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. $y=x^2$)?

• 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. Jan 19, 2013 at 16:03
• 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). Jan 19, 2013 at 16:12
• 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). Jan 19, 2013 at 16:57

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.