1
$\begingroup$

As I understand, for an external observer nothing even enters a black hole because it takes infinite time in his frame of reference for something to actually reach the event horizon. I also read that for an actual object falling into a black hole his proper time is different so he actually falls into the black hole. but the question is still there for light: Does an external observer far from the black hole see that light slows down as it approaches the event horizon? If so, why is this so, because the only reason I understand for a change in the apparent speed of light is the creation of space in between the path, but $c$ will always remain a constant with respect to space itself.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
3
$\begingroup$

c will always remain a constant with respect to space itself.

We can't define a speed relative to space itself, only relative to an observer. In GR, we can only define speed relative to a nearby observer. To a nearby observer, the speed of light is always $c$, because GR is locally the same as SR.

the only reason I understand for a change in the apparent speed of light is the creation of space in between the path

No, GR simply doesn't have a well-defined notion of velocity as measured by a distant observer.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ "No, GR simply doesn't have a well-defined notion of velocity as measured by a distant observer." It does have a notion of coordinate velocity relative to whatever coordinate system you want to use though, and in this sense, light's coordinate velocity needn't be constant (though this is just as true for non-inertial coordinate systems in flat spacetime, like Rindler coordinates, as it is for coordinate systems in curved spacetime like Schwarzschild coordinates). $\endgroup$
    – Hypnosifl
    Commented Nov 20, 2014 at 15:12
-1
$\begingroup$

For the first part you are correct but it is a bit of an illusion. If say a person fell into a black hole, then there are two perspectives:

  1. From the perspective of the person, they fall straight in (ignoring whether they survive to observe it or not).

  2. From an outside observer the person appears to slow down as they approach the event horizon of the black hole (and will never actually pass it). This is not due to any actual slowing down, simply the light bouncing off them (so we can observe them) takes longer and longer to get back to us as the person approaches the black hole. Consider exactly at the event horizon bounced light is stationary, trying to escape upwards (away from the black hole at the speed of light) but pulled downwards by an exactly equal force of gravity so it will never reach us. Before the event horizon the light can just escape (slowly) so it takes a long time to reach us and after it passes the event horizon it can't escape (gets pulled back into the black hole).

Does an external observer far from the black hole see that light slows down as it approaches the event horizon?

Well, no. If you fire light at a black hole you will see nothing (you can't see light traveling away from you, just like if you shine a torch up at a clear nights sky).

But if we assume a laser firing at the black hole and there just happened to be dust falling in (everywhere) that reflected some laser light then we could view the reflected laser light from an angle to the side of where the laser is fired from.

This situation however behaves exactly the same as the person falling in. The light that isn't reflected by dust falls straight in (no problems and nothing special happens) and the reflected light takes longer and longer to make the return trip the closer it gets to the event horizon (due to the increasing gravity) after which it never makes it back.

So in reality the laser fires straight into the black hole at the speed of light and what we observe (reflected light from dust) from outside is the laser firing in and travelling slower and slower towards the event horizon until it appears to stop.

The stopping is just an illusion due to the immense gravity's effect on the reflected light (the non-reflected light has already passed through a lot earlier) and there is nothing else untoward happening like "creation of space in between the path".

Improve this answer
$\endgroup$
2
  • $\begingroup$ "the reflected light takes longer and longer to make the return trip" - wrong, -1. It takes exactly the same time always, because the speed of light is constant. It just gets red-shifted. $\endgroup$
    – Anixx
    Commented Jan 1, 2015 at 12:09
  • 1
    $\begingroup$ @Anixx - The speed of light is not constant in a general non-inertial coordinate system, it would only be constant as measured in local inertial frames at different points along the light ray's worldline. Take a look at the diagrams of light ray worldlines in Eddington-Finkelstein coordinates around a non-rotating black hole here for example--light rays traveling in one direction have a uniform speed, in the other direction they move slower closer to the horizon. $\endgroup$
    – Hypnosifl
    Commented Jan 1, 2015 at 14:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.