# Lower Limit of central pressure in Star

1. Sep 20, 2009

### godzilla5002

Hello,

I was given a question in school that says use the hydrostatic equilibrium equation and mass conservation equation to come up with a lower limit of the central pressure of a star at it's centre. Here is how I think:

1) Refer to the density in both equations as the same and sub one into the other.
2) After I did I get: dp = -[G*M(r)]/[4*pi*R^4]dm,

Now from here, I want to integrate using dm and M(r),, but M(r) is a equation with respect to r... so I can't right. Is there some insight you can give me to derive a lower limit for the pressure.

2. Oct 15, 2009

### motoroller

Assuming M(r) is independent of r seems to give the right answer but I can't figure out why. That's the only way I can see of doing the problem.

3. Oct 27, 2009

### paco_uk

I think the trick is to use the fact that we only want a lower limit so if we replace the function with something we can integrate and is definitely lower we will be rigorously correct.

Starting with:

$$dp = -\frac{G m(r)}{4 \pi r^4} dm,$$

Integrate from the centre to the surface (I've multiplied by -1 to get PC positive as that is what we are interested in. We will assume PS the surface pressure is 0.

$$P_C - P_S = \int_0^{M} \frac{G m(r)}{4 \pi r^4} dm,$$

Now, we replace r4 (the variable) with R4, the constant radius of the star. Since r4 $$\leq$$ R4 over the whole range of integration we can be sure that the answer we get is smaller than the true answer.

$$P_C - P_S > \int_0^{M} \frac{G m(r)}{4 \pi R^4} dm,$$

Do the integral and get:

$$P_C> \frac{G M^2}{8 \pi R^4} dm,$$

This is a rigorous lower limit for the central pressure of the star.