If all stars rotate, why was there a theory developed that requires non-rotating stars?

Question

According to Penrose's research, a non-rotating star would end up, after gravitational collapse, as a perfectly spherical black hole. However, every star in the universe has some kind of angular momentum.

Why even bother doing that research if that won't ever happen in the universe and does it have any implications for the future of astrophysics?

Would you mind providing more information about the research, e.g. linking to a paper about it? — HDE 226868, Aug 21 '19 at 13:58
I suppose it's the solution to a simplified model of reality as a first step? That's not unusual in science... — Peter - Reinstate Monica, Aug 22 '19 at 05:47
"However, every star in the universe" You've checked them all have you? — TripeHound, Aug 22 '19 at 13:03
@TripeHound Every star in the universe has an absolute spin of at least zero. — wizzwizz4, Aug 22 '19 at 13:56
FWIW, there's a graph at the end of this answer of the spin of 19 supermassive black holes. As you can see, they have spin speeds that are a significant fraction of the speed of light. — PM 2Ring, Aug 23 '19 at 02:20
Ever solved a problem by treating macroscopic objects as point masses? None of those objects are actually mathematical points, and yet, you can make certain assumptions, disregard certain details, and use the resulting model to understand what is going on, and make predictions - as long as you keep in mind that there are scenarios where such a model is less applicable, or not applicable, because the assumptions you made no longer apply and the details you disregarded start to produce significant effects. There's always a domain of applicability. Everything in science is like that. — Filip Milovanović, Aug 24 '19 at 00:12
You usually have to start simple before doing the more complicated things. Turns out that rotation is kinda complicated. — Daddy Kropotkin, Jun 27 '21 at 16:09

score 53 · Answer 1 · answered Aug 21 '19 at 14:35

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In a similar way, we could ask...

No beams can be exactly 1 meter long. No beams can be exactly straight. The material making up a beam cannot be truly isotropic. So why should we bother calculating the stress in a 1 meter straight beam having isotropic material?

Because knowing how to perform this calculation is a building block for doing more complex calculations.

The non-rotating black hole calculation also provides a limiting solution. The solution for a spinning star's collapse will approach this solution as the spin approaches zero.

Similarly, Newton told us that as external forces approach zero, the path of a moving object will approach a straight line. This is useful to know even though there is no place in our universe that doesn't have gravitational influence.

answered Aug 21 '19 at 14:35

James

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Assume a spherical cow... – RonJohn Aug 22 '19 at 03:23
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I'm not sure if the metre is still defined against a standard, but if so, there is one stick that is exactly 1 metre long (by definition). Perhaps not entirely relevant to your point though. – Roland Heath Aug 22 '19 at 05:39
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@RolandHeath It hasn't been since 1960. – Graipher Aug 22 '19 at 06:50
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+1, but is it obvious that the non-rotating case is a limiting solution? A priori there might be global (topological?) effects that come into play as the angular momentum density grows towards infinity just before a singularity forms. – hmakholm left over Monica Aug 22 '19 at 12:54
@HenningMakholm: I might be misunderstanding your point (I'm not a physicist). I meant that it is the limit of low-spin, not the limit of high spin. – James Aug 22 '19 at 14:11
@James In a very simple handwaving "model", the collapse to a black star is also a limit size $\to 0$, and during the collapse the angular momentum remains constant. This might not play nicely with our limit of angular mometum $\to 0$. (Conceptually, think of $\lim_{x\to 0} \lim_{y\to 0} \arctan(x/y)y/x = 0$ vs $\lim_{y\to 0} \lim_{x\to 0} \arctan(x/y)y/x = 1$.) – JiK Aug 22 '19 at 18:44
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@James: My point is that a collapsing star with low but nonzero angular momentum has to go through a phase where its angular momentum density diverges to infinity during the collapse -- whereas a star with zero angular momentum can have zero angular momentum density during its entire collapse. That might (at least a priori) give rise to a qualitative difference that is not respected by the limiting process. – hmakholm left over Monica Aug 23 '19 at 12:24
@HenningMakholm: Thank you! I understand your point now. – James Aug 23 '19 at 12:26

score 47 · Answer 2 · answered Aug 21 '19 at 12:19

All models are approximations, we judge a model on how useful it is.

Understanding the collapse of a non-rotating star to a black hole gives insight into the nature of gravitational collapse. Much of the physics of collapse does not depend on spin. The formation of an event horizon, for example.

Models can be refined, and in this case, considering rotation leads to further insight, and a non-spherically symmetric structure with multiple singular horizons.

All models are necessarily simplifications. But the non-rotating model is still useful.

score 20 · Accepted Answer · answered Aug 21 '19 at 22:55

Another consideration is that the physics that describe a rotating black hole was much harder to develop.

The maths describing the Schwarzschild (uncharged, non-spinning) black hole was developed in 1916. This was expanded to charged, non-spinning black holes in 1918 (The Reissner–Nordström metric)

It wasn't until 1963 that the Kerr metric for uncharged spinning black holes was developed. Two years later, the most general form, the Kerr-Newman metric was found.

I wouldn't fancy waiting 47 years for a more accurate black hole model to be developed before doing any meaningful work in the field.

Also note that the pure Schwarzschild solution is static: it's eternal, not formed by collapse, and it's the only object in an otherwise empty universe. But it's still a useful solution, despite these unnatural simplifications. — PM 2Ring, Aug 23 '19 at 02:15

score 3 · Answer 4 · answered Aug 22 '19 at 18:50

Our sun's rotation period is 24.47 days at the equator and almost 38 days at the poles, our planet's rotational period is 23h 56m 4.098,903,691s. Use of Schwarzschild equations for either case isn't exact.

If you used the equation for non-rotating objects to calculate the time at the altitude of GPS satellites (~ 20,200 km or 12,550 miles) then you would be off by 38,636 nanoseconds per day. A Julian year is defined as 365.25 days of exactly 86,400 seconds (SI base unit), totalling exactly 31,557,600 seconds in the Julian astronomical year. The Gregorian calendar year (400 year average) is 365.2425 days.

Multiplying 365.2425 x 38,636 = 14,111,509.23 nanoseconds, that's 0.0141 seconds per year. If being off by that amount isn't of any concern to you then you can use the easier equation, such as for calculations involving the star HR 1362 which has a rotational period that is 306.9 ± 0.4 days.

score -1 · Answer 5 · answered Aug 21 '19 at 12:21

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You're right: all stars rotate. The only reason I can think of why astrophysicists make calculations for a non-rotating star or black hole is that it makes their calculations a bit easier. Although all stars rotate, some rotate much faster than others, and their masses vary too, so there is a wide degree of uncertainty which is reduced by calculating for a star that does not rotate.

answered Aug 21 '19 at 12:21

Michael Walsby

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How certain can we be that *all* stars rotate? There are a lot of stars and many many possible (theoretical) interactions that would slow rotation. – Valorum Aug 21 '19 at 22:11
No one has found one yet. I suspect it would cause a sensation if one were discovered. – Michael Walsby Aug 21 '19 at 22:26
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@Valorum Yeah, I was thinking about a stellar collision where the stars are rotating in opposite directions. If the rotational energy is exactly opposite you'll get a non-rotating result. Very unlikely, not utterly impossible--thus it probably will happen somewhere, someday. – Loren Pechtel Aug 22 '19 at 03:41
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@LorenPechtel The rotational momentum needs to be exactly equal. I think that counts as utterly impossible. – Martin Bonner supports Monica Aug 22 '19 at 07:28
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@Valorum Because the chance for "zero" angular momentum approaches 0 much faster than the amount of stars grow with "sample size". – paul23 Aug 24 '19 at 17:44
@MartinBonner - Surely it doesn't need to be exact, just very similar. A star that is hit with multiple objects could (theoretically) eventually reverse in rotation which means that for at least a moment in time, the rotation would be zero. Hit it with multiple smaller objects and the rotation might slow to approximately zero. – Valorum Aug 24 '19 at 17:49
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@Valorum Nobody has any problem with approximately zero; that's obvious. For a star whose rotation is reversed: obviously "the system" (star+impactor) has angular momentum in the opposite direction to the star alone. The star + impactor will mix in complex ways, so I don't think it is helpful to talk about "the star" reversing direction. – Martin Bonner supports Monica Aug 26 '19 at 08:29

If all stars rotate, why was there a theory developed that requires non-rotating stars?

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