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A problem regarding to Lagrangian in Classical Mechanics

  1. Dec 30, 2011 #1
    1. The problem statement, all variables and given/known data
    I have a problem regarding to lagrangian.

    If L is a Lagrangian for a system of n degrees of freedom satisfying Lagrange's equations, show by direct substitution that

    [itex] L' = L + \frac{d F(q_1,...,q_n,t)}{d t}[/itex]

    also satisfies Lagrange's equations where F is any ARBITRARY BUT DIFFERENTIABLE function of its arguments.



    2. Relevant equations
    Lagrange's equations:
    [itex] \frac{\partial L}{\partial q_i} - \frac{d}{d t}\frac{\partial L}{\dot{\partial q_i}} =0[/itex]


    3. The attempt at a solution
    Equivalently we have to find
    [itex] \frac{\partial F}{\partial q_i} - \frac{d}{d t}\frac{\partial F}{\partial \dot{q_i}} =0[/itex]
    It is obvious that [itex]\frac{\partial F}{\partial \dot{q_i}}=0[/itex].
    But how can I get [itex] \frac{\partial F}{\partial q_i}=0[/itex] ?

    Thank you.
     
    Last edited: Dec 30, 2011
  2. jcsd
  3. Dec 30, 2011 #2

    I like Serena

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    Homework Helper

    Welcome to PF, iiternal! :smile:

    This looks like an exercise in multivariable differentiation.

    Let's start with your relevant equation.
    You seem to have dropped a partial derivative there...

    I'm afraid that since F is an arbitrary function of ##q_i##, you won't get ##\frac{\partial F}{\partial q_i}=0##.

    I believe you have to substitute L' in Lagrange's equation and expand everything.
    This means expanding dF/dt into partial derivatives.
    Do you know how to do that?
     
  4. Dec 30, 2011 #3
    Thank you very much! You are absolutely right!

    let ## G = \frac{dF}{dt} = \sum\frac{\partial F}{\partial q_i}\dot{q_i}+\frac{\partial F}{\partial t}##;
    Substitute it into Lagrange's Equation
    ## \frac{\partial G}{\partial q_i} - \frac{d}{dt}\frac{\partial G}{\partial \dot{q_i}}
    = (\frac{\partial^2F}{\partial q_i^2}\dot{q} + \frac{\partial^2F}{\partial q_i\partial t}) - \frac{d}{dt}\frac{\partial F}{\partial q_i} ##

    I believe the last term can somehow cancel both the first two terms.
    When I tried to expand the last term, I faced another problem
    ## \frac{d}{dt}\frac{\partial F}{\partial q_i} = \frac{\partial }{\partial q}\frac{\partial F}{\partial t} + ???##
    How can I take derivative of the denominator of a differentiation?

     
  5. Dec 30, 2011 #4

    I like Serena

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    Good! :)

    I'm afraid you should not use the index i everywhere and keep the summations.

    Your equations should read:
    ## G = \frac{dF}{dt} = \sum\limits_j\frac{\partial F}{\partial q_j}\dot{q_j}+\frac{\partial F}{\partial t} ##

    Substitute it into Lagrange's Equation
    ## \frac{\partial G}{\partial q_i} - \frac{d}{dt}\frac{\partial G}{\partial \dot{q_i}}
    = \sum\limits_j ... ##

    And:
    ## \frac{d}{dt}\frac{\partial F}{\partial q_i} = \sum\limits_j {\partial^2 F \over \partial q_j\partial q_i} \dot q_j + {\partial^2 F \over \partial t\partial q_i}##
     
  6. Dec 30, 2011 #5
    Great !!!
    Thank you very much!
    Happy New Year.

     
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