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I have a question about a specific step in this problem that comes from 'University of Chicago Graduate Problems in Physics', Cronin, Greenberg, and Telegdi.

A long hollow cylindrical conductor of radius $a$ is divided into two parts by a plane through the axis, and the parts are separated by a small interval. If the two parts are kept at potentials $V_1$ and $V2$, show that the potential at any point within the cylinder is given by $$ V=\frac{\left(V_{1}+V_{2}\right)}{2}+2 \frac{\left(V_{1}-V_{2}\right)}{\pi} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2 n-1)}\left(\frac{r}{a}\right)^{2 n-1} \cos (2 n-1) \theta $$

When cross-referencing my solution with that of the textbook's, I see that I have the correct approach. Using the Poisson equation,

$$ \nabla^{2} V=0=\frac{\partial^{2} V}{\partial r^{2}}+\frac{1}{r} \frac{\partial V}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2} V}{\partial \theta^{2}} $$

and separation of variables yields:

$$V=\sum_{m=0}^{\infty}\left(\frac{r}{a}\right)^{m}\left[A_{m} \cos (m \theta)+B_{m} \sin (m \theta)\right]$$

Now

$$V(r=a, \theta)=\sum_{m=0}^{\infty}\left[A_{m} \cos (m \theta)+B_{m} \sin (m \theta)\right]$$

$B_m=0$ by symmetry since $V(\theta) = V(-\theta)$. $A_m$ is calculated using the orthogonality of trig. functions, and the boundary conditions of $V_1$ and $V_2$,

$$ \int_{-\pi / 2}^{\pi / 2} V_{1} \cos n \theta d \theta+\int_{\pi / 2}^{3 \pi / 2} V_{2} \cos n \theta d \theta=\frac{2\left(V_{1}-V_{2}\right)}{n} \sin \left(\frac{n \pi}{2}\right)=\pi A_{n} $$

but I don't understand how to correctly find the $A_0$ term. The solutions have the following statement:

$$ \pi\left(V_{2}+V_{1}\right)=2 \pi A_{0} $$

I'm not sure how this comes about.

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  • $\begingroup$ It would help if you published your starting equation, as well as the boundary conditions. If so, I'd gladly look at your solution. $\endgroup$
    – Gert
    Sep 18, 2020 at 17:04
  • $\begingroup$ I added the starting Poisson equation. The boundary conditions I refer to are the $V_1$ and $V_2$ of the two pieces of the conductor. $\endgroup$
    – user244484
    Sep 18, 2020 at 17:10
  • $\begingroup$ Thanks. Let me have a look at it now. $\endgroup$
    – Gert
    Sep 18, 2020 at 17:12
  • $\begingroup$ I don't think yu want $A_0$ (for $m=0$) I think your 'counter should start at $1$: look at the solution you quoted: it starts at $n=1$ $\endgroup$
    – Gert
    Sep 18, 2020 at 18:12
  • $\begingroup$ @Philip Tbh Philip, I'm not finding OP's result at all. My separ. of var. yields something different altogether.. So I need to solve that discrepancy first. $\endgroup$
    – Gert
    Sep 18, 2020 at 21:07

1 Answer 1

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This is pretty much exactly how you do a Fourier decomposition. The term $A_0$ is usually called the "DC" component of the decomposition, and is obtained by simply integrating the function that you're decomposing as a Fourier sum (in your case, it's $V(a,\theta)$).

You should be able to show that $$A_0 = \frac{1}{2\pi}\int_0^{2\pi} V(a,\theta)\text{d}\theta,$$

since all the sines and cosines integrate out to zero, leaving only a constant term behind. But this integral is quite trivial to do in this case, since $$V(a,\theta) = \begin{cases}V_1 \quad 0 < \theta < \pi \\ V_2 \quad \pi< \theta < 2\pi\end{cases},$$

and so $$2 \pi A_0 = \int_0^\pi V_1 \text{d}\theta + \int_{\pi}^{2\pi} V_2 \text{d}\theta = \pi\, (V_2 + V_1).$$

A much nicer (and quicker!) way to do it would be to realise that $A_0$ is the value of the potential when $r = 0$, which is exactly in between the two halves of the cylinder, at its centre. By symmetry, the potential at that point must be the average of the potential on either side, by cylindrical symmetry about the axis.

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    $\begingroup$ Very clear, thanks. +1 from me. $\endgroup$
    – Gert
    Sep 18, 2020 at 21:35
  • $\begingroup$ Thanks, the symmetry argument is what I had originally, but I haven't been able to convince myself that it works. I can see that the "potential must change smoothly from $V_1$ to $V_2$" if you are on the rim, $r=a$, but is this necessarily the case for all $r$? Is $V(r,0) = (V_1+V_2)/2$? Because if it isn't the case, then all the symmetry argument gives you is for $r=a$: $(V_1+V_2)/2 = \sum_{n=0}^\infty (1)^n A_n cos(0) = \sum_{n=0}^\infty A_n$. And I don't think this gives you $A_0$. $\endgroup$
    – user244484
    Sep 19, 2020 at 14:32
  • $\begingroup$ You're completely right, I was wrong. $A_0$ is the potential when $r=0$, i.e. $A_0 = V(0,\theta)$, the potential at the center of the cylinder, which, by symmetry, must be $\frac{V_1 + V_2}{2}$, since all the other contributions must cancel out. I'm changing it :) $\endgroup$
    – Philip
    Sep 19, 2020 at 16:44
  • $\begingroup$ I can finally confirm the whole solution to be correct. $\endgroup$
    – Gert
    Sep 20, 2020 at 0:52

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