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Imagine a Schwarzschild black hole with mass $10^6 M_\odot$ in (almost) empty space. I remain at rest at a great distance and drop a robust clock. The time keeping mechanism of the clock is the radioactive decay of a suitable $\beta$ emitting isotope, insensitive to gravity. As the clock approaches the event horizon of the black hole, it "ticks" at an ever slower pace according to my wrist watch. I will never see the clock pass the even horizon, due to gravitational time dilation. So far, so good.

What about the clock at the instant it passes the event horizon in free fall? Does it tick just like it always did? I would say it ticks at its normal pace (although it's hard to imagine an experiment to verify this) because the event horizon isn't a special place for the clock. I've read quite a few related questions here, but some experts seem to disagree.

Answering the question "Equivalence principle and the meaning of the coordinate speed of light" John Duffield states

Even for the gedanken observer at the event horizon with his optical clock. Gravitational time dilation goes infinite. His clock is stopped, and he's stopped too. Contrary to what Kruskal-Szekeres coordinates suggest, the stopped observer doesn't see his stopped clock ticking normally "in his frame".

Duffield's answer appears in a strange grey on my monitor, and Javier disagrees. That's confusing for me.

My question: does the clock tick normally at the event horizon, according to an indestructible observer who also passes the event horizon in free fall, next to the clock?

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    $\begingroup$ Answers appear in gray when they have been downvoted to a score of -3 or less, which was an appropriate response to that answer. $\endgroup$ – J. Murray May 20 '18 at 18:56
  • $\begingroup$ The distant observer never sees the falling object reach the event horizon. $\endgroup$ – StephenG May 20 '18 at 18:59
  • $\begingroup$ @StephenG Agreed, that's why I wrote "If I still would be able to see the clock". The question is not what the distant observer sees (or doesn't see). $\endgroup$ – gamma1954 May 20 '18 at 19:09
  • $\begingroup$ It appears answers need to address what the observer measures as proper time when falling through the event horizon $\endgroup$ – Triatticus May 20 '18 at 21:43
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    $\begingroup$ @RobJeffries Are you suggesting John Duffield might like to read one? ;-) $\endgroup$ – gamma1954 May 21 '18 at 22:43
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does the clock tick normally at the event horizon, according to an observer right next to the clock?

The problem here is that even a gedanken experiment can not disregard physical laws. It is not possible to have something with rest mass stationary at the event horizon of a black hole. The gravitational acceleration at $r = 2M$ approaches infinity and things are crossing the horizon with lightspeed. This means nothing can be hold here by a rope, it will brake. - For the same reason there is no observer next to the clock at the event horizon.

Only particles with zero rest mass can "stay" at $r = 2M$, e.g. a photon emitted there radially outwards. Such a photon has a frequency (which can be understood as a kind of clock) according to it's energy. It moves locally with light speed and therefor has no rest frame because nothing with rest mass can move with light speed. In other words any observer crossing the event horizon will measure locally the speed of light c as everywhere else.

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    $\begingroup$ In the question there is no clock or observer stationary at the event horizon. Both are moving in free fall through the event horizon. $\endgroup$ – gamma1954 May 20 '18 at 20:14
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    $\begingroup$ @gamma1954: If that's the case, then edit the question to say so. $\endgroup$ – Ben Crowell May 21 '18 at 2:31
  • $\begingroup$ @BenCrowell I have edited the question according to your suggestion. $\endgroup$ – gamma1954 May 21 '18 at 18:45
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If you stay at any fixed distance from the event horizon and keep a bunch of clocks around you, you will see them all run at the same, "normal" rate. This is just a basic consistency requirement; a clock is a device that keeps track of time, but anything can work as a clock, including your body and your brain. If you saw the clock running differently, not matter the nature of the clock, it would mean that there is something special either about the clock or about the human body. Neither of these possibilities is contemplated within our current laws of physics.

There is a problem with your question, which is that you can't stay still at the event horizon. But if you hold a clock while in free fall you will see it running as normal, at least until the tidal forces start to break things.

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  • $\begingroup$ Thank you for explaining that the clock will tick normally. I didn't mean (and didn't write) that the observer remains at the event horizon, so there is no problem. Both the observer and the clock are in free fall. $\endgroup$ – gamma1954 May 20 '18 at 20:11
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Yes, he could edit to make sure everybody sees it and the question is clear, but he's already made comments that he does not mean an observer or clock at rest in the horizon, and as he states, he never said that. With that assumption the answer is clear, and I think that was all he was asking.

So the simple answer is yes, the observer falling with the clock will see the clock just ticking at the observer's proper time rate, I.e., the observer's and the one near him falling at the same rate, measure the same time. As is said by Susskingd and others, GR says there is no drama at the horizon, as has been understood for a long time.

Unless you go by the firewall hypothesis arguments, which have not been disproven, that's it. If you consider quantum effect and thus the black hole information paradox (which was not the question, the question is inherently about GR, and not quantum info), the firewall destroys the clock and the observer at or as they get very close to the horizon.

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  • $\begingroup$ "No drama at the event horizon" Well, @timm stated that acceleration at the event horizon approaches infinity. That to me sounds like serious drama. $\endgroup$ – flippiefanus May 21 '18 at 4:24
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    $\begingroup$ @flippiefanus the acceleration needed to hover at fixed $r$ goes to infinity at the horizon. But the free falling observer isn't hovering and so they feel no acceleration - at the horizon or anywhere else. $\endgroup$ – John Rennie May 21 '18 at 4:37
  • $\begingroup$ @John Rennie: Yes, I understand that, but I don't think that is the acceleration that timm is refering to. As I understand it, even at a short distance from the event horizon, I would see the objects at the event horizon moving at the speed of light away from me. That implies that when I free-fall toward the event horizon, my acceleration would have to approach infinity. $\endgroup$ – flippiefanus May 21 '18 at 6:14
  • $\begingroup$ @flippiefanus the shell observer at $r$ records the radial acceleration of a free-fall object as $g_{shell} = Mr^{-2}/\sqrt(1-2M/r)$ which approaches infinity as r approaches $2M$. From this it follows what John Rennie explained. $\endgroup$ – timm May 21 '18 at 7:59
  • $\begingroup$ @timm: Thanks for the details. What precisely is a "shell observer"? $\endgroup$ – flippiefanus May 22 '18 at 4:27

protected by Qmechanic May 21 '18 at 3:47

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