Breaking the speed of light?

  • #1
jamesbolt
5
0
Ok I have a question regarding black holes and light. If a photon is travelling directly towards the centre of a black hole, when it comes within its gravitational pull, theoretically it is being given GPE therefore it must go faster than the speed of light because a black hole has enough energy to overcome a photon travelling in the opposite direction. IF this is true then surly it breaks some of the laws of physics? IF not why not?
 

Answers and Replies

  • #2
Nugatory
Mentor
13,984
7,548
Ok I have a question regarding black holes and light. If a photon is traveling directly towards the centre of a black hole, when it comes within its gravitational pull, theoretically it is being given GPE therefore it must go faster than the speed of light because a black hole has enough energy to overcome a photon traveling in the opposite direction. IF this is true then surely it breaks some of the laws of physics? IF not why not?

Giving energy to light doesn't change its speed, it changes its frequency/wavelength. Light is blue-shifted as it falls into a gravity well, red-shifted as it climbs out. You don't need a black hole to do this, any gravitating object works. The black hole case is interestingly different because the red shift (increase in wavelength, decrease in frequency) becomes infinite for light that is traveling outwards from the event horizon.

Note that all this discussion of redshift and blueshift is relative to some observer, just as the speeds of ordinary objects would be. You don't need a black hole or gravitational time dilation to see why this must be: if two observers are traveling towards or away from a light source, they will measure different frequencies and wavelengths because of the Doppler effect.
 
  • #3
jamesbolt
5
0
If giving energy to light does not change its speed, how does a photon travelling directly away from a black hole not detectable? While its wavelength is changing it will still be readable no? IF not then why not? and if the wavelength change is infinite surely it becomes either a flat line, not a wave of a solid block because the wave lengths are infinitely close together?
 
  • #4
PAllen
Science Advisor
8,948
2,158
If giving energy to light does not change its speed, how does a photon travelling directly away from a black hole not detectable? While its wavelength is changing it will still be readable no? IF not then why not? and if the wavelength change is infinite surely it becomes either a flat line, not a wave of a solid block because the wave lengths are infinitely close together?

The outgoing light trapped at the horizon, its speed is c relative to any local inertial frame. Speed of at a distance is ill defined in GR; further speed of light in non-inertial coordinates is easily greater than c even in special relativity. A static observer outside the horizon is a non-inertial observer.

(Note that speed of light can be greater than c in pure SR in Fermi-Normal coordinates (the most physical) for a uniformly accelerating observer).
 
  • #5
jamesbolt
5
0
The outgoing light trapped at the horizon, its speed is c relative to any local inertial frame. Speed of at a distance is ill defined in GR; further speed of light in non-inertial coordinates is easily greater than c even in special relativity. A static observer outside the horizon is a non-inertial observer.

(Note that speed of light can be greater than c in pure SR in Fermi-Normal coordinates (the most physical) for a uniformly accelerating observer).

Thank you, this is teaching me a lot, however, it does not explain why the light would not change into either a block or a straight line. Also if the wavelength is being constantly shortened at an infinite rate how does the light go anywhere as the speed of light is not infinite
 
  • #6
yuiop
3,962
20
... The black hole case is interestingly different because the red shift (increase in wavelength, decrease in frequency) becomes infinite for light that is traveling outwards from the event horizon.

Note that all this discussion of redshift and blueshift is relative to some observer, just as the speeds of ordinary objects would be. You don't need a black hole or gravitational time dilation to see why this must be: if two observers are traveling towards or away from a light source, they will measure different frequencies and wavelengths because of the Doppler effect.

Light emitted 'outward' exactly from the event horizon stays at the event horizon, so any light that actually leaves the event horizon must be less than infinitely red shifted. I guess there might be some oddball situations such as an evaporating black hole with a receding event horizon where a photon trapped at the event horizon might escape, but I am not sure what the redshift would be in that case.
 
  • #7
jamesbolt
5
0
Light emitted 'outward' exactly from the event horizon stays at the event horizon, so any light that actually leaves the event horizon must be less than infinitely red shifted. I guess there might be some oddball situations such as an evaporating black hole with a receding event horizon where a photon trapped at the event horizon might escape, but I am not sure what the redshift would be in that case.

Why does the light get trapped in the event horizon, there must be enough energy to do so and therefore enough to break the speed of light
 
  • #8
yuiop
3,962
20
Why does the light get trapped in the event horizon, there must be enough energy to do so and therefore enough to break the speed of light
In Kruskal-Szekeres coordinates it can be seen that the event horizon is (very loosely speaking) moving outwards at the speed of light, so outgoing light can only keep pace with it but not overtake it.
stworm.gif


In the above image the event horizon of the black hole is the diagonal red line going from bottom left to top right. The outgoing light rays (orange) are parallel to the event horizon, so if emitted at the event horizon, they cannot escape.

P.S. To provide some context, the above coordinate system is the point of view of an observer moving towards the black hole and to such an observer, the event horizon appears to be coming towards him at the speed of light.
 
Last edited:
  • Like
Likes 1 person
  • #9
Simon Bridge
Science Advisor
Homework Helper
17,874
1,657
When talking about relativity, be sure to always include the observer.
Explanations for how things happen tend to depend on where the observer is and what the have been doing.
Consider the explanations for the twins paradox for example.
 
  • #10
WannabeNewton
Science Advisor
5,829
547
Why does the light get trapped in the event horizon, there must be enough energy to do so and therefore enough to break the speed of light

What's breaking the speed of light? Nothing is. Light trapped at the event horizon still travels at ##c## relative to an observer falling into the event horizon at the very instant this observer passes by the event horizon. The event horizon is a null surface! It is not an object in space as you may be imagining in your head.

What you should do is solve for the radially ingoing/outgoing radiation in the Schwarzschild geometry and sketch the light cones defined by the worldlines of the ingoing/outgoing radiation inside the event horizon in terms of Eddington-Finkelstein coordinates. Even though the local light cone structure is trivial, globally there is a big difference due to the highly non-trivial space-time structure inside of the event horizon. Recall that future-directed time-like curves and null geodesics must pass through the future lobes of light cones from event to event.
 
Last edited:
  • #11
39,049
16,800
If giving energy to light does not change its speed, how does a photon travelling directly away from a black hole not detectable?

Giving energy to light doesn't change its speed, but light moving in curved spacetime will travel on paths that don't match your intuitions, which are based on flat spacetime. The light at the event horizon of a black hole is moving radially outward at c (relative to local observers): but spacetime is curved enough at the horizon that radially outgoing light stays at the horizon.

One way of visualizing this is to think of light as traveling along the local light cones, and gravity as "tilting" the light cones inward towards the gravitating body. (WannabeNewton gave similar advice.) Far away from a black hole, spacetime is nearly flat, so the light cones go up and to the left, and up and to the right, at close to 45 degrees (in perfectly flat spacetime, they would be at exactly 45 degrees, viewed on a spacetime diagram). But as you get closer to the hole, the light cones tilt inward: more precisely, the outer sides of the light cones get closer and closer to vertical. At the event horizon, the outer sides of the light cones are exactly vertical, meaning that radially outgoing light stays at the same radius.

This page has some diagrams that show what I described above:

http://asymptotia.com/2008/03/10/tipping-the-light-cone-black-holes/

There are better diagrams in relativity textbooks, such as MTW, but unfortunately they aren't easily visible online.
 

Suggested for: Breaking the speed of light?

  • Last Post
Replies
25
Views
843
  • Last Post
Replies
18
Views
579
  • Last Post
Replies
22
Views
619
  • Last Post
Replies
29
Views
634
Replies
7
Views
545
Replies
13
Views
560
  • Last Post
Replies
34
Views
1K
  • Last Post
Replies
1
Views
239
Replies
45
Views
806
Replies
1
Views
375
Top