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Foreword.

Since light can't escape from a black hole, we can assume that light is influenced by gravity.

So I can imagine that light that would otherwise escape from a black hole is in fact restrained/slowed from the gravity in the black hole.

A light ray/photon that is going in direction of a black hole can also be accelerated from black hole gravity? In this case is light going faster that light speed?

Argue each point if i'm wrong.

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  • $\begingroup$ Related: physics.stackexchange.com/q/249026, physics.stackexchange.com/q/221991 $\endgroup$
    – Kyle Kanos
    Commented Nov 21, 2016 at 16:21
  • $\begingroup$ I'm not sure the reason for the downvote, it's a perfectly reasonable question to ask (even if it's been asked before) $\endgroup$
    – Kyle Kanos
    Commented Nov 21, 2016 at 16:23
  • $\begingroup$ Hi Rudi, I added in the word otherwise to try and clarify your question and reduce down votes based on that misreading . If it changes the meaning of your question, please edit it out. $\endgroup$
    – user108787
    Commented Nov 21, 2016 at 16:56
  • $\begingroup$ @JohnRennie in your answer i found this, that clarify me:" dr/dt=(1−(2GM/rc^2))c So in this case the speed of light is not cc. In fact it's less than cc for any value of rr less then infinity." so i can assume that light never goes at light speed apart in vacuum? So, for example here on earth light speed tends at maximum light speed. exatly? $\endgroup$
    – RudiDudi
    Commented Nov 21, 2016 at 17:03
  • $\begingroup$ @RudiDudi: If you measure the speed of light at your location then you'll always get the value $c$ regardless of where you are. If you measure the speed of light at a distance location you may get a value different from $c$, as I describe in the linked question. This speed is known as the coordinate speed to distinguish it from the local speed. Assuming you are on the Earth's surface then as measured by you the coordinate speed increases with altitude. So for example you would measure the coordinate speed of light at the altitude of a GPS satellite to be faster than $c$. $\endgroup$
    – John Rennie
    Commented Nov 21, 2016 at 17:07

1 Answer 1

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A black hole is a construct in General Relativity. In GR light follows geodesics with velocity c, these geodesics are very distorted close to the black hole:

black hole

The five geodesics drawn on the embedding surface of space-time outside the black hole represent the possible trajectories of a free-falling body passing at different distances from the black hole. The geodesics 1,2 and 3 are increasingly affected by the curvature. Geodesic 4 falls into the gravity well and intersects itself as it comes out. Geodesic 5 falls straight into the hole and is trapped.

In the case of a photon it has energy, and it follows the geodesic with c velocity, which cannot increase due to the theory of GR. Its energy ( frequency) can increase and this is what happens.

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  • $\begingroup$ Are there any implications as to what frequency photons can experience a change in energy? For instance, a high energy cosmic ray photon following the geodesic. If it noticed a growth in energy, would this growth just be smaller as it is near the end of the spectrum? Or does something else happen that we would not expect? $\endgroup$
    – bleuofblue
    Commented Nov 21, 2016 at 18:30
  • $\begingroup$ It follows the rule E=h*nu. it is the opposite of the red shift, called blue shift en.wikipedia.org/wiki/Geodesics_in_general_relativity $\endgroup$
    – anna v
    Commented Nov 21, 2016 at 20:09

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