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Hamilton's principle: why dA rather than dA/dt?

  1. Feb 1, 2012 #1
    Hamilton's principle is described as

    [itex]\delta[/itex]I=[itex]\delta[/itex][itex]\int[/itex]L dt = 0

    so as the action is stationary.

    This does not seem to be the same as dI/dt = 0, which is how I understand the condition for a function being stationary.

    Am I misinterpreting the equation?
  2. jcsd
  3. Feb 2, 2012 #2
    The action is a *functional* of the trajectory. That is to say, its input is a function (position as a function of time) and its output is a number. This is contrasted with a regular function whose input is a number and whose output is a number. So when we say we want the action to be stationary we mean that if we modify the input to the action functional slightly, the output should be unchanged. The input, though, is a function, not a time. So here "stationary" means: if we replace the trajectory x(t) with a slightly different trajectory x(t)+a(t), where a(t) is a function that is always small, then the action of the new trajectory should be the same as the action of the old trajectory.
  4. Feb 2, 2012 #3
    Thanks. Your explanation was much clearer than the textbooks I've been reading.
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