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?
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:
- Inside a black hole, can I see my reflection in a mirror (is the law of reflection still valid)?
Best Answer
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.
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.