Reference: Moore, Thomas A., *A General Relativity Workbook*, University Science Books (2013) – Chapter 12; Problem 12.7.

We’ve seen what the view of black hole looks like for a stationary observer at various distances from the black hole. We can do similar calculations for an observer falling in radially from infinity. For such an observer, we’ve already worked out the four-velocity components:

For a particle starting at rest at infinity and moving radially inwards, , and , so these equations reduce to

These are the components of the basis vector in the Schwarzschild frame:

Note that this vector is already normalized, since

From this, we can work out the three spatial basis vectors. For , we know that and, since the axis is aligned (by definition) with the direction, , so we get

We also have the normalization condition which gives us

We choose for since the axis is aligned with the direction. Thus:

Since , the condition tells us that , so the normalization condition then says that and are the same as before, namely

We can now follow the same procedure as before to calculate the critical angle at which a photon emitted by the observer is absorbed by the black hole. We have the photon’s four-momentum:

The 3-velocity components as measured by the observer are

Using our basis vectors from above, we get

The sine of the emitted angle is, as before

As a check, we can also calculate the cosine:

After a bit of algebra, it can be confirmed that which is reassuring.

The critical angle occurs when the impact parameter , so

Unlike the case for a stationary observer which is defined only for , this formula is actually well-defined for all , although for . We’ll plot versus (in units of ) for both signs, with the plus sign in red and the minus sign in blue. We see that something odd happens at :

I’m not totally sure of the interpretation, but if we look at the analysis of the stationary observer that we did earlier, we see that at , the critical emission angle is . That is, for , the photons at the critical angle are emitted such , while for , they are emitted with . This means that we should take the minus sign for in the former case, and the plus sign in the latter. Since the observer is stationary, it is in the same frame as the global Schwarzschild frame.

When dealing with a moving observer, we are still doing the calculations in the global frame (that is, the frame of the central mass ); it is only the local basis vectors that have changed due to the motion of the observer. Therefore, the switch from positive to negative still happens at , since we are using as calculated in the global frame. In the plot, therefore, we should use the red curve for and the blue curve for .

To the *moving* observer, however, the angle subtended by the black hole is different from that for a stationary observer at the same distance from the black hole. For example, the maximum of occurs at so it is at that radius that the critical angle *relative to the moving observer* is .

The plot of allows us to determine the quadrant of :

At other radii, we have:

- ;
- ;
- ;

I’m not sure what to make of the last result, since I’d imagine is inside the black hole, where presumably the Schwarzschild metric breaks down.

From the formula we can work out the observed energy of a photon fired radially from infinity at the observer. If the photon is coming in on a radial line, the impact parameter and so we get

where is the photon’s energy as observed at infinity. Since for all , the light is always red-shifted. The fractional change in wavelength is

Pingback: Circular orbit: appearance to a falling observer | Physics tutorials

Pingback: Local flat frame for a circular orbit | Physics tutorials

Pingback: Black holes: are they really black? | Physics pages

Pingback: Painlevé-Gullstrand coordinates: derivation using a local flat frame | Physics pages

Pingback: Riemann tensor in the Schwarzschild metric: observer’s view | Physics pages

growescienceFrom Asher Weinerman….

Hi – I didn’t notice you replied to my previous comment – sorry!

OK – I finally found your mistake in problem 12.7. Here is the reason you must use your red curve outside 3GM and your blue curve inside 3GM: In equation 29, when you calculate negative ot (dot) p, your plus or minus E term must change to minus or plus E as you multiply the negative sign through the equation. In other words, ot (dot) p has a plus or minus E term, so negative ot (dot) p must have a minus or plus E term (reverse the meaning of the plus and minus). So now minus corresponds to outgoing and plus corresponds to incoming. So use the red curve outside 3GM and the blue curve inside 3GM.