Do light experience Doppler shift along and against frame dragging?

Question

Imagine 2 photons traveling along the equator in opposite directions, of course I'm not suggesting light goes into orbit I'm just wandering if these photons will be blue and red shifted ever so slightly as massive body such as Earth spins in general relativity?

Answer 1

To answer this we have to start with the equation that describes the geometry around a spherically symmetric rotating mass. This is the Kerr metric. I'll write this out in full below, which is going to look pretty scary, but for our purposes we find the equation simplifies a great deal:

$$\begin{align} ds^2 &= -(1 - \frac{r_s r}{\rho^2})dt^2 \\ &+ \frac{\rho^2}{\Delta}dr^2 \\ &+ \rho^2d\theta^2 \\ &+ (r^2 + \alpha^2 + \frac{r_s r\alpha^2}{\rho^2}\sin^2\theta)\sin^2\theta d\phi^2 \\ &+ \frac{2r_sr\alpha\sin^2\theta}{\rho^2}dt d\phi \end{align}$$

Where:

$$\begin{align} r_s &= 2M \\ \alpha &= \frac{J}{M} \\ \rho^2 &= r^2 + \alpha^2\cos^2\theta \\ \Delta &= r^2 - r_sr + \alpha^2 \end{align}$$

In the equation $J$ is the angular momentum of the black hole, $r$ is the distance from the centre of the black hole, $\theta$ is the latitude, $\phi$ is the longitude and $t$ is time.

This simplifies because we can assume all motion is in the equatorial plane so $\theta$ is constant and therefore $d\theta=0$ and $\rho=r$. We'll also consider the moment that the light is travelling tangentially to our circle of radius $r$, so at this point $dr\approx 0$. Lastly for light $ds=0$, and all this simplifies the metric to:

$$\begin{align} 0 &= -(1 - \frac{r_s}{r})dt^2 \\ &+ \left(r^2 + \alpha^2 + \frac{r_s \alpha^2}{r}\right) d\phi^2 \\ &+ \frac{2r_s \alpha}{r}dt d\phi \end{align}$$

And finally $d\phi/dt$ is just the angular velocity $\omega$ so if we divide through by $dt^2$ and rearrange we get:

$$ 0 = \left(r^2 + \alpha^2 + \frac{r_s \alpha^2}{r}\right) \omega^2 + \frac{2r_s \alpha}{r}\omega - (1 - \frac{r_s}{r}) \tag{1} $$

And that's the equation we need. Let's do a quick sanity check and take the non-rotating limit i.e. set $\alpha = 0$. Equation (1) immediately gives us:

$$ r \omega = v = \pm \sqrt{1 - \frac{r_s}{r}} $$

Which is the right answer! We get two values for $\omega$ with equal magnitudes and opposite signs corresponding to the two light beams going in opposite directions. So the two beams have identical speeds $v = \sqrt{1 - r_s/r}$ (note we are using units where $c=1$). The speed of the light is reduced by a factor of $\sqrt{1 - r_s/r}$ due to the time dilation.

OK, thus reassured lets go back to equation (1) and solve it for non-zero rotation i.e. $\alpha \ne 0$. The equation is just a quadratic and the quadratic formula gives us:

$$ \omega = \frac{-\frac{r_s \alpha}{r}}{\left(r^2 + \alpha^2 + \frac{r_s \alpha^2}{r}\right)} \pm \frac{\sqrt{ \left(\frac{r_s \alpha}{r}\right)^2 + \left(r^2 + \alpha^2 + \frac{r_s \alpha^2}{r}\right)(1 - \frac{r_s}{r})}}{\left(r^2 + \alpha^2 + \frac{r_s \alpha^2}{r}\right)} $$

I don't propose to take the algebra any farther, but it should now be immediately obvious that our two values for the angular velocity are not equal. They differ by:

$$ \omega_+ - \omega_- = 2\frac{-\frac{r_s \alpha}{r}}{\left(r^2 + \alpha^2 + \frac{r_s \alpha^2}{r}\right)} $$

And this difference is the effect of the frame dragging.

Answer 2

Arguably, no. Here's my reasoning:

First, note that when talking about frequency shifts in a generic general-relativistic setting, you need to specify your observers (emitter and absorber) as well as the photon trajectory. The generalized Doppler shift is then nothing but the difference in the time-component of photon momentum as measured in the two reference frames. This will yield special-relativistic Doppler shift as well as gravitational and cosmological redshift (which can be phenomenologically attributed to time dilation and spatial expansion, respectively).

Now, there are two different circular photon orbits in the equatorial plane of a Kerr black hole (cf Wikipedia), one prograde and one retrograde. An observer sitting at fixed Boyer–Lindquist coordinates along either of the orbits will always measure the same photon energy, no matter how often the photon has circled the black hole.

However, when you consider more complicated trajectories (eg when starting at an arbitrary radius and shooting photons in opposite directions around the hole), this story will generally change - but you will no longer compare apples to apples due to the inequivalent positions of the absorbers.