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I posted my questions in a previous post MO, but it seems that a more refined version for question on the Euler-Lagrange equation is needed. So, I post my question again.

In standard symplectic geometry, we assume the "canonical coordinates" $(p_i,q_i)$ at least locally, under with the symplectic form $\omega$ is written as \begin{equation} \omega=\sum_i dp_i \wedge dq_i \ . \end{equation}

Then the Hamilton's equations are $\dot{p_i}= -\frac{\partial H}{\partial q_i}$ and $\dot{q_i}= \frac{\partial H}{\partial p_i}$ while the Euler-Lagrange equation is \begin{equation} \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i}=\frac{\partial L}{\partial q_i} \end{equation}

However, I am curious about how to write the Hamilton's equations and Euler-Lagrange equation in "arbitrary" coordinates $(p'_i,q'_i)$ where $\omega \neq \sum_i dp'_i \wedge dq'_i$.

In "Classical Dynamics : A Contemporary Approach" (1998) by Jose Saletan, the Euler Lagrange equation is expressed in a coordinate-free way by \begin{equation} \mathcal{L}_\Delta \theta_L = d L \end{equation} where $\Delta$ is the vector field generating time translation and $\theta_L$ is the canonical $1$-form corresponding to the Lagrangian $L$. However, he only defines these quantities in terms of canonical (local) coordinates as \begin{equation} \theta_L : =\sum_i \frac{\partial L}{\partial \dot{q}_i} dq_i \text{ while } \Delta:= \sum_i \dot{q_i} \frac{\partial}{\partial q_i} + \ddot{q_i} \frac{\partial}{\partial \dot{q_i}} \end{equation}

So, I don't think it "truly" coordinate-free...

Could anyone please provide the general formula for the Euler-Lagrange equation either

  1. In a "truly" coordinate-independent way (which is more desired for me)
  2. In "arbitrary" local coordinates $(p'_i,q'_i)$ in which $\omega \neq \sum_i dp'_i \wedge dq'_i$

I would deeply appreciate any help...

Isaac
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    Hamilton's equations are a special case of Euler-Lagrange equations derived from the variational principle $S[p,q] = \int (\omega(p,\dot{q}) - H(p,q)) dt$. You can take the dynamical variables $z = (p,q)$, rewrite them in any set of coordinates you please and derive the EL equations with respect to them. This kind of question comes up from time to time. See some discussion and references here. – Igor Khavkine Jan 12 '24 at 16:31
  • @IgorKhavkine Thank you. I will read your link very carefully. – Isaac Jan 12 '24 at 16:38
  • @IgorKhavkine Just to check my understanding : in your formula above, $z=(p,q)$ is NOT necessarily the canonical coordinates under which $\omega=\sum_i dp_i \wedge dq_i$. Am I correct? – Isaac Jan 12 '24 at 16:38

1 Answers1

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The answer is in this short paper by Erik Curiel. The Geometry of the Euler-Lagrange Equation

The key idea is to express the Euler-Lagrange equation in the tangent bundle, in terms of a structure tensor that the tangent bundle automatically comes equipped with.

Let $Q$ be the coordinate manifold. Pick coordinate chart $q_i$ of $Q$. Then $q_i, \dot{q}_i$ is a chart of the tangent bundle $TQ$.

Define a $(1,1)$ tensor $J$ on $TQ$ by

$$ J \begin{bmatrix} \vec{\dot{q}} \cr \vec{\ddot{q}} \end{bmatrix} = \begin{bmatrix} 0 \cr \vec{\dot{q}} \end{bmatrix} $$

You can check that this is well defined and doesn't depend on the coordinate chart.

Let $\xi$ be a vector field in $TQ$, which is to be the solution of the Euler-Lagrange equation. Then the equation can be expressed as:

$$ \mathscr{L}_\xi ({dL} \cdot J) = dL $$

Lawrence D'Anna
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  • Thank you. So, your $q_i$'s depend on time? How do you know that $q_i, \dot{q}_i$ are linearly independent so that they comprise a local chart for $TQ$? These issues have always confused me... – Isaac Jan 13 '24 at 19:05
  • $q_i$ is a chart of $Q$ by definition, so $q_i$ does not directly depend on time. $q_i, \dot{q}_i$ is the associated chart of $TQ$, where a vector at $\vec{q}$ is expressed as $V = V^i \dot{q}_i $.

    The way time comes into it is though $\xi$. The time evolution of the system is the flow through $TQ$ generated by $\xi$. So $\mathcal{L}_\xi$ is $d/dt$

     – Lawrence D'Anna Jan 14 '24 at 14:57
  • ah, sorry that should be $V = \dot{q}^i \partial / \partial q_i $ – Lawrence D'Anna Jan 14 '24 at 15:09
  • by the way, I just picked up Manuel de Leon’s Methods of Differential Geometry in Analytical Mechanics and skimmed some of the chapters last night, and it has some pretty good coverage of these concepts – Lawrence D'Anna Jan 14 '24 at 18:01
  • So, $\dot{q}_i$ is defined by $\xi = \dot{q}_i \frac{\partial}{\partial q_i}$? – Isaac Jan 14 '24 at 18:11
  • If you could let me know which chapters cover the concepts, I would be very grateful. – Isaac Jan 14 '24 at 18:12
  • @Isaac I wouldn't say that $\dot{q}_i$ is defined that way. You can think of any vector field on any manifold as expressing a first-order ODE. The equation you wrote corresponds to $\xi$ being "second order", that is it's expressing a second-order ODE on $Q$.

    A path $f(t)$ though $TQ$ induces a path $\pi(f(t))$ though $Q$. That path has a differential $(\pi \circ f)'(t)$ which is again a path on $TQ$. You want that to equal $f$. ξ is second order iff its flow paths have that property.

    The E-L equation is a 2nd order ODE, so its solutions are required to have this property.

     – Lawrence D'Anna Jan 15 '24 at 20:22
  • ... er, sorry the actual second order property is $ξ = \dot{q}_i \frac{∂}{∂q_i} + a_i \frac{∂}{∂\dot{q}_i}$ where $a_i$ is arbitrary. The vertical part of ξ is constrained but the horizontal part is allowed to be anything. This can also be written $Jξ = Δ$, where $Δ = \dot{q}^i \frac{∂}{∂\dot{q}_i}$ – Lawrence D'Anna Jan 15 '24 at 20:28
  • OK. Thank you. I am looking into the index of that Manuel de Leon's book, but having difficulty because I do not know the keywords... I would deeply appreciate if you tell me the keywords or any relevant chapters! – Isaac Jan 15 '24 at 22:10
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    @Isaac Section 7.1. Equation (7.2) is equivalent to the version I gave here, you can derive it by applying Cartan's magic formula to expand the Lie derivative, and then applying the 2nd order condition on for $\xi$ – Lawrence D'Anna Jan 16 '24 at 19:00