2

So I'm doing some work from Taylor's mechanics book. He says for the problems in the book, we require the system to be holonomic - that is the number of generalized coordinates = number of Deg. of freedom. Why does this have to be the case?

I have been looking through his proof for a single particle, where he proves the Lagrangian, for the correct path taken by the particle, minimizes the action integral, but he doesn't say 'for this step in the proof to be true, we require the system to be holonomic'.

So why does this feature have to be true?

Qmechanic
  • 220,844
milanios
  • 85

1 Answers1

3

  1. Actually, that the constraints are holonomic is not always sufficient to obtain a Lagrangian formulation. E.g. there could still be sliding friction.

  2. What is needed in the derivation of Lagrange equation's from Newton's laws is d'Alembert's principle, which we will write in the form$^1$ $$\sum_{i=1}^N {\bf F}^{(c)}_i\cdot \delta {\bf r}_i~=~0, \tag{1} $$ cf. Ref. 1, i.e. that the total virtual work of the constraint forces ${\bf F}^{(c)}_i$ on $N$ point particles at positions ${\bf r}_1,\ldots,{\bf r}_N, $ is zero.

  3. It is possible to show that broad classes of constraint forces of holonomic types satisfy d'Alembert's principle, see e.g. this Phys.SE post and links therein.

  4. Concerning OP's specific question about where Ref. 1 implicitly uses the assumption of only holonomic constraints, it is in eq. (7.49), which is essentially eq. (1).

References:

  1. J.R. Taylor, Classical Mechanics, 2005; eq. (7.49).

--

$^1$It is tempting to call eq. (1) the Principle of virtual work, but strictly speaking, the principle of virtual work is just d'Alembert's principle for a static system.

Qmechanic
  • 220,844