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It is known that a Kerr Black Hole (=BH) isn't the same as the Schwarzschild BH viewed from a rotating frame of reference (see here and here). The geometry due to it's "intrinsic rotation" seems to be different from the passive transformation of coordinates viewed with respect to an orbiting test particle around Schwarzschild space-time.

Can we say something similar about a moving BH? There has been perturbative solutions like Aichelburg-Sexl spacetime, Boosted Schwarzschild spacetime , or non-perturbative like in H.Bondi 1962, which involves doing diffeomorphism transformations mimicking Lorentz-boost at infinity. However, they just look like a result of passive transformations w.r.t. observers far away from the source and its not clear what this transformations represent near the massive body.

A related post already exists, but I'm still not sure how to define a "moving BH". It can be wrt observers at infinity (just like in above solutions). However, even a Schwarzschild BH will look like a moving mass wrt orbiting test particle, yet its still Schwarzschild. If the moving particle observes anything similar to the above solutions then qualitatively it should detect gravitational radiations which would constantly decrease the mass of the source. Probably I'm mistaken, but there seems to be some kind of "absoluteness" of a body's state of rest or motion unless the curvature effect is turned off

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    $\begingroup$ If the black hole is moving with constant velocity you can just transform into its rest frame and give all the velocity to the orbiting particle, velocity is relative. You don't need a black hole or relativity for that, that is also valid for regular masses and charges. Only if the black hole is accelerated it will radiate gravitational radiation, but if the particle in orbit has neglible mass compared to the black hole the latter will have only neglible acceleration and unaccelerated objects don't produce gravitational radiation. $\endgroup$
    – Yukterez
    Commented Aug 15, 2023 at 0:16
  • $\begingroup$ @Yukterez Even if the mass of the test particle is not negligible, the radiation would still be the same in the boosted or transformed cases. The relativity principle of Poincare is fundamental - there is no difference between test and uniform linear motion +1 $\endgroup$
    – safesphere
    Commented Aug 15, 2023 at 2:50
  • $\begingroup$ @Yukterez and safesphere thank you for the comments. However, I didn't get your points. The boosted schw solution for e. g. shows a relatively non accelerating BH, but it is still radiating. Acceleration doesn't seem to be a necessary component. Lorentz boost that these authors have done only makes sense near infinity. Relative velocity has a "global meaning" near infinity, but it becomes ambiguous in presence of curvature. How do we know that a moving BH won't have any additional properties apart from viewing it from a test particle POV? $\endgroup$
    – KP99
    Commented Aug 15, 2023 at 6:35
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    $\begingroup$ The velocity is not ambiguous, but can be calculated anywhere through a coordinate change. If a boosted solution radiates, then a non-busted one radiates too, because the boost does not change physics per the relativity principle. Spacetime is invariant to uniform linear boosts the same way it is invariant to spatial translations. Either changes only the coordinate system, but not physics. Deriving a boosted solution or boosting the Schwarzschild solution is a mathematical choice for the same physics. The result is the same, because math is consistent. $\endgroup$
    – safesphere
    Commented Aug 15, 2023 at 14:17
  • $\begingroup$ Relative velocity becomes ambiguous depending on how we compare the velocity in different tangent planes (see here). Boost isn't an isometry in general, nor is translation. So I didn't get what you mean by spacetime is invariant. Its hard to make sense of such transformations in presence of curvature. If you look at the above papers , you will see that boosted solution has grav. rad but the original metric doesn't. When curvature is turned off, the ambiguity disappears and boosting a particle doesn't change the physics. $\endgroup$
    – KP99
    Commented Aug 15, 2023 at 14:41

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