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In Lipschitzs Classical Mechanics a Lagrangian is defined as:

$L(q,q',t)$ for some trajectory $q(t)$ of a particle

And the action is defined as:

$S:=\int^a_b L(q,q',t) dt$

How does one express this on the language of forms? Is the following possible?

Consider:

$L:TM \rightarrow \mathbb{R}$

so that $L$ is a 1-form; ie $L \in \Omega^1 M$ which is neccessary to integrate over a trajectory $q$ considered as curve, a 1-dimensional manifold:

$q:I \rightarrow J \subseteq M$.

Then defining the action:

$S:=\int_J L$

we have $S=\int_{qI} L=\int_I q^*L=\int_I Lq_*$

Which we define as the Lagrangian $\int_I L(q,q',t) dt$ Introduced by Lipschitz.

Does this work?

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You may know about it already, but you can find an excellent account of Lagrangian Mechanics on manifolds in the book Mathematical Methods of Classical Mechanics by V. Arnold.

Also to specifically address your question:

$L:TM \rightarrow \mathbb{R}$

so that $L$ is a 1-form; ie $L \in \Omega^1 M$ which is neccessary to integrate over a trajectory $q$ considered as curve, a 1-dimensional manifold:

$q:I \rightarrow J \subseteq M$.

This part is not normally true, consider for example the free Lagrangian (evaluated at a point $p$ as a function $L_p : T_pM \to\mathbb{R}$) $L_p(v) = <v,v>_p$. This is definitely not linear in $v$, which is necessary for a 1-form. (Here < , > is some Riemannian metric on the manifold).

Also it may be that the set traced out by the curve is not a submanifold, this occurs in the case of a self-intersecting curve.

You do not need to have a 1-form to integrate over the curve. Since $L$ is a function $TM \to \mathbb{R}$, for a differentiable curve $\gamma : [a,b] \to M$ the composition: $L_{\gamma(t)}(\dot\gamma(t))$ is a function $[a,b] \to \mathbb{R}$ over which one can integrate.

Defining an action $S$ by $S[\gamma(t)]=\int_a^b L_{\gamma(t)}(\dot \gamma(t))\, dt$ one can recover the old equations of motion under the condition that $S$ is extremal for a physical path: By choosing any coordinate chart $\phi$ on a subset $U \subseteq M$, there is an induced set of coordinates on $TU \subseteq TM$ given by

$\tilde \phi (p,v) = (\{\phi(p)^i\},\{v_i\})$ where $v=\sum_i v_i\, v(\phi(p)^i) \equiv \sum_i v_i \frac{\partial}{\partial\phi^i}$

On these coordinates $L$ gets the old form of a function $L(q,q')$ acting on two tuples of numbers and then e.o.m. $\frac{d}{dt}\frac{\partial L}{\partial q'}- \frac{\partial L}{\partial q} =0$ are recovered.

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  • $\begingroup$ Thanks; the second objection can by considering immersed sub-manifolds; but the first shows it isn't possible to use forms at all; your final expression is what Lipschitz is obviously considering... $\endgroup$ – Mozibur Ullah May 23 '15 at 18:07
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    $\begingroup$ If you only allow paths that are immersed sub-manifolds you will discard paths that would otherwise extremise the action. An example of a physically realised self intersecting path would be a masspoint on a circle in the x-y-plane that can roll along x-axis. $\endgroup$ – s.harp May 23 '15 at 18:11
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    $\begingroup$ I'm not so sure now that this is correct; a volume form $\omega$ be typically expressed as $\omega=\omega' (dx^1 \wedge dx^n)$, for some function $\omega'$; so your function $L_p$ is actually the coefficient of a volume form and not one itself; as a form it should be written as $L_p dt$ which is obviously linear, since $dt$ is. $\endgroup$ – Mozibur Ullah May 23 '15 at 23:02
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    $\begingroup$ An immersed manifold, at least the definition I'm acquainted with, is an injection $:I \rightarrow M$ whose tangent spaces are also injective; this means that it can allow self-crossings; since typically everything is smooth the only paths I see that could be missing are non-smooth paths (for example a link) - is this what you were getting at? $\endgroup$ – Mozibur Ullah May 23 '15 at 23:10
  • $\begingroup$ You are correct, I was mistaken with the definition of a immersed submanifold. $\endgroup$ – s.harp May 24 '15 at 13:38

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