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With all this situation, it’s also given that the gravitational influence is strong enough to turn the photon back. Now, the photon is being attracted directly from behind the path it is travelling on, so it can’t turn either way and can just go directly behind.

The speed of light is constant, so will it just switch directions in an instant? Or will something else happen? Please enlighten me.

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  • $\begingroup$ If you treat the photon as a classical massless particle, then it's equation of motion will be given by the geodesic equation. en.wikipedia.org/wiki/… $\endgroup$
    – Bruce Lee
    Commented Apr 30, 2018 at 11:17
  • $\begingroup$ @BruceLee That looks to be more along the lines of an answer than a comment. Granted, it probably wouldn't make a very good answer unless you expand it a bit, but it's best not to use the comments for posting even partial answers. $\endgroup$
    – David Z
    Commented Apr 30, 2018 at 11:22
  • $\begingroup$ @DavidZ IMO it is not an answer, it is meant to be a hint by which the given questions can be solved. $\endgroup$
    – Bruce Lee
    Commented Apr 30, 2018 at 11:28
  • $\begingroup$ It sounds like you mean a photon inside a black hole, rather than outside. Can you confirm that please. $\endgroup$
    – StephenG - Help Ukraine
    Commented Apr 30, 2018 at 11:29
  • $\begingroup$ Photon outside a black hole @StephenG $\endgroup$
    – Bhanu Kumar
    Commented Apr 30, 2018 at 11:35

3 Answers 3

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If the photon has been emitted outside the black hole, directly away from the black hole, then the hole's gravity will not turn it back and it will escape to infinity. The only effect gravity will have on it is redshifting its wave length.

If you emit a photon inside the black hole, then it is no longer possible for it to go away from the center. Every direction either a photon or a massive particle could go is towards the center; in a sense, trying to avoid the singularity once you are below the horizon is like trying to avoid tomorrow when you are outside.

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  • $\begingroup$ The other answer also states this, but can you please tell me why the photon will not turn back? $\endgroup$
    – Bhanu Kumar
    Commented Apr 30, 2018 at 11:55
  • $\begingroup$ I haven't really studied general relativity. But intuitively speaking, the black hole horizon is precisely the surface where the escape velocity is equal to the speed of light. Therefore, above the horizon the escape velocity is less than the speed of light. Therefore, above the horizon a photon (or a massive particle with a speed greater than escape velocity) can escape away from the black hole. Likewise, gravity can't change the speed of light, and there's no reason why it should affect its direction. (It would have an effect if the gravity and the photon's trajectory were not aligned.) $\endgroup$
    – Mike Rosoft
    Commented Apr 30, 2018 at 12:16
  • $\begingroup$ Thank you very much. I can see how stupid i was to ask this question without thinking about this before. The explanation is deeply appreciated. $\endgroup$
    – Bhanu Kumar
    Commented Apr 30, 2018 at 12:51
  • $\begingroup$ Photons are not attracted by gravity. Photons simply move in space curved by gravity. The space curvature near a black hole becomes critical at the event horizon. This is the point of no return for photons or anything else. The curvature is not critical outside the horizon, so photons simply fly away by moving in space as they normally do. However, time moves slower at the horizon. So for a remote observer the speed of light near the horizon is very slow and it would tale a long time for the photons to reach us from there. $\endgroup$
    – safesphere
    Commented Apr 30, 2018 at 13:10
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The photon would lose energy, and hence increase in wavelength, but its velocity would not change in either magnitude or direction.

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  • $\begingroup$ Can you please state why the direction will not change? It's told that black holes are able to influence light's direction. I have a really faint idea why, but an explanation would be appreciated. $\endgroup$
    – Bhanu Kumar
    Commented Apr 30, 2018 at 11:53
  • $\begingroup$ Because, according to the question, it's directly behind the direction of travel of this photon $\endgroup$
    – RogerJBarlow
    Commented Apr 30, 2018 at 12:36
  • $\begingroup$ To put it another way, because the photon travels on a radial path outwards, otherwise its direction would change. $\endgroup$
    – timm
    Commented Apr 30, 2018 at 13:44
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As stated in a previous answer, the photon emitted in a radial direction will not come back to the black hole and will move away in the same direction with speed $ c $ to infinity, with a decreasing measured frequency.

This can be completed with the effect of general relativity on the photon geodesic which will be different from that of classical mechanics.

Applying the Schwarzschild metric to an outgoing geodesic $ \frac{dr}{dt}>0 $, it is written as: $$ ct=r+R_s\ln{(\frac{r}{R_s}-1)}+constant\ \ \ \ \ [A]$$ with $ r $ radial coordinate of the photon, $ c $ speed of the light in vacuum and $ R_s $ Schwarzschild radius $ = \frac{2GM}{c^2} $.

When $ r\gg R_s $, the logarithm term is negligible and $ [A] $ turns into: $$ ct\simeq r+constant $$ which is the equation for radial light beams in a flat space-time (classical mechanics).

In the hope of having usefully completed the answer,

Best regards.

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