The question is related to Sean Carroll's Spacetime and Geometry ex 6.6.
Consider a Kerr black hole with an accretion disk of negligible mass. Particles in the disk follow geodesics. Some iron in the disk emits photons at constant frequency $\nu_0$ in the rest frame. Then what is the observed frequency from the edge of the disk at the edges and center of the disk? Consider both cases where the disk and the black hole rotate in the same and opposite directions. Limit everything in the equatorial plane so $\theta=\pi/2$.
By edge of the disk, imaging we are looking at the accretion disk edge on, then we source from one edge to be approaching and the other recessing.
I'll use -+++ as the signature.
It immediately came to my mind that we can use $\omega=-g_{\mu\nu}\frac{dx^\mu}{d\lambda} U^\nu$, where $x$ is a null geodesics connecting emitter and detector parametrized by $\lambda$, $U^\nu = \frac{dX^\nu}{d\tau}$ which describes the iron and observer. With some algebra, we should find the frequency to be
$$\omega = EU^t – LU^\varphi + g_{rr}\frac{dr}{d\lambda}U^r$$
where E and L are energy and angular momentum obtained via Killing vectors. The redshift is, therefore,
$$
1 + z = \frac{\omega_e}{\omega_d} = \frac{EU_e^t – LU_e^\varphi + g_{rr}\frac{dr}{d\lambda}U_e^r}{EU_d^t – LU_d^\varphi + g_{rr}\frac{dr}{d\lambda}U_d^r}
$$
where I use e and d to denote emitter and detector respectively.
This is where I got stuck. We may consider the iron atoms to move in a circular orbit so that we ignore the $U_e^r$ term. In addition, for the observer to be stationary far away, we can also take $U_d^r = U_d^\varphi = 0$.
$$
1 + z = \frac{EU_e^t – LU_e^\varphi}{EU_d^t}
$$
This formula seems to be only dependent on radius r. As an observer, we should expect the photon to have the same redshift at a fixed radius r. This conclusion is not making much sense. Wouldn't we expect different redshifts due to the motion of the source on opposite edges of the disk? What am I missing here?
Many thanks for any help in advance!
Best Answer
The thing you are missing is that the null geodesic connecting the emitter to the detector (and therefore $E$ and $L$ ,or more specifically the ratio $b=L/E$) will depend on the location of the emitter along its orbit. When the emitter is approaching the value of $b$ will be different than when it is receding.
Consequently, the formula you find doesn't just depend on $r$.