I've had some discussions recently about black holes, and the issue of infalling bodies taking forever to reach the event horizon. That's essentially what Einstein said in his 1939 paper on a stationary system with spherical symmetry consisting of many gravitating masses. He said "it is easy to show that both light rays and material particles take an infinitely long time (measured in “coordinate time”) in order to reach the point r = μ/2 when originating from a point r > μ/2”.

Now, I'm a big fan of Einstein. But there seems to be two issues with this:

  • One is that Einstein concluded that black holes cannot form, but we have good evidence that there are black holes out there. The obvious example is Sagittarius A*. That's where's there’s something with a mass 4.28 million times the Sun, with a diameter of less than 44 million kilometres, and we can't see it. Surely it's just got to be a black hole.

  • The other issue is that falling bodies don't slow down. Imagine you drop a body at elevation A and it falls down to elevation B. The "force" of gravity relates to the first derivative of gravitational potential. Hence the bigger the difference in gravitational time dilation between elevations A and B, the faster the body falls past B. Then if you drop a body at elevation B it falls down to elevation C. Again the bigger the difference in gravitational time dilation between elevations B and C, the faster the body falls past C. In a typical gravitational field the force of gravity at B is greater than at A. Hence as the body descends, the acceleration increases as well as the falling speed.

Imagine a gedanken spaceship from which we've suspended a cable. We have clocks at different elevations, so we can measure the gravitational time dilation at each elevation. We can also release test bodies at each elevation and record the clock readings as they fall past other elevations:

enter image description here

At the end of the experiment we can reel in the cable and upload the recorded measurements to ascertain how our test bodies behaved. My understanding is that we will always find that time dilation always increases as we descend, that the falling body always accelerate downwards, and that both the acceleration and the falling speed always increases as the body descends. Is this correct? Or do falling bodies somehow stop accelerating? And do falling bodies ever slow down?

  • 4
    $\begingroup$ It depends on frame of reference. As you know. $\endgroup$
    – ProfRob
    Feb 1, 2019 at 0:22
  • $\begingroup$ Since information cannot be retrieved once it has crossed the event horizon, I assume that the lowest data-collecting point (clock C) is outside the horizon. That being the case, and since the observer is local, is there a reason the motion of the test bodies would not be as expected (i.e. accelerating)? $\endgroup$ Feb 1, 2019 at 6:17
  • $\begingroup$ @chappo : yes, the lowest collecting point is outside the event horizon. I don't think there's any reason why the motion of the test bodies wouldn't be as expected. $\endgroup$ Feb 1, 2019 at 11:17
  • $\begingroup$ @Rob Jeffries : I don't know. And as far as I do know, references frames have no objective existence, but all observers will agree that the lower the clock, the slower it's going. Following your comment here I'd be grateful if you'd answer this question. $\endgroup$ Feb 1, 2019 at 11:23
  • 1
    $\begingroup$ @JohnDuffield I'm afraid most people who have studied this are confident that relativity is pretty clear and consistent on these points and that there isn't a question, at least until you enter the parts where quantum mechanics starts to be significant, which this question doesn't. $\endgroup$ Feb 1, 2019 at 17:37

1 Answer 1


According to the standard interpretation of General Relativity (e.g. as presented in "Exploring black holes" by Taylor & Wheeler chapter 3, or "Black holes, white dwarfs & neutron stars" by Shapiro & Teukolsky, pp 343-345) then yes they do. But it depends on the frame of reference of the observer - there is no absolute answer.

According to an observer far from the black hole, the rate of change of radial coordinate with time (for an object that started falling radially inwards far from a non-rotating black hole) is given by $$\frac{dr}{dt} = -\left(1 - \frac{r_s}{r}\right)\left(\frac{r_s}{r}\right)^{1/2}$$ where $r_s$ is the Schwarzschild radius and $r$ and $t$ are Schwarzschild coordinates.

If we call this the infalling speed as measured by a distant observer then we can see by differentiation that it goes through a maximum at $r=3r_s$ and that $dr/dt \rightarrow 0$ as $r \rightarrow r_s$.

However, an observer accompanying the falling particle would totally disagree. To them, their velocity is given by $dr/d\tau$, the rate of change of $r$ with respect to the time on their clock. $$\frac{dr}{d\tau} = -\left( \frac{r_s}{r}\right)^{1/2}$$ which continues to increase up to and below the event horizon.

The latter appears to admit the possibility of faster than light travel, but no more so than me (correctly) saying that if you travel at close to the speed of light you can get to a star 10 light years away in much less than 10 years (as measured on your clock).

Finally we could have the point of view of stationary "shell" observers at fixed radial distances (outside the event horizon, because no stationary observer is possible below the event horizon). They would measure the speed of objects falling past them to be $$\frac{dr_{\rm shell}}{d\tau_{\rm shell}} = -\left(\frac{r_s}{r}\right)^{1/2}.$$ This means that the reports of stationary observers (which is the gist of your question I think) at increasingly lower heights, is indeed that the velocity of the falling object is increasing as it falls.

There is no paradox to these apparently contradictory points of view. Measurements of non-local events and phenomena are not bound to agree in General Relativity, where there isn't even agreement on what is meant by "now" or whose coordinate system in what frame of reference should be used in any particular circumstance.

  • $\begingroup$ Thanks for the answer Rob, +1. Let's keep this simple by staying outside the event horizon. So, falling bodies increase their speed according to some observers, but slow down and stop after r=3r$_s$ according to others. According to Taylor & Wheeler. That sounds like a paradox to me. Surely something is badly wrong here? The infalling body does what it does, and what it does doesn't depend on some observer. I've got a great idea. Why don't you ask a question pointing this out, and then ask if anybody can resolve this paradox? $\endgroup$ Feb 1, 2019 at 13:04
  • 2
    $\begingroup$ Nothing's wrong, different observers always measure different speeds. Even Galileo understood that. All that is wrong is the now-dead concept that speed is a function of the object rather than a function of the coordinates used to analyze the object. $\endgroup$
    – Ken G
    Feb 1, 2019 at 13:56
  • $\begingroup$ Everyone would agree on actual events. For example if your observer C had a measuring rod and a clock, everyone might agree that the falling object passed the top end of the measuring rod when the clock said 0 and the bottom end when it said 1. But the spaceship. A, B and C would all have different ideas about how long it took the clock to go from 0 to 1 and consequently how fast the falling object was going. The falling observer might also disagree about the length of the measuring rod. $\endgroup$ Feb 1, 2019 at 14:05
  • $\begingroup$ If you hauled A, B and C up afterwards you would genuinely find that their clocks had recorded different amounts of total elapsed time, so this isn't just some kind of illusion. $\endgroup$ Feb 1, 2019 at 14:06
  • 1
    $\begingroup$ Not every observer agrees the time dilation increases, time dilation is a coordinate issue. The only, yes only, situation in which elapsed time is an invariant observable is when it is proper time, i.e., the time registered on a clock that is present at two events, and then the clock registers the proper time between those events along that path connecting them. That is the sole, and only, kind of time interval that is not a matter of coordinates. $\endgroup$
    – Ken G
    Feb 4, 2019 at 3:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .