Wikipedia states:

holonomic constraints are relations between the position variables (and possibly time1) which can be expressed in the following form: $$f(q_{1},q_{2},q_{3},\ldots ,q_{n},t)=0$$

where $\{q_{1},q_{2},q_{3},\ldots ,q_{n}\}$ are the $n$ coordinates which describe the system. For example, the motion of a particle constrained to lie on the surface of a sphere is subject to a holonomic constraint, but if the particle is able to fall off the sphere under the influence of gravity, the constraint becomes non-holonomic. [...] the second non-holonomic case may be given by: $$r^{2}-a^{2}\geq 0.$$

Is this really a non-holonomic constraint? Consider the following function $f(r)=\min(r^2-a^2,0)$. Then we have $$r^{2}-a^{2}\geq 0\quad\text{iff}\quad f(r)=0.$$

Doesn't this mean that the constraint is in fact holonomic?


2 Answers 2


OP's example is clever. However in physics, the constraint function $f$ is often implicitly assumed to obey various regularity conditions, e.g. differentiability, which OP's example fails. For details, see e.g. this related Phys.SE post.


No. A system with $n$ degrees of freedom is said to be under a holonomic constraint if imposition of the constraint reduces the number of degrees of freedom by $1$ , to $n-1$. Every holonomic constraint can be captured by some relation of the form $f(q_1...q_n,t)=0$, but not every constraint that can be captured in this form is necessarily holonomic. You have provided an example.

When $(x^2+y^2+z^2) = a^2$ is the constraint satisfied, transforming to spherical coordinates (where one d.o.f is explicitly $r= \sqrt{x^2+y^2+z^2}$ ) allows us to describe the motion constrained on the sphere using $\theta$ and $\phi$, ie, reduction of the number of d.o.f from $3$ to $2$. Sometimes, it may not be straightforward to find this coordinate transform ; in such cases, we automate the process of describing the resultant $n-1$ dimensional motion by using the Lagrange Multiplier method.

So, calling a constraint on a system with $n$ d.o.f holonomic implies that imposing this constraint turns the system into one with n-1 d.o.f. This is by definition (see comments below)

What you have constructed indeed has the functional form described above, but it does not describe a situation with $1$ d.o.f less. The motion is very clearly still $3$-dimensional.

  • $\begingroup$ I am confused by this answer, because it seems to directly contradict both wikipedia and Goldstein et al (Classical mechanics). I am not saying that your answer is wrong, but I am confused about it. $\endgroup$
    – user56834
    Mar 21, 2020 at 14:46
  • $\begingroup$ Read the text after eq (38.2) in Mechanics, Landau (3rd ed), where he uses precisely the same definition. (Paraphrasing) his words, "A constraint is said to be holonomic if it can be reduced to a relation of coordinates only, which could be used to express the position of the bodies in terms of fewer co-ordinates, corresponding to the actual number of degrees of freedom.". Operational phrase - fewer co-ordinates $\endgroup$
    – insomniac
    Mar 21, 2020 at 15:12
  • $\begingroup$ The point is this : the reason one marks out a subset from the set of all possible constraints, and give it a special name , holonomic, is because there is something special (geometrically) about them : namely, they all reduce the number of degrees of freedom (or reduce the dimension of the configuration space, if you prefer) by $1$. $\endgroup$
    – insomniac
    Mar 21, 2020 at 15:29
  • 2
    $\begingroup$ @user56834 It's worth noting that the regularity conditions Qmechanic mentions (smoothness, full rank) allow one to deduce (via regular value theorem) that $f^{-1}(0)$ - the subspace of configurations allowed by the constraint - is a submanifold of the configurations space of one dimension less, so imposition of the regularity conditions is actually very much the same as demanding that one can use the constraint to eliminate a degree of freedom. $\endgroup$
    – ACuriousMind
    Mar 21, 2020 at 15:38

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