3
$\begingroup$

Wikipedia and other sources define holonomic constraints as a function

$$ f(\vec{r}_1, \ldots, \vec{r}_N, t) \equiv 0, $$

and says the number of degrees of freedom in a system is reduced by the number of independent holonomic constraints.

I could take multiple such constraints $f_1, \ldots, f_m$ and formulate them as single one that is fulfilled if and only if all $f_i$ are fulfilled:

$$ f = \sum_{i=1}^{m}{\lvert f_i \rvert}. $$

This combined $f$ would obviously reduce the number of degrees of freedom by $m$ instead of $1$.

Alternatively, to avoid the absolute value, I could use a sum of squares

$$ f = \sum_{i=1}^{m} f_i^2 $$

instead. Where is my error in reasoning?

$\endgroup$
4
$\begingroup$

Well, in the definition of holonomic constraints $f_1, \ldots, f_m$, there are also two technical regularity conditions (which OP's counterexamples do not fulfill):

  1. The functions $f_1, \ldots, f_m,$ should be continuously differentiable with $m\leq 3N$.

  2. The $m\times 3N$ rectangular Jacobian matrix $$\frac{\partial(f_1, \ldots, f_m)}{\partial(\vec{r}_1, \ldots, \vec{r}_N)}$$ should have rank $m$.

The regularity conditions 1 & 2 are imposed to ensure the local existence of generalized coordinates $q_1, \ldots, q_n$, in some open neighborhood, where $n:=3N-m$, via the inverse function theorem.

See also this related Phys.SE post.

References:

  1. M. Henneaux & C. Teitelboim, Quantization of Gauge Systems, 1994; Subsection 1.1.2, p. 7.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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