- #1
- 9,495
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In another thread, Dalespam posted this paper:
http://arxiv.org/abs/gr-qc/0311038
The paper is, overall, very nice and introduces a very convenient unification of all coordinate systems for the spherically symmetric vacuum GR solution that manifest the static character ( of the exterior geometry) as well as the spherical symmetry.
However, there is a discussion of redshift on page 6, that deserves a cautionary statement. The authors sum up a calculation as follows:
"The redshift formula is directly obtained from this ratio;
since it doesn’t depend on the choice of time coordinate,
it holds for all variations on the general line element.
Thus, the redshift is infinite if the emitter is located at
r1 = 2M, even if the coordinate system allows for a crossing
of the event horizon in finite time."
I have the following issues with this summary:
1) The calculation only shows the well known result that redshift between two static observers approaches infinite as you pick one of them closer and closer to the horizon (if you imagine one of them moving, you no longer have a static observer, and the given formula no longer suffices, by itself).
2) There is no such thing as redshift from light emitted at the horizon and received by a static observer. The light is trapped, pure and simple.
3) Between two static observers, the formula is reversible and indicates blue shift for radially ingoing light. Again, though, there can be no static observer to detect infinite blueshift.
4) Instead, any observer receiving light at the horizon from an exterior static observer must be following a time like path, and, depending on the path, will receive a finite shift that can be any amount in the red or blue direction.
In sum, I find this statement by the authors exceedingly misleading, even dead wrong. I think they know this, and it is just careless editing.
http://arxiv.org/abs/gr-qc/0311038
The paper is, overall, very nice and introduces a very convenient unification of all coordinate systems for the spherically symmetric vacuum GR solution that manifest the static character ( of the exterior geometry) as well as the spherical symmetry.
However, there is a discussion of redshift on page 6, that deserves a cautionary statement. The authors sum up a calculation as follows:
"The redshift formula is directly obtained from this ratio;
since it doesn’t depend on the choice of time coordinate,
it holds for all variations on the general line element.
Thus, the redshift is infinite if the emitter is located at
r1 = 2M, even if the coordinate system allows for a crossing
of the event horizon in finite time."
I have the following issues with this summary:
1) The calculation only shows the well known result that redshift between two static observers approaches infinite as you pick one of them closer and closer to the horizon (if you imagine one of them moving, you no longer have a static observer, and the given formula no longer suffices, by itself).
2) There is no such thing as redshift from light emitted at the horizon and received by a static observer. The light is trapped, pure and simple.
3) Between two static observers, the formula is reversible and indicates blue shift for radially ingoing light. Again, though, there can be no static observer to detect infinite blueshift.
4) Instead, any observer receiving light at the horizon from an exterior static observer must be following a time like path, and, depending on the path, will receive a finite shift that can be any amount in the red or blue direction.
In sum, I find this statement by the authors exceedingly misleading, even dead wrong. I think they know this, and it is just careless editing.
It's obvious from their line element that it is *not* static, or even stationary, at r <= 2M, since the [itex]\partial / \partial u[/itex] Killing vector field is no longer timelike, but they never discuss that or even point it out.