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

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?

• Frictionless spherical cows are useful abstractions too... Aug 22 '19 at 5:13
• I suppose it's the solution to a simplified model of reality as a first step? That's not unusual in science... Aug 22 '19 at 5:47
• "However, every star in the universe" You've checked them all have you? Aug 22 '19 at 13:03
• "All models are wrong, but some are useful" Aug 22 '19 at 22:14

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.

• Assume a spherical cow... Aug 22 '19 at 3:23
• 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. Aug 22 '19 at 5:39
• @RolandHeath It hasn't been since 1960. Aug 22 '19 at 6:50
• +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. Aug 22 '19 at 12:54
• @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. Aug 23 '19 at 12:24

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.

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. Aug 23 '19 at 2:15

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.

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.

• 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. Aug 21 '19 at 22:11
• @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. Aug 22 '19 at 3:41
• @LorenPechtel The rotational momentum needs to be exactly equal. I think that counts as utterly impossible. Aug 22 '19 at 7:28
• @Valorum Because the chance for "zero" angular momentum approaches 0 much faster than the amount of stars grow with "sample size". Aug 24 '19 at 17:44
• @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. Aug 26 '19 at 8:29