Proof that the event horizon of a black hole is a "null surface" (meaning a normal vector to the hypersurface $r=2m$ is a null vector)

Question

I want to show that a normal vector to the event horizon $r=2m$ is null. In general, if a hypersurface is defined by $$F(x^\mu)=0 \,\,,$$ then a normal vector is $\nabla F = F_{,\mu}$. I will use Lemaitre coordinates $(τ,ρ,θ,ϕ)$ (see Dirac's "General Theory of Relativity" chap. 19, or Landau & Lifshitz "Classical Theory of Fields" 4th Ed., sec. 102), where the metric is $$ds^2 =dτ^2 - \frac{2m}{r}dρ^2 - r^2dθ^2 - r^2(\sin{θ})^2 dϕ^2 \,\, .$$ The event horizon is described by the equation $$ρ-τ = \frac{4m}{3} \,\, .$$

Taking $F(τ,ρ,θ,ϕ) = ρ-τ-4m⁄3$, we have $F_{,μ} = (-1,1,0,0)$. So, $$F_\mu F^\mu = g^{μν} F_{,μ} F_{,ν}=0$$ on the surface; therefore, $F_{,μ}$ is a null vector on the event horizon hypersurface $r=2m$.

Is this correct?

I originally wanted to show this by first showing that a vector $dx^\mu$ lying on the event horizon is null, and then showing that any vector perpendicular to a (non-zero) null vector must be null, but this assumption turns out to be false.

My proof does not seem to require the knowledge that a vector lying on the event horizon is null.

Answer 1

First of all, not every vector tangent to the event horizon is a null vector. Second, you seem to conflate vectors and covectors way too much (granted Dirac and L&L do so, and it’s definitely good to learn directly from the masters but not everything…).

Anyway, consider the function $F=\rho-\tau$. Then, $dF=-d\tau+d\rho$, i.e $F_{\mu}=(-1,1,0,0)$, so yes \begin{align} g_{\mu\nu}F^{\mu}F^{\nu}= g^{\mu\nu}F_{\mu}F_{\nu}=1 \cdot(-1)^2 + \left(-\frac{r}{2M}\right)\cdot (1)^2=1-\frac{r}{2M}, \end{align} which indeed vanishes on the event horizon. This shows $F^{\mu}$ does indeed constitute a null vector when $r=2M$, and since the hypersurface $r=2M$ is a certain level set of $F$, it follows that the event horizon is indeed a null hypersurface.