The interior Schwarzschild metric

$$c^2 {d \tau}^{2} = \frac{1}{4} \left( 3 \sqrt{1-\frac {r_s}{r_g}}-\sqrt{1-\frac{r^2 r_s}{r_g^3}} \right)^2 c^2 dt^2 - \left( 1-\frac{r^2 r_s}{r_g^3} \right)^{-1} dr^2 - r^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right)$$

approximates the metric inside a planet or star of some homogeneous material (see Wikipedia). Here, $r_s$ is the Schwarzschild radius of its mass and $r_g>r_s$ its radius, $r$ the radius at which we want to know the metric.

To derive the speed of light at the surface $(r=r_g$), I follow the derivation shown in this answer and set $d\tau=0$ to get

$$0 = (1-\frac{r_s}{r_g})c^2 dt^2 - (1-\frac{r_s}{r_g})^{-1}dr^2 -r_g^2d\Omega^2$$

where $d\Omega = \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right)$. Rearranging gives:

$$\left(\frac{dr}{dt}\right)^2 + (1-\frac{r_s}{r_g})r_g^2 \left(\frac{d\Omega}{dt}\right)^2 = (1-\frac{r_s}{r_g})^2c^2 $$

Taking the square root on both sides, on the left side we have the speed of light on the surface of the object according to my watch (read: coordinate time) limited by a constant depending on its mass and radius being smaller than the speed of light $c$ far away from any gravitational source.

Now comes the head scratching:

1) I assume that if the speed of light is like that, any massive particle near $r_g$ has to obey the same speed limit. In particular the atoms of this object. This is, with regard to coordinate time, i.e. what I would measure when watching it with my telescope and timing with my watch, so to say.

2) Now start reducing $r_g-r_s$ by either increasing the mass or reducing its radius such that $r_s/r_g \to 1$ such that the speed of light nicely approaches zero. The speed of the surface particles (and by extension the whole object) has to go to zero too.

This looks as if the object has to come to a halt and does not move anymore.

I am rather confident in the interpretation of $dt$ as the time my watch shows, so to say. But clearly black holes ($r_s\to r_g$) are not nailed to the sky.

In which coordinate system does the object slow down, to a halt in the limit, and how does this coordinate system relate to me observing it with a telescope (undisturbed by any nearby gravitational sources).


1 Answer 1


I believe you are referring to the fact that, in the exterior Schwarzschild metric, the speed of light at the event horizon of a black hole is zero. The Schwarzschild solution is static, so the Schwarzschild coordinates are those used by a remote observer who is stationary relative to the black hole. This solution does not apply to a black hole that is moving in your coordinates. To derive a solution for a moving black hole one could solve the gravitational equations for this case.

However, a simpler approach is to use the relativity principle. Consider a stationary black hole in my coordinates as a Schwarzschild observer while you pass by me with a constant velocity. This is equivalent to the black hole moving with the opposite velocity in your coordinates. Thus we obtain the Schwarzschild solution for a moving black hole where all velocities at the horizon are non-zero, but a constant value, so the black hole is not "nailed" to the sky.

Having said this, I must mention that the concepts of distance and remote velocity don't have a unique definition in General Relativity, so one must be careful using these concepts. This of course doesn't change the fact that countless galaxies are in motion relative to us and many of them have supermassive black holes in the center definitely not "nailed" to the sky.


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