# Checking my understanding of $T^*M$ as a symplectic manifold and the links between the classical description of Lagrangians vs this invariant way.

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:



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 $L_x(x, v) - \frac{d}{dt} L_v (x, v) = 0$?

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You start with a manifold $M$. A Lagrangian function is just an element $L\in C^\infty(T M)$. Call the pair $(M,L)$ a Lagrangian system. A motion in the Lagrangian system is just a curve $\gamma:\mathbb{R}\to M$ whose lift, $(\gamma(t),\gamma^\prime(t))\in T_{\gamma(t)}M$, satisfies the Euler-Lagrange equations.

A Hamiltonian system is a triple $(X,\omega, H)$ where $(X,\omega)$ is a symplectic manifold and $H\in C^\infty(X)$ is called the Hamiltonian function. The integral curves of $X_H$, the Hamiltonian vector field, are characterized by solving Hamilton's equations. Note that you can always turn $T^\ast M$ into a symplectic manifold in the way you described above.

Given the Lagrangian system as above, we get the associated Legendre transform $\Phi_L: T M\to T^\ast M$. Under certain conditions on $L$, we get that $\Phi_L$ is a diffemorphism. The Legendre transform relates the two formulations via the following theorem:

A curve $\gamma:\mathbb{R}\to M$ is a motion in $(M,L)$ (i.e. satisfies the Euler-Lagrange equations) if and only if $\Phi_L(\gamma(t),\gamma^\prime(t))$ is an integral curve for $X_H$ (i.e. satisfies the Hamilton equations)

Here $H=L\circ \Phi_L^{-1}$. That is we are considering the Lagrangian system $(M,L)$ and the Hamiltonian system $(T^\ast M,\omega, H=L\circ\Phi_L^{-1})$ with $H\in C^\infty (T^\ast M)$ and $X_H\in \Gamma(T(T^\ast M))$.

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Thank you for your incredibly clear answer :) –  muzzlator Jul 7 '14 at 6:08
I think it should be that $H = E (\phi^{-1}_L)$ where $E:TM \rightarrow \mathbb{R}$ is the function $E(v) = (\phi_L(v))(v)-L(v)$. See below for the correct explanation. –  Sina Jan 14 at 17:37

Simply because you pull back the equation $\omega(X,\cdot) = dH$ to an equation on $T^*(TM)$ using the Legendre transformation. Indeed denote $\phi_L: TM \rightarrow T^*M$ the Legendre transformation and denote the Hamiltonian as $H(p) = p(\phi_L^{-1}(p)) - L(\phi^{-1}_L(p))$ (this is how you define an Hamiltonian from the Lagrangian in the classical setting). Then if you define the quantity "energy" as you defined above that is $E(v)= (\phi_L(v))(v) - L(v)$ it is easy to see that $E(v) = H(\phi_L(v))$

Now applying the pull back to the equation you get:

$$\phi^*_L (\omega(X_H,\cdot) = d(\phi_L^*H)$$

$$(\phi^*_L\omega)(D\phi^{-1}_L X_H, \cdot) = dE$$

$$\Omega_L(X_L,\cdot) = dE$$

Just one remark on how you can also pass to Lagrangian mechanics from Hamiltonian in a "natural way". As you have remarked you can build the canonical 1-form on $T^*M$ as $\Theta_{x,p}(W) = p\circ d\pi^*(W)$. Now if $\pi_*: TM \rightarrow M$ is the canonical projection, given a Lagrangian one can define a 1-form on $TM$ with almost the same procedure as follows: $(\Omega_L)_{x,v}(U) = \phi_L(v) ( d\pi^*(U))$ for $U$ in $T(TM)$. If you want the differential of this 1-form to be symplectic, you need the non-degeneracy condition $det(\frac{\partial^2L}{\partial v_i \partial v_j}) \neq 0$. This is the same symplectic form $\Omega_L$ above. You just use an isomorphism between $TM$ and $T^*M$ to transport the symplectic structure on $T^*M$ to $TM$. Infact in the cases where $L(v) = g(v,v) - V(\pi_*(v))$ then Lagrangian transformation is simply the isomorphism between $TM$ and $T^*M$ given by the metric (the musical isomorphism).

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