Are there forces which do not involve a change in momentum?

I am familiar with the equation $$\vec{F}=m \vec{a}$$ I am wondering as to whether it is possible for something to exert a force on another object without changing the momentum of said object. My question is: is force defined as a change of momentum? If that is the case, my question is pointless. But if it is not. What is the underlying framework?

Also, when some phenomena causes the spins or angular momentum of subatomic particles to change, does it mean that the phenomena caused the particle to accelerate or decelerate? What is force in the framework of quantum mechanics? Does the definition of force only apply in classical mechanics (Sidenote: I am not very familiar with quantum mechanics. I apologize beforehand...)

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Do you want your frame to be inertial? If not then consider having some force exerted on you, from your own perspective your momentum won't change. (This can of course be explained away by fictitious forces.) –  user21433 Mar 6 '14 at 17:39

Actually Newton's second law is better stated as $$F=\frac{dp}{dt}$$ and this is even valid in relativity, both SR and GR, expressed in the right way $$f^\mu = \frac{dp^\mu}{d\tau}=m\frac{du^\mu}{d\tau} = m u^\nu\nabla_\nu u^\mu$$ (for massive particles) so classically forces are always imply a change in momentum. In QFT the concept of a force is no more useful and we talk about interactions but generally they also change momentum.

And answering to your question I believe a force is not defined as the change of the momentum but as the cause of that change.

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You can have multiple forces exerted on an object that add to zero. Then there will be no momentum change. Think of the two of us leaning against opposite sides of a door with the same force. The door does not change momentum, nor does either of us. I am exerting a force on my chair as I sit here.

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IMHO this is the most appropriate answer, although the others give good insight into the general force equations. –  Carl Witthoft Mar 6 '14 at 18:48
I think that in this case we would say that, while there may be multiple forces acting on an object, or even that there are multiple forces exerted on the object, that there is no net force exerted on the object. –  Bob Gilmore Mar 6 '14 at 20:26
@BobGilmore I've always been Intrigued by this idea of a completely inanimate object, suddenly applying force to something. When I sit down on my chair, why does the force start applying force now, and not before? It feels almost arbitrary. I mean I know it's the case, and that its a law, I just don't understand why its so necessary –  Cruncher Mar 6 '14 at 21:25
I'm not entirely sure this is appropriate just because the net force of the objects in your example is zero. Individually, each of the forces would cause a change in momentum, so it's not as if they have the special property of breaking Newton's second law. –  EtaZetaTheta Mar 6 '14 at 21:35
@Cruncher The chair does undergo a minute change when it begins to exert upward force - it's being squished under your weight, like a spring. –  Brilliand Mar 6 '14 at 21:40

No, all forces involve a change in momentum.

In classical mechanics force is defined as a change in momentum.

In quantum field theory particles interact via exchanging one or more bosons (see feyman diagrams). These bosons always have momentum and therefore the momentum of the interacting particles changes as well.

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All unbalanced forces - c.f. Ross Millikan's answer. –  Mister Mystère Mar 6 '14 at 21:36

A universal relation is that the force exerted on an object equals the time derivative of momentum. No force, no momentum change, vice versa.

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Not exactly what you were asking for, but any force exerted perpendicular to the direction of motion does not change the magnitude of momentum -- though it does change the direction.

Two examples are the force exerted by a uniform magnetic field on a moving charged particle, and the force of gravity on a satellite in a perfectly circular orbit.

But momentum is inherently a vector quantity, so this does not really answer your question.

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Absolutely not true! As you add velocity in a perpendicular direction the magnitude of the momentum will change. Momentum is a vector. –  kleineg Mar 7 '14 at 16:04
I don't understand why the -1. I explicitly said in my answer that "momentum is inherently a vector quantity". Also, I had been led to believe that in order to have uniform circular motion (constant magnitude of momentum but varying direction) there had to be a force perpendicular to the direction of motion -- can you please explain how that's wrong? –  Snowbody Mar 12 '14 at 21:30
In uniform circular motion, since the magnitude of the momentum is constant we know that as momentum increases in one direction it must decrease in another. So in this case the force is not perpendicular but actually includes a portion of the force in a direction opposite the motion, and the direction of the force must change constantly. If you add a perpendicular force the magnitude of the momentum will increase, and since the force stays perpendicular it will keep increasing the momentum in that direction. –  kleineg Mar 13 '14 at 12:45
Okay, I see the assumptions you're making, and I think I know why you're wrong. You're breaking up the momentum into two perpendicular components, neither of which is guaranteed to be parallel or perpendicular to the current direction of motion. Both the x and y components of momentum are changing, so the force must have both x and y components. But if you do the math, you'll discover that instantaneously, the force to maintain uniform circular motion is always perpendicular to the instantaneous velocity. I'm not talking about "adding" a perpendicular force; I'm talking about maintaining. –  Snowbody Mar 16 '14 at 4:11
@kleineg The answer is right. An easy way to see this is noting that such a force doesn't do work. Thus the kinetic energy doesn't change, so speed remains constant, the same applies to magnitude of momentum. –  Ruslan Dec 24 '14 at 12:58

As I interpret your question, there is one (and only) situation where "forces do not involve a change in momentum" and that is when the vector sum of the forces is zero. However, if you meant "net force," then the answer is no.

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