Frequency of the photon in general relativity

Question

There is an observer at ($r,\theta, \phi$) outside of the Schwarzschild blackhole. A beacon is falling into the black hole along $r$ coordinates of the metric and is emitting radiation. At $r_{em}$, it emits a photon of frequency $\omega_{em}$, what will be the frequency of the light which is observed by the observer (assuming only radial trajectories)? In Sean Caroll, Chapter-5 eq. (5.100), he defines the frequency of the photon to be $$\omega = -g_{\mu \nu} U^{\mu} \frac{dx^{\nu}}{d\lambda}.\tag{5.100}$$However, I am not sure how he got this expression? Is there some intuitive way to get the same expression.

I was thinking about this and it seems that the $\omega$ would change according to the time dilation factor because of the curvature components of the metric.

Please correct me if I am wrong and kindly guide me.

Answer 1

To answer your question, we need a few calculations that I will try to make short.

Using the Schwarzschild metrics, and considering that the photon is emitted in a radial direction, the energy of the photon measured by a static observer is: $$ E = -\frac{c}{\sqrt{1-\frac{2GM}{c^2r}}}\vec{ξ(0)}.\vec{p} $$ with $ r $ radial coordinate, $ \vec{ξ(0)} $ vector of Killing and $ \vec{p} $ four-momentum of the photon.

Along a light geodesics, the quantity $ \vec{ξ(0)}.\vec{p} $ is maintained, then you can write for the emission and for the reception of the photon $ (\vec{ξ(0)}.\vec{p})_{em}=(\vec{ξ(0)}.\vec{p})_{rec} $ which means: $$ E_{rec}=\sqrt{\frac{1-\frac{2GM}{c^2r_{em}}}{1-\frac{2GM}{c^2r_{rec}}}}E_{em}\ \ \ \ \ [1] $$
In addition, because $ E=\frac{hc}{\lambda} $ with $ \lambda $ wave length of the photon, you have with $ \omega $ frequency of the photon: $$ \frac{E_{rec}}{E_{em}}=\frac{\lambda_{em}}{\lambda_{rec}}=\frac{\omega_{rec}}{\omega_{em}} $$ which gives according to $ [1] $: $$ \omega_{rec}=\sqrt{\frac{1-\frac{2GM}{c^2r_{em}}}{1-\frac{2GM}{c^2r_{rec}}}}\omega_{em} $$

This shows that the reception frequency of the photon does indeed change with respect to its emission frequency.

Hoping to have answered your question,

Best regards.

Answer 2

The RHS of eq. (5.100) is the RHS of $$E~=~-p_{\mu}U^{\mu},\tag{3.63}$$ where the affine parameter $\lambda\to a\lambda+b$ can be scaled such that $$p^{\mu}~=~\frac{dx^{\mu}}{d\lambda}\tag{3.62}$$ is the 4-momentum of the photon. Here Carroll is using units with $\hbar=1$ so that $E=\hbar\omega=\omega.$

In eq. (3.89) and in a paragraph between eqs. (5.62-63) Carroll argues that the scalar invariant $\omega$ can be thought of as the observed frequency of the photon if we use local initial/normal coordinates $\left. g_{\mu\nu}\right|_p=\eta_{\mu\nu}$ for a static observer $U^{\mu}=(1,0,0,0)$.