We have seen that the energy-momentum of a cloud of dust, according to an observer with 4-velocity $\vv$, is given by 1) \begin{equation} \Tvec_{\vv} = \rho\,\uu \,(-\uu\cdot\vv) \end{equation} If the dust is moving at speed $v=\tanh\beta$ with respect to the observer $\vv$, then the 4-velocity $\uu$ of the dust takes the form \begin{equation} \uu = \begin{pmatrix} 1\\ \us\\ \end{pmatrix} \cosh\beta \label{ucomp} \end{equation} where we have introduced the notation $\us$ for the 3-velocity of the dust. In this frame, $\vv$ is the time direction, so \begin{equation} \uu\cdot\vv = -\cosh\beta \end{equation} and we have \begin{equation} \Tvec_{\vv} = \rho \begin{pmatrix} 1\\ \us\\ \end{pmatrix} \cosh^2\beta = \rho_{\hbox{obs}} \begin{pmatrix} 1\\ \us\\ \end{pmatrix} \end{equation}

We expect a classical conservation law along the lines of \begin{equation} \dot\rho_{\hbox{obs}} + \grads\cdot(\rho_{\hbox{obs}}\,\us) = 0 \label{Mconserve} \end{equation} where $\grads$ denotes the 3-dimensional gradient, which says that the matter density can only change if the matter flow “diverges”. This conservation law is essentially the requirement that \begin{equation} \grad\cdot\Tvec_{\vv} = 0 \end{equation} where this divergence operator is 4-dimensional (and therefore includes a $t$-derivative). But \begin{equation} \Tvec_{\vv} = - T^i{}_j \,v^j \,\ee_i \end{equation} and conservation should hold for any $\vv$. It therefore seems reasonable to expect a conservation condition of the form \begin{equation} \grad\cdot (T^i{}_j \,\ee_i) = 0 \label{divvec} \end{equation} This is a divergence on the vector part of $\Tvec$, but the symmetry of the energy-momentum tensor allows us to regard this as a divergence on the 1-form part of $\Tvec$ instead, that is, to expect a conservation condition of the form \begin{equation} {*}d{*} T^i = {*}d{*} (T^i{}_j\,\sigma^j) = 0 \label{divform} \end{equation} The equivalence between ($\ref{divvec}$) and ($\ref{divform}$) in special relativity (and Minkowski coordinates) can be easily checked, since both notions of divergence involve simply taking the $x^i$-derivative of the $i$th component; the symmetry of the energy momentum tensor allows us to perform this sum on either the first (vector) or second (1-form) “slot” of $T^i{}_j$. In curved space (and/or curvilinear coordinates), the above argument fails. However, there is an obvious alternative, which turns out to be correct: Demand conservation in the form \begin{equation} {*}d{*} \Tvec = \zero \label{Tconserve} \end{equation} where $d$ now acts on both the 1-form and the vector parts of $\Tvec$.

It is straightforward to run this argument backward and recover the expected conservation law ($\ref{Mconserve}$) for dust in Minkowski space. First of all, in Minkowski coordinates, the basis vectors are constant, so that ($\ref{Tconserve}$) implies \begin{equation} {*}d{*} T^i = 0 \end{equation} for each energy-momentum 1-form \begin{equation} T^i = \rho\,u^i\,u_j\,\sigma^j \end{equation} But the components of $\uu$ are given by ($\ref{ucomp}$), so that \begin{align} T^t &= \rho_{\hbox{obs}} \> (-dt + \bar{u}_i\,dx^i) \\ T^j &= \bar{u}^j T^t \end{align} where $i$, $j$ run only over the three spatial indices. Setting \begin{equation} {*}d{*} T^t = 0 \end{equation} leads immediately to ($\ref{Mconserve}$), which can be thought of as conservation of energy. Inserting this result into \begin{equation} {*}d{*} T^j = 0 \end{equation} leads to \begin{equation} \rho_{\hbox{obs}} \left( \dot{\bar{u}}^i+\bar{u}^j\,\Partial{\bar{u}^i}{x^j} \right) = 0 \end{equation} or equivalently \begin{equation} \dot\us + (\us\cdot\grads)\us = \zeros \end{equation} which is the well-known Navier-Stokes equation, and can be thought of as a statement of conservation of momentum.

We conclude that ($\ref{Tconserve}$) does indeed provide a reasonable notion of conservation.