Self Gravity

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Physics

Poisson Equation of Self Gravity

The electric field generated by a point chargeis

=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}

On the other hand, the gravitational acceleration by a point mass of is

=-G\frac{M}{r^{2}

The Poisson equation for electrostatic potential phi_{e is

nabla^{2}\phi=-\frac{\rho}{\epsilon_{0}

Analytically we can get the Poisson equation for the gravity

nabla^{2}\phi=4\pi G\rh

where rh is the mass density. This equation describes how the potential ph is determined by the mass density distribution.

For example, consider the uniform density distribution

begin{equation}\label{eq:uniDensity}\tag{1} \rho=\begon{cases} \rho_0 & r < R \\ 0 & r \ge R \end{cases} \end{equation}

\rho = \begin{cases} \rho_0 & r<R
0 & r \ge R \end{cases}

Use spherical coordinates, the Poisson equation for the density distribution Equation (1) can be written as

begin{equation}\label{eq:poissonSphCoord}\tag{2} \frac{1}{r^{2}}\frac{\partial }{\partial r}\left(r^{2}\frac{\partial\phi}{\partial r} \right)=4\pi G\rho \end{equation}

Or

begin{equation}\label{eq:poissonSphCoord1}\tag{3} \frac{\partial ^{2}\phi }{\partial r^{2}}+\frac{2}{r}\frac{\partial\phi}{\partial r}=4\pi G\rho \end{equation}

Solving Equation (3) with the density Equation (2), we obtain the solution for the potential for the uniform density sphere:

begin{equation}\label{eq:solUniSphere}\tag{4} \phi(r)=\begin{cases} -\frac{4 \pi}{3} G\rho \frac{R^{3}}{r} & r>R\\ -\frac{2\pi G\rho}{3}\left[R^{2}-r^{2}\right]-\frac{4\pi}{3}G\rho R^{2} &r\ge R \end{cases} \end{equation}

Free-fall Time

The free-fall time is the characteristic time that would take a body to collapse under its own gravitational attraction, if no other forces existed to oppose the collapse. Using Gauss's theorem, the gravitational acceleration _{r is given by

begin{equation}\label{eq:graAcce}\tag{5} g_{r}=-\frac{GM_{r}}{r^{2}} \end{equation}

where _{r is the mass inside

begin{equation}\label{eq:insideMass}\tag{6} M_{r}=\int_{0}^{r}\rho4\pi r^{2}dr \end{equation}

So the equation of motion for a gas molecule under a control of the self-gravity can be written as

begin{equation}\label{eq:motion}\tag{7} \frac{d^{2}r}{dt^{2}}=-\frac{GM_{r}}{r^{2}} \end{equation}

Equation (7) can be written as the form of the conservation of mechanical energy

begin{equation}\label{eq:conserE}\tag{8} \frac{1}{2}\left(\frac{dr}{dt}\right)^{2}-\frac{GM_{r}}{r}=E \end{equation}

Suppose the gas molecule is initially located at =. So the total energy

begin{equation}\label{eq:totalE}\tag{9} E=-\frac{GM_{r}(R)}{R} \end{equation}

Now consider gas collapse. So _{r}= M_{r}(R. Equation (8) gives

begin{equation}\label{eq:approx1}\tag{10} \frac{dr}{dt}=-[2GM_{r}(R)]^{\frac{1}{2}}\left(\frac{1}{r}-\frac{1}{R} \right)^{\frac{1}{2}} \end{equation}

Let

begin{equation}\label{eq:Etadef}\tag{11} r=R\cos^{2}\eta \end{equation}

From the initial state we have eta(t=0)=. So Equation (10) reduces to

begin{equation}\label{eq:reduced}\tag{12} \cos^{2}\eta\frac{d\eta}{dt}=\left(\frac{GM_{r}(R)}{2R^{3}}\right)^{\frac{1}{2}} \end{equation}

Solve Equation (12) we have

begin{equation}\label{eq:sol}\tag{13} t=\left(\frac{R^{3}}{2GM_{r}(R)} \right)^{\frac{1}{2}}\left(\eta+\frac{\sin2\eta}{2} \right) \end{equation}

Let = or eta=\frac{\pi}{2. we have the free-fall time

begin{equation}\ref{eq:freeFall}\tag{14} \begin{aligned} t_{ff}&=\left(\frac{R^{3}}{2GM_{r}(R)}\right)^{\frac{1}{2}}\frac{\pi}{2}\\ &=\left(\frac{3\pi}{32G\rho_{0}}\right)^{\frac{1}{2}} \end{aligned} \end{equation}

where

begin{equation}\label{rhoBar}\tag{15} \rho_{0}=\frac{M_{r}(R)}{\frac{4\pi R^{3}}{3}} \end{equation}

is the average density of the gas.

Implementation

Uniform Density Cloud

In this section we consider the problem of the collapse of a pressureless (=), uniform, initially motionless cloud which has an analytic solution.

The definition of et in Equation (11) can also be written as

begin{equation}\tag{16} \cos\eta=\left(\frac{\rho}{\rho_{0}} \right)^{-\frac{1}{6}} \end{equation}

From Equations (13) and (14) we have

begin{equation}\label{eq:trho}\tag{17} \frac{t_{\rho}}{t_{ff}}=\frac{2}{\pi}\left[\eta + \frac{1}{2}\sin(2\eta) \right] \end{equation}

Equation (17) describes the time it takes for the cloud to collapse to a density rh.

Boundary Conditions

Present Considerations

Some notes on periodic BC's and particles

With periodic boundary conditions, we would like the total potential to be unchanged under the conversion of gas to particle. This requires adjusting the particle potential so that it has the same constant of integration as well as being periodic. When hypre solves for the potential, the constant of integration is a free parameter and I believe hypre adjusts it so that \phi>=. To calculate the potential of a point charge, we need the Greens function corresponding to

nabla_D^2 G(x,y) = \delta^D(x-y

where D is the dimenesion of the problem. For 3D,

(x,y)=\frac{-1}{4\pi|x-y|

and for 2D,

(x,y)=\frac{\ln(|x-y|)}{2\pi

and for 1D,

(x,y)=\frac{|x-y|}{2

Now with periodic BC's, the solution to the gas potential is given by

nabla^2 \phi = 4\pi G (\rho-\bar{\rho}, so the solution to the potential would actually be given by phi(y)=-\int_V{\frac{M\delta(x-y)-M/V}{|x-y|}dx where is the mass of the particle. Note this is not the same as using mirror versions of the particle… An easier way to solve this however, is to use fourier transforms.

^2\hat{\phi}(k)=4\pi G\hat{\rho}(k with hat{\phi}(0)=. Now hat{\rho}(k is just the fourier transform of a delta function times the mass, which is , so hat\phi(k)=\frac{4\pi GM}{k^2 and phi(x)=4 \pi G M\int{\frac{1}{k^2}e^{ikx} dk

This can be discretized as

phi(l)=\frac{4 \pi GM}{\Delta k^2} \displaystyle \sum_{j=-N/2}^{N/2}{\frac{1}{j^2}e^{2\pi ijl/N} \frac{\Delta k}{2\pi}}=\frac{4 \pi GM}{(2 \pi/L)^2}\left(\frac{2 \pi}{L}\right) \displaystyle \sum_{j=1}^{N/2}{\frac{1}{j^2}\cos(2\pi jl/N)

This still requires summation over many wave numbers - although the higher wave numbers have less impact because of the ^{-2 dependence. However in 2D this is lessened because there are more wave vectors in each anuli, and in 3D there is equal power in each shell. This then, becomes an order N^{6 operation}

Solutions:

• Don't ever use sink potential?
• Don't use sink potential to update gas potential
• Solve for gas potential after accretion?