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

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?

• Are you familiar with functional derivatives?
– fqq
Jul 18, 2020 at 9:47
• Why do you think $\displaystyle\frac{\partial \varepsilon'}{\partial \varepsilon}=0$?
– user258881
Jul 18, 2020 at 10:13
• @FakeMod Because that is the way we treat time derivatives when dealing with partial derivatives? For example, the partial derivative of the Lagrangian $\frac{1}{2}mv^2 - V(x)$ with respect to x is $- dV/dx$ because, obviously, the kinetic energy term is treated as a constant. Jul 18, 2020 at 13:44

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.

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

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 $$\epsilon$$.