# Variation of path in the derivation of the Euler-Lagrange equation

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).

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.
• 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? Commented Nov 25, 2019 at 17:10