To simulate light scattering on the rotating black hole we have used this paper and this code. First, we made animation for light beam scattering in the equatorial plane

Figure 1

For not equatorial plane the animation looks like

Figure 2

This animation shows difference of light orbits in the direction of rotation and in the counterrotating direction

Figure 3

Gravitational redshift is determined by the equation $$1+z=\frac {\lambda _0}{\lambda _1}=\frac {\Sigma}{\rho \sqrt {\Delta}} (1)$$
where $\Sigma , \rho , \Delta $ - parameters of the Kerr metric are expressed in terms of radial and angular coordinates, $$\rho ^2=r^2+ a^2 \cos ^2\theta, \Sigma ^2=(r^2+a^2)^2-a^2 \Delta \sin ^2\theta , \Delta = r^2+a^2- 2 M r$$
This question comes from consideration of gravitational red shift picture on the Wikipedia page, which is widely used. My animation is the analog of this picture in the Kerr metric. Is it correct to show redshift from red to blue when light coming to the border of ergosphere computed with equation (1)? Second question comes from the answer on Do light experience Doppler shift along and against frame dragging? and answer on Gravitational lensing redshift around a Kerr black hole. How we can compute frame dragging effect on the light in the Kerr metric?

Update 1. Taken into account answer @Void the last picture has been recalculated. The new animation looks very different from above.

Figure 4

  • $\begingroup$ Who are these experts, and what are they saying exactly? Your question is a bit vague. $\endgroup$
    – Javier
    Commented May 12, 2021 at 17:13
  • $\begingroup$ @Javier Experts preferer state anonymous. But they ask two questions: 1) For what reasons will the photon turn blue or red along the way? 2) And most importantly, where will photon take energy to turn blue, or where to give it when photon turns red? $\endgroup$ Commented May 12, 2021 at 17:39
  • 1
    $\begingroup$ @AlexTrounev Nice figures! If you read the Gravitational redshift wiki, how does that not answer your question? Note that the frequency of the photon is observer-dependent, so there is some nuance to the statement of "photon changing frequency". $\endgroup$
    – Void
    Commented May 15, 2021 at 19:28
  • $\begingroup$ @Void Thank you for you comment. I understand, that this animation is just an illustration to the gravitational red shift. I read wiki page and some other pages connected to the picture discussed. What about explanation on this page en.wikiversity.org/wiki/Astronomy_college_course/… ? $\endgroup$ Commented May 15, 2021 at 21:13

1 Answer 1


Redshift of photons is an observer-dependent notion. That is, if we want to know how photons get redshifted from point to point, we need to ask who is measuring their frequency. If we do not specify the observers, the question has no meaning.

When one speaks of gravitational redshift, one typically implicitly refers to a class of observers that play a special role with respect to the space-time. For a wave-vector $k^\mu$ at a given point, the observer with four-velocity $u^\mu$ will see the photon to have a frequency $\omega = -u^\mu k_\mu$. If you have a field of observers with velocities $u^\mu(x^\nu)$, you can formally assign a frequency to every value of the affine parameter along the lightray as $\omega(\lambda) = -u^\mu(x^\nu(\lambda)) k_\mu(\lambda)$. This is how I interpret the pictures in your post.

In Schwarzschild space-time, one would usually choose the static observers with four-velocity $$u^\mu = \frac{1}{\sqrt{-g_{tt}}} \delta^\mu_t$$ In Kerr space-time, things are a little bit more complicated. Generally, it is natural to choose observers that have a non-zero $u^t$ and $u^\varphi$. The redshift formula is then also dependent on the impact parameter of the photon $-k_\varphi/k_t$. (Note that $k_\varphi,k_t$ are conserved along null geodesics.) The only way to avoid this is to choose the static observers with the same formula for four-velocity as given above. These will not exist within the ergosphere. However, outside the ergosphere you can show that you will have the redshift (blueshift) with respect to infinity given as $$\frac{\omega}{\omega_\infty} = \frac{1}{\sqrt{-g_{tt}}} = \sqrt{\frac{r^2 + a^2 \cos^2 \vartheta}{1 - 2 Mr}}$$ This does not match your formula (1). Instead, your formula (1) corresponds to $\sqrt{-g^{tt}}$, which is incorrect (it could be completed by an impact parameter term to correspond to the so-called zero-angular momentum observers), even though the qualitative behaviour is OK.

As for your question about frame-dragging. Frame-dragging is hard to quantify locally, it refers once again to observer-dependent comparison between points. You observe frame-dragging in your plots when the photons make "tiny loops" at the edges of the ergosphere. These correspond to the fact that the photons are forced to corotate with the black hole when sufficiently close even though they counter-rotated up to that point.

Come to think of it, there may be an issue in your code. You should see this phenomenon to some degree, but some photons should definitely end up in the black hole.

There is no unique way to locally characterize frame-dragging. I would recommend to instead use global outcomes such as the outgoing angles of the photons as its measure.

  • $\begingroup$ What about method used in this paper iopscience.iop.org/article/10.1088/0067-0049/218/1/4/pdf ? $\endgroup$ Commented Jun 2, 2021 at 17:58
  • 1
    $\begingroup$ It seems to be correct. They use the so-called zero-angular momentum observers to define redshift in that case. $\endgroup$
    – Void
    Commented Jun 2, 2021 at 19:26
  • $\begingroup$ Thank you very much for your answer. Actually there is no issue in my code since I used reflective boundary condition for the light approaching ergosphere, while usually it should be nonreflective condition. $\endgroup$ Commented Jun 2, 2021 at 21:54
  • $\begingroup$ I guess that is fine as long as you keep in mind that the real physical light-rays will never actually reflect at the boundary of the ergosphere. $\endgroup$
    – Void
    Commented Jun 3, 2021 at 9:16
  • $\begingroup$ In theory yes, you are right. But in practice we need to keep in a mind all possibilities including reflection of light approaching ergosphere or maybe some region close to it. $\endgroup$ Commented Jun 3, 2021 at 10:33

Your Answer

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

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