# Mass of star from Lane-Emden equation

Suppose we have equation of state $p=K\rho^{1+\frac{1}{n}},$ where $\gamma=1+\frac{1}{n}$ for some star. Then by standard calculations we obtain equation for enthalpy $h$: $$\Delta h+4\pi G\left(\frac{\gamma-1}{K\gamma}\right)^nh^n=0.$$ We can transform the above the equation to the following form $$w''(x)+\frac{2}{x}w'(x)+w^n(x)=0,$$ where $w(x)=\alpha h(x), \ r=\lambda x$ are some rescaled function and variable, i.e. we obtain Lane-Emden equation.

Let $x_0:=\inf\{x>0\ |\ w_n(x)=0\}.$ Then one can easily check that the radius $R$ of star is $\lambda x_0$.

My question is : How to show ( in the simplest way ) that the mass $M$ of a star is proportional to $\lambda^3x_0^2w_n'(x_0)$ ?

• There are some inconsistencies in your equations; you introduce $\alpha$ but don't define it, and you introduce $r$ but don't use it. Please correct this. It seems to me that the the mass $M$ of a star is proportional to volume $R^3=\lambda^3 x_0^3$, so you probably need to find out where the additional $w_n'(x_0)/x_0$ comes from. – nluigi May 25 '16 at 6:10
• $\alpha$ is some constant, which is used for make new enthalpy (i.e. $w$) dimensionless. Analogously for $x$. Equation $r=\lambda x$ is a definition for $x$ not for $r$ - radius $r$ is in the previous version of equation. It is contained in Laplace operator (in spherical coordinates). – mikis May 25 '16 at 7:38
• I'm confused about what $w_n$ is... is that supposed to be $w^n$? Or have I forgotten some bit of standard stellar astrophysics notation? – Kyle Oman Jun 16 '16 at 8:08
• Lane-Emden equation has a parameter $n$. This equation defines a family of special functions denoted by $\{w_n(x)\}$. In our case $w_n(x)$ means exactly that it is a solution of Lane-Emden equation in which last term has an exponent equal to $n$. – mikis Jun 16 '16 at 12:24

Let us first define $\lambda$ by the equation $$\lambda^2=\frac{4\pi G}{K(n+1)\rho_c^{\frac{1-n}{n}}}$$ for a constant $K$ and central pressure $\rho_c$, which gives us a scaled radius leading to the Lane-Emden equation. Now, we know that for a spherically symmetric body, $$M=\int_0^R 4\pi r^2\rho(r)\mathrm{d}r$$ In our polytropic relation, though, $$\rho=\rho_cw^n,\quad \mathrm{d}r=\lambda \mathrm{d}x,\quad r^2=\lambda^2x^2,\quad R=\lambda x_0$$ Therefore, $$M=\int_0^{x_0}4\pi\lambda^3\rho_cx^2w^n\mathrm{d}x=4\pi\lambda^3\rho_c\int_0^{x_0} x^2w^n\mathrm{d}x$$ Substituting in from the Lane-Emden equation yields $$M=4\pi\lambda^3\rho_c\int_0^{x_0}\left[-\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\frac{\mathrm{d}w}{\mathrm{d}x}\right)\right]\mathrm{d}x=4\pi\lambda^3\rho_c\left[-x^2\frac{\mathrm{d}w}{\mathrm{d}x}\right]_{x=x_0}$$ This shows that $M\propto \lambda^3x_0^2w_n'(x_0)$, completing the proof.
The derivation is somewhat common, and can be found in e.g. this text. Different variables are used; as far as I know, some of the ones you're used ($w$ and $x$) are non-standard, and are generally replaced by $\theta$ and $\xi$, respectively.