# Coulomb potential in 2D

I know that the Coulomb potential is logarithmic is two dimensions, and that (see for instance this paper: http://pil.phys.uniroma1.it/~satlongrange/abstracts/samaj.pdf) a length scale naturally arises:

$$V(\mathbf{x}) = - \ln \left( \frac{\left| \mathbf{x} \right|}{L} \right)$$

I can't see what's the physical meaning of this length scale, and, most of all, I can't see how this length scale can come up while deriving the 2D Coulomb potential by means of a Fourier transform:

$$V(\mathbf{x}) = \int \frac{\mathrm{d^2 k}}{\left( 2 \pi \right)^2} \frac{e^{\mathrm{i} \mathbf{k} \cdot \mathbf{x}}}{\left| \mathbf{k} \right|^2}$$

I would appreciate some references where the two-dimensional Fourier transform is carried out explicitly and some insight about the physical meaning of L and how can it arise from the aforementioned integral.

• – Qmechanic May 18 '12 at 21:51

## 3 Answers

The length scale $L$ has to be present in the denominator for dimensional reasons – only logarithms of dimensionless quantities are really "well-defined" unless one wants to introduce bizarre units such as the "logarithm of a meter".

On the other hand, the dependence on $L$ is largely trivial and unphysical for most purposes. Replace $L$ by $K$ and you will get $$V_K(x,y) = -\ln \left( \frac{|\vec x|}{K} \right) = V_L(x,y) +\ln(K/L)$$ which only differs by the additive shift, $\ln(K/L)$ from the original potential you mentioned. Shifts of potentials by a constant are largely inconsequential. In particular, the gradient of $V$, $\nabla V$, isn't changed at all. To derive the simple claim about the shift above, I only used $\ln(A/B) = \ln(A)-\ln(B)$ a few times.

The Fourier transform of your potential may be derived by realizing what the Laplacian of the potential is. The Laplacian is the two-dimensional delta-function. In the momentum basis, it's equivalent to the identity $$(p_x^2+p_y^2) \tilde V(p_x,p_x) = 1$$ which is easily solved by $\tilde V =1/(p_x^2+p_y^2)$. However, the behavior of $\tilde V$ isn't quite well-defined for the point $p_x=p_y=0$ where one can add a multiple of a delta-function. This is because $$(p_x^2+p_y^2) \delta(p_x) \delta (p_y) = 0$$ so $\tilde V \to \tilde V + K \delta(p_x)\delta(p_y)$ transforms a solution into another solution. Of course, the two-dimensional delta-function in the momentum space is nothing else than the Fourier transform of the constant term $\ln(K/L)$ we discussed in the position basis so the two ambiguities are the same.

Now, you could think that the momentum-basis form of the potential, $1/(p_x^2+p_y^2)$, is unique because it has no length scale in it and no delta-function in it while we don't see a corresponding unique form of the position-basis potential – because the expressions with any length scale are equally good. But this is really an illusion. As a distribution, $1/(p_x^2+p_y^2)$ is ill-defined (in the very same sense as $\ln(x^2+y^2)$ would be in the position basis) and we must specify what its behavior near the origin is.

This ambiguity is the two-dimensional generalization of the subtleties connected with the one-dimensional "principal value" of $1/x$ as a distribution. $1/x$ multiplied by a test function is well-defined if we agree that the symmetric region $x\in(-\epsilon,+\epsilon)$ is removed. That's what we mean by the principal value.

On the other hand, if you compute the two-dimensional integral of $1/(p_x^2+p_y^2) f(p_x,p_y)$ where $f$ is continuous near the origin, you may switch to the polar coordinates where $r$ in $r\,dr\,d\phi$ is beaten by $1/r^2=1/(p_x^2+p_y^2)$ so you still have a divergent integral that has to be regulated. A way to regulate it is to cut if off and remove the disk $r<p_{\rm min}$ for some small $p_{\rm min}$. Such a cutoff induces an additive shift dependence that is logarithmic in the cutoff. For the same dimensional reasons as before, one has to take the logarithmic dimensionless so what we need to subtract (or add?) to erase most of the dependence on the cutoff is something like $$f(0) \ln(p_{\rm min} / L_P)$$ where $L_P$ is the counterpart of $L$, the length scale you started with. Sorry if I omitted some dimensionless coefficients. Clearly, the change of $L_P$ is equivalent to redefining the distribution by an additive shift by $\delta^{(2)}(\vec p)\times L_P$ and $L_P \sim 1/L$ plays the same role of the scale we had before, in the position basis.

• Thanks Luboš, very thorough answer! I'd have one more question...You say that the choice of the length scale is largely inconsequential, however (maybe I should have written that in the original question) I'm struggling with the calculation of the Coulomb self-energy for a 2D object: $$\int \mathrm{d}^3 x \int \mathrm{d}^3 y V\left( \left| \mathbf{x} - \mathbf{y} \right| \right)$$ as in this case the choice of the scale length seems to affect a physical quantity. Can you please explain how your excellent discussion about the unphysicalness of $L$ relates with the integral above? Thank you! – zakk May 18 '12 at 19:36
• Thanks for your interest, zakk. The additive shift in the self-energy is unphysical whenever you turn off gravity (and, at the same moment, one must be unable to compare the energy with the energy of a state without the self-interacting object). The reason is that all the equations of motion - Lagrange, Hamilton - depend on derivatives of L,H with respect to fields and their derivatives and the derivative of a constant is zero. When gravity is turned on, the additive shift matters and gravitates but the corresponding scales L are determined in a complete theory. – Luboš Motl May 19 '12 at 4:12

## I can't see what's the physical meaning of this length scale, and, most of all, I can't see how this length scale can come up while deriving the 2D Coulomb potential by means of a Fourier transform:

zakk, the constant L in the logarithm is the integration constant which is present for any differential equation and means that the Poisson equation used to derive V has infinity of solutions. It can be fixed by prescribing the value of potential V(r_0) at some distance r_0.

The same happen in 3D; there we have

V = 1/r + C

and we choose C = 0.

In your case we cannot choose L = 0, but L = 1 meter is good.

If V is supposed to be potential of electric force, the value of L is otherwise completely arbitrary.

The Fourier transform method can only find one particular solution of the equation. It cannot supply the integration constant L. You still have to add a constant - ln L to your Fourier integral to get the general solution.

Your integral for self-energy does not seem right. What is your starting point?

I don't know if you have solved it or not, but my suggestion is to add a convergence factor $$e^{-a\,r}$$ and get the integral, and then let $a \to 0$ (the same trick with yukawa potential) You can do this integral by the mathematica.