Why is the density in GR equal to $\rho_0\dfrac{dx^0}{ds}\sqrt{-g}$?

Question

In general relativity, the continuity equation says $$ \partial_{\mu}\left(\rho_0c\dfrac{dx^{\mu}}{ds}\sqrt{-g}\right) = 0 $$

with $\rho_0$ being the proper density, as seen by an observer who is at rest with respect to the matter.

This is a general-relativistic version of $$ \partial_{\mu}\left(\dfrac{\rho_0}{\sqrt{1-\tfrac{v^2}{c^2}}}\dfrac{dx^{\mu}}{dt}\right) = 0 $$

which in turn, is a lorentz-covariant version of $$ \partial_{\mu}\left(\rho_0\dfrac{dx^{\mu}}{dt}\right) = 0. $$

I can understand the physical intuition behind the second one. An observer who looks at a given volume element and sees matter moving at speed $v$ will perceive a lorentz contracted piece of matter with density equal to $$ \dfrac{\rho_0}{\sqrt{1-\tfrac{v^2}{c^2}}}. $$

However, I don't see the motivation for the first equation (apart from imposing general covariance on the second one). The matter density there seems equal to $$ \dfrac{\rho_0c}{\sqrt{\dfrac{dx^{\mu}}{dt}g_{\mu\nu}\dfrac{dx^{\nu}}{dt}}}\sqrt{-g} $$

but I am at a loss as to why.

Answer 1

In addition, to the answer form Navid, I would like to add something regarding the below comment:

OP asked: However, I don't see the motivation for the first equation (apart from imposing general covariance on the second one). The matter density there seems equal to $$ \dfrac{\rho_0c}{\sqrt{\dfrac{dx^{\mu}}{dt}g_{\mu\nu}\dfrac{dx^{\nu}}{dt}}}\sqrt{-g} $$

I can write your Eq. 1 $$ \partial_{\mu}\left(\rho_{0} \frac{d x^{\mu}}{d s} \sqrt{-g}\right)=0$$ as $$ \partial_{\mu}\left(\rho_{0} \frac{d x^{\mu}}{d t} \frac{d t}{d s} \sqrt{-g}\right)=0$$

where $t$ is the coordinate time, and $s$ is the proper time (or the affine parameter).

If you compare this with your Eq. 2, then matter density then is equal to $$\rho_{0} \frac{dt}{ds}. $$ For a flat spacetime, you will find that $\frac{dt}{ds} = 1/\sqrt{1 - v^2/c^2}$ which agrees with your Eq. 2, while for a generic spacetime, it will depend on the form of the metric.

Answer 2

The first equation is actually $$\nabla_{\mu}J^{\mu}=\frac{1}{\sqrt{-g}}\partial_{\mu}(\sqrt{-g}J^{\mu})=0$$ where $J^{\mu}=\rho_0 u^{\mu}$. Instead of trying to express this like your second equation, I would write it as $$\partial_{\mu}\big((\rho_0\sqrt{-g})\frac{dx^{\mu}}{d\tau}\big)=0$$ This is similar to your second equation if we take $$\rho_0\sqrt{-g}=\bar{\rho}_{0}\,\,\,\rightarrow\,\,\,\partial_{\mu}\bigg(\frac{\bar{\rho}_0}{\sqrt{1-v^2/c^2}}\frac{dx^{\mu}}{dt}\bigg)=0$$ where $\bar{\rho}_0$ can be taken to be the proper density at the asymptotically flat region of the spacetime. Now this simply tells us that intuition behind the GR version of the equation is that the proper density would depend on local geometry: $$\rho_{0}=\frac{\bar{\rho}_0}{\sqrt{-g}} $$ For example in the expanding universe $ds^2=-dt^2+a^2(t)dx_i^2$: $$\rho_0=\frac{\bar{\rho}_0}{a^3(t)}$$ Hope this helps.