# From Newtonian systems to Lagrange mechanics using Euler - Lagrange equations

I'm asking the same question here: https://physics.stackexchange.com/questions/206758/from-newtonian-systems-to-lagrange-mechanics-using-euler-lagrange-equations Shoud I delete the question from physics?

I am reading some notions about Lagrangian Mechanics from Holm's book.(page 12).

There is:

Every Newtonian system, $$m_i \ddot{q_i}=\frac{\partial V}{\partial q_i}, \mbox{ } i=\overline{1,N},$$ is equivalent to the Euler _ Lagrange equations,

$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q}\right)-\frac{\partial L}{\partial q}=0,$$

for the Langrangian $L$ defined by:

$$L(q, \dot{q})=\sum^{N}_{i=1}{\frac{1}{2}m_{i}\|\dot{q_{i}}\|^{2}}-V(q).$$

Proof For $L$ defined in the theorem,

$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q_{i}}}\right)-\frac{\partial L}{\partial q_{i}}=\ldots$$

I don't know how to derive:

$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q_{i}}}\right)$$

What I've tried:

$$\frac{d}{dt}{v(q)}=0$$ so I have to compute only $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q_{i}}}\right)$$

First of all I must to compute:

$$\frac{\partial L}{\partial \dot q_{i}}=\frac{1}{2}m_{i}\frac{\partial}{\partial \dot{q_i}}\|\dot q_{i}\|^2$$ So I have to compute the derivative for a dot product

So:

$$\frac{\partial}{\partial}\|\dot{q_i}\|^2=\frac{\partial}{\partial \dot {q_i}}\langle\dot{q_i},\dot{q_i}\rangle=\langle 1, \dot{q_i} \rangle + \langle \dot{q_i}, 1 \rangle$$

From here I got stuck... It is ok what I have done?

• It will help a lot if you explain notation. Is $q_{i}$ a component of a vector? or one of many quantities? I assume the dot above a variable means a time derivative of that variable? if I recall that was a convention. Do your angle brackets mean inner-product, or is it a vector notation. This question seems impossible to answer without a lot of context (in the form of explanation of what the quantities are and what the notation means). – TravisJ Sep 14 '15 at 23:58