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How to prove Momentum = Gradient of Action?

  1. May 27, 2015 #1
    How can show that momentum is the gradient of the action for the free particle? I tried it like this for one dimensional case:

    [tex] s=\int Ldt [/tex]
    [tex] ds=Ldt [/tex]
    [tex] ds=\frac{mv^2}{2}dt\:[/tex]
    Velocity is constant right? So I should be able to to this:
    [tex] \frac{ds}{dx}=\frac{mv^2}{2}\frac{dt}{dx} [/tex]
    I'm not sure about the following step where I use: [tex] \frac{dt}{dx}=\frac{1}{v} [/tex] is this correct?
    [tex] \frac{ds}{dx}=\frac{mv^2}{2}\frac{1}{v} [/tex]
    And finally I get:
    [tex] \frac{ds}{dx}=\frac{mv}{2} [/tex]
    It differs from momentum by the factor of 1/2, where did I make the mistake?
    Last edited: May 27, 2015
  2. jcsd
  3. May 27, 2015 #2


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    The Lagrangian is to be written as a function of t, q, and qdot (, or t, x, and v in your 1D example). The weird thing about it is that x and v are supposed to be treated as formally distinct independent variables, and v cannot be replaced by dx/dt.

    I'm actually not sure what "gradient of the action" means, since action is a functional, depending on the path, rather than depending on the position like a normal function.
    Last edited: May 27, 2015
  4. May 27, 2015 #3


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    Right. You need to differentiate the potential energy part of the Lagrangian
  5. May 27, 2015 #4
    Suppose I have no potential, just the kinetic energy of a free particle wandering around (actually moving at a straight line with a constant velocity), the Lagrangian will be equal to kinetic energy only. I've found out a hint why my derivation might be wrong, there's a thing called abbreviated action and it differs from ordinary action which is just an integral of Lagrangian (if it's in units of energy) with time. But I'm clueless what does that mean and how or why it differs from action.


    [tex] S_0=\int pdq [/tex]
    [tex] dS_0=pdq [/tex]
    [tex] \frac{dS_0}{dq}=p [/tex]
    For more general case I see it's a gradient but what the hell is abbreviated action? I came across this thing in quantum mechanics textbook (in introductory chapter, where classical mechanics was discussed)
    Last edited: May 27, 2015
  6. May 27, 2015 #5


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    I'm rusty on these things but one key element is that you're looking at variations on a path that are supposed to be independent. That is p and x are assumed independent in partial derivatives. You can look at http://www.physics.usu.edu/torre/6010_Fall_2010/Lectures/02.pdf
  7. May 28, 2015 #6


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    Free particle moves with constant velocity [tex]v = \frac{x - x_{0}}{t - t_{0}} = \mbox{const.}[/tex] So the action is [tex]S = \frac{m}{2} ( \frac{x - x_{0}}{t - t_{0}} )^{2} \int^{t}_{t_{0}} d t = \frac{m}{2} \frac{(x - x_{0})^{2}}{t - t_{0}} .[/tex] Now, you can calculate its partial derivative with respect to [itex]x[/itex] [tex]\frac{\partial S}{\partial x} = m \frac{x - x_{0}}{t - t_{0}} = m v = p.[/tex] Notice, you also get the Hamiltonian from [tex]\frac{\partial S}{\partial t} = - \frac{m}{2} ( \frac{x - x_{0}}{t - t_{0}} )^{2} = - H.[/tex] Now you can back check as follow [tex]S = \int d S = \int \left( \frac{\partial S}{\partial x} d x + \frac{\partial S}{\partial t} d t \right) = \int \left( p \dot{x} - H \right) \ d t = \int L d t .[/tex] Of course, to be accurate you must think of the action as function of the end points [itex]S = S ( x , t ; x_{0} , t_{0} )[/itex]. For the general case, you can use the variation principle to show that [tex]\delta S = \int_{0}^{t} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \delta q \right) = p \delta q |_{0}^{t} .[/tex] Assuming [itex]\delta q (0) = 0[/itex] we find [tex]\delta S = p(t) \delta q , \ \ \Rightarrow \ p(t) = \frac{\partial S}{\partial q} .[/tex]

  8. Jun 22, 2015 #7
    Oh, thanks, now I understand. I didn't realize that velocity doesn't depend on time and so Lagrangian goes out of the integral here
    [tex]S = \frac{m}{2} ( \frac{x - x_{0}}{t - t_{0}} )^{2} \int^{t}_{t_{0}} d t [/tex]
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