6
$\begingroup$

The Euler-Lagrange equation states that the time flow is given by a vector field such that the vector field contracted with the symplectic form gives dL, where L is the Lagrangian function on the tangent bundle.

$\endgroup$
0

3 Answers 3

4
$\begingroup$

I don’t think your characterization of the vector field is quite correct; I would recommend Cartan (1922, §§184-185), Godbillon (1969, Proposition XI.3.7), or Souriau (1970, Theorem 7.29). Or this answer conveniently linked in this page’s “related” list $\to$.

$\endgroup$
2
  • 1
    $\begingroup$ The last link does not give a complete answer but it is helpful. Other links however like Cartan and souriaou are just completely wrong , they are as far from coordinate free as possible i dont know why you posted that. $\endgroup$
    – Tyson
    Dec 12, 2017 at 21:43
  • 6
    $\begingroup$ @Tyson 1) I posted “that” because as I said, it corrects your erroneous characterization of the vector field. Its contraction with the 2-form is not $dL$ but $-dH$ where $H$ is the Hamiltonian. 2) What matters in differential geometry is that results be coordinate-free. Better come to terms with the fact that not all proofs can or need to be. (E.g.: existence of the exterior derivative.) 3) That said, I think Souriau’s formula (7.28) (= 6th displayed in Cartan’s §184) gives about as close to an intrinsic argument as you’ll get. Call them “completely wrong” at your own risk :-) $\endgroup$ Dec 13, 2017 at 12:17
3
$\begingroup$

Usually one thinks of the symplectic structure $\omega=d\lambda$ as living on the cotangent bundle $T^*N$ of a manifold $N$, where it is the exterior derivative of the Lioville 1-form $\lambda$. Hence, contracting it with a vectorfield will require that the vectorfield lives on $T^*N$ and the result will be a 1-form on $T^*N$. Since your Lagrangian $L$ lives on $TN$, $dL$ is a 1-form on $TN$ so something needs to be corrected.

The way one usually looks at this is the following: If the Lagrangian $L$ on $TN$ is Tonelli then it gives rise to a Legendre transformation $$ l:TN \to T^*N, $$ which is a diffeomorphism. This allows one to transfer the dynamics of $L$ to the cotangent bundle. The way this is done is by defining a Hamiltonian function $H:T^*N \to \mathbb{R}$ via $$H(l(q,v))=\partial_vL(q,v)\cdot v-L(q,v), $$
where $\partial_vL(q,v)$ denotes the fiber derivative of $L$ and $(q,v)$ are coordinates on $TL$, $q$ being the base-coordinate and $v$ the fiber-coordinate. The Hamiltonian flow of $H$ is then defined exactly as you say: Namely it is the flow generated by the vectorfield $X_H$ whose contraction with $\omega$ equals $dH$. The vectorfield $X_L$ on $TN$ which generates the dynamics dictated by $L$ is now obtained by $$dl \cdot X_L=X_H.$$ Of course one can also derive the fomula for $X_L$ using the principle of least action.

Deriving all the above is a bit tedious, but it is carried out in great detail in the first chapter of Mazzucchelli's book "Critical Point Theory for Lagrangian Systems" http://www.springer.com/gp/book/9783034801621.

$\endgroup$
3
$\begingroup$

Nester, James M. "Invariant derivation of the Euler-Lagrange equation." Journal of Physics A: Mathematical, Nuclear and General 21.21 (1988): L1013-L1017.

Note that this is a physics paper, so the derivation is not completely rigorous. I am convinced it can be made rigorous.

Edit. Here is a math reference:

Connections, Sprays and Finsler Structures by Jozsef Szilasi, Rezso L Lovas, David Cs Kertesz ISBN: 9789814440097

It's in section 5.2.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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