4

Sagittarius A* is the Milky Way's own supermassive black hole. Up till recently, I used to think that Sagittarius A* was a static black hole, i.e. It had no rotational speed at all.

But after doing a bit of research on black holes, I found out that Sagittarius A* is actually a Kerr-Newman black hole, i.e. it rotates. But apart from that, there was no available data on its rotational speed. What vague information I have found suggests that it rotates only slowly, but even then, there is no information about its rotational speed.

What exactly is the rotational speed of Sagittarius A?

Alastor
  • 2,639
  • 1
  • 9
  • 35
  • 1
    That looks like an answer, @PM2Ring. – David Hammen Oct 26 '22 at 15:32
  • 1
    The odds that an object has zero rotation rate with respect to local inertial statistically is a space of measure zero. A Schwarzschild black hole (or static black hole) is a spherical cow. A very nice and very early spherical cow (Schwarzschild came up with the idea just a year after Einstein published his general theory of relativity), but a spherical cow nonetheless. – David Hammen Oct 26 '22 at 19:47
  • Still, proving the possibility of spherical cows does prove the possibility of more general cows. :) – paul garrett Oct 26 '22 at 22:26

1 Answers1

4

According to An Upper Limit on the Spin of SgrA* Based on Stellar Orbits in Its Vicinity, by Giacomo Fragione and Abraham Loeb (2020) ApJL 901 L32:

The spin of the massive black hole (BH) at the center of the Milky Way, SgrA*, has been poorly constrained so far.

Using normalised units where 0 is no spin and 1 is the maximum possible spin, Fragione and Loeb give an upper limit of 0.1 for the spin of Sagittarius A*.

That corresponds to a spin speed at the event horizon of $0.1c$, which sounds rather fast. However, it's common for SMBHs (supermassive black holes) to have spins greater than 0.5.

Here's a graph of SMBH spins, from zephyr's answer to Maximum spin rate of a black hole?

SMBH spin

That graph comes from a paper by E. Samuel Reich, Spin rate of black holes pinned down, Nature 500, 135 (2013).

As zephyr's answer mentions, we expect stellar mass black holes, which are the remnants of core-collapse supernovae, to have substantial spin, due to conservation of angular momentum. Similarly, neutron stars formed via core collapse have high spin. However, as Tim Rias says in a comment on that answer,

the merging black holes observed by LIGO and Virgo generally seem to have had very little or no spin (with a couple of recent exceptions).

Those mergers involve intermediate mass black holes (under 100 solar masses), which are (presumably) formed by the merger of smaller black holes and other accreted matter. Conceivably, the merger of two black holes of roughly equal but opposite spin results in a merged BH of low spin. Also, some angular momentum is shed during the merger via gravitational radiation. However, merging can also increase spin, by converting orbital angular momentum to spin.

We still don't have a good theory to explain how SMBHs managed to become so large in the early universe. That theory also needs to explain why their spins tend to be large.

PM 2Ring
  • 13,836
  • 2
  • 40
  • 58
  • FWIW, the Nature Scientific Reports article Divergent reflections around the photon sphere of a black hole by Albert Sneppen discusses photon trajectories around Schwarzschild and Kerr BHs, with useful equations, and nice diagrams. – PM 2Ring Oct 26 '22 at 23:40
  • 1
    Which means that Sgr A* rotates at 10% the speed of light. – Alastor Oct 27 '22 at 07:47
  • @Furious That's an upper limit, but it would be unusual if its actual spin rate was substantially lower than 0.1. – PM 2Ring Oct 27 '22 at 07:56
  • Is the limit of the rotational speed of a black hole given by simply not exceeding the speed of light at the event horizon? From reading elsewhere it seems the limit is given by some relation between angular momentum and mass, but maybe the two are equivalent. Also, recent studies puts the actual spin of Sagittarius A* at around 0.9 of the maximum value, so to find the actual rotational period, do I just calculate the the circumference and divide it by c, and then multiply that by 0.9? – Outis Nemo Mar 11 '24 at 10:55
  • @OutisNemo The spin parameter of a BH is $a=J/Mc$, where $J$ is the angular momentum & $M$ is the mass. That has units of length, and a limiting value of $r_s/2=GM/c^2$. So a dimensionless form of the spin parameter is $A^\ast=\frac{Jc}{GM^2}$, which must be <1. At the limit, the event horizon breaks down & you get a naked singularity. See https://physics.stackexchange.com/a/350349/123208 & https://physics.stackexchange.com/a/804601/123208 – PM 2Ring Mar 11 '24 at 14:06
  • So, does that mean I can calculate the actual rotation rate of Sagittarius A* like I did above, or is it more complicated? I'm just trying to find the actual rotation rate, but I can't find it anywhere. – Outis Nemo Mar 11 '24 at 16:52
  • @OutisNemo I think so, but I'm not sure: my knowledge of the Kerr metric is very patchy. Maybe ask a question on Physics.SE. However, the spin parameter is probably more useful, because it's an invariant. The "actual" rotation rate isn't a straightforward thing because time dilation is intense, different observers see different things, and there are various choices of space & time coordinates. – PM 2Ring Mar 11 '24 at 17:18
  • @PM2Ring: I see. Well, I'm mostly interested in what the rotation rate would be as observed from Earth. The context for it is that I'm speculating about whether or not Sagittarius A* could possible cause ripples in the galactic magnetic field like our sun does in the heliomagnetic field as it rotates. But other than that I'm also just curious about it. – Outis Nemo Mar 11 '24 at 23:02