14
$\begingroup$

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 $L_2(\upsilon)=V(\tau_Q\upsilon)$?

$\endgroup$
5
  • $\begingroup$ @joriki I'm not a native speaker. My apologies, I didn't know the word was nonstandard. $\endgroup$
    – akater
    May 31, 2011 at 14:01
  • 2
    $\begingroup$ @Akater: No apologies required -- I hadn't heard the word before and thought it was a nice and plausible analogy to "somewhere" :-) $\endgroup$
    – joriki
    May 31, 2011 at 14:34
  • $\begingroup$ @Joriki, Akater: It's not really a word. The expression you are looking for is "for some reason". $\endgroup$
    – Glen Wheeler
    May 31, 2011 at 14:45
  • $\begingroup$ @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. $\endgroup$
    – joriki
    May 31, 2011 at 14:57
  • $\begingroup$ @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. $\endgroup$
    – Glen Wheeler
    May 31, 2011 at 17:27

1 Answer 1

Reset to default
4
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .