# How Lagrange equations imply Newton equation

Sorry for my English (it's not my mother language). I'm self-studying Mechanics and I'm have a little problem :

We can see that in Landau's book or in Wikipedia that when we inject the Lagrangian in Euler-Lagrange equation the term $\frac{\partial v²}{\partial q}$ vanishes. So we get $$\frac{\partial L}{\partial q}= - \frac{\partial U}{\partial q}$$

here more details :

We want to proof that Euler-lagrange equation imply Newton's second law :

Euler-Lagrange equation state that

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = \frac{\partial L}{\partial q}$$

And we set that, for conservative potential fields $L= T-V(q)= \frac{1}{2}mv² - V(q)$

But if we inject L in Euler-Lagrange equation we will get

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = m\frac{dv}{dt} - \frac{d}{dt}\frac{\partial V}{\partial \dot{q}}$$

And

$$\frac{\partial L}{\partial q} = \frac{1}{2}m\frac{\partial v²}{\partial q} + F$$

In Landau's book the terms $\frac{\partial V}{\partial \dot{q}}$ and $\frac{1}{2}m\frac{\partial v²}{\partial q}$ vanishes without any explanation! Then I'm asking:

Why these terms vanish?

• Welcome to Mathematics SE. The question as you've currently written it is hard to answer. Could you provide a complete context? It would help if you describe the problem from the beginning. – Simon S Nov 15 '14 at 20:16

This is quite simple. To better visualize Newton's law let's not use generalized coordinates, instead, let's use Cartesian ones. Euler-Lagrange equations hence stands: $$\frac{d}{dt}\frac{\partial L}{\partial\dot x} - \frac{\partial L}{\partial x} = 0$$
Since $L = T - V$, it is clear to know that: $$\frac{\partial L}{\partial x} = \frac{\partial T}{\partial x} - \frac{\partial V}{\partial x} = -\frac{\partial V}{\partial x} \quad\Longrightarrow\quad \frac{\partial L}{\partial x} = -\frac{\partial V}{\partial x} = F$$
Now, on the other term, the potential doesn't depends on the velocity. Hence: $$\frac{\partial L}{\partial \dot x} = \frac{\partial T}{\partial \dot x} - \frac{\partial V}{\partial \dot x} = \frac{\partial T}{\partial \dot x} \quad\Longrightarrow\quad \frac{\partial L}{\partial \dot x} = \frac{\partial T}{\partial \dot x} = p$$
This can be easily seen: $$T = \frac{1}{2}m\dot x^2 \Longrightarrow \frac{\partial T}{\partial \dot x} = m\dot x = p$$
Taking those results and plugging into Euler-Lagrange equations, we have: $$\frac{dp}{dt} - F = 0$$