Calculate Pressure in a Star at Center

[Irid]

Homework Statement

Given the density function \rho = \rho(r) calculate the pressure at the center of a star.

Homework Equations

F = \frac{GMm}{r^2}

P = \frac{\Delta F}{\Delta A}

The Attempt at a Solution

Choose some radius r. Then the gravitational attraction there is

\Delta F = \frac{GM(r) \Delta m}{r^2}

and the resulting pressure is

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

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

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

Near the center

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

and so

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

Where's my error?

 

[dynamicsolo]

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

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

 

[Irid]

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 \rho_c, so the mass out to a small radius \Delta r is

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

If you plug this in, there remains one more power of a small \Delta r which causes the infinity.

 

[Dick]

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.

 

[dynamicsolo]

I can confirm that an approach working from

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

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 G\rho^{2}r^{2}, which has units of pressure.

The structure equation

\frac{dP}{dr} = -\frac{GM(r)\rho(r)}{r^2}

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.