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If light can be bent by gravity, does a mass as dense as a star pull any fraction of photons back towards itself?

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    $\begingroup$ No. The photons just lose part of their energy. $\endgroup$ – CuriousOne Oct 10 '14 at 9:08
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For visible stars, the answer is no. In Newtonian physics, a star that would pull something travelling at light speed back to itself, i.e. a star for which the escape velocity were $c$, was called a dark star and seems to have been first postulated by the Rev. John Mitchell in a paper to the Royal Society in London in 1783. The great Simon Pierre de Laplace postulated the same idea some years later. It is important to take heed that in theory there was nothing stopping something escaping a dark star by climbing a rope let down by a helpful spaceship, nor was there any known lightspeed limit in those days.

In General Relativity, the analogous concept is a black hole. By definition, if a star is not a black hole, light shone upwards will escape the star's gravitational field, although light is red-shifted in doing so, heavily so if the star is massive. Moreover, in GTR there is no faster than light communication, and gravity is not thought of as a force. In GTR, a black hole is no longer something that a friendly spaceship dangling a rope could help you escape from. A black hole curves space and time such that the futures of anything within the Schwarzschild horizon must lie wholly within the black hole. You can no more escape from a black hole than you can go backwards in time; indeed these two deeds are the same thing in the "curved" spacetime of the black hole.

Edit As CuriousOne points out, a quantum mechanical treatment of the black hole shows that photons can be emitted as Hawking Radiation. This theoretical foretelling was made by Stephen Hawking in 1974: the theory is piecemeal and ad hoc, but it is very simple and fundamental, so I don't believe many physicists seriously believe Hawking radiation will be absent from a full quantum theory of gravity. For "normal sized" black holes formed from collapse of stars, this radiation is exquisitely faint, but microscopic black holes emit much stronger Hawking radiation.

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    $\begingroup$ Nice explanation! You could still add that quantum mechanics may allow photons to escape, after all. $\endgroup$ – CuriousOne Oct 10 '14 at 9:20
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    $\begingroup$ @CuriousOne See edit: I'm afraid my theoretical comfort zone ends at the end of the Big Black Book, but of course, if you're talking about Hawking radiation, then this should be part of every physicist's general knowledge. $\endgroup$ – Selene Routley Oct 10 '14 at 9:29
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    $\begingroup$ "The big black book" - what a lovely name! I agree, Hawking radiation is closer to the speculative end of the pond. $\endgroup$ – CuriousOne Oct 10 '14 at 9:36
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    $\begingroup$ And of course, even if the rope problem wasn't unsolvable due to space-time geometry, you would still need a rope of infinite strength - which can't exist in a universe with a finite speed of light. And a universe with infinite speed of light would obviously not have black holes. Black holes are... funny :D $\endgroup$ – Luaan Oct 10 '14 at 12:57
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    $\begingroup$ I think it's important to note that the radiation emitted from black holes does not originate inside the event horizon. Any photon that originates inside the event horizon will eventually make its way to the center. $\endgroup$ – Gabe Oct 10 '14 at 15:48

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