1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question on Lagrangian and Action

  1. May 21, 2006 #1
    I'm reading my textbook and trying to follow the math on how to minimize the action for an arbitrary Lagrangian. The author states that the action is:
    [tex]
    S[x(t)] = \int^{t_B}_{t_A} dt L( \dot x(t),x(t))
    [/tex]

    Then the author goes on to talk about finding the extrema for the action by computing [itex] \delta S[x(t)] [/itex]. The author says to compute this by substituing [itex] x(t) + \delta x(t) [/itex] into the definition for the action, expanding to 1st order and integrate by parts. The text then shows this:
    [tex]
    \delta S[x(t)] = \int^{t_B}_{t_A} dt [\frac {\partial L}{\partial \dot x(t)} \delta \dot x(t) + \frac {\partial L}{\partial x(t)} \delta x(t)] = [ \frac {\partial L}{\partial \dot x(t)} \delta x(t) ]^{t_B}_{t_A} + \int^{t_B}_{t_A} dt [ - \frac {d}{dt} \frac {\partial L}{\partial \dot x(t)} + \frac {\partial L}{\partial x(t)} ] \delta x(t)
    [/tex]

    What I don't understand is how the author got this. If all he did was substitute and expand like he said to do, what happened to the 0th order term from the expansion? I'm also unsure how he got the right most form of the equation using integration by parts.
     
    Last edited: May 21, 2006
  2. jcsd
  3. May 21, 2006 #2

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Given a function [itex]X(t)=x(t)+\delta{x}(t)[/itex], (where x(t) is the assumed solution of the resulting diff. eq.) he is looking at the first order term of:
    [tex]\bigtriangleup{S}\equiv{S}(X(t))-S(x(t))[/tex]
    That first order expression is proportional to [itex]\delta{x}(t)[/itex], and is called [itex]\delta{S}(x(t))[/itex]
     
  4. May 21, 2006 #3
    Ah, ok. That makes sense, I just wish the author had made that clear.

    Thanks
     
  5. May 21, 2006 #4
    One other question that's sorta related to the first. In this case, the author talked about a value [itex] \delta S[x(t)] [/itex]. I notice that if I simpily take the first order derivative, I get something that looks exactly the same except for all the little deltas changing to d's. Is there any real difference? Or is the delta notation just to keep you from being confused as to what is being integrated?
     
  6. May 21, 2006 #5

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Well, the variation can be regarded as a differential on A FUNCTION SPACE, rather than, say, on the real line of numbers.

    It is the type of space we're working with that is different; the basic "idea" is the same.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Question on Lagrangian and Action
  1. Lagrangian question (Replies: 1)

  2. Lagrangian question (Replies: 11)

Loading...