Approximation of solution for Laplace's equation in 5d (Kaluza-Klein)

Question

I apologize for the following question because it will seem like a cheap please help me with my homework one.

I just want a hint as to what direction to follow. Suppose we have a 5d space where the 5th dimension is periodic. It models a circle of radius R. We will take R to be very small, this setup reminded me of Kaluza-Klein theory. A solution for the Laplace equation is

$$V=-Q \sum\limits_{n=-\infty}^{n=\infty} \frac{1}{r^2+(y+2\pi n R)^2},$$ $r$ is the length for the other 4 dimensions. The exercise asks me to simplify the problem by approximating the sum to an integral with respect to y, assuming $R \to 0$. My question is how to do this approximation systematically, should I use something like Euler-Maclaurin formula and ignore the error? My problem here is that we sum with respect to n, so the integral won't be with respect to y. Another idea a had is to first expand the function $\approx \frac{1}{r^2+y^2}-\frac{4 \pi n y R}{{(r^2+y^2)}^2}$, but this seems to make things worse when I try to take the sum (first term is infinite). Maybe there is a trick I am unaware of, my intuition says that to first order we will still have $V=-Q/r$ but I can't replicate it.

Answer 1

You can build an intuition using basic electrostatics. Instead of thinking in terms of periodised space, imagine instead that you have real charges spaced by $2\pi R$ along a direction. As $R\to0$, if each charge goes to zero appropriately, you recover a continuous, uniform linear distribution of charge and should recover the Green's function of dimension minus one.

Mathematically, yes Euler-Maclaurin does the trick, but it's a bit of an overkill as it's just the first term of the expansion. Your sum is essentially a Riemann sum, on an unbounded domain, so depending on how you construct the integral, by definition you get the integral. From the physical insight, you rescale the charge as: $$ Q = 2\pi R\lambda $$ with $\lambda$ the uniform linear charge density. You can think in $D$ dimensions, you have the Green's solution (in your case $D=5$, you made a mistake in the formula, you were using the formula for $D=4$ without the numerical prefactor): $$ V = \frac{\Gamma(D/2-1)}{4\pi^{D/2}}\sum_{n\in\mathbb Z}\frac Q{(r^2+(y+2\pi nR)^2)^{D/2-1}} $$ As $R\to0$, you therefore get: $$ \begin{align} V &\sim \frac{\Gamma(D/2-1)}{4\pi^{D/2}}\int_{-\infty}^{+\infty}\frac{\lambda dt}{(r^2+(y+t)^2)^{D/2-1}} \\ &= \frac{\Gamma(D/2-1)}{4\pi^{D/2}}\frac\lambda{r^{D-3}}B\left(\frac{D-1}2-1,\frac12\right) \\ &= \frac{\Gamma(D/2-1)}{4\pi^{D/2}}\frac\lambda{r^{D-3}}\frac{\Gamma\left(\frac{D-1}2-1\right)\sqrt\pi}{\Gamma\left(\frac D2-1\right)} \\ &= \frac{\Gamma\left(\frac{D-1}2-1\right)}{4\pi^{(D-1)/2}}\frac\lambda{r^{D-3}} \end{align} $$ which is consistent with the physical insight. Conversely, when $R\to\infty$, you recover the Green's function for the entire space, so $R$ serves to interpolate between two dimensions.

Hope this helps.