θ = π/2 choose ony vaues of =, 3, 5,. V = n=0 A n r 2n+ P 2n+ (cos(θ)) Finay, match the boundary condition, F(x) with x = cos(θ) when r = a, by projecting out of the sum the coefficients, A n A n = 2n + 2a (2n+) dxf(x) P 2n+ (x) On the other hand suppose we seek a soution for r > a. In this case, the soution shoud have the form; V = B r (+) P (x) As above, we wish the soution finite on the z axis, azimuthay symmetric and finite as r. We wi need to change the boundary conditions sighty to find a stabe soution in spherica coordinates as the appication of Cauchy (Neumann) conditions must be appied on a cosed surface. The boundary probem is reay due to a mixture of boundary conditions on spherica and cyindrica surfaces. Thus choose to put the potentia on the surface at z = 0 to a constant vaue which we can choose as V = 0. There wi be another potentia, V = V 0 on a pane which we take as r. Then the boundary conditions are ; V = 0 for z = 0, and r = a V = E z z as r The boundary condition V = E z z is reay a Neumann condition specifying the eectric fied, E = V. Then we must have a soution of the form; V = E z [r cos(θ)] + n B n r (2n+2) P 2n+ (x) 5 Exampe5 As another exampe, consider a dieectric sphere of dieectric constant, ǫ, in a uniform fied in the ẑ direction, E = E z ẑ. The boundary conditions are; V = E z z = E z r cos(θ) as r Tangentia E is continuous at r = a 6
V = E_{z} z z V = E_{z} z a x V = E_{z} z V = E_{z} z Figure 7: The potentia of a dieectric sphere in a uniform eectric fied Perpendicuar D = ǫ E is continuous at r = a The boundary conditions at r = a are derived for a dieectric without surface charge or currents from Gauss and Ampere s aws. The soution must have the form after appying azimutha symmetry and requiring a finite soution on the z axis (x = cos(θ)); V = A r P (x) + B r (+) P (x) To match the boundary condition as r require the soution to be; V = E z r cos(θ) + B r (+) P (x) Then match the boundary conditions at r = a. The boundary condition that tangentia E is continuous is the same as requiring a continuous potentia at r = a. A a P = E z r P + For D r to be continuous; ǫ V r in = ǫ V r out B a (+) P ǫ r A a P = B ( + ) a (+2) P = E z P Sove for the coefficients; A 0 = 0 B 0 = 0 A = 3 ǫ r + 2 E z 7
B = (ǫ r )a 3 ǫ r + 2 E z A other coefficients vanish. The potentia is; Inside r < a V = 3E z ǫ r + 2 r cos(θ) Outside r > a V = E z r cos(θ) + [ ǫ r + ǫ r + 2 ] E z a3 r 2 cos(θ) 6 Mutipoe EM fieds We ook for soutions to Maxwe s equations with the time component suppressed. Let k = ω/c. The Hemhotz equation is; 2 B + k 2 B = 0 The operator, 2 in the above is the vector Lapacian. We want a set of spherica harmonic soutions (mutipoes) for B ( or for E. Each component of B satisfies the Hemhotz equation as we as div B = 0 and we want to find soutions in terms of the spherica Hanke functions as these have the correct boundary conditions as r. The soution is then expected to have the form; B a m h L Y m In the above h is the spherica Hanke function giving the radia dependence as r and the anguar term is the vector spherica harmonic χ m whose divergence vanishes. This soution then satisfies the Hemhotz equation has appropriate boundary conditions as r, and has B = 0. The soution for the eectric fied is simiar, and to get a fuy compete set of functions we need to combine the fieds, (ie the magnetic fied as above and the magnetic fied component due to the eectric fied E = iωb, and of course the eectric component due to the magnetic fied). Therefore the compete soution takes the form; B = m A(E) m f χ m (i/k) A(M) m g χ m E = (i/k)a(e) m f χ m m + A(M) m g χ m 8
For a wave moving radiay outward, f, g are the Hanke functions of the first kind. The coefficient A(E) [A(M)] represents the amount of eectric [magnetic] mutipoe in the soution. A(E) means the source is due to a charge density and A(M) is due to a magnetic moment density. 9