Are rigid body collisions elastic or inelastic?

Question

Suppose two rigid bodies collide head-on in a vacuum. Will the collision be elastic or inelastic?

In most rigid body computer systems you have to specificy a coefficient of restitution for the collision, so the answer would depend on that. But this is a post-hoc mechanism used so we can better model real world systems; it's not a natural consequence of our assumption that bodies are rigid.

For elastic collisions in the real world bodies would deform the relax over some amount of time which provides the mechanism for impulses to change the directions of the bodies. But in a world of only rigid bodies they can't do that.

If collisions are inelastic, I can't see a plausible mechanism that would absorb the excess energy. The bodies can't deform by definition, so it can't be that. We could say the bodies warm up, but temperature is not normally included as a property in rigid body systems and I'm not even sure if it would be a well defined property in such a system without introducing a whole slew of other properties and phyical laws.

Perhaps the rigid body assumption violates conservation of energy? But that doesn't seem right since conservation of energy comes from time symmetry (from Noether's Theorem), which should still hold for a rigid body system.

Perhaps the result of collisions between rigid bodies is ill-defined because the concept of rigid bodies is non-physical? But there are lots of non-physical simplifications used in physics and mathematics that are still self-consistent. Maybe what happens when two bodies collide is another assumption you have to explicitly make for the system to be self-consistent?

Lots of questions but I'm not sure how to approach looking for an answer.

Answer 1

Suppose two rigid bodies collide head-on in a vacuum. Will the collision be elastic or inelastic?

A collision between two rigid bodies is elastic. A rigid body is an idealization of a body that does not deform or change shape. A inelastic collision always involves some degree of deformation. At the macroscopic level, there is no such thing as a rigid body. Consequently, at the macroscopic level all collisions are inelastic.

Hope this helps.

Answer 2

The rigid body model is quite useful to model many interesting physical systems, and, for example, robotics heavily hinges on such a model.

The rigidity constraint simply means that all distances of the body parts keep constant during the motion. The physical consequence is that many dissipation processes, like the transformation of motional kinetic energy into sound waves or plastic deformations, can be excluded and disappear as possible mechanisms of inelasticity.

Does the rigid body dynamics is equivalent to a constraint of elasticity? Not really. The lack of macroscopic deformations is not enough to exclude the possibility of friction when two rigid body surfaces get in contact.

In such a case, friction forces may do work and dissipate macroscopic kinetic energy. For this reason, when dealing with rigid body dynamics, people speak about smooth frictionless surfaces when the elastic dynamics model should be enforced in the model.

Notice that, in a way, the presence of friction in rigid body interactions may be considered a matter of definition. Microscopic models for friction imply microscopic atomic dynamics. Therefore, if the rigidity constraint is maintained down to the atomic scale, one could argue that friction should be excluded. However, the way people work in the field of modeling the motion of macroscopic bodies is to assume a continuum mechanics approach: the rigidity constraint applies to small regions, but not so small to break a continuous media description in terms of strain-stress tensors. Therefore, there is room for friction without disrupting the rigid body idealization.

Answer 3

"But there are lots of non-physical simplifications used in physics and mathematics that are still self-consistent"

Simplification of that sort is called modeling, or abstraction. Any abstraction "forgets about" (ignores) details that are deemed not relevant. In physics, these abstractions model aspects of the real world. So while an abstraction should be well-behaved within the context of the formal system it's embedded in, when you map it back to the real world, if the assumption that these "forgotten" details can be ignored for your purposes is incorrect, then the model isn't applicable (at least, not without modification), or doesn't make sense.

What I'm saying is that the rigid body model explicitly ignores any kind of elastic deformation (though not some of the other effects associated with it) and assumes that there is no observable/relevant/significant plastic deformation (and no disintegration). The exact mechanism of the collision is not part of the model, there's just an abstract parameter that influences the outcome. The rigid body can just do that - exactly how is outside of its scope.

So when you start looking for a microscopic mechanism while still clinging to the rigid body abstraction, you break these assumptions and run into contradictions.

"But this is a post-hoc mechanism used so we can better model real world systems; it's not a natural consequence of our assumption that bodies are rigid"

Yeah! It's a part of the particular abstraction you're working with, put in there so that it captures an aspect of the real world. It's not something that's a derived consequence, it's just a given. It's part of the bedrock (in this particular context). In other words, as soon as you remove that, you have a model different from the one you started with, and one that is less suitable to represent the situations you're trying to understand.