At the beginning of the Lagrangian Mechanics Wikipedia page, it gives a $\delta$ function on the stationary point of the action $\cal S$:

Given the time instants $t_1$ and $t_2,$ Lagrangian mechanics postulates that a smooth path $x_0: [t_1,t_2] \to M$ describes the time evolution of the given system if and only if $x_0$ is a ''stationary point'' of the action functional $$ {\cal S}[x]\, \stackrel{\text{def}}{=}\, \int^{t_2}_{t_1} L(x(t),{\dot x}(t),t)\, dt. $$ If $M$ is an open subset of $\mathbb{R}^n$ and $t_1,$ $t_2$ are finite, then the smooth path $x_0$ is a stationary point of ${\cal S}$ if all its directional derivatives at $x_0$ vanish, i.e., for every smooth $\delta : [t_1,t_2] \to \mathbb{R}^n,$ $$ \delta {\cal S}\ \stackrel{\text{def}}{=}\ \frac{d}{d\varepsilon}\Biggl|_{\varepsilon=0} {\cal S}\left[x_0+\varepsilon \delta\right] = 0. $$

What is $\epsilon$ here?

  • $\begingroup$ its a big article can you plz quote the relevant part? $\endgroup$
    – lineage
    Commented Jun 16, 2021 at 4:25
  • 1
    $\begingroup$ It's a small dimensionless variable (a number) that gives different variations (when acted upon the variation). When you differentiate wrt to this variable, you get the variation itself. $\endgroup$
    – user304539
    Commented Jun 16, 2021 at 4:41

1 Answer 1


$\epsilon$ is a real, positive, small number that modulates the strength of the perturbation $\delta$ (remember that $\delta$ is a function $\delta:[t_1,t_2] \rightarrow R^n$ in this context). When $\epsilon$ goes to zero, the total perturbation goes to zero, no matter what $\delta$ is.

This variable is a way to translate the "magnitude" of a multidimensional perturbation (on a position) to a single real number.


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