Pressure Inside an Ideal Gas

Question

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

$p = \frac13 \rho \langle v^2 \rangle$

where $\rho$ is the mass density of the gas and $\langle v^2 \rangle$ is the average square of the Maxwell velocity distribution. This is the relation for the pressure at the boundary of the vessel containing the ideal gas.

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

$T_{\mu \nu} = \left( \begin{array}{cccc} \rho & 0 & 0 & 0 \\ 0 & p & 0 & 0 \\ 0 & 0 & p &0 \\ 0& 0 & 0 & p \end{array} \right) $

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?

Answer 1

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.

Answer 2

[see my correction in the comments. Sorry, Ian.]

Ian is missing the fact that the area segment that he is considering is a directed area. In our imagination we think of the area segment as a tiny transparent, infinitely thin rectangle, and if we flip it 180 degrees it looks the same. We can't be imposing that limitation on our area. We can only reasonably insist that it returns to its original state after a rotation of 360 degrees. Imagine that the little area has an arrow pointing perpendicularly from the surface. Now the momentum that crosses the surface has to be reckoned according to whether or not your atom is crossing in the positive direction or the negative direction. When two atoms cross, one from the left the other from the right, the momentum that crosses the directed area is 2mv. Suddenly the pressure is no longer zero.

A directed area is something we don't ordinarily think about, but we do use them without naming them when we use the integral form of Gauss' Law. On the other hand, a directed line segment is quite familiar; we call it a vector instead of "directed line segment". The directed area is more naturally described mathematically by an even more unfamiliar concept: the bivector, which naturally accounts for the directionality of an area segment without having to glue perpendicular arrows on to it.