http://SaturnianCosmology.Org/ mirrored file
   For complete access to all the files of this collection
        see http://SaturnianCosmology.org/search.php
  ==========================================================



   Discover Magazine | RSS Feed Bad Astronomy 


The Cosmic Hand of Destruction
.......

   47. 47.   Tom Marking Says:
       April 9th, 2009 at 3:20 pm
       I've heard it claimed before that the energy output of the pulsar
       nebula equals the loss in rotational kinetic energy of the
       rotating neutron star. I've never done the calculation before, but
       it's never too late to start so...
       The best data I could find is for the Crab Nebula pulsar (PSR
       B0531+21) which is still one of the most rapidly spinning pulsars
       known:
       http://en.wikipedia.org/wiki/Crab_Pulsar
       So we have the following equation for rotational kinetic energy:

       Er = 0.5 * I * omega^2

       where

       Er is rotational kinetic energy in joules,
       I is the moment of inertia in kilogram - meter^2, and
       omega is the angular velocity in radians per second

       For a sphere of uniform density we have:

       I = (2/5) * m * r^2

       where

       m is the mass of the neutron star in kilograms, and
       r is the radius of the neutron star in meters

       Combining the previous two equations yields:

       Er = (1/5) *m * r^2 * omega^2

       But omega = 2*pi / T

       where
       pi = 3.14..., and
       T is the period of rotation in seconds

       So,

       Er = 4*pi^2 * m * r^2 / (5 * T^2)

       Take the derivative with respect to time, t:

       dEr/dt = (-8*pi^2 * m * r^2 / (5 * T^3)) * (dT/dt)

       where
       dT/dt is the rate of period increase per second

       Plugging in some numbers for the Crab Nebula pulsar we have:
       m = 1.4 to 2.0 solar masses = 2.78E30 to 3.98E30 kg
       r = 12.5 km = 1.25E4 meters
       T = 33 milliseconds = 3.3E-2 seconds
       dT/dt = 38 nanoseconds per day = 4.40E-13 seconds per second
       So dEr/dt = -8.40E31 to -1.20E32 watts = -218,000 to -312,000 suns
       The thinking is that this 218,000 to 312,000 solar luminosity
       decrease in rotational kinetic energy of the pulsar goes into the
       radiation emission of the Crab Nebula. From various sources it
       appears that the total luminosity of the Crab Nebula is 75,000
       suns, so the calculated number for loss of rotational kinetic
       energy is 2.9 to 4.1 times too big. It's in the right ball park
       but not particularly close.
       Perhaps the difference is accounted for by imperfect efficiency in
       the conversion of rotational energy into radiation. If so the
       efficiency would be from about 24 percent to 34 percent, which is
       actually slightly better than your car's efficiency in converting
       chemical energy into mechanical work. Still, I'm a bit puzzled by
       such low efficiency numbers.
.......... other:

 This will alter the formula for moment of
       inertia by reducing it substantially. The formula for a sphere of
       uniform density is not valid. This may account for the factor of
       2.9 to 4.1 between the computed loss of rotational energy and the
       emitted radiation.