This is my third question from a series were I progressively refine my thought experiment. The others are:

Assume I am stationary over a Schwarzschild black hole of mass $M$. Some force is keeping me at a distance $d$ over the event horizon (or the black hole center, whatever is easier to answer). At time $0$ I drop a flashlight blinking at frequency $f$ into the black hole. At what time $T_{M, d}(f)$, in my own frame of reference, will I see the last blink of the flashlight?

Some assumptions (non-comprehensive list):

  • Assume only general relativity (the world is not quantized, no Hawking radiation);
  • The flashlight is not destroyed/spaghettified by the tidal forces;
  • I can always detect the flashlight as long as it has not crossed the event horizon;
  • I can set $f$ to any arbitrarily big frequency.

It would be nice if the answer was the actual function/algorithm describing $T_{M, d}(f)$, so I could graph for increasing values of $f$, but the real information I want to know is:

Is it true that $$\lim_{f \rightarrow \infty} T_{M, d}(f) = \infty \;\;\;\; \text{?}$$

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    $\begingroup$ Why are you more interested in what outside observers see than in what actually happens to the flashlight, which is that it crosses the horizon and soon after reaches the singularity? $\endgroup$
    – G. Smith
    Commented Aug 29, 2019 at 21:58
  • $\begingroup$ Because if the limit actually tends to infinity, and I plug back Hawking radiation to the world, I can definitely say nothing ever falls into the black hole, and just before reaching the event horizon, the black hole explodes, making the issue of what happens to the flashlight after the event horizon a moot point. $\endgroup$
    – lvella
    Commented Aug 29, 2019 at 22:09
  • $\begingroup$ Good luck with that. $\endgroup$
    – G. Smith
    Commented Aug 29, 2019 at 22:11
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    $\begingroup$ It is not even a fringe idea, it is simply based on not understanding black holes. $\endgroup$
    – TimRias
    Commented Aug 30, 2019 at 7:26
  • 2
    $\begingroup$ You cannot however properly graph your function, because the time to the last flash can be arbitrary long at any frequency of flashing. What matters is not the frequency, but how close to the horizon the last flash happens. It can happen by chance very close even at a very low frequency. $\endgroup$
    – safesphere
    Commented Aug 30, 2019 at 8:59

2 Answers 2


The coordinate acceleration by proper time for a radially infalling observer is simply the Newtonian expression

$$ \ddot{r} = \frac{{\rm d}^2 r}{{\rm d} \tau^2} = -\frac{G M}{r^2}$$

which integrates to

$$ \dot{r} = \frac{{\rm d} r}{{\rm d} \tau} = -\sqrt{\frac{2 G M (r_1-r)}{r_1 \ r}} $$

so if you start at rest at $r_1$ and freely fall to $r_2$ the elapsed proper time is

$$ \tau = \int_{r_1}^{r_2} \frac{1}{\sqrt{2 G M \left(\frac{1}{r}-\frac{1}{r_2}\right)}} \, \text{d}r$$

The time dilation relative to the coordinate bookkeeper far away from the black hole is

$$ \dot{t} = \frac{{\rm d} t}{{\rm d} \tau} =\sqrt{\frac{c^2 \ r-c^2 \ r_s +r \ \dot{r}^2}{c \ (r- r_s ) \ (1-r_s/r)}} $$

For example let's plug in some numbers; if you fall from the photon sphere at $r_1=1.5 \ r_s$ you reach the horizon at $r_2=r_s=2G M/c^2$ when your proper time is $\tau=3.993468$ so if you set the intervall of your flashlight to give $100$ pulses per $1GM/c^3$ proper time, the last received pulse is that given at $\tau=3.99 \ G M/c^3$ when the flash light is at $r=2.001999 \ G M/c^2$, which is at coordinate time $t=18.43309 \ G M/c^3$.

If you want to know when this last pulse of light is observed you have to add the light travel time; the shapirodelayed coordinate velocity of a radially outgoing light ray is

$$ r' = \frac{{\rm d} r}{{\rm d} t} = c-\frac{2 G M}{r c} $$

so the light travel coordinate time from where that signal was emmitted to the observer at $r_1=3 G M/c^2$ is $t_{\rm \ light} = 13.42798 \ G M/c^3$, so the last signal which is emmitted outside of the horizon will be received at $t_{\rm obs}=t+t_{\rm \ light}=31.86107 \ G M/c^3$, or in terms of the proper time of the stationary shell observer at $r=r_1$ that $399 \rm th$ and last signal would be received at ${_{T}}_{\rm obs}=t_{\rm obs} \sqrt{1-r_s/r_1}=18.395 \ G M/c^3$.

If you set the frequency of your flashlight to the limit of infinitely many pulses per unit proper time, the last received signal will also be seen only after an infinite coordinate- or shell-time.

  • $\begingroup$ Can I just ask the reference for the following? $$\dot{t} = \frac{{\rm d} t}{{\rm d} \tau} =\sqrt{\frac{c^2 \ r-c^2 \ r_s +r \ \dot{r}^2}{c \ (r- r_s ) \ (1-r_s/r)}}$$ $\endgroup$ Commented Aug 31, 2019 at 14:32
  • $\begingroup$ The total time dilation in natural units and in terms of the local velocity and the lapse function is 1/√(1-v²)/√(1-2/r), so if you set the tangential velocity component to 0 and dr/dτ=p_r·√g_rr, where p_r is the specific covariant linear radial momentum v_r/√(1-v²), you get this expression, see notizblock.yukterez.net/viewtopic.php?p=554#p554 or en.wikipedia.org/wiki/… $\endgroup$
    – Yukterez
    Commented Aug 31, 2019 at 16:16

There is nothing like some badly drawn paint art to explain relativity. So here we go.

Below we see a spacetime diagram of the experiment. This type of diagram (a Penrose diagram) is drawn is such a way that light always travels on diagonal lines.

enter image description here

Between the time that you (the green line) drop the flash light (the purple line) and that the flashlight crosses the event horizon (blue diagonal) a finite amount of proper time will pass for the flashlight. Let call that $T$.

If the flashlight flashes with a frequency $f$ the flashlight will emit $n =\lfloor T/f\rfloor$ flashes before crossing the event horizon.

Each of these flashes (and only these flashes) will reach you. However, for each subsequent flash it will take longer to reach you.

So yes, in the limit that $f\to 0$ the number of seen flashes $n$ will go to infinity. From the diagram it is also immediately clear that this answer does not depend on whether the flashlight is freely falling or rocket power (this will change $T$ but it will always be finite if the flashlight crosses the event horizon), or whether you keep position, orbit the black hole, or return to the distant universe.

EDIT: In the limit $f\to 0$ it also follows that the last flash is emitted arbitrarily close to the event horizon. Since this flash follows the same diagonal slope as the event horizon, this means that this flash intersect the green line (you) arbitrarily close to the top right corner of the diagram, which represent $t\to\infty$ for your clock. So the arrival time of the last flash on your clock will also tend to infinity. Note that this is not because it took such a "long time" for the last flash to be emitted, but because it took a "long time" for the signal to reach you.

EDIT2: I've add another diagonal red line. Once you (the green line) has crossed this redline, there is no way for you the ever send a signal to the flashlight again. In particular, you could no longer dive into the black hole after your flashlight and reunite with it.

*The astute reader will wonder what "long time" means, and rightly so. For the purpose of this statement, lets just think about it as far away in the diagram.

  • 1
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Chris
    Commented Aug 30, 2019 at 10:33
  • $\begingroup$ 'Note that this is not because it took such a "long time" for the last flash to be emitted, but because it took a "long time" for the signal to reach you.' This is in direct contradiction with the other answer. According to it, $\frac{dt}{d \tau}$ approaches infinity as the object approaches $r_s$. $\endgroup$
    – lvella
    Commented Sep 7, 2019 at 23:15
  • $\begingroup$ @Ivella It only is, if you insist on using the Schwarzschild time as your time coordinate. Is the diagram show, this is not necessarily a good idea. $\endgroup$
    – TimRias
    Commented Sep 9, 2019 at 7:42
  • $\begingroup$ @mmeent Why it is not a good idea? The objects crosses the event horizon in a finite amount of $\tau$, but in a infinite amount of $t$, that much I get it. But considering the hawking radiation, the black hole evaporates in a finite amount of $t$, thus $t$, not $\tau$, seems much more important to me. $\endgroup$
    – lvella
    Commented Sep 16, 2019 at 15:03
  • $\begingroup$ @Ivella Because most of the evaporation happens in a region of spacetime well-distinct from where the object crosses the horizon. In terms of measures of time that make sense in the vincinity of the horizon, the object crosses the horizon long before the black hole starts to evaporate. $\endgroup$
    – TimRias
    Commented Sep 16, 2019 at 15:41

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