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Unformatted text preview: Electromagnetic Theory I Problem Set 7 Due: 10 October 2011 25. J, Problem 3.23. Note that you are actually obtaining various representations of the Dirichlet Green function (just set q = ) for the interior of a cylinder when you do this problem. In particular, the form in part c) is an expansion of the Green function in a complete set of functions in all three coordinates as we discussed in class. Solution : The potentials of this problem are just q/ , ( q/ 4 ) times my (Jacksons) Green functions for this geometry. The first two expansions a) and b) are adaptations, to a finite range in : 0 < < a , of the Green functions for the region between two parallel planes, that we obtained in problems 22a) and 23c) in Problem Set 6 respectively. The easiest one to obtain is the second form which differs from that in 23c) by the following extra term in the radial factor 1 L I m ( n < /L ) I m ( n > /L ) K m ( na/L ) I m ( na/L ) = 1 L I m ( n/L ) I m ( n /L ) K m ( na/L ) I m ( na/L ) . (1) The equality follows because the left side is symmetric in < > . This term has no singularity at = and satisfies Laplaces equation throughout the region. Its role is to make the potential vanish at > = a . Since the new term has no singularity at = , it doesnt contribute to the delta function. The first (discontinuous) term is exactly the same as in problem 23c) so it produces the proper delta function. Thus the correctness of the second expansion is proved. The first expansion differs from that in 22a) by the replacement of the integral over continuous k by a sum over the discrete values k mn = x mn /a , so that the Bessel functions in the expansion vanish at , = a . We must only check that the new (constant) factors are such as to produce the proper delta function. From the orthogonality condition, Jackson (3.95). it follows that the completeness condition for discrete k is 2 a 2 summationdisplay n =1 J m ( x mn /a ) J m ( x mn /a ) J 2 m +1 ( x mn ) = 1 ( ) (2) Comparing this to the continuous k completeness relation, 1 ( ) = integraldisplay kdkJ  m  ( k ) J  m  ( k ) (3) we see that the extra constant factors 2 / ( a 2 k mn J 2 m +1 ( x mn )) = 2 / ( ax mn J 2 m +1 ( x mn )) in the first expansion, compared to those in 22a), are precisely what is needed to give the correct normalization of the delta function. 1 Finally the third expansion involves a triple sum of eigenfunctions of 2 with eigenval ues ( x mn /a ) 2 + ( k/L ) 2 . Each term satisfies the boundary conditions in , z and periodicity in . Applying 2 to the expansion leaves the representation for the product of delta functions 1 ( ) ( ) ( z z ) already discussed. To obtain either of the first two expansions from this triple sum, either the sum over k or the sum over...
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This note was uploaded on 01/08/2012 for the course PHY 6346 taught by Professor Staff during the Spring '08 term at University of Florida.
 Spring '08
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