Partial derivative of Lagrangian with respect to velocity

  • #1
604
13

Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
stevendaryl
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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.
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]
 
  • #3
604
13
that is true.
 
  • #4
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,401
2,577
that is true.
Well, if [itex]p = m\dot{q}[/itex], then [itex]\frac{\partial}{\partial \dot{q}} p = m[/itex]
 

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