0
$\begingroup$

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.

  1. $n=0$ $$ \theta(\xi) = 1-\frac{1}{6}\xi^2$$
  2. $n=1$ $$\theta (\xi)= \frac{\sin \xi}{\xi}$$
  3. $n=5$ $$\theta(\xi) = \frac{1}{\sqrt{1+\xi^2/3}}$$

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

  1. 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.
  2. 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?

$\endgroup$
  • 1
    $\begingroup$ What would $\xi>\xi_0$ even mean? (aside from you've extended your model to a region where it is necessarily invalid) $\endgroup$ – Kyle Kanos Apr 26 '15 at 14:49
  • $\begingroup$ While I agree with you, what I wanted to ask is how to rigorously formulate that $\xi>\xi_0$ violates something that it should be replaced with proper condition to be something. $\rho>0$ would be a physical condition, and I love it, but the form of the Poisson's Eq. can also be applied to the electrostatics, which deals with negative source terms, so constraining with that seemed unnatural to me. For example, for the case of $n=1$, $\theta >0$ again after $\xi > \xi_1$. Why is this not a physical solution? I think the answer for my question can also explain this. $\endgroup$ – Hojin Cho Apr 26 '15 at 17:49
  • $\begingroup$ A volume of gas can be negative if the volume of liquid allow it. Both together shoudn't be under 0. $\endgroup$ – user78946 Apr 28 '15 at 1:09
  • $\begingroup$ Well, in the polytrope, we're really only considering one fluid (and truthfully a plasma), so there isn't the ability to "donate" pressure. $\endgroup$ – Kyle Kanos Apr 28 '15 at 2:17
1
$\begingroup$

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.

This is generally understood as an implicit restriction/constraint of the hydrostatic equilibrium (HSE) condition for starting the Lane-Emden equation: HSE is for the star and not all space.

Thus, your domain is $D\in[0,\,\xi_0]$ not $D\in\mathbb R$ because the star extends to a "radius" of $\xi_0$, not $\infty$. This domain eliminates the possibility of $\xi>\xi_0$.

$\endgroup$
  • $\begingroup$ OK. I think I finally get it. So, what I understood is: Outside of the star is just vacuum, which does not necessarily satisfy HSE, since there is no entity to experience acceleratoin. So the equation loses its constraint on the pressure, and that is replaced with the physical condition where density should not be negative. Is this correct? $\endgroup$ – Hojin Cho Apr 26 '15 at 18:37
  • $\begingroup$ That sounds about right to me. I think I'd argue on grounds of the density of the star being zero for $r>r_\star$ (though there is a density in the ISM) & therefore the pressure of the star would be zero as well; but it seems to be a viable explanation both directions. $\endgroup$ – Kyle Kanos Apr 26 '15 at 18:45
1
$\begingroup$

I would go with the hydrostatic equilibrium condition being the source of the issue here. When there is no polytropic gas when $\xi \gt \xi_0$, there is no need for any hydrostatic equilibrium to be in place - the gravitational potential is free to take on any value, determined (together with boundary conditions) by Laplace's equation: $$\nabla^2\Phi = 0$$ Whereas the Lane-Emden equation does not hold for $\xi \gt \xi_0$, trying to extrapolate its solutions to this region is what is leading to the negative densities.

However, deciding where to "switch" between the two equations is more or less based on a physical argument, that we cannot have negative densities - and the only way out is to decide that one of our starting conditions (i.e. the assumption that a gas in equilibrium is actually present) has failed in this region.

$\endgroup$
0
$\begingroup$

The assumption

$$ P=K\rho^{1+1/n} $$

is a very simplistic mathematical model and should be applied with care. When the model predicts matter has negative density at some point, it is a sign that it is too simple to get it right there.

The model neglects lots of things - radiation, magnetic field, solar wind, etc.

$\endgroup$

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.