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In most physics books I've read, such as Goldstein's Classical Mechanics, the explanation of Hamilton's principle took into consideration the equation (1) known as Action:

$$\displaystyle I = \int_{t_1}^{t_2}f(q(t), \dot{q}(t),t)dt\tag1$$

From it, we can arrive at the following Euler-Lagrange equation:

$$\displaystyle\frac{\partial f}{\partial y} - \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{\partial f}{\partial \dot{y}} \right ) = 0\tag2$$

I'm familiar with the steps that lead from (1) to (2). But I don't understand why the functional used in Action took this configuration space, $(q, \dot{q}, t)$, generalized coordinates, generalized velocies and time. The way it is usually explained, it seems that the Action as it is in (1) was pulled out of thin air.

I believe that derivation of the Euler-Lagrange equation (2), was formulated first using the d'Alembert principle and the concept of virtual work, so, Hamilton found the configuration space that, after some steps, lead to (2). Would this be it or is there another reason?

Edited part: In an effort to clarify my question, I will give some more examples. For instance, as suggested in comments, why no $\ddot{q}$, $\dot{q}$,... ? Or, why not only (q, t)? There may be other options that I haven't considered. I don't understand the rationale behind that choice in the functional.

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    $\begingroup$ What, beyond $t$, $q(t)$ and $q'(t)$, would you expect to find in there? $\endgroup$
    – WillO
    Apr 6 at 2:56
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    $\begingroup$ Hi Daniel. Welcome to Phys.SE. Is your question Why no $\ddot{q}$, $\dddot{q}$, $\ldots$? $\endgroup$
    – Qmechanic
    Apr 6 at 8:09
  • $\begingroup$ Hi, @Qmechanic .Yes, but this would also be a particular case, such as 'Why not only (q, t)?' There may be other options that I haven't considered. I don't understand the rationale behind that choice in the functional. $\endgroup$
    – Daniel
    Apr 6 at 16:25

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Well, the function $f$ in the action integral is always identical to the Lagrangian in Hamilton's principle: $$\delta I=\delta\int_1^2L(q,\dot q,t)\;dt=0,\;\;\text{For Some Path in Configuration space}.$$ Since the Lagrangian is defined as the difference between the kinetic and potential energies $L=T-V$, then a configuration space consisting of the generalized coordinates and velocities is natural, since these are the variables found in the Lagrangian. As to being pulled out of thin air, since Hamilton's principle is a principle, it needs not proof or derivation, only affirmation as rightly reflecting experiment.

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  • $\begingroup$ Thank you, Albertus MAgnus.But saying like this, seems that Lagrangian function (L) in Hamilton's principle is always equivalent to the difference between kinetic and potential energy, (L = T - V), it's important to note that this is not universally true. While it's a common choice, particularly in classical mechanics, there are scenarios where the Lagrangian may take different forms, such as in systems with non-conservative forces. Therefore, while the (L = T - V) formulation is often appropriate and useful, it's not a strict rule that applies to all physical systems $\endgroup$
    – Daniel
    Apr 6 at 17:37
  • $\begingroup$ Your point regarding the fact that the principle wouldn't require proof or derivation from another idea is well taken. $\endgroup$
    – Daniel
    Apr 6 at 17:59
  • $\begingroup$ @Daniel I see what you are saying. $\endgroup$ Apr 6 at 21:41

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