# How to derive a logarithmic potential from Newtonian?

Suppose we believe that the formula for Newtonian potential in $R^3$ is correct:

$\varphi(\bar{x}) = \frac{1}{|x|} = \frac{1}{r}$, disregarding the constant.

What is the justification of the fact that on $R^2$ this potential reduces to $\varphi(\bar{x}) = -\log( |x| ) = -\log( r )$?

Essentially, I could imagine two reductions: a potential of a point-mass or a potential of a string-mass.

The point one:

$$\int_{x \in R}\delta(x-x_0)\frac{1}{\sqrt{x^2+y^2+z^2}} = \frac{1}{\sqrt{{x_0}^2+y^2+z^2}}$$ or if we want to renormalize it, say, $$\frac{1}{\sqrt{{x_0}^2+y^2+z^2}} - \varphi(x_0) = \frac{1}{\sqrt{{\frac{1}{\phi(x_0)}}^2+y^2+z^2}} - \varphi(x_0)$$

This is a projection of a Newtonian potential on a 2-D plane $(x_0, y, z)$. Doesn't look like a logarithm...

If we look at the string, lying on the X axis:

$$\delta(y,z)*\frac{1}{\sqrt{x^2+y^2+z^2}} = \int_{-\infty}^{\infty}\frac{dx}{\sqrt{{x}^2+y^2+z^2}}$$

I don't really understand how to compute the last expression. Something like: (I am not sure about this formula.)

$$\int_{-\infty}^{\infty}\frac{dx}{\sqrt{{x}^2+y^2+z^2}} = v.p \int\frac{dx}{\sqrt{{x}^2+y^2+z^2}} + \int\frac{\delta(x)dx}{\sqrt{{x}^2+y^2+z^2}}$$

The right summant is no logarithmic, and the left doesn't even converge!

However, just blindly plugging it into the table of integrals would give me something like

$\log| x + \sqrt{ x^2 + (r=y^2+z^2)^2}|/_{-\infty}^{\infty}$

Which is somehow related to the logarithm, but since it has an essential singularity at 0, it is not correct to use the Newton-Leibniz formula.

Could someone enlighten me or point to a right direction (m.b. book)?

• The potential is defined as a "primitive" of a vector field. In $\mathbb{R}^2$, such a "primitive" is a logarithm, while in higher dimension it is a pure power of $r$. – Siminore Jul 23 '16 at 7:48

## 1 Answer

Your second attempt, the string-mass, is on the right track. If you started in $\mathbb{R}^n$ with $n\ge 4$, it would work out in a straightforward way: at distance $r$ from an infinite string, the potential is a constant multiple of
$$\int_{-\infty}^\infty \frac{1}{(r^2+z^2)^{(n-2)/2}}\,dz =\int_{-\infty}^\infty \frac{1}{(1+u^2)^{(n-2)/2}}\,\frac{du}{r^{n-3}} = C/r^{n-3}$$ using the substitution $u=z/r$.

The transition from $\mathbb{R}^3$ to $\mathbb{R}^2$ is more difficult because the potential theory has a different flavor in $\mathbb{R}^2$ (the potential is unbounded). As you noted, the potential of a uniformly charged infinite string is infinite: the integral given above diverges when $n=3$. In such a situation, one often uses regularization: subtract off the largest asymptotic term and extract information from the rest.

Consider the finite string from $(0,0,-L)$ to $(0,0,L)$. Its potential at the point $(r,0,0)$ is equal to $$\int_{-L}^L \frac{1}{\sqrt{r^2+z^2}}\,dz = \log\frac{\sqrt{r^2+L^2}+L}{\sqrt{r^2+L^2}-L} = 2 \log(\sqrt{r^2+L^2}+L) - 2\log r$$ This expression tends to infinity at the rate $2\log L$. The latter quantity doesn't depend on $r$, so we may as well subtract it. (Recall that potential is defined up to an arbitrary constant; this subtraction is just a way of normalizing it). After subtracting $2\log L$ we get
$$2 \log\left(\sqrt{(r/L)^2+1}+1\right) - 2\log r\to 2\log 2-2\log r$$ i.e., the logarithmic potential.