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I am reading Goldstein's Classical Mechanics book, and I came across that:

In a rigid body the internal forces do no work

Is this statement based on the assumption that the internal forces are central? Or, is it true even when the internal forces not central?

Goldstein defines a rigid body as a system of particles in which the distance between the particles remain constant, and does not require that the internal forces to be central.

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  • $\begingroup$ If the distance between the particles remains constant, how could there be any internal work? $\endgroup$
    – garyp
    Feb 23, 2019 at 14:32
  • $\begingroup$ On pg 56, in Appendix A of "Physically Based Modeling - Rigid Body Simulation" by David Baraff of Pixar: [pdfs.semanticscholar.org/902c/… he gives a derivation which he describes as follows: "The derivation in this appendix is (we feel) much shorter and considerably more elegant than the one found in traditional sources such as Goldstein." I am finding these notes by Baraff very useful. $\endgroup$
    – Simon
    Feb 23, 2019 at 14:45

3 Answers 3

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Recall the difference between the weak and strong Newton's third law, cf. e.g. this Phys.SE post.

  1. If the internal forces satisfy the weak Newton's third law (but not the strong Newton's third law, i.e. without the collinarity assumption), then it is not guaranteed that the internal forces do no work, cf. e.g. Fig. 1.

    ^ F
    |
    |                  2
    x------------------x
    1                  |
                       |
                       v F
    

    $\uparrow$ Fig. 1: A rigid body consisting of 2 point-particles with a pair of non-collinear internal forces $F$. Note that there in principle could be other internal forces holding the 2 point-particles together.

  2. If the internal forces satisfy the strong Newton's third law, then the internal forces do no work, cf. e.g. this Phys.SE post.

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  • $\begingroup$ Then, the definition of a rigid body as used in Goldstein's assumes that the strong Newton's third law holds for internal forces. $\endgroup$
    – user74261
    Jul 22, 2016 at 18:49
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    $\begingroup$ $\uparrow$ Yes, apparently. $\endgroup$
    – Qmechanic
    Jul 22, 2016 at 18:51
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In a rigid body, according to Goldstein's definition, the distance between any two constituent particles does not change. Work done is force times distance moved in the direction of the force. There is no relative movement in the direction of any force. Therefore, regardless of the form of the internal forces, no work is done by or against them.

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As @sammy gerbil says, for a rigid body the internal forces do no work since the distance between constituent particles is fixed.

Also, since the distance between particles is fixed for a rigid body, the internal energy is constant and there can be no "heating" effects. This is why physics mechanics texts can evaluate the work done by friction on a rigid body as affecting only the kinetic energy of the body, not "heating" the body.

For a system of particles in general (not rigid) the internal forces do no work only if Newton's third law holds. And the internal energy of the system is not necessarily constant. The work/energy development of physics mechanics must be extended to the first law of thermodynamics to consider heat and change in internal energy.

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