Lecture02_part2_V1.2

Lecture02_part2_V1.2 - V1.2 PHY501 Lecture II-Part 2...

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Unformatted text preview: V1.2 PHY501 Lecture II-Part 2 Boundary Value Problems (continued) Spherical Harmonics The Laplacian in spherical polar coordinates is 2 1 1 = 2 r2 + 2 r r r r sin sin 1 + 2 2 r sin 2 2 (64) We separate variables as before, but this time retain the possibility of azimuthal dependence Once again (r, , ) = R(r)P( )Q( ) R(r) = A r + B r +1 (66) (65) And in azimuth we have 1 d 2Q = m2 Qd 2 Q( ) = e im (67) If the problem spans the full range in azimuth, then we have a periodic boundary condition--i.e., Q( + 2 ) = Q( ) In we are left with m = integer (68) d 2 dP (1 x ) dx + dx ( + 1) m2 P =0 2 (1 x ) (69) Note that this reduces to the Legendre Equation if m=0 (no azimuthal dependence). Spherical Harmonics The solutions to Eq. 69 are called the Associated Legendre Functions and are denoted dm P (x) = ( 1) (1 x ) P (x) m dx m d +m 2 ( 1) m = (1 x 2 ) 2 dx +m ( x 1) 2 ! m m m 2 2 (70) There is an orthogonality relation 1 I ' The usual practice is to combine P and Q into a single overall function called the Spherical Harmonic 2 ( + m)! P ' ( x ) P ( x ) dx = 2 + 1 ( m)! 1 m m ' (71) Y( ) = Y m( , ) = 2 + 1 ( m)! m P (cos )e im 4 ( + m)! (72) Where the coefficient under the radical has been chosen such that d Y m'* ( , )Y m ( , ) = ' 4 ' m' m (73) It is left as an exercise to show that Y m* ( ', ')Y m ( , ) = = 0 m= ( ') ( ') (74) Spherical Harmonics The first few spherical harmonics Y00 = Y11 = Y22 = 0 2 1 4 3 sin e i 8 1 15 sin 2 e 2i 4 2 3cos 2 ( 1) Y10 = Y21 = 3 cos 4 15 sin cos e i 8 (75) 1 5 Y = 2 4 Since the Y m ( , ) are an orthogonal set, we can expand any function of direction in terms of them--i.e., f( , )= = 0 m= A mY m ( , ) d Y m* ( , ) f ( , ) (76) (77) where Am = 4 Thus the general solution to a spherical polar coordinate problem can be written (r, , ) = = 0 m= ( A mr + B mr ( +1) )Y m ( , ) (78) Addition Theorem for Spherical Harmonics The addition theorem for spherical harmonics states 4 that m* m P (cos ) = 2 + 1 m= Y ( ', ')Y ( , ) z x (79) where the angles in Eq. 79 are shown the in the figure to the right. For a proof see Jackson Sec. 3.6. Inserting this into Eq. 63, we obtain x y x 1 x x' = = 1 r> =0 r< r> P (cos ) r< m* Y ( ', ')Y m ( , ) (80) 2 + 1 r> +1 ,m 4 Expansion of Green Functions in Spherical Coordinates In Sec. 2.6 of Jackson it is shown that the Green function for a sphere of radius a can be written G'( x, x ') = 1 x x' r' x a a x' r' 2 r= x where (81) r'= x ' This function was derived with the assumption that we have a physical charge outside the sphere and an image charge inside the sphere. This is the "exterior problem." To tackle the "interior problem" we must modify Eq. 81 slightly--i.e., G'( x, x ') = 1 x x' r x' b b x r 2 (82) b and r r' to be consistent with Jackson where a 3.9. If we then insert Eq. 80 into Eq. 82 we obtain G( x, x ') = r< 2 + 1 r> +1 ,m 4 (rr') b2 +1 Y m* ( ', ')Y m ( , ) (83) where in the second term in the brackets, we have 2 taken r< = rr' and r> = b (for a more rigorous derivation of Eq. 83, see Jackson Sec. 3.9). Expansion of Green Functions in Spherical Coordinates z As an example, consider a loop of total charge q of radius a. We describe the charge density as b a y q ( x ') = (r' a) (cos ') (84) 2 2 a x (You can check this to confirm that V ( x )dV '= q .) The general expression for the potential is (from Eq. 12) ( x) = 1 4 0 V G'( x, x ') ( x ')dV ' 0 S ( x ') G'( x, x ') da' (85) n' where ( x ') S = 0 and ( x ') and G( x, x ') are given. Thus 1 q 4 ( x) = 4 0 2 a 2 ,m 2 + 1 (cos ')Y m* ( ', ')Y m ( , )d(cos ')d 4 ( r' a) r' 1 r +1 r b 2 +1 r'2 dr' (86) Expansion of Green Functions in Spherical Coordinates Remarks on evaluating Eq. 86 1. All but the m=0 terms integrate to zero. 2. The azimuthal integral is 2 . 0* 3. (cos ') "selects" Y ( 2 , ') = 2 +1 P (0) 4 2 +1 P (cos ) 4. Y ( , ) becomes Y ( , ) = 4 5. (r' a) "selects" r' = a 0 0 Putting this all together Eq. 86 becomes ( x) = a q 4 0 +2 1 4 1 r +1 0 q 2 a2 r b 2 +1 2 +1 P (0)P (cos )2 2 +1 4 (87) 1 r +1 4 = a P (0)P (cos ) r b 2 +1 for r > a and r< = r. One can obtain the result for r > a by taking r> = r' Laplace's Equation in Cylindrical Coordinates We start with the Laplacian in cylindrical coordinates 2 2 ( , ,z) = 2 + 1 + 1 2 2 2 2 + z 2 = 0 (88) and follow the usual procedure*--i.e., we write which leads to ( , ,z) = R( )Q( )Z(z) k 2Z = 0 2 (89) (90) and d 2Z dz 2 Z(z) = e kz d 2Q + 2 d Q=0 Q(q) = e i (91) k to get for the R equation we substitute x 2 d 2 R 1 dR + + 1 2 R = 0 (92) 2 dx x dx x We could solve this using power series (see Jackson 3.7) but instead choose to write the solution in terms of Bessel functions--i.e., R( ) = AJ (k ) + BY (k ) *filling in the details is left to the reader (93) where J and Y are respectively Bessel and Neumann functions of the first kind. Bessel Function Digression Bessel Functions Note that J 0 = 1 and J m = 0 m 1 Ym (x 0) Neumann Functions Zeros of Bessel Functions: Laplace's Equation in Cylindrical Coordinates The general solution is then ( , ,z) = m,n [ Amn Jm (kmn ) + BmnYm (kmn )] e im e kmn z (94) where kmn corresponds to the n'th zero (or extremum) of the Bessel or Neumann function of order m. The details of the expansion will depend on the problem at hand. Orthogonality: if kmna is the n'th zero of a Bessel function of order m, then a J m ( kmn 0 ) Jm ( kmn' ) a2 2 d = [ J m +1 (k mn a)] 2 nn' (95) We can thus express an arbitrary function satisfying the boundary conditions as f( )= where n Amn J m (k mn ) (96) a 2 0 Amn = 2 a [ J m +1 (k mn a)] 2 f ( )J m ( k mn' ) d (97) ...
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This note was uploaded on 02/04/2012 for the course PHY 501 at Princeton.

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