On Mass-radius Relation & Lane-Emden Equation

• Introduction
• Lane-Emden Equation
• Analytical Solution to Lame-Emden Equation
  • Boundary conditions
  • n = 0
  • n = 1
  • n = 5

Introduction

​ In physics, matter is said to follow a relation called equation of state (EOS), that is, the internal pressure P is a function of temperature T and density $\rho​$ (or volume V). This relation describe a lot of useful thing of a matter. Using the knowledge of thermodynamics, one can easily derive many properties using such equation. One of the most important properties is the mass radius relation.

​ Why is this relation so important? This is a measurable quantity thanks to the scientists in NASA. It can be derived from the observation of the planet’s orbit. It defines the intrinsic property of a planet. For instance, Earth is consist of surface, mantle and core. A coarse model to describe the earth can be a three-layered model: a water surface, a silicon mantle and an iron core. Use the EoS of $H_2O$, $MgSiO_3$ and Fe, we can derive a mass radius relation for earth. If we can find some planets outside the solar system with a similar mass radius relation like Earth, it is likely to be a habitable planet. (Well, of course not that simple)

​ One way derive a mass radius relation from EoS is to solve the Lane-Emden equation.

Lane-Emden Equation

Consider the mass of a spherical shell

$$\frac{dm(r)}{dr} = 4 \pi r^2\rho(r)\label{mass}\tag{1}$$

the equation of hydrostatic equilibrium (HSE)

$$\frac{dP(r)}{dr} = \frac{-Gm(r)\rho(r)}{r^2}\label{HSE}\tag{2}$$

and a special form of EOS

$$P = K\rho^{(n+1)/n}\label{EOS}\tag{3}$$

where $K$ and $n$ are constants.

This problem, toward to solution of which fundamental contributions have been made by Lane, Ritter, Kelvin, Emden, and Fowler, is also of considerable physical interest.

Use Eq.$\ref{mass}$ and Eq.$\ref{HSE}$,

$$\frac{1}{r^2} \frac{d}{dr}(\frac{r^2}{\rho}\frac{dP}{dr}) = -4\pi G\rho\label{origin}\tag{4}$$

according to Eq.$\ref{EOS}$, we can write

$$\rho = \lambda\theta^n\tag{5}$$ $$P = K\rho^{(n+1)/n} = K\lambda^{(n+1)/n}\theta^{n+1}\label{EoSrho}\tag{6}$$

Substitute Eq.$\ref{EoSrho}$ into Eq.$\ref{origin}$,

$$[\frac{(n+1)K}{4\pi G}\lambda^{(1/n)-1}]\frac{1}{r^2}\frac{d}{dr}(r^2\frac{d\theta}{dr}) = - \theta^n\tag{7}$$

Now we introduce a dimensionless variable $\xi$, which is defined by

$$r = a \xi;\ a = [\frac{(n+1)K}{4\pi G}\lambda^{(1/n)-1}]^{1/2}\tag{8}$$

Plug in

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

This star equation is referred as the “Lane-Emden equation of index $n$.”

Analytical Solution to Lame-Emden Equation

The Lane-Emden equation has analytical solution for n = 0, 1 and 5. For other situations, the equation must be solved numerically. In this post, only analytical solutions will be discussed.

Boundary conditions

To solve the Lane-Emden equation, only complete polytropes should be considered, which means Eq.$\ref{EoSrho}$ should be valid throughout the entire mass. So $\lambda$ is equal to the central density $\rho_c$. Further, $d\theta/d\xi$ must vanish at origin. So the boundary conditions can be written as:

$$\theta = 1\ \&\ \frac{d\theta}{d\xi} = 0\ at\ \xi=0\label{BC}\tag{9}$$

n = 0

Now the Lane-Emden equation is

$$\frac{1}{\xi^2}\frac{d}{d\xi}(\xi^2\frac{d\theta}{d\xi}) = -1\tag{10}$$

after first integration,

$$\xi^2\frac{d\theta}{d\xi}= -\frac{1}{3}\xi^3 - C\tag{11}$$

the second integration,

$\theta = D + \frac{C}{\xi} - \frac{1}{6}\xi^2\tag{12}$ where C, D are integration constant.

Add the boundary condition $\theta =1$,

$$\theta_0 = 1 - \frac{1}{6}\xi^2\tag{13}$$

n = 1

Introduce a useful transformation,

$$\theta = \frac{\chi}{\xi}\tag{14}$$

and reduce the Lane-Emden equation to

$$\frac{d^2\chi}{d\xi^2}=-\frac{\chi^n}{\xi^{n-1}}\label{tf1}\tag{15}$$

then for $n = 1$

$$\frac{d^2\chi}{d\xi^2} = -\chi\label{n=1}\tag{16}$$

The general solution for Eq.$\ref{n=1}$ is

$$\begin{align} \chi = C \sin(\xi - \delta)\\ \theta = \frac{C\sin(\xi-\delta)}{\xi}\tag{18} \end{align}$$

Restrict the solution equals to zero at the origin, $\delta = 0$. Then make $\theta =1$, $C = 1$

$$\theta_1 = \frac{sin\xi}{\xi}\tag{19}$$

n = 5

This situation is a little bit more complicated. I will make an individual post on it.

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