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 Lagrangian $L$ 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


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

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

  • 1
    $\begingroup$ Comment to the title (v2): To derive Lagrangian mechanics from Newtonian mechanics, we are supposed to derive EL equations (as opposed to assuming or using them). Did you mean the opposite way: From Lagrangian mechanics to Newtonian mechanics using EL equations? $\endgroup$ – Qmechanic Sep 13 '15 at 11:39
  • $\begingroup$ Crossposted to math.stackexchange.com/q/1433397/11127 $\endgroup$ – Qmechanic Sep 13 '15 at 17:21

If I understood well, you want to find the second law of Newton using the Lagrangian and the Euler-Lagrange equation?

To make it easy, I'll only take one direction "$x$". Even if you can work with more degrees of freedom, it's the same application of Euler-Lagrange equation, so like that the writing will be lighter.

So you have the Lagrangian $$L_{x}=\frac{1}{2}m\dot{x}^{2}-V(x)$$ and the EL equation : $$\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})=\frac{\partial L}{\partial x}$$

Now you just apply the EL to the Lagrangian, so :

$$\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})=\frac{d}{dt}(m\dot{x})=m\ddot{x}$$


$$\frac{\partial L}{\partial x}= -\frac{\partial{V}}{\partial x}=F_{x}$$

If the force is conservative and derive from a potential energy :

$$F_{x}= -grad(V(x))$$

So we find :

$$ m\ddot{x}=F_{x}$$

Second law of Newton. Like the EL works regardless of the coordinate system, we put "$q$" for expressed a general coordinate. Hope it was what you were looking for!

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Given the Lagrangian $L(q, \dot{q})=\sum^{N}_{j=1}{\frac{1}{2}m_{j}\dot{q_{j}}^{2}}-V(q)$, $$\frac{\partial L}{\partial \dot q_{i}} =\sum^{N}_{j=1}{\frac{1}{2}m_{j}\frac{\partial (\dot{q_{j}}^{2})}{\partial \dot q_{i}}} = \sum^{N}_{j=1}{\frac{1}{2}m_{j} 2 \dot{q}_{j}\delta_{ij}} = m_i\dot{q}_i.$$

I guess you are stuck because of the dot product, but the index $i$ is not for the particle, but for a single degree of freedom, i.e. if you have one particle in a 3D space, your index is $i=1, 2, 3$, whilst with $N$ particles $i=1,2, \dots ,3N$.

In the case $\dot{q}_i = v_i$, your kinetic energy for one particle is $$T = \frac{1}{2}m (v_x^2+v_y^2+v_z^2),$$ for $N$ particles $$T =\sum_{i=1}^N \frac{1}{2}m_i (v_{ix}^2+v_{iy}^2+v_{iz}^2) = \sum_{j=1}^{N'}\frac{1}{2}m_j v_j^2$$ where $N'=3N$ and $j$ is the new index.

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