Stress-energy-momentum tensor

Question

In Wald's General Relativity, he writes on pg 61

For an observer with 4-velocity $v^a$, the component $T_{ab}v^a v^b$ is interpreted as the energy density, i.e. the mass-energy per unit volume, as measured by the observer.

However, if we use this with the stress tensor for dust $$T_{ab}=\rho_0 u_a u_b,$$

we get $$T_{ab}v^a v^b= \rho_0 u_a u_b v^a v^b = (\rho_0 u_a v^a) (\rho_0 u_b v^b)/\rho_0 = U^2/ \rho_0$$

where I assumed $U=- \rho_0 u_b v^b$ is the energy density (since $E= - m_0 u_a v^a $).

What am I missing here?

Answer 1

I henceforth assume $c=1$. Consider a 3-volume $\Delta \Sigma_0$ at rest with the dust. For the rest observer (this is its definition) $u^\mu$ has only (unit) temporal component.
The energy (i.e the mass) associated with that portion of system is $\Delta \Sigma_0 \rho_0$. The 4-momentum of that portion is therefore $\Delta p^\mu := \Delta \Sigma_0 \rho_0 u^\mu$.

Now consider another observer with four velocity $v^\mu$. The energy of that portion of system is the temporal component of $\Delta p^\mu$ computed in his/her reference frame. In other words it is: $$\Delta p^\mu v_\mu = \rho_0 \Delta \Sigma_0 u^\mu v_\mu$$ However we are interested in the density of energy computed in that reference frame. It is defined as: $$\frac{\Delta p^\mu v_\mu}{\Delta \Sigma} = \rho_0\frac{\Delta \Sigma_0}{\Delta \Sigma} u^\mu v_\mu\:.$$ $\Delta \Sigma$ is the volume, as measured in the considered reference frame, of the portion of dust defined by $\Delta \Sigma_0$ at rest with the dust. As is well known if a solid body has a volume $\Delta \Sigma_0$ in its rest frame, it has volume $$\Delta \Sigma = \Delta \Sigma_0\sqrt{1-v^2} $$ in a reference frame where it is seen to have a speed $v$. The above relation can be re-written $$\Delta \Sigma_0 = \Delta \Sigma u^\nu v_\nu $$ where $u^\nu$ is the $4$-velocity of the body and now $v^\nu$ the $4$-velocity of the other reference frame. Coming back to our dust system with field of $4$-velocity $u^\nu$, we can conclude that the energy density for a generic observer with $4$-velocity $v^\nu$ is, as correctly stated in Wald's book: $$\frac{\Delta p^\mu v_\mu}{\Delta \Sigma} = \rho_0 \frac{\Delta \Sigma_0}{\Delta \Sigma} u^\mu v_\mu = \rho_0 u^\nu v_\nu u^\mu v_\mu = T^{\mu\nu} v_\mu v_\nu\:.$$

Answer 2

We can evaluate it from the observer's rest frame. Then $v^\mu = (1, 0, 0, 0)$ and $u_\mu = (\gamma , -\gamma\mathbf v)$ where $\gamma = \frac{1}{\sqrt{1-v^2}}$ and $\mathbf v$ is the relative velocity. (We put $c = 1$.) Then $$T_{\mu\nu}v^\mu v^\nu = \rho_0 \gamma^2.$$ However, $\rho_0$ is the density in the rest frame of the fluid. In the observer's rest frame the density is $\rho = \rho_0\gamma$ because of length contraction! So the energy density according to our observer is $$E = \rho\gamma = \frac{\rho}{\sqrt{1-v^2}} = \rho + \frac{\rho v^2}{2} + O(v^4).$$

By the way, I think everyone could benefit from reading about the stress-energy tensor in Misner, Thorne and Wheeler, Chapter 5.