Changes between Version 61 and Version 62 of SelfGravityDevel

Timestamp:

07/09/13 21:17:43 (12 years ago)

Author:

Jonathan

Comment:

—

SelfGravityDevel

@@ -6 +6 @@
 === Poisson Equation of Self Gravity ===
 
-  The electric field generated by a point charge[[latex(Q)]]is
-
- [[latex(E=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}})]]
-
- On the other hand, the gravitational acceleration by a point mass of [[latex(M)]] is
-
- [[latex(g=-G\frac{M}{r^{2}})]]  [[BR]]
-
- The Poisson equation for electrostatic potential [[latex(\phi_{e})]] is 
-
- [[latex(\nabla^{2}\phi=-\frac{\rho}{\epsilon_{0}})]]
+  The electric field generated by a point charge[[latex($Q$)]]is
+
+ [[latex($E=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}$)]]
+
+ On the other hand, the gravitational acceleration by a point mass of [[latex($M$)]] is
+
+ [[latex($g=-G\frac{M}{r^{2}}$)]]  [[BR]]
+
+ The Poisson equation for electrostatic potential [[latex($\phi_{e}$)]] is 
+
+ [[latex($\nabla^{2}\phi=-\frac{\rho}{\epsilon_{0}}$)]]
 
  Analytically  we can get the Poisson equation for the gravity
 
- [[latex(\nabla^{2}\phi=4\pi G\rho)]]
-
- where [[latex(\rho)]] is the mass density. This equation describes how the potential [[latex(\phi)]] is determined by the mass density distribution.
+ [[latex($\nabla^{2}\phi=4\pi G\rho$)]]
+
+ where [[latex($\rho$)]] is the mass density. This equation describes how the potential [[latex($\phi$)]] is determined by the mass density distribution.
 
 For example, consider the uniform density distribution
@@ -66 +66 @@
 
  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 [[latex(g_{r})]] is given by
+ Using Gauss's theorem, the gravitational acceleration [[latex($g_{r}$)]] is given by
 {{{
 #!latex
@@ -73 +73 @@
 \end{equation}
 }}}
- where [[latex(M_{r})]] is the mass inside [[latex(r)]]
+ where [[latex($M_{r}$)]] is the mass inside [[latex($r$)]]
 {{{
 #!latex
@@ -94 +94 @@
 \end{equation}
 }}}
- Suppose the gas molecule is initially located at [[latex(r=R)]]. So the total energy
+ Suppose the gas molecule is initially located at [[latex($r=R$)]]. So the total energy
 {{{
 #!latex
@@ -101 +101 @@
 \end{equation}
 }}}
- Now consider gas collapse. So [[latex(M_{r}= M_{r}(R))]]. Equation (8) gives
+ Now consider gas collapse. So [[latex($M_{r}= M_{r}(R)$)]]. Equation (8) gives
 {{{
 #!latex
@@ -115 +115 @@
 \end{equation}
 }}}
- From the initial state we have [[latex(\eta(t=0)=0)]]. So Equation (10) reduces to
+ From the initial state we have [[latex($\eta(t=0)=0$)]]. So Equation (10) reduces to
 {{{
 #!latex
@@ -129 +129 @@
 \end{equation}
 }}}
- Let [[latex(r=0)]] or [[latex(\eta=\frac{\pi}{2})]]. we have the free-fall time
+ Let [[latex($r=0$)]] or [[latex($\eta=\frac{\pi}{2}$)]]. we have the free-fall time
 {{{
 #!latex
 \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}}
+ 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}
@@ -151 +151 @@
 == Implementation ==
 === Uniform Density Cloud ===
- In this section we consider the problem of the collapse of a pressureless ([[latex(T=0)]]), uniform, initially motionless cloud which has an analytic solution.
-
- The definition of [[latex(\eta)]] in Equation (11) can also be written as
+ In this section we consider the problem of the collapse of a pressureless ([[latex($T=0$)]]), uniform, initially motionless cloud which has an analytic solution.
+
+ The definition of [[latex($\eta$)]] in Equation (11) can also be written as
 {{{
 #!latex
@@ -167 +167 @@
 \end{equation}
 }}}
- Equation (17) describes the time it takes for the cloud to collapse to a density [[latex(\rho)]].
+ Equation (17) describes the time it takes for the cloud to collapse to a density [[latex($\rho$)]].
 
 === Boundary Conditions ===
@@ -176 +176 @@
 
 === 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 [[latex(<\phi>=0)]].  To calculate the potential of a point charge, we need the Greens function corresponding to 
-
- [[latex(\nabla_D^2 G(x,y) = \delta^D(x-y))]] 
+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 [[latex($<\phi>=0$)]].  To calculate the potential of a point charge, we need the Greens function corresponding to 
+
+ [[latex($\nabla_D^2 G(x,y) = \delta^D(x-y)$)]] 
 
  where D is the dimenesion of the problem.  For 3D, 
 
- [[latex(G(x,y)=\frac{-1}{4\pi|x-y|})]]
+ [[latex($G(x,y)=\frac{-1}{4\pi|x-y|}$)]]
 
  and for 2D, 
 
- [[latex(G(x,y)=\frac{\ln(|x-y|)}{2\pi})]]
+ [[latex($G(x,y)=\frac{\ln(|x-y|)}{2\pi}$)]]
 
  and for 1D,
 
- [[latex(G(x,y)=\frac{|x-y|}{2})]]
+ [[latex($G(x,y)=\frac{|x-y|}{2}$)]]
 
 Now with periodic BC's, the solution to the gas potential is given by 
 
- [[latex(\nabla^2 \phi = 4\pi G (\rho-\bar{\rho}))]], so the solution to the potential would actually be given by [[latex(\phi(y)=-\int_V{\frac{M\delta(x-y)-M/V}{|x-y|}dx})]] where [[latex(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.  
-
- [[latex(k^2\hat{\phi}(k)=4\pi G\hat{\rho}(k))]] with [[latex(\hat{\phi}(0)=0)]].  Now [[latex(\hat{\rho}(k))]] is just the fourier transform of a delta function times the mass, which is [[latex(M)]], so [[latex(\hat\phi(k)=\frac{4\pi GM}{k^2})]] and [[latex(\phi(x)=4 \pi G M\int{\frac{1}{k^2}e^{ikx} dk})]]
+ [[latex($\nabla^2 \phi = 4\pi G (\rho-\bar{\rho})$)]], so the solution to the potential would actually be given by [[latex($\phi(y)=-\int_V{\frac{M\delta(x-y)-M/V}{|x-y|}dx}$)]] where [[latex($M$)]] 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.  
+
+ [[latex($k^2\hat{\phi}(k)=4\pi G\hat{\rho}(k)$)]] with [[latex($\hat{\phi}(0)=0$)]].  Now [[latex($\hat{\rho}(k)$)]] is just the fourier transform of a delta function times the mass, which is [[latex($M$)]], so [[latex($\hat\phi(k)=\frac{4\pi GM}{k^2}$)]] and [[latex($\phi(x)=4 \pi G M\int{\frac{1}{k^2}e^{ikx} dk}$)]]
 
  This can be discretized as 
 
- [[latex(\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 [[latex(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 :|  
+ [[latex($\phi(l)=\frac{4 \pi GM}{\Delta k^2} \displaystyle \sum_{j=-N/2}^{N/2}{\frac{1}{j^2}e^{2\pi ijl/N} \frac{\Delta k}{2\pi}}=\frac{4 \pi GM}{(2 \pi/L)^2}\left(\frac{2 \pi}{L}\right) \displaystyle \sum_{j=1}^{N/2}{\frac{1}{j^2}\cos(2\pi jl/N)}$)]]
+
+ This still requires summation over many wave numbers - although the higher wave numbers have less impact because of the [[latex($k^{-2}$)]] 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: