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Suppose we have a manifold $M$ and a Lagrangian function $L:TM\rightarrow \mathbb{R}$. I want to prove that a curve is a critical point of the action if and only if it satisfies the Euler-Lagrange equations.

So we have $S(c(s))=\int_{a}^{b}L(c(s,t),c'(s,t))dt$, and $\frac{d}{ds}S(c(s))|_{s=0}=\frac{d}{ds}\int_{a}^{b}L(c(s,t),c'(s,t))dt$ . Now all the proofs I have seen then go one to use the Leibniz rule and the derivative of the composition of functions and this is where I have my doubts on why we can use since $L$ is a function from the tangent bundle to $\mathbb{R}$. Now I guess there is something probably hapenning using charts that is not mentioned but I am not quite convinced and I can't seem to do with myself. Any help with this is aprecciated. Thanks in advance.

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  • $\begingroup$ Does the approach you describe involve the term $\frac{\partial L}{\partial c}(c,c')$? $\endgroup$
    – Kajelad
    Oct 3 '20 at 19:10
  • $\begingroup$ Yeah it involves that I am just a bit confused why we can do that since $L$ is a function from the Tangent Bundle. $\endgroup$
    – Something
    Oct 3 '20 at 20:32
  • $\begingroup$ Comment to the post (v1): Why does $c$ sometimes has 1 argument and sometimes has 2 argument? $\endgroup$
    – Qmechanic
    Oct 3 '20 at 20:35
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The standard derivation of the Euler-Lagrange equation on manifolds uses coordinates. Without coordinates (or some other splitting of $TTM$), one can't take a "horizontal derivative" analogous to $\partial L/\partial x$. The fact that the Lagrangian is written as a function of two arguments suggests that this is already being done in your case, at least implicitly.

Let $x^1,\cdots,x^n:U\to\mathbb{R}$ be a set of local coordinates on an open set $U\subseteq M$. These induce coordinates $(x^1,\cdots,x^n,v^1,\cdots,v^n)$ on $\pi^{-1}(U)\subseteq TM$ in the standard way. We can write the local representative of the Lagrangian in these coordinates as $L(x,v)=L(x^1,\cdots,x^n,v^1,\cdots,v^n)$.

Using local coordiantes, the derivation of the EL equation proceeds almost exactly like it does in $\mathbb{R}^n$, with the caveat that one must work locally. To show that a stationary path $c$ satisfies the EL-equations in local coordinates, it is sufficient to restrict attention to local variations such that $c$ only changes inside of (a compact subset of) the coordinate domain. For the converse, you can cover the path with coordinate charts and use a partition of unity.

Edit: Here's how to "localize" the first variation formula in a particular set of coordinates.

The action integral is perfectly well defined independently of coordinates $$ S(\gamma)=\int_a^bL(\dot{\gamma}(t))dt $$ Where $\gamma:[a.b]\to M$ is a curve and dots represent derivatives with respect to $t$. Given coordinates $x^i$ on $U$ which intersect with the image of $\gamma$, we may choose an interval $[a',b']\subseteq[a,b]$ with $\gamma([a',b'])\subset U$, and a variation $\gamma_s$ such that $\gamma_0=\gamma$ and $\gamma_s(t)=\gamma(t)$ for $t\notin(a',b')$. Before imposing coordinates, we may commute the integral and derivative $$ \frac{d}{ds}S(\gamma_s)|_{s=0}=\int_a^b\frac{d}{ds}L(\dot{\gamma}_s(t))|_{s=0}dt $$ However, since $\gamma_s$ is constant with respect to $s$ outside of $(a',b')$, we have $$ \frac{d}{ds}S(\gamma_s)|_{s=0}=\int_{a'}^{b'}\frac{d}{ds}L(\dot{\gamma}_s(t))|_{s=0}dt $$ and now, since $\gamma_s(t)\in U$ for $t\in[a',b']$, $s\in(-\epsilon,\epsilon)$, the integral can be passed to local coordinates without issue.

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  • $\begingroup$ Yes I am trying to do it using looking local corrdinates , I am just not sure why we are able to pass the derivative to inside the integral and then use the formula for the derivative of the composition since for the most part it seems to me we won't be working with functions defined over the euclidean space , maybe I am forgetting some chart somewhere but that is what is not clear to me. $\endgroup$
    – Something
    Oct 3 '20 at 21:25
  • $\begingroup$ You can exchange the integral and derivative before passing to coordinates. I've added a section on how to do this. $\endgroup$
    – Kajelad
    Oct 3 '20 at 23:05
  • $\begingroup$ Ah ok I think I got it , I forgot that despite $L$ being a function on the tangent bundle we have that $L \circ c(s,t)$ we will be a function from $(-\epsilon,\epsilon)$ to $\mathbb{R}$. Then we just use local coordinates to use the derivative of the composition. $\endgroup$
    – Something
    Oct 4 '20 at 6:33

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