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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?

Brendan Darrer
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agaminon
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1 Answers1

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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.
Qmechanic
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