Suppose I am at rest at a great distance $r_0$ from a black hole with a mass $M$ without rotation or charge.
During my free fall in vacuum from $\tau=0$ and $r=r_0$, I will pass the event horizon in finite proper time and the increment of my proper time in Schwarzschild coordinates is $$\text{d}\tau=(2M/r-2M/r_0)^{-1/2}\,\text{d}r$$ (disregarding my destruction by tidal forces etc.)
If it is correct to view $$\frac{\text{d}r}{\text{d}\tau}=\sqrt{2M/r-2M/r_0}$$ as the locally measured speed, I would travel at $v>c=1$ shortly after passing the event horizon and before reaching $r=0$. But the locally measured speed of light is always $c=1$. This seems contradictory.
Where do I go wrong? Is it correct to say that $$\frac{\text{d}r}{\text{d}\tau}=\sqrt{2M/r-2M/r_0}$$ is the locally measured speed even within the event horizon and can this be greater than $c=1$?
I've read quite a few Q&A's in stackexchange, but I couldn't find and answer to this question.

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    $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$
    – David Z
    Commented Jun 5, 2020 at 0:08

2 Answers 2


The problem is that in GR coordinates do not necessarily have a physical meaning. They are just a way of labelling points in spacetime. For example the Schwarzschild coordinate $r$ is not a radial distance. It's actually the circumference of the circle centred on the black hole and passing through your point divided by $2\pi$. That is, it's what the radial distance would be if space were flat.

This means that the coordinate velocity $dr/dt$ also does not have a physical significance. You can certainly calculate $dr/dt$, e.g. for the observer at infinity, and you would find the coordinate velocity for that observer is:

$$ v = \left(1 - \frac{r_s}{r}\right)\sqrt{\frac{r_s}{r}}c \tag{1} $$

giving the notorious result that the infalling object slows to a halt at the horizon.

Alternative you could ask what an observer hovering some distance $r$ (these are known as shell observers) would observe i.e. what speed would the falling object pass them. And in that case the result is:

$$ v = \sqrt{\frac{r_s}{r}}c \tag{2} $$

and now we find that the speed the falling object passes the shell observer tends to $c$ as the shell observer approaches the horizon. The difference between the two results is due to the relative time dilation between the shell observer and the observer at infinity.

More generally the shell observer and the observer at infinity will always observe different velocities. If you're interested I go into this in detail in my answer to the question Does light really travel more slowly near a massive body? As explained in that answer the coordinate velocity can be greater than $c$ even outside the horizon, and this is because the coordinate velocity is not a physically meaningful quantity.

Now you are asking about the velocity inside the horizon, but this is even harder to discuss in any meaningful way. For any observer outside the horizon no object ever passes the horizon, so there is no velocity inside to observe. And inside the horizon it's impossible to stay at fixed $r$ so we cannot have any shell observers to watch the falling object flash past them. I guess the best we could do is ask how fast the infalling observer observes the singularity to be approaching, though note that this is theoretical since no light from the singularity could ever reach the observer's eye. And the answer that the speed would indeed exceed $c$ inside the horizon, though I must emphasise again that you shouldn't assign any physical sigificance to this.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    Commented Jun 5, 2020 at 0:08

You can use Gullstrand–Painlevé coordinates to calculate the speed of an object falling into a black hole both inside and outside of the event horizon. To be clear the speed would be that observed by a distant observer. It involves a coordinate transform that removes the singularity in Schwarzchild metric that occurs at the event horizon. For a "raindrop" falling in from rest at infinity:

$$\frac{dr}{dt_r} = -\sqrt{\frac{2M}{r}}$$

This is the negative of the Newtonian escape velocity FYI. It equals the speed of light at the event horizon and is faster within it.

Just FYI I found the above on the wikipedia article about Gullstrand–Painlevé coordinates

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    $\begingroup$ That's not the speed measured by a distant observer, that's the speed measured by a local and stationary observer. $\endgroup$
    – Yukterez
    Commented Dec 17, 2023 at 22:58
  • $\begingroup$ And there is no stationary observer inside the event horizon. $\endgroup$
    – ProfRob
    Commented Dec 17, 2023 at 23:02
  • $\begingroup$ This is what would be observed by the distant observer if they could communicate with the stationary observers, if those could exist. $\endgroup$
    – dllahr
    Commented Dec 18, 2023 at 17:04

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