Time Dilation and Redshift for a Static Black Hole - Comments

[SlowThinker]

Regarding the hovering case 1 (above) and free fall case 2

I've realized there are too many things going on to keep track of, without a very careful analysis: Lorentzian time dilation, length contraction and simultaneity shift, plus these same effects caused by gravity.
Ideally, we'd name 5 points (top and bottom of a hovering ship, top and bottom of a falling ship, and a distant observer), and describe how each sees the other 4.
I gave up

 

[Bernard McBryan]

I tried to make some simplifying assumptions to simplify this scenario to focus just on the time dilation aspects:
1) assumed ship (or very small clock) small enough to ignore the tidal forces: thus the ship is small enough in length (and width) so that the tidal forces can be ignored. ( e.g. head to feet height less than .25 inches). Alternately, one can assume that the black hole is larger (e.g. 1 Billion or 1 Trillion solar masses, such that the g force and tidal forces are weaker and can be neglected).
2) To further simplify and eliminate velocity based terms, the falling ship can reverse course after a very short time, before it reached relativistic speeds. I eliminated the orbiting case 3, so velocities can be kept at non-relativistic speeds.
3) I also intentionally did all this outside the event horizon to avoid its entanglements.
I guess I did not even need a black hole, but the question would be true outside any sun or planet (but with time slowdowns at the microsecond level).
The basic question: Does a clock in free fall, and a hovering clock at the same gravitational altitudes experience the same gravitational slowdowns from the three perspectives and how do they explain the twin paradox from the three perspectives (hovering, free fall, far away outside observer).

 

[PeterDonis]

Regarding the hovering case 1 (above) and free fall case 2 gravitational time dilation being equal seems to possibly violate the equivalence principle.

No, it doesn't. The EP doesn't say that free fall and proper acceleration can't yield the same local experimental results in some particular cases. It only says that free fall compared to free fall, or proper acceleration compared to proper acceleration of the same magnitude, can never yield different experimental local results.

Case 2 is in free fall, and assuming small enough to ignore the tidal forces, does not feel the gravity, nor her acceleration downward.

Correct.

I suppose she sees the hovering case 1 time dilate due to the gravity, or perceived acceleration away from her.

No. Remember we are talking about the instant where the free faller is momentarily at rest, right next to the hoverer. They are motionless at that instant with respect to each other, and they are both at the same altitude, so neither one sees the other as time dilated at that instant.

Of course, after that instant, the free faller and the hoverer will start moving relative to each other, and consequently they will start seeing the effects of time dilation (due to both relative motion and being at different altitudes). But I specifically restricted attention, in my example, to the instant where they are momentarily at rest relative to each other, to eliminate those effects.

Does a clock in free fall, and a hovering clock at the same gravitational altitudes experience the same gravitational slowdowns from the three perspectives

At the instant they are at rest relative to each other, yes. Otherwise no. See above.

how do they explain the twin paradox from the three perspectives (hovering, free fall, far away outside observer)

First somebody needs to state a "twin paradox" scenario--that is, a scenario where two observers start out together, separate for a while, then come back together and compare the elapsed times on their clocks. Nobody has yet done that in this thread. Once a specific scenario is stated, explaining how it works from the three perspectives will be straightforward.

 

[Bernard McBryan]

Mentor said: "First somebody needs to state a "twin paradox" scenario--that is, a scenario where two observers start out together, separate for a while, then come back together and compare the elapsed times on their clocks. Nobody has yet done that in this thread. Once a specific scenario is stated, explaining how it works from the three perspectives will be straightforward."

The scenario is a simple extension of the above case (or original cases) where two twin ships start at the same altitudes in a gravitational field, both initially hovering. The first continues to hover in place, but the second case momentarily turning their engine off, and then free falls long enough to create some additional time dilation from one (or both perspectives), but not long enough to cross the event horizon or reach excessive speeds. Then, the engines are re-engaged, first to stop the downward fall, and then slowly to rise to the original altitude. To simplify the scenario, a larger 1 billion or 1 trillion solar mass black hole could be used to reduce the tidal gravitational forces across the top/bottom of each ship, and even across the change in altitudes of the two ships. The second twin ship, as she returns may only be slightly different due to the small distance and time traveled (rather than years in special relativity based twin paradox). Would there be a difference? Who would be "younger"? And what would the external observer record?

Thanks for your help thus far. It has been very helpful.
Bernie.

 

[PeterDonis]

The first continues to hover in place, but the second case momentarily turning their engine off, and then free falls long enough to create some additional time dilation from one (or both perspectives), but not long enough to cross the event horizon or reach excessive speeds. Then, the engines are re-engaged, first to stop the downward fall, and then slowly to rise to the original altitude.

Ok, this is a good scenario. We'll call the two ships A (the hovering ship) and B (the traveling ship, that falls down and then climbs back up).

To simplify the scenario, a larger 1 billion or 1 trillion solar mass black hole could be used to reduce the tidal gravitational forces across the top/bottom of each ship, and even across the change in altitudes of the two ships.

Yes, this is fine; nothing will depend on there being measurable tidal gravity involved, so we can assume it's negligible.

Would there be a difference? Who would be "younger"?

Yes, there would be a difference. The person on ship B would be younger when the two ships meet up again. This is easily seen from the fact that ship B does two things relative to ship A, both of which create increased time dilation relative to A: ship B goes to a lower altitude (so ship B has more gravitational time dilation), and ship B moves while ship A remains stationary (so ship B has additional time dilation due to motion, which ship A does not have).

And what would the external observer record?

The external observer records just what I described above. The difference in aging between ships A and B when they meet up again is an invariant; all observers must agree on it.

 

[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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