The use of $x_\varepsilon (t) = x(t) + \varepsilon (t)$ and $x_\varepsilon (t) = x(t) + \varepsilon \eta (t)$ in proving Hamilton's principle

Question

The following Wikipedia page uses $x_\varepsilon (t) = x(t) + \varepsilon (t)$ in the proof.

https://en.wikipedia.org/wiki/Hamilton%27s_principle#Mathematical_formulation

But in my mechanics book (by David Morin), the author uses something similar to $x_\varepsilon (t) = x(t) + \varepsilon \eta (t), \ \varepsilon \in \mathbb{R}$.

Since the proof using the concept of $\delta \mathcal S$ is written on the Wikipedia page I won't go through that.

Consider the following partial derivative of the action functional (whichever epsilon it is).

$$\frac{\partial \mathcal S}{\partial \varepsilon} = \int_{t_1}^{t_2} \frac{\partial \mathscr L}{\partial \varepsilon} \ dt = \int_{t_1}^{t_2} \frac{\partial \mathscr L}{\partial x_\varepsilon}\frac{\partial x_\varepsilon}{\partial \varepsilon} + \frac{\partial \mathscr L}{\partial x'_\varepsilon}\frac{\partial x'_\varepsilon}{\partial \varepsilon} \ dt $$

Note that I used the notation $x'$ to indicate the time derivative.

If I put $x_\varepsilon (t) = x(t) + \varepsilon \eta (t)$ here then it works and I get the Euler-Lagrange equation.

But if I substitute $x_\varepsilon (t) = x(t) + \varepsilon (t)$ into the integral, I obtain

$$ \frac{\partial x_\varepsilon}{\partial \varepsilon} = 1 $$ and $$ \frac{\partial x'_\varepsilon}{\partial \varepsilon} = \frac{\partial \varepsilon '}{\partial \varepsilon} = 0$$

which lead to an absurd result.

Why?

Answer 1

Recall that we are looking for a function $x(t)$, such that first order changes in it will induce second order changes in $S[x(t)]$. These changes in $x(t)$ can be expressed in two different ways, as you have identified:

1. $$x_{\epsilon}(t) = x(t) + \epsilon \eta(t)$$ where $\epsilon$ tracks the "smallness" and $\eta(t)$ is any (suitable) arbitrary function. In this set-up, we seek a condition on $x(t)$ such that $\delta S = 0 + \mathcal{O}(\epsilon^2)$ for all $\eta(t)$.

2. $$x_{\epsilon}(t) = x(t) + \epsilon(t)$$ where now $\epsilon(t)$ is any suitable arbitrary function and is used to track the "smallness" (i.e. the size of the variation of $x'(t)$ from $x(t)$). Here we also seek a condition on $x(t)$ such that $\delta S = 0 + \mathcal{O}(\epsilon^2)$ but the "for all $\eta$" is incorporated into the nature of $\epsilon$.

Viewing $S$ as a function of $\epsilon$ for the purpose of variation, $S(\epsilon) = S(0) + \frac{\mathrm{d}S}{\mathrm{d}\epsilon}\bigr|_{\epsilon = 0} \epsilon + \mathcal{O}(\epsilon^2)$ implies that we require $\frac{\mathrm{d}S}{\mathrm{d}\epsilon}\bigr|_{\epsilon = 0} = 0$.

1. $$S(\epsilon) = S(0) + \int \mathrm{d}t \left(\frac{\partial L}{\partial x} \eta + \frac{\partial L}{\partial x'} \eta' \right) \epsilon + \mathcal{O}(\epsilon^2)$$. Therefore, we require:

$$ \frac{\mathrm{d}S}{\mathrm{d}\epsilon} = \int \mathrm{d}t \left(\frac{\partial L}{\partial x} \eta + \frac{\partial L}{\partial x'} \eta' \right) = 0$$ as in Morin's derivation.

2. $$S(\epsilon) = S(0) + \int \mathrm{d}t \left(\frac{\partial L}{\partial x} \epsilon + \frac{\partial L}{\partial x'} \epsilon' \right) + \mathcal{O}(\epsilon^2)$$

Like before, we also require the integrand in the second term to vanish, but in its current form, $\frac{\mathrm{d}S}{\mathrm{d}\epsilon}$ doesn't naturally drop out like in case 1, since we have an $\epsilon'$. However, using integration by parts (like we would need to in case 1) will reduce this integrand to a term proportional to $\epsilon$ again, so like in case 1, we'll find that $\frac{\mathrm{d}S}{\mathrm{d}\epsilon}\bigr|_{\epsilon = 0} = 0$ gives us the Euler-Lagrange equations.

Side Note:

I'm not quite sure here how to justify why I've written $\mathcal{O}(\epsilon^2)$ over $\mathcal{O}(\epsilon^2, \epsilon^{'2})$ - i.e. I've made the assumption that terms of order $\epsilon'$ are of order $\epsilon$.