Suppose that you're doing electrostatics in two dimensions (or, equivalently, in 3D with translation invariance along $z$) and you're studying the response to a point charge in an angle sector of opening angle $\alpha$, i.e. in the domain $$ \mathcal D=\{(r\cos(\theta),r\sin(\theta))\in\mathbb R^2:0<\theta<\alpha\}, $$ under Dirichlet boundary conditions, i.e. between two conducting planes at dihedral angle $\alpha$.

Now, if you're actually faced with such a problem, the natural response is to say "ah, yeah, you can probably get the Green's function of the problem in terms of some expansion in terms of fractional-order Bessel functions something something something", and nobody can really remember the details, but luckily we have Jackson's Classical Electrodynamics, which informs us in §2.11 and in problem 2.25 that the Green's function for the problem is \begin{align} G(\mathbf r,\mathbf r') = 4 \sum_{n=1}^\infty \frac1n \frac{r_<^{n\pi/\alpha}}{r_>^{n\pi/\alpha}} \sin\left(\frac{n\pi}{\alpha}\theta \right)\sin\left(\frac{n\pi}{\alpha}\theta' \right), \tag 1 \end{align} satisfying $\nabla^2_\mathbf r G(\mathbf r,\mathbf r') = \delta(\mathbf r-\mathbf r')$ (up to signs and factors of $\pi$), where the arbitrary-real-number nature of the corner angle $\alpha$ manifests itself in the non-integer powers of the radii involved in the expansion.

On the other hand, for some special angles $\alpha$ of the form $\alpha = \pi/k$, where $k$ is an integer, one can just ditch all of that and simply use the method of image charges, by judiciously planting copies of our source charge at angles between $\alpha$ and $2\pi$ in such a way that the potential vanishes at the boundary of $\partial \mathcal D$ like it needs to, giving us the Green function in the form $$ G(\mathbf r,\mathbf r') =\sum_{i}q_i\ln(\|\mathbf r- \mathbf r_i\|) \tag 2 $$ where one of the $\mathbf r_i$ is $\mathbf r'$ and the rest are image charges.

OK, so after all that setup, here's my question: is it possible to directly reduce the general result in $(1)$ to the method-of-images result in $(2)$ through some means or another? Intuitively, it feels like that should be the case (though there's no promises that it'd be pretty), because $(1)$ contains all the information about the problem's solution and it should be possible to whittle it down to simpler versions where those exist; more prosaically, when $\alpha=\pi/k$ those exponents now read $r^{n\pi/\alpha} = r^{nk}$, which looks rather more manageable.

Along these lines, Jackson does add one tack to that problem 2.25, saying

(b) By means of complex-variable techniques or other means, show that the series can be summed to give a closed form, $$ G(\mathbf{r},\mathbf{r}') = \ln\left[\frac{ r^{2\pi/\alpha}+r'^{2\pi/\alpha}-2(rr')^{\pi/\alpha}\cos(\pi(\theta+\theta')/\alpha) }{ r^{2\pi/\alpha}+r'^{2\pi/\alpha}-2(rr')^{\pi/\alpha}\cos(\pi(\theta-\theta')/\alpha) }\right], $$

which looks like a step in the right direction, but (i) I can't fully see how you'd sum that series, and more importantly (ii) I'm not fully sure how you'd reduce that simplified series to the method-of-images solution.

So: is this possible? and if so, how? Also, a bit more ambitiously: does that still hold in 3D, where the Green's function goes up into Bessel expansions but the method of images still works?


For Question (i), note that $$ \sin \left( \frac{n \pi}{\alpha} \theta \right) \sin \left( \frac{n \pi}{\alpha} \theta' \right) = \frac{1}{2} \left[ \cos \left(\frac{n \pi}{\alpha} (\theta - \theta') \right) - \cos \left(\frac{n \pi}{\alpha} (\theta + \theta') \right)\right] \\ = \frac{1}{2} \Re \left[ e^{n \pi i (\theta - \theta')/\alpha} - e^{n \pi i (\theta + \theta')/\alpha} \right] $$ and so $$ G(\mathbf r,\mathbf r') = 2 \Re \left\{ \sum_{n=1}^\infty \frac1n \left[\left(\frac{r_<}{r_>} e^{i (\theta - \theta')}\right)^{n\pi/\alpha} - \left(\frac{r_<}{r_>} e^{i (\theta + \theta')}\right)^{n\pi/\alpha} \right]\right\} $$ The summation can now be seen to be equivalent to two power series for natural logarithms, since $$ \ln(1 - z) = \sum_{n=1}^\infty \frac{1}{n} z^n. $$ After much algebra (which I won't go into here, since this kind of qualifies as a "homework-like question"), this should reduce to the form given in the Jackson exercise.

I'll think a bit more about Question (ii) and see if I can come up with anything. It's probably worth noting that if the original charge $q$ is at coordinates $(r', \theta')$, then the angular positions of the image charges in the corresponding problem are $\theta', \theta' + 2\alpha, \theta' + 4 \alpha, ... \theta' + 2(k-1)\alpha$ for the positive charges, and $-\theta', -(\theta' + 2\alpha), -(\theta' + 4 \alpha), ... -(\theta' + 2(k-1)\alpha)$ for the negative charges. This means that (2) can be written as $$ G(\mathbf r,\mathbf r') = \frac{1}{2} q \sum_{n=0}^{k-1} \ln \left[ \frac{r^2 + (r')^2 - 2 r r' \cos( \theta - (\theta' + n \alpha)) }{r^2 + (r')^2 - 2 r r' \cos( \theta + (\theta' + n \alpha))}\right], $$ which looks like a step in the right direction, but it's still not clear to me how you'd go from this equation to the form written in the Jackson exercise.

  • $\begingroup$ Yeah, that's probably plenty in terms of getting between the two forms in Jackson, and the work back from the method-of-images result is in the right direction, but there's still some more bridges to build there. $\endgroup$ – Emilio Pisanty Aug 2 '17 at 21:09

Michael Seifert's answer is basically correct, and I'll just fill in a few steps missing in the answer to Question (ii).

In terms of Jackson's Green function, the potential due to the charge is: $$V(\mathbf r, \mathbf r') = \frac{q}{4\pi\epsilon_0}G_{Jackson}(\mathbf r, \mathbf r')$$ To obtain the same result using the method of images, the expression for the Green's function must be properly normalised. The original charge is actually a line charge (in 3D), and so will be the image charges. The potential is therefore the sum of potentials of line charges: $$V(\textbf r, \textbf r')=-\frac{1}{2\pi\epsilon_0}\sum_{i}q_i\ln(\|\mathbf r- \mathbf r_i\|)$$ Arranging the image charges as suggested by Michael Seifert, that is placing positive charges at locations $\theta'+2n\alpha$ and negative charges at $-(\theta'+2n\alpha)$ for $n = 0, \ldots, k-1$ we obtain the following expression: $$V(\mathbf r,\mathbf r') = \frac{q}{4\pi\epsilon_0} \sum_{n=0}^{k-1} \ln \left[ \frac{r^2 + (r')^2 - 2 r r' \cos( \theta + \theta' + 2n \alpha)}{r^2 + (r')^2 - 2 r r' \cos( \theta - \theta' - 2n \alpha) }\right]$$ This differs from the one obtained in the other answer because the normalisation is different, numerator and denominator in the logarithm have been swapped, and we have $2n\alpha$ instead of $n\alpha$ in the argument of the cosine (which was probably only a typo).

At this point, to show that the expression in Jackson and that obtained with the method of image charges agree, it is sufficient to show that:

$$\prod_{n=0}^{k-1} \left[z^2 + 1 - 2 z \cos( \beta + 2n \pi/k)\right]=\left[z^{2k} + 1 - 2 z^k \cos(k\beta)\right]\tag{*}$$ where we've taken $z=r/r'$ and $\beta = \theta +\theta'$. The same proof works for the denominator as well.

Let $u = z e^{i\beta}$ and $v_n = e^{2in\pi/k}$. Then we can write the left hand side of ($*$) as: $$\prod_{n=0}^{k-1} |u-v_n|^2=\left| \prod_{n=0}^{k-1} (u-v_n) \right|^2=\left| u^k-1 \right|^2$$ which is the expression on the right hand side of ($*$). The second equality follows because the $v_n$ are the $k^{\mathrm{th}}$ roots of unity, so the polynomial in $u$ is cyclotomic.


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