# The Lagrange equations from mechanics

aclaret
I was having a doubt about the Lagrangian mechanics. Possible we can derive the lagranges equations of by extremisation principle of action, that is assume we already guess what is the lagrangian of the systeme. I say that minimisation procedure rely on assume a lagrangian, and then show it derive correct motions. that okay, but backward. can derive the lagranges equation from mechanics?

Not sure if I understood the question completely, but I think the answer is yes! You can derive those equations just using d'Alembert's principle. You have a system ##\mathcal{S}## with ##k## degrees of freedom, described by generalised co-ordinates ##\mathbf{q} = (q^1, \dots, q^k)## and generalised velocities ##\dot{\mathbf{q}} = (\dot{q}^1, \dots, \dot{q}^k)##. Acting on any particle ##\mathcal{P}_a \in \mathcal{S}## is a total force ##\mathbf{F}_a## which may be decomposed into the sum of an specified force ##\mathbf{F}^{(s)}_{a}##, which includes known external and internal forces, as well as an unknown constraint force ##\mathbf{F}^{(c)}_{a}##.

d'Alembert's principle states that if the constraint forces do zero work, then the specified force (alone) satisfies$$\sum_a m_a \ddot{\boldsymbol{x}}_a \cdot \mathrm{d}\boldsymbol{x}_a^* = \sum_a \mathbf{F}^{(s)}_a \cdot \mathrm{d}\boldsymbol{x}_a^*$$where the ##\mathrm{d}\boldsymbol{x}_a^*## is any virtual infinitesimal displacement of ##\mathcal{P}_a##. It follows that$$\sum_a m_a \ddot{\boldsymbol{x}}_a \cdot \frac{\partial \boldsymbol{x}_a}{\partial q^i} = \sum_a \mathbf{F}^{(s)}_a \cdot \frac{ \partial \boldsymbol{x}_a}{\partial q^i}$$You can define the generalised force ##Q_i## corresponding to ##q^i##$$Q_i := \sum_a \mathbf{F}^{(s)}_a \cdot \frac{ \partial \boldsymbol{x}_a}{\partial q^i}$$Defining ##T := \sum \frac{1}{2} m_a \dot{\boldsymbol{x}}_a^2##, the left-hand side equals ##\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial T}{\partial \dot{q}^i} - \frac{\partial T}{\partial q^i}##; to show this notice that by the chain rule ##\dot{\boldsymbol{x}}_a = \left( \partial \boldsymbol{x}_a / \partial q^i \right) \dot{q}^i## from which it follows that$$\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial T}{\partial \dot{q}^i} = \frac{d}{dt} \sum_a m_a \dot{\boldsymbol{x}}_a \cdot \frac{\partial \boldsymbol{x}_a}{\partial q^i} = \sum_a m_a \ddot{\boldsymbol{x}}_a \cdot \frac{\partial \boldsymbol{x}_a}{\partial q^i} + \sum_a m_a \dot{\boldsymbol{x}}_a \cdot \frac{\partial^2 \boldsymbol{x}_a}{\partial q^i \partial q^j} \dot{q}^j$$and similarly$$\frac{\partial T}{\partial q_i} = \sum_a m_a \dot{\boldsymbol{x}}_a \cdot \frac{\partial \dot{\boldsymbol{x}}_a}{\partial q^i} = \sum_a m_a \dot{\boldsymbol{x}}_a \cdot \frac{\partial^2 \boldsymbol{x}_a}{\partial q^i q^j} \dot{q}^j$$which proves the result; you end up with Lagrange's equation$$\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial T}{\partial \dot{q}^i} - \frac{\partial T}{\partial q^i} = Q_i$$If the applied forces ##\mathbf{F}^{(s)}_a## are all conservative, it follows that the generalised forces themselves can be written as ##Q_i = - \partial \varphi / \partial q^i ## for some function ##\varphi = \varphi(\mathbf{q})##. Notice also that since ##\varphi## does not depend on the velocities ##\dot{\mathbf{q}}##, we have ##\partial \varphi / \partial \dot{q}^i = 0## and thus$$Q_i = - \frac{\partial \varphi}{\partial q^i} + 0 = - \frac{\partial \varphi}{\partial q^i} + \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial \varphi}{\partial \dot{q}_i}$$Inserting this into Lagrange's equations and defining ##\mathscr{L} := T - \varphi## gives$$\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial \mathscr{L}}{\partial \dot{q}^i} - \frac{\partial \mathscr{L}}{\partial q^i} = 0$$Is that sort of what you were after?

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wrobel
andresB
What etotheipi says is basically correct (though it has to be modified for situations with variable masses https://aapt.scitation.org/doi/10.1119/1.4885349), but I want to address the conceptual issue the OP arises.

"guessing" the Lagrangian is not that different than "guessing" the correct forces to put into Newton 2nd law. They both requires a talk between the mathematics and the experience/experiments. So, I see nothing backwards with the Lagrangian approach.

aclaret
aclaret
$$\sum_a m_a \ddot{\boldsymbol{x}}_a \cdot \mathrm{d}\boldsymbol{x}_a^* = \sum_a \mathbf{F}^{(s)}_a \cdot \mathrm{d}\boldsymbol{x}_a^*$$where the ##\mathrm{d}\boldsymbol{x}_a^*## is any virtual infinitesimal displacement of ##\mathcal{P}_a##. It follows that$$\sum_a m_a \ddot{\boldsymbol{x}}_a \cdot \frac{\partial \boldsymbol{x}_a}{\partial q^i} = \sum_a \mathbf{F}^{(s)}_a \cdot \frac{ \partial \boldsymbol{x}_a}{\partial q^i}$$
how does result follow? you make error here

"guessing" the Lagrangian is not that different than "guessing" the correct forces to put into Newton 2nd law. They both requires a talk between the mathematics and the experience/experiments. So, I see nothing backwards with the Lagrangian approach.
thank, yes that what i ask. i am satisfied

how does result follow? you make error here
You start from $$\sum_a m_a \ddot{\boldsymbol{x}}_a \cdot \frac{\mathrm{d}\boldsymbol{x}_a^*}{dt} = \sum_a \mathbf{F}^{(s)}_a \cdot \frac{\mathrm{d}\boldsymbol{x}_a^*}{dt}$$For each generalised coordinate in turn, we'll prescribe a certain virtual motion with ##\dot{q}^{\xi} = 1## and ##\dot{q}^i = 0## for ##i \neq {\xi}##. Then, by the chain rule, ##d \boldsymbol{x}_a^* / dt = (\partial \boldsymbol{x}_a / \partial q^i) \dot{q}^i = \partial \boldsymbol{x}_a / \partial q^{\xi}## and$$\sum_a m_a \ddot{\boldsymbol{x}}_a \cdot \frac{\partial \boldsymbol{x}_a}{\partial q^{\xi}} = \sum_a \mathbf{F}^{(s)}_a \cdot \frac{\partial \boldsymbol{x}_a}{\partial q^{\xi}}$$Just replace the free index ##{\xi} \mapsto i## and you've got the next line.

vanhees71