Let's say that I'm hovering in a rocket at constant spatial coordinates outside a Schwarzschild black hole.
I drop a bulb into the black hole, and it emits some light at a distance of $r_e$ from the center, with a wavelength of $\lambda_{e}$ in the rest frame of the bulb.
What would the wavelength of the light be when it reaches me, at $r_{obs}$ in terms of the radius at which it is emitted, $r_e$?
This is a subquestion from Sean Carroll's Spacetime and Geomtery. Earlier in the chapter, Carroll asserts that any stationary observer $(U^i= 0)$ measures the frequency of a photon following a null geodesic $x^{\mu}(\lambda)$ to be
$$\omega= -g_{\mu\nu}U^\mu\frac{dx^\nu}{d\lambda}$$
I don't understand where this expression comes from.
How does one even conceptualize things like wavelength and frequency of light in terms of general relativistic quantities like $U, g_{\mu\nu}, ds^2$, etc?
Best Answer
Here are some ideas re your question:
Let us consider the path taken by the torch in the presence of a black hole, and let us assume the observer is outside the horizon. For the sake of simplicity let us assume the light source (the torch) is falling in a rectilinear way.
Some algebra and quite a bit of physics, combining the principle of equivalence and some aspects of special relativity, can show that the geometry of the path of the torch is given by the equation (taking only rectilinear motion into account):
$ds^2=\left(1-\frac{2GM/c^2}{r}\right) c^2dt^2-\frac{dr^2}{\left(1-\frac{2GM/c^2}{r}\right)}$.
Where r the distance of the torch from the centre of the BH and M is the mass of the BH. The coefficients of $dt^2$ and $dr^2$ are the metric “tensor components” of space-time geometry for this particular question. For the light of the torch, however, the path is a geodesic curve – line of shortest path taken by light in a hugely curved space-time, and the above equation becomes $ds^2=0$ hence:
$\left(1-\frac{2GM/c^2}{r}\right) c^2dt^2-\frac{dr^2}{\left(1-\frac{2GM/c^2}{r}\right)}=0$.
The latter equation gives the speed of the light source, the torch, as it falls towards the BH from outside the horizon, and observed by the observer at some very large distance away from the BH
$v(r)=c(1-2GM/c^2r)$.
The frequency shift $z=\frac{f-f_0}{f_0}$ relates to the speed of the light source via the equation
$v(r)=c\frac{z^2+2z}{z^2+2z+2}.$
The last equation gives the way the frequency shift varies as a function of $r$, and how it is affected by the mass, M, of the BH. Here, $f$ is the frequency received by the observer, while $f_0$ is the actual frequency emitted by the light source (the torch.)