# Partial derivative of Lagrangian with respect to velocity

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
Staff Emeritus
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:

$\frac{\partial}{\partial \dot{q}} \frac{1}{2} \dot{q} p = (\frac{\partial}{\partial \dot{q}} \dot{q}) \frac{1}{2} p + \frac{1}{2} \dot{q} (\frac{\partial}{\partial \dot{q}} p)$

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