# Mass conservation in spherical coordinate

See four velocity $$u^\alpha = \gamma(1,\beta,0,0)$$ in a spherical coordinates $$(ct,r,\theta,\phi)$$,

The mass conservation is $$\begin{equation} \nabla_\mu(\rho u^\mu) = 0 \end{equation}$$

Then how it can be expressed as something like (I'm not sure if it is correct expression)

$$\begin{equation} \frac{\partial\rho}{\partial t} + \frac{1}{r^2}\frac{\partial(\rho r^2 \gamma\beta)}{\partial r} = 0 \end{equation}$$

• When you say 'in spherical coordinates', do you mean that it is flat space with spherical coords (metric $ds^2=-dt^2+dr^2+r^2d\theta^2 + r^2sin^2(\theta)d\phi^2$)? The covariant divergence depends on the metric. – jacob1729 Apr 16 '19 at 17:23
• @jacob1729 you're right, the length is exactly what you proposed. My question is also how does the covariant divergence depends on the metric? – Paul Chen Apr 16 '19 at 17:28
• See eg the first answer here or the early chapters of most GR books have a section on tensor analysis that derives these things. – jacob1729 Apr 16 '19 at 17:34

$$\begin{equation} g_{\mu\nu} = \begin{bmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & r^2 & 0\\ 0 & 0 & 0 & r^2\sin^2\theta \end{bmatrix} \end{equation}$$
$$\begin{equation} \nabla_\mu V^\mu = \frac{1}{\sqrt{g}}\partial_\mu(\sqrt{g}V^\mu) \end{equation}$$
So the conservation of the mass should be like below $$\nabla_\mu(\rho u^\mu) = 0 \\ \frac{\partial(\rho\gamma)}{\partial ct} + \frac{1}{r^2}\frac{\partial(r^2\rho u_r)}{\partial r} = 0 \\ \frac{1}{r^2c}\partial_t(r^2\rho\gamma) + \frac{1}{r^2}\partial_r(r^2\rho\gamma\beta) = 0 \\ \frac{1}{r^2c}\partial_t(r^2\rho\gamma) + \frac{1}{r^2}\beta\partial_r(r^2\rho\gamma) + \rho\gamma\partial_r\beta = 0 \\ \frac{1}{r^2ct}\frac{\rm d}{{\rm d}t}(r^2\rho\gamma) + \rho\gamma\partial_r\beta = 0$$
• I think you dropped a $\partial_t$ in that last line. – Kyle Kanos Apr 19 '19 at 10:59