EDIT: Judging by the low number of upvotes, I guess not many people are convinced by the logical argument I presented earlier, so I have decided to present a case based on calculations instead:
Calculation: Pressure is force per unit area (F/A
). Using the definition of force in terms of momentum we can write the equation for pressure as:
$$ P = \frac{\Delta p}{\Delta t \cdot A} = \frac{m \Delta v}{A \Delta t} $$
Considering just the right hand wall of the vessel with area A and with velocity to the right defined as positive, the pressure on the right hand wall of the vessel when stationary is:
$$ P = \frac{N m \Delta v}{A \Delta t} = \frac{N m (v_0 - v_1)}{A \Delta t} $$
where N is the number of particles that collide with the wall in a given time interval, $v_0$ is the velocity of a given particle before it impacts the wall and $v_1$ is the velocity of the particle after the collision. Assuming elastic collisions, the velocity after the impact is equal in magnitude and opposite in direction to the velocity before the collision, so $v_1=−v_0$ and the above equation can be written as:
$$ P = \frac{N m (v_0 - v_1)}{A \Delta t} = \frac{N m (v_0 + v_0)}{A \Delta t} = \frac{N m (2 v_0)}{A \Delta t} \tag{1}$$
Now if we consider the situation when the vessel is moving with constant velocity u to the right, the calculation becomes:
$$ P = \frac{N m ((v_0 + u) - (v_1 + u))}{A \Delta t} = \frac{N m (v_0 + u + v_0 - u)}{A \Delta t} = \frac{N m (2 v_0)}{A \Delta t}, $$
which is the same as equation (1) so the motion of the vessel has had no effect on the pressure on the right hand wall. All the same applies to any other wall of the vessel, so we can conclude that relative motion of a vessel containing a gas under pressure has no effect on the pressure in the vessel in any reference frame. In other words, gas pressure can be considered an invariant.
which is the same as equation (1) so the motion of the vessel has had no effect on the pressure on the right hand wall. All the same applies to any other wall of the vessel, so we can conclude that relative motion of a vessel containing a gas under pressure has no effect on the pressure in the vessel in any reference frame. In other words, gas pressure can be considered an invariant.
Logical argument: Imagine the two equal containers have flat sides that slide alongside each other. Each container has a hole in its sliding side that is sealed by the surface of the other container most of the time. When the holes coincide, gas can move from one container to the other if there is a pressure difference between the two containers. Which way does the gas move?
POV 1) If in frame A, we say container A is stationary and container B is moving and has greater pressure due to its movement, we would have a net flow from container B to container A, and we would expect to end up with more gas molecules in container A.
POV 2) In frame B, the observer considers container B to be stationary and container A to be moving. If movement causes the pressure to increase, in this reference frame, we would expect a net gas flow from A to B and to end up with more gas molecules in container B.
These two points of view contradict each other and obviously cannot happen at the same time. Both containers cannot end up with more gas molecules than the other. That is a physical impossibility. If the movement of a container causes the pressure to change, we end up with a contradiction, so the logical conclusion is that the pressure does not change with relative velocity.
Relativistic calculation: A naïve analysis of the situation in Special Relativity would suggest the moving container should length contract and so the pressure in the moving container should increase, but that is not the correct conclusion.
In SR the Lorentz transformations of force are:
$F'_{\parallel} = F$ and
$F'_{\perp} = F\gamma^{-1}$
The Lorentz transformations of area are:
$A'_{\parallel} = A$ and
$A'_{\perp} = A\gamma^{-1}$
where the $\parallel$ and $\perp$ subscripts indicate parallel and perpendicular measurements respectively and the unprimed quantities are the proper measurements when the vessel is at rest with respect to the observer. It follows that the pressures measured when the vessel is moving relative to the observer are:
$$P'_{\parallel} = \frac{F'_{\parallel}}{A'_{\parallel}} = \frac{F}{A} = P$$ and
$$P'_{\perp} = \frac{F'_{\perp}}{A'_{\perp}} = \frac{F\gamma^{-1}}{A_{\perp}\gamma^{-1}} = \frac{F}{A} = P$$
so it turns out even in Special Relativity, the relative motion has no effect on the pressure on the walls of the container in any direction.
We could do more involved calculations involving the Lorentz transformation of momentum and relativistic velocity addition etc. but we still end up with the same result. Gas pressure is invariant.
Also, take another case in which the container, instead of moving with
constant velocity, is moving with constant acceleration. In that case,
what will be the relation between the pressures observed in the two
containers?
In this case, we end up with a pressure and density gradient in the accelerating container, while the pressure and density in the stationary container remain uniform. Unlike the purely relative inertial case, we can always tell the difference between a frame with acceleration and an inertial frame.
