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I am not asking what happens to light that is emitted from a flashlight. I am specifically asking what happens to light that hits a mirror inside a black hole.

I have read this question:

So inside the horizon even a light ray directed outwards actually moves inwards not outwards.

How does light behave within a black hole's event horizon?

And this one:

Light cannot move outwards inside the event horizon. I would guess you're thinking that an outgoing light ray might leave you in the outgoing direction, then slow to a halt and return - hence you would see yourself. However this doesn't happen. The light leaving you moves inwards not outwards, but since you fall inwards faster than the light does, the light still leaves you (at velocity c) and never returns. The briefest glance at the diagram shows that your worldline and the worldlines of the light rays can only intersect at one point, i.e. the point you shine the light rays inwards and outwards.

Would the inside of a black hole be like a giant mirror?

enter image description here

If you shoot a light beam behind the event horizon of a black hole, what happens to the light?

So basically these ones say that the light that would hit a mirror, would actually leave the mirror, because the mirror itself (and the observer with it) is moving faster then the light, so the light could only interact with the mirror once, and never return to the mirror.

Now imagine this setup, the observer is holding a flashlight and a mirror, the observer is facing inwards, and the mirror that the observer is holding is facing outwards (towards the observer), so that normally you could see yourself in the mirror, because light from the flashlight would bounce back from the mirror into the observer's eyes. But, based on these answers, the way light moves is different inside the black hole.

I am asking about the case when some light would hit the mirror.

Now this is the part I am asking about, what happens to the light that actually hits the mirror. The law of reflection states that the angle of incidence must be equal to the angle of reflection.

But based on the answers, this cannot happen, because when the light hits the mirror(facing outwards), it cannot move outwards, it must move inwards (even light must move towards the singularity). But that is not allowed, because the mirror is facing outwards. Based on the law of reflection, the inwards moving light (that is hitting the mirror) should be reflected outwards (which is not allowed because that would mean moving away from the singularity). Does that mean that the law of reflection is not valid inside the black hole?

Question:

  1. Inside a black hole, can I see my reflection in a mirror (is the law of reflection still valid)?
Árpád Szendrei
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    Tell me the truth. Is this question from your wife? – Xfce4 Sep 20 '21 at 22:39
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    Ultimately we can never know what goes on past the event horizon (by definition), so there's no meaningful way to answer the question. We could speculate, but those speculations cannot be tested, at least not in any meaningful way (you could jump into a black hole to satisfy your curiosity, but you'll never be able to tell anyone else what you found). – Eric Smith Sep 21 '21 at 02:29
  • @safesphere I think I understand. But do you think the mirror reflection would still work though? – Árpád Szendrei Sep 21 '21 at 18:23
  • Impossible? I see reflections from my mirror's past and send signals to my own future every day. – g s Sep 21 '21 at 19:07

5 Answers5

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The mirror works more or less like you'd expect it to anywhere else. If the mirror and the emitter/observer are separated by a distance sufficient that there is a tidal difference in the gravitational field between them such that the acceleration due to gravity is different by $\Delta g$, then it looks like the observer is accelerating away from the mirror at $a=\Delta g$.

Below, a MSPaint graph of the path of the emitter / observer, the photon, and the mirror, for an emitter and mirror separated by a distance $\Delta r << r$, with the mirror directly below the emitter. The origin is an arbitrary point inside the event horizon below the emitter, $r_s>r>>0$.

The purple curve (mirror path) should be $\Delta r$ distance away from the blue curve (emitter path) at all time. It isn't, because my MSPaint skills are not that good.

enter image description here

Note how the grey path, which is what our observer/emitter sees, is the exact path we would expect for a light beam in no gravitational field, going out at c, bouncing off of a mirror, and coming right back to where it started at c.

If we had tidal effects (or if the emitter is on a rocket boosting away from the center while the mirror is in free fall), the second $\Delta r$ would be larger, so the photon would take longer to get back to the emitter. The observer/emitter measures a time dilation appropriate to a gravitational field of $\Delta g$, exactly as if they were accelerating away at $\Delta g$ far away from any gravitational fields.

Edit with correction for the graph: the grey path is emitter path minus photon path, not the other way around.

g s
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  • Thank you so much! Are you saying that the photon would actually bounce back from the mirror? – Árpád Szendrei Sep 21 '21 at 01:42
  • Yes, from the frame of a massive object inside the event horizon. – g s Sep 21 '21 at 04:21
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    As a side note, for human-scaled things, you need a really, really big black hole before you can start to approximate $\Delta g \approx 0$. I'm not even sure if the kinds of black holes found at the centers of galaxies would be big enough. – g s Sep 21 '21 at 05:02
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    The tidal acceleration at the event horizon are on the order of $d c^6/(GM)^2$, where $d$ is the size of the object falling in and $M$ is the mass. For $d \approx 1$ m, this works out to be about $4 \times 10^{8} \text{ m/s}^2$ for $M \approx 10 M_\odot$ (which is obviously no good) but only about $3 \times 10^{-2} \text{ m/s}^2$ for Sagittarius A*, $M \approx 4 \times 10^6 M_\odot$ (which is harmless.) – Michael Seifert Sep 21 '21 at 15:24
  • @gs stellar-sized black holes are incompatible with human-sized observers even before you reach the event horizon. Supermassive black holes should be fine, as long as they're not currently eating... assuming they obey raw general relativity on the inside, which they probably don't, but the question assumes they do, so w/e – John Dvorak Sep 21 '21 at 15:32
  • @MichaelSeifert 3e-2 m/s^2 is actually ... much larger than I expected. – John Dvorak Sep 21 '21 at 15:34
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There's a sense in which you can't define an inwards/outwards direction in the interior of a black hole, but this doesn't mean that mirrors suddenly start misbehaving.

A similar thing happens in cosmology. We know the spatial curvature of the universe is close to zero. If it's slightly positive (which is within the error bars) then the universe is spatially closed and the total number of galaxies is finite, but if it's zero or slightly negative then the total number of galaxies is (or at least can be) infinite. In a spatially closed universe, you can argue that it's impossible to go "away" from Earth because you're also going "toward" Earth along the other segment of the great-circle geodesic that encircles the whole universe. As soon as the curvature reaches zero, that is no longer true. But nothing visibly changes locally when the curvature passes that threshold. We can't even tell the difference with our best instruments.

Similarly, when you cross the event horizon, the spacetime curvature passes a threshold and there's technically no longer an outward direction, but this isn't something that you'll notice unless you look at the black hole geometry on a scale that's large compared to the Schwarzschild radius. Your reflection will be essentially unchanged. The mirror just isn't technically closer to the center of the black hole any more.

benrg
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The statement

So inside the horizon even a light ray directed outwards actually moves inwards not outwards.

is somewhat misleading because “inwards” and “outwards” are defined relative to a distant observer - but a distant observer cannot see what happens inside the event horizon.

To see how “inwards” and “outwards” can be relative terms and depend on the reference frame of the observer, here is an analogy.

Imagine you are on a high speed train travelling at $100$ mph. You bounce a ball off the front wall of your carriage. From your point of view the ball travels forwards at, say, $30$ mph, bounces off the wall, and travels back to you at $30$ mph.

But from the point of view of an observer on a station platform the ball travels forward at $130$ mph, bounces off the wall, and then continues to travel forwards at $70$ mph. The ball never moves backwards relative to the platform observer - instead you catch up with the ball.

gandalf61
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The thing to remember is that inside a black hole, time has stopped from our reference here on Earth. But if you are inside the black hole, time moves normally for you. So, assuming that you did not get crushed into a singularity as you went into the black hole, (and you and the mirror certainly would be crushed into a singularity) then for you everything would proceed exactly as normal. You would look into the mirror and see your reflection. You could eat your dinner and live your life.

But remember that in any real sense, this would only occur at the end of the universe for those of us outside. Inside the black hole time runs, but only for those inside, which is just impossible. But I hope you get the idea.

foolishmuse
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  • This has some errors. He is not crushed at all, but stretched to death by tidal forces. Nor would he finish his dinner; he falls to the center in finite (and quite short) proper time. Nor is there anything unusual going on implying a weird eternal moment. We're accustomed to "what you see is what is real" in relativity because, if the astronaut ever returned from near the event horizon, we know we'd agree with him about what he saw, after doing a bit of math to adjust for relativity of simultaneity.... – g s Sep 21 '21 at 00:06
  • Likewise, if we do the math to adjust for relativity of simultaneity for the astronaut who crosses the horizon and never comes back, we'll agree with him all the way down: he fell through the horizon, reflecting light that was dimmer and dimmer and of longer and longer wavelength. – g s Sep 21 '21 at 00:08
  • If the time stops inside the black hole from the point of view of an external observer what kind of transformation would you get a "normal" time flow for the internal observer? Do you multiply with infinity? – Andrei Sep 21 '21 at 06:48
  • @Andrei of course where time is actually stopped, the math gets unrealistic and we are looking at infinity. I guess the best way to think of it is if you are sitting down to dinner at 6:00pm. For you inside a black hole, it would be exactly 6:00pm and you would be in the act of sitting down to dinner until the end of the universe. But you would not experience anything different than if you were sitting down to dinner at 6:00pm here on Earth. – foolishmuse Sep 21 '21 at 15:04
  • @gs Yes, I should not have said "as you went into the black hole" because you would be stretched. I was thinking of already being inside the black hole and crushed by the gravitational pull in the centre. But for the astronaut inside the black hole, I don't think he would see light that was dimmer or of longer wavelength, because for him time has dilated to the same degree. – foolishmuse Sep 21 '21 at 15:07
  • In order to feel something you need some time to flow, isn't it? So, the inside observer would perceive nothing, he is frozen. If we also accept the fact that black holes evaporate there is no chance for an inside observer to stay until the end of the universe, it will evaporate together with the black hole before being able to move a finger. – Andrei Sep 22 '21 at 10:10
  • @Andrei of course it gets rather esoteric at this point. I guess you could say what are you experiencing right NOW. You might have plans for the rest of your day, including what you are having for dinner. From your point of view it remains NOW for ever, but you don't have any idea that you are stuck. – foolishmuse Sep 22 '21 at 15:40
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The position of you and that of the mirror define an inertial coordinate system (you are still in inertial motion remember!).

So, if the mirror is close enough to you to neglect the local curvature it will still work on you, even within the event horizon.

m4r35n357
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