A non-relativistic ideal gas exerts a pressure at the surface of its container H

Question

A non-relativistic ideal gas exerts a pressure at the surface of its container

However, if one were to place an infinitesimal "test" area inside the boundary of the vessel the momentum flux across that area would be 0 since the distribution of velocities is symmetric. That is, as many particles would cross one way across the area as cross the opposite way. This suggests that the pressure inside an ideal gas is 0.

In general relativity an ideal gas is usually presented as an example of a perfect fluid, that is one with stress energy tensor equal to

Since the stress energy tensor should be a local function it seems strange to assign to p (in the frame in which the velocity distribution is isotropic) in the above equation the value of the pressure at the boundary of the vessel.

I can see why this is done: one wants the stress energy tensor to be a smooth function, and also it should be that a gas at finite temperature should gravitate more than a gas at zero temperature (dust)... however this is point is not articulated, in for instance MTW.

Perhaps I am missing something elementary?

Explanation / Answer

The stress is the current of the conserved momentum. If momentum is going from one place to another, you have a stress, because the momentum has to go through all the places inbetween. If your intuition comes from action-at-a-distance physics, this can be confusing, because in Newtonian gravity, momentum can jump across empty space between two gravitating objects, leading to a nonlocal force. But this is no longer true in relativity.

When you have a gas compressed by a container, the x-momentum flows from one x-end of the container, where a positive x-force is applied (x-momentum pumped into the gas) to the other x-end, going in the x-direction. This means that there is a flow of x-momentum in the x-direction, a pressure.

If you place an x-wall somewhere, the force exerted by the left half of the gas on the right half is equal and opposite to the force exerted by the right half on the left half. But this does not mean that there is zero stress. The rightward force is a flow of x momentum in the positive x direction, and the leftward force is also a flow of x momentum in the positive x direction.

This is a bit confusing because of the vector nature of momentum, and the tensor nature of stress, so consider a scalar conserved charge. Suppose you have a wire with a nonzero flow of electric charge, a nonzero current, and you cut the wire somewhere with a plane. The current through the plane is the flow of charge from left to right, and if 8 coulombs per second are flowing from left to right, then it is obvious that -8 coulombs per second are flowing from right to left, in order for charge to be conserved. But does this mean that the current is always zero? Obviously not--- the current is defined by one or the other of these quantities, not the sum.

Similarly, in a gas, the momentum currents are defined by the force one part of the gas exerts on another, both through particles leaving the region carrying a certain amount of momentum with them, and collisions right by the boundary transferring momentum from particles inside to particles outside. The net momentum flow is defined by the force of one half on the other, or by the negative of the force the other way, not by the sum of the two.

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