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