I have read this answer from John Rennie.


Where he says that to a Schwarzschild observer, anything reaching the event horizon will seem to be frozen there forever.

Now if we send mirrors from all angles towards the black hole so that they reach the event horizon horizontally (perpendicular to the center of the black hole) with their reflective side outwards, then we can practically wrap the black hole in mirror gift-wrapping.

If these mirrors will seem to be frozen at the event horizon forever in fact, then they will reflect light.

This is where I am getting confused. If the mirrors are visible to an outside observer, then do they see their own actions in the mirror, or do they see an image from the past in the mirror (from the past moment when the mirror reached the horizon)?

This is a contradiction, because if they see a frozen image, then that image needs to be made of photons that are reflected by the mirror only once when the mirror reached the horizon. How can these photons keep being reflected for all time? That needs more and more photons coming from the horizon from the mirror surface. If the image is frozen, then the new photons cannot be ones reflected from our present (outside). Because in that case, the image would change with us. If the image is frozen, it must be made of photons from the past. Now those photons reflect our past once, understandable. But how can photons keep coming from the mirror with the same image? Those new photons with the same image must be reflected from somewhere originally? The mirror cannot keep producing new and new amounts of photons to create the same image? The mirror will either reflect those photons or will receive and reflect new photons from outside and keep reflecting them with new images.

If they see an interactive, not frozen image, then it is a contradiction too, because the interactive image means photons are being reflected all the time from the mirror, making our present life visible in an interactive way. This needs the mirror to be actually there physically frozen (from an outside observer's view). In this case, the mirror receives photons from outside and reflects them back, but to do that, the mirror must physically be there (in the observer's frame).

The only logical solution might be, that the mirror viewed from the outside observer's frame is physically there frozen at the horizon. In that case, the mirror will reflect an interactive, not frozen image.


  1. Will they see their present being reflected from the mirrors interactively going on?

  2. Or will they see their image frozen from the past (moment in past when the mirror reached the horizon)?

  • $\begingroup$ The mirror never reaches the horizon, from the faraway observer's point of view. It never stops approaching the horizon, though it does so more and more slowly over time. It also redshifts more and more as time passes, so there will eventually be a time when the radiation from the mirror will be at a lower frequency than your detector can detect. So nothing is "frozen" in this situation. Also, if you put an ordinary mirror a light-year away in flat space, then the image you see in the mirror will be you from two years ago. $\endgroup$ Aug 2, 2018 at 19:42

1 Answer 1


First of all, only if we can neglect the mass of the mirror, can we assume that it will be forever frozen at the event horizon by the clock of an outside observer. If the mirror has a small but finite mass $m$ the event horizon would expand to account for a new mass and the mirror would thus cross the horizon on a timescale $\frac{r_s}{c}\ln \frac Mm$ (where $M$ is the mass of a black hole) from, say, the moment the mirror crossed photon sphere, $r=\frac32 r_s$. If we double that time for the reflected light to climb back to a reasonable distance away from a black hole or be redshifted beyond any threshold of detection, after that there is no trace of the mirror at all, we have only a black hole without any observable outside features. For stellar mass black holes the time interval would be quite small from a macroscopic point of view: for a black hole of $1 M_\odot$ the $\frac {r_s}{ c} \approx 10^{-5}\,\text{s}$ and assuming that "mirror" is made of single sheet of graphene (specific surface area $2630\,\text{m}^2/\text{g}$) the $\ln$ factor would be about $70$, so within about $10^{-3}\,\text{s}$ after the mirror fall reasonably close to the black hole horizon there would be nothing left to see.

But, even if we assume that the mirror is absolutely weightless and by the clock of an outside observer it never crosses the horizon, still the reflections from it would be observable only for a finite time. The main reason is that there is only finite time interval during which the light sent from a static observer some distance away from the horizon chasing the falling mirror would be able to catch up with it before it enters the horizon (remember that the horizon crossing occurs in a finite time by the moving mirror own clocks). So there is a moment in time past which nothing is reflected. (And that moment is not the moment when the mirror crosses the horizon but instead, some earlier moment from an outside light source when the photon sent after the mirror would catch up to it just as it crosses the horizon).

As the amount of energy that could be reflected is finite and the time of photon's return trip back to an outside observer will increase indefinitely the closer to the horizon the mirror gets, the power of reflected radiation would be dropping exponentially to zero and would get below the threshold of observation in a finite time (again the characteristic timescale for this process is Schwarzschild crossing time $\frac{r_s}c$).

More than that, from the energy that gets to the mirror before the horizon the fraction that could actually get back to outside observers, is also decreasing and drops to zero as the mirror approaches the horizon. This is due to two effects. Fist, the normal component of photons momentum is redshifted if reflected from the mirror moving away from it. Second, a photon may be able to escape the near-horizon region only if it does not have too large angular momentum to energy ratio. In other words, the photon momentum must be within an "escape cone" around a radial normal to the mirror, which gets narrower the closer to horizon the mirror gets (for example for $\frac{r-r_s}{r_s}=10^{-5}$ the angle would be about $1^\circ$). Since reflection does not change the angular momentum of a photon but lowers its energy for a mirror near the horizon only photons moving near radially would reflect on the trajectory escaping the black hole and their energy (as measured by a static observer) would be significantly redshifted by this reflection. The rest of the reflected photons would be moving along the trajectories that would fall back into the horizon.

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    $\begingroup$ Would it be possible to show how one can compute the time scale $r_s/c \ln(M/m)$; or perhaps give a reference where one can find this calculation? Thanks $\endgroup$ Aug 7, 2018 at 6:51
  • $\begingroup$ @flippiefanus: My answer here gives an outline of calculations and a reference to the book by Frolov & Novikov. $\endgroup$
    – A.V.S.
    Aug 7, 2018 at 14:33

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