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I've always heard that black holes are black because light can't escape its intense gravity. If photons are massless how can they be influenced by gravity?

Or is it time that stops (relative to a distant observer) at the event-horizon that prevents photons to move towards the observer?

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marked as duplicate by Alfred Centauri, Rococo, DilithiumMatrix, John Rennie black-holes May 21 '17 at 6:19

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  • $\begingroup$ Black holes are back because nothing can escape from them which enters them, not even light. $\endgroup$ – Wrichik Basu May 20 '17 at 19:29
  • $\begingroup$ Does time stop at the event-horizon or at the singularity? If it stops at the horizon, can anything enter the black hole? Maybe there is no singularity? I'm just curious. I don't know exactly what is proven and what is hypothesis. $\endgroup$ – Frode May 20 '17 at 19:45
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    $\begingroup$ Gravity couples to everything, not just to things that have mass. Anything that has non zero rest mass and/or non zero momentum is affected by gravity. $\endgroup$ – Avantgarde May 20 '17 at 20:05
  • $\begingroup$ @Alfred I don't think it's a duplicate. There are differences. $\endgroup$ – Wrichik Basu May 20 '17 at 20:55
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    $\begingroup$ Gravity is space-time curvature in GR, so it is not clear to me that this question has a meaningful answer. $\endgroup$ – Rococo May 20 '17 at 21:06
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Black holes are black because as has been mentioned previously light cannot escape from within the event horizon.

In general relativity light always follows a straight path through space which is referred to as a geodesic. However around massive objects where space is curved, it no longer behaves as a Euclidian geometry. This means to an observer assuming a Euclidian space, the light appears to curve in presence of massive objects.

In a black hole the curvature becomes so great that the light cannot move outwards from the region inside the event horizon any more, hence making the black hole "black".

The escape velocity at the event horizon can actually be derived using classical Newtonian gravity even for light. Escape velocity is calculated from

$\frac{1}{2}mv^2=\frac{GMm}{r}$

It becomes quite clear that the mass of the smaller object cancels out and that the final escape velocity

$v_{escape}=\sqrt{\frac{GM}{r}}$

only depends on the Mass of the large object and the distance from its center. If you want to find the Schwarzschild radius of a black hole of mass M, then you simply set the escape velocity = c and rearrange for r (really this equation should not be used for black holes but if you are new to physics it is a basic approximation).

The effect that the time seems to stop when something reaches the event horizon is linked to this effect of the photon not being able to escape the potential well around the black hole. If the light is regarded as a wave traveling away from the black hole, it is gravitationally red shifted, which through the invariance of c makes the time appear to slow down for the object in the proximity of the black hole. As soon as the object passes the event horizon, light coming from it becomes infinitely red shifted making it look like time stops to an outside observer. At the same time, through the red shift, the photons lose all there energy and hence technically have not escaped from the black hole.

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  • $\begingroup$ Thank you for a good explanation. Easier to connect the dots now $\endgroup$ – Frode May 20 '17 at 22:01
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    $\begingroup$ The calculation of escape velocity using Newtonian gravity should not be applied to a black hole near the event horizon. You need general relativity to describe a black hole. It's purely a coincidence that using an escape velocity of $c$ in the Newtonian formula will recover the Schwarzschild radius. $\endgroup$ – StephenG May 21 '17 at 2:12
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    $\begingroup$ Jaywalker. Red shifting does not cause time dilation. The other way around. Time dilation derives from the metric, for an observer far away. The time between light wave peaks thus increases, that's the period, so frequent decreases. That's how you figure redshifts. Also, loosing all their energy through redshifts means they have not escaped is a somewhat baroque way to explain that lightlike geodesics inside or at the horizon can only go inwards. You're making things more mysterious than they are. $\endgroup$ – Bob Bee May 21 '17 at 5:59

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