(c) 2009 Andisheh Mahdavi
Physics 725: Special and General Relativity
You are orbiting at coordinate radius r=1000km above a solar mass black hole. You drop a green flare (rest wavelength = 520 nm) into the black hole with an initial velocity of 10 km/s. Describe what you see.
The Schwarzschild radius of the black hole is 2.95km; therefore, first thing to realize here is that the observer is at a radius r = 339 M. For this observer, there isn't going to be too much of a difference between the wavelength observed at r = 339 M and r = Infinity. According to equation 9.20 in Hartle, the two wavelength should just differ by a factor of 
(for a stationary observer, which is what we are). Plugging this in, we get that
Therefore, there is less than 0.2% difference between the wavelength of light at r=1000 km and at infinity. As a result, we are justified to just find λ(t), the wavelength at infinity as a function of Schwarzschild coordinate time.
The general approach is to solve for t(r) and λ(r), and plot each on one axis of an XY plot, therefore yielding the desired function. Let's first solve for t(r).
Because we have no angular momentum, the effective potential is identical to the Newtonian potential. In fact, as we'll see the only relativistic effects that come into play are gravitational time dilation and the doppler effect. Here's the equation of motion, Hartle 9.29. Note that E is dimensionless, so we need to put a 
in front of it to make the units work out:
Now let's solve for r'[τ]. The [[1]] picks out the negative (infalling) solution:
Now, we're looking for t(r), so we also need a formula for t'[τ]. This is given by Hartle equations 9.22 and 9.27:
So, putting these together, we have that dt/dτ = dt/dτ ×dτ/dr, so that we have
Mathematica is too dumb to figure it out, but this integrand can be recast in the form:
where rs is the radius in units of the Schwarzschild radius, k = -1/(2E), and C is a bunch of constants. This gives us, with a little manipulation, the form
Now we can put back in the constants we took out before:
OK, so that was not pretty---the indefinite integral is definitely horrendous looking. Let's see what the function looks like.
Since we've been told that the initial velocity in the rest frame of the flare is -10 km/s, we can use the the r'[τ] formula to solve for E. The following expression solve for the energy given that r'[τ]= -10 km/s and r[τ] = 1000km:
Now we have everything we need to plot t(r). to do this, we evaluate the indefinite integral at a given point r, and subtract from it the indefinite integral at r = 1000 km (where we set our clock to t=0).
The plot here shows the infall from r=1000 km to r=5km. This is essentially Newtonian infall, and getting to two Schwarzschild radii takes essentially 0.1 seconds.
Let's look at the infall from 5 km to 3 km... should be interesting:
As expected, t(r) starts cusping into a straight vertical line as you head towards the event horizon---time freezes from the point of view of the observer. The flare gets within a few meters of the horizon very quickly but then appears to freeze.
Now that we have t(r), let's work out λ(r), so that we can get the color as a function of time. As you remember, the observed energy of a photon is -p·u, where p is the photon's four-vector and u is the observer's four-vector. Now we know that the rest-frame color of the flare is green, i.e. it is emitting photons with energy 
. Since we are dealing with pure radial infall, only t and r are relevant, so
Note the use of the "dot" to carry out matrix multiplication in Mathematica. Next, we have the ubiquitous and useful p·p=0 (for photons) and u·u=-1 (for timelike particles):
Our final constraint comes from the fact that we already know 
=r'[τ]:
OK---let's solve all four equations for four unknowns. Because the quantities are squared, there will be more than one unique solution, and the [[3]] below picks out the ones where pt and ut have the right sign:
So here's the key part. We want to know the wavelength seen at infinity. Well, the world line of an observer at rest at infinity is just u = cξ = (c,0). Therefore the energy observed at infinity is 
=-c p·ξ. In other words, the dot product of the time Killing vector and the photon's four-vector is always the energy observed at infinity. Let's find it:
OK, now let's plug in our constants again:
That was the wavelength as a function of r. The limit of human vision is about 650nm, which means that you stop seeing the flare with your naked eye (because it switched to the infrared) once its coordinate is r=75km or so. When does this happen? We can now obtain our final answer by plotting the wavelength as a function of time.
There's our final answer. At around 0.13 seconds after you launch the flare, it disappears from view. Note how steeply the wavelength turn up as a function of time.
