I understand that to an outside observer, the light from a star that is collapsing into a black hole will become more and more red-shifted as the surface of the star appears to approach the black hole event horizon. The outside observer will never actually see the surface of the star cross the black hole event horizon. This applies to all outside observers: at infinity, in orbit around the star/black hole or those using a rocket to hover above the black hole.
Conversely, I know that for someone on the surface of the star that is collapsing to form a black hole it will appear quite different. The observer on the surface will not see anything unusual happen as they cross the event horizon and in a finite time they will reach the singularity at the center of the black hole where we do not know what will happen since general relativity breaks down in a singularity.
So, now consider an observer that starts at a great distance from the star who is continually falling directly into the star that has formed a black hole. Assume that he is falling in an exactly radial direction with no angular momentum. While the observer is still very far from the black hole, he will see the (original) visible light of the star get red-shifted to infrared, to microwaves and then to longer and longer radio waves. But as he approaches the black hole, he starts to fall faster and faster, so I assume he starts to see that red-shifted photons from the star surface will begin to be blue-shifted by his increasing speed as he falls. So, I think that the red-shift photons will be blue-shifted such that when the observer crosses the event horizon he will see the "visible light" photons that were sitting there at the horizon waiting for him. Since he is falling at the speed of light (in some sense) when he crosses the horizon these photons will become "visible light" again.
So the first question is this true? When he crosses the event horizon will the surface of the star be back to having no net red or blue shift? Will a visible light photon emitted by the star as it crosses the horizon be once again a visible light photon?
The second part of my question is what about the number density of photons? Will it look as "bright" as it would have looked for an observer falling with the surface of the black hole or will it look "dimmer" as if the surface was further away?
Finally, what happens as the falling observer continues his fall past the event horizon of the black hole. Will the photons form the original in-falling star be red or blue shifted. What will the observer see during the short time before he himself hits the singularity?
This is a follow on to my previous question since this was not answered there.
Answer
You need to be a lot more careful when you use the phrase red-shift, due to how frequency is measured in general relativity. Roughly speaking, a photon is characterised by its wave vector $k$, which is a light-like four vector. The frequency measured by an observer is $g(\tau,k)$ where $g$ is the metric tensor, and $\tau$ is the unit vector tangent to the observer's world-line.
A little bit of basic Lorentzian geometry tells you that for any given photon $k$, instantaneously there can be observers seeing that photon with arbitrary high and arbitrarily low frequency.
So: start with your space-time. Fix a point on the event horizon. Fix a photon passing through that space-time event. For any frequency you want to see, you can choose a time-like vector at that space-time event that realises that frequency. Now, since the vector is time-like and the event horizon is null, the geodesic generated by that vector must start from outside the event horizon and crosses inside. Being a geodesic, it represents a free fall. So the conclusion is:
For any frequency you want to see, you can find a free falling observer starting outside of the black hole, such that it crosses the event horizon at the given space-time event and observes the frequency you want him to see.
So you ask, what is this whole business about gravitational red-shift of Schwarzschild black holes? I wrote a longer blog post on this topic some time ago and I won't be as detailed here. But the point is that on the Schwarzschild black holes (and in general, on any spherically symmetric solution of the Einstein's equations), one can break the freedom given by local Lorentz invariance by using the global geometry.
On Schwarzschild we have that the solution is stationary. Hence we can use the time-like Killing vector field* for the time-translation symmetry as a "global ruler" with respect to which to measure the frequency of photons. This is what it is meant by "gravitational redshift" in most textbooks on general relativity (see, e.g. Wald). Note that since we fixed a background ruler, the frequency that is being talked about is different from the frequency "as seen by an arbitrary infalling observer".
(There is another sense in which redshift is often talked about, which involves two infalling observers, one "departing first" with the second "to follow". In this case you again need the time-translation symmetry to make sense of the statement that the second observer "departed from the same spatial point as the first observer, but at a later time.)
It turns out, for general spherically symmetric solutions, there is this thing called a Kodama vector field, which happens to coincide with the Killing vector field on Schwarzschild. Outside of the event horizon, the Kodama vector field is time-like, and hence can be used as a substitute for the global ruler with respect to which to measure red-shift, when the space-time is assumed to be spherically symmetric, but not necessarily stationary. Again, this notion of redshift is observer independent. And it has played important roles (though sometimes manifesting in ways that are not immediately apparently related to red-shift, through choices of coordinates and what-not) in the study of dynamical, spherically symmetric gravitational collapse in the mathematical physics literature.
To summarise:
If you just compare the frequency of light measured (a) at its emission at the surface of the star in the rest frame associated to the collapse and (b) by an arbitrary free-falling observer, you can get basically any values you want. (Basically because the Doppler effect depends on the velocity of the observer, and you can change that to anything you like by choosing appropriate initial data for the free fall.)
One last comment about your last question:
You asked about what happens in the interior of the black hole. Again, any frequency can be realised by time-like observers locally. The question then boils down to whether you can construct such time-like observers to have come from free fall starting outside the black hole. By basic causality considerations, if you start with a time-like vector at a space-time event inside the black hole formed from gravitational collapse, going backwards along the time-like geodesic generated by the vector you will either hit the surface of your star, or exit the black hole. Though precisely how the two are divided depends on the precise nature of the gravitational collapse.
I should add that if you use the "global ruler" point of view, arguments have been put forth that analogous to how one expects red shift near the event horizon, one should also expect blue shifts near any Cauchy horizon that should exist. This has been demonstrated (mathematically) in the Reissner-Nordstrom (and similar) black holes. But as even the red-shift can sometimes run into problems (extreme charged black holes), one should not expect the statement about blue shifts near the Cauchy horizons to be true for all space-times.
No comments:
Post a Comment