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Euler Lagrange equation of motion

  1. Jun 1, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the equations of motion for both r and [itex]\theta[/itex] of
    5E%7B2%7D%20%5Cright%29%5Cdot%7Br%7D%5E%7B2%7D+r%5E%7B2%7D%5Cdot%7B%5Ctheta%7D%20%5Cright%29-mgH.png

    2. Relevant equations
    My problem is taking the derivative wrt time of

    ex?%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20%5Cdot%7B%5Ctheta%7D%7D=mr%5E2%5Cdot%7B%5Ctheta%7D.png
    and
    [itex]\dfrac{\partial\mathcal{L}}{\partial\dot{r}}=m \dot{r} \left( 1 + \left( \dfrac{\partial H}{\partial r}\right)^2 \right)[/itex]

    3. The attempt at a solution
    So
    %7B%5Ctheta%7D%7D%20%5Cright%29=2mr%5Cdot%7Br%7D%5Cdot%7B%5Ctheta%7D+mr%5E2%5Cddot%7B%5Ctheta%7D.png
    I think this is correct. I have the solution to the second expression but I am unsure how it was found
    [itex]
    m \ddot{r} \left( 1 + \left( \dfrac{\partial H}{\partial r}\right)^2 \right) + 2 m \dot{r}^2 \dfrac{\partial H}{\partial r}\dfrac{\partial^2 H}{\partial r \partial t}
    [/itex]
    The first part of this expression is fine it is the second part I am unsure of.
     
  2. jcsd
  3. Jun 1, 2015 #2

    Orodruin

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    Please specify what ##H## is a function of.
     
  4. Jun 1, 2015 #3
    HI, sorry H(r).
     
  5. Jun 1, 2015 #4

    Orodruin

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    So if H is only a function of r, the only possible derivative of H is the total derivative wrt r. Since r depends on t, the derivative wrt t is given by the chain rule. This also goes for dH/dr, which is also a function of r. The partial derivative of H wrt t does not make sense if H does not depend explicitly on t.
     
  6. Jun 1, 2015 #5

    Ray Vickson

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    So, if we write ##K(r) = (\partial H(r) /\partial r)^2## (which is just some computable function of ##r##) you have
    [tex] {\cal{L}} = \frac{1}{2} m \left((1+ K(r)) \dot{r}^2 + r^2 \dot{\theta} \right) - mgH(r) [/tex]
    Thus,
    [tex] {\cal{L}}_r \equiv \partial {\cal{L}} /\partial r = \frac{1}{2}m \left( K'(r) \dot{r}^2 + 2 r \dot{\theta} \right) - m g H'(r) [/tex]
    so
    [tex] \frac{d}{dt} {\cal{L}}_r = \frac{1}{2} m \left( K''(r) \dot{r}^3 + K'(r) 2 \dot{r} \ddot{r} + 2 \dot{\theta} \right) - mg H''(r) \dot{r} [/tex]

    If ##H## does not depend explicitly on ##t## there should be no ##\partial H/\partial t## anywhere.
     
  7. Jun 1, 2015 #6

    Orodruin

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    This, although correct, is not relevant for the Euler-Lagrange equations which contain what you would call ##d\mathcal L_{\dot r}/dt##. Also, please allow the OP to try to do the derivatives for himself.
     
  8. Jun 1, 2015 #7

    Ray Vickson

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    You are right. Anyway, the OP did try to obtain the derivatives for himself, but made some errors. I admit maybe I should have not written so much detail (which does not matter anyway, since I did the wrong calculation!).
     
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