Calculating four-momentum from energy-momentum tensor How does one calculate four-momentum from the energy-momentum tensor $T_{\mu\nu}$? Would it be the eigenvector of the four-momentum tensor? If so, which eigenvector should be considered (in case of multiple eigenvectors)? If the answer is the time-like eigenvector, would this eigenvector be unique or it's possible there might be more than one time-like eigenvector? Also, what is the physical interpretation of the other eigenvectors?
 A: You extract an energy-momentum vector from the stress-energy-momentum density by integration over a space-like 3-surface (whereas integration over a time-like surface would yield momentum fluxes). If your space-like surface is defined by the condition $t=\text{const}$, that means integrating the first column $T_{\mu0}$.
It might help to consider the invariant geometric objects involved, instead of just their coordinate representations:
For point particles, momentum is a covector
$$
p_\mu\mathrm dx^\mu
$$
For continuous media (fields and fluids), we have to promote this to a momentum density
$$
\rho_\mu\mathrm dx^\mu\otimes\mathrm dV
$$
where $$\mathrm dV = \mathrm dx\wedge\mathrm dy\wedge\mathrm dz$$ is the 3-dimensional volume element. That is no longer an invariant object, as a change of frame will mix-in $$\mathrm dt\wedge\mathrm dx\wedge\mathrm dy\qquad \mathrm dt\wedge\mathrm dx\wedge\mathrm dz\qquad \mathrm dt\wedge\mathrm dy\wedge\mathrm dz$$ By purely geometric reasoning, the energy-momentum vector thus has to be promoted to a tensor
$$
T_{\mu\nu}\mathrm dx^\mu\otimes\star\mathrm dx^\nu
$$
where we have represented the surface elements via their normal vectors by Hodge duality.
