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According to Prof. Hawking, light rays will 'hover' on the edge of a black hole. If this is true, and the light 'stops' on the edge, how can the electric/magnetic fields which, constitute the light, continue their self-perpetuating state?

What does Hawking mean? His quote is,

the boundary of the black hole, the event horizon, is formed by rays of light that just fail to getaway from the black hole. Instead, they stay forever, hovering on the edge of the black hole.

The Theory Of Everything

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RaSullivan
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    In whose coordinate system? –  May 10 '16 at 21:30
  • ??! What. In spacetime..... GR – RaSullivan May 10 '16 at 21:39
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    The sense in which "a ray of light hovers on the edge of a black hole" is that the world-line of a particle $(ct,R,0,0)$ (as the coordinate $t$ advances and $R$ is constant right on the edge of the black hole, the Schwarzschild radius) covers zero proper time with the Schwarzschild metric. This is a very precise statement and it's very easily misunderstood by an audience looking for popular science! –  May 10 '16 at 22:03
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    (My point being: your question has, built into it, the assumption of a coordinate system, and the notion of a coordinate suggestively labeled $t$.) –  May 10 '16 at 22:10
  • Light doesn't "hover" at the event horizon but it decays exponentially. Not sure what Hawking said, but your interpretation of it is certainly 100% false. – CuriousOne May 10 '16 at 23:45
  • S.W.Hawking's Theory of Everything 2003 New Millennium Press ISBN 978-1-59777-611-0. Quote Instead, they /light rays/ stay forever, hovering on the edge of the Black hole. End quote. The question is about the B and H waves that constitute the self propagating light ray hovering. – RaSullivan May 11 '16 at 00:11
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    Additionally. I am making a serious inquiry and you should not preassume my intellectual status in your comments, ThankYou! – RaSullivan May 11 '16 at 00:18
  • Perhaps Mr Hawking misinterpreted himself. I would really appreciate honest attempts to enlighten me on this question, It is a foundational intersection in many ways regarding numerous issues, such as self-propagation for instance. – RaSullivan May 11 '16 at 01:02
  • He is referring to the fact that photons emitted directly above the event horizon will for a short time (until the Shwarzschild radius fluctuates, either via radiation or absorption) be stuck there, infinitely red-shifted to a distant observer. The event horizon is the point where the black hole's escape velocity reaches C. – Drunken Code Monkey May 11 '16 at 03:18
  • Relevant: http://physics.stackexchange.com/questions/21319/how-can-anything-ever-fall-into-a-black-hole-as-seen-from-an-outside-observer?rq=1 – DilithiumMatrix May 11 '16 at 17:12
  • Let's say we dip a LC-oscillator into a very deep gravity well. The oscillator includes a counter that counts the oscillations. From the reading of the counter we can deduce that the oscillator kind of tends to freeze at low altitudes. – stuffu May 12 '16 at 04:16
  • Seems well posed to me; there's a quote, properly attributed, about physics, and a request for an explanation. Voting to re-open. – Kyle Oman May 12 '16 at 20:57

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Hawking, I believe, is referring to a more metaphorical 'hovering'. As light, or anything, approaches the event horizon, it becomes more and more redshifted---it's motion appearing to go slower and slower and slower, approaching zero apparent velocity to an outside observer (approximately) infinitely far away. Anything falling into a BH, thus appears to end up 'hovering' just outside of it.

From the perspective of the object falling into the BH, or an observer traveling nearby/similarly to it, nothing special happens. To the infalling observer, time still seems to pass normally... everything is the same. So there is no problem with the electromagnetic wave itself behaving (basically) normally as it approaches the blackhole.

There are lots of questions and material about this and related subjects on physics.stackexchange which might be helpful.

Aside: Apologies for the extremely pedantic and unhelpful comments you received on your question.

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DilithiumMatrix
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