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 $\lambda^3x_0^2w_n'(x_0)$ ?

  • $\begingroup$ 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. $\endgroup$ – nluigi May 25 '16 at 6:10
  • 1
    $\begingroup$ $\alpha$ 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). $\endgroup$ – mikis May 25 '16 at 7:38
  • 1
    $\begingroup$ 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? $\endgroup$ – Kyle Oman Jun 16 '16 at 8:08
  • $\begingroup$ 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 $n$. $\endgroup$ – 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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.