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emsol08 - Electromagnetic Theory I Solution Set 8 Due 24...

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Electromagnetic Theory I Solution Set 8 Due: 24 October 2011 29. J, Problem 4.4 Solution : a)The 2 l moment about a point r 0 has the form Q i 1 ··· i l r 0 = A l integraldisplay d 3 r [( r r 0 ) i 1 · · · ( r r 0 ) i l Traces ] ρ ( r ) = Q i 1 ··· i l 0 r i 1 0 Q i 2 ··· i l 0 r i 2 0 Q i 1 i 3 ··· i l 0 + · · · (1) where the dots signify terms of higher order in r 0 as well as terms with different indices on r 0 . But if the coefficients of the positive powers of r 0 are proportional to a lower multipole moment than the 2 l moment and l is the lowest nonvanishing moment all these terms are zero and we would have Q i 1 ··· i l r 0 is independent of r 0 . It is clear that those coefficients involve an integral of powers of r i times the charge density, but it is not obvious that the coefficients are traceless in all pairs of indices, as the multipole moments must be. To see that they are actual multipole moments it is better to consider alternative multipole expansions of the potential due to the charge distribution: φ = 1 4 πǫ 0 integraldisplay d 3 x ρ ( r ) | r r | = 1 4 πǫ 0 integraldisplay d 3 x ρ ( r ) | r r 0 ( r r 0 ) | summationdisplay n =0 r i 1 · · · r i n Q i 1 ··· i n n ! r 2 n +1 = summationdisplay n =0 ( r r 0 ) i 1 · · · ( r r 0 ) i n Q i 1 ··· i n r 0 n ! | r r 0 | 2 n +1 (2) Now take r → ∞ on both sides. Then if n = l is the lowest nonvanishing moment, the equality tends to r i 1 · · · r i l Q i 1 ··· i l l ! r 2 l +1 = r i 1 · · · r i l Q i 1 ··· i l r 0 l ! r 2 l +1 (3) which implies Q i 1 ··· i l = Q i 1 ··· i l r 0 . By expanding both sides of the equality in powers of r r 0 , one sees that the higher nonvanishing multipole Q i 1 ··· i n r 0 is equal to Q i 1 ··· i n + powers of r 0 times lower multipole moments.
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