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Consider the following Lagrangian (Exercise 3.6B from Abraham and Marsden's Foundations of Mechanics): $ L(\upsilon)=\frac12g(\upsilon,\upsilon)+V(\tau_Q\upsilon)+g(\upsilon,Y(\tau_Q\upsilon)) $ ($V \colon Q \rightarrow \mathbb{R}$ is a smooth function; $Y \colon Q \rightarrow TQ$ is a vector field; $\tau_Q\colon TQ\rightarrow Q$ is the tangent bundle). My aim is to calculate the corresponding Legendre transform $FL \colon TQ \rightarrow T^*Q$. It was easy to deal with the first term $L_1(\upsilon)=\frac12g(\upsilon,\upsilon)$: $ \begin{aligned} \langle FL_1(\upsilon) |\, w\rangle &= \left.\frac{d}{ds}\left[\vphantom{\frac{d}{ds}}L_1(\upsilon+sw)\right]\right|_{s=0}=\left.\frac{d}{ds}\left[\vphantom{\frac{d}{ds}}\frac12g(\upsilon+sw,\upsilon+sw)\right]\right|_{s=0}\\ &=\frac12\left.\frac{d}{ds}\left[\vphantom{\frac{d}{ds}}g(\upsilon,\upsilon)+g(\upsilon,sw)+g(sw,\upsilon)+g(sw,sw)\right]\right|_{s=0}\\ &=\frac12\left.\frac{d}{ds}\left[\vphantom{\frac{d}{ds}}s(g(\upsilon,w)+g(w,\upsilon))+s^2g(w,w)\right]\right|_{s=0}\\ &=g(\upsilon,w), \end{aligned} $ but I'm stuck with the other two terms for the reasons I myself do not fully understand. I must be able to calculate $FL$ using simple chain rule, but the differential $T\tau_Q$ of the bundle somewhy confuses me.

So, in case it's appropriate for Math.SE,

could someone provide an example of calculation for $L_2(\upsilon)=V(\tau_Q\upsilon)$?

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    @Jor I take your point, but would like to also point out that if Akater is indeed still learning English and used the word inadvertently, then it is much better to become aware of standard (and dare I say it, correct) usage. I have yet to hear "somewhy" spoken.2011-05-31

1 Answers 1

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The Legendre Transformation of $L$, is the fiber-preserving smooth map $\mathbb{F}L:TQ\to T^\ast Q$ defined by $\langle\mathbb{F}L(v),w\rangle=\left.\frac{d}{dt}\right|_{t=0}L(v+tw),\quad \forall x\in Q,v,w\in T_xQ.$

For the bilinearity of $g$, you have got:

$\left.\frac{d}{dt}\right|_{t=0}\ \frac{1}{2}g(v+tw,v+tw)=g(v,w).$

Remembering that $u$ and $w$ lie on the same fiber of $\tau_Q$, we get:

$\left.\frac{d}{dt}\right|_{t=0}\ V(\tau_Q(v+tw))=0,$

$\left.\frac{d}{dt}\right|_{t=0}\ g(v+tw,Y(\tau_Q(v+tw))=g(w,Y(\tau_Q(v)).$

Summarizing $\mathbb{F}L(v)=g(v+Y(\tau_Q(v)),\cdot).$