# Deriving the hydrostatic equilibrium equation.

1. May 24, 2006

### Brewer

A question asks for me to show that the minimum pressure of a star is given by:
$$P_{min}=\frac{GM^2}{8\pi R^4}$$

where M and R are the mass and radius of the star.

Gravitational force towards center = $$\frac{GM(r)\delta M(r)}{r^2}$$
Pressure force outwards = $$(P(r)-P(r+\delta r))\delta A = \delta P (r)\delta A$$
$$\delta M(r) = \delta A\delta r \rho (r)$$

In equilibrium Gravitational up + Pressure down = 0

:. $$\frac{GM(r) \delta A \delta r \rho (r)}{r^2} + \delta P (r)\deltaA = 0$$
$$\delta P(r) = - \frac{GM(r) \rho (r) \delta r}{r^2}$$
as $$\rho (r) = \frac{M(r)}{\frac{4}{3}\pi r^3}$$
:. $$\delta P(r) = \frac{-GM(r)M(r) \delta r}{\frac{4}{3} \pi r^5}$$

after reducing $$\delta$$ to zero and integrating, I end up with:
$$P(r) = \frac{3G(M(r))^2}{16\pi r^4}$$

The minimum pressure will be when at r=R, so the variables will check out, but I still have incorrect constants. Can you spot anywhere I've gone wrong, or an incorrect assumption in this derivation?

Thanks guys

Last edited: May 25, 2006
2. May 24, 2006

### vsage

You might want to go back and edit some of your LaTeX as some of it is hidden or didn't show up right.

3. May 25, 2006

### Brewer

I think this is it. Well I hope so anyway.

4. May 25, 2006

### Curious3141

I am very sure that there's a coefficient of "3" missing in the expression for core pressure you're supposed to derive, please check the question.

You have the right concept, but your derivation assumes that m is independent of r. This is not correct. Assume a uniform density $$\rho$$

Then the mass m enclosed by a spherical shell of radius r measured from the core is given by :

$$m = \frac{4}{3}\pi \rho r^3$$ --(1)

Differentiate that (you'll see why in a minute)

$$dm = 4\pi \rho r^2 dr$$ --(2)

The inward gravitational force at a point distance r from the core is given by :

$$F_G = \frac{Gmdm}{r^2} = \frac{16}{3}\pi^2 G \rho^2 r^3 dr$$

which you get by substituting (1) and (2). Note that use is made of the fact that the uniform spherical shell beyond r contributes nothing to the gravitational field at r, a well-known result that is not proved here.

You have a pressure p that decreases uniformly as you travel outward from the center. The nett outward pressure exerted on the spherical shell at distance r from the core is given by (-dp). The area of the spherical shell is simply $$4\pi r^2$$

Hence the nett force outward due to pressure is given by

$$F_P = -Adp = -4\pi r^2 dp$$

For equilibrium, $$F_P = F_G$$ giving

$$-4\pi r^2 dp = \frac{16}{3}\pi^2 G \rho^2 r^3 dr$$
$$dp = -\frac{4}{3}\pi G \rho^2 r dr$$

The pressure at the exterior of the star (radius R) will be zero, hence

$$\int_0^P dp = -\int_R^0 \frac{4}{3}\pi G \rho^2 r dr$$

which will yield

$$P = \frac{2}{3}\pi R^2 G \rho^2$$

since you know that $$M = \frac{4}{3}\pi R^3 \rho$$, do the algebra to get

$$P = \frac{3GM^2}{8\pi R^4}$$

which is the stellar pressure at the core.

Last edited: May 25, 2006
5. May 25, 2006

### Brewer

Thank you.

6. May 25, 2006

### Curious3141

You're welcome.