Is this momentum and if so how is it derived?

So I'm reading Susskind's "The Theoretical Minimum" and on pages 128 and 129, he has the following equations:

First he starts with the Lagrangian for the system:

$$L=\frac{1}{2}(\dot{q_1}^2+\dot{q_2}^2) -V(q_1-q_2)$$

Then he shows the following two derivations:

$$\dot{p_1} = -V'(q_1-q_2)$$ $$\dot{p_2} = V'(q_1-q_2)$$

Where $V'$ is stated to be the derivative of $V$.

The derivation itself is left as an exercise to the reader and I can't figure it out.

In the previous chapter, he had defined the momentum as:

$$p_i=\frac{\partial L}{\partial \dot{q_i}}$$ on page 124.

But since $V$ does not depend upon either $\dot{q_1}$ or $\dot{q_2}$, equations 2 and 3 don't seem to follow from equation 1. Could someone show me what this derivation looks like? Or maybe $p$ means something else here.

• Are you familiar with the Hamiltonian formalism? You might want to look that up ... Nov 25 '16 at 0:10
• The equation you're missing is the Euler-Lagrange equation $\dot{p}_i = \partial L / \partial q_i$. Nov 25 '16 at 0:11
• Does he give classical mechanics lectures on YouTube, as well as the QM and others that I know are up there?
– user108787
Nov 25 '16 at 0:33
• Sanya, that is the next chapter. knzhou, thank you. I had not recognized it in that form. I can accept your answer if you post it. CountTo10 yes he does. Nov 25 '16 at 0:51

Given the action $S=\int L~dt$, defined in terms of the Lagrangian $L(q,\dot{q})$, we have the following relationships: