Time Dilation and Redshift for a Static Black Hole - Comments

[Nugatory]

Would there be a difference? Who would be "younger"? And what would the external observer record?

The only thing I might add to PeterDonis's excellent answer is that this problem responds well to the method described in the "Doppler Shift analysis" in the twin paradox FAQ. Imagine that both spaceships carry a strobe light that flashes once a second while they're separated. Clearly the total number of times each ship's light flashes is the amount of time that elapsed on that ship while they wrere separated. Furthermore, both ships can see and count the number of flashes from the other ship's strobe. Thanks to time dilation, gravitational effects, and light travel time it may not be clear when the flashes will get to the other ship, but it is clear that they will get there and be counted eventually (and before the reunion).

Thus, if I see my light flash eight times while we separated, and I count ten flashes from your ship... I know that you aged ten seconds while I only aged eight. All observers every must agree about the number of times each strobe flashed; we can put a counter on the strobe just to be sure.

 

[Mike Holland]

For someone free-falling into the hole along a radial trajectory, the "time dilation" factor isn't really well-defined, because this observer is not at rest relative to an observer at infinity, so they don't have a common standard of simultaneity. However, the "redshift factor" for light emitted by the free-falling observer and received by an observer at infinity is well-defined, and can be calculated as the redshift factor for a static observer at altitude r r, which is what you wrote down, combined with the Doppler redshift for an observer falling inward at velocity v=2GM/r − − − − − − √ v = \sqrt{2GM/r} relative to the static observer.

Peter, I think Kip Thorne gets this wrong in his book "Black Holes and Time Warps". After his fable about ants pushing balls out of a hole, he concludes that what we see is the Doppler effect for successive photons from a falling body taking longer and longer to escape, and he concludes "that it appears to freeze as seen from far away is an illusion." (p249). As you point out above, what we would see is the combination of the two effects.

Mike

 

[PeterDonis]

After his fable about ants pushing balls out of a hole, he concludes that what we see is the Doppler effect for successive photons from a falling body taking longer and longer to escape

IIRC, that's because he is not using the concept of gravitational time dilation in that passage; he is viewing the entire redshift as "Doppler" (though you still have to be careful with that term because it's not the ordinary SR Doppler shift in an inertial frame--there is no inertial frame that covers both the free-faller near the horizon and the distant observer). The splitting that I describe, into gravitational and Doppler parts, is not the only possible way of describing what happens; it just happens to be a useful one. Thorne's way, which is a different way, can be useful too.

he concludes "that it appears to freeze as seen from far away is an illusion."

Which is true, regardless of whether you split the observed redshift into gravitational and Doppler parts or not. Nothing is "frozen" when viewed locally; that is an invariant, independent of how we describe what the distant observer sees.

 

[Boing3000]

they will see the contraction slow down and effectively come to a stop.

Yes, this is what remote observers will see, in the sense of the light signals they receive.

I don't agree. A remote observer will never see anything "effectively coming to a stop". Only the event horizon is at time infinity. Everything outside is at finite time and is not stopped. Beside the object will effectively be out of "sighting" range way before due to the extreme red shifting of the signals

 

[Dale]

@PeterDonis do you know if the technical definition of "event horizon" excludes the Rindler horizon for any reason? If it does not, then I would say that we are always passing through event horizons.

 

[PeterDonis]

do you know if the technical definition of "event horizon" excludes the Rindler horizon for any reason?

Yes, it does. The technical definition of an event horizon is that it is the boundary of the causal past of future null infinity. The Rindler horizon does not meet that definition; all of Minkowski spacetime, regardless of which side of any Rindler horizon it is on, is in the causal past of future null infinity.

 

[PeterDonis]

A remote observer will never see anything "effectively coming to a stop". Only the event horizon is at time infinity. Everything outside is at finite time and is not stopped.

This is a fair point. A more precise way of saying it would be that a remote observer will see the apparent "speed" (I put this in scare-quotes because it's a coordinate speed and doesn't really have a physical meaning) of an infalling object get closer and closer to zero, without ever quite reaching it, and he will see the apparent position of the object get closer and closer to the horizon, without ever quite reaching it.

Beside the object will effectively be out of "sighting" range way before due to the extreme red shifting of the signals

In practical terms this will happen fairly quickly, yes. Discussions of the theory involved usually ignore this point, but it will certainly come into play if we ever try to run any actual experiments along these lines.

 

[PeterDonis]

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