# Black hole light pulses

1. Sep 29, 2011

### keithdow

Here is a simple one. An observer far away from a black hole emits a light pulse every second. A second observer has just crossed into the event horizon and received one of his pulses. How many additional pulses will the second observer receive before he is crushed at pi*m proper time.

Thanks.

2. Sep 30, 2011

### Phrak

Infinitely many.

3. Sep 30, 2011

### keithdow

Thanks!

Now how do I show that?

4. Sep 30, 2011

### PAllen

I don't think this is right from analyzing it in Kruskal map. It looks like a finite number. Just need to set up a somewhat messy integral.

5. Oct 1, 2011

### Phrak

I might agree if you are talking about an interior solution. I dont know exaclt what "just inside the event horizon" was meant by the OP. I assumed R minus some infinitessimal.

Otherwise, at the horizon the frequency ratio 0:1 just as the ratio of proper times are 1:0.

6. Oct 1, 2011

### George Jones

Staff Emeritus
I am not sure what you are trying to say, but I don't think that it is correct.

Suppose observer that A hovers at a great distance from a black hole, and that observer B hovers very close to the event horizon. The light that B receives from A is tremendously blueshifted. Now suppose that observer C falls freely from a great distance. C whizzes by B with great speed, and, just past B, light sent from B to C is tremendously Doppler reshifted. What about light from A to C? The gravitation blueshift from A to B is less that the Doppler redshift from B to C. As C crosses the event horizon, C sees light from distant stars redshifted, not blueshifted.

Last edited: Oct 1, 2011
7. Oct 1, 2011

### Naty1

I doubt that.

If anyone has an idea about how close to the
singularity a photon can get, that would be interesting.
Seems like the big "observer" will be crushed before
the tiny photon anyway.

8. Oct 1, 2011

### PAllen

The photon reaches the singularity along a fixed path, just like always, a null geodesic. An observer of light pulses can be idealized: an elementary particle. Then both can reach the singularity (along different spacetime paths).

Anyway, it graphically obvious looking at a Kruskal chart, that for any timelike path from the horizon to the singularity (doomed observer), and light pulses coming from any hovering source outside the horizon, that a finite number will reach the doomed observer on the way from the horizon to the singularity.

9. Oct 1, 2011

### George Jones

Staff Emeritus
Yes, the freely falling observer encounters the singularity in a finite amount of proper time, and the frequency seen by the freely falling observer goes to zero as the freely falling observer approaches the singularity. I have done the calculation in a Painleve-Gullstrand chart (which has the same r coordinate as Schwarzschild), and

$$f' = \frac{f}{1 + \sqrt{\frac{2M}{r}}},$$

where $f'$ is the frequency received by the freely observer when the observer is at $r$, and $f$ is the frequency emitted by the very distant observer.

10. Oct 1, 2011

### pervect

Staff Emeritus
The answer is definitely finite, and related to the fact that one doesn't see the entire history of the universe play out when one falls into a Shwarzschild black hole.

As to how to obtain it, there are several ways, but I'd suggest using ingoing Eddington Fiknlestein coordinates

http://en.wikipedia.org/w/index.php?title=Eddington–Finkelstein_coordinates&oldid=332272883

just because the null coordinate 'v' is constant along ingoing null geodesics (i.e. infalling light rays). So you can think of v being defined by a timestamped signal from an observer "at infinity". i.e pulse 0 is sent out at 0 seconds, pulse 1 is sent out at 1 second, etc.

The process of finding the infalling geodesic is somewhat complex, but managable. I'd suggest first writing the geodesic equations from the Eddington Finklestein metric, looking at how ugly they are to solve without further input, then using the Killing vector in Schwazschild coorfdinates to find the Killing vector in E.F. coordinates and using the fact that the dot product of the Killing Vector and the tangent vector (four velocity) is constant to simplify the problem.

Then go back and check that said solution does in fact solve the geodesic equations.

11. Oct 2, 2011

### Passionflower

We can actually split the Doppler effect from the perspective of a free falling (from infinity) observer receiving signals from a far away observer in two factors, a gravitational part and a velocity based part.

Below are is a graph illustrating the effect for rs=1 in Schwarzschild coordinates.

As you can see the gravitational part (yellow) contributes to the blue shift factor while the velocity based part (red) contributes to the red shift factor, and the velocity based part wins overall except near the singularity as you can see in the resulting green line.

[PLAIN]http://img812.imageshack.us/img812/3936/doppler.png [Broken]

The gravitational factor is basically:
$$\Large {\frac {1}{\sqrt {1-{\frac {{\it rs}}{r}}}}}$$
While the velocity factor is simply the relativistic Doppler formula:
$$\Large \sqrt {{\frac {1-v_{loc}}{1+v_{loc}}}}$$
The local velocity for a free falling (from infinity) observer can be obtained by:
$$\Large \sqrt {{\frac {{\it rs}}{r}}}$$

Last edited by a moderator: May 5, 2017