3
$\begingroup$

I would like to resolve a few problems I'm having regarding the exact procedure of Lagrangian mechanics when formulated as the tangent bundle of configuration space. These problems are not overly technical but I will list definitions below so that we are all on the same page and in the event that my confusion is arising from incorrect definitions that will be apparent from the start to the reader.


Definitions:

By configuration space I mean $\Bbb R^N:=M$, the $N$-dimensional manifold whose points are associated with the positions of our particles. This question does not relate to the generalisation to field theories. The starting point for Lagrangian mechanics is the tangent bundle of this manifold $\Bbb R^{2N}:=TM$. The Lagrangian is then a scalar function on the tangent bundle: $$\mathcal L:TM\rightarrow \Bbb R \tag{1}.$$

The action is a functional which a priori is a functional of trajectories through $TM$ (in particular, not through configuration space, in general, since $q(t)$ and $\dot q(t)$ are independent parameters unless we are explicitly on a solution to the equations of motion (EOM)).

When we say on-shell we are restricting our discussion such that the domain of the action functional is those trajectories through $TM$ such that the Euler-Lagrange equations are satisfied. In this specific case it is enough to specify a curve through configuration space since on-shell the $q(t)$ and $\dot q(t)$ coordinates are related through $\dot q(t)=\frac{dq(t)}{dt}$.

When we say off-shell we mean we are considering all possible trajectories through $TM$ with no guarantee that the Euler-Lagrange equations are satisfied and in particular $\dot q(t)\neq \frac{dq(t)}{dt}$ in general.

We assume throughout that the Lagrangian has no explicit time dependence.

Note: I use the term "symmetry" here when perhaps the more correct term is "quasi-symmetry" as explained by QMechanic here and in various other places around the site.


Now, Noether's theorem states that for every continuous off-shell symmetry of the action there exists a corresponding on-shell conserved quantity $f(q(t),\dot q(t))$, often called a "Noether charge". However, the Physics SE question here essentially asks if symmetries of the Lagrangian and symmetries of the action are equivalent, to which the answer (modulo some technicalities) is yes.

My problem is then how to define "symmetries of the action" and "symmetries of the Lagrangian". Since we are working on a manifold it seems natural to ask if a symmetry of the Lagrangian simply means that we perform a coordinate change on $TM$ (i.e. we work with passive transformations) such that the transformed Lagrangian function is of the form:

$$q(t)\rightarrow q'(t),\quad \dot q(t)\rightarrow \dot q'(t)\quad \Rightarrow\quad \mathcal L'(q'(t),\dot q'(t))=\mathcal L(q(t),\dot q(t))+\frac{dF}{dt}+\mathcal O(\epsilon^2) \tag{2}.$$

In words, our coordinate transformation induces a change in the Lagrangian to first order such that the difference is expressible as a total time derivative of some function $F(q(t),\dot q(t))$. If this is incorrect I would appreciate some guidance towards a more correct answer.

On the other hand, the action is a functional, and so a "symmetry" of the action (may?) require a different construction and so exactly how one would go about inducing a change in the action is another point of confusion. Since we are talking about off-shell symmetries of the action I assume we must talk about making some generic change to all admissible (smooth) curves through the tangent bundle and ask whether for such a change:

$$\gamma (q(t),\dot q(t))\rightarrow \gamma'(q(t),\dot q(t))\quad \Rightarrow\quad S[\gamma]=S[\gamma'] \tag{3}.$$

Again, I am unsure if this is an accurate description of what we are doing, and if it is not I would appreciate a push in the right direction here too. Perhaps the symmetries of the action also simply involve a change in coordinate on $TM$, and we don't have to go to the trouble of talking about transformations on the space of curves $\gamma:[t_1,t_2]\rightarrow TM$?

I hope this isn't considered too much of an information dump to be answerable, however I wasn't able to find an answer on the site that explicitly talks about the differential geometric structure and solves my problems.

$\endgroup$
0
2
$\begingroup$

First, if I understand you correctly, I believe you do have the definitions correct up to things which possibly happen at the boundaries of the integration region (time, here). Though if you are worried about the comparison between the statement in terms of the action and in terms of the Lagrangian, then perhaps this is not a negligible point.

To comment on the transformations in (2) and (3), note that these ideas can be unified by instead considering the action of a vector field $\xi$ on TM as defined by its flow (and hence the infinitesimal transformations are given by the Lie derivatives...indeed the variation of the Lagrangian and action would now also be given by the Lie action of $\xi$ since they are scalars over TM). The advantage of using flows is the removal of coordinate dependence since, at the end of the day, if all you're doing is defining coordinate transformations, then any good, coordinate independent quantity on TM must be blind to such changes.

Let me know if this doesn't quite get at what you're looking for.

This may not directly address your question as-posed because it deals with field theory and abandons this tangent bundle description, but perhaps this sort of thing will hit a chord if you haven't come across it before.

$\endgroup$
1
$\begingroup$
  1. Notation: Be aware that unlike OP, a dot denotes time-differentiation $$\dot{q}~\equiv~ \frac{dq}{dt}$$ in this answer.

  2. OP correctly identifies that generalized position $q$ and generalized velocity $v$ are independent variables of the Lagrangian $L(q,v,t)$, cf. e.g. this Phys.SE post.

  3. However, given an infinitesimal vertical transformation $\delta q$, the definition that $\delta$ is an infinitesimal quasi-symmetry of the Lagrangian $L$ is that $\delta L(q,\dot{q},t)$ is a total time derivative, not that $\delta L(q,v,t)$ is a total time derivative.

  4. The issue at stake is similar to this related Phys.SE post.

$\endgroup$
2
  • $\begingroup$ Thanks for providing an answer to this question too I appreciate your time. Just as a follow up on point 3, the implication here is that during any transformation in the domain of the action $(q(t)\rightarrow q'(t))$ the induced change to the curve in the tangent bundle (along which we integrate the Lagrangian function) is such that $\dot q(t)=dq(t)/dt$ $\forall t$? In other words, the only point in our discussion of classical mechanics at which $q$ and $\dot q$ (or $q$ and $v$ as you notate in your answer) are independent quantities is in their role as coordinate functions on $TM$. $\endgroup$
    – Charlie
    Dec 16 '20 at 20:45
  • 1
    $\begingroup$ Yes, with $\dot{q}$ in your notation being $v$ in my notation :) $\endgroup$
    – Qmechanic
    Dec 16 '20 at 20:51
0
$\begingroup$

After doing a fair amount of reading over the past 24 hours I think I am in a position to self-answer this question now.

The answer here given by ACuriousMind addresses my question about the domain of the action functional. In summary, while $q$ and $\dot q$ are of course independent functions at the manifold level on the tangent bundle $TM$, we do not consider arbitrary curves through the tangent bundle as the domain of the action. Rather, curves through $M$ (configuration space), $\gamma(q(t))$, naturally induce curves on $TM$, $\bar \gamma(q(t),\dot q(t))$, such that the action is defined by:

$$S[\gamma(q(t))]=\int_{t_0}^{t_1}dt\text{ }\mathcal L(\bar \gamma(q(t),\dot q(t)) \tag{1}.$$

So at all times while varying our path (be it in the derivation of the Euler-Lagrange equations or in Noether's theorem) we are dealing with curves through configuration space, as this is the domain of the action.

My definition of on-shell and off-shell are also incorrect. Off-shell refers to the curves $\bar \gamma(q(t),\dot q(t))$ (in which $\dot q=dq(t)/dt$) that do not solve the Euler-Lagrange equations, on-shell refers to the subset of these curves that do solve the Euler-Lagrange equations.

$\endgroup$

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.