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?

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    $\begingroup$ Why does a net momentum flux of zero for some area suggest a pressure of zero? You completely lost me at that point. $\endgroup$ – Alan Rominger Oct 19 '11 at 5:30
  • $\begingroup$ Pressure is defined to be the force per unit area. If when the particles collide with the test area they do so elastically, the force (integrated over a small time window) exerted will be the momentum flux (integrated over that time window). Basically as many particles bounce off on one side as they do on the other... the net force is zero... the pressure is zero $\endgroup$ – Ian Oct 19 '11 at 5:42
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    $\begingroup$ Your wording is proposing a closed area integral - a Gaussian surface. The integral over any such surface for an idealized gas model should give zero, of course. If I put any object into a gas, I do not expect a net force (aside from buoyancy, drag, etc). In order to define "pressure" you need to take a once-sided integral, and only count those particles going in one direction, like the case of a boundary surface. If I have a jar in a vacuum with a gas in it, the basic idea of pressure is that I count the collisions on one side of the wall only. $\endgroup$ – Alan Rominger Oct 19 '11 at 6:19
  • $\begingroup$ I think Z is missing the more general definition of pressure: perpendicular momentum transported across an area. In the case where the area sits on the wall of the container, that definition can be restated as the usual "particles bouncing off of the wall" argument. But his question still stands in the case that the area is within the gas. An interstellar gas has no containing wall. Is pressure a sensible concept in this case? Of course it is! Is the pressure zero? Certainly not! I'll give an answer below... $\endgroup$ – garyp Oct 20 '11 at 1:43
  • $\begingroup$ The momentum flux is not zero. If it were, there would be no stress. @Zassounotsukushi: this is the definition of stress. $\endgroup$ – Ron Maimon Oct 20 '11 at 6:01

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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  • $\begingroup$ Somebody gets a kick from downvoting all my answers. This answer is the correct resolution of the confusion, regardless. $\endgroup$ – Ron Maimon Oct 20 '11 at 9:33
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    $\begingroup$ Well, I up-voted it. $\endgroup$ – garyp Oct 20 '11 at 12:17

[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.

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  • $\begingroup$ I don't believe I was missing that fact.... the test area is of course directed. Consider the test area to have unit normal $\hat{n} = (1,0,0)$. A sensible definition of the momentum flux would be $\vec{p} \cdot \hat{n}$. A particle traveling from the "left" would have momentum $\vec{p} = (mv ,0 ,0)$ and would contribute a flux $mv$. A particle traveling from the "right" would have momentum $\vec{p} = ( -mv ,0,0)$ and would contribute a flux of $-mv$. These cancel. $\endgroup$ – Ian Oct 20 '11 at 2:40
  • $\begingroup$ The source of my confusion was in how the Stress tensor (or Stress energy tensor) is defined. It turns out that it is defined such that given a test area one only considers the force that would be exerted on the area from the side opposite the direction of the unit normal. This means that the pressure inside the gas is identical to the pressure at the boundary. $\endgroup$ – Ian Oct 20 '11 at 3:50
  • $\begingroup$ Yes. (lesson learned: I posted an answer because I struggled with the same question some years ago, and came up with a somewhat satisfying answer. I forgot to make sure I correctly remembered the answer.) What I neglected was that in considering the momentum added by the two particles ... one from the right, the other from the left ... one has to ask how much momentum the two particles carry to the right (for instance). They both carry $mv$ to the right, giving $2mv$. I'll go out on another limb and suggest that the factor of two is important. But Ron's answer is better anyway. $\endgroup$ – garyp Oct 20 '11 at 12:45

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