From the Schwarzschild metric $$ds^2=(1-2m/r)dt^2-(1-2m/r)^{-1}dr^2-r^2(dθ^2+\sin^2⁡θ dϕ^2)$$ on the surface $r=2m$ (setting $dr=0$) we have $$ds^2=-r^2(dθ^2+\sin^2⁡θ dϕ^2).$$

This looks spacelike ($ds^2<0$) rather than null ($ds^2=0$). How does one get $ds^2=0$ if $dθ$ and $dϕ$ are not zero?

Edit: Poorly worded question. I should have said how does one show that the normal vector to the horizon (as a hypersurface) is null?


2 Answers 2


Two wrongs apparently do make a right. First, you’re using bad coordinates for the event horizon, and second, as a result, you divided by $0$ by plugging in $r=2m$. But, somehow you got that the induced metric tensor is $g=-r^2\,g_{S^2}$, which actually is the correct induced metric on the event horizon of Schwarzschild. But unfortunately, you made a third error and concluded that this is spacelike, when in fact this says the induced metric is degenerate, and thus the hypersurface is null (keep in mind a hypersurface is 3-dimensional in this case, so a non-degenerate metric would at the very minimum see three coordinates, but you only see two, so it’s definitely degenerate).

Correct Way to Analyze the Spacetime.

As mentioned already, you should use a good set of coordinates which cover the event horizon. One example is by the Eddington-Finkelstein coordinates $(v:=t+r^*,r,\theta,\phi)$, in which the metric is (with my signature, which is the opposite of yours, because I don’t want to type something wrong later on out of habit) \begin{align} g&=-\left(1-\frac{2m}{r}\right)\,dv^2+2\,dv\,dr+r^2g_{\Bbb{S}^2}. \end{align} This is valid for $(v,r)\in \Bbb{R}\times (0,\infty)$ and $(\theta,\phi)$ any spherical coordinates on $\Bbb{S}^2$. In particular, this covers the Schwarzschild exterior and also the black hole region. Hence, now we can set $r=2m$ (i.e consider the pullback of this metric to the hypersurface $\mathcal{H}_+:=\{(v,r,\theta,\phi)\,:\, r=2m\}$) to get \begin{align} g_{\mathcal{H}_+}&=0+0+4m^2\,g_{\Bbb{S}^2}=4m^2g_{\Bbb{S}^2}. \end{align} This is a degenerate metric on the 3-dimensional manifold $\mathcal{H}_+$, thereby proving that the event horizon is a null hypersurface.

By the way, another fun feature to notice is that the coordinate vector field $T:=\left(\frac{\partial}{\partial v}\right)_{(v,r,\theta,\phi)}$ in the $(v,r,\theta,\phi)$ coordinate system agrees with the vector field $\left(\frac{\partial}{\partial t}\right)_{(t,r,\theta,\phi)}$ from our original $(t,r,\theta,\phi)$ coordinate system (see here for the detailed calculation); in fact it is the unique analytic extension (but analyticity isn’t really the way to judge things; the key is that this $T$ is Killing, and timelike for large $r$, in fact for $2m<r<\infty$). Now, we can observe that $g(T,T)=g_{vv}=-\left(1-\frac{2m}{r}\right)$, so with my sign convention, $T$ is spacelike, null, timelike according as to whether $r\in (0,2m), r=2m, r\in (2m,\infty)$ respectively. So, our usual notion of time is getting more and more null as we approach the event horizon (geometrically, the timecones are “tilting”, until their boundary, which is the null cone, becomes tangent to the event horizon).

Equivalence of Definitions of Null Hypersurface

There are actually (atleast two) ways to describe a null hypersurface:


Let $(V,g)$ be a Lorentzian inner product space and $H\subset V$ a codimension-1 subspace. Then, $H$ is degenerate if and only if the orthogonal complement $H^{\perp}$ is spanned by a non-zero null vector (i.e the normal is null).

In either of these cases, we say $H$ is a null hyperplane.

To prove the $(\implies)$ direction, the meaning of degeneracy is that there exists a non-zero vector $\zeta\in H$ such that $g(\zeta,\cdot)|_H=0$. In particular, this says $\zeta$ is orthogonal to $H$, and since $\zeta\in H$, it says $g(\zeta,\zeta)=0$, and so $\zeta$ is null, i.e $H$ has a null normal.

Conversely, suppose $\zeta$ is a (non-zero) null normal. By virtue of being null, we have $g(\zeta,\zeta)=0$, meaning $\zeta\in (H^{\perp})^{\perp}=H$ (the inclusion $H\subset H^{\perp\perp}$ is true for any subspace by simple definition unwinding; the reverse inclusion follows from the rank-nullity theorem which implies these two subspaces have the same dimension, and thus must be equal). Thus, we have a non-zero vector $\zeta\in H$ which is orthogonal to $H$, i.e $g(\zeta,\cdot)|_H=0$, thus proving that $H$ is a degenerate subspace relative to $g$.

If you apply this argument at each tangent space, then you of course get the corresponding statement at the level of Lorentzian manifolds and null hypersurfaces. So, above, we saw directly that $g_{\mathcal{H}_+}$ was a degenerate metric and concluded from here that the hypersurface is null. Alternatively, we can see that the vector field $T=\frac{\partial}{\partial v}$ is such that

  • it is obviously nowhere-vanishing (it’s a coordinate vector field afterall)
  • $g_{vv}=0$ (i.e it is null). Said again, $T$ is null and orthogonal to $\frac{\partial}{\partial v}$.
  • $g_{v,\theta}=g_{v,\phi}=0$, so $T$ is orthogonal to $\frac{\partial}{\partial\theta}, \frac{\partial}{\partial\phi} $.

Since $\frac{\partial}{\partial v}, \frac{\partial}{\partial\theta} , \frac{\partial}{\partial\phi} $ are tangent to the hypersurface $\mathcal{H}_+=\{r=2m\}$, and span the whole tangent spaces of the hypersurface, it follows that $T$ is orthogonal to the hypersurface. Thus, we see according to our second definition that $\mathcal{H}_+$ is a null hypersurface.

A third way to see/calculate that the hypersurface is null is to again use the Eddington-Finkelstein coordinates as above, and explicitly compute the normal vector, $\text{grad}_g(r)$ given by the gradient of the function $r$, or simply the inner product with itself (since we know that is the normal, and we only want to know its causal nature): \begin{align} g(\text{grad}(r),\text{grad}(r))&=g^{ab}\frac{\partial r}{\partial x^a}\frac{\partial r}{\partial x^b}=g^{rr}, \end{align} since we’re dealing with coordinate vector-fields. So, we just need to invert the suitable matrix and extract this $rr$ component; a short calculation reveals it is $1-\frac{2m}{r}$, so when $r=2m$, it is a null vector field. Hence, the normal to $\mathcal{H}_+$ is null, so once again proving the hypersurface is null.


Before anything else, I should point out that Schwarzschild coordinates $t,r,\theta,\phi$ are not good to discuss the event horizon, as they are ill-defined there. You might prefer advanced Eddington–Finkelstein coordinates instead.

That being said, points in a null hypersurface may be spacelike related. Being a null hypersurface only means the normal vector to the surface is null. For example, notice you don't need to go to curved spacetime to have this sort of behavior. Pick the $x = t$ plane in Minkowski spacetime. Two points at the same value of $x$, but different values of $y$, are spacelike related. There is nothing wrong with that, and the hyperplane is still null.

  • $\begingroup$ Yes I was thinking of phrasing it other coordinates (eg Gullstrand-Painleve) to eliminate the coordinate singularity at $r=2m$ but didn't think, strictly speaking, it was necessary. Thank you -- yes of course all that I really need is to show that the normal vector is null (such as on a light cone, as in your example). $\endgroup$
    – Khun Chang
    Commented May 25 at 20:17
  • $\begingroup$ I will switch to better coordinates, express the surface as $F(x^\mu)=0$ and calculate the norm of the normal vector $g^{\mu\nu}F_{,\mu}F_{,\nu}$ and see if I get zero. $\endgroup$
    – Khun Chang
    Commented May 25 at 20:32
  • 1
    $\begingroup$ @KhunChang That should work! $\endgroup$ Commented May 25 at 20:39
  • $\begingroup$ @KhunChang “Eliminating the coordinate singularity” is a popular textbook lie. This cannot be done by any mathematically valid coordinate transformation. The transformation used for this purpose is singular, which is a mathematical nonsense like dividing infinity by infinity. The result of this operation is not unity, but undefined, the singularity remains. Also, the very same procedure of applying an invalid transformation can be used to create a singularity anywhere, such as in your kitchen, that is a physical nonsense. Using coordinate transformations for removing singularities is absurd. $\endgroup$
    – safesphere
    Commented May 27 at 5:13
  • 1
    $\begingroup$ @NíckolasAlves Yep, it absolutely did! Piece of cake with Gullstrand-Painleve coordinates. $\endgroup$
    – Khun Chang
    Commented May 28 at 6:03

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