AdS black brane

An anti de Sitter black brane is a solution of the Einstein equations in the presence of a negative cosmological constant which possesses a planar event horizon. This is distinct from an anti de Sitter black hole solution which has a spherical event horizon. The negative cosmological constant implies that the spacetime will asymptote to an anti de Sitter spacetime at spatial infinity.

Math development
The Einstein equation is given by"$   R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}+\Lambda g_{\mu\nu}=0,$"where $$R_{\mu\nu}$$ is the Ricci curvature tensor, R is the Ricci scalar, $$\Lambda$$ is the cosmological constant and $$g_{\mu\nu}$$ is the metric we are solving for.

We will work in d spacetime dimensions with coordinates $$(t,r,x_1,...,x_{d-2})$$ where $$r\geq0$$ and $$-\infin<t,x_1,...,x_{d-2}<\infin$$. The line element for a spacetime that is stationary, time reversal invariant, space inversion invariant, rotationally invariant

and translationally invariant in the $$x_i$$ directions is given by,

$$ds^2=L^2\left(\frac{dr^2}{r^2h(r)}+r^2(-dt^2f(r)+d\vec{x}^2)\right)$$.

Replacing the cosmological constant with a length scale L

$$\Lambda=-\frac{1}{2L^2}(d-1)(d-2)$$,

we find that,

$$f(r)=a\left(1-\frac{b}{r^{d-1}}\right)$$

$$h(r)=1-\frac{b}{r^{d-1}}$$

with $$a$$ and $$b$$ integration constants, is a solution to the Einstein equation.

The integration constant $$a$$ is associated with a residual symmetry associated with a rescaling of the time coordinate. If we require that the line element takes the form,

$$ds^2=L^2\left(\frac{dr^2}{r^2}+r^2(-dt^2+d\vec{x})\right)$$, when r goes to infinity, then we must set $$a=1$$.

The point $$r=0$$ represents a curvature singularity and the point $$r^{d-1}=b$$ is a coordinate singularity when $$b>0$$. To see this, we switch to the coordinate system $$(v,r,x_1,...,x_{d-2})$$ where $$v=t+r^*(r)$$ and $$r^*(r)$$ is defined by the differential equation,"$\frac{dr^*}{dr}=\frac{1}{r^2h(r)}$."The line element in this coordinate system is given by,"$ds^2=L^2(-r^2h(r)dv^2+2dvdr+r^2d\vec{x}^2)$,"which is regular at $$r^{d-1}=b$$. The surface $$r^{d-1}=b$$ is an event horizon.