Charged Black Hole in 1+2 dimensions without cosmological constant Does a charged black hole in 2+1D exist? I am interested in a theory of the form:
$$ S = \int d^3 x \sqrt{-g}\big(\cfrac{R}{2} - \cfrac{1}{2}F^2\big) $$
where $F^2 = F_{ab}F^{ab}$.
The field equations are:
$$G_{ab} = 2T_{ab} $$
$$\nabla^{a}F_{ab}=0$$
Imposing a one degree of freedom metric:
$$ds^2 = -f(r)dt^2 + f^{-1}(r)dr^2 + r^2 d\theta^2 $$
and an ansatz for the $U(1)$ field:
$$A_{a} = (-\phi(r),0,0).$$
Maxwell's equations can be immediately integrated to yield:
$$ \phi(r) = c_1\ln(r) +c_2 $$
I cannot find any mistakes in my calculation, neither any paper on 2+1D charged black holes without cosmological constant. If i am correct what does $c_1$ and $c_2$ represent? Why the potential does not vanish at infinity?
 A: My apologies for reading/answering too quickly. In the absence of a cosmological constant, the vacuum Einstein equations do not support a black hole solution in 2+1 dimensions, but the electrovacuum Einstein equations do.
The solution you're looking for is described in this paper by Deser and Mazur.  The metric due to a point source with charge $e$ and mass $m$ takes the form
$$ds^2 = \left(1-\frac{Ge^2}{1-4Gm} \log(r)\right)^2 dt^2 +\exp\left[Ge^2\log(r)^2 - 8Gm\log(r)\right](dr^2+r^2d\theta^2)$$
which exhibits a Killing horizon (not an event horizon) at $r=\exp\left[\frac{1-4Gm}{Ge^2}\right]$

The electrostatic potential you refer to is the same as one obtains for an infinitely long wire in 3+1 dimensions.  In $d+1$ dimensions, the electric field from a point charge goes like $1/r^{d-1}$, which means that the potential goes like $1/r^{d-2}$ if $d\neq 2$, and $\log(r)$ for $d=2$.
Physically, this simply means that if you started a test charge at rest and let it be pushed out to infinity by the electric field, then its kinetic energy would grow without bound as it traveled further and further away. The reason this feature does not emerge in 3 or more spatial dimensions is that the electric field falls off sufficiently quickly that the kinetic energy of the aforementioned test charge remains bounded.
A: The electric field is given by the derivative of the potential $\phi(r)$ which indeed vanishes at infinity.
