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We understand that we derive Lagrange equations, $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_j}\right)-\frac{\partial L}{\partial q_j}=0, \tag{1}$$ starting from d'Alembert's principle of virtual work being zero, which is, $$({\bf F}_i-\dot{{\bf p}_i})\cdot\delta {\bf r}_i=0, \tag{2}$$ with the help of generalized coordinates.

In the same spirit, from the calculus of variations, starting from optimizing the action (Hamilton's principle), and for a generic function, $f(y,\dot{y},x)$, we obtain the differential equation, $$\frac{d}{dx}\left(\frac{\partial f}{\partial \dot{y}}\right)-\frac{\partial f}{\partial y}=0. \tag{3}$$

In the spirit of Goldstein's Classical Mechanics, we simply identify Euler-Lagrange equations as follows,

$$x \rightarrow t,$$ $$y \rightarrow q_j,$$ $$f \rightarrow L.$$ How do we recognize this last identification? Is it from the semblance of the eqs. (1) and (3)? Or is there any other reasoning involved? Like, if we didn't know the d'Alembert's principle, then, how do we know for sure that "Lagrangian is the function that extremizes the action functional"?

Qmechanic
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user31694
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2 Answers2

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In a general problem involving an integral of the form

$$I=\int_{x_1}^{x_2}f(x,y(x),y'(x)) \ dx $$

we know that we can obtain the generic Euler-Lagrange equations. The "identification" is simply defining the same problem involving the action integral, and therefore the relevant mechanical variables

$$ S = \int_{q_1}^{q_2} L(t,q,\dot{q}) \ dt,$$

which should result in the Euler-Lagrange equations of lagrangian formalism because of the identical mathematical structure. Alternatively, one can just start with the action integral and find the equations that already involve the lagrangian, in the same way one would find the generic Euler-Lagrange equations.

agaminon
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  1. If there is a d'Alembert's principle, then we can show Lagrange equations, cf. e.g. this Phys.SE post. If Lagrange equations are of the form of Euler-Lagrange equations, then the corresponding action satisfies a stationary action/Hamilton's principle.

    However, it may be not unique, cf. e.g. this example.

  2. If there is not a d'Alembert's principle, then OP seems to ask: "Why should there be a stationary action/Hamilton's principle?" Well, it may not exists, cf. e.g. this Phys.SE post. See also e.g. this & this related Phys.SE posts.

Qmechanic
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