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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?

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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.

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  • $\begingroup$ 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 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 at 17:28

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