0
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – jacob1729 Apr 16 '19 at 17:23
  • $\begingroup$ @jacob1729 you're right, the length is exactly what you proposed. My question is also how does the covariant divergence depends on the metric? $\endgroup$ – Paul Chen Apr 16 '19 at 17:28
  • $\begingroup$ See eg the first answer here or the early chapters of most GR books have a section on tensor analysis that derives these things. $\endgroup$ – jacob1729 Apr 16 '19 at 17:34
2
$\begingroup$

I now find the answer!

\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}

According to Section 4.5 in Blau 2018 GR lecture notes, the covariant divergence can be written compactly as

\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 $$ \nabla_\mu(\rho u^\mu) = 0 \\ \frac{\partial(\rho\gamma)}{\partial ct} + \frac{1}{r^2}\frac{\partial(r^2\rho u_r)}{\partial r} = 0 \\ \frac{1}{r^2c}\partial_t(r^2\rho\gamma) + \frac{1}{r^2}\partial_r(r^2\rho\gamma\beta) = 0 \\ \frac{1}{r^2c}\partial_t(r^2\rho\gamma) + \frac{1}{r^2}\beta\partial_r(r^2\rho\gamma) + \rho\gamma\partial_r\beta = 0 \\ \frac{1}{r^2ct}\frac{\rm d}{{\rm d}t}(r^2\rho\gamma) + \rho\gamma\partial_r\beta = 0 $$

$\endgroup$
  • $\begingroup$ I think you dropped a $\partial_t$ in that last line. $\endgroup$ – Kyle Kanos Apr 19 '19 at 10:59
  • $\begingroup$ @KyleKanos Thank you! it should be a total derivation sign $\endgroup$ – Paul Chen Apr 19 '19 at 13:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.