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For:

$$\mathcal{L}=\mathcal{L}(q_j,\dot{q_j},t)=T-V$$

the Euler-Lagrange equation is simply:

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

Now, using the above definition for the Lagrangian:

$$\frac{d}{dt}\left(\frac{\partial T}{\partial \dot{q_j}} -\frac{\partial V}{\partial \dot{q_j}}\right)-\frac{\partial T}{\partial q_j}=-\frac{\partial V}{\partial q_j}$$

Defining:

$$Q_j=-\frac{\partial V}{\partial q_j}$$

as a "generalized force", and:

$$\Pi_j=\frac{\partial T}{\partial \dot{q_j}}$$

as a "generalized momentum" and assuming that $V$ does not depend on the generalized velocities, we have:

$$\frac{d\Pi_j}{dt}-\frac{\partial T}{\partial q_j}=Q_j$$

Which is very similar to the common:

$$\frac{dp}{dt}=F$$

except for an added factor. This rearrangement of the Euler-Lagrange equations to me, makes it more clear that they are equivalent to Newton's second law.

Now, first of all, have I made any mathematical mistakes? Secondly, while this is a "discussion" question which seemingly is less recommended, what do you think about it? Does it make the equivalence shine more clearly?

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    $\begingroup$ Well, in classical mechanics the E-L equations are derived from the second law, using D'Alembert principle and supposing that $F$ can be taken as the gradient of a scalar function. $\endgroup$
    – Claudio Saspinski
    Commented Jun 19, 2022 at 23:46
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    $\begingroup$ @ClaudioSaspinski Yes, but the final result is rather warped from the starting point. It's hard to see intuitively that the equations are essentially equivalent. $\endgroup$
    – agaminon
    Commented Jun 19, 2022 at 23:49
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    $\begingroup$ As a matter of fact, this can be seem precisely as a proof of Newton's second law. This being said, don't forget that which laws are fundamentals changes with the era, and with the field of research. Newton's laws came long before Euler-Lagrange equations, but in a modern approach of physics, the former is considered a consequence of the latter. During my PhD, it was a game between students to prove any fundamental laws this way, building a lagrangian and applying Noether's theorem. Newton's law, Schrödinger's equation, Maxwell's equations... Tedious, but fun. $\endgroup$
    – Miyase
    Commented Jun 20, 2022 at 0:18
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    $\begingroup$ I notice that you placed quotes around generalized force. I surmise you are concerned that maybe a notion of generalized force is not admissible. In my opionion such a concern is not necessary. Given that position and velocity can be expressed in terms of generalized coordinates: it follows logically that acceleration can also be expressed in terms of generalized coordinates. If acceleration can meaningfully be expressed in generalized coordinates, so can force. See also: connection between work-energy theorem and EL-equation with L=T-V $\endgroup$
    – Cleonis
    Commented Jun 20, 2022 at 16:01

1 Answer 1

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Yes, one may use d'Alembert principle to rewrite Newton's 2nd Law as Lagrange equations $$ \sum_{i=1}^N\underbrace{\left(\dot{\bf p}_i-{\bf F}_i\right)}_{\text{Newton's 2nd Law}}\cdot \delta {\bf r}_i ~=~ \sum_{j=1}^n \underbrace{\left(\frac{d}{dt} \frac{\partial T}{\partial \dot{q}^j} -\frac{\partial T}{\partial q^j}-Q_j\right)}_{\text{Lagrange equations}} \delta q^j,\tag{1}$$ and vice-versa, cf. e.g. Ref. 1 and this related Phys.SE post. Here $\delta$ denotes an infinitesimal virtual displacement.

References:

  1. H. Goldstein, Classical Mechanics, Chapter 1.
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