I am observing (theoretically) an object falling into a black hole from a safe distance away. My understanding is that from far away it appears as if the body will asymptotically approach the event horizon, but I will never see it cross it. Although the object will flatten itself and encompass the entire event horizon its center of mass will slow down.

I deduce that there is a radially repulsive force to counter act gravity from my point of view. Is there a different way to do $\sum F=m \ddot{r}$ where I do not have this fictitious force, and if not what do we call this force? This is equivalent to a centripetal force needed to keep something in orbit, although in this case it is the opposite.


From the responses I see the free-fall EOM is

$$\sum F=ma=\frac{GMm}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}}$$

So the local gravitational acceleration is $$g = \frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}}$$

and any motion besides a free fall is $F_{ext} = m (\ddot{r}-g)$ or

$$ \ddot{r} = \frac{r_S c^2}{2 r^\frac{3}{2} \sqrt{r-r_S}} $$ with $r_S = \frac{2 G M}{c^2}$.

So how can this produce $\ddot{r} \approx 0$ near the event horizon?

  • 1
    $\begingroup$ The title of the question does not correspond to the question. Your question seems to correspond to an observer which is at a fixed Schwarzschild radial coordinate ("outside" the black hole), so it is not an inertial frame. Only free falling observers correspond to an inertial frame. $\endgroup$
    – Trimok
    Commented Nov 1, 2013 at 20:34
  • $\begingroup$ Ok if I am watching from really far away then. $\endgroup$ Commented Nov 1, 2013 at 22:52
  • $\begingroup$ This is not helping in answering the question, but do you have a working definition of what the expression "falling into a back hole" means? For example, does it imply the radial distance between the object and the black hole shrinks in time? $\endgroup$
    – Nikolaj-K
    Commented Nov 1, 2013 at 23:07
  • 2
    $\begingroup$ I find the discussion of what far away means a bit odd. perhaps I should not make this assumption, but it seems obvious to me that ja72 wants to use the Schwarzschild coordinates. The difference between Schwarzschild and shell coordinates can be made negligable by hovering far enough away from the black hole. $\endgroup$ Commented Nov 2, 2013 at 10:07
  • 1
    $\begingroup$ this description for "what happens" cosmology.berkeley.edu/Education/BHfaq.html#q4 might help $\endgroup$
    – anna v
    Commented Nov 3, 2013 at 11:57

3 Answers 3


In Newtonian mechanics the equation you use to describe the trajectory of the falling particle is the usual $F = ma$. In General Relativity you shouldn't be surprised that you use a more sophisticated equation that gives the Newtonian equation as the low curvature limit. This equation is called the geodesic equation:

$$ \frac{d^2x^\gamma}{dq^2} + \Gamma^\gamma_{\mu\nu} \frac{dx^\mu}{dq} \frac{dx^\nu}{dq} = 0 $$

The variable $q$ parametrises the trajectory - it's proportional to the proper time i.e. the time measured by the falling particle. The equation gives you $t(q)$, $r(q)$, $\theta(q)$ and $\phi(q)$ in whatever coordinate system you wish to work in.

You can calculate the EOM of the falling particle then invent a fictional force that is a function of distance to make Newton's equation work, but this would be an intellectual exercise only and wouldn't have any physical significance.

If you're interested, Phil's answer to What is the weight equation through general relativity? analyses a similar problem.


Re your comment to Stan's answer, there's nothing physically opposing the motion of the falling body. What you're seeing is analogous to the length contraction and time dilation you get in special relativity. The radial coordinate $r$ is defined as the circumference of a circle drawn round the black hole divided by $2\pi$. It is not the same as the length you'd measure if you let down a tape measure towards the event horizon. So you shouldn't be surprised that the speed you get by dividing $dr$ by $dt$ turns out a bit odd.

To see this look at this diagram:


The $dr$ you're using to calculate the speed is the distance on the flat plane, and it's immediately obvious that this isn't the same as the proper distance, $ds$, you get by integrating $dr$ along the trajectory. You can calculate the proper distance simply by holding the other parameters constant and integrating the metric:

$$ \int ds = \int_{r_1}^{r_2} \frac{dr}{\sqrt{1 - r_s/r}} $$

As $r_2 \rightarrow r_s$ this integral goes to infinity. So your result is that the falling body takes an infinite time to move an infinite distance.

But this doesn't mean that the distance to the event horizon is infinite any more than it means time stops there. All it means is that $r$ and $t$ are not the simple quantities that we dwellers in low curvature spacetime think they are.

  • $\begingroup$ Thanks for the link. I still don't see how a free falling particle stops moving without the influence of an external force, other than gravity. $\endgroup$ Commented Nov 2, 2013 at 19:41
  • $\begingroup$ The first $=$ should be a $+$. $\endgroup$
    – Stan Liou
    Commented Nov 2, 2013 at 21:28
  • $\begingroup$ @StanLiou: thanks :-) Please feel free to edit any answer of mine if you spot such egregious mistakes! $\endgroup$ Commented Nov 3, 2013 at 7:18
  • $\begingroup$ @JohnRennie Nice graph. So yes I will be observing time dilation somehow and in order for the laws of physics to be valid for all reference frames maybe I interpret the free fall as an increase in mass $m= m_0/\sqrt{1-\left(\frac{v}{c}\right)^2}$ or something just like in special relativity. Am I correct? $\endgroup$ Commented Nov 5, 2013 at 17:11
  • $\begingroup$ @ja72: that is a truly horrible way to interpret the free fall. Any self respecting physics teacher will throw you from the class for mentioning relativistic mass as it's a concept that causes nothing but confusion, and indeed has done so for generations of unfortunate physics students. You should not attempt to force fit GR into matching your intuition. Just go with the maths. My point above was just to show that the $r$ coordinate is not what you intuitively think it is. $\endgroup$ Commented Nov 5, 2013 at 17:16

I will assume a Schwarzschild black hole with mass $M$ and Schwarzschild coordinates, $-+++$ sign convention, and units of $G = c = 1$.

Because the metric is independent of $t$, $\partial_t$ is a Killing vector field, and hence for any geodesic with tangent vector $u$, there is a constant of motion $$\epsilon = -\partial_t\cdot u = \left(1-\frac{2M}{r}\right)\frac{dt}{d\tau}\text{.}$$ Similarly, $\partial_\phi$ is another Killing vector corresponding to angular momentum conservation, but for radial freefall we don't care about that. Also, I assumed the geodesic is timelike, and hence the affine parameter can be taken to be the proper time $\tau$ along it.

Together with the timelike condition $u\cdot u = -1$, the constant of motion above implies that $$\frac{1}{2}\dot{r}^2 - \frac{M}{r} = \frac{\epsilon^2-1}{2} = e$$ is constant, where $e$ is the relativistic analogue of specific mechanical energy of the orbit, and overdot represents differentiation with respect to $\tau$. Differentiating this, we get: $$\ddot{r} = -\frac{M}{r^2}\text{.}$$ Since we actually want to get the equation of motion with respect to Schwarzschild coordinate time, $$\frac{dr}{dt} = \dot{r}\frac{d\tau}{dt}\text{,}$$ $$\frac{d^2r}{dt^2} = \frac{d\tau}{dt}\frac{d}{d\tau}\left[\dot{r}\frac{d\tau}{dt}\right]\text{,}$$ and all that remains is plugging and chugging.

But it is obvious that these quantities go to $0$ as $r\to 2M$: $\epsilon>0$ if the particle starts anywhere outside the horizon, and since $\dot{r}$ is finite is well-behaved there while $\frac{d\tau}{dt} = \left(1-\frac{2M}{r}\right)\epsilon^{-1}\to 0$, the claim follows.

  • $\begingroup$ I am following, but what is the conclusion? How would a falling particle appear from "far away" if it is in free-fall? If the apparent motion stops, what is stopping it from my point of view? $\endgroup$ Commented Nov 3, 2013 at 0:29
  • $\begingroup$ @ja72: I think I see what you're getting at, and I've edited my answer accordingly. Have a look and see if it helps. $\endgroup$ Commented Nov 3, 2013 at 10:41
  • $\begingroup$ I'm not sure what you mean. I interpreted your question as asking how to derive $d^2r/dt^2$ and showing that it is $0$ near the horizon. If that's not it, could you clarify? For a fast enough object in radial freefall, this quantity can even be always positive, as the object moves to regions of greater gravitational time dilation. $\endgroup$
    – Stan Liou
    Commented Nov 3, 2013 at 22:44

As other answers pointed out, you can interpret it as either dilation of time, increase in distance, increase in inertial mass, depending on the system of coordinates and terminology you use.

I also can point out that you can interpret it as frame-dragging, a second-order gravitational force.

  • $\begingroup$ You seem to be saying that those are all mutually exclusive but shouldn't time dilation and proper distance increase be considered together? Wouldn't an external observer who watches someone fall into a black hole see both their time being dilated and their proper distance increasing (which would look like length contraction) and see that both effects contribute to the apparent slowing of the fall? $\endgroup$ Commented Apr 4, 2022 at 21:16

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