How can any spatially extended object have 4-momentum assigned? We know that the 4-momentum of a point particle is of the form $p^{\mu} = (E/c, p^{i})$, whose transformations across different inertial reference frames are given by Lorentz transformations. However, a spatially extended object is not a point particle and therefore has an energy density. The covariant way to talk about energy densities is by invoking the stress-energy-momentum tensor $T^{\mu\nu}$.
Despite this fact, I still see people talk about extended objects (like spaceships and planets) as if they are point particles with a covariant 4-momentum. What I want to know is, how can we possibly talk about the 4-momentum of an object in a covariant manner if it is an extended body with some $T^{\mu\nu}$? Is there some way to start with $T^{\mu\nu}$ and get to 4-momentum $p^{\mu}$ of an object in a covariant manner? If so, can anyone show this in excruciatingly explicit terms?
 A: Consider a particular inertial reference frame with coordinates $\{t, x, y, z\}$, and let $t^a$, $x^a$, $y^a$, and $z^a$ be their corresponding orthonormal basis vectors.  The set of events in spacetime with a given $t = t_0$ is known as a "timelike hypersurface".  This hypersurface is three-dimensional, and it corresponds to all of the events at $t = t_0$ in that particular reference frame.  We can then construct an integral
$$
p^a = \iiint_{t = t_0} T^{a b} t_b \, d^3x
$$
over the $t = t_0$ hypersurface.  This defines the total four-momentum as seen in the given reference frame at the time $t_0$.
However, this construction involved picking a particular reference frame, and it is not obvious that our result is consistent across reference frames.  To see that it is, consider the same integral defined on a surface of constant $t'$ (with corresponding volume element $d^3x'$):
$$
{p'}^a = \iiint_{t' = t'_0} T^{a b} t'_b \, d^3x'
$$
We wish to show that $p^a = {p'}^a$.
Consider the region of spacetime $\mathcal{V}$ bounded by the surfaces $t = t_0$ and $t' = t'_0$.  Assume further that the region of spacetime in which $T_{ab} \neq 0$ intersects each of the hypersurfaces in a compact set, so that we can add timelike hypersurfaces spanning between the surfaces $t = t_0$ and $t' = t'_0$ to make a closed compact region of spacetime.  Let $n^a$ be the unit normal vector pointing outward from this spacetime volume.

From the assumption above, the integral of $T^{ab} n_b$ over the timelike "sides" of this spacetime region is zero;  and we have $n^a = t^a$ on the $t = t_0$ surface and $n^a = - {t'}^a$ on the $t' = t'_0$ surface.  Thus, we have
$$
p^a - {p'}^a = \iiint_{t = t_0} T^{a b} t_b \, d^3x - \iiint_{t' = t'_0} T^{a b} t'_b \, d^3x' = \iiint_{\partial \mathcal{V}} T^{a b} n_b d^3 x
$$
But by the spacetime version of Gauss's Law, we have
$$
\iiint_{\partial \mathcal{V}} T^{a b} n_b d^3 x = \iiiint_\mathcal{V} \partial_b T^{ab} \, d^4 x = 0
$$
since the stress-energy tensor is conserved.  Thus, $p^a = {p'}^a$, and the four-vector $p^a$ is well-defined between different reference frames.  (The components of $p^a$ as observed in different reference frames are different, of course, but the vector itself is well-defined and constant regardless of the reference frame we choose for the purposes of performing the integral.)
This argument can also be extended to situations where $T_{ab}$ vanishes "sufficiently rapidly at infinity" (rather than being confined to a compact intersection with the timelike hypersurfaces), and it can also be extended to completely arbitrary (non-flat) hypersurfaces in flat spacetime as well.
