{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# emsol07 - Electromagnetic Theory I Problem Set 7 Due 10...

This preview shows pages 1–3. Sign up to view the full content.

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 = ǫ 0 ) 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/ǫ 0 , ( q/ 4 πǫ 0 ) times my (Jackson’s) Green functions for this geometry. The first two expansions a) and b) are adaptations, to a finite range in ρ : 0 < ρ < 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 πL I m ( nπρ < /L ) I m ( nπρ > /L ) K m ( nπa/L ) I m ( nπa/L ) = 1 πL I m ( nπρ/L ) I m ( nπρ /L ) K m ( nπa/L ) I m ( nπa/L ) . (1) The equality follows because the left side is symmetric in ρ < ρ > . This term has no singularity at ρ = ρ and satisfies Laplace’s equation throughout the region. It’s role is to make the potential vanish at ρ > = a . Since the new term has no singularity at ρ = ρ , it doesn’t 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 ρ, ρ = 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 ρ/a ) J m ( x mn ρ /a ) J 2 m +1 ( x mn ) = 1 ρ δ ( ρ ρ ) (2) Comparing this to the continuous k completeness relation, 1 ρ δ ( ρ ρ ) = integraldisplay 0 kdkJ | m | ( ) J | m | ( ) (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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Finally the third expansion involves a triple sum of eigenfunctions of −∇ 2 with eigenval- ues ( x mn /a ) 2 + ( kπ/L ) 2 . Each term satisfies the boundary conditions in ρ, z and periodicity in ϕ . Applying −∇ 2 to the expansion leaves the representation for the product of delta functions 1 ρ δ ( ρ ρ ) δ ( ϕ ϕ ) δ ( 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 n must be explicitly performed. Since all forms have been proved, we can infer the result of that summation by comparison.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 8

emsol07 - Electromagnetic Theory I Problem Set 7 Due 10...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online