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

${\frac {dP}{dr}}=-{\frac {GM_{r}}{r^{2}}}\rho ,{\frac {dM_{r}}{dr}}=4\pi r^{2}\rho ,$

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

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

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 $n=3/2$ .