Deriving the conservation of mass in a perfect fluid I have a problem on a homework assignment. I will write the question and then what I have so far. I just want some guidance on what I am doing wrong (or right). Any help would be greatly appreciated:
In these exercises we study the equations of motion of a perfect fluid. These are the conservation laws:
\begin{eqnarray} 
                    \nabla_{\mu}T^{\mu\nu} = 0 \space (1) \\
                    \nabla_{\mu}I^{\mu} = 0 \space (2)
\end{eqnarray}
Where $T^{\mu\nu}$ is the energy-momentum-stress tensor, and $I^{\mu}$ the particle current:
\begin{eqnarray}
                    T^{\mu\nu} = \rho{u}^{\mu}u^{\nu} + p((g^{-1})^{\mu\nu} + u^{\mu}u^{\nu}) \space (3) \\
                    I^{\mu} = nu^{\mu} \space (4)
\end{eqnarray}
Here $\rho$ is the energy density, $u^{\mu}$ the material velocity (a future-directed unit time-like vectorfield), and p denotes the pressure while n is the [article density. 
Exercise 1 [Nonrelatvisitic limit of the conservation laws]: Let us first study the quasi-Newtonian limit of the conservation laws. We consider (1) and (2) in Minkowski space and assume that with respect to a system of stationary observers:
\begin{eqnarray}
                u^{0} = \frac{1}{\sqrt{1-\frac{|v|^2}{c^2}}}, u^i = \frac{\frac{v^i}{c}}{\sqrt{1-\frac{|v|^2}{c^2}}} \space (5) 
\end{eqnarray}
Moreover, we assume that 
\begin{eqnarray}
              \rho = \mu{c^2} + h \space (6)
\end{eqnarray}
where $\mu$ is the mass density and $h$ is the internal energy density.
(i) Derive in the nonrelativistic limit $c\rightarrow\infty$ the conservation of mass law,
\begin{eqnarray}
                \frac{\partial\mu}{\partial{t}} + \nabla\cdot(\mu{v}) = 0. \space (7)
\end{eqnarray}
Hints: Express the coponents of (3) explicitly in orders of c using (5) and (6), and derive the conservation law from the 0-component of (1).
So I have done a little work on this but I don't see how I can derive (7). Maybe there is some math trick I need to use or maybe this is a lot easier than I am making it out to be. So, I put (5) and (6) into (3) and I get the following:
\begin{eqnarray}
T^{\mu\nu} = (p + \mu{c^2} + h)\frac{v^{\mu}v^{\nu}}{c^2-|v|^2} + p(g^{-1})^{\mu\nu}
\end{eqnarray}
I am not sure what I am supposed to do from this point. I am not sure if I derived this properly and am I supposed to take the derivative of this in terms of $t$ and what do they mean by expressing things as orders of $c$. Thank you and have a great day!
Well after doing some work on my own in this problem I did get somewhere but not this question. I will show only the result of my derivation for at least one part and maybe more detail on the derivation of another part that is giving me trouble, but just to recall what I want to derive:
\begin{eqnarray}
 &&\nabla_{\mu}T^{\mu0} = \partial_0T^{00}+\partial_iT^{i0} = 0
\end{eqnarray}
I got the first term just fine (I think):
\begin{eqnarray}
  \partial_0T^{00} = \frac{d\mu}{dt}c^2
\end{eqnarray}
but I am not getting something nice for the second term, so I will show my derivation in more detail:
\begin{eqnarray}
  &&\partial_iT^{i0} = \partial_i((p+\rho)u^{0}u^{i}+p(g^{-1})^{i0} \\
  &&\partial_iT^{i0} = \partial_i((p+\rho)u^{0}u^{i}) \\
  &&\partial_iT^{i0} = \partial_i((p+\mu{c^2}+h)u^{0}u^{i}) \\
  &&\partial_iT^{i0} = \partial_i((p+\rho)(\frac{\frac{v^i}{c}}{1-\frac{|v|^2}{c^2}})) \\
\end{eqnarray}
If I play around with this I somehow get:
\begin{eqnarray}
  \partial_iT^{i0} = \nabla\cdot(\mu{v})c
\end{eqnarray}
Which I don't think is quite right... any help would be nice :)
 A: Since it's homework I'm not going to solve this for you, but here's what they mean:
1) Taking the 0-th component of (1) means looking at
\begin{align}
\nabla_\mu T^{\mu 0} = 0 \implies -\partial_0 T^{00} + \partial_i T^{i 0} = 0.
\end{align}
Remember that $\partial_0 = \frac{1}{c} \partial_t$ and $\partial_{i=1,2,3} = \partial_x, \partial_y, \partial_z$. So you can see the $\partial_t() + \nabla \cdot () = 0$ form of the equation coming out already.
2) In the limit $c \to \infty$, the term $v/c$ is small. That means we can Taylor expand the Lorentz factor in $v/c$. Recall how this is done: $(1 - x)^{-1/2} = 1 + \frac{1}{2} x + \cdots$. So this is what they mean by expanding it in orders of $c$.
So you should get stuff like
\begin{align}
(\cdots) + \frac{1}{c} (\cdots) + \frac{1}{c^2} (\cdots) + \cdots = 0,
\end{align}
Look at the term without any factors of $c$, that being the lowest order, and since $c \to \infty$ the other terms $\to 0$, the lowest order term should give you the continuity equation.
