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

  1. Apr 17, 2012 #1
    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
     
  2. jcsd
  3. Apr 17, 2012 #2
    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.
     
  4. Apr 17, 2012 #3
    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.
     
  5. Apr 18, 2012 #4
    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).


    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.
     
  6. Apr 18, 2012 #5
    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.
     
  7. Apr 18, 2012 #6

    Nabeshin

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    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!
     
  8. Apr 18, 2012 #7
    Haha, sorry Nabeshin---great response though ;)

    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
     
  9. Apr 18, 2012 #8

    Nabeshin

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    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.
     
  10. Apr 18, 2012 #9
    I definitely agree. Especially considering the roll of tectonics, I think the unknowns in the equation of state make it a pretty *shaky* subject
     
  11. Apr 18, 2012 #10

    mfb

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  12. Apr 20, 2012 #11
    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
     
  13. Apr 20, 2012 #12

    Chronos

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    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] [Broken]
    For more technical details on rotation limits see: http://arxiv.org/abs/0704.0799
     
    Last edited by a moderator: May 5, 2017
  14. Apr 20, 2012 #13

    Ich

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    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.
     
  15. Apr 20, 2012 #14
    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.

    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.


    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?
     
  16. Apr 20, 2012 #15
    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)
     
  17. Apr 21, 2012 #16

    Chronos

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    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?
     
  18. Apr 21, 2012 #17
    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)
     
  19. Apr 22, 2012 #18

    Chronos

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    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.
     
  20. Apr 22, 2012 #19
    But we're concerned with pressure, which is related to the internal energy density.
     
  21. Apr 23, 2012 #20

    Chronos

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