By switching to a different set of coordinates, can you make problem with semi holonomic constraints into a problem with holonomic constraints? If so, then when can you do this? I wold like to know if this is possible for all semi-holonomic problems, some semi-holonomic problems or no semi-holonomic problems.

My intuition is that it probably works for some specific cases, but not in general. However, I don't know the reason why. Would be very nice with some sort of proof.

(Constraints on the form: $f=f(q_i,\dot q_i,t)$.)


1 Answer 1

  1. A non-holonomic$^1$ constraint is by definition a constraint that is not holonomic, e.g. on the form $f(q,\dot{q},t)=0$ or an inequality.

  2. A semi-holonomic constraint $$ \omega~\equiv~\sum_{j=1}^na_j(q,t)~\mathrm{d}q^j+a_0(q,t)\mathrm{d}t~=~0 $$ is equivalent to a holonomic constraint iff there exist an integrating factor $\lambda(q,t)\neq 0$ and a one-form $\eta$ such that $$ \lambda\omega+ f\eta~\equiv~\mathrm{d}f ,$$ cf. my related Phys.SE answer here.


$^1$ If you are using the 3rd edition of Goldstein, be aware of erratum.

  • $\begingroup$ I have Pearson New International edition Third edition from 2014. But the last correction made on the errata homepage was in 2010. ( astro.physics.sc.edu/Goldstein ) I checked one of the revisions ( a $\delta$ corrected to $\partial$) and it was indeed corrected in my edition. Guess it is updated then. $\endgroup$
    – Vebjorn
    Sep 13, 2022 at 6:21
  • 1
    $\begingroup$ Eq. (2.26) is still wrong/misleading at best. $\endgroup$
    – Qmechanic
    Sep 13, 2022 at 14:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.