0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $3.5$.

$\endgroup$
1
$\begingroup$

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: $$V \propto \log{r}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.