So, I know that photons do not travel fast enough to escape a black hole once it passes the event horizon. Also, I know that the photons themselves aren't affected by the gravity, but rather their path instead. My question is, if the photons are stuck in-between the singularity and the event horizon, where do they go? Do they build up around the singularity, and they just haven't built up enough to pass the event horizon, or do they somehow escape and just don't emit light?
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asked Mar 3, 2015 at 0:38
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3 Answers
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The photons do not get "stuck" at the event horizon, from their own reference frame they are still traveling at C. The event horizon is simply the point where gravity from the singularity is strong enough that the escape velocity exceeds C. Just above the event horizon, photons can still escape the black hole, and just below it they sink into the singularity. Because the singularity is theorized to be infinitely small and of infinite density, the photons and everything else just keep falling into themselves forever.
EDIT 1: Classically the singularity at the centre of the black hole is infinitely small and the tidal forces at the singularity are infinite. This means that at some point the tidal forces will overcome the strong force and the atom will be torn apart into protons and neutrons, then quarks, then ???... Exactly what happens no one knows because we don't have a theory of quantum gravity to model it.
EDIT 2: It IS possible for a photon to get "paused" at the event horizon for a short amount of time. A photon emitted exactly at the event horizon will be in equilibrium right up until the black hole emits some Hawking radiation or swallows something, which would fluctuate the Schwarzschild radius and leave the photon either inside or outside the event horizon. If it is left outside the event horizon and escapes, it would appear as infinitely redshifted to a distant observer (it would be invisible).
answered Mar 3, 2015 at 3:47
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1$\begingroup$ What do you mean by "falling into itself forever"? Slower-than-light particles falling through the horizon are predicted to reach the singularity in finite proper time, at least that's theoretically true in pure GR (as opposed to some more accurate theory of quantum gravity that we don't have yet). $\endgroup$ Commented Mar 3, 2015 at 3:55
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$\begingroup$ what do you mean by ''their own reference frame they are still traveling at C''? does photon frame exist? $\endgroup$– PaulCommented Mar 3, 2015 at 4:19
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1$\begingroup$ For your Edit2: You forget that a lot of massive stuff is also falling through the horizon and the photon has a probability of hitting it and interacting changing its trapped state or be absorbed if of the correct energy for a spectrum. $\endgroup$– anna vCommented Mar 3, 2015 at 5:09
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$\begingroup$ I'm not sure I agree with your statement that something could be at event horizon and later find a way to escape. The very definition of the event horizon (in terms of future behaviour) implies that something at the event horizon will take an infinite amount of time to escape. If it takes a finite time then that means you misidentified the location of the event horizon. $\endgroup$– NaniteCommented Mar 3, 2015 at 8:44
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1$\begingroup$ If a photon is emitted right as the last particle the black hole will ever feed on crosses the event horizon, on the outside of the event horizon, and if the black hole only ever shrinks from that point on until infinity, the photon will eventually overcome the gravity holding it at the horizon and escape. Theoretically anyway. But of course nothing PAST the event horizon will ever make it out. $\endgroup$ Commented Jul 30, 2016 at 4:27
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In a classical Schwarzschild black hole, inside the event horizon all things, whether they be massless photons or bodies with mass, will travel towards smaller radial coordinate. This applies even to light that is emitted outwards from inside the event horizon.
That is, both light and mass are inevitably compelled to move inwards and will ultimately encounter the singularity, so there is no build up of anything between the event horizon and the singularity.
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According to the second postulate of special relativity, light is always observed as moving at c. - However, the proper time of a photon is zero. That means, from the hypothetical point of view of the photon, no time is passing for the photon which is infalling through the event horizon towards the singularity, and afterwards emitted as Hawking radiation.
For the observers (whatever observation means in this case, because there is no direct way of measuring of photons before crossing the event horizon), in general relativity the velocity of the photon c becomes strongly decelerated in front of the event horizon, the photon is approaching (nearly) eternally the event horizon without ever reaching it.
But the photon will not approach eternally the event horizon. It will change direction when emitted as Hawking radiation. That means, from its current point near the event horizon it will be ejected outwards as radiation (from the observers' point of view).
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$\begingroup$ So what you're saying, is that the black hole doesn't trap light, rather the event horizon is acting like a shield repelling any photons that come near it? $\endgroup$ Commented Mar 3, 2015 at 10:14
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$\begingroup$ No, everything is trapped by the black hole. There is no direct connection between the infalling process and Hawking radiation, we can only say: what comes in must come out, at the latest when the black hole is perishing. $\endgroup$ Commented Mar 3, 2015 at 11:02
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$\begingroup$ Now black holes can perish? $\endgroup$ Commented Mar 3, 2015 at 21:04
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$\begingroup$ Black holes are shrinking due to Hawking radiation $\endgroup$ Commented Mar 3, 2015 at 21:10
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$\begingroup$ F*ck, what's next imaginary numbers? Can you explain (or send a link that does) how black holes are shrinking? $\endgroup$ Commented Mar 3, 2015 at 22:36