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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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6  
+1 for "somewhy" :-) –  joriki May 31 '11 at 13:33
    
@joriki I'm not a native speaker. My apologies, I didn't know the word was nonstandard. –  Akater May 31 '11 at 14:01
1  
@Akater: No apologies required -- I hadn't heard the word before and thought it was a nice and plausible analogy to "somewhere" :-) –  joriki May 31 '11 at 14:34
    
@Joriki, Akater: It's not really a word. The expression you are looking for is "for some reason". –  Glen Wheeler May 31 '11 at 14:45
    
@Glen: That's a prescriptivist view. Descriptively speaking, it's a word because people use it as a word: urbandictionary.com/define.php?term=somewhy. I wouldn't go so far as to infer that Akater is looking for a different expression -- to the contrary, becoming familiar with non-standard language use is an important part of getting to know a language for non-native speakers. –  joriki May 31 '11 at 14:57

1 Answer 1

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

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