Say you start with the Vlasov-Poisson Equation
$\frac{\partial f}{\partial t} + v \frac{\partial f}{\partial x} + a \frac{\partial f}{\partial v} = 0$
where
$a=-\frac{\partial \phi}{\partial x}$
and
$\frac{\partial^2 \phi}{\partial x^2}= 4 \pi G \rho$
Now say you take a moment of this eqn... multiply by $v$ and integrate (which ultimately gets you to one of Jeans' equations):
$\int dv \left[ v \frac{\partial f}{\partial t} + v^2 \frac{\partial f}{\partial x} + va \frac{\partial f}{\partial v}\right] =0$
$\frac{\partial}{\partial t} \left[\left< v \right> \rho \right] +\frac{\partial}{\partial x} \left[\left< v^2 \right> \rho \right]+ \int dv \frac{\partial}{\partial v} \left[ a f v \right] - \int dv \,\, af = 0$
$\frac{\partial}{\partial t} \left[\left< v \right> \rho \right] +\frac{\partial}{\partial x} \left[\left< v^2 \right> \rho \right]+ \left a f v \right|_{-\infty}^{\,\, \infty} - a\rho = 0$
$\left a f v \right|_{-\infty}^{\,\, \infty} \rightarrow 0$ as long as f vanishes at $\infty$ (kindof strongly).
It's not 100% clear that this is okay, but if we assume we have reached a stationary solution then we can set $\frac{\partial f}{\partial t}=0$ and if we define $\sigma^2=\left< v^2 \right> $ then we are left with the following:
$\frac{\partial}{\partial x} \left( \sigma^2 \rho \right) - a \rho =0$
This is just a statement that for steady state, the gravitational force inward has to balance the pressure force outward. But here, things get a little weird.
Suppose we assume that our attractor solution is a steady state solution of this type. Then we know $\rho=Ax^a$. We can try plugging this in (this weird thing came up because I was trying to solve for the scaling of $\sigma$ that we should observe in the simulations if the system is obeying the V-P limit).
$\frac{1}{\rho}\frac{\partial}{\partial x} \left( \sigma^2 \rho \right) = a = -\frac{\partial \phi}{\partial x}$
$\frac{\partial}{\partial x} \left[\frac{1}{\rho}\frac{\partial}{\partial x} \left( \sigma^2 \rho \right) \right] = -\frac{\partial^2 \phi}{\partial x^2}= -4 \pi G \rho$
$\frac{1}{\rho}\frac{\partial}{\partial x} \left( \sigma^2 \rho \right) = -4 \pi G \frac{1}{a+1} A x^{a+1}$
$\frac{\partial}{\partial x} \left( \sigma^2 \rho \right) = -4 \pi G \frac{1}{a+1} A^2 x^{2a+1}$
This is a pretty strange place to be at.
Observation 1) It turns out, there actually
is something special and different about the power law scaling $\rho \propto x^{-1/2}$. The LHS is the gradient of the pressure, and the RHS is the gravitational force. If $a=-1/2$ then the gravitational force is constant as a function of position from the center of the system.
Observation 2) There is the distressing fact that the sign of the coefficient is negative. I have looked and looked, but I haven't found an error yet. When I showed this to Jerry he pointed out that probably the constant of integration in the next step cannot be ignored, it's the pressure at the center. If you integrate this guy one more time you get:
$\left( \sigma^2 \rho \right) = -4 \pi G \frac{1}{2(a+1)^2} A^2 x^{2a+2} + C$
Where here it's not clear if C is even finite, but if it is, I have no idea how to get it... it's $\rho(0)\sigma^2(0)$, whatever that is.
Observation 3) For $a=-1/2$, $\phi\propto x^{3/2}$ and is therefore growing with $x$. This is opposite from the 3D case, where usually $\phi$ decreases with radius, e.g. $\phi \propto r^{-1}$. Consider the force due to pressure:
$-\frac{\partial}{\partial x} \left( \sigma^2 \rho \right) = \rho \frac{\partial \phi}{\partial x}$
The RHS has a different sign in 1D and in 3D. It looks like the pressure in 1D points in the same direction as the force, which is probably not news to anybody else but it does mean I am completely confused as to how we managed to arrive at a steady state in the first place.
Observation 4) Turns out, sigma is not a power law in 1D as previously thought, it's a sum of power laws
$\sigma =C_1 x^{-a} - C_2 \frac{1}{(a+1)^2}x^{a+2} $
where $C_1$ and $C_2$ are positive constants and $C_1$ might formally be infinite.
Incidentally, modulo this weird pressure stuff, I have an idea that the gravitational force being constant is the reason that the $a=-1/2$ scaling is the dividing line between power law initial conditions that go to the attractor solution, and those that don't (and indeed probably the reason that $a=-1/2$ is the attractor). When $a$ is shallower than $-1/2$ the force increases with radius, whereas if it's steeper the force decreases with radius. Somehow if the force is increasing with radius it has enough umph to shove the profile toward the attractor solution, but if it's decreasing with radius it can't push it in any farther (especially since the object is already more "collapsed" than an $x^{-1/2}$ object). Anyhow, I am dinking around trying to find a way to say this thought with math rather than this annoying handwavy argument. Jerry is too... he will probably beat me (if it can be done at all).
Alexia
