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When deriving the Euler-Lagrange equation in one dimension the "correct" path, $f(x)$, is the path along which the action is stationary upon infinitesimal modifications of the path, $\epsilon\eta(x)$.

$$f^*(x)=f(x)+\epsilon\eta(x)$$

My question is, is $\epsilon$ just a scalar? And if so how can the following two infinitesimal path modifications that have completely different shapes be related by just a scalar? (Apologies for crudely draw diagrams).

enter image description here

enter image description here

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The (infinitesimal) parameter $\epsilon$ you have introduced is a scalar and introduced in such a way that for $\epsilon = 0$, the new path $f^*(x)$ is equal to the original path $f(x)$ for which the action is stationary.

The function $\eta(x)$ has the property that it vanishes at $A$ and $B$, i.e. $\eta (A) = \eta(B) = 0$. And the two different path which deviate from $f(x)$ are different due to different $\eta(x)$ for each $f^*(x)$. And therefore you can not relate the two modified path by just a scalar.

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    $\begingroup$ So $\epsilon\eta(x)$ is effectively an infinite set of modifications of some specified shape $\eta(x)$, which for instance could be the modified shape in the first diagram. But in the second diagram $\eta(x)$ is a different modification and $\epsilon\eta(x)$ is another infinite set of all possible scaled variations of the second shape? $\endgroup$
    – Charlie
    Commented Nov 25, 2019 at 17:10
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    $\begingroup$ Yes, that is a good summary! $\endgroup$
    – MST
    Commented Nov 25, 2019 at 17:17
  • $\begingroup$ Thanks for your help this is a lot clearer now :) $\endgroup$
    – Charlie
    Commented Nov 25, 2019 at 17:24

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