Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $L$.


$\newcommand{\rn}{\mathbb{R}} \newcommand{\fn}{\mathbb{F}} \newcommand{\into}{\rightarrow}$ 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.

share|improve this question

1 Answer 1

up vote 5 down vote accepted

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 $X_H\in \Gamma(T(T^\ast M))$.

share|improve this answer
Thank you for your incredibly clear answer :) –  muzzlator Jul 7 at 6:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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