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Pressure in a star

  1. Jun 30, 2008 #1
    1. The problem statement, all variables and given/known data
    Given the density function [tex]\rho = \rho(r)[/tex] calculate the pressure at the center of a star.

    2. Relevant equations

    [tex]F = \frac{GMm}{r^2}[/tex]

    [tex]P = \frac{\Delta F}{\Delta A}[/tex]

    3. The attempt at a solution
    Choose some radius [tex]r[/tex]. Then the gravitational attraction there is

    [tex]\Delta F = \frac{GM(r) \Delta m}{r^2}[/tex]

    and the resulting pressure is

    [tex]P = \frac{\Delta F}{\Delta A} = \frac{GM(r)}{r^2} \frac{\Delta m}{\Delta A}[/tex].

    We can interpret [tex]\Delta m[/tex] as the total mass above radius [tex]r[/tex] and [tex]A[/tex] as the area of the sphere at that radius. Then

    [tex]P = \frac{GM(r) [M_0-M(r)]}{4\pi r^4}[/tex].

    Near the center

    [tex]\frac{M(0)}{4\pi r^3} \approx \rho_c/3[/tex]

    and so

    [tex]P(0) \approx \frac{G\rho _c M_0}{3r} \rightarrow \infty[/tex].

    Where's my error?
  2. jcsd
  3. Jun 30, 2008 #2


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    Homework Helper

    Don't you need to bring in the density function [tex] \rho(r)[/tex] into this in order to deal with M(r)? It seems a bit mysterious how you got a result out of

    P = \frac{GM(r) [M_0-M(r)]}{4\pi r^4}[/tex]
  4. Jun 30, 2008 #3
    I didn't get a result, I got an infinity, which I don't like. I skipped a limiting procedure in my derivation. Since I'm looking for the pressure at the center, very near to the center I can assume that density is constant with value [tex]\rho_c[/tex], so the mass out to a small radius [tex]\Delta r[/tex] is

    [tex]M(\Delta r) \approx \rho_0 4\pi/3 (\Delta r)^3[/tex]

    If you plug this in, there remains one more power of a small [tex]\Delta r[/tex] which causes the infinity.
  5. Jun 30, 2008 #4


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    The relevant equations are dP/dr=-GM(r)*rho(r)/r^2, where dM(r)/dr=4*pi*r^2*rho(r) and M(r) is the total mass contained within radius r. So you can see immediately that dP/dr -> 0 as r -> 0. So there's no singularity at the origin. But to get the central pressure you have to integrate the M(r) first and then do the integral to get the pressure. I don't think you get get it by using only the total mass and properties at the center.
  6. Jun 30, 2008 #5


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    I can confirm that an approach working from

    [tex]P = \frac{GM(r) [M_0-M(r)]}{4\pi r^4}[/tex]

    confuses the issue. The appearance of a singularity is only due to the fact that the M_0-M(r) is concealing a factor of r^3 (M_0 is not just a constant), so the dimensions for the quotient are [tex]G\rho^{2}r^{2}[/tex], which has units of pressure.

    The structure equation

    [tex]\frac{dP}{dr} = -\frac{GM(r)\rho(r)}{r^2}[/tex]

    can only be integrated exactly for a handful of cases, unfortunately, so there are only a few simple models where you can calculate a straightforward result for the central pressure. In realistic situations, the integration is performed numerically.
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