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Lecture02_part2_V1.2

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

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PHY501 Lecture II-Part 2 Boundary Value Problems (continued) ±²³´µ¶ ·¸¹¹ ºµ¶¶ »¼½¾¿ ºÀ ÁÂÃÄÅ¸ÆÇ È¸¹Ã¼ ±ÆÂÉ¹¼ÊË V1.2

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±²³´ !µ ¶·¸¹º»¼½ ¾»¿¸² À¼·Á¿²ÂÃ ! Spherical Harmonics The Laplacian in spherical polar coordinates is ± 2 ² = 1 r 2 ³ ³ r r 2 ³ ² ³ r + 1 r 2 sin ´ ³ ³´ sin ´ ³ ² ³´ µ · ¸ ¹ º + 1 r 2 sin 2 ´ ³ 2 ² ³» 2 (64) ± ( r , ² , ³ ) = R ( r ) P ( ² ) Q ( ³ ) (65) We separate variables as before, but this time retain the possibility of azimuthal dependence R ( r ) = A ± r ± + B ± r ± + 1 (66) Once again And in azimuth we have 1 Q d 2 Q d ± 2 = ² m 2 ³ Q( ± ) = e ± im ± (67) If the problem spans the full range in azimuth, then we have a periodic boundary condition—i.e., Q( ± + 2 ² ) = Q( ± ) ³ m = integer (68) In we are left with d dx 1 ± x 2 ( ) dP dx + ± ± + 1 ( ) ± m 2 1 ± x 2 ( ) ² ³ ´ ´ µ · · P = 0 (69) ± ÄÅ² Æ²¹²¼»¿ÇÈ²º ±²É²¹º¼² ÊË¸»Ì·¹ Note that this reduces to the Legendre Equation if m=0 (no azimuthal dependence).
±²³´µ ¶· ¸¹º»¼½¾¿ À½Áº² Â¾¹ÃÁ²ÄÅ Æ Spherical Harmonics The solutions to Eq. 69 are called the Associated Legendre Functions and are denoted P ± m ( x ) = ( ± 1) m 1 ± x 2 ( ) m 2 d m dx m P ± ( x ) = ( ± 1) m 2 ± ± ! 1 ± x 2 ( ) m 2 d ± + m dx ± + m x 2 ± 1 ( ) ± (70) There is an orthogonality relation The usual practice is to combine P and Q into a single overall function called the Spherical Harmonic Where the coefficient under the radical has been chosen such that It is left as an exercise to show that I ±± ' ± P ± ' m x ( ) P ± m x ( ) ² 1 1 ³ dx = 2 2 ± + 1 ± + m ( ) ! ± ² m ( ) ! ´ ± ' ± (71) Y ( ± ) = Y ± m ( ² , ³ ) = 2 ± + 1 4 ´ ± µ m ( ) ! ± + m ( ) ! P ± m cos ² ( ) e im ³ (72) d ± 4 ² ³ Y ± ' m '* ( ´ , µ ) Y ± m ( ´ , µ ) = ± ' ± m ' m (73) Y ± m * ( ± ', ² ') Y ± m ( ± , ² ) m = ³ ± ± ´ ± = 0 µ ´ = ¶ ² ³ ² ' ( ) ¶ ± ³ ± ' ( ) (74)

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Spherical Harmonics ±²³´µ¶ ·¸¹¹ ºµ¶¶ »¼½¾¿ ºÀ ÁÂÃÄÅ¸ÆÇ È¸¹Ã¼ ±ÆÂÉ¹¼ÊË Ì Y 0 0
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Lecture02_part2_V1.2 - V1.2 PHY501 Lecture II-Part 2...

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