What do a lifeguard, an ant and a photon have in common? – PhotoNick

By CoperNick

Alessandro Santoni

January 27, 2020

Let’s think for the next few minutes that we are in the Australian hemisphere, where at the moment is summer: we are enjoying some free time at the beach, sunbathing, until we hear someone screaming. The lifeguard notices the situation and starts running towards the drowning person (hopefully not in slow-mo like in the Baywatch series!). But which path allows the lifeguard to be as fast as possible? This problem is known as the Lifeguard’s paradox, and actually it’s not trivial as it seems!

For sure, he should go through a stretch of road $L_{sand}$ on the sand and another one $L_{water}$ into water, and clearly his velocity $v_{sand}$ on the sand will be greater than the velocity $v_{water}$ in water. Simplifying our scheme, we should imagine something like this:

Figure 1: geometric representation of a rescue in the sea.

Where $a$ is the distance of the lifeguard from the sea, $b$ is the distance of the drowning person from the shore, $d$ is the distance from the lifeguard at the point $A$ and the swimmer in $B$ , and $c$ the distance from the projections of the point $A$ and $B$ on the shore – which we can easily find through Pitagora’s theorem

$c= \sqrt{d^2-(a+b)^2}$

Now let’s introduce the variable $x= \overline{CK}$ , so that $\overline{KD} = c-x$ , where $K$ was the intersection point between the segment $\overline{CD}$ and the trajectory of the lifeguard

Figure 2: introduction of the variable x in our scheme

Then, taking into account the right triangles $\triangle CKA$ and $\triangle DKB$ , we can write

$L_{sand}=\sqrt{a^2+x^2} \, ; \,\,\, L_{water} =\sqrt{b^2+ (c-x)^2}$

At this point, we can calculate the total time $\Delta t _{TOT}$ needed to save the person in the sea in function of the length $x$ : this will be

$\Delta t _{TOT} (x) = \frac{L_{sand}}{v_{sand}} +\frac{L_{water}}{v_{water}} = \frac{\sqrt{a^2+x^2}}{v_{sand}}+\frac{\sqrt{b^2+(c-x)^2}}{v_{water}}$

which is the function that we have to minimize in function of $x$ to find the path of shorter duration. In order to do so, we need some infinitesimal calculus: in particular we have to find the critical point, by deriving our expression and putting it to zero. Don’t mind about the mathematics now: let’s only notice that, after the calculations, we find the following condition for the desired trajectory

$\frac{1}{v_{sand}}\frac{x}{\sqrt{a^2+x^2}}=\frac{1}{v_{water}}\frac{c-x}{\sqrt{b^2+(c-x)^2}}$

Okay, now you are maybe asking yourself what all this story is really about… We are getting there! Defining the two angles $\alpha$ and $\beta$ made by the trajectory and the vertical, through the point $K$

Figure 3: display of the diffraction scheme with angles.

and replacing

$\sin{\alpha} = \frac{x}{\sqrt{a^2+x^2}} \, ;\,\,\, \sin{\beta} = \frac{c-x}{\sqrt{b^2+(c-x)^2}}$

in the previous relation, we obtain the new equation:

$\frac{\sin{\alpha}}{v_{sand}} = \frac{\sin{\beta}}{v_{water}}$

Which is the analogue of Snell’s Law which describes the refraction of light when passing through a boundary between two materials

$n_1 \sin{\alpha} = n_2 \sin{\beta}$

where $n_1$ and $n_2$ are the refraction indices of the two different medias, and can be expressed as the ratio among the velocity of light in the vacuum and the velocity in the physical media: $n = c/v$

This mean that, if we could imagine photons as small, non-quantum marbles of light, when moving from a point to another they would “choose” always the faster path available, exactly as the lifeguard!

It is also worth mentioning that our observation coincides (almost) completely to the Fermat’s principle of least time, which was formulated in the 17th century, and can be thought as the theoretical link between geometrical or ray optics and the later developed wave optics. [1]

Nevertheless, as you can imagine, light is “brainless” and it cannot really “think” of what pathway to take; whilst the lifeguard would probably let the person drown during the calculations.

However, what I discovered while writing this article have blown my mind and would probably do for you: in 2012, a team of German scientists have demonstrated [2] that ants (yes, those tiny insects with six legs and antennas) seem to be able to complete such calculations and find the faster path while moving through different surfaces. This means that ants, in some mysterious way, follow the Fermat’s principle!

Figure 4: “Refracted” trail of a group of ants at the medium border between smooth (white) and rough (green) felt

We have discovered, at least, that a bunch of ants is better at math than a lifeguard… and that’s maybe something that you could keep in mind the next time you go swimming!

In conclusion, it is also interesting to note that optimization problems, like the one we discussed, have some analogies with the theory of optimal transport: a relatively new frontier in mathematics. In the references you can find the link [3] to an introductive speech given by mathematician Alessio Figalli, where he also explain some concepts from the research that earned him the Fields Medal (the equivalent of Nobel Prize, for Mathematics) in 2018.

Note: all the geometric drawings in this article were made with GeoGebra [4], a free software that allows you to visualise lots of different and useful operations regarding mathematics: give it a try!

References

[1] https://journals.plos.org/plosone/article?id=10.1371/journal.pone.0059739

[3] https://www.youtube.com/watch?v=jyL6ZECM7oM