4-momentum through a spacelike-othogonal hypersurface for a perfect fluid in special relativity My understanding of the stress-energy tensor in special relativity (or in general relativity), is that it gives you the flux density of 4-momentum flowing through an oriented 3D hypersurface. So at some point (event) $P$ in spacetime in an inertial frame in SR (or using geodesic coordinates in GR), such that the metric is simply $\eta_{\mu\nu}$ at $P$, and working in relativistic units ($c=1$), I'd expect the 4-momentum $dp^{\alpha}$ flowing through a hypersurface with 3-volume $dV$, oriented along the normal 1-form $n_{\beta}$ to be:
$$dp^{\alpha}=T^{\alpha\beta} n_{\beta} dV$$
Now, I'd like to consider a perfect fluid with $T^{\alpha\beta}=(\rho_0+p_s)u^{\alpha}u^{\beta}+\eta^{\alpha\beta}p_s$ (with rest mass-energy density $\rho_0$, static pressure $p_s$ and 4-velocity flow field $u^{\alpha}$). In particular, I want to consider the situation in a locally inertial co-moving rest frame at $P$, i.e. where $u^{\alpha}=\delta^{\alpha}_0$. If I now consider an infinitesimal hypersurface at $P$, oriented along the $x^1$ coordinate axis, i.e. $n_{\beta}=\delta^1_{\beta}$ and $dV=dx^0 dx^2 dx^3$, the total 4-momentum flowing through the 2D spatial surface with dimensions $dx^2 dx^3$, during time $dx^0$ should be:
$$dp^{\alpha}=[(\rho_0+p_s)\delta^{\alpha}_0\delta^{\beta}_0+\eta^{\alpha\beta}p_s]\delta_{\beta}^1dx^0 dx^2 dx^3$$
$$=[(\rho_0+p_s)\delta^{\alpha}_0\delta^1_0+\eta^{\alpha 1}p_s]dx^0 dx^2 dx^3$$
$$=\eta^{\alpha 1}p_s dx^0 dx^2 dx^3$$
Since the metric is diagonal, this implies that the total 4-momentum, flowing through $dx^0 dx^2 dx^3$ has only one component, $dp^1=p_s dx^0 dx^2 dx^3$, i.e. it is entirely spacelike. How can this be? I would have expected any 4-momentum to be timelike, for a material fluid such as the one I'm considering here. Am I fundamentally misunderstanding the nature of the stress-energy tensor? I'd be grateful for any inisghts into my dilemma :)
 A: Short answer: Pressure is flow of momentum.
For your energy-momentum tensor, spacelike flows are $\sim p$, timelike flow is $\sim\rho$. This makes intuitive sense: If you take a small 3-volume $[t,t+\epsilon]\times [x^2,x^2+\epsilon]\times[x^3,x^3+\epsilon]$ in your hypersurface, your pressure is transmitting a momentum (call momentum $P$ to distinguish from pressure $p$, and somewhat heuristically)
$$\int\limits _t^{t+\epsilon}\int\limits _{x^2}^{x^2+\epsilon}\int\limits _{x^3}^{x^3+\epsilon} p \;\text{d}x^2 \text{d}x^3 \text{d}t = \int\limits _{t}^{t+\epsilon} F \;\text{d}t = \int\limits _{t}^{t+\epsilon} \dot P \;\text{d}t = \Delta P\,.$$
Here I've used that $\int p\, \text{d}A$  is a force $F$, and force is change in momentum, $F=\dot P$.
A: I guess I should modify my understanding of the momentum-flow represented by the stress-energy tensor: I should take $T^{\alpha\beta} n_{\beta} dV$ to be the net 4-momentum flowing out through the hypersurface normal to $n_{\beta}$. That is to say, all the 4-momentum flowing out of the hypersurface minus all the 4-momentum flowing in from the other side:
$$dp^{\alpha}_{net}=dp^{\alpha}_{out}-dp^{\alpha}_{in}=T^{\alpha\beta} n_{\beta} dV$$
If $n_{\beta}$ is timelike, then $dp^{\alpha}_{in}=0$ since no 4-momentum associated with matter flows backwards in time. Likewise, for dust without any static pressure ($T^{\alpha\beta}=\rho_o u^\alpha u^\beta$) in a frame where the fluid moves in the positive $x^1$ direction ($u^1 > 0$), momentum flows out of a hypersurface perpendicular to the $x^1$ direction (both $dp^0_{out}$ and $dp^1_{out} \ne 0$), but none flows in ($dp^{0,1}_{in}=0$). If the fluid moves in the negative $x^1$ direction, then no momentum flows out through a hypersurface perpendicular to positive $x_1$ ($dp^\alpha_{out}=0$), but momentum does flow in. In this case, $dp^0_{in}>0$ and $dp^1_{in}<0$, so I should expect $dp^0_{net}=-dp^0_{in}<0$ and $dp^1_{net}=-dp^1_{in}>0$. This is indeed born out by $dp^{\alpha}_{net}=T^{\alpha\beta} n_{\beta} dV=\rho_o u^{\alpha}u^{\beta} n_{\beta} dV$ for $u^0>0$, $u^1<1$ and $n_1=+1$.
Only in the case of a non-zero static pressure accross a spacelike hypersurface, is there a situation with both outflowing momentum and inflowing momentum. On average, for each fluid particle flowing out of the surface with positive $x^1$ momentum (and positive $x^0$ momentum, i.e. energy), there'll be a particle flowing into the surface with equal positive $x^0$ momentum (i.e. energy), but opposite, negative $x^1$ momentum. So in the net momentum, the $x^0$ momenta cancel out (no net gain in energy), but the $x^1$ momenta add up, leading to a purely spatial net momentum transfer.
