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this one's rarer but more paradoxial than the twin paradox, although the possibilities are either Newtonian or Einsteinian.
well it goes thus:
Newton said that orbits are reversible. That means light striking on a mirror normally will strike back at the same speed normal to the mirror. The confrontation comes when we theoritically assume a light source near to a black hole, but not so near, emitting a photon ray which hits a mirror kept on the surface of the black hole. According to simple non relativistic motion, i.e. Newtonian motion, the photon which bounce back leaving the black hole to reach the source again. But this aint possible according to Einstein relativity. accrd. to that, the photon will not accelarate more than c in its journey upto the mirror, but if it bounces back, the gravity will decelarate it down and it will never leave a certain sphere with finite radius.

2007-05-12 23:11:45 · 5 answers · asked by Anonymous in Science & Mathematics Physics

my solution for the paradox is that no reflecting surface can exist on a black hole, and that all has to do with its name, black hole. the black hole is thought to be a dense mass which absorbs everything falling on it. that means the light will never be reflected back in the first place. what do you think guys?

2007-05-12 23:11:57 · update #1

5 answers

(About my English, I'm a Russian)

Photons don't accelerate or decelerate (this is the clue). They always move with constant speed. Their energy determined by their frequency
E = h*nu, h - Planck constant, nu - frequency. Gravitational field affects photons like they have mass m = E / c^2.
So, when photons travel towards a black hole, their energy rises, frequency also rises as a consequence. If the mirror is near to the event horizon, and the mirror is still able to reflect (our photons may become roentgen or even gamma ones) photons will reflect back and will travel from the black hole loosing their high energy and decreasing frequency. They will get back their initial frequency, when far away from the black hole again. This situation is the same no matter how close the mirror to the event horizon is.
If the mirror sinks into the event horizon, photons will not come back. And when the mirror is just on the event horizon it's just the moment when one situation changes to another.

2007-05-13 02:22:52 · answer #1 · answered by Anonymous · 1 0

I think you're basically right. A reflective surface can't exist at the event horizon of a black hole, because light cannot move away from that surface due to the escape velocity being greater than the speed of light. Newton had never contemplated the existence of such a thing as a black hole, and that's why his theories could lead to this seeming paradox.

2007-05-13 00:20:05 · answer #2 · answered by DavidK93 7 · 0 0

Well, since black holes don't have a surface, the question is moot. However, if you are referring to the point above the Event Horizon (the point where all events mathematically describable stop), then the question does have an answer. The photon of light doesn't accelerate. Light travels at the speed of light, neither faster nor slower. The only effect noticeable due to the gravity well of the black hole would be a reduction in its frequency.

2016-05-17 05:53:13 · answer #3 · answered by luella 4 · 0 0

A reflective surface probably could not exist on the surface of a black hole it would be crushed beyond recognition but if light was reflected from someting on its surface that light would cirle the black hole for eternity.

2007-05-20 06:54:48 · answer #4 · answered by johnandeileen2000 7 · 0 0

Well, A black hole is probably a good example of a black body.

so, if all the wavelengths of the electromagnetic spectrum are absorbed by the black body, what remains to be reflected?

2007-05-13 02:07:05 · answer #5 · answered by Anonymous · 0 0

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