[phkahler]

Now that all the physicists are reading, I have a question: Are the trajectories of light beams following the curvature of space-time reversible? In other words, if I reflect a light beam back on itself, does it return to its source?

Because if that's true (and my feeble understanding is that it is true), then light can't enter a black hole or the reverse path would allow light to escape. What am I missing here?

[lisper]

IANA physicist, but I have a pretty good intuitive understanding of relativity. Yes, light paths are reversible. The reason you can't get a beam of light out of a black hole by reflecting an incoming beam is that from the point of view of an observer outside the event horizon, and inbound beam never makes it past the horizon inbound because of time dilation. At the horizon, time stops (relative to frames of reference outside the hole).

[Ynought]

Yes, to the best of my knowledge, in general light beams following the curvature of space-time are reversible so that you would be correct for everything EXCEPT for the special case of black holes.

In the case of black holes, the escape velocity of the black hole is simply too great to allow light to escape at it's velocity if it originates from within the event horizon.

It would be like if you were rolling tennis balls to a friend across a trampoline. If you place a bowling ball in the middle of the trampoline, suddenly rolling the balls straight to your friend won't work anymore, because you need to roll them angled to account for the bowling ball in the middle of the trampoline causing the trampoline to be bent a bit so the balls will curve around the bowling ball and a person on the other side of the trampoline can then grab them and roll them back to you with the exact same speed and angle that you rolled them to that person.

However, if you tried to roll a tennis ball instead from the MIDDLE of the trampoline (where the bowling ball is) to your friend you would find that no matter what angle you chose you could not get the tennis ball out if you used the same speed to roll it as before. On the other hand, your friend will of course have no problem rolling tennis balls from the side of the trampoline inward to you.

This is the situation you find light in. Because the absolute maximum velocity that ANYTHING including light can have in a vacuum is equal to c if the black hole is bending the trampoline of space time so much that that velocity doesn't surpass the curvature then it doesn't matter what angle the light faces, it's just going to end up heading right back towards the black hole.

[antognini]

If the spacetime is static, then yes, the trajectory of a light beam (or anything else) is reversible. The question you pose is an excellent one and is one of the reasons that many physicists believed that black holes could not exist after Karl Schwarzschild discovered the Schwarzschild metric that describes spacetime around a spherically symmetric black hole.

So what happens if you direct a beam of light radially into a black hole? A distant observer will calculate that as the light beam descends into the black hole it will be (according to a distant observer) progressively blueshifted until it reaches the event horizon, at which point it will have blueshifted to infinite energy. In fact, according to the calculations of a distant observer, the light beam will never cross the event horizon, so the statement that the light beam is reversible remains true. The reason that the light beam (according to the distant observer) doesn't cross the horizon is that if you look at the metric, it states that ds^2 = (1 - R_Schw / r) c^2 dt^2 - 1 / (1 - R_Schw / r) dr^2, where R_Schw is the radius of the event horizon. Notice that at the event horizon r = R_Schw and the dr^2 term diverges. The distant observer therefore concludes that there is a singularity at the event horizon. This analysis lead physicists to initially conclude that black holes were unphysical.

A more careful analysis reveals that this isn't the case, however. We have been using this metric which describes the shape of spacetime according to a distant observer. But what we're really concerned with is not what a distant observer calculates, but how a local observer perceives spacetime. If we now switch to a coordinate system in which we are free falling into the black hole along with the beam of light, we find that nothing remarkable happens when we cross the event horizon. The light is blueshifted relative to its initial energy, but it doesn't have an infinite amount of energy. Thus there is no true singularity at the event horizon. This singularity that the distant observer found must therefore have been an artifact of the particular coordinate system that the distant observer was using. The free falling observer can then hold up a mirror and send off a light beam and have it return back to him even after they have crossed the event horizon.

[lisper]

You have this backwards. To a distant observer, light going into a black hole is red-shifted, not blue-shifted. If this were not the case, sending even a single photon into a black hole would produce infinite energy.

[antognini]

No, any photons that are emitted from matter close to the event horizon and travel toward the distant observer will be redshifted. Photons falling into the potential well of the black hole will be blueshifted. Of course, the distant observer is not able to directly detect photons falling into the black hole because he is far away from it. All he can do is calculate what he would expect the energy of the photons to be. This calculation gives an infinite energy as the photon approaches the event horizon, but it's not a "real" energy, so the black hole doesn't gain an infinite amount of energy.

An analogous effect occurs for an observer stationed at rest, hovering above the event horizon. Such an observer will observe the photons from afar to be blueshifed in agreement with the distant observer. Moreover, as this stationary observer gets closer and closer to the event horizon, the photons will be blueshifted to infinitely large energies.

[lisper]

Ah. OK, I misunderstood what you were saying. This is right.

[lutusp]

> Are the trajectories of light beams following the curvature of space-time reversible? In other words, if I reflect a light beam back on itself, does it return to its source?

Yes, it would, assuming you have a way to reflect a beam back the way it came, like the kind of retro-reflectors the Apollo astronauts left on the moon. If you were an observer and saw a laser beam emanating from some distant place, and you wanted to assure that your laser beam would be most likely to be seen there, you would point it at the exact opposite heading -- assuming neither the source nor destination were in motion.

> Because if that's true (and my feeble understanding is that it is true), then light can't enter a black hole or the reverse path would allow light to escape.

That's a different question with a different answer. If you point a laser beam at a surface that absorbs most or all the radiation that falls on it, it's possible that none of the initial radiation will be returned to the source. But -- very important -- all the energy is accounted for. If the initial light beam has energy E, then the energy absorbed and reflected by the target will also equal E, not necessarily the same wavelengths but the same total energy. This applies to black holes as well -- they conserve energy.

[aroberge]

Physicist here. Imagine a fish swimming down a river. Down the river, there is a water fall, with water falling faster than the fish can swim. Past the point where the water is falling faster than the fish can swim, the fish can not turn back and swim up the waterfall, back into the upper part of the river.

The same thing happens in spacetime. As you get passed the event horizon, there is no way to move through spacetime so as to emerge back through it.

[raphaelj]

I'm not into Physics, but should not the space stretch itself to infinity so the light will never actually reach the mirror ?

[sampo]

I think, even if light beam trajectories were reversible (to which I cannot answer) in normal space, the singularity breaks the rules of the normal space.