For this video, I’m doing something a little different. I got the chance to sit down with Steven Strogatz and record a conversation. For those of you who don’t know, Steve is a mathematician at Cornell. He’s the author of several popular math books and a frequent contributor to, among
other things, Radiolab and The New York Times. To put it shortly, he’s one of the great mass communicators of math in our time.
In our conversation, we talked about a lot of things. But it was all centering around this one very famous problem in the history of math,
the brachistochrone. And for the first two-thirds or so of the video, I’m just gonna play some of that
conversation. We lay out the problem, talk about some of its history, and go through the solution
by Johann Bernoulli from the 17th century. After that, I’m gonna show this proof that Steve showed me. It’s by a modern mathematician, Mark Levi. And it gives a certain geometric insight to Johann Bernoulli’s original solution. And at the very end, I have a little challenge for you.
Grant: We should probably start off by just defining the problem itself.
Steve: Okay, alright, you want me to take a crack at that?
Grant: Yeah, go for it.
Steve: Okay, yeah, so it’s this complicated word, first of all, brachistochrone, that
comes from two — gee, I have to check. Are those Latin or Greek words, I think?
Grant: I’m pretty sure they’re Greek.
Steve: Okay, so Greek words for “the shortest time”. And it refers to a question that was posed by one of the Bernoulli brothers, by
Johann Bernoulli. If you imagine like a chute and there’s a particle moving down a chute, being pulled
by gravity. What’s the path of the chute that connects two points so that it goes from point A to
point B in the shortest amount of time?
Grant: I think what I like most about this problem is that it’s relatively easy to
describe qualitatively what you’re going for. You know, you want the path to be short, something like a straight line. But you want the object to get going fast, which requires starting steeply. And that adds length to your line. But making this quantitative and actually finding the balance with a specific
curve? It’s not at all obvious and makes for a really interesting problem.
Steve: It is! It’s a really interesting thing. I mean most people when they first hear it assume that the shortest path will give
the shortest time, that the straight line is the best. But-but as you say it, it can help to build up some steam by rolling straight down at
first, or-or not necessarily rolling. I mean, you could picture it sliding. That doesn’t really matter, how we phrase it. But, so Galileo had thought about this himself much earlier than Johann Bernoulli in
1638. And Galileo thought that an arc of a circle would be the best thing. So he had the idea that a bit of curvature might help.
Grant: And it turns out that the arc of the circle is not the right answer. It’s good, but there are better solutions. And the history of real solutions starts with Johann Bernoulli posing this as a
Steve: So that’s then in June of 1696. And he posed it as a challenge really to the mathematical world at that time. For him, that meant the mathematicians of Europe. And in particular, he was very concerned to show off that he was smarter than his
brother, you know. So he had a brother, Jacob. And the two of them were quite bitter rivals actually, both tremendous
mathematicians. But Johann Bernoulli fancied himself the greatest mathematician of his era, not just
better than his brother. But, you know, I think he thought that he might be better than-than Leibniz, who was
alive at the time, and Isaac Newton, who was by then sort of an old man. I mean, more or less retired from doing math. Newton was the Warden of the Mint, be something like the secretary of the treasury
Grant: And Newton shows him up, right? He stays up all night and solves it even though it took Johann Bernoulli two weeks to
Steve: That, right. That’s the great story, that Newton was shown the problem, wasn’t really pleased to
be challenged, especially by somebody that he considered beneath him. I mean, he considered pretty much everybody beneath him. But-but yeah, Newton stayed up all night, solved it, and then sent it in anonymously
to The Philosophical Transactions, the journal at the time. And it was published anonymously. And, so Newton complained in a letter to a friend of his. He said, “I do not love to be dunned and teased by foreigners about mathematical
things.” So he didn’t enjoy this challenge, but he did solve it. The famous legend is that Johann Bernoulli on seeing this anonymous solution said, “I
recognize the lion by his claw.” I don’t know if that’s true, but it’s a great story. Everyone loves to tell that story.
Grant: And I suspect part of the reason that Johann was so eager to challenge other
mathematicians like Newton is he secretly knew that his own solution was unusually
clever. Maybe we should start going into what he does.
Steve: Yes, he imagines that to solve the problem, you let light take care of it for
you. Because Fermat in the early 1600s had shown that you could-you could state the way
that light travels, whether bouncing off of a mirror or refracting from air into
water. Where it bends or going through a lens. All the motion of light could be understood by saying that light takes whatever path
gets it from point A to point B in the shortest time.
Grant: Which is a really awesome perspective when you think about it. Cause usually you think very locally in terms of what happens to a particle at each
specific point. This steps back and looks at all possible paths and says, nature chooses the best
Steve: Yes, it is. It’s a beautiful and-and, as you say, really an awe inspiring mental shift. For some people, literally are inspiring in the sense that it had religious
overtones, that somehow nature is imbued with this property of doing the most
Grant: Oh, interesting!
Steve: But leaving that aside, you know, you could just say it’s an empirical fact
that that is how light behaves. And so, Johann Bernoulli’s idea was to then use Fermat’s principle of least time. And say, let’s pretend that instead of a particle sliding down a chute, it’s light
travelling through media of different index of refraction. Meaning that the light would go at different speeds as it-as it successively went,
sort of, down the chute.
Grant: And-and I think before we dive into that case, we should look at something
So, at this point in the conversation we talked for a while about Snell’s law. This is a result in physics that describes how light bends when it goes from one
material into another, where its speed changes. I made a separate video out of this, talking about how we can prove it using Fermat’s
principle, together with the very neat argument using imaginary constant tension
springs. But for now, all you need to know is the statement of Snell’s law itself.
When a beam of light passes from one medium into another, and you consider the angle
that it makes with a line perpendicular to the boundary between those two
materials. The sine of that angle divided by the speed of light stays constant as you move from
one medium to the next. So what Johann Bernoulli does is find a neat way to take advantage of that fact, this
sin-of-𝜃-over-𝑣-stays-constant fact, for the brachistochrone problem.
Steve: When he thinks about what’s happening with the particle sliding down the
chute. He notices that by conservation of energy, the velocity that the particle has will be
proportional to the square root of the distance from the top.
Grant: And just to spell that out a little bit more, the loss in potential energy is
its mass times the gravitational constant times 𝑦, that distance from the top. And when you set that equal to the kinetic energy, one-half times 𝑚𝑣 squared, and
you rearrange, the velocity 𝑣 will indeed end up being proportional to the square
root of 𝑦.
Steve: Yes, so-so that then gives him the idea about, let’s imagine, glass of many
different layers, each with a different velocity characteristic for the light in
it. The velocity in the first one is 𝑣 one and the next one is 𝑣 two and the next one
is 𝑣 three. And these are all gonna be proportional to the square root of 𝑦 one or 𝑦 two or 𝑦
Grant: And, in principle, you should be thinking about a limiting process where you
have infinitely many infinitely thin layers. And this is kind of a continuous change for the speed of light.
Steve: And so-so then his question is, if light is always instantaneously obeying
Snell’s law, as it goes from one medium to the next, so that 𝑣 over sin 𝜃 is
always a constant, as I move from one layer to the next. What is that path where, you know, such that these tangent lines are always
instantaneously obeying Snell’s law?
Grant: And, for the record, we should probably just state exactly what that property
Grant: So the conclusion that Johann made was that if you look at whatever the
time-minimizing curve is and you take any point on that curve. The sine of the angle between the tangent line at that point and the vertical divided
by the square root of the vertical distance between that point and the start of the
curve. That’s gonna be some constant independent of the point that you chose. And when Johann Bernoulli first saw this, correct me if I’m wrong, he just recognized
it as the differential equation for a cycloid, the shape traced by the point on the
rim of a rolling wheel.
But it’s not obvious, certainly not obvious to me, why this
sin-of-𝜃-over-square-root-𝑦 property has anything to do with rolling wheels.
Steve: It’s-it’s not at all obvious, but this is again the-the genius of Mark Levi to
Grant: You wanna say a few words about Mark Levi?
Steve: Yeah, well Mark Levi is a-a very clever as well as a very nice guy who is a
friend of mine and a terrific mathematician at Penn State who has written a book
called the Mathematical Mechanic. In which he uses principles of mechanics and, more generally, physics to solve all
kinds of math problems. That is rather than math in the service of science, it’s science in the service of
math. And as an example of the kinds of clever things that he does, he recently published a
little note, very short. Showing that if you look at the geometry of a cycloid, just drawing the correct lines
in the right places, that-that this principle of velocity over sin 𝜃 being constant
is-is built-in to the motion of the cycloid itself.
So in that conversation, we never actually talked about the details of the proof
itself. It’s kind of a hard thing to do without visuals. But I think a lot of you out there enjoy seeing the math and not just talking about
the math. It’s also a really elegant little piece of geometry. So, I’m gonna go through it here.
Imagine a wheel rolling on the ceiling. And picture a point 𝑃 on the rim of that wheel. Mark Levi’s first insight was that the point where the wheel touches the ceiling,
that I’ll call 𝐶, acts as this instantaneous center of rotation for the trajectory
of 𝑃. It’s as if, for that moment, 𝑃 is on the end of a pendulum whose base is at 𝐶. Since the tangent line of any circle is always perpendicular to the radius, the
tangent line of the cycloid path of 𝑃 is perpendicular to the line 𝑃𝐶. This gives us a right angle inside of the circle. And any right triangle inscribed in a circle must have the diameter as its
hypotenuse. So from that, you can conclude that the tangent line always intersects the bottom of
the circle. Now, let 𝜃 be the angle between this tangent line and the vertical. We get a pair of similar triangles, which I’ll just show on the screen.
You can see that the length of 𝑃𝐶 is the diameter times sin of 𝜃. Using the second similar triangle, this length times sin of 𝜃 again gives the
distance between 𝑃 and the ceiling. The distance that we were calling 𝑦 earlier. Rearranging this, we see that sin of 𝜃 divided by the square root of 𝑦 is equal to
one divided by the square root of the diameter. Since the diameter of a circle, of course, stays constant throughout the rotation,
this implies that the sin of 𝜃 divided by square root of 𝑦, it’s constant on a
cycloid. And that’s exactly the Snell’s law property that we’re looking for. So, when you combine Johann Bernoulli’s insight with this little geometry proof,
that’s the cleverest solution of the brachistochrone that I’ve ever seen.
And I could call it done here. But given that the whole history of this problem started with a challenge that Johann
Bernoulli posed, I wanna finish things off with a little challenge of my own. When I was playing around with the equations of a cycloid, something interesting
popped out. Consider an object sliding down the cycloid due to gravity. And think about where it is along the curve as a function of time. Now think about how the curve is defined, as this trajectory of the point on a rim of
a rotating wheel. How might you tweak the rate at which the wheel rotates so that when the object
starts sliding, the marked point on the rim of the wheel always stays fixed to that
Do you start rotating it slowly and increase its speed? If so, according to what function? It turns out, the wheel will rotate at a constant rate, which is surprising. This means that gravity pulls you along the cycloid in precisely the same way that a
constantly rotating wheel would. The warm-up part of this challenge is just, confirm this for yourself. It’s kind of fun to see how it falls out of the equations.
But this got me thinking. If we look back at our original brachistochrone problem, asking about the path of
fastest decent between two given points, maybe there is a slick way to reframe our
thinking. How about it look, if instead of describing the trajectory of a sliding object in
terms of its 𝑥- and 𝑦-coordinates, we described it in terms of the angle that the
velocity vector makes as a function of time. I mean, you can imagine defining a curve by having an object start sliding then
turning a knob to determine the angle at which it’s sliding at each point in time,
always being pulled by gravity.
If you describe the angle of the knob as a function of time, you are in fact uniquely
describing a curve. You’re basically using a differential equation, since what’s given is the slope as a
function of some other parameter. In this case, time. So what’s interesting here is that when you look at the solution of the
brachistochrone problem, not in the 𝑥𝑦-plane but in the 𝑡𝜃-plane — where 𝑡 is
time, 𝜃 is the angle of the path — all of the brachistochrone solutions are
straight lines. That is to say, 𝜃 increases at a constant rate with respect to 𝑡. When the solution of a curve-minimization problem is a straight line, it’s highly
suggestive that there is some way to view it as a shortest-path problem.
Here, it’s not so straight forward. Since the boundary conditions that your object started at point A and ended at point
B in the 𝑥𝑦-space doesn’t just look like going from one point to another in the
𝜃𝑡-space. Nevertheless, my challenge to you is this. Can you find another solution to the brachistochrone problem by explaining why it
must be the case that a time-minimizing trajectory, when represented in 𝑡𝜃-space,
looks like a straight line?