Template:ASTR508/Polytropes

Polytropes

Reading Assignment

Chapter 4 of Stellar-Astrophysics Notes

What are polytropes?

They are a simple model for a star where the pressure is proportional to some power of the density:

$P=K{\rho }^{1+1/n}.$

Why did we use such an odd exponent? Let's look at the equations of hydrostatic equilibrium and gravity

$\frac{dP}{dr}=-\frac{G{M}_r}{r^2}\rho ,\frac{d{M}_r}{dr}=4\pi r^2\rho ,$

and we can combine these

$\frac{1}{r^2}\frac{d}{dr}\left(\frac{r^2}{\rho }\frac{dP}{dr}\right)=-4\pi G\rho .$

Now let's define some dimensionless quantities

$\rho ={\rho }_c{\theta }^n,P=P_c{\theta }^{n+1},r=\alpha \xi$

where

${\alpha }^2=\frac{K(n+1)}{4\pi G}{\rho }_c^{1/n-1}.$

And now we have a dimensionless equation of the hydrostatic equilibrium with gravity

$\frac{1}{{\xi }^2}\frac{d}{d\xi }\left({\xi }^2\frac{d\theta }{d\xi }\right)=-{\theta }^n$

as known as the Lane-Emden equation. We have to solve this with two conditions at $\xi =0$: $\theta =1,d\theta /d\xi =0$, so for each value of $n$ there is a single solution that we can scale to the size of the star that we are interested in. The solution is only valid where $\theta >0$ so the surface of the star lies where $\theta$ vanishes, called ${\xi }_1$. There are three closed-form solutions.

$n=0,\theta =1-\frac{{\xi }^2}{6},{\xi }_1=\sqrt{6},$

$n=1,\theta =\frac{\sin \xi }{\xi },{\xi }_1=\pi$

and

$n=5,\theta ={(1+\frac{{\xi }^2}{3})}^{-1/2},{\xi }_1=\mathrm{\infty }$

What do these three solutions mean physically?

Assignment

Calculate the function $\theta (\xi )$ for various values of $n$ between 0 and 5. Some important ones in addition to the ones here are $n=3$ and $n=3/2$.