Gravitational Redshift around a Schwarzschild Black Hole

Question

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?

Answer 1

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.)

Answer 2

It would have been helpful if the OP could have given a reference to where this material occurs in Carroll. I looked through the free arxiv version (there is also a free html version) and couldn't find it. The question is also a little broad and vague as to what the OP's background is. There are basically two parts to the question: (1) justification and interpretation of the form given in Carroll for the redshift factor, and (2) application of this to the problem of the falling lightbulb.

Re #2, I think the following works as a heuristic derivation of the form of the relation given by Carroll, and may help to explain its physical content and what the math means. A particle's energy-momentum four-vector $p$ is proportional to its velocity four-vector $v$. This is basically because what other direction would there be for it to point? (It also makes sense in terms of corresponding correctly to the Newtonian relation $p=mv$.) Now a ray of light doesn't actually have a normalizable velocity four-vector, but that's OK, because we're just working with proportionalities. We don't need normalization. We also know that the frequency four-vector $\omega$ is proportional to $p$. There are purely classical reasons for this, but for people in the modern era this is easier to justify simply because the proportionality constant is Planck's constant in the case of a single photon.

Stated in index notation, we have $\omega^a \propto v^a$. This is in terms of the contravariant form of the frequency, but ordinarily we work with its covector form, which is defined in such a way that for an observer with velocity $U$, the rate $\omega$ (scalar) at which wavefronts arrive is $\omega=U^a\omega_a$. We therefore have $\omega \propto U_a v^a$, which is essentially what Carroll seems to be saying (although the context is lacking).

The $-$ sign in Carroll is just a proportionality constant, so we can ignore it. Presumably he has it there because he's working in the $-+++$ metric and he wants this to come out positive. The $g_{\mu\nu}$ is there because he's lowering an index. The fact that the observer is stationary doesn't actually affect the proportionality argument in any way, but presumably for the case he's considered (perhaps a Doppler shift referred to a stationary observer at infinity), that assumption is necessary in order to make the proportionality constant be the one he gives. The choice of the affine parameter $\lambda$ is arbitrary, and again it's a little hard to know how Carroll intends that ambiguity to have been resolved when he states the constant of proportionality, since any redefinition of the affine parameter $\lambda\rightarrow a\lambda+b$ will change the result by a factor of $a$. (And because this is a null geodesic, there is no natural choice for the affine parameter, such as a proper time.)

Re #1, I think we would need to see the actual statement of the problem, because some things seem to be left out. Specifically, it's not clear to me whether the flashlight was dropped from rest at infinity or from rest at $r_\text{obs}$. In either case, you probably need to find its velocity four-vector when it reaches $r_e$, but that would be a separate calculation. Also, it's not stated whether the observer is at rest or moving relative to the black hole. If at rest, then the observer has to be outside the horizon, and the flash cannot be observed if it was emitted from inside the horizon.