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DS/dt = -H?

  1. Dec 10, 2006 #1
    I'm trying to understand how [tex]\frac{\partial S}{\partial t} = -\mathcal{H}[/tex]. I put the simplest/one dimensional Lagrangian ([tex]mv^2/2-V[/tex]) and tried to derive it, but I failed:

    [tex]\frac{\partial S}{\partial t} = \frac{\partial }{\partial t} \int_{t_i}^{t_f} Ldt[/tex]

    noting that x and [tex]\dot{x}[/tex] is a function of t, expanding the integrand into power series and collecting only the first terms:

    [tex]= \int_{t_i}^{t_f} \left( \frac{\partial (m\dot{x}^2)}{\partial \dot{x}}\frac{\partial \dot{x}}{\partial t} - \frac{\partial V}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial L}{\partial t} \right) dt[/tex]

    since [tex]-\frac{\partial V}{\partial x} = F[/tex] and because Lagrangian has no direct time dependence, the last term should drop. Also, [tex]\frac{\partial x}{\partial t} = \frac{dx}{dt}[/tex] since it's a function of t only (same goes for [tex]\dot{x}[/tex])

    [tex]= \int_{\dot{x}(t_0)}^{\dot{x}(t_f)} m\dot{x} d\dot{x} + \int_{x(t_0)}^{x(t_f)} F dx[/tex]
    which's not something like [tex]-\mathcal{H}[/tex]. Any ideas what has gone wrong? And could someone work out full derivation?
     
    Last edited: Dec 10, 2006
  2. jcsd
  3. Dec 10, 2006 #2

    vanesch

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    The dS/dt = - H applies when you consider S a function of the endpoint S(q_f, t_f), so it is actually [tex]\partial S/\partial t_f[/tex] if I remember well.
     
  4. Dec 11, 2006 #3

    dextercioby

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    S is a functional, not a function, it doesn't have partial derivatives, but functional derivatives, like Ga^teaux or Frechet...

    So your [itex] \frac{\partial S}{\partial t} = - H [/itex] is what physicists call "bull****".

    Daniel.
     
  5. Dec 11, 2006 #4

    vanesch

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    No, you should check Landau and Lif****z. At a certain moment (just before they introduce the minimum action principle of Maupertuis - I have the 3rd edition from Mir, in French), they consider the action, for a given initial point (q1,t1), as a function of the final point (q2,t2), using only the correct path:
    [tex] S(q_1,t_1,q_2,t_2) = \int_{t_1}^{t_2} dt L(q(t),\dot{q}(t),t) [/tex] in which we already substituted the correct [tex]q(t)[/tex] which is the solution to the variational problem. As such, S is simply a function of the begin and end points, and by convention, we fix the begin point, which makes S then an ordinary function of the end points.
    Turns out that [tex]\frac{\partial S}{\partial q_i}[/tex] equals [tex]p_i[/tex] (or minus ? I don't remember...) and [tex]\frac{\partial S}{\partial t}[/tex] equals [tex]-H[/tex]. The partial derivative is by changing the end time of the end point, without changing the point itself: this will slightly modify the dynamical solution (as now the trajectory between q_1 and q_2 will have to take a bit more time), and hence the action integral for that solution.
     
    Last edited: Dec 11, 2006
  6. Dec 11, 2006 #5

    vanesch

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    There is a simple way to see how this comes about in all generality.
    Consider the solution to the dynamical problem q(t), which is such that [tex]q_1 = q(t_1)[/tex] (point (1) ) and [tex]q_2 = q(t_2)[/tex] (point (2)). In other words, it is the extremal problem, solved:
    [tex]
    \delta \int_{t_1}^{t_2} dt L(q,\dot{q},t) = 0
    [/tex]

    The corresponding action corresponds to:
    [tex]
    S_1 = \int_{t_1}^{t_2} dt L(q(t),\dot{q}(t),t)
    [/tex]

    We can say that the point (q1,t1) and the point (q2,t2) are "on the path" q(t).

    Now, consider that we lengthen the dynamical solution q(t) a bit further. In order to do so, there has to be continuity of the derivatives at point 2. We consider that we lengthen the path a bit to (q2 + dq, t + dt). We call this point, point (3). Note that we don't really have a choice for dq ! dq will be [tex]\dot{q} dt[/tex], because it is determined by the "initial" conditions of the arrival of the trajectory at point (2).

    We can consider the dynamical problem between (1) and (3), but we know already that the path between (1) and (2) will be part of it. So our second action will be:
    [tex]
    S_2 = \int_{t_1}^{t_2+dt} dt L(q(t),\dot{q}(t),t) = S_1 + L(q_2,\dot{q}_2,t) dt
    [/tex]

    We can write this as:
    [tex]
    \frac{dS}{dt} = L
    [/tex]

    which reads: the derivative of the action (along the correct path) equals the lagrangian (at the end point we're considering).

    But this is not what we are after. We wanted the solution of the dynamical problem, with t shifted from t2 to t2 + dt, but without changing the end point.

    So we have to work back a bit. We can say that:
    [tex]\frac{dS}{dt} dt = \frac{\partial S}{\partial q} dq + \frac{\partial S}{\partial t}dt [/tex]
    from which we derive:

    [tex]
    \frac{\partial S}{\partial t} = \frac{dS}{dt} - \frac{\partial S}{\partial q} \frac{dq}{dt}
    [/tex]

    where we have to remember that the increments dq and dt were along a genuine dynamical path. If we now accept (can be shown too) that [tex]\frac{\partial S}{\partial q} = p[/tex], then we have:

    [tex]
    \frac{\partial S}{\partial t} = L - p \dot{q}
    [/tex]

    which is nothing else but - H.
     
    Last edited: Dec 11, 2006
  7. Dec 11, 2006 #6

    dextercioby

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    I see now, it's another matter than i had been thinking of in the first place.

    A suggestion of notation would be to consider changing [itex] dt,dq [/itex] with [itex] \Delta t, \Delta q [/itex].

    Keeping the d's means using differential forms where they shouldn't belong.

    Daniel.
     
  8. Dec 11, 2006 #7

    vanesch

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    Yes, that's a good idea...
     
  9. Dec 11, 2006 #8
    Thanks vanesch, that leaves yo show [tex]\frac{\partial S}{\partial q} = p[/tex], but it appears to be as troublesome as the first question... (I'm a newbie to such things) I'll give it a try in the evening.

    This's simply a flame and I wonder where the admins are.

    Daniel, thank you for your warm, enlightening comment. I'm glad that I don't fit into category of people whom you call physicist.
     
  10. Dec 11, 2006 #9

    vanesch

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    Well, you can work out the variation of the action:
    [tex] q \rightarrow q + \Delta q [/tex], as you do to derive the E-L equations in the first place, keeping the time boundaries fixed:

    [tex] \Delta S = \left[\frac{\partial L}{\partial \dot{q}} \Delta q \right]_{t_1}^{t_2} + \int_{t_1}^{t_2}\left(
    \frac{\partial L}{\partial q}- \frac{d}{dt}\frac{\partial L}{\partial \dot{q}}\right) \Delta q dt [/tex]

    Usually, we say that the variation [tex] \Delta q [/tex] has to vanish on the end points, which makes the integrated term disappear. If otherwise Delta is a random small variation, then the only way to get this expression to 0, is by requiring the E-L equation to be valid. That's usually how the E-L equation is derived from the variational principle.

    However, this time, we do things differently. We consider that the E-L equation IS satisfied (we are working with a solution to the dynamical problem), but we consider a variation which does NOT vanish at the end point (2) (although it does, at the end point (1)). We say that [tex] \Delta q (t_2) = \Delta q[/tex]
    As such, the thing that remains is:
    [tex] \Delta S = \frac{\partial L}{\partial \dot{q}} \Delta q \right]_{t_2} [/tex]
    from which we can deduce that [tex]\partial S / \partial q_2 = p [/tex]
     
    Last edited: Dec 11, 2006
  11. Dec 12, 2006 #10

    vanesch

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    Let's not get into a shouting contest here. Daniel misunderstood the original question, no need to get nervous over it.
     
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