6
$\begingroup$

Light from a source passes by a Kerr black hole on two sides at the equator and converges at the observer. The axis of rotation of the black hole is perpendicular to the direction of light. Two rays of light pass through the spacetime regions of a significant frame dragging, on one side along and on the other side against the direction of light.

Would frame dragging cause a red shift of one ray and a blue shift of the other? Or would both rays come to the observer with the same frequency?

Improve this question
$\endgroup$
1

1 Answer 1

Reset to default
4
$\begingroup$

Kerr metric is stationary and has a Killing vector field $\xi_t=\partial _t $, time-like outside of the black hole, representing stationarity. This means that the quantity: $$ E = g_{\mu\nu }\xi^\mu_t p ^\nu ,$$ where $p^\mu$ is 4-momentum (of, say, a photon), is conserved along the geodesics.

So, if both the source of light and observer are at rest relative to the black hole and are far away from it, then the observed frequency of light would be the same as the emitted frequency, for all light rays reaching the observer.

Improve this answer
$\endgroup$
11
  • 1
    $\begingroup$ That is really non-intuitive, given that light rays can get strongly deflected and even forced into prograde orbits. On the other hand, accretion disks and orbiting objects will show a lot of gravitational redhshift plus the redshift/blueshift due to rotation velocity. I guess this shows the power of using Killing vector fields. $\endgroup$
    – Anders Sandberg
    Jan 31, 2019 at 17:02
  • $\begingroup$ @AndersSandberg: if your intuition is better developed for EM phenomena, then consider the following analogy. A charge enters the region with stationary electric and magnetic fields. Its trajectory could be quite complex, but we know that if the charge escapes the region of EM fields its energy would be the same as before it entered it (ignoring the radiation). $\endgroup$
    – A.V.S.
    Jan 31, 2019 at 17:28
  • $\begingroup$ @A.V.S. Does your answer imply that energy cannot be extracted from a rotating black hole by using the EM radiation? This old paper for example claims that energy can be extracted: pdfs.semanticscholar.org/07ef/… - However, this seems strange, because, for example, the magnetic field of a permanent magnet cannot be a source of energy, even if the magnet is made of a superconducting ring with electrons rotating in a circle and thus sinewhat conceptually similar to a Kerr black hole. Is the claim in the paper correct? Thanks! $\endgroup$
    – safesphere
    Sep 5, 2019 at 20:01
  • 1
    $\begingroup$ @safesphere: Energy can be extracted, but this is a wave effect called superradiance that classically corresponds to the scattered waves around rotating BH having larger overall energy than the incident wave but still the same frequency (away from the BH). On a quantum level, for single photons we have the nonzero probability of black hole producing a second photon (coherent with the first one), or vacuum state producing outgoing photon using the energy of black hole (the last effect smoothly interpolates into pure Hawking radiation for nonrotating BHs)… $\endgroup$
    – A.V.S.
    Sep 7, 2019 at 5:03
  • $\begingroup$ … cont. See the book arXiv:1501.06570 for more examples of superradiance (not only in black holes). $\endgroup$
    – A.V.S.
    Sep 7, 2019 at 5:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.