Missing step: Euler-Lagrange equations for the action integral

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Homework Help Overview

The discussion revolves around the application of the Euler-Lagrange equations in the context of the action integral, specifically focusing on the transition from a variation of the action to its first-order expansion using Taylor's theorem in a one-dimensional case.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand the derivation of the variation of the action integral and expresses confusion regarding the use of Taylor's theorem for multiple variables.
  • Some participants provide insights into Taylor's theorem and its application to the problem, while others seek further clarification on the steps involved in the derivation.

Discussion Status

Participants are actively engaging with the concepts, with some providing explanations and others requesting additional help. There is an indication that some have found external resources helpful, but the original poster still seeks a clearer understanding of the derivation process.

Contextual Notes

The original poster expresses uncertainty about specific terminology and concepts related to Taylor's theorem and its application in the context of the problem. There is a mention of a desire for a simplified explanation or step-by-step guidance.

alfredblase
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Hi its me again, stuck once more. Sorry guys and gals :P
Ok a problem I found on http://en.wikipedia.org/wiki/Action_(physics)
In a 1-D case how do we get from:
[tex]\delta S = \int_{t_1}^{t_2} [L(x + \varepsilon, \dot{x} + \dot{\varepsilon})-L(x,\dot{x})]dt[/tex]
to:
[tex]\delta S = \int_{t_1}^{t_2} \left(\varepsilon \frac{\pd L}{\pd x} + \dot{\varepsilon} \frac {\pd L} {\pd \dot{x}}\right)dt[/tex]
where [tex]\varepsilon = x_1(t) - x(t)[/tex]
and where the first order expansion of L in ε and ε′ is used? I don't even know what that last phrase means, so if someone could explain that to me too, that would be great.
Thankyou very much.
 
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Consider a real-valued function of 2 variables, say [tex]f = f(x,y)[/tex]. Taylor's theorem for multivariable calculus gives

[tex] f(x + \Delta x , y + \Delta y) - f(x,y) = \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y + ...[/tex]

In your variation principle, [tex]L[/tex] is a function of the "independent variables" [tex]x[/tex] and [tex]\dot{x}[/tex], so

[tex] L(x + \varepsilon , \dot{x} + \dot{\varepsilon}) - L(x,\dot{x}) = \frac{\partial L}{\partial x} \varepsilon + \frac{\partial L}{\partial \dot{x}} \dot{\varepsilon} + ...[/tex]

Neglecting the ... (i.e., the terms higher than first-order) gives the desired result.

Note: the "independent variables" actually are functions themselves.

Regards,
George
 
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Thanks George! =)

Edit: just posting a version of the last formula that shows correctly;

[tex]L (x + \varepsilon , \dot{x} + \dot{\varepsilon}) - L (x, \dot{x}) = \frac{\partial L}{\partial x} \varepsilon + \frac{\partial L}{\partial \dot{x}} \dot{\varepsilon} + ...[/tex]
 
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need more help on this one sorry...

I still don't understand how to get from one to other. I thought boning up on the Taylor theorem for multiple variables wouldn't be too hard, but I was wrong.
Can some one post a step by step, dummies guide to getting from:
[tex]f(x + \Delta x , y + \Delta y) - f(x,y)[/tex]
to:
[tex]\frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y + ...[/tex]
please?
thanx
38
 
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