Pressure in Stress Tensor

Question

In the stress energy tensor $T^{\mu \nu}$ the flux of $i$-component of momentum in the $i$-direction is $T^{ii}$. This is identified as the pressure. However for a system in equilibrium and at rest there is no net momentum flux in any direction, although there is still a positive pressure. It seems to me that the momentum flux must represent the net pressure in a particular direction rather than the absolute pressure. Where is my misunderstanding?

Pressure is force, and if there is force then momentum is moving from one place to another (because of Newton's second law). It's just that the density of momentum doesn't change because the incoming flux is equal to the outgoing flux. — Javier, Apr 18 '17 at 18:35
Related: Intuitive understanding of the elements in the stress-energy tensor — John Rennie, Apr 18 '17 at 19:25
I agree with you. I never liked the idea of interpreting the stress tensor as a momentum flux. But, with the pressure present, even if you interpret the stress tensor that way, then when you couple with the stress equilibrium equation, you still get the right answer. — Chet Miller, Apr 18 '17 at 20:21

score 1 · Answer 1 · answered Apr 18 '17 at 19:59

Note that the conservation law for energy and momentum is formally written as:

$\partial_\mu T^{\mu \nu} = 0$.

The momentum density is given by the vector $T^{0i}$. Even if this quantity is zero the pressure $T^{ii}$ can be nonzero. If off-diagonal components of the stress tensor are also zero it holds

$\partial_i T^{ii} = 0$ (no summation convention used here!)

equivalently $T^{ii} = const.$

Pressure is a momentum flux directed normal to a boundary of an elementary cell.

For constant pressure it holds $\oint_\Sigma T^{ii}n d^3x = 0$ with unit normal vector $n$ for the boundary $\Sigma$, because pressure integrated on some face of $\Sigma$ is equal to pressure integrated on the opposite face (with opposite normal vector direction). Therefore the net flux for constant pressure is zero.

Holds for all surfaces $\Sigma$.

score 1 · Answer 2 · answered Apr 19 '17 at 18:51

I think I've realised where my misunderstanding is. I was imagining that the momentum-flux of the particles flowing in one direction is exactly counterbalanced by momentum-flux of the particles flowing in the opposite direction, and so there is zero net flux of momentum. However the momentum fluxes in opposite directions do not cancel. To see this, observe that contribution to the momentum flux (in the $i$-direction) from a single particle is $v_i p_i$ where $v_i$ is the velocity in the $i$-direction. Since $p_i = mv_i$, $p_i$ and $v_i$ have the same sign. If the particle is moving in the positive direction, both $v_i$ and $p_i$ are positive and so $v_i p_i$ is positive. But if they are both negative, the minus signs cancel and so there is still a positive contribution to the momentum flux.

As the other commenters have pointed out, the average momentum can stay zero everywhere (and so the systems remain at rest) even with a positive momentum flux so long as the divergence of the momentum flux is zero. That is, each point accumulates no momentum because the momentum flowing in to a point is counterbalanced by the momentum flowing out. This reminds me of the counterintuitive nature of the heat equation in which the steady state solution (i.e. constant temperature) is not, in general, one of zero heat flux.

Pressure in Stress Tensor

2 Answers2