Can internal forces do work?

Work is a subtle concept that can be approached in various ways. There are ambiguities that can be reduced somewhat through the careful use of language. Ultimately, however, these ambiguities can only be eliminated by specifying a theory of physics and stating the definition within that theory.

Suppose our theory is Newton's laws applied to frictionless gear trains. Then the forces are normal forces exerted by a gear tooth A on another gear tooth B. By Newton's third law, the forces are equal in magnitude and opposite in direction. Since these are normal forces, the displacements of A and B are equal. Work can be defined either as $dW=F\cdot dx$ or as the transfer of energy by a mechanical force; since these are normal forces, the two definitions are equivalent. By Newton's third law, A's work on B cancels B's work on A.

Suppose instead that our theory is Maxwell's theory applied to oppositely charged particles A and B moving in a vacuum. We release A and B at some distance from one another. Their subsequent motion is that they oscillate about their common center of mass, radiating electromagnetic waves (and periodically passing through each other). There is clearly no way we can define work so as to make A's work on B always cancel B's work on A. Energy is going out into the electromagnetic waves, and there is no way to count this energy into the definition of work, since it's nonmechanical.

Well, I think the issue must be that all those references that claim that internal forces don't do work must refer to RIGID BODIES or PARTICLES. If that's the case, they're right. But in general, for deformable bodies, that's not true and your textbook is right.

• In a rigid body, all the external work done to the body is transformed into kinetic energy. The work of the internal forces is zero, since any couple of internal forces will act in opposite directions (the former fact is known as the work-energy theorem)

• In a general body (think of continuum mechanics), part of the external work will be done to compensate for some internal work, and not all of it will become kinetic energy. For simplicity, let's assume that the internal forces come from a potential and let's see an example:

Let our system be a spring that is compressed by an external force F, until it reaches a steady position. There is an external work on the system because the force creates a displacement, but in the end the spring is steady, so there is no increase in the kinetic energy between the initial and final states. Where did all this energy go? It compensated the work of the internal forces of the spring. Because these internal forces derive from a potential, the work of the internal forces amounts exactly the change in the potential.

In general we have, with usual sign conventions (but that changes a lot, it is better to think carefully on the problem than to just substitute) $$ - W_e = dK - W_i = dK + dU $$ Where $W_e$ is the external work, $W_i$ the internal, $dK$ the change in kinetic energy and $dU$ the change in the internal energy, that in the example is just the change in the potential.