In the variational calculus, we find the Euler-Lagrange equation by using the following:

$S[x(t) + \epsilon f(t)] = S[x(t)] + \delta S + O(\epsilon^2)$ (1)

$S[x(t) + \epsilon f(t)] = S[x(t)] + \frac{d}{d\epsilon}S[x(t) + \epsilon f(t)]\Bigr|_{\epsilon=0}\epsilon + O(\epsilon^2)$

Now, I understand that in (1), the LHS's value has to be less (over all - including all orders) than the RHS's value because $x(t)$ is what minimizes the action, but I also understand that $\delta S$ must be $0$. So, even if $\delta S = 0$, (1) still obeys the principle such that the LHS is less than the RHS because the RHS also has $O(\epsilon^2)$. So we say actions are equal in first order.

Now, if I look at i.e $\delta S = \frac{d}{d\epsilon}S[x(t) + \epsilon f(t)]\Bigr|_{\epsilon=0}\epsilon$, what this tells me in words is how much action changes between our paths, so since we set this to $0$, then we're officially saying that action values are the same on both paths, even though, on true path, the value has to be less. What's going on? Can this be explained in layman's terms?


2 Answers 2


The term $$\frac{d}{d\epsilon} S[x(t)+\epsilon f(t)]\vert_{\epsilon=0}\epsilon$$ does not tell you how much the action changes between the paths $x(t)$ and $x(t)+\epsilon f(t)$. That would simply be the difference

$$S[x(t)+\epsilon f(t)] - S[x(t)].$$

What it does tell you is how much the action, $S[x(t)]$, changes (either increases or decreases) if you start to change the path $x(t)$ by adding a multiple of $f(t)$.

As $x(t)$ is a minimizer, this quantity should only be able to increase. However, if $\delta S$ is nonzero, then for a small positive $\epsilon$ we would either see the quantity increase or decrease, and for $-\epsilon$ we would observe the other behavior by linearity. The only logical conclusion is that this quantity must be 0.

  • $\begingroup$ Thanks for the great answer. aren't they really the same ? $\endgroup$
    – Giorgi
    Dec 1, 2023 at 14:16
  • $\begingroup$ No, but the distinction has nothing to do with calculus of variations. Even in one dimension, there is a difference between $f(x+h)-f(x)$ and $f'(x)$. At a minimum $f'(x)=0$, but $f(x+h)-f(x)$ will be positive for sufficiently small $h$. $\endgroup$
    – podiki
    Dec 1, 2023 at 14:20
  • $\begingroup$ I think I got it. imagine $f(x) = x^2$. $f'(3) = 6$ while $f(3+h) - f(3) \approx 0.0001$. So this is the idea right ? One is the actual change, another is the rate of change. The same happens for my original question then. $\endgroup$
    – Giorgi
    Dec 1, 2023 at 14:41
  • $\begingroup$ Hm, but we also multiply it by $\epsilon$, which basically means it's not the rate of change, but the change itself. So that's what's confusing me now $\endgroup$
    – Giorgi
    Dec 1, 2023 at 14:46
  • $\begingroup$ The two quantities will be very close when $\epsilon$ is small, but they are still fundamentally different. The whole idea is that the first one must be zero even for small nonzero $\epsilon$. While the other will in general not be zero and determined by the higher order terms in the expansion. $\endgroup$
    – podiki
    Dec 1, 2023 at 14:48

In this answer I will discuss a specific instance: the case of the surface area of a soap film stretching between two coaxial rings. The idea is to discuss in such a way that it readily generalizes to other cases.

The webpage: The shape of soap bubbles has several pictures of a film stretching between two coaxial rings.

The surface area of that soap film is given by the following integral:

$$ I = 2 \pi \int_{x_0}^{x_1} y \ \sqrt{1 + (y')^2} \ dx \tag{1} $$

As we know, the shape of this minimal surface is called catenoid. We can think of the shape as a surface of revolution. Take a catenary curve (hyperbolic cosine) and rotate it.

The integration obtains the surface area by summing the areas of adjacent circular strips.

The area of each strip is a function of the circumference of the strip, and the width of the strip; the surface area is proportional to the product of these two factors.

It pays for the soap film to contract in the middle; the contraction reduces circumference, hence less surface area. But that works only up to a point. The steeper the slope the wider the strips are, which tends to increase the surface area.

In the variation space I will call the minimal surface area the 'sweet spot'.

Now start from a state of not contracted enough, and as you allow the curve to move to the sweet spot look at the two factors that determine the surface area.

Track how the circumference factor affects the surface area, and how the slope factor affects the surface area.

As the soap film contracts from not-contracted-enough to the sweet spot: in the multiplication the circumference factor outperforms the slope factor; the surface area decreases.

Continue the contraction past the sweet spot.
Now: in the multiplication the slope factor outperforms the circumference factor; the surface area increases again.

Again: the value of a surface area is obtained by multiplication; because of that there is no meaningful way to compare values directly. Instead what is compared is (contribution to) rate of change; a derivative. Here that derivative is derivative with respect to variation.

We have: as you sweep out variation the rate of change coming from each of the contributing factors can be expressed as a function of the variation sweep; as you sweep out variation there is a single spot where the two rates of change from each of the contributing factors match each other.

The Euler-Lagrange equation expresses that balance of contributions.

$$ \frac{\partial L}{\partial y} - \frac{d}{dx}\left(\frac{\partial L}{\partial y'}\right) = 0 \tag{2} $$

The Euler-Lagrange equation is not about comparing values; the Euler-Lagrange equation is about comparing contributions to rate of change.

The purpose of the helperfunction (in your question notated as $f(t)$) is to facilitate sweeping out variation. It is convenient to sweep out the variation by multiplying the helper function with a factor $\epsilon$, and then evaluate the derivative with respect to $\epsilon$.

In the notation the following expresses that as we sweep out variation we need that derivative with respect to $\epsilon$ at a single point only: at the point $\epsilon=0$

$$ \Bigr|_{\epsilon=0} \tag{3} $$

Incidentally: about the reason why the higher orders (of Taylor series expansion) do not affect the result:
For finite values of $\epsilon$ the higher order terms do make a difference, of course. However, that is inconsequential; we need the evaluation to be deviation-free only at a single point in variation space: the point $\epsilon=0$. At $\epsilon=0$ retaining only the linear term is sufficient.


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