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

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


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

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?

  • $\begingroup$ 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. $\endgroup$ – 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: $$ \frac{dp}{dt} - F = 0 $$

Which are, Newton's law.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.