Goldstein says that when a system of $N$ particles is subject to $k$ holonomic constraints, the positions $\mathbf{r}_1, \dots, \mathbf{r}_N$ can be parameterized by $3N - k$ independent coordinates $q_1, \dots, q_{3N - k}$ and time. He then says that:

It is always assumed that we can also transform back from the ($q_i$) to the ($\mathbf{r}_l$) set, i.e., that [the parameterizations] combined with the $k$ equations of constraint can be inverted to obtain any $q_i$ as a function of the ($\mathbf{r}_l$) variable and time.

My question: Why would we need the $k$ equations of constraint? It seems to me that all of the constraint information is stored in the parameterizations of $\mathbf{r}_1, \dots, \mathbf{r}_N$. No?


It is a fundamental result$^1$ in the theory of embedded differentiable submanifolds that they can equivalently be described

  • locally$^2$ as a parametrized submanifold/graph,

  • or locally as a constrained submanifold.

Example: An ellipse in the 2D plane can either be described by a parametrization $(x,y)=(a\cos\theta,b\sin\theta)$ or via a constraint $(x/a)^2+(y/b)^2=1$.

Depending on application, both descriptions can be useful. If one of the descriptions fails, it means that some of the technical regularity conditions (which are mostly implicitly assumed in Goldstein) are not fulfilled, cf. e.g. this & this Phys.SE posts.


$^1$ This result is included in any decent textbook on differential geometry (DG). (See e.g. Proposition 3.2.1 in Ben Andrews, Lectures on DG.) The main tool in its proof is the inverse function theorem.

$^2$ The word "locally" here means "in an open neighborhood".

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  • $\begingroup$ This is the answer but I feel like "either" conveys one or the other, maybe I'd suggest "equivalently". $\endgroup$ – Ryan Thorngren Mar 6 '19 at 19:07
  • $\begingroup$ @Ryan Thorngren: $\uparrow$ Adapted. $\endgroup$ – Qmechanic Mar 6 '19 at 19:08
  • $\begingroup$ I probably should've pointed this out earlier, but this is beyond my level of understanding. I'm aware, however, that this answer may come in handy to someone else. $\endgroup$ – user113773 Mar 9 '19 at 2:14
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Mar 10 '19 at 20:28

I understand it like this:

Example: one particle with sphere coordinates (parameter $r\,,\theta\,,\varphi$)

$$\vec{R}=\left[ \begin {array}{c} x\\ y\\ z\end {array} \right] =\left[ \begin {array}{c} r\cos \left( \theta \right) \sin \left( \varphi \right) \\ r\sin \left( \theta \right) \sin \left( \varphi \right) \\ r\cos \left( \varphi \right) \end {array} \right]\tag 1$$

we can solve equation (1) for $r\,,\theta\,,\varphi$

$$r=\sqrt{x^2+y^2+z^2}\tag 2$$ $$\theta=\arctan(x/y)\tag 3 $$ $$\varphi=\frac{z}{x^2+y^2}\tag 4$$

the constraint equation

$$r^2-l^2\cos^2(\varphi)\cos(\theta)=0\tag 5$$

now we can choose the generalized coordinates:

if we solve equation (5) for $r$ then we get (3) $\quad q_1(x,y)=\theta$ and (4) $\quad q_2(x,y,z)=\varphi$

if we solve equation (5) for $\theta$ then we get (2) $\quad q_1(x,y,z)=r$ and (4) $\quad q_2(x,y,z)=\varphi$


if we solve equation (5) for $\varphi$ then we get (2) $\quad q_1(x,y,z)=r$ and (3) $\quad q_2(x,y)=\theta$

we get always unique result for $q_1(x,y,z)$ and $q_2(x,y,z)$ and we used the "inverse" of the position vector and the constrained equation.

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I think that the author is just referring to "they" as the numbers that you have for the different variables $Q=\{q_i\}$ whereas you are correct that these numbers only make sense in this context via their parameterizations $\mathbf r_i(Q)$ and if you understand the sentence in that way it is expressing a tautology and no further information is needed.

With that said if you had a really simple system there are probably some degenerate-ish cases where just the equations for those parameters are not invertible without fully knowing the constraints. For example we might have two particles living in 2D, one is constrained to live on the line $x=0$ and the other is constrained to live on the line $y=0$, and let's say that they are joined by a spring with rest length $\ell$. We know that we can describe this system with two variables $(x, y)$ and the mapping from $(\mathbf r_1, \mathbf r_2) \mapsto (x, y)$ is going to be $$\big([x_1, y_1], [x_2, y_2]\big) \mapsto (x_1, y_2)$$ but discovering that $x_2=0, y_2=0$ is not directly possible just given this function; it is not invertible.

But with that said this is indeed somewhat of an unnatural way to describe the description. The more natural $\mathbf r_i(Q)$ way is indeed to specify $$\mathbf r_1(x, y) = [x, 0]\\\mathbf r_2(x, y) = [0, y]$$and this indeed does embody the constraints and therefore no further reference to the constraints is needed to use the $x,y$ to determine the positions.

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We need k constraint equations because we need to find 3N variables and we do not have 3N independent coordinate equations but only 3N-K ( from q3N-k ), the k restrictions (remembering that constraints are relations between the position variables (and possibly time) Goldstein Classical Mechanics 3ED p. 12) will provide a set of k equations that together with the 3N-K equations of independent coordinates will form a set that makes the system uniquely determined (number of equations equal to the number of unknown variables).


the local space is Minkowsk’s space-time thus requiring the coordinate of time. Our problem is reduced to discovering 3 coordinates of local space for each particle.

how are we dealing with two systems of space (local and generalized) we will need to guarantee the inverse transform, the transform back is ensured in the holonomic system.

Holonomic system is defined in contrast of the definition for non-holonomic system given by Hertz (Non-holonomic system are systems in which the kinematic conditions give only relations between differentials of the coordinates and not as finite relations between the coordinates themselves) therefore, holonomic systems always have integrable differential functions ∂F/(∂qi )dqi (i = 1 to 3N-k) allowing the inverse transform.

Remembering - In the holonomic system k constraints leads to the reduction to 3N-k independent coordinates or equivalently to rN functions with k implicit constraints (Goldstein Classical Mechanics 3ED p. 13), Therefore, rN functions not linearly independent, but equivalent to 3N-k set of linearly independent equations in generalized space qi. The k constraints allow you to eliminate the dependent variables.

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