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Evaluating the non-relativistic Action

  1. Feb 9, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider the classical action [tex](m/2) \int dt \dot{x}^{2}[/tex] where m is the mass of the non-relativistic point particle, and [tex]\dot{x} = dx/dt [/tex] .

    This classical path starts at [tex](x_{1}, t_{1})[/tex] and ends at [tex](x_{2}, t_{2})[/tex]. Could someone help me evaluate this classical action explicitly please?


    2. Relevant equations



    3. The attempt at a solution

    Thanks a lot!
     
  2. jcsd
  3. Feb 9, 2010 #2

    Simply add the time bounds of the path to the integral to get

    [tex](m/2) \int^{t_2}_{t_1} \dot{x}^{2}dt[/tex].

    Now turn this into a position-bounded integral as follows:

    [tex](m/2) \int^{t_2}_{t_1} \dot{x}^{2}dt=(m/2) \int^{x_2}_{x_1} \dot{x}dx[/tex].

    Finally take the time derivative out of integrand and solve the integral!

    AB
     
  4. Feb 9, 2010 #3
    Hi there,

    Could you kindly explain why one can turn the time-bounded integral into a position-bounded integral?

    Thanks a lot!
     
  5. Feb 10, 2010 #4
    As in the classical theory of action integrals, the bounds must be either spatial or temporal, so we cannot do the analysis over the dynamics of particles along a bounded path with both elements of time and space simultaneously. So in the integral in question, you can use the chain rule to kill two of time differentials dt by expanding one of [tex]\dot{x}[/tex]'s and cancelling out dt's to only have a dx left behind. The remaining is straightforward to be treated. Since now integral element is dx, so the bound must be spatial and then you can easily factor the operator d/dt out of the integral to get

    [tex]\frac{m}{2}\frac{d}{dt}\int^{x_2}_{x_1} x dx[/tex].

    Is it all clear now?

    AB
     
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