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All explanations of jet propulsion that I've seen are formulated as "due to conservation of momentum, air with momentum coming out of one end means the rocket must gain momentum in the opposing direction".

However it can't be the case that particles simply exiting some area with a momentum means all of a sudden something else must gain opposing momentum. If we had a straight pipe open on both ends, particles suddenly exiting one side (maybe a chemical exploding inside or the like) does not mean the pipe will all of a sudden start moving in the opposing direction. You need one side of the pipe to be closed in order for the pipe to move opposingly. Similarly it can't be the case that particles suddenly exiting an inflated balloon or rocket means the object must propel in the opposing direction due to some 'rule'.

I'm assuming what is actually happening is a case of 'particle collision averages'. In the case of a balloon the air inside is constantly bouncing off the walls in all direction. So when I let go of an inflated balloon I have disturbed this average. There are now particles hitting a side of the balloon with no opposing force, as the particles opposing that wall are now simply exiting the balloon.

Assuming my above formulation is correct, does this mean all statements regarding conservation of momentum are simply a rule of thumb for this above 'particle collisions average' explanation?

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    $\begingroup$ Does this answer your question? Conservation of Momentum $\endgroup$
    – Farcher
    Commented Jun 6, 2023 at 7:46
  • $\begingroup$ One issue is that you are conflating "How does conservation of momentum work?" with "What makes a solid object behave like a solid object?" $\endgroup$
    – RBarryYoung
    Commented Jun 6, 2023 at 14:03
  • $\begingroup$ When you say "it can't be the case....," in fact that is exactly the case. In the open pipe example, the pipe may not move, but the gas particles must exit both sides and not just one side. In order for any of the particles or objects to acquire net motion they must "push off" something else in the opposite direction. $\endgroup$
    – RC_23
    Commented Jun 7, 2023 at 23:20
  • $\begingroup$ "However it can't be the case that particles simply exiting some area with a momentum means all of a sudden something else must gain opposing momentum." It's not the exiting with momentum that is important here. This phrasing implies that momentum has already been imparted. It's the imparting of the momentum (which will eventually cause the exit) that is the part you should focus on. $\endgroup$
    – Flater
    Commented Jun 7, 2023 at 23:25

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An ion thruster works by accelerating particles using an electromagnetic field. No collisions are necessary.

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  • $\begingroup$ I am actually not sure how to reconcile my explanation with an ion thruster, I was actually looking to see if there was some type of thruster like the ion thruster but hadn't found one. I'll think on it more $\endgroup$
    – one two
    Commented Jun 6, 2023 at 0:32
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    $\begingroup$ The is still an interaction - a collision at a distance. $\endgroup$
    – Farcher
    Commented Jun 6, 2023 at 7:42
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    $\begingroup$ @Farcher All collisions are at a distance $\endgroup$
    – Some Guy
    Commented Jun 6, 2023 at 17:30
  • $\begingroup$ Since all non-nuclear "collisions" are accelerations due to electromagnetic interactions you may as well consider the quite literally "particular" electromagnetic interaction of the ions with the accelerating field a large-scale collision. If it was an ion trap and not open on one side, the "collisions" would cancel each other out very much like in a conflagration in a closed chamber without the nozzle exit. $\endgroup$
    – Peter - Reinstate Monica
    Commented Jun 6, 2023 at 22:37
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    $\begingroup$ @Peter-ReinstateMonica The repulsive force in non-nuclear collisions is the Pauli force, not electromagnetic. $\endgroup$
    – John Doty
    Commented Jun 6, 2023 at 22:45
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The key part of how propulsion of any kind works is that there must be a non-zero momentum flux. That is, the momentum of that which enters can not be the same as that which leaves if you want the apparatus itself to gain any momentum from the operation.

The amazing thing about the conservation of momentum is that this is true for any apparatus you can imagine, regardless of how it internally functions. It doesn't matter if there are particles bouncing off the walls or being accelerated by fields - all that actually matters is that there is a non-zero momentum flux through the apparatus.

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The conservation of momentum is not a new physical force that you introduce alongside the ideas you already understand of particles whacking into one another. It is a zoomed out way of describing those same whacks without worrying about the specifics.

You are certainly correct that the particle collisions are playing an important role, just like your balloon example. We can imagine a situation where at t=0 we have a straight segment of pipe floating in space with a bullet passing through the opening (not touching the walls). The pipe does not change speed at all as the bullet exits it.

With the jet the air gets sucked in at a low speed, and it exits at a much higher speed. So we know the air changed velocity, from Newton we know a force was needed to do that. Also from Newton we know their must have been an equal and opposite force on something else. The only other thing is the engine itself, and we get the answer. The conversation of momentum is just a cleaner and faster way of giving that Newtonian argument. Its the interaction force between the two things (air and engine) that is important, whether the air is inside the engine at any particular time is not relevant.

The application of the conservation of momentum principle to this engine is indeed a rule of thumb, because it is carrying several implicit assumptions, for example that the air was not moving (or at least not so fast) when it went into the engine. Its also assuming that the acceleration actually came from the engine, not from some other device that imposed a force at a distance on the air (eg. electrical or gravitational) when the air just happened to be inside the engine.

However, more the specific statement "momentum is conserved" is not a rule of thumb, it is a proper rule that (in current physics) has no exceptions.

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Conservation of momentum still works when only non-contact forces are at play - clearly, these scenarios cannot rely on particle collisions, since there is no physical contact between objects whatsoever. Momentum may be conserved in the case of gravity or electromagnetic forces, without requiring any particles to physically "touch" at all.

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This is an interesting question. I guess there are several steps to point out here:

  • First, it's not just about bouncing particles, since non-contact forces like electromagnetic forces also conserve momentum. (Then again, in QFT you model forces with exchange particles. Those don't bounce but are emitted and absorbed, but the general idea is similar. So you may think of those as generalised bouncing.)
  • But the important point is that all (bouncing or otherwise) interactions do conserve momentum. That's a rather fundamental property of nature. Note that it's not a rule of thumb, and is not only true on average.
  • Thus, if you see some ejected air, you can of course try to look at the individual bounces and average those. That can be instructive, because you actually get a better understanding of the fundamental processes, and it's presumably necessary to see what's going on where in the balloon or rocket or whatnot.
  • However, if you want to know the thrust of the balloon, there is a shortcut: Momentum is conserved, so no matter what the individual bounces or electromagnetic interactions do, the only way to eject something at one end with some momentum is to impart the opposite momentum to something else, and that can only be the balloon in this case.
  • So in these cases, momentum conservation really is a result of many small bounces which individually conserve momentum.
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However it can't be the case that particles simply exiting some area with a momentum means all of a sudden something else must gain opposing momentum.

Why not?

If we had a straight pipe open on both ends, particles suddenly exiting one side (maybe a chemical exploding inside or the like) does not mean the pipe will all of a sudden start moving in the opposing direction. You need one side of the pipe to be closed in order for the pipe to move opposingly.

You just need there to be imbalance between exiting one side and the other. Consider that the explosion inside the pipe is very near one end and far from the other end, then it can happen that the gas exiting one end is much more than the gas exiting the other end. Then the pipe will move. It can be as simple as that there is viscosity between the air and the pipe, that pulls the pipe to move the other direction.

I'm assuming what is actually happening is a case of 'particle collision averages'.

The correct and proper idea is actually not that far off, so I think it is better for you to just learn the fully correct version: Momentum is simply a conserved quantity, and if you had an arbitrarily defined system that initially had a certain amount of momentum, and gained or lost some certain another amount of momentum, then the final amount of momentum must be the sum of them.

A consequence of this is that, if you started with zero average momentum, then the centre of mass of the system must be fixed. If any one part of it moves a certain direction by a certain amount, then the rest of the system must move the same amount backwards the other direction. If the average momentum is non-zero, it is the same behaviour but with additional translation of the average momentum.

Your thing with the walls obviously increased the effect because you forbid some motion, leading to even more lopsidedness in the motion. But it is not necessary to have walls to get this behaviour.

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  • $\begingroup$ I see what you mean by the explosion being near one end. However the pipe would move like you said due to viscous forces etc. If the pipe was in a vacuum I don't see why the open ended pipe would move. Which is what I mean when I say "is it all reducible to particle collisions", because even in your viscous forces case its due to particles colliding and pushing, but its the air. $\endgroup$
    – one two
    Commented Jun 6, 2023 at 0:30
  • $\begingroup$ Well, that is because you have only considered scenarios whereby random motion being suddenly restricted, is the source of motion. What I am trying to impress upon you is that any transfer of momentum is sufficient, and the motion of any part of a whole, by conservation of the location of the centre of mass, is sufficient to move the rest. Ion thrusters is one example of something completely else causing the motion. $\endgroup$
    – naturallyInconsistent
    Commented Jun 6, 2023 at 0:35
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No

If you were aboard the ISS, and you grabbed a fellow astronaut with roughly the same mass, then pushed them away, what do you think will happen to the two of you, and why?

Hopefully, you will not say: "Well, there were no collisions involved, so nothing will happen. We will just sit there happily." Instead, you should say the obvious: "We will go flying in opposite directions, at roughly equal speed, just like momentum conservation laws tell us."

Suppose you are standing on a frozen lake wearing hockey skates, and you have a spring-loaded gun that launches heavy bricks wherever you aim. If you point that straight in front of you and pull the trigger, what do you expect to happen? Hopefully, you will predict that it will push you backwards at the same time the gun launches the projectile forwards. Is that a "collision"? If you want to call it that, sure.

Now, what if instead of shooting bricks, suppose it shoots ping-pong balls. Will that change the ultimate outcome? And then what if we change the ping-pong balls to peas? Will you stop moving backwards from the recoil? Will momentum stop being conserved when you reduce the peas to dust grains? To air molecules?

If you were floating in space, with your oxygen supply about to run out, but a space station is located just a few hundred meters away, stationary relative to you, and all you have loose is a wrench, what are you going to do? Are you going to hit yourself with the wrench as hard as you can so that it bounces off and sends you towards the station? Or will you just, you know...throw it in the opposite direction?

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