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There are various theoretical models for accretion disks and relativistic jets from black holes that considers the black hole to be either rotating (Kerr geometry) or non-rotating (Schwarzschild geometry). But when astronomers detect a source, they detects whether the source is rotating or non-rotating. One such example is the GRS 1915+105 which is an X-ray binary star system featuring a regular star and a black hole is observed to rotate at close to 1,150 times per second with a spin parameter value between 0.82 and 1.00 (maximum possible value).

So my question is the following:

How do astronomers detect whether a black hole is rotating or non-rotating? And if the black hole is rotating, how do they calculate the spinning rate?

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    $\begingroup$ Walter Lewin explains this questions nicely in this lecture youtube.com/watch?v=313C1zo9pyE $\endgroup$ Sep 15, 2020 at 7:52
  • $\begingroup$ Rotating black holes produce gravitational waves. These waves can be accurately measured here on earth. $\endgroup$
    – Sam
    Sep 15, 2020 at 12:25
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    $\begingroup$ An ideal rotating black hole does not produce gravitational waves. Gravitational waves are produced during a binary merger, but not after the resulting black hole settles to a steady state. $\endgroup$ Sep 15, 2020 at 12:49
  • $\begingroup$ @physicopath it's a great lecture by Lewin, but it does not examine the spin of a black hole, so your comment is incorrect and should be deleted. :-) $\endgroup$ Sep 18, 2020 at 4:36

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Here is from the abstract of a paper about theory and observation of the spin of a black

..non-zero spin leaving an indelible imprint on the space-time closest to the black hole. As a consequence of relativistic frame-dragging, particle orbits are affected both in terms of stability and precession, which impacts on the emission characteristics of accreting black holes both stellar mass in black hole binaries (BHBs) and supermassive in active galactic nuclei (AGN). Over the last 30 years, techniques have been developed that take into account these changes to estimate the spin ....

It is not a simple formula, but the paper goes into the mathematical details.

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