Why pressures in opposite directions don't cancel out effects of each other in stress-energy tensor?

Question

here is how I think: pressure can cause to flow of momentum without flow of matter. a simple example is newton's cradle. now consider pressure witch by same mechanism and intercepting of many particles. consider perfect fluid at rest. at an event in space-time flow of momentum in all directions is same and I think so the net flow of 4-momentum must be zero, but, $T$ don't give zero and I confused why is so.

Answer 1

Pressure does NOT have a direction. It is scalar.

Answer 2

I think this is basically covered in Intuitive understanding of the elements in the stress-energy tensor, but maybe it's worth specifically focussing on the pressure terms in the stress-energy tensor.

If we consider our system to be made up from point particles then the diagonal terms look like:

$$ T^{ii} = \sum \gamma m (v^i)^2 \delta(x - x^i) $$

where $v^i$ is the coordinate velocity (not the four-velocity) of the point particle of mass $m$. The $\delta(x - x^i)$ term just makes the contribution to the stress-energy tensor zero everywhere apart from the position of the particle. The $\gamma m (v^i)^2$ term is basically a relativistic kinetic energy, and the kinetic energy of the particles in a gas is proportional to the pressure, which is why the diagonal terms are a pressure.

The contributions from the individual particles don't cancel because they sum $v^2$, and that is always positive regardless of the direction of $\mathbf v$. But what I want to do here is offer a simple intuitive reason why the diagonal terms can't cancel out.

Suppose we consider some ball of gas and the gravitational field it generates. If we want to increase the pressure we need to increase the particle kinetic energies, and that means increasing the temperature. So we have to add energy (as heat) to the system. But in the rest frame of the ball of gas the mass and energy are related by Einstein's famous equation:

$$ E = mc^2 $$

So if we have added energy we must have increased the mass, and that will have increased the gravitational field. So the increased velocities of the particles cannot cancel out.

Answer 3

Tensor of energy-momentum $T$ isn't made just of and does not reflect just densities of energy and momentum; in $T^{\mu \nu}$ there are also spatial-spatial components, each of which gives some momentum flux (rate of transfer of momentum per unit area). $T^{ik}$ is rate of transfer, per unit area, of $i$-th cartesian component of momentum, from one side of an oriented surface with surface vector aligned with the axis $k$. This often has to be assigned non-zero value because we want the matrix $T^{\mu\nu}$ to be such a function of position and time that it obeys the law of local conservation of momentum.

These spatial-spatial components $T^{ik}$ can be non-zero at some point of space even if momentum density there is zero. This is easy to see in the case of isotropic incoherent randomly moving dust (or ideal gas in ballistic regime, where collisions are negligible). Any region of space then is associated with zero momentum, because probability density in momentum space is isotropic too; probability density for $\mathbf p$ is the same as that for $-\mathbf p$. But for any imaginary planar element in real space, particles do move through the element from one side to the other, in both directions, thus there is momentum flux both from left-to-right and from right-to-left. This is different from a situation where there is no particle motion and thus no momentum transfer at all. It is this non-zero transfer of momentum, happening at equal rate in both directions, that manifests as non-zero components of $T^{ik}$.

Similar situation can happen for a fluid, or EM field in vacuum; there can be non-zero transport of momentum through a planar element in both directions, compatible with a stationary state where density of net momentum is zero. In case of fluid, this happens for a static fluid under pressure, as matter on one side of the planar element pushes matter on the other side and vice versa; in case of EM field, this happens for a static electric field or static magnetic field, or when both fields are parallel to each other. Momentum flux in the case of EM field in vacuum is hard to visualize as there is nothing moving there, but $T$ is still a perfectly well-defined theoretical quantity, obeying the law of local conservation, and in general, one has to have non-zero elements $T^{ik}$ in static electric or magnetic field, free of net momentum, to satisfy that.