alfredblase
- 228
- 0
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.
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.
Last edited: