Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose a large neutron star were to be spun-up by a particular pattern of mass-accretion. The increased centrifugal force could presumably mitigate the increased gravity thus delaying gravitational collapse. How far could this process be taken?

share|cite|improve this question
Two things: neutron stars are already spinning very rapidly. Do you mean faster than they are usually observed? Second, the neutron stars has already collapsed gravitationally as far as it will go with its mass. Do you mean "can rotation prevent a neutron star with enough mass to become a black hole from actually becoming a black hole?" – spencer nelson Jan 29 '11 at 22:01
Just for clarity. Yes, it's already spinning pretty fast. The question is, can we add both mass and angular momentum in such a way as to stave off gravitational collapse to a BH and for how long, indefinitely? Note that as the neutron star spins up, it will distort into a very flat ellipsoid with the circumferential velocity approaching that of light. – Nigel Seel Jan 30 '11 at 8:48
up vote 2 down vote accepted

Excerpt from a report I once wrote on this topic:

Gravitationally bound compact objects --- such as neutron stars --- cannot have pulsar periods less than 0.3msec. The pulsar period is of course radius and mass dependent, but is bound from below by a neutron star just on the verge of forming a black hole singularity.

Strange quark stars are hypothetical objects qualitatively different from other compact stars such as neutron stars. First, they are composed of strange quark matter, a hypothetical form of matter consisting of equal part u, d and s quarks. This phase must be of lower energy than hadronic matter for a strange star to be stable to quark fusion. Second, unlike neutrons stars which are held together by gravity, strange stars are bound by the strong interaction. As strange stars are not bound by gravity, they are free to occupy the sub-millisecond pulsar region.


The limiting angular velocity for a gravitationally bound star goes as


This comes from Newtonian arguments (centrifugal vs. gravitation force). The GR version turns out to have the same form, just a different multiplicative constant ("accurate to better than 10%" according to Glendenning). So, to maximise the angular velocity, you maximise M and minimize R. But, if you do that, you make the start more compact, and eventually it forms a black hole! Quantitatively, there's your limit.

But note that the limit is not one number it's actually a function mass M. To get rid of R, use the fact that according to Einstein's equations of stellar structure (Oppenheimer-Volkoff equations for hydrostatic equilibrium)

$M/R < 4/9$.

In the end the formula for the rotational period is:

$P > 0.167 M / M_{sun} [msec]$

For more please check the book Compact Stars by N. K. Glendenning, it's very good.

share|cite|improve this answer
Interesting report: it wasn't clear whether you derived the 0.3 ms minimal period. Could you just comment qualitatively as to where it comes from? Thanks. – Nigel Seel Jan 30 '11 at 21:18
Edited per your request. Advise to check out book if really interested! – mtrencseni Jan 30 '11 at 23:01

Here is a link to a paper where rotationally delayed collapse is seriously considered:
It is found likely to occur in some cases, but the delays are of order 10 milliseconds,
(which is significant for gravitational wave detection.)

share|cite|improve this answer
This paper, about the coalescence of binary neutron stars, is interesting and has some great figures from the numerical simulation. However, the kind of spin-up the question thought-experiment is considering is intended to impart a greater angular momentum to mass ratio than I think the paper deals with. – Nigel Seel Jan 30 '11 at 19:24

The maximum rotation rate of a neutron star is: $\Omega=(M/R^3)^{1/2}$ where M is the non-rotating mass and R the Radius. I admit that I havent worked this out, but I found it in a thesis I have been studying in connection with another Stack question "Geometrical and Physical Aspects of Rotating Neutron Star Models". Around equation (4.5), there is further discussion of conditions and variants.

Note that White Dwarfs are similar objects which are held together by electron degeneracy pressure - and may obey similar equations.

share|cite|improve this answer
And above the limit, what? Mass loss? – dmckee Jan 30 '11 at 0:38
One possible outcome of the OP scenario may be neutron star stabilisation against gravitational collapse with increasing mass. Another might even be break-up due to net outward acceleration at the equator... – Nigel Seel Jan 30 '11 at 8:52
Your link is to a pretty impressive PhD thesis. The question is even more non-trivial than I thought when I posed it. – Nigel Seel Jan 30 '11 at 21:24
Thanks. I have just been reviewing your more recent answers and reckon I understand the question better now. The appropriate limit is the Tolman-Oppenheimer-Volkov limit at < 3 Sun masses. It is the Neutron equivalent of Chandrasekhar (white dwarf). See also…. – Roy Simpson Jan 30 '11 at 21:34
@dmckee : Yes there is a mass shredding due to the high centrifugal force above that limit. – Roy Simpson Jan 31 '11 at 16:51

The problem with this idea is that you get a fairly modest reduction of the pressure at the center of the star because the poles of rotation don't feel the effect. The detailed calculation would have to be relativistic and assume an equation of state for neutronium.

Last I heard theory and observation could put only fairly broad limits on the real equation of state.

share|cite|improve this answer
The surface of the rotating neutron star must be an equipotential. So under very high spin the shape must distort to something like a very squashed ellipsoid. The polar diameter is then far less than the equatorial diameter and hence the polar gravity is reduced to the same degree as the equatorial gravity. – Nigel Seel Jan 30 '11 at 8:45
@Nigel Seel: Newtonian concepts like a well-defined equipotential surface are not particularly well-defined in strong field general relativity. – Jerry Schirmer Jan 30 '11 at 15:30

Neutron stars are stable end points of stellar evolution. So long as their mass is less than whatever the modern, updated version of the Chandrasekhar mass is (something like 1.5-2.2 times the mass of the sun), they won't collapse to a black hole.

If they're over the Chandrasekhar mass, then the core will be the most compact part of the NS, and centrifugal forces will be less important, because the mass will be close to the center of rotation, and the core will collapse first, and you'll get a NS supernova.

I don't get the premise of this question.

share|cite|improve this answer
The premise of the question: from the point of view of a reference frame rotating with the neutron star, the neutron star is experiencing a variable gravitational field outwardly-directed and orthogonal to the axis of rotation. This would serve to counteract the star's tendency to gravitational collapse. If we carefully add mass to the star while continuing to spin it up is there a point when gravitational collapse happens anyway and why (you answered yes with a good reason). if you're happy that your answer addresses the intent of the question I'll mark it accepted. – Nigel Seel Jan 30 '11 at 19:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.