# Intuition for why/how $\delta \int_A^B n(\mathbf{r}) \ ds = 0$?

My textbook, Fundamentals of Photonics, 3rd edition, by Teich and Saleh, says the following:

Fermat's Principle. Optical rays travelling between two points, $$A$$ and $$B$$, follow a path such that the time of travel (or the optical pathlength) between the two points is an extremum relative to neighboring paths. This is expressed mathematically as

$$\delta \int_A^B n(\mathbf{r}) \ ds = 0, \tag{1.1-2}$$

where the symbol $$\delta$$, which is read "the variation of," signifies that the optical pathlength is either minimized or maximized, or is a point of inflection. ...

This probably doubles as a mathematics question, but I'm going to ask it here anyway.

How does the fact that the optical rays follow a path such that the time of travel (optical pathlength) between two points is an extremum relative to neighboring paths imply the result $$\delta \int_A^B n(\mathbf{r}) \ ds = 0$$? I'm struggling to develop an intuition for why/how the "variation of" optical pathlength would be $$0$$ in this case.

I would greatly appreciate it if people could please take the time to clarify this.

• Do you understand that the integral corresponds to the time taken for light to travel that path? – jacob1729 Nov 6 at 16:59
• @jacob1729 Yes, but it doesn't make sense to me that the time would be equal to $0$. That implies that the light ray travels the path instantaneously. – The Pointer Nov 6 at 17:02
• @The Pointer. $\delta \dots$ means variation. The variation of the action with respect to path change is zero, i.e. (electrical) path length is minimum (not zero) – Cryo Nov 6 at 17:10
• $\delta$ isn't a number, it's expressing the difference in time, much like $df$ isn't the product of $d$ and $f$ but the difference in $f$. – jacob1729 Nov 6 at 17:10
• Intuition, given Fermat's principle: v = ds/dt so dt =ds/v In this case, n is the index of refraction = c/v . Putting this in gives dt = n ds/c where n is a function of position and c is a constant which can be ignored when minimizing the total time. Keep in mind that to minimize a function, you find the derivative and set it equal to zero. – R.W. Bird Nov 6 at 19:01

A variation is a fancy derivative. If you start with the integral $$I=\int_A^B f(x,x')dx$$ one first makes this into a parametrized integral $$I(\epsilon)=\int_A^B f(x(\epsilon),x'(\epsilon))dx$$ with $$x(\epsilon)$$ and $$x'(\epsilon)$$ "parametrized path" so that the true path is at $$\epsilon=0$$. Then $$\delta I=\frac{d}{d\epsilon}I(\epsilon)=0$$.

When looking for points where a function $$g$$ is extremal, the condition $$d g/dx=0$$ provides an algebraic equation to find the points $$x_0$$ where $$g$$ is extremal.

For the integral $$I$$, we're not looking points where the integral is extremal; instead the variation $$\delta I=0$$ provides a differential equation to be satisfied by the function $$f(x,x')$$ (here this function is the path of light) that produces an extremum of the integral.

So Fermat's principle states the path to travel from $$A$$ to $$B$$ will be such that the total time $$\int_A^B dt= \int_A^B n ds$$ is extremal, i.e. if you pick any neighbouring path, the time will be longer (assuming the extremum is a minimum).

This is certainly true when the index $$n$$ is constant: the path is then a straight line between $$A$$ and $$B$$; since the straight line is the path between two points, the time taken for light travelling at constant speed (since $$n$$ is constant) is minimum, i.e. $$\delta I=0$$.

In more general cases the index will not be constant so the more general integral $$\int_A^B n(s) ds$$ is the general way of obtaining the total travel time.

For an excellent discussion see: Boas, Mary L. Mathematical methods in the physical sciences. John Wiley & Sons, 2006.

• Thanks for the answer. What do you mean by "the true path is at $\epsilon=0$"? And how did you get $\delta I=\frac{d}{d\epsilon}I(\epsilon)=0$? – The Pointer Nov 6 at 18:03
• $\delta I=\frac{d}{d\epsilon}I(\epsilon)$ is the definition of the variation. Also the family of path is set up so that $\epsilon=0$ is the true path. Highly recommend you read chapter on calculus of variation in Boas. – ZeroTheHero Nov 6 at 18:33
• I will begin reading the chapter now. – The Pointer Nov 6 at 18:41
• Ok, so what you meant by "the true path is at $\epsilon = 0$" is that, on page 475 of the aforementioned text, for the equation $Y(x) = y(x) + \epsilon \eta(x)$, where $y(x)$ is the desired extremal and $\epsilon$ is a parameter, when $\epsilon = 0$, we get the desired extremal $Y(x) = y(x)$, as distinguished from the alternative, arbitrary paths. – The Pointer Nov 7 at 11:02
• Ahh, ok, I think I understand now. So $\delta \int_A^B n(\mathbf{r}) \ ds$ is equivalent to $\delta I$, where $I = \int_A^B n(\mathbf{r}) \ ds$. And so, as you said, the variation $\delta I=0$ provides a differential equation to be satisfied by the function $f(x,x')$ (here this function is the path of light) that produces an extremum of the integral. So it is a condition that we are imposing (that is, we are imposing that we must have $\delta \int_A^B n(\mathbf{r}) \ ds = 0$) in order for the statement to be true. So, in other words, it is true by definition. – The Pointer Nov 7 at 11:27