# Does the potential of a charged ring diverge on the ring?

I know that the density and potential (in spherycals) of a charged ring is, respectively,:

$$\rho(\textbf{r}) = \frac{\lambda}{a} \delta(r-a)\delta(\theta-\tfrac{\pi}{2})$$

$$\varphi(\textbf{r})= \frac{2\pi a \lambda}{r_>} \left[ 1+ \sum_{n=1}^\infty (-1)^n \frac{(2n-1)!!}{(2n)!!}\left(\frac{r_<}{r_>}\right)^{2n}P_{2n}(\cos\theta) \right]$$

Where $$P_{2n}$$ is the $$2n$$-th Legendre Polynomial, and $$r_<=\min\{a,r\},r_>=\max\{a,r\}$$. If I evaluate $$\mathbf r$$ in the ring ($$r=a,\theta=\tfrac{\pi}{2}$$):

$$\varphi(\mathbf r)\,\propto\, \left[ 1+ \sum_{n=1}^\infty (-1)^n \frac{(2n-1)!!}{(2n)!!} \right] \to\infty$$

So this is a problem (I suppose).

• Actually, I'm rethinking my answer. Are you sure that your sum for the energy does not also have a term like $P_{2n}(0)$? My answer was assuming that the sum was as you stated, but in trying to derive your result, I seem to have gotten something different. Jun 4, 2020 at 12:38
• @Philip Thank you so much for your answer, you have a recurrence relation for the Polynomials Legendre $(n+1)P_{n+1}=(2n+1)xP_n(x) - nP_{n-1}(x)$, and $P_0(0)=1,\quad P_1(0)=0$, so you can say $P_{2n}(0)=(-1)^n\frac{(2n-1)!!}{(2n)!!}$ and $P_{2n+1}(0)=0$ for $n>0$ Jun 4, 2020 at 17:22
• Oh great! I never paid as much attention as I should have to recurrence relations. But in that case, would the formula for the potential $\varphi(\mathbf{r})$ also have the same double factorial terms in it? I'm afraid that given your formula for $\varphi$, I can't derive your formula for $U_E$. It might be a typo? Also, does my answer answer your question? :) Jun 4, 2020 at 17:43
• It is the continous form of the energy for discrete charges. You answer my question, thank you. Jun 4, 2020 at 21:41

This is actually a fun question, I learnt something new about double factorials while trying to answer it!

I don't see why that term diverges. Using the identities on Wikipedia for the "double factorial", we have that for even integers $$k$$,

$$\int_0^{\pi/2} \sin^{k}(x)\text{d}x = \frac{(k-1)!!}{(k)!!}\frac{\pi}{2}.$$

We can use this to calculate the sum term you have explicitly.

$$\sum_{n=1}^\infty (-1)^n \frac{(2n-1)!!}{(2n)!!} = \frac{2}{\pi}\sum_{n=1}^\infty (-1)^n \int_0^{\pi/2} \sin^{2n}(x)\text{d}x = \frac{2}{\pi}\int_0^{\pi/2} \text{d}x \sum_{n=1}^\infty (-1)^n \sin^{2n}(x).$$

Where in the last step I've interchanged the sum and the integral. This particular sum is quite easy to do, and I'll leave it as an exercise to show that

$$\sum_{n=1}^\infty (-1)^n \sin^{2n}(x) = -\frac{\sin^2(x)}{1+\sin^2(x)}.$$

We can now perform the integral and show that $$-\frac{2}{\pi}\int_0^{\pi/2} \frac{\sin^2(x)}{1+\sin^2(x)} \text{d}x = \frac{-2 + \sqrt{2}}{2}.$$

Thus, $$\sum_{n=1}^\infty (-1)^n \frac{(2n-1)!!}{(2n)!!} = \frac{-2 + \sqrt{2}}{2} < \infty,$$

• @Elborito I think it might be better to just edit this question, perhaps? I think repeating questions is frowned up. Also, I am not sure how you got the $$\left(\frac{(2n-1)!!}{(2n)!!}\right)^2$$ term, since I only got one which came from the Legendre Polynomial recurrence relations. Perhaps you could edit the question with a little bit more of your working? Jun 9, 2020 at 15:07