Question: What does it mean by when we obtain negative density for specific region in space as a solution of Poisson's equations?
I will explain the situation as detailed as possible, so that anyone who understands Newtonian mechanics can get the question without any background in astronomy. So, you may skip the blockquoted part if you know what it is.
Background Physics
In the fields of astronomy, spherically symmetric Poisson's equation for polytropic gas is called Lane-Emden Equation.
Polytropic gas is an ideal gas whose equation of state is given by $$ P = K \rho ^{1+1/n} $$ though it is denoted by $n$, polytropic index can be any positive real number, but most likely between $3/2$ and $3$ in case of stellar physics.
When we assume hydrostatic equilibrium, that is, density configuration does not change over time, and without any other forces but gravity and gas pressure, the following condition should be satisfied $$ \nabla \left(\Phi + P \right ) = 0 $$ since net force should be 0 everywhere. Here, $\Phi$ denotes the gravitational potential.
If we assume spherical symmetry, the Poisson's equation is, $$ \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d\Phi}{dr}\right) = 4\pi G \rho $$ along with the hydrostatic condition, can be written solely in terms of density, $\rho$. In dimensionless form, the equation can be written as $$\frac{1}{\xi^2}\frac{d}{d\xi}\left(\xi^2\frac{d\theta}{d\xi}\right) +\theta^n = 0 $$ where $\xi$ denotes dimensionless distance factor, and $\theta$ denotes dimensionless density factor. Note that $\theta=1$ and $d\theta/d\xi=0$ at the center, $\xi=0$,.
There are 3 known analytic solutions.
- $n=0$ $$ \theta(\xi) = 1-\frac{1}{6}\xi^2$$
- $n=1$ $$\theta (\xi)= \frac{\sin \xi}{\xi}$$
- $n=5$ $$\theta(\xi) = \frac{1}{\sqrt{1+\xi^2/3}}$$
In general, all solutions with $n<5$ has $\theta=0$ at finite $\xi=\xi_0$. Beyond that point, the solution for density becomes negative. In astronomy, we define this $0$ to be the (dimensionless) radius of the star given the polytropic index. But, I am irritated that the solution goes negative and I cannot come up with good logical explanation preventing that.
Of course, physically, when density approaches to $0$ the pressure drops more quickly in case of $n>1$. It is not that strange that after some point pressure and density vanishes. But I want mathematical support for this issue, and I think there is an unused condition that can solve this problem. What did I miss? What I could think of is
- Pressure gradient: When we assume hydrostatic equilibrium, the pressure gradient is taken. Maybe the pressure should be 0 when $\xi>\xi_0$, so that constraint should be explicitly applied.
- Polytropic assumption failure: Maybe the polytropic equation of states cannot be applied near the surface of the star, $\xi \simeq \xi_0$, that this problem solely arises from assuming the polytrope, and cannot be resolved by providing additional constraints.
Does anyone has a good answer for this problem?
