Is there a limit on the rotational speed of fast spinning pulsars?

[tigerstef]

And if so, what is the limit?

Some pulsars spin very fast. more than 700 times per second, or even faster?

So the centrifugal force would be very strong at the equator, threatening to rip them apart.

Is there a limit on how fast a pulsar can spin before the centrifugal force would destroy them?

What holds pulsars together? Their gravity or their density, or both?

Sorry if my questions are dumb.

TIA

 

[ImaLooser]

And if so, what is the limit?

Some pulsars spin very fast. more than 700 times per second, or even faster?

So the centrifugal force would be very strong at the equator, threatening to rip them apart.

Is there a limit on how fast a pulsar can spin before the centrifugal force would destroy them?

What holds pulsars together? Their gravity or their density, or both?

Sorry if my questions are dumb.

TIA

They are held together by gravity. The fastest pulsars rotate at the maximum. Above that the star becomes ovoid (egg shaped) and radiates gravitational waves. Such waves consume a great deal of energy, so the star can't rotate that fast.

Actually, I have been told that most of the gravity comes from the pressure, not the mass.

 

[ImaLooser]

And if so, what is the limit?

Some pulsars spin very fast. more than 700 times per second, or even faster?

So the centrifugal force would be very strong at the equator, threatening to rip them apart.

Is there a limit on how fast a pulsar can spin before the centrifugal force would destroy them?

What holds pulsars together? Their gravity or their density, or both?

Sorry if my questions are dumb.

TIA

They are held together by gravity. The fastest pulsars rotate at the maximum. Above that the star becomes ovoid (egg shaped) and radiates gravitational waves. Such waves consume a great deal of energy, so the star can't rotate that fast.

Actually, I have been told that most of the gravity comes from the pressure, not the mass.

 

[zhermes]

Some pulsars spin very fast. more than 700 times per second, or even faster?
So the centrifugal force would be very strong at the equator, threatening to rip them apart.
Is there a limit on how fast a pulsar can spin before the centrifugal force would destroy them?
What holds pulsars together? Their gravity or their density, or both?

Great questions tigerstef. There is a limit, called the 'breakup velocity' at which the centrifugal force outweighs the pulsar's (neutron star's) gravity. Gravity is what holds all stars together---density (or specifically mass) is required for that gravity.

Its easy to approximate the breakup limit by assuming that the centrifugal force (at the equator) exactly cancels out gravity:
[tex] \frac{v^2}{R} = G\frac{M}{R^2} [/tex]
where the velocity at the surface can be written as [itex]v = \omega R[/itex] where [itex]\omega[/itex] is the angular velocity. You can easily solve this equation for a star of mass equal to approximately the sun, and a radius of about 10 km (characteristic of a neutron star).

They are held together by gravity. The fastest pulsars rotate at the maximum. Above that the star becomes ovoid (egg shaped) and radiates gravitational waves. Such waves consume a great deal of energy, so the star can't rotate that fast. Actually, I have been told that most of the gravity comes from the pressure, not the mass.

The term generally used for the distended shape is 'oblate' --- which is very different from 'egg shaped'. Additionally, the oblate-distortion does not produce gravitational waves as it is only a dipole distortion, and a quadrapolar distortion is required to produce gravitational waves. Finally, the dominate contribution to gravity is the rest-mass... not the pressure.

 

[Steely Dan]

Its easy to approximate the breakup limit by assuming that the centrifugal force (at the equator) exactly cancels out gravity:
[tex] \frac{v^2}{R} = G\frac{M}{R^2} [/tex]
where the velocity at the surface can be written as [itex]v = \omega R[/itex] where [itex]\omega[/itex] is the angular velocity. You can easily solve this equation for a star of mass equal to approximately the sun, and a radius of about 10 km (characteristic of a neutron star).

Keep in mind though, that general relativitistic corrections to gravity are important on the surface of a neutron star. As it is, if you do the calculation you'll see that the answer for those parameters is a substantial fraction of the speed of light, which means you know you've left the realm of ordinary physics.

 

[Nabeshin]

They are held together by gravity. The fastest pulsars rotate at the maximum. Above that the star becomes ovoid (egg shaped) and radiates gravitational waves. Such waves consume a great deal of energy, so the star can't rotate that fast.

Actually, I have been told that most of the gravity comes from the pressure, not the mass.

Well, if the stars were symmetric this wouldn't really be a problem. A rotating oblate spherioid will not produce a changing quadrupole moment, so no radiation will take place. In practice, neutron stars usually have small 'mountains' (only need to be a few cm high) which, as the stars spin, do produce a changing quadrupole moment and hence radiate. I'm not sure of the power of this type of radiation, but I suspect effects with the magnetic field will likely be equally important.

Edit: Zhermes beat me to it!

 

[zhermes]

Edit: Zhermes beat me to it!

Haha, sorry Nabeshin---great response though ;)

In practice, neutron stars usually have small 'mountains' (only need to be a few cm high) which, as the stars spin, do produce a changing quadrupole moment and hence radiate. I'm not sure of the power of this type of radiation, but I suspect effects with the magnetic field will likely be equally important.

Just to be clear, there's no existing evidence for those mountains, and very very minimal constraints based on pulsar spin-down times. The magnetic field effects are incredibly dominant in all pulsars; it would only be low-magnetic field neutron stars (if such things exist) in which GW-spindown of the star might be noticeable... but low-magnetic field NS's aren't really observable :/

P.S. I did see a paper on the arXiv today about quark-mountains on pulsars... Over my head, but hey, cool stuff

 

[Nabeshin]

Just to be clear, there's no existing evidence for those mountains, and very very minimal constraints based on pulsar spin-down times. The magnetic field effects are incredibly dominant in all pulsars; it would only be low-magnetic field neutron stars (if such things exist) in which GW-spindown of the star might be noticeable... but low-magnetic field NS's aren't really observable :/

Interesting. I figured B-field effects would dominate, but wasn't really sure. I know the mountains are one thing LIGO is looking for, and I see no reason a-priori to preclude their existence. With all of the tectonic activity one might expect on a neutron star, it seems only a question of degree.

 

[zhermes]

I definitely agree. Especially considering the roll of tectonics, I think the unknowns in the equation of state make it a pretty *shaky* subject

 

[mfb]

Here is that article:
http://arxiv.org/abs/1204.3781]Gravitational[/PLAIN] waves from color-magnetic `mountains' in neutron stars[/quote]

The idea to use a gravitational wave detector to measure properties of quantum chromodynamics is just awesome.

 

[stevebd1]

And if so, what is the limit?

Some pulsars spin very fast. more than 700 times per second, or even faster?

So the centrifugal force would be very strong at the equator, threatening to rip them apart.

Is there a limit on how fast a pulsar can spin before the centrifugal force would destroy them?

What holds pulsars together? Their gravity or their density, or both?

Sorry if my questions are dumb.

TIA

Here's a paper that looks at the maximum spin for various neutron stars, fig.'s 5, 6 & 7 depending on mass and composition-

Neutron Star Interiors and the Equation of State of Superdense Matter
http://arxiv.org/abs/0705.2708v2

 

[Chronos]

The breakup speed is about 2000 Hz whereas the most rapidly rotating neutron star known is about 716 Hz - re: http://www.int.washington.edu/talks/WorkShops/int_11_2b/People/Wasserman_I/Wasserman.pdf]
For more technical details on rotation limits see: http://arxiv.org/abs/0704.0799

 

[Ich]

The term generally used for the distended shape is 'oblate' --- which is very different from 'egg shaped'. Additionally, the oblate-distortion does not produce gravitational waves as it is only a dipole distortion

Just for the record: "egg-shaped" = prolate is the opposite of oblate = "wafer shaped". Both are pure quadrupole distortions, though. It's just that when the rotation is around the symmetry axis, there is no change in quadrupole momentum, hence no radiation.

 

[ImaLooser]

They are held together by gravity. The fastest pulsars rotate at the maximum. Above that the star becomes ovoid (egg shaped) and radiates gravitational waves. Such waves consume a great deal of energy, so the star can't rotate that fast. Actually, I have been told that most of the gravity comes from the pressure, not the mass.

The term generally used for the distended shape is 'oblate' --- which is very different from 'egg shaped'. Additionally, the oblate-distortion does not produce gravitational waves as it is only a dipole distortion, and a quadrapolar distortion is required to produce gravitational waves. Finally, the dominate contribution to gravity is the rest-mass... not the pressure.

That is why I used the word ovoid, not oblate. The star would become ovoid well before the breakup limit. It can't become ovoid because the energy required to do so escapes in gravitational waves instead.

Finally, the dominate contribution to gravity is the rest-mass... not the pressure.

Captain Renault: "Why did you come to Casablanca?"
Rick: "For the waters."
"There are no waters in Casablanca."
"I was misinformed."

So...what is the percentage of the gravitation that comes from pressure?

 

[zhermes]

That is why I used the word ovoid, not oblate. The star would become ovoid well before the breakup limit. It can't become ovoid because the energy required to do so escapes in gravitational waves instead.

This doesn't make sense. Things don't become 'egg-shaped' when they spin. And the 'egg-shape', if spinning along the axis of symmetry, still has no quadrupole moment.

So...what is the percentage of the gravitation that comes from pressure?

About 10% (http://www.physics.drexel.edu/~bob/Term_Reports/Whitehead_hw1.pdf)

 

[Chronos]

Pressure, in and of itself, makes zero contribution to gravitational field strength. It's strictly a function of mass. For an analogy consider this - does a rock on a mountain top have stronger gravity than the same rock at the bottom of the sea?

 

[zhermes]

A rock on a mountain top at a high temperature, does have stronger gravity than a cold rock on a mountain top. No?
(Note, mountain top superfluous)

 

[Chronos]

Yes, a hot rock has energy content, and energy has mass equivalence. A rock on a mountain top has more potential energy than a rock on the sea floor, but, this is not the same as actual energy content.

 

[zhermes]

But we're concerned with pressure, which is related to the internal energy density.

 

[Chronos]

Does a cylinder of compressed air weigh more that a cylinder of the same mass containing the same mass of uncompressed air?

 

[Ich]

Does a cylinder of compressed air weigh more that a cylinder of the same mass containing the same mass of uncompressed air?

There's tension in the cylinder = negative pressure. So there's no net effect (see here for expample).

Pressure, in and of itself, makes zero contribution to gravitational field strength.

It does, assuming an ideal fluid the source term for gravity is [itex]\rho + 3p[/itex], see here.

 

[Steely Dan]

Does a cylinder of compressed air weigh more that a cylinder of the same mass containing the same mass of uncompressed air?

How could you put the same mass of compressed air inside the same cylinder without decompressing it? You would need more of it to fill the cylinder and maintain the pressure.

 

[zhermes]

Does a cylinder of compressed air weigh more that a cylinder of the same mass containing the same mass of uncompressed air?

Yes, it does. Because pressure does contribute to the stress-energy tensor---as Ich eloquently points out.

If the pressurized cylinder contains the same rest-mass of gas, it must be at a higher temperature to maintain the higher pressure. As I tried to elucidate with the 'hot rock', that higher temperature increases the gravity.

 

[Chronos]

Let's avoid the mess of wading into the stress energy tensor pool by trying another approach. Let us suppose we have a planet in a stable orbit around an unbelievably massive star. The star runs out of fuel and begins collapsing. Due, however, to its incredible mass, it collapses directly into a black hole without any mass loss. We now have an identical mass but under increasing pressure as it collapses towards a singularity. Does this alter the orbit of the planet? If not, is it safe to conclude pressure does not contribute to gravitational field strength at any fixed distance from the center of mass?

 

[zhermes]

Of course the gravity stays the same in that case---because the energy is conserved. If instead you are considering the difference between two scenarios, one with more energy and the other with less, then obviously the former will have stronger gravity. Period. If you have a neutron star's rest-mass of dark matter, it would have weaker gravity than an actual neutron star with the same rest-mass. Its the exact same principle as why the reactants and products before and after fusion/fission have a difference in rest-mass.

Metaphors and analogies are often not an effective means of scientific discussion; if you have a reason why the pressure should not be included in the stress-energy tensor of einstein's equations, I would be very interested to hear it.

 

[Chronos]

I agree, conservation of energy is the issue here. A star is, for all practical purposes, a closed system with a total energy content defined by the aggregate mass of its constituent particles. It cannot create more energy via its own gravitational field. Yes, pressure contributes to the stress energy tensor, but, so does the gravitational field causing this pressure. The gravitational field has negative energy, so, the contribution of pressure to the stress energy tensor is exactly canceled out by the negative energy contribution of gravity. I trust you will agree this is not a metaphor or analogy.

 

[ImaLooser]

This doesn't make sense. Things don't become 'egg-shaped' when they spin. And the 'egg-shape', if spinning along the axis of symmetry, still has no quadrupole moment.


About 10% (http://www.physics.drexel.edu/~bob/Term_Reports/Whitehead_hw1.pdf)

Yes they do. When the rate of spin exceeds a certain amount the shape of a non-rigid oblate spheroid becomes asymmetrical. You are correct in that the asymmetry is not along the axis of rotation. It is equatorial, whatever that would be called. Here is my source.

It seems that as a body spins faster at some point it stops being an oblate spheroid and turns into something more like an egg spinning on its side.Yes, that's the transition from Maclaurin spheroid to Jacobi ellipsoid. The oblate spheroid becomes unstable when the polar radius drops below 0.58 of the equatorial radius, according to my copy of Lang's Astrophysical Formulae.
This happens when ω = [0.37423πGρ]1/2. Plugging in a middle-of-the-road neutron star density of 4.5x1017 kg.m-3 gives me ω = 6000 rad.s-1 (please check my working! , implying that a true millisecond pulsar would be right on the cusp of transition. Of course, there are all sorts of GR considerations which may make that observation nonsense, but it's interesting.

Grant Hutchison

 

[Ich]

The gravitational field has negative energy, so, the contribution of pressure to the stress energy tensor is exactly canceled out by the negative energy contribution of gravity.

I agree with that, though I don't know the details. But obviously you don't expect monopole radiation, so something like that has to happen if pressure changes.

But a canceled contribution is still a long way from

Pressure, in and of itself, makes zero contribution to gravitational field strength. It's strictly a function of mass.

Unless you mean special situations/definitions where the Mass ist defined by its gravitational field, like Komar mass.
Doesn't matter, as long as we agree that there are situations where the gravitation of pressure, in and of itself, becomes important. For example Dark Energy (repulsion despite positive energy density) or the radiation dominated era.

 

[Chronos]

My 'pressure is irrelevant' remark was only intended for closed geodesics, like neutron stars, which was the topic of discussion. It is a bit more complicated for an unbounded geodesic, like the universe.