Time-Dependent Classical Lagrangian with variation of time

1. Jun 12, 2014

586

Hello everyone!

I was reading the following review:

http://relativity.livingreviews.org/open?pubNo=lrr-2009-4&amp;page=articlesu23.html [Broken]

And I got stuck at the first equation; (10.1)

So how I understand this is that there are two variations,

$\tilde{q}(t)=q(t)+\delta q(t) \hspace{1cm} \text{and} \hspace{1cm} \tilde{t}=t+\delta t$

Further we also have a `total variaton' for q at first order:
$\tilde{q}(\tilde{t})=q(t)+\delta q(t)+\dot{q}(t)\delta t$

and its derivative,
$\dot{\tilde{q}}(\tilde{t})=\dot{q}(t)+\delta\dot{q}(t)+\ddot{q}(t) \delta t$

So now how is $\delta L(q,\dot{q},t)$ defined?

Last edited by a moderator: May 6, 2017
2. Jun 13, 2014

UltrafastPED

If the latter, see the attachment at https://www.physicsforums.com/showthread.php?t=752726#2

It contains a leisurely introduction to the calculus of variations, followed by derivations of Lagrangians & Hamiltonians.

3. Jun 13, 2014

586

The question is about the fundamentals of calculus of variations. I know how to derive the usual Euler-Lagrange equations without the extra variation in t $\tilde{t}=t+\delta t$. But I am having trouble incorporating this extra variation.

if i define:
$\delta L = \frac{\partial L}{\partial q} \delta q + \frac{\partial L}{\partial q} \frac{\partial q}{\partial t}\delta t + \frac{\partial L }{\partial \dot{q} } \delta \dot{q} +\frac{\partial L}{\partial \dot{q}} \frac{\partial \dot{q}}{\partial t}\delta t + \frac{\partial L}{\partial t} \delta t$

then it gets a similar result as (10.1) but everywhere there is $\dot{q}$ they have $-\dot{q}$.

Its driving me pretty crazy. Any help would be greatly appreciated.

4. Jun 13, 2014

UltrafastPED

5. Jun 13, 2014

586

Unfortunately the attachment does not treat coordinate variations.

6. Jun 13, 2014

UltrafastPED

Then you will need to find a specialized text book.

7. Jun 15, 2014