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Initial-Value Version of Hamilton's Principle?

  1. Sep 2, 2013 #1
    As far as I understand, Hamilton's principle (a.k.a. "least action" or "stationary action") requires that you know both the initial and final location of a particle. Then, based on the requirement that the action must be stationary along any "possible" path, it will tell you what path(s) the particle possibly could have taken between those two points.

    What I'm wondering is, is there some sort of "initial value" form of this principle? That is, if I know the initial position and the initial velocity of a particle, can I determine the path it's going to take by minimizing (stationary-izing) something? Is that something, in fact, the Lagrangian?
     
  2. jcsd
  3. Sep 2, 2013 #2
    Hamilton's principle says that at equal times the variation in the paths must be zero (this eliminates the boundary/surface term), and out all paths left we chose the one which extremizes the action.

    If you want to solve initial value problems, then you should start with Newton's second law and proceed. Your last sentence confuses me. The Lagrangian is the functional that appears inside the integral for the expression of the action, S=∫L dt.
     
  4. Sep 2, 2013 #3

    UltrafastPED

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    Hamilton's principle specifies the starting and ending times for the action integral.
     
  5. Sep 2, 2013 #4
    Sorry, you're right, the last sentence was a mistake. I should have said something like "some functional of the Lagrangian" rather than just "Lagrangian." What I was getting at is I'm wondering whether we can solve the initial-value version of the problem by extremizing something related to the Lagrangian. I'm certainly aware that it can be done with Newton's second law (solving a vector differential equation), but I'm wondering if it can also be done through the extremization of some quantity.

    I'm not sure what you mean by "at equal times the variation in the paths must be zero."

    In every statement of Hamilton's principle that I've seen, it's assumed that we know the initial and final times, and that we know the position of the particle at the initial and final time. If you don't know the initial and final positions, then the typical derivation of the Euler-Lagrange equations using integration by parts doesn't work.
     
  6. Sep 2, 2013 #5
    That is tricky wording on my part. Say we were to do the integration from t1 to t2, then the distance (variation) between two paths should be zero at those two values of the time.

    EDIT: I meant to respond to the part about other possible extremizations. I'm not aware of any such construction. This is a pretty heuristic comment, but the dynamics of the system is really contained in the shape of the curves. It's like dropping a ball from different heights; up to a constant the equation for the position is identical. It decreases as t2. That t2 behavior comes from the form of the potential.
     
    Last edited: Sep 2, 2013
  7. Sep 3, 2013 #6

    UltrafastPED

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    IIRC, Cornelius Lanczos goes over this in "The Variational Principles of Mechanics", but I'm having trouble finding it ... so from memory I think it goes like this:

    The perfect cancellations that we see in the "usual" development of the Lagrangian equations of motion are very convenient, but not the only possibility. They correspond to the "weak variational principle". If you allow one of them to be non-zero it produces a "boundary term"; this boundary term has an interesting structure when transformed to the Hamiltonian: it is an integral of the form [p dq].

    I will look further in Lanczos today and see what he actually had to say.

    See Lanczos, chapter 5, especially sections 3 and following. This is consistent with Voko's comment.
     
    Last edited: Sep 3, 2013
  8. Sep 3, 2013 #7
    In calculus of variations, there is a (lacking a better term) most general problem, which minimizes a functional where neither the range of the integral nor the values of the function at the limits of integration are specified.

    It is shown that even in this case the Euler-Lagrange must be satisfied, plus there is an additional condition: $$

    \left[ \sum_{i = 0}^n p_i \delta q_i - H \delta t \right]_{t = t_0}^{t = t_1} = 0

    $$ where ## q_i, \ p_i, \ H ## are the canonical variables and the Hamiltonian corresponding to the functional, and ##t_0, \ t_1 ## are the unknown limits of integration (i.e., initial and final time). If there are no additional constraints (such as that the boundary values must be on certain surfaces), this yields ## 2n + 2 ## additional equations, which matches the number of unknown constants.

    Now, if the initial time is given then ## \delta t |_{t = t_0} = 0 ##; if the initial positions are given, ## \delta q_i |_{t = t_0} = 0 ##, we are left with ##n + 1 ## additional equations, which also matches the number of still unknown constants.

    But if we now also specify initial velocities, we may over-determine the problem. The solution will then exist only if the initial velocities are compatible with the solution that one obtains without specifying the velocities. This is analogous to specifying initial velocities when the end position is specified: they cannot be arbitrary.

    Note the above applies to any functional, not necessarily to the classical least action functional.
     
  9. Sep 3, 2013 #8
    Thanks, that's exactly what I was looking for!
     
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