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How is the density of a white dwarf star calculated?

  1. Apr 7, 2004 #1

    JJ

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    I have no idea. And how is the pressure/force/intensity of gravitational collapse known?
     
  2. jcsd
  3. Apr 8, 2004 #2

    Kurdt

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    Basically its just a case with balancing the outward pressure due to the forces between atoms and molecule and the inward pull of the gravity on the mass of all those atoms. Once you equate them you can gain the radius of the star and therefore the density and internal pressure etc.
     
  4. Apr 8, 2004 #3

    JJ

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    How is the inward pull of gravity calculated?
     
  5. Apr 8, 2004 #4

    Kurdt

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    The self gravitation of a star can be approximated by ust using the normal law for two bodies attracting to one another through gravity or you can set up an integral using shells of the star as differential elements of mass.
     
  6. Apr 8, 2004 #5

    chroot

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    JJ,

    Are you asking to see derivation of hydrostatic equlilbrium, the Chandrasekhar limit, and so on?

    - Warren
     
  7. Apr 9, 2004 #6

    JJ

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    I don't know what half of those things are, but I was looking to be able to understand the Chandrasekhar limit better. I think I'll google it.
     
  8. Apr 9, 2004 #7

    jon

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    By observing a star we notice that it has a constant surface temperature (T)and radiates a constant amount of energy into space (L-luminosity). From the equation:

    L = 4*pi*R^2 * T^4 * a constant

    We can assume that the radius must remain constant as well. The fact that there is no overall accelerations upon any individual shell of matter within the star tells us that it is in a state of hydrostatic equilibrium. Otherwise matter could drift outward or inward causing the radius to change, which we know cannot happen.

    Hydrostatic equilibrium is also a differential equation that states how pressure changes with radius within the star. As you go deeper into the star the pressure force must increase to balance the increasing weight above.

    A rough calculation can be obtained by integrating the equation using the radius at the surface and radius at the centre (0) as endpoints of analytic integration. To be more exact you need to write a computer program and perform numerical integration with many more equations and steps, taking into account the unusual conditions within a star of that kind.

    Don't know if that's any help. Good luck on your search
     
  9. Apr 9, 2004 #8

    chroot

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    Borrowing heavily from Carroll and Ostlie's "Intro to Modern Astrophysics"...

    The differential equation describing hydrostatic equilibrium is

    [tex]\frac{dP}{dr} = -G\frac{M_r \rho}{r^2} = -\rho g[/tex]

    where [itex]M_r[/itex] is the mass interior to radius [itex]r[/itex] and [itex]\rho[/itex] is the density.

    This equation "clearly indicates that in order for a star to be static, a pressure gradient [itex]dP/dr[/itex] must exist to counteract the force of gravity. It is not the pressure that supports the star, but the change in pressure with radius."

    Assuming that the star (a white dwarf) has a uniform density (this is an approximation, but it's close to being correct),

    [tex]\frac{dP}{dr} = -G \frac{(\frac{4}{3}\pi r^3 \rho) \rho}{r^2} = -\frac{4}{3} \pi G \rho^2 r[/tex]

    Solving the integral to find the pressure as a function of radius, using P=0 at the surface, one obtains

    [tex]P(r) = \frac{2}{3} \pi G \rho^2 (R^2-r^2)[/tex]

    You can use this expression to find the pressure at the center of a white dwarf star by setting r=0:

    [tex]P_c = \frac{2}{3} \pi G \rho^2 R^2[/tex]

    In the extreme relativistic limit (i.e. when degenerate electrons are moving in the limit of v->c), the electron degeneracy pressure can be shown to be

    [tex]P = \frac{(3 \pi^2)^{1/3}}{4} \hbar c \left[ \left(\frac{Z}{A}\right) \frac{\rho}{m_H} \right]^{4/3}[/tex]

    where [itex]m_H[/itex] is the mass of a hydrogen atom, Z is the number of protons, and A is the number of total nucleons.

    Now, equating the central pressure as derived from hydostatic equlibrium arguments with the pressure at the extreme limit of electron degeneracy lets us calculate the maximum permissable mass of a white dwarf star. This is only an estimate, however, because it again assumes that [itex]\rho[/itex] is uniform.

    [tex]M_{Ch} \approx \frac{3 \sqrt{2 \pi}}{8} \left( \frac{\hbar c}{G} \right)^{3/2} \left[ \left( \frac{Z}{A} \right) \frac{1}{m_H} \right]^2[/tex]

    This winds up being [itex]0.44 M_{\odot}[/itex].

    A more precise calculation, with the density modelled more accurately, produces the accepted value [itex]M_{Ch} = 1.44 M_\odot[/itex].

    - Warren
     
  10. Apr 11, 2004 #9

    Nereid

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    Although this doesn't seem to be directly relevant to your question JJ, you may be interested to to know how the average density of a white dwarf star is determined observationally. In a few words: from its mass, radius (diameter), and surface gravity! The first two can be found (with varing degrees of accuracy) for white dwarfs in binaries (esp eclipsing binaries); the last from analysis of the spectral lines.
     
  11. Apr 13, 2004 #10

    JJ

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    Warren, that's nuts, something I was wondering about. And Nereid, I hadn't approached the problem from such a simple direction. Though couldn't you know the rough density from only two of those variables?

    Thanks for a thread with a goldmine of information I'll take a week to decode.
     
  12. Apr 13, 2004 #11

    Nereid

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    Forget not the magic words "with varing degrees of accuracy" :wink:

    The more methods, the merrier. :biggrin:
     
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