Let's summarize the discussion in the preceding section, but without the restriction to Minkowski space. Matter is described by the energy-momentum 1-forms \begin{equation} T^i = T^i{}_j\,\sigma^j \end{equation} which can be combined into a single vector-valued 1-form \begin{equation} \Tvec = T^i \ee_i \end{equation} Consider an observer with velocity vector $\vv$ satisfying \begin{equation} \vv \,d\tau = d\rr \end{equation} and drop the arrow to write as usual \begin{equation} v=\vv\cdot d\rr \end{equation} for the 1-form physically equivalent to $\vv$. The observer's 4-velocity $\vv$ is a future-pointing, timelike unit vector, so $\vv\cdot\vv=-1$ and therefore $v=-d\tau$.

Given an energy-momentum tensor $\Tvec$ and an observer with velocity $v$, it is reasonable to ask how to obtain the energy density seen by this observer. Not surprisingly, this involves some sort of dot product between $\Tvec$ and $v$, but it requires a dot product both on the vector part of $\Tvec$, and on its 1-form part. Begin by using the metric $g$ to perform the latter operation, defining 1) \begin{equation} \Tvec_{\vv} = -g(\Tvec,v) \end{equation} which we can write in terms of components using \begin{equation} g(\Tvec,v) = T^i{}_j\ee_i \,g(\sigma^j,\sigma^k) \,v_k = T^i{}_j \,v^j \,\ee_i \end{equation} since $v=v_k\,\sigma^k$. The observed energy density is then given by \begin{equation} \rho_{\hbox{obs}} = -\Tvec_{\vv} \cdot \vv = g(\Tvec\cdot\vv,v) \label{rhoobs} \end{equation} where in components \begin{equation} \Tvec\cdot\vv = \left(T^i{}_j\sigma^j\ee_i\right)\cdot \left(v^k\,\ee_k\right) = T^i{}_j\sigma^jv_i \end{equation} and where the last equality in (\ref{rhoobs}) reflects the ability to perform these two types of dot products in either order. The condition that the energy-momentum tensor be symmetric can be expressed in the form \begin{equation} g(\Tvec,v)\cdot d\rr = \Tvec \cdot \vv \end{equation} (for all spacetime vectors $\vv$), which we will henceforth assume, and which is equivalent to $T^{ij}=T^{ji}$.

The energy-momentum tensor for dust with velocity vector $\uu$ is given by \begin{equation} \Tvec = \rho\,\uu\,u \end{equation} where $u=\uu\cdot d\rr=-d\lambda$, with $\lambda$ denoting the proper time of the dust particles in order to avoid confusion with the proper time $\tau$ of the observer. Equivalently, \begin{align} \Tvec_{\uu} &= \rho\,\uu \\ \Tvec \cdot \uu &= -\rho\,u \end{align} The energy density in the rest frame of the dust particles is then \begin{equation} -\Tvec_{\uu}\cdot\uu = \rho \end{equation} as expected, and the observed energy density in some other frame is given by \begin{equation} -\Tvec_{\vv}\cdot\vv = \rho\,\uu\,g(u,v) \cdot\vv = \rho\,(\uu\cdot\vv)^2 \end{equation} since $g(u,v)=\uu\cdot\vv$. We again recover the expected result, since $\uu\cdot\vv=\cosh\beta$ (and $\rho=mn$ in the example considered in the previous section).

Dust is not very interesting, since the dust particles do not interact with each other. A slightly more general model is a perfect fluid, in which there is also a pressure density $p$. The energy-momentum tensor for a perfect fluid with velocity vector $\uu$ is given by \begin{equation} \Tvec = (\rho+p)\,\uu\,u + p\,d\rr \end{equation} for which \begin{equation} \Tvec\cdot\uu = -(\rho+p)\,u + p\,u = -\rho\,u \end{equation} as was the case for dust, so the energy density (at rest) is still $\rho$. In (spacelike) directions $\vv$ orthogonal to $\uu$ we have instead \begin{align} -\Tvec_{\vv}\cdot\vv &= g(\Tvec\cdot\vv,v) \nonumber\\ &= g(p\,v,v) \nonumber\\ &= p \vv\cdot\vv \nonumber\\ &= p \end{align} where we have used $\uu\cdot\vv=0$ and further assumed $|\vv|=1$. Thus, $\rho$ and $p$ correspond to the energy and pressure densities 2) in the rest frame of the fluid, and $p$ is the same in all (spatial) directions in this frame.