# Problem in Euler-Lagrange imply Newton

I'm self-studying Mechanics and I 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 $\frac{\partial v²}{\partial q}$ vanish. 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 law :

Euler lagrange equation state that $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = \frac{\partial L}{\partial q}$

And $L= T-V(q)= \frac{1}{2}mv² - V(q)$

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 $\frac{1}{2}m\frac{\partial v²}{\partial q}$ Vanish without any explanation

Why these terms vanish?

The first term vanishes because you have assumed that the potential is independent of $\dot{q}=v$ (as you have said, $V(q)$). The second term vanishes because speed and possition are independent.

• But suppose for exemple that we deal with a simple pendulum. If the solve de the DE, we get : $\theta = acos(wt+\varphi )$ , so we can find a function f such that : $t = f(\theta )$ and $\frac{d\theta }{dt}= - awsin(wt+\varphi )$ Thus $\dot{\theta}= - awsin(wf(\theta)+\varphi )$ and thus we will have $\frac{\partial L }{\partial \theta} = \frac{1}{2}ml\frac{\partial \dot{\theta²} }{\partial f}\frac{\partial f}{\partial \theta} -mglsin(\theta)$ . So why the term $\frac{1}{2}ml\frac{\partial \dot{\theta²} }{\partial f}\frac{\partial f}{\partial \theta}$ vanish ? – Med Saâd Alami Nov 15 '14 at 21:22
• @MedSaâdAlami Newton's equations are specifically written as $m\ddot{x}=-U'(x)$, where $x$ is a coordinate in your Euclidean system. If you want to prove Newton's equations, then, you'd better choose to write $L$ in terms of Euclidean coordinates first. (The pendulum adds an extra degree of complication, therefore, because there is now a force of constraint when you move back to Euclidean coordinates!) – user12029 Nov 15 '14 at 21:58
• @MedSaâdAlami, so certainly that term doesn't vanish. But, if it did vanish, what would that get you? (Also, $\theta=a \cos(\omega t)$ does not solve the simple pendulum problem, it solves the simple harmonic oscillator problem.) For reference I think the proof of Newton's laws from the principle of least action is given in L&L as equation 5.3. It's so simple because the first step in this would be writing $L$ in the form 5.1. – user12029 Nov 15 '14 at 22:04

The Lagrangian is defined in the most simple case as a function of $q$, $\dot{q}$ and $t$: $L(q,\dot{q},t)$. This notation implies that $q$, $\dot{q}$ and $t$ are by definition independent variables. This is how you have to interpret the partial derivatives to $q$ and $\dot{q}$, it doesn't make sense to write: $q = f(\dot{q})$, because both are considered independent variables in the lagrangian. Its just a function of three variables, so $\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}}$ is mathematically just the same as $\frac{\partial L(x,y,t)}{\partial y}$ where we replaced $q$ by $x$ and $\dot{q}$ by $y$: it is the derivative to the second variable of the function.

Ofcourse when you solve the problem and find a $q(t)$ you can find a relation for this solution between $q$ and $\dot{q}$: $q = f(\dot{q})$, but this doesn't alter the fact that you should consider the variables of the $L$ above as independent, this solution has nothing to do with the independent variables of the lagrangian.

So when using the Lagrange formula: $$\frac{d}{dt}\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}} = \frac{\partial L(q,\dot{q},t)}{\partial q}$$ the idea is that for the partial derivatives you just look at $q$ and $\dot{q}$ as independent variables, neglecting possible solutions you may have in mind. Calculating these derivatives will result in new functions of $q$, $\dot{q}$ and $t$, f.i.: $F(q,\dot{q},t) =\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}}$. Then you have to calculate a total derivative to $t$ of $F$: to do this you have to substitute a possible solution $q(t)$, i.e. $F \circ q = F(q(t),\dot{q} (t),t) = G(t)$, the result is a function of only $t$, only when calculating this total derivative you can use dependency between $q$ and $\dot{q}$ because you have to insert a possible solution.

Why these terms vanish?

If

$$L = \frac{mv^2}{2} - V(q) = \frac{m(\dot q)^2}{2} - V(q)$$

Then

$$\frac{\partial L}{\partial \dot q} = m \dot q$$

Why? Because $V(q)$ is a function of $q$ thus the partial derivative of $V$ with respect to $\dot q$ is zero.

Similarly

$$\frac{\partial L}{\partial q} = -\frac{\partial V}{\partial q}$$

Why? Because $\frac{m(\dot q)^2}{2}$ is a function of $\dot q$ thus the partial derivative of $\frac{m(\dot q)^2}{2}$ with respect to $q$ is zero.

• Ok i understood that... but the question is not as simple as you think : Why $q$ and $\dot{q}$ have to be independant ? – Med Saâd Alami Nov 15 '14 at 23:38
• @MedSaâdAlami, they're not independent but, as I wrote and emphasized, these are partial derivatives. – Alfred Centauri Nov 15 '14 at 23:46
• So if they're not independant we can write for exemple $q=f(\dot{q})$ and we will have $\frac{\partial V(q)}{\partial \dot{q}}= \frac{\partial V(f(\dot{q}))}{\partial q} = \frac{\partial V(f(\dot{q}))}{\partial f} \frac{\partial f}{\partial \dot{q}}$... So ? how to conclude ? (i think that their is somthing about calculus that i don't understand) – Med Saâd Alami Nov 16 '14 at 0:01