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Leonard Susskind : Classical Mechanics

  1. Jun 19, 2008 #1


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  3. Oct 10, 2009 #2
    isn't the derivative of the first derivative squared:
    d/dt (x')^2 = 2x'x''? why does susskind claim it is 2x'', in his classical lecture 3?
  4. Oct 11, 2009 #3
    he assumes X(t) so 'x' is a function of time not only of x , he is using the chain rule
  5. Oct 11, 2009 #4
    Ithink he is not using the chain rule properly. if x is a function of time only, d/dt (dx/dt)^2 = 2dx/dt * d^2x/dt^2
  6. Oct 11, 2009 #5


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    You've already asked this question, and had it answered in another thread. I don't know where abouts in the video you've seen this, but I'm guessing it has to do with the Euler-Lagrange equations:

    \frac{d}{dt}\Big(\frac{\partial\mathcal{L}}{\partial\dot{x}}\Big)=\frac{\partial\mathcal{L}}{\partial x}[/tex]

    So, the LHS is not taking the time derivative of the Lagrangian, but is instead the time derivative of the derivative of the Lagrangian with respect to the coordinate velocity. It is important to treat the coordinate velocity as a variable; that is [tex]\mathcal{L}\equiv\mathcal{L}(x, \dot{x})[/tex].

    If this doesn't clear things up, let me know the exact time in the video that you're confused with, and I'll try and look at it.
  7. Oct 11, 2009 #6
    I had the thread deleted because i decided I wanted to post it here.
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