# Fermat's Principle: no first-order change in time?

I was reading the chapter on Fermat's principle in the Feynman lecture series. The principle is stated along these lines:

"The correct statement is the following: a ray going in a certain particular path has the property that if we make a small change (say a one percent shift) in the ray in any manner whatever, say in the location at which it comes to the mirror, or the shape of the curve, or anything, there will be no first-order change in the time; there will be only a second-order change in the time. In other words, the principle is that light takes a path such that there are many other paths nearby which take almost exactly the same time"

Could someone please explain what "no first-order change in the time" means here?

Optics: The principle of least time

• – Frobenius Sep 21 at 19:15

Any continuous and differentiable function $$f(x)$$ can be expressed as a Taylor series: $$f(x_0+\delta x) = f(x_0) + \frac{\mathrm{d}f}{\mathrm{d}x}\bigg|_{x_0}\delta x + \frac{1}{2}\frac{\mathrm{d} x^2f}{\mathrm{d}^2x}\bigg|_{x_0}\delta x^2 + \dots + \frac{1}{n!}\frac{\mathrm{d}^nf}{\mathrm{d}^nx}\bigg|_{x_0}\delta x^n .$$

Each of these terms are called of the $$n^{\mathrm{th}}$$ order.

If there is no "first-order" contribution, then $$\frac{\mathrm{d}f}{\mathrm{d}x}\bigg|_{x_0} = 0$$, i.e. $$x=x_0$$ is a stationary point. In the limit of infinitesimal $$\delta x \rightarrow \mathrm{d}x \rightarrow 0$$, all $$\mathcal{O}(n)$$ contributions to th Taylor series tend to zero too. But anything that goes like $$\delta x^{n>1}$$ goes to $$0$$ faster than the first-order correction. Which means that for a minuscule $$\delta x$$, the only "correction" would be given by $$\frac{\mathrm{d}f}{\mathrm{d}x}\bigg|_{x_0}$$. If that's also $$0$$, then there is no correction and the function is stationary.

In this case, your $$f$$ is actually the time $$t$$.

• so essentially is he saying this? $$f(x_0+\delta x) = f(x_0)$$ – Eliza Sep 19 at 18:11
• In the limit of $\delta x \rightarrow 0$ yes. It means the function has a maximum/minimum there, so it’s “flat” at that point. – SuperCiocia Sep 19 at 19:06
• So Feynman is implying that light rays not only travels in the direction which takes least time but also in the direction at which it has many choices of the same kind (ie same transit time) nearby? – Eliza Sep 19 at 19:23
• Feynman is just saying that light takes the path that takes the least time. So time is minimised. – SuperCiocia Sep 19 at 20:18

By no first-order changes Feynman means that the first-order functional derivative vanishes, or equivalently, the path is stationary.

By the way, no first-order changes is a common talking point of Feynman. Listen e.g. to 46:48-48:48 in the talk The Character of Physical Law, part 4, where he makes similar remarks about the principle of least action.