1
$\begingroup$

The configuration space of a system of particles $(m_i,x_i)$, $i=1,\dots,n$, subject to constraints $$\Phi (x)=0,\qquad \Phi\colon \mathbb R^{3n}\to \mathbb R ^{3n-k},\qquad x=(x_1,...,x_n),$$ if the constraint is nice enough (i.e. if $0$ is a regular value of $\Phi$), can be described as a $k$-dimensional submanifold of $\mathbb R ^{3n}$, and this clearly has some advantages. Those systems are called “autonomous” (and if someone could throw some light on the terminology I'd also be grate).

Now, suppose that $\Phi$ depends on time, i.e. the constraints are:$$\Phi(x,t)=0.$$ In this case it isn't immediately obvious to me what would be the most natural way to describe the configuration space. For example, one might define $g^t(x)=\Phi(x,t)$, suppose that $0\in \mathbb R ^{3n-k}$ is still a regular value of $g^t$ and define the configuration space at time $t$ as $$M^t=(g^t)^ {-1}(0),$$ and describe the position of the system at time $t$ with $k+1$ parameters $(q_1,\dots,q_k,t)$. This works good if, for example, the manifolds $M^t$ are essentially the same: for example, if $M^t\subset \mathbb R ^3$ is a ring that rotates about the $z$ axis, the manifold is simply $S^1$. But is this description always possible? I mean, $M^t$ could possibly change in time so that the coordinates $(q_1,\dots,q_k,t)$ don't mean a thing for some $t$.

Another conceivable way, I suppose, would be to consider the ($k+1$-dimensional) manifold: $$M=\Phi ^{-1}(0)\subset \mathbb R ^{3n+1}\ni (x_1,...,x_n,t).$$ For example, in this post, I sketched a proof of the Noether's theorem for non-autonomous systems following a similar idea. The main problem that occurs to me is that, in this way, we make the statement of propositions like, for example, d'Alembert principle more complicated, because we can no more consider virtual displacements as tangent vectors to the configuration space.

So, to summarize: given a constraint of the form $\Phi(x,t)=0$, what is the most natural way to describe the configuration space of such a system?

$\endgroup$
1
$\begingroup$

Regardless of the form of whichever (holonomic) constraints you may have, non-autonomous systems are most naturally understood from a field theoretical viewpoint. More precisely, one should understand Lagrangian mechanics as Lagrangian field theory in $0+1$ dimensions (that is, the space-time manifold is just the real time line). There, coordinates over each time instant constitute the fibers (which could well be manifolds instead of vector spaces) of a fiber bundle over the real line, and the pairing of positions and velocities are the fibers of the corresponding first-order jet bundle of this fiber bundle. Histories, on their turn, are then understood as field configurations. Hence, the Noether theorem for non-autonomous systems is just the Noether theorem for classical field theory. One advantage of this viewpoint is that the apparent distinction in the description of autonomous and non-autonomous systems disappears.

Holonomic constraints can still be understood, as in the autonomous case, as conditions on initial data which are preserved under the dynamics. In other words, there should be an one-to-one correspondence between solutions of the equations of motion and initial data at a certain instant of time (= "initial-data surface") satisfying such constraints. In particular, there can be no "topology change" among fibers, in compliance with the fiber bundle picture. Notice as well that, since the base manifold is 1-dimensional, all fiber bundles over it are trivial, so the total space of the bundle is always of the form (position space) $\times$ (time). With these considerations in mind, one concludes that holonomic constraints of the kind $\Phi(x,t)=0$ should specify a sub-bundle of the above fiber bundle. More generally, if the constraints involve the veolcities as well, one should instead consider sub-bundles of the first-order jet bundle.

$\endgroup$
  • $\begingroup$ Hi Pedro, I'm sorry but I know nothing about field theory, and I can understand little from the second part of your answer. Could you possibly try to state it in simpler terms? From what I get, you are saying that the world line of the system always lies in a $k+1$-dimensional manifold of the form $\mathbb R \times M$, where $k=$ number of constraints? Also, by saying that “there should be a 1-1 correspondece...“, do you mean the uniqueness of the solution of the equations of motion? I'm sorry, but my “vocabulary” is a bit limited. Thank you. $\endgroup$ – pppqqq Feb 9 '14 at 12:24
  • $\begingroup$ Indeed, all possible worldlines live in a manifold of the kind $\mathbb{R}\times M$, where $M$ can be thought of as the "position manifold". However, this includes both worldlines which are solutions of the equations of motion and those which are not. If there were no constraints in the dynamics, the dimension $k$ of $M$ would be the number of degrees of freedom of the system. Once you have constraints as above, you have to count the number of physical degrees of freedom, which is $k$ minus the number of independent constraints. This is the dimension of the so-called constraint manifold. $\endgroup$ – Pedro Lauridsen Ribeiro Feb 10 '14 at 2:01
  • $\begingroup$ As for the "1-1 correspondence" I've mentioned, you got it right, it refers to uniqueness of the solutions of the equations of motion for given initial data. $\endgroup$ – Pedro Lauridsen Ribeiro Feb 10 '14 at 2:02
  • $\begingroup$ (continued from my first comment) The constraint manifold is just the submanifold of $M$ where the "physical" (i.e. constrained) trajectories live. One can see this viewpoint as a "field theoretical" one because each worldline can be thought of as a "field configuration" over the time line. The value of the field configuration at a given time $t$ is just the position of the trajectory at that time. $\endgroup$ – Pedro Lauridsen Ribeiro Feb 10 '14 at 2:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.