What I am going to ask is probably a result of unrigorous treatment of the submanifold in question.

Radial Null Geodesics of Schwarzschild

So start with Schwarzschild spacetime. The metric tensor is in Schwarzschild coordinates

$$g=-f(r)dt^2+f(r)^{-1}dr^2+r^2(d\theta^2+\sin^2\theta d\phi^2),\quad f(r)=1-\frac{2M}{r}.$$

We study then radial null geodesics. These are curves $\gamma : I\subset \mathbb{R}\to M$ satisfying three conditions

  1. They are radial, $\theta\circ\gamma,\phi\circ\gamma$ are constant.

  2. They are null, $g(\gamma',\gamma')=0$.

  3. They satisfy the geodesic equation.

Putting together (1), (2) and (3) we can prove that these curves are very simple in Eddington-Finkelstein coordinates. They divide in two classes, ingoing and outgoing.

For the ingoing we use coordinates $(v,r,\theta,\phi)$ with $v = t+r_\ast$. Then (1), (2) and (3) implies that the curve is $\lambda \mapsto (v_0,\lambda,\theta_0,\phi_0)$.

Likewise, for outgoing we use coordiantes $(u,r,\theta,\phi)$ with $u = t-r_\ast$. Then (1), (2) and (3) implies the curve is $\lambda\mapsto (u_0,\lambda,\theta_0,\phi_0)$.

Generators of the Horizon

Now we wish to study the surface $r = 2M$. In Schwarzschild coordinate this is not possible because this locus isn't in the domain of the chart. But it is in the domain of the Eddington-Finkelstein coordinates. We use the $(v,r,\theta,\phi)$ ones.

The horizon is $r = 2M$. Its normal one-form is $dr$. The dual vector is $\ell =\partial_v + f(r) \partial_r$. At $r=2M$ we have $\ell=\partial_v$ because $f(r)$ vanishes.

Since $\ell$ is the normal vector to the surface and since it is null at the surface, its integral lines are the generators of the null surface.

The generator thus are specified as $(r,\theta,\phi)$ constant, with $r = 2M$ which defines the surface. Plugging this into the geodesic equation we can find that the generators are $$\lambda\mapsto(4M\ln \lambda, 2M, \theta_0,\phi_0).$$

These are of course curves inside the null surface $\mathcal{H}$ but since $\mathcal{H}\subset M$ they can be seen as curves in $M$.

These curves satisfy:

  1. They have constant $\theta$ and $\phi$;

  2. They are null;

  3. They are geodesics;

Now wait a moment, these are the three conditions for the radial null geodesics.

So it seems that the horizon generators are radial null curves.

Still, these curves have constant $r$ and have not the form which we found for radial null curves. Still, it is a fact that the three conditions implies the form given previously.

So what is the inconsistency here? My guess is that satisfying (1), (2) and (3) inside a submanifold doesn't imply satisfying (1), (2) and (3) on the full manifold. In particular intuitively I imagine the issue is that a geodesic on the surface is not a geodesic on the ambient space due to the constraint. But I don't know how to make this precise.

  • $\begingroup$ $\theta$ and $\varphi$ being constant doesn't exclude the possibility of $r$ being constant too. $\endgroup$
    – Javier
    Commented Mar 8, 2019 at 21:03
  • $\begingroup$ Sure, but that's not the issue I have. For the first solutions $r$ itself is the affine parameter, which is clearly not constant, while for the horizon generators $r$ is constant. So the horizon generators doesn't seem to be a subset of the first solutions. What am I missing here ? $\endgroup$
    – Gold
    Commented Mar 8, 2019 at 21:49
  • $\begingroup$ Oh, I see. Well, I'll check the math later, but the most obvious explanation is that you missed some solutions when working in EF coordinates. Possibly some implicit extra assumption snuck in. $\endgroup$
    – Javier
    Commented Mar 8, 2019 at 22:11
  • 1
    $\begingroup$ They divide in two classes, ingoing and outgoing A horizon generator is neither ingoing or outgoing but a separatrix of those classes. $\endgroup$
    – A.V.S.
    Commented Mar 9, 2019 at 7:47
  • $\begingroup$ Thanks to both for the comments. With them I believe I was able to solve the issue. In fact, supposing $r$ varies is the extra assumption which made me lost a third class of solutions - those with $r$ constant. I added one answer with this worked out in case anyone ends up with the same doubt. If anything is found to be wrong, corrections are welcome. $\endgroup$
    – Gold
    Commented Mar 9, 2019 at 13:14

1 Answer 1


With the help of the comments I was able to find what is the issue. I'm posting it here, in the case anyone has the same doubt. In fact horizon generators are radial null geodesics. They are, in fact, a third class, separate from ingoing and outgoing. So let's work this out.

In Schwarzschild coordinates, the geodesic equation follows from the Lagrangian

$$L=\frac{1}{2}\left(-f(r)\dot{t}^2+f(r)^{-1}\dot{r}^2+r^2(\dot{\theta}^2+\sin^2\theta \dot{\phi}^2)\right).$$

One needs to work out the Euler-Lagrange equations. The $t$ equation is

$$\dfrac{d}{d\lambda}\frac{\partial L}{\partial \dot{t}}=0$$

Which actually yields one constant of motion


Now, the null condition is $$-f(r)\dot{t}^2+f(r)^{-1}\dot{r}^2=0.$$

Using the constant of motion we get the $r$ equation

$$\dot{r}^2=E^2\Longrightarrow \dot{r}=\pm E$$

From which it follows immediately that $r=\alpha \lambda + \beta$. Here comes the analysis, however. We have two situations here:

  1. The case $\alpha\neq 0$. In this one we can reparameterize in order to get a new parameter on which $r = \lambda$. Once we substitute in the first equation we get the ingoing/outgoing null geodesics which are simpler in Eddington-Finkelstein coordinates.

  2. The case $\alpha =0$. Obviously is a valid solution, but now we obviously cannot reparameterize so that $r = \lambda$. Getting back to either the null condition or the $t$ equation, we see that they are satisfied if and only if at the constant $r = r_0$ one has $f(r_0)=0$. This happens only at $r = 2M$ which is not in the domain of the chart, so these solutions simply can't be studied in Schwarzschild coordinates.

    We go to Eddington-Finkelstein coordinates which covers this region and make the equations simple. I belive that it is obvious that any other chart covering this region would do. In $(v,r,\theta,\phi)$ coordinates the metric becomes $$ds^2=-f(r)dv^2+2dvdr+r^2(d\theta^2+\sin^2\theta d\phi^2)$$

    The geodesic equation will follow from the Lagrangian $$L=\frac{1}{2}\left(-f(r)\dot{v}^2+2\dot{r}\dot{v}+r^2(\dot{\theta}^2+\sin^2\theta \dot{\phi}^2)\right)$$

    Working out the $v$ and $r$ Euler-Lagrange equations we get for the $v$ equation a constant of motion $$-f(r)\dot{v}+\dot{r}=E$$

    and for the $r$ equation we get $$\ddot{v}+\frac{1}{2}f'(r)\dot{v}^2-r(\dot{\theta}^2+\sin^2\theta\dot{\phi}^2)=0,$$

    which for radial simplifies $$\ddot{v}^2+\frac{1}{2}f'(r)\dot{v}^2=0.$$

    Now recall we set out to study the solutions which have $r$ constant. Recall these are null only if the constant is $r = 2M$. So the first equation is trivially satisfied because the LHS is zero, which is of course constant. The second equation gives the important one. Computing $f'(2M)$ we get


    Solving this equation gives $v = 4M \ln \lambda$. So we get the missing solutions with $r$ constant.

    We can finally get the remaining ones working in $(u,r,\theta,\phi)$ coordinates as well.

So in conclusion: radial null geodesics divide not in two, but three categories as pointed out in comments. Supposing that in the $r$ solution $\alpha\neq 0$ is the implicit extra assumption that snuck in as also pointed out in comments. Treating the $\alpha = 0$ case yields the remaining solutions, which clearly cannot be parameterized by $r$.

  • $\begingroup$ Could you please clarify what the expression for $\lambda$ is? $\endgroup$
    – safesphere
    Commented May 21, 2019 at 16:41
  • $\begingroup$ In my notation $\lambda$ is an affine parameter for the geodesic under consideration. The relation between coordinates along the geodesic and $\lambda$ are found by solving the autoparallel equation. For ingoing/outgoing radial null geodesics it is always possible to reparameterize so that $r = \lambda$. For generators of the future horizon we have instead $v = 4M\ln \lambda$ or $\lambda = e^{v/4M}$. For generators of the past horizon there is a similar formula. Does it clarify the matter? $\endgroup$
    – Gold
    Commented May 21, 2019 at 17:33
  • $\begingroup$ Yes (I assume $v$ in the exponent is the EF time coordinate). Let me ponder this. Thank you! $\endgroup$
    – safesphere
    Commented May 21, 2019 at 17:40

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