Why does the quantity $P^i=\int d^3x T^{0i}$ represent the momentum of a field? For a classical field $\phi$, the space integral of $T^{0i}$ denoted by $P^i=\int d^3x T^{0i}$ where $T^{\mu\nu}$ is the energy-momentum tensor is called the momentum of the field. This is not very obvious. It almost looks like a dictum to me. How do I go about verifying that this quantity indeed has all the properties that a momentum should have without quantizing the theory? 
 A: Momentum is defined via the Noether theorem. It is the conserved charge correspoding to translations. If you consider $\delta x^{\mu} = a^{\mu}$ where $a^{\mu}$ is just a constant four-vector with real components, you can run the Noether procedure to compute 
$$
\delta S = \int d^{4} x \, a^{\mu} \partial^{\nu} T_{\mu \nu}
$$
Now given a conserved current $j^{\mu}$ -- that is a four-vector that satisfies $\partial_\mu j^{\mu} = 0$ -- you can define a conserved charge by
$$
Q = \int d^{3} x \, j^{0}
$$
It is conserved because the time derivative gives a space derivative if we use $\partial_\mu j^{\mu}$ = 0, which then gives the integral of a total derivative. If we consider that this current decays sufficiently fast at infinity, the boundary of our $3d$ space, then Stokes theorem establishes $\dot{Q} = 0$.
If you ask why do we need to consider $j^{0}$, it is because an observer at rest has four-velocity given by $u = (1,0,0,0)$. So, the projection of $j^{\mu}$ onto his spacetime trajectory (which is essentially a line through the time axis) is given by $ - g(u, j) = j^{0}$. This gives the density of energy-momentum that this observer will measure in his spacetime trajectory. 
And there you have it:
$$
Q = \int d^{3} x \, j^{0} = \int d^{3} x \, a^{\mu} T^{0}_{\mu} = \int d^{3} x \,
a^{i} T^{0}_{i}
$$
In the last line I wrote $a^{\mu} = a^{i}$ because momentum is the Noether current corresponding to space translations.
