# Inverse square law in 2+1 dimensional universe from a Yukawa coupling?

There is a nice result that in 3+1 space time, a Yukawa coupling leads to an inverse square law force as the mass of the scalar field goes to zero. I was wondering what the corresponding force in a 2+1 dimensional Universe would be? Geometrically, I would guess that it should be $1/r$ rather than $1/r^2$, but my explicit calculation for a Yukawa coupling gives that it is $1/r^2$ in 2+1 space as well. I was wondering if that was correct?

## 1 Answer

It's log(r) potential in 2d space, so a 1/r force in the massless limit. The Yukawa potential is computed here: https://physics.stackexchange.com/questions/32337/yukawa-potential . The reduction to 2+1 dimensions is the Fourier transform of the 2 dimensional projection of the propagator for $\omega=0$,

$$V(r) = \int {e^{ik_x r} \over k_x^2 + k_y^2 + m^2 } {dk_x dk_y\over (2\pi)^2}$$

Where I have made the $2\pi$ factors explicit (they are implicit in the k-measure in the linked answer) and I have placed the separation vector between the two sources in the x-direction. Although this requires special functions when m is nonzero, it becomes elementary when you plug in m=0. It's the solution to Laplace's equation in 2d:

$$\nabla^2 V(x) = \delta^2(x)$$

And this means, by Gauss's law in 2d that the magnitude of the radial gradient of V (using a Gaussian circle) is given by

$$|\nabla V| = {1\over 2\pi r}$$

Which integrates to a log potential

$$V(r) = {1\over 2\pi} \log(r)$$

This potential is divergent at both small and large r, which makes the Fourier integral a little tricky to do explicitly from the integral. But you can do it also, if you know that

$$\lim_{k_{max}\to\infty}\int_0^{k_{max}} \sin(kx) dk = \lim_{k_{max}\to\infty}{1-\cos(k_{max}x)\over x}={1\over x}$$

This integral is oscillatory and ill defined in the Riemann sense, but it oscillates around the right-hand side value, which is the value one should use. This can be justified with lattices, but it is best to define the propagator directly as solution to the associated equation of motion.

So the massless propagator in d Euclidean dimensions is always the solution to Laplace's equation in d-dimensions, and the Yukawa potential in d+1 dimensions is always the Euclidean reduction in time, so it is the propagator is d Euclidean dimensions. The propagator in d dimensions is found by Gauss's law using a d-1 dimensional Gaussian surface, and this gives the gradient of the propagator:

$$|\nabla V(r)| = {1\over S_{d-1} r^{d-1}}$$

So that in any spatial dimension d, the Euclidean propagator/Yukawa potential is

$$V(r) = {1\over S_{d-1} r^{d-2}}$$

Where $S_{d-1}$ is the prefactor for the volume of a d-1 dimensional ball.

• cannot open the link in the answer above, can anyone show the details? – Chuan Chen Apr 2 '16 at 7:31