Self Gravity

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Physics

Poisson Equation of Self Gravity

The electric field generated by a point charge Q is

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

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

g=-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\rho

where \rho is the mass density. This equation describes how the potential \phi is determined by the mass density distribution.

For example, consider the uniform density distribution

%%(latexrender) [tex] \rho=\begin{cases} \rho_{0}, &r<R

0, & r\ge R

\end{cases}$)]]

[\tex]

Implementation

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>=0. 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,

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

and for 2D,

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

and for 1D,

G(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 M 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.

k^2\hat{\phi}(k)=4\pi G\hat{\rho}(k) with \hat{\phi}(0)=0. Now \hat{\rho}(k) is just the fourier transform of a delta function times the mass, which is M, 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 k^{-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?