Say you dropped a mirror into a black hole while observing at a distance and holding a clock such that the clock's face was pointing to the black hole. What is the latest time you would view on the reflection of the clock? What time would the reflection of the clock show after waiting for a long period of time.
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After some time, when the mirror is close to the black hole, you wouldn't be able to see anything because the visible light would be redshifted. So that means that the latest time you would be able to see is the one where at the time, the light reflected would still be in the visible range. – twisted manifold May 18 '22 at 14:33
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3@twistedmanifold, Light returning from the mirror to the observer's eye would be red-shifted, but wouldn't light reaching the mirror from the face of the clock be equally blue-shifted? Wouldn't those two shifts cancel each other? (supposing some magic force is holding the mirror at a fixed location W.R.T. the black hole. I realize that the question becomes more complicated when the mirror is in free fall.) – Solomon Slow May 18 '22 at 16:33
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The latest time you would view on reflection would the moment the mirror has crossed the event horizon, I suppose. – JanG May 18 '22 at 18:14
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If the mirror (magically) makes periodic stops for the light from your clock to catch up, then you’d see the reflections forever, they would not be redshifted, but increasingly delayed. – safesphere May 19 '22 at 04:10
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Good point, @SolomonSlow, however, as well as the gravitational blue & red shifts, the reflected light also gets red shifted because the mirror is moving away from the clock. But I'm not quite sure how to calculate this correctly... – PM 2Ring May 19 '22 at 04:13
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1BTW, you probably need a SMBH to do this experiment. With a stellar mass BH, you'll get spaghettified if you're within 100 km or so of the BH. See https://physics.stackexchange.com/a/631427/123208 – PM 2Ring May 19 '22 at 04:25
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@SolomonSlow You are correct, energy is conserved in the Schwarzschild spacetime. If the mirror is stationary and massive enough not to move under the pressure of the blueshifted light, then the reflections are not redshifted. – safesphere May 19 '22 at 04:57
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@safesphere, thanks for correcting me. I have mixed the observers. The one traveling with the mirror will see reflection until he crossed the event horizon. But for the outside observer the mirror will never cross the event horizon so he will see the reflection forever. – JanG May 19 '22 at 05:53
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The first and second inequality/equation in ProfRob's answer here is the answer to your question. I'm posting this as an answer instead of a suggested duplicate because the questions are distinct and parts of the answers are unrelated to your question.
The image reflected in the mirror asymptotically approaches $\Delta t$ as $t$ goes to infinity for the observer, during which time the image is also infinitely dimmed and redshifted. If you tune the clock so that it ticks for the last time at exactly $\Delta t$, you never see that last tick in the reflection. Given arbitrarily sensitive instruments, the last image you measure reflected by the mirror (after a potentially very long wait, depending on how fast your clock ticks) is one tick before $\Delta t$.
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Very good answer. The main take away, I believe is to consider that an infalling object will take an infinite amount of time to pass through the event horizon as observed by an observer far away. – twisted manifold May 19 '22 at 10:54
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The clock will measure the proper time of an observer falling into the black hole. The observer will reach the horizon in a finite proper time; you will observe the clock reaching this value asymptotically (you will never see the clock actually reach the value it has when it reaches the horizon but you will see it get closer and closer).
The clock's light will also be redshifted as it approaches the horizon, meaning that the light will have longer and longer wavelengths (and smaller and smaller frequencies) until eventually you are not able to detect the light (assuming there's a maximum wavelength of light you can detect).
Andrew
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1No, I'M holding the clock, it's the mirror which is being dropped into the black hole. – blademan9999 May 19 '22 at 01:48
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@blademan9999 I see. Well, it's pretty much the same idea. You can trace a null geodesic from you to the mirror and back. There will be a critical null geodesic that hits the mirror as it crosses the horizon which is the last one that is reflected back. Photons emitted before the critical one will reflect and return to you. Photons emitted after will not. So, the time on your clock at the moment the critical null geodesic is emitted, is the time you will see, asymptotically. – Andrew May 19 '22 at 01:55
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If you know about penrose diagrams, the critical null geodesic will be a 45 degree line coming in from scri minus that intersects a timelike trajectory (the mirror) at the horizon. – Andrew May 19 '22 at 01:57
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I don’t think the downvote is deserved. The answer is in the comment and could be more elaborate, but it is not wrong. So +1. – safesphere May 19 '22 at 16:30