1
$\begingroup$

Consider the first derivation of the Euler-Lagrange equation shown in this Wikipedia article. We begin by considering the path $f(x)$ that yields stationary action. We then consider a new path with the addition of a small perturbation given by:

$$g_\epsilon(x) \ = \ f(x) \ + \ \epsilon \eta(x)$$

Here is my concern: At no point in this proof do we actually make use of the fact that $\epsilon$ is small. It would seem as though this proof would hold for any size of $\epsilon$. I do understand that $\epsilon$ terms of order greater than or equal to $2$ in the Taylor expansion of the action (in terms of $\epsilon$) will become negligible if $\epsilon$ is very small. Along with the absence of a first-order $\epsilon$ term, this makes the action at the original and perturbed path essentially the same, but this doesn't seem to have anything to do with the proof itself.

$\endgroup$
4
  • $\begingroup$ Someone may correct me but $\epsilon$ in this case can take any value. $g_\epsilon(x)$ defines a set of curves. $\endgroup$ – Charlie Apr 20 '20 at 0:54
  • $\begingroup$ Oh interesting, would saying that $\epsilon$ is small be incorrect then? (I've seen this on Wikipedia and a few other places) $\endgroup$ – Jack Ceroni Apr 20 '20 at 1:00
  • 1
    $\begingroup$ Someone more qualified could give a better answer than me I'm afraid. $\endgroup$ – Charlie Apr 20 '20 at 1:42
  • 1
    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Vivek Apr 20 '20 at 2:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.