Suppose a point particle is constrained to move on the curve $y=x^2$. This would then be a non-holonomic geometric constraint since the particle has one degree of freedom and requires two coordinates ($x$ and $y$) to describe its position fully (clarification about aforementioned statement is needed). My question is (if) this constraint is non-holonomic, how come $y$ can be expressed in terms of $x$ as a finite relation (i.e. $y=x^2$)?

  • $\begingroup$ Hi Sumukh Atreya. Welcome to Physics.SE. Here, we use an unique TeX markup called MathJax, same as Math.SE. The markup is very much helpful in understanding equations, etc. Please have a look here for an introductory, or atleast have a look at our FAQ for an overview. For now, I'll help revising your post. $\endgroup$ Jan 19, 2013 at 16:03
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    $\begingroup$ It seems hard to deal with the question because it is based on two major propositions. If one of them were wrong, it would already be bad - but all of them seem to be wrong. First, if $y=x^2$, it's enough to know $x$ to determine the exact position $(x,y)$. We don't need two positions. Second of all, $y=x^2$ doesn't give a nonholonomic constraint. A non-holonomic constraint is written in terms of differentials and this constraint can't be integrated (it's impossible to reduce it to differentials-free form). $\endgroup$ Jan 19, 2013 at 16:12
  • $\begingroup$ You seem to be mistaking the usual Cartesian coordinate system for a required coordinate system. That's not the case: indeed in the Lagrangiana and Hamiltonian formulation of mechanics we explicitly use "generalized coordinates", but even in a Newtonian formulation we can use non-Cartesian coordinates like spherical and polar cylndrical coordinates (albeit at the cost of sticking Jacobians all over the usual formulas). $\endgroup$ Jan 19, 2013 at 16:57

1 Answer 1


Using a Cartesian coordinate system, the particle has two degrees of freedom because you need two variables to define its position. You use the holonomic contraint to eliminate a coordinate and so reduce the coordinates and degree of freedom by one, but transforming the coordinates in such a way to make this obvious. I.e along the path it's constrained to move along! Clever eh? $$\begin{align*}dt &= \sqrt{\left(\frac{dy}{dx}\right)^2 + 1}= \sqrt{\left(2x\right)^2 + 1}\\ t &= \frac 1 4\left(2\sqrt{4x^2 + 1}x+ \text{sinh}^{-1}(2x)\right) \end{align*}$$

Holonomic was introduced by Hertz in his book on mechanics, and comes from Greek to mean whole law. A holonomic constraint is therefore expressed in terms of the coordinates and not differentials that usually can't be integrated. This is equivalent to $f(q_1,q_2...,q_n,t) = 0$ making your constraint $y - x^2 = 0$ also holonomic.

  • $\begingroup$ Thanks for all the help everyone. This issue is clear to me now. $\endgroup$ Jan 20, 2013 at 2:05
  • $\begingroup$ @SumukhAtreya if you think my answer is useful, then hit the up icon beside the zero. If you think it's also correct, then hit the right answer icon also. As you can see, I'm begging for some points here ;) $\endgroup$ Jan 20, 2013 at 14:24

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