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I am working through a book titled "An introduction to mechanics and symmetry" by Marsden and Ratiu. I have written up a brief summary trying to solidify my understanding of the general principles.

Two questions:

i) Is my understanding of the ideas sound? I realise I should probably know if I understand the material once I understand it but I think it takes some time to internalize this stuff.

ii) How does one actually prove that the flows of the vector field $X_L$ correspond to the Euler Lagrange equations of $L$.

Notes:

$\newcommand{\rn}{\mathbb{R}} \newcommand{\fn}{\mathbb{F}} \newcommand{\into}{\rightarrow}$ Let $T^*M$ be the cotangent bundle of $M$. The Liouville 1-form $\Theta$ on $T^*M$ is defined as follows. Let $(x, \omega) \in T^*M_x$ and then let $W$ be in $TT^*M_{(x, \omega)}$. Then $\Theta_{(x, \omega)} (W) = \omega(d \pi^* (W))$ where $\pi^* : T^*M \into M$ is the projection map. With this, we can define a 2-form $\Omega$ on $T^*M$ by taking the exterior derivative, $\Omega = -d\Theta$.

For a function $H : T^*M \into \rn$, the Hamiltonian vector field $X_H$ on the cotangent bundle is the vector field satisfying the property that $dH = \iota_{X_H} \Omega$. Flows under this vector field represent solutions to the classical Hamiltonian equations under a suitable Hamiltonian function $H$.

To switch between Hamiltonian formalism and the Lagrangian formalism, we use the Legendre transform. Given a map $L : TM \into \rn$, define $\fn L : TM \into T^*M$ by $\fn L(v)(w) = \frac{d}{ds}\big|_{s=0} L(v + sw)$. We have the corresponding forms on $TM$ given by $\Theta_L = (\fn L)^* \Theta$ and $\Omega_L = (\fn L)^* \Omega$. $\Theta_L$ is called the Lagrangian 1-form and $\Omega_L$ the Lagrangian 2-form.

Then let $X_L$ be a vector field on $TM$ such that $\iota_X \Omega_L = dE$ where the energy $E$ is defined by $E(x,v) = \mathbb{F}L(v)\dot(v) - L(x,v) = \Theta_L(X)(v) - L(v)$

Then the integral curves (I think) of $X_L$ are solutions to the Euler Lagrange equations. Suppose a curve $v(t) \in TM$ satisfies $v^{(1)}(t) = X(x, v)$. Then does $v^i(t) = x^{i(1)}(t)$ and $L_x(x, v) - \frac{d}{dt} L_v (x, v) = 0$?

Thanks for your time.

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