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B Partial derivative of Lagrangian with respect to velocity

  1. May 1, 2016 #1
    I came across a simple equation in classical mechanics,
    $$\frac{\partial L}{\partial \dot{q}}=p$$
    how to derive that?
    On one hand,
    $$L=\frac{1}{2}m\dot{q}^2-V$$
    so, $$\frac{\partial L}{\partial \dot{q}}=m\dot{q}=p$$
    On the other hand,
    $$L=\frac{1}{2}m\dot{q}^2-V=\frac{1}{2}m\dot{q}\dot{q}-V=\frac{1}{2}\dot{q}p-V$$
    $$\frac{\partial L}{\partial \dot{q}}=\frac{1}{2}p$$
    which is half value from the first derivation.
     
  2. jcsd
  3. May 1, 2016 #2

    stevendaryl

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    Well, writing [itex]\frac{1}{2}m \dot{q}^2[/itex] as [itex]\frac{1}{2}\dot{q} p[/itex] doesn't change anything; you have to use the product rule:

    [itex]\frac{\partial}{\partial \dot{q}} \frac{1}{2} \dot{q} p = (\frac{\partial}{\partial \dot{q}} \dot{q}) \frac{1}{2} p + \frac{1}{2} \dot{q} (\frac{\partial}{\partial \dot{q}} p)[/itex]
     
  4. May 1, 2016 #3
    that is true.
     
  5. May 1, 2016 #4

    stevendaryl

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    Well, if [itex]p = m\dot{q}[/itex], then [itex]\frac{\partial}{\partial \dot{q}} p = m[/itex]
     
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