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I'm trying to understand how it's possible to always reduce the degrees of freedom of a system represented by $n$ coordinates $q_{i}$ for which holds $f(q_{1},...,q_{n})=0.$

For example, let's consider the system represented by $x+\cos y=5$. How can i show that this system has only 1 degree of freedom?

Qmechanic
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AleWolf
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2 Answers2

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Your system is currently represented by two coordinates, $x$ and $y$. Without any constraints, any point $(x,y)$ on the plane could be a valid state of your system. But you have a constraint: valid states must obey the equation $x+\cos y=5$. This restricts the space of valid states for your system from a two-dimensional plane to a one-dimensional curve, specifically this curve:

enter image description here

Since your state space is constrained to be one-dimensional, you now only need one parameter to distinguish between different states. This parameter is your one degree of freedom. For example, you might choose to express the state of the system as solely a function of the single parameter $w$, where:

$$y=w$$ $$x=5-\cos w$$

probably_someone
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  • Thanks. How can i prove formally that it's possible to always reduce the degrees of freedom of a system represented by $n$ coordinates $q_{i}$ for which holds $f(q_{1},...,q_{n})=0.$ ? – AleWolf Nov 25 '19 at 12:17
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    @AleQuercia Simply solve the equation $f(q_1,...,q_n)=0$ for $q_1$, so that you have $q_1=g(q_2,...,q_n)$. (If for some reason such a solution is proven not to exist for $q_1$, simply solve for some other coordinate; if this is true for all coordinates, most likely you have an unphysical constraint. See Qmechanic's answer and link for formal requirements on constraints.) Since $q_1$ is completely determined by $q_2,...,q_n$, it's not a free parameter (in other words, simultaneously choosing an arbitrary $q_1$ and choosing $q_2,...,q_n$ would lead to a contradiction). – probably_someone Nov 25 '19 at 14:25
  • Let's consider a physical system X described by a set of coordinates, is it possible to say (informally) if the system X is holonomic or no only considering his physical behaviour ? For example a rolling disk on a plane picture – AleWolf Nov 25 '19 at 16:49
  • @AleQuercia What do you mean by the system's "physical behavior"? Are you saying that you for some reason have chosen coordinates but don't know the constraints your system is subject to? Because if you know the constraints, you have your answer: if the constraints are all expressible as $f(q_1,...,q_n,t)=0$ (note the absence of any time derivatives of the generalized coordinates), then the system is holonomic. – probably_someone Nov 25 '19 at 17:28
  • For example, the method shown in the first reply here . I don't understand how to apply it for every possible system – AleWolf Nov 25 '19 at 17:38
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    @AleQuercia The method shown there won't apply to every possible system. – probably_someone Nov 25 '19 at 17:39
  • Can you show me why ? – AleWolf Nov 25 '19 at 17:40
  • @AleQuercia For one thing, not every system has periodic coordinates (this is how $\theta$ and $\phi$ are eliminated in that argument). – probably_someone Nov 25 '19 at 17:43
  • Thanks so much :) – AleWolf Nov 25 '19 at 17:59
  • I have another doubt, if i take the system $cosx+cosy=5$ and i fix the $x$ then $cosy$ can assume infinite values. To me this system seems as holonomic as the rolling disk example, where if i fix $x$,$y$ and $\theta$ then $\phi$ can assume infinite values ... – AleWolf Nov 25 '19 at 18:09
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Technically, given a $3N$-dimensional position manifold, it is implicitly assumed that the zero-locus $$\bigcap_{\ell=1}^m\{f_{\ell}=0\}$$ for the $m$ independent holonomic constraints is an embedded submanifold, which necessarily have dimension $3N-m$. Hence the constrained system has $3N-m$ generalized coordinates/degrees of freedom.

See also this related Phys.SE post.

Qmechanic
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