It is know that in 1+1 dimensions, the Coulomb potential is of linear form: $$ V(x) = Cx$$ and in 1+2 dimensions, of the form: $$V(x,y) = - \ln\left(\frac{L}{\sqrt{x^2 + y^2}}\right).$$ And I am wondering how these results are derived? In 1+d dimensions, the equation that takes the form $\nabla \cdot \vec{E}$ results from the equation ($\nu =0$) (source free): $$ \partial^\mu F_{\mu \nu}=0 $$ Where $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ and then we try to find a solution to the equation $\nabla V = \vec{E}$. In 1+1 dim, we can solve: $$ \partial^\mu F_{\mu 0} = \frac{\partial E}{\partial x } = 0$$ Which gives $E= C$, $C$ a constant, and then: $$ V(x) = \int E\: \mathrm{d}x = Cx$$ But in 2-d I am unsure how to get the logarithmic form of the Coulomb potential. Thanks.


2 Answers 2


You can look at this problem from a different direction. Take Gauss' law

$$\int\boldsymbol{\rm E}\cdot\boldsymbol{{\rm d}S}=\frac{Q_{in}}{\varepsilon_{0}}$$

and lets try to generalize it for arbitrary number of dimensions $d$. Assuming a point charge $q$ at the origin, from symmetry considerations one has

$$\left|\boldsymbol{\rm E}\right|S_{d-1}r^{d-1}=\frac{q}{\varepsilon_{0}}$$

where $S_{d-1}r^{d-1}$ is the surface area of a $d-1$-sphere in $d$ dimensions. Thus the general form of the electric field is

$$\boldsymbol{\rm E}=\frac{q}{\varepsilon_{0}S_{d-1}r^{d-1}}\hat{r}$$

In particular, for $d=2$

$$\boldsymbol{\rm E}=\frac{q}{2\pi\varepsilon_{0}r}\hat{r}$$

and thus the potential is

$$V=-\int_{r_{0}}^{r}\boldsymbol{E}\cdot\boldsymbol{{\rm d}\ell}=-\frac{q}{2\pi\varepsilon_{0}}\ln\left(\frac{r}{r_{0}}\right)$$

This analysis is based on the book "A First Course in String Theory" by Barton Zwiebach. You can look it up in section $3.5$.


You basically just use the same method you use in $3D$. Your mistake is assuming that the covariant Maxwell equation is source-free. More generally (one of) the Maxwell equations is: $$\partial_\mu F^{\mu\nu}=-\mu_0 J^\nu$$ Some people use different conventions for the constant in front of the current. For a static point charge at the origin in $d+1$ dimensions, the current is given by $$J^\nu = \begin{pmatrix} c q \delta^{(d)}(\textbf{x}) \\ \textbf{0}\end{pmatrix}$$ where $\delta^{(d)}$ is the $d$-dimensional delta function. We can identify the elements in the first row of $F^{\mu\nu}$ with the electric field. Therefore taking the $\nu=0$ component of the equation above, we get: $$\nabla \cdot \textbf{E}=\frac{q}{\epsilon}\delta^{(d)}(\textbf{x})$$ where both $\nabla$ and $\textbf{E}$ have $d$ components. Assuming that $\textbf{E}$ has "spherical" symmetry, we can solve this by using the same method we used in $3+1$ dimensions, that is by integrating both expressions on the volume of a $d-1$ sphere. Using the properties of the delta function and the divergence theorem we get: $$\textbf{E} A_{d-1}=\frac{q}{\epsilon_0} \implies \textbf{E} \propto \frac{1}{r^{d-1}}$$ where $A_{d-1}$ is the area of a $d-1$ dimensional sphere of radius $r$.

From this, we see that in $1+1$ dimensions, $$\textbf{E} \propto 1$$ so integrating we get for the $1+1$ dimensional Coulomb potential: $$V\propto r$$ while in $2+1$ dimensions, $$\textbf{E} \propto \frac{1}{r}$$ and integrating we obtain the $2+1$ dimensional Coulomb potential: $$V \propto \log{r}$$


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