Partial derivative of Lagrangian with respect to velocity

[Adel Makram]

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.

 

[stevendaryl]

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 \frac{1}{2}m \dot{q}^2 as \frac{1}{2}\dot{q} p doesn't change anything; you have to use the product rule:

\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)

 

[Adel Makram]

that is true.

 

[stevendaryl]

that is true.

Well, if p = m\dot{q}, then \frac{\partial}{\partial \dot{q}} p = m