17 | | [[latex($\Delta p=-\frac{\Delta t}{\Delta x} \left(F_{i+1/2}-F_{i-1/2} \right )$)]] which is just [[latex($\dot{p} = f_{grav}$)]] where [[latex($f_{grav}=-\nabla F$)]] which requires [[latex($\nabla F=\rho \nabla \phi$)]]. Since [[latex($\nabla^2 \phi = 4\pi G (\rho-\bar{\rho})$)]] we can substitute for [[latex($\rho$)]] and we have [[latex($\nabla F = (\frac{\nabla^2\phi}{4 \pi G}+\bar{\rho}) \nabla \phi$)]] which reduces to [[latex($\nabla F=\nabla\left [\frac{1}{2} \frac{\left (\nabla \phi\right )^2}{4 \pi G} \bar{\rho} \phi \right ]$)]] where we can identify the equivalent momentum flux as [[latex($F=\frac{1}{2} \frac{\left (\nabla \phi\right )^2}{4 \pi G} + \bar{\rho} \phi$)]] |
| 17 | [[latex($\Delta p=-\frac{\Delta t}{\Delta x} \left(F_{i+1/2}-F_{i-1/2} \right )$)]] which is just [[latex($\dot{p} = f_{grav}$)]] where [[latex($f_{grav}=-\nabla \cdot F$)]] which requires [[latex($\nabla \cdot F=\rho \nabla \phi$)]]. Since [[latex($\nabla^2 \phi = 4\pi G (\rho-\bar{\rho})$)]] we can substitute for [[latex($\rho$)]] and we have [[latex($\nabla \cdot F = (\frac{\nabla^2\phi}{4 \pi G}+\bar{\rho}) \nabla \phi$)]] which is equivalent (in 1D) to [[latex($\nabla \cdot \left [\frac{1}{2} \frac{\left (\nabla \phi\right )^2}{4 \pi G} + \bar{\rho} \phi \right ]$)]] where we can identify the equivalent momentum flux tensor as [[latex($F=\frac{1}{2} \frac{\left (\nabla \phi\right )^2}{4 \pi G} + \bar{\rho} \phi$)]] |
| 18 | |
| 19 | * In more than 1D, [[latex($\nabla \cdot \left [\frac{1}{2} \frac{\left (\nabla \phi : \nabla \phi \right )}{4 \pi G} + \bar{\rho} \phi I \right ]_i =\partial_j (\partial_j \phi \partial_i \phi) +\bar{\rho} \partial_i \phi$)]] |
| 20 | |