General proof of $m^2 = E^2 - p^2$ All proofs I know of, starting from Einstein's famous 1905 article, only deal with special cases (actually, yes, I know that formula like $E=mc^2$ were known before Einstein in the context of electromagnetism). Would there be a more general proof? One that would work on curved spacetime for added bonus. 
I am thinking of something along the following lines. Starting from the energy-momentum tensor $T^{\mu\nu}$, we could integrate $T^{00}$ and $T^{i0}$ over some region to get an energy and a momentum but do they always form a 4-vector in the simple Minkovskian case?
 A: 
I am thinking of something along the following lines. Starting from the energy-momentum tensor $T^{\mu\nu}$, we could integrate $T^{00}$ and $T^{i0}$ over some region to get an energy and a momentum but do they always form a 4-vector in the simple Minkovskian case?

This is correct, when properly interpreted.
Working in flat spacetime (Minkowski metric), suppose that we have a tensor field $T^{AB}(x)$. In general, the quantities
$$
    P^A = \int dx^1\,dx^2\,dx^3\ T^{0A}(x)
\tag{1}
$$
will be functions of the remaining coordinate $x^0$. But if the tensor $T^{AB}(x)$ satisfies 
$$
\partial_A T^{AB}(x)=0,
\tag{2}
$$
then the quantities (1) are independent of $x^0$, and in this case we can show that the quantities $P^A$ are the components of a four-vector. To do this, consider two frames, $x$ and $\tilde x$, that are related to each other by a Lorentz transformation
$$
   \tilde x^A=\Lambda^A_B x^B
\hskip2cm
\text{with }\det\Lambda=1.
\tag{3}
$$
Let $v$ and $\tilde v$, respectively, be the timelike vectors with components $(1,0,0,0)$ in each of the two frames. These are related by
$$
   \tilde v^A=\Lambda^A_B v^B.
\tag{4}
$$
Now consider the quantities
\begin{align}
    P^F 
   & = \int \epsilon_{ABCD}\, dx^A\wedge dx^B\wedge dx^C\ v^D \,v_E T^{EF}(x)
\\
    \tilde P^F 
   & = \int \epsilon_{ABCD}\, d\tilde x^A\wedge d\tilde x^B\wedge d\tilde x^C\ 
  \tilde v^D \,\tilde v_E \tilde T^{EF}(\tilde x)
\tag{5}
\end{align}
where $\epsilon_{ABCD}$ is completely antisymmetric in its indices, normalized so that
$$
\det\Lambda = \epsilon_{ABCD}\Lambda^A_1\Lambda^B_2\Lambda^C_3\Lambda^D_4.
\tag{6}
$$
The integrals are written this way, instead of writing them as in (1), so that we can use the identity
$$
    \epsilon_{ABCD}\, dx^A\wedge dx^B\wedge dx^C\ v^D
   =  \epsilon_{ABCD}\, d\tilde x^A\wedge d\tilde x^B\wedge d\tilde x^C\ 
 \tilde v^D,
\tag{7}
$$
which is an obvious consqeuence of equations (3)-(4) and (6). Use (7) in (5) to get
\begin{align}
    \tilde P^F
    = \int \epsilon_{ABCD}\, dx^A\wedge dx^B\wedge dx^C\ v^D
  \,\tilde v_E \tilde T^{EF}(\tilde x).
\tag{8}
\end{align}
The assertion that $v$ and $T$ are tensors implies
$$
\tilde v_E \tilde T^{EF}(\tilde x)
=
v_E T^{EG}(x)\Lambda_G^F,
\tag{9}
$$
and using this in (8) gives
$$
\tilde P^F=\Lambda^F_G P^G,
$$
which is the desired result. In summary, if the condition (2) is satisfied, then the quantity (1) is a four-vector in the sense that if the construction on the right-hand side of (1) is repeated in two different frames, then the resulting quantities $P^A$ and $\tilde P^A$ are related (as four-vectors) by the same Lorentz transformation that relates those two frames.
By the way, the quantity $P_A P^A$ is invariant under Lorentz transformations, and it is interpreted as the invariant mass of the system.
For confirmation, here are a few on-line sources where the four-vector character of (1) is mentioned:


*

*Equations (4.35) and (4.67) in http://edu.itp.phys.ethz.ch/hs12/qft1/Chapter04.pdf

*Equation (8.138) in  http://users.physik.fu-berlin.de/~kleinert/b6/psfiles/Chapter-7-conslaw.pdf

*Equation (1.43) in http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf
