So I will preface this by saying the following is simply a thought experiment. However, in my field, I have heard seemingly legitimate arguments for different answers.

The thought experiment goes as follows:

Suppose we have a closed shock tube in the vacuum of space and free of any gravitational or frictional forces. As many of you are probably aware, the shock tube has a high pressure side termed the driver section (state 4), and a low pressure side termed the driven section (state 1). The two sides are initially separated by a diaphragm until at a given instance, the diaphragm is ruptured. A shock wave begins propagating into the driven section and simultaneously a rarefaction begins propagating into the driver section. The post-shock and rarefaction state are denoted by states 2 and 3 in the schematic below.

enter image description here

Now the physical value of these states can be determined in any classical text in gas dynamics, however, for the purposes of this thought experiment they are unimportant. As the gas goes into motion within the shock tube, a series of shock, rarefaction, and contact surface reflections and interactions will take place until eventually the gas inside the tube becomes stationary and reaches a spatially uniform equilibrium. The question then becomes, after all of the internal motion of the gas has resided, does the shock tube itself experience any net acceleration in any direction?

A control volume analysis of the shock tube system would indicate that the external forces acting on the control volume are defined by,

$$ \sum F_{ext} = \frac{\partial}{\partial t} \int_V \rho \vec{V} \ dV + \int_S \rho \vec{V} (\vec{V} \cdot \hat{n}) \ dS$$

However, since the system is closed, there is no momentum flux through the surrounding control volume and we are simply left with,

$$ \sum F_{ext} = \frac{\partial}{\partial t} \int_V \rho \vec{V} \ dV $$

Now this would be the force on the control volume, and from the dynamics of the fluid, it is obvious that there will be instances during the gas motion for which $\frac{\partial}{\partial t} \int_V \rho \vec{V} \ dV \neq 0 $. This is usually enough for some people, to which they claim, obviously the system will experience some form of a net acceleration in a given direction. Then the second crowd will come along and suggests, it is impossible for the shock tube system to experience any net acceleration simply because all of the motion from the gas within the shock tube acts as internal forces, and only external forces can provide a net acceleration to the system.

So I am curious, what are your thoughts as to which is the correct answer?


1 Answer 1


It is true that in the absence of external forces, the center of mass of a system does not change position. That said, components of the system can shift around all they want, assuming that the center of mass stays in the same place. In this case, the gas is initially distributed unevenly, as higher-pressure (meaning higher-density) gas is on one side of the wall than the other. Hence, the center of mass of the system will not be in the geometric center.

However, once the gas reaches equilibrium, being distributed evenly, the center of mass should be at the geometric center of the system (as the system is now completely symmetric across its midline). Since the center of mass's position relative to an outside observer does not change, then the system must accelerate and decelerate in such a way that the geometric center of the system occupies the original center of mass's position once the gas reaches equilibrium.

  • $\begingroup$ So I suppose if the gasses were selected such that the system had mass evenly distributed initially (say high pressure helium on one side and low pressure air on the other), then it is technically possible to have the system accelerate and then decelerate back to its original position. That is an interesting result. $\endgroup$
    – TRF
    Dec 30, 2016 at 5:01
  • 1
    $\begingroup$ That is exactly correct. The system can only ultimately move as far as the displacement of its center of mass, as seen in the frame moving with the object. $\endgroup$ Dec 30, 2016 at 5:02
  • $\begingroup$ As a little anecdote, as a student I once visited an installation that operated a very large shock tube. As I remember it had a diameter of the order of maybe five or six feet, and was at least 60 feet long (perhaps longer, this was many years ago). The whole tube was mounted on rails, and the engineers running the thing told us that the whole tube was moving by more than a foot during each "shot". $\endgroup$
    – Pirx
    Dec 30, 2016 at 17:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.