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In a local universe empty of any matter except a Kerr black hole and an observer, that observer is spinning at the same rate as the black hole and observes it from a great distance directly above its north pole. Will he deem it to be a Schwarzschild black hole?

If the observer is spinning at a different rate, will he see a Kerr black hole with different angular momentum?

Edouard
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LePtC
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2 Answers2

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No. A coordinate transform cannot change the angular momentum (or mass) of a Kerr black hole. In particular, the Schwarzschild black hole is spherically symmetric. The Kerr black hole is not.

By the way, observers and coordinate systems are independent concepts. The "frame of reference" language tends to suggest some kind of connection between them, but they are independent, even though they are conflated with each other in many introductions to special relativity (where we tend to consider only a very limited class of coordinate transformations). In any case, whichever of these concepts the question might mean by "frame of reference," the answer is still no: we cannot make a rotating black hole look like a nonrotating black hole just by changing the frame of reference.

Chiral Anomaly
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  • Frame of reference and coordinate systems are distinct, but they aren't independent. The central issue is frame of reference is not inertial. – Acccumulation Sep 19 '21 at 07:08
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    FWIW, something analogous goes in classical mechanics (Newtonian mechanics), too. If you could hover above the north or south pole of the Earth and rotate along with it, you'd still note it has an equatorial bulge. – The_Sympathizer Sep 19 '21 at 11:37
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    @Acccumulation Any observer can be described using any coordinate system (better: any coordinate atlas, and choosing a coordinate system does not require choosing any observers. In that sense, they are independent. – Chiral Anomaly Sep 19 '21 at 13:24
  • Given an observer, there are some coordinate systems in which their space coordinates are identically (0,0,0). So there is a relationship between coordinate system and observer. – Acccumulation Sep 19 '21 at 21:10
  • It is unclear how you intend to prove that coordinate transformations cannot change the angular momentum by calculating the Kretschmann scalar. Nor is it clear how this implies that the geodesic will be different. As it stands this answer seems to be trying to prove that the sky is blue by observing that water is wet and boiling an egg. – TimRias Sep 19 '21 at 21:31
  • @mmeent I removed the Kretschmann-scalar statement from the answer, because the approach I had in mind was indeed flawed. I also removed the statement about geodesics, because even though the difference in symmetry does prove that the patterns of geodesics are different, it doesn't by itself prove that the patterns of lightlike geodesics are different, so it doesn't by itself justify my statement about the light from distant stars -- which was irrelevant anyway, given my interpretation of the question. – Chiral Anomaly Sep 19 '21 at 22:22
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    @mmeent The conclusion is unchanged, though. I'm interpreting the question to be "Is a spinning black hole physically distinct from a non-spinning black hole," not a question about the appearance of a black hole (which is a more difficult issue). The answer asserts that we can't un-spin a black hole by changing coordinate systems, and the fact that one spacetime has spherical symmetry and the other does not is enough to prove this. – Chiral Anomaly Sep 19 '21 at 22:23
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You can't really evaluate how a black hole in our universe works by comparing it to a thought experiment of a universe containing just the black hole. There is a phenomenon called "frame dragging" in which the rotating black hole alters the spacetime around it through its rotation. However, in the reference frame of an observer rotating with the black hole, the rest of the universe is rotating, and from this observer's perspective, its rotation causes frame dragging in the opposite direction. The effect of the rest of the universe is that the black hole's frame dragging is partially cancelled out (and the net frame dragging decreases as one goes away from the black hole). So while the Mach principle does assert that a universe containing nothing but a rotating black hole would be indistinguishable from a universe containing nothing but a non-rotating black hole, that doesn't mean that a rotating black hole with our surrounding universe is indistinguishable from a non-rotating black hole with our surrounding universe.

Relativity says that the rules for what things do don't depend on the reference frame, but doesn't mean that reference frames can't differ in how things are affected by the rest of the universe. A car traveling at the same velocity as the Earth will experience different phenomena than a car traveling at a different velocity, but that's not because there are different physics operating on them, it's because the surrounding matter is interacting with them differently.

Acccumulation
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  • @safesphere According to the Mach principle, in a universe with just the black hole, the universe would be indistinguishable from a universe with a non-rotating black hole. So the rest of the universe must be cancelling out the effects of the rotation in some way. – Acccumulation Sep 19 '21 at 21:02