# What is $\epsilon$ in the $\delta$ smooth action functional of the Lagrangian?

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?

• its a big article can you plz quote the relevant part? Commented Jun 16, 2021 at 4:25
• 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.
– user304539
Commented Jun 16, 2021 at 4:41

$$\epsilon$$ 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.