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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: $f=f(q_i,\dot q_i,t)$.)

Qmechanic
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Vebjorn
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1 Answers1

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  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.

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

Qmechanic
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  • I have Pearson New International edition Third edition from 2014. But the last correction made on the errata homepage was in 2010. ( http://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. – Vebjorn Sep 13 '22 at 06:21
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    Eq. (2.26) is still wrong/misleading at best. – Qmechanic Sep 13 '22 at 14:26