About calculus of variation and lagrangian formulation

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The discussion centers on the principle of least action and its derivation of Newton's second law. A variation in the path is introduced, where x(t) is perturbed by a small function a(t). The integral for action S is transformed into S' through a Taylor series expansion, leading to an expression that includes terms involving the perturbation a(t). The key point is that the calculations simplify to show how the action changes under this variation, ultimately confirming the relationship S' = S + δS. The participants clarify that a first-order Taylor series approximation is employed in the derivation.
physlad
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I was reading about the principle of least action and how to derive Newton's second out of it.

at a certain point I didn't follow the calculations,

so the author defines a variation in the path, x(t) \longrightarrow x'(t) = x(t) + a(t), a \ll x

a(t_1) = a(t_2) = 0

Now, S \longrightarrow S' = \int_{t_1}^{t_2} (m/2 (\dot{x} +\dot{a})^2 - V(x +a)) dt

= \int_{t_1}^{t_2} {1/2 m\dot{x}^2 + m\dot{x}\dot{a} - [V(x) + aV'(x)]} dt + O(a^2)

(what happened exactly here? could anybody tell me??)

then

= S + \int_{t_1}^{t_2} [m\dot{x}\dot{a} - aV'(x)] dt

\equiv S + \delta{S}
 
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I assume you meant

<br /> S \longrightarrow S&#039; = \int_{t_1}^{t_2} (m/2 (\dot{x}(t) +\dot{a}(t))^2 - V(x(t) + a(t))) dt<br />

? It looks like they simply used a differential approximation. (i.e. a first-order Taylor series)
 
Yes, I got it. Thanks
 
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