If space craft nears a black hole we see it getting slower and slower to the point it would appear to stop moving due to red shift. Then how fast are these black holes moving?

enter image description here

  • 1
    $\begingroup$ Another question by OP about the same video: physics.stackexchange.com/q/225673/2451 $\endgroup$
    – Qmechanic
    Commented Jun 29, 2016 at 22:47
  • 1
    $\begingroup$ Please don't cut and paste random stuff on the internet without attribution. It's rude. Who is the person who made the video? Also, there's no way we can possibly give you a quantitative answer to this question based on a video. There is nothing that sets a time or distance scale. What is the point of offering a bounty on the question when you haven't even provided any of the information that would be needed in order to answer it? $\endgroup$
    – user4552
    Commented Feb 12, 2019 at 22:17
  • $\begingroup$ @BenCrowell Well the video is at least watermarked. Hence we know it is made by the SXS collaboration. $\endgroup$
    – TimRias
    Commented Feb 13, 2019 at 8:30
  • $\begingroup$ How can a user with a reputation of 1 offer any bounty at all? $\endgroup$
    – D. Halsey
    Commented Feb 14, 2019 at 1:20
  • $\begingroup$ @D.Halsey The bounty amount is subtracted from the offerer's rep as soon as the bounty is started, so you're seeing the post-subtraction rep. See meta.stackexchange.com/questions/16065 and physics.stackexchange.com/users/148704/muze?tab=topactivity $\endgroup$ Commented Feb 14, 2019 at 4:13

1 Answer 1


Let me begin by stressing that this question does not have a physically meaningful answer in any precise way. This is due to three reasons:

  1. There are no global frames of reference in general relativity (see e.g. this Q&A: How do frames of reference work in general relativity, and are they described by coordinate systems?). Hence there is no meaningful way of assigning a velocity relative to the observer (or the center-of-mass for that matter) to the black holes. In particular, if the black holes were pointlike objects (but see 2.) they would be at rest in their own local reference frame.

  2. The black holes are not physical objects with any precise localisation. Their individual event horizons are extended (and deformed). Hence we cannot assign a precise position to the black holes, let alone a velocity.

  3. The spacetime of a binary black hole merger is highly dynamical. This excludes many approximations one could make to mitigate the above two issues, strengthening their consequences.

That being said we can have a go at giving an approximate answer. Based on the relative sizes of the two photon spheres in the video, we can estimate the ratio of the two black hole masses to be roughly 1:4. Furthermore, lets assume the individual black holes are not spinning. From simulation data it can be read off (e.g. from the peak of the gravitational wave frequency) that the merger happens at angular velocity $\omega = 0.16 c^3/(GM)$, where $G$ is the gravitational constant, $c$ the speed of light, and $M$ the total mass of the binary. From Effective One Body descriptions it is known that the merger happens there and about a radial separation from the "center of mass" $r = 3 GM/c^2$. We can thus roughly estimate the orbital velocity as $r\omega =0.6\,c$.

However, I stress again that this is (and fundamentally can be) only a "back of the envelope" calculation.


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.