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Lagrangian classical action for particle with constant force

  1. Feb 23, 2014 #1

    micole

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    1. The problem statement, all variables and given/known data
    for particle with lagrangian L = m/2 dx/dt^2 + fx where x is constant force, what is ScL (classical action)


    2. Relevant equations
    d/dt (∂L/∂(dx/dt)) = ∂L/∂x
    ScL = ∫m/2 dx/dt^2 + fx dt from ti to tf

    3. The attempt at a solution
    d/dt (∂L/∂(dx/dt)) = ∂L/∂x implies f = md2x/dt2 (f = ma)

    on aside, from this you can calculate the familiar equations
    dx/dt (t) = f/m(t) + dx/dt(0) or v(t) = at + v0
    x(t) = 1/2 f/m t^2 + dx/dt(0) t + x(0)

    Question: is it helpful to define ti as zero so Xi = x(ti) = X(0) = 0? And if I do this, is my
    solution generalizable? I think yes

    Scl = ∫m/2 dx/dt^2 + fx dt
    for ∫m/2 dx/dt^2 I integrate by parts with
    u =dx/dt --> du = (d2x/dt2) dt
    and
    dv = dx/dt dt --> v = x

    Scl = m/2 (dx/dt)(x) from ti to tf -m/2∫fx dt + ∫fx dt
    Scl = m/2 {(xf )(f/m tf +dx/dt(0)) - xi(f/m ti + dx/dt(0))} + m/2 ∫fx dt
    where xi(f/m ti + dx/dt(0)) goes away if its okay to set xi = x(0) = 0

    so potentially im left with
    Scl = m/2 (xf )(f/m tf +dx/dt(0)) + fm/2 ∫x dt
    I have X as a function of t so i can integrate the right hand term ( ∫x dt) but I end up with a big equation thats a function with terms t^3/3 and many other terms. Most of the solutions to these problems have ended up simple and elegant so im assuming im either making an error or there is a slick math trick that simplifies by function of ti, tf and xf, xi into something lovely.
    Thats the part i would like help on please

    Thanks very much in advance
     
  2. jcsd
  3. Feb 24, 2014 #2

    micole

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    Someone please help (; 67 views and no replies! It's lonely
     
  4. Feb 24, 2014 #3

    hilbert2

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    Well, the action is just the time integral of the Lagrangian over the time interval that is being considered:

    ##S=\int_{t_{1}}^{t_{2}}L(t)dt=\int_{t_{1}}^{t_{2}}\left[\frac{m}{2}\left(\frac{dx(t)}{dt}\right)^{2}+Fx(t)\right]dt##, where ##F## is the constant force.

    As it's basic physics that the trajectory of the particle is ##x(t)=x_{0}+v_{0}t+\frac{F}{2m}t^{2}##, it's easy to calculate the integral provided that the force ##F##, the initial position and velocity ##x_{0},v_{0}## and the time interval ##[t_{1},t_{2}]## are known.
     
  5. Feb 25, 2014 #4

    micole

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    Thanks hilbert. None of those are known. If you take a look at my work you see that what we are being asked is to simplify this integral from a long function of t (as you and I both indicate, it's basic physics to write x as a function of t) into a succinct statement. The point is not to substitute numbers (see final paragraph of my question)
    Thanks for any help
     
  6. Feb 25, 2014 #5

    micole

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    And I see an error in my statement. The coefficient of integral of Fx is (2-m)/2, not m/2
     
  7. Feb 25, 2014 #6

    hilbert2

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    How is that possible? You are subtracting a dimensional quantity ##m## from a bare number ##2##.

    I'm not sure how the integral ##\int_{t_{1}}^{t_{2}}\left[\frac{m}{2}\left(\frac{dx(t)}{dt}\right)^{2}+Fx(t)\right]dt## could be simplified without solving for ##x(t)## with the initial conditions ##x_{0}##, ##v_{0}## known. Integration by parts doesn't seem to help.
     
  8. Feb 25, 2014 #7

    micole

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    Me neither, exactly my question
     
    Last edited: Feb 25, 2014
  9. Feb 25, 2014 #8
    Hi.
    Is this the exact problem statement? It doesn't suggest that you should get a particularly succinct (or enlightening) result... You are given a specific potential, so you're indeed supposed to solve for x(t) in terms of a (= the constant acceleration f/m), v0 and x0, then integrate.
    You can always make abbreviations like L[x(t0)] in your final result, but you'll always get something with at least four terms and no immediate connection to something familiar.

    The thing is, this kind of Lagrangian is defined as L = T – V, where T and V are determined by your choice of coordinates; then it's the principle of least action that gives you the familiar equations of motion upon varying S and you would be going in circles by trying to get something edifying from S. It's just what it's supposed to be: its specific form may look uneventful but that's just how things are sometimes... :)
     
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