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  • You may be aware that I’m a huge fan of the YouTube channel 3blue1brown, run by Grant

  • Sanderson. Grant makes excellent videos about math, and mathy aspects of other topics, so

  • I’m letting him take over my channel for the day. Grant, take it away.

  • (Grant) A week ago, I put out a tweet showing a peculiar place where an ellipse arises,

  • but what I didn’t mention is that this arbitrary-seeming construction is highly relevant to a once

  • lost lecture by Richard Feynman on why planets orbit in ellipses.

  • The construction starts by drawing a circle, and choosing some point within the circle

  • which is not the center, what I’ll call aneccentricpoint.

  • Then draw a bunch of lines from this eccentric point to the circumference of the circle.

  • For each of those lines, rotate it 90 degrees about its midpoint. Once you do this for all

  • the lines, an ellipse emerges in the middle. Out of context, this a mildly pleasing curiosity,

  • but there’s a much deeper form of satisfaction on its way once you understand the full story

  • surrounding this.

  • Front and center in that story is Richard Feynman, whose famous along a number of dimensions.

  • To scientists, he’s a giant of 20th century physics, winner of the Nobel prize for his

  • foundational insights for Quantum Electro Dynamics among many other things.

  • To the public, he’s a refreshing contradiction to stereotypes about physicists: A safe-cracking,

  • bongo-playing, mildly-philanderous non-conformist whose heavily brooklyn-accented voice youve

  • probably heard either relaying some bit of no-nonsense pragmatic wisdom about the only

  • sensible way to view the world, or else some wry joke told through a crooked smile.

  • But to physics students, he was an exceptionally skillful teacher, both for his charisma and

  • his uncanny ability to make complicated topics feel natural and approachable.

  • Many of the lectures he gave as a CalTech freshman course are immortalized in the now

  • famousFeynman lectures”, whose three volumes are available for free online.

  • But not all of the lectures he gave made it into this collection.

  • One in particular, a guest lecture lecture given in March 13, 1964 entitledThe motion

  • of planets around the sun”, survived only as an unpublished partial transcript with

  • a smattering of notes buried in the office of one of Feynman’s colleagues until it

  • was eventually dug up by Caltech archivist Judith Goodstein.

  • Despite the absence of some crucial blackboard drawings to follow what Feynman was saying,

  • her husband David eventually reconstructed the argument of the lecture, which the two

  • of them published in a book titledFeynman’s lost lecture”, conveying both the lecture

  • itself and the surrounding story in a really beautiful way.

  • Here, I’d like to give a more animated and slightly simplified retelling of the argument

  • Feynman presented.

  • The lecture itself is about why planets and other astronomical objects orbit the sun in

  • ellipses. It ultimately has to do with the inverse square

  • law, the fact that the gravitational force pulling an object towards the sun is inversely

  • proportional to the square of the distance between the orbiting object and the sun. But

  • why? How exactly does that give rise to an ellipse, of all shapes?

  • Of course, the gravitational attraction between planets, moons comets and such means that

  • no one orbit is a perfect ellipse, but to a very good approximation this is the shape

  • of an orbit. You can solve this analytically, setting up

  • the appropriate differential equation and seeing the formula for an ellipse pop out,

  • but Feynman’s goal was not to rely on any heavy mathematical machinery. In fact, let’s

  • take a listen to him articulate his own goal. I am going to give what I will call an elementary

  • demonstration. But elementary does not mean easy to understand. Elementary means that

  • very little is required to know ahead of time in order to understand it, except to have

  • an infinite amount of intelligence. There may be a large number of steps that

  • hard to follow, but to each does not require already knowing the calculus or Fourier transforms.

  • Yeah, that’s all, infinite intelligence. I think youre up to that, don’t you?

  • I’ve done what I can to simplify things down further, but that’s not to say a good

  • deal of focus won’t be required.

  • First thing’s first, we need some definition of an ellipse, otherwise there’s no hope

  • of proving that theyre the shape of orbits. Some of you may be familiar with the classic

  • way to construct an ellipse using two thumbtacks and a piece of string. Use the thumb tacks

  • to fix the ends of a small length of string in place, then pull the string taut with a

  • pencil, and trace out a curve while keeping the string taut.

  • It’s similar to how you might use a single pushpin to construct a circle, where the fixed

  • length of string guarantees that every point you trace is a constant distance from the

  • thumbtack. But in this case, with two thumbtacks, what property are you guaranteeing about each

  • point you trace? Well, at every point, the sum of the distances

  • from that point to each of the thumbtacks will be the full length of the string, right?

  • So the defining property of this curve is that when you draw lines from any point on

  • the curve to these two special thumbtack locations, the sum of the lengths of those lines is a

  • constant, namely the length of the string. Each of these points is called a “focus

  • of your ellipse, collectively calledfoci”. Fun fact, the word focus comes from the latin

  • forfireplace”, since one of the first places ellipses were studied was for orbits

  • around the sun, a sort of fireplace of the solar system, sitting at one of the foci of

  • a planet’s orbit. Making up a bit of terminology, let’s call

  • this constant sum of the distances from any point on the ellipse to the two foci thefocal

  • sumof the ellipse.

  • Well get to orbital mechanics in a moment, but first let’s turn back to that construction

  • I showed at the beginning, which will come up again later in the story.

  • Remember, we take all these lines from an eccentric point of a circle to its circumference,

  • and rotate each 90 degrees about its center, why on earth should an ellipse pop up?

  • You could just take my word for it, but I think youll be much more satisfied in the

  • end if we take the time now for a brief sidestep into geometry proof land.

  • First off, there are really only two special points in this diagram, the eccentric point

  • from which all the lines emerge, and the center of the circle, so you might guess that each

  • of these is a focus of the ellipse. Given the defining property of an ellipse,

  • you know youre going to want to look at the sum of the distances from these two points

  • to...something. Also, if youre doing a geometry problem

  • involving a circle, youll very likely want to draw a radius of that circle, and at some

  • point use the fact that this radius has a constant length no matter where you draw it.

  • I mean, that’s what defines a circle, so youll probably need to incorporate that

  • somewhere. With those two thoughts in the back of our

  • mind, let’s limit our attention to just one of these lines, touching some point P

  • on the circle. Remember what happens in our construction:

  • You rotate this line from the eccentric point 90 degrees about its center, and the geometry

  • enthusiasts in the room might fancifully call this a “perpendicular bisectorof the

  • original line. Take a moment to think about the sum of the

  • distances from our two proposed focus points to any point Q along this perpendicular bisector.

  • The key insight here is that you can find two similar triangles to conclude that the

  • distance from the eccentric point to Q is the same as the distance from Q to P.

  • So, that means adding the distances to each focus is the same as adding the distances

  • from the center to Q, then Q to P. Now there are two key things I want you to

  • notice: First, at the point where this perpendicular bisector intersects the radius, that sum is

  • clearly the radius of the circle. Since that radius is a constant no matter where we draw

  • it, the focal sum at that intersection point stays constant, which by definition means

  • it traces out an ellipse, specifically an ellipse whose focal sum equals the radius

  • of this circle. Isn’t that neat? Second, because the sum of these two lengths

  • at every other point on this perpendicular bisector is larger than the radius, meaning

  • the sum of the distances to the foci from those points are bigger than the ellipse’s

  • focal sum, all other points of this line must lie outside the ellipse.

  • What this means, and this will be important, is that this perpendicular bisector, the line

  • we got after our special 90 degree rotation, is tangent to the ellipse.

  • So the reason all the lines we drew earlier make an ellipse appear is because were

  • drawing a bunch of that ellipse’s tangent lines.

  • The reason this will be important, as youll say later, is that this tangency direction

  • will correspond to the velocity of an orbiting object.

  • Okay, geometry proofiness done, onto some actual physics and orbital mechanics!

  • The first fact to use is Kepler’s (very beautiful) second law, which says that as

  • an object orbits around the sun, the area it sweeps out during a given amount of time,

  • like 1 day, will be a constant, no matter where you are in the orbit.

  • For example, think of a comet whose orbit is very skewed. Close to the sun, it’s getting

  • whipped around very quickly, so it covers a larger arc length during a given time interval.

  • Farther away, itll move slower, so covers a shorter arc length during that same time.

  • And this trade off between radius and arc length balances in just such a way that the

  • swept area is the same. A quick way to see why this is true is to

  • leverage conservation of angular momentum. For a tiny time step, delta t, the area swept

  • out is essentially a triangle. In principle you should think of this as a

  • small sliver for a tiny time step, but I’ll draw it thicker so we can better see all the

  • parts. The area is ½ base times height, right?

  • The base is the distance to the sun, and the height will be this little length here, which

  • you can think of as the component of the object’s velocity perpendicular to the line to the

  • sun, which I’ll call v_perp, multiplied by the small duration of time.

  • So the area is ½ R * (v_perp) * (delta t). Conservation of angular momentum with respect

  • to a given origin point, like the sun, tells us that this radius time the component of

  • velocity perpendicular to it will remain constant, so long as all forces acting on the object

  • are directed towards that origin. Well, specifically it says this quantity times

  • the mass of the object stays constant, but the mass of an orbiting object won’t be

  • changing. So! Our expression for the area swept out

  • depends only on the amount of time that has passed, delta t.

  • Historically, this went the other way around, and Kepler’s second law is one of the empirical

  • facts that led to an understanding of angular momentum.

  • I should emphasize, this law does not assume that the orbit is an ellipse. Heck, it doesn’t

  • even assume the inverse square law, the only thing needed for this to hold is that the

  • only force acting on the orbiting object is directed straight towards the sun.

  • This is a fact Feynman spent much more time showing, recounting an argument by Newton

  • in his Principia, but it kind of distracts from our main target, so I figure assuming

  • conservation of angular momentum is good enough for our purposes here, albeit at some loss

  • of elementarity.

  • At this point we don’t know the shape of an orbit; for all we know it’s some wonky

  • non-elliptical egg shape. The inverse square law will help pin down

  • that shape precisely, but the strategy is a little indirect. Before showing the shape

  • of the path traced by the orbiting object, well show the shape traced out by the velocity

  • vectors. Here, let me show you what I mean by that.

  • As the object orbits, its velocity will be changing, always tangent to the curve of the

  • orbit, longer at points where the object moves quickly, and shorter at points where it moves

  • more slowly. What well show is that if you take all

  • these velocity vectors, and collect them together so that their tails all sit a single point,

  • their tips actually trace out a perfect circle. This is a pretty awesome fact, if you ask

  • me. The velocity spins around and gets faster and slower at various angles, but evidently

  • the laws of physics cook things up just right so that these trace out a perfect circle.

  • The astute among you might have a little internal lightbulb starting to turn on at the sight

  • of this circle with an off-center point.

  • Now, why on earth should this be true? Feynman describes being unable to easily follow

  • Newton at this point, so instead he comes up with his own elegant line of reasoning

  • to explain where this circle comes from. He starts by looking at the orbit, and slicing

  • it up into little pieces which all cover the same angle with respect to the sun.

  • Alright, now think about how the amount of time it takes the orbiting object to traverse

  • one of these equal-angle slices changes as it gets farther away.

  • Well, by Kepler’s 2nd law, it’s proportional to the area swept out, right? And because

  • these slices have the same angle, as you get farther away from the sun, not only does the

  • radius increase, but the component of arc length perpendicular to that radial line goes

  • up in proportion to that radius. So the area of one of these slices, and hence the time

  • it takes the object to traverse it, is proportional to the distance away from the sun squared.

  • In principle, well ultimately be considering very small slices, so there won’t be ambiguity

  • in what I mean by the radius from the planet to the sun on a given slice, and the relevant

  • bits of arc length will be effectively straight. Alright, now think about how the inverse square

  • law comes into play. At any given point, the force the sun imparts on the object is proportional

  • to 1/(the radius)^2, but what does that really mean? What force is is the acceleration on

  • the object, the amount that it’s velocity changes per unit time, multiplied by that

  • object’s mass. This is enough to give us a super useful bit

  • of information about how the velocity of our orbiting object changes from one slice to

  • the next. The change in velocity is acceleration times change in time, right? Which means its

  • proportional to the change in time over the radius squared.

  • But since the time it takes to traverse one slice is proportional to the radius squared,

  • these terms cancel, so the change in velocity as the object traverses a given slice is actually

  • some constant that doesn’t depend on the slice at all.

  • In other words, if you look the velocity at the start of the slice, and at the end of

  • the slice, then directly compare them by joining their tails, looking at the difference between

  • the two, the vector joining their tips, this difference has the same length no matter which

  • slice of the orbit you were looking at. Also! Since the force vector is always pointing

  • towards the sun, as we go from the start of one slice to the next, that force vector is

  • turning by a constant angle. In geometry lingo, you might say that all theexternal angles

  • of this polygon that has formed will be equal. I know this is a little tricky, but hang in

  • there! Remember that all you need to follow along is infinite intelligence.

  • Take a moment to make sure it’s clear what’s happening without velocity diagram: The change

  • from one vector to the next, the little difference vector joining one tip to the next, will always

  • have the same length, which was a consequence of the perfect cancelation that happens when

  • mixing Kepler’s second law with the inverse square law.

  • Now, because those constant-length change vectors rotate by a constant angle each time,

  • it means they form regular polygon. As we consider finer and finer slices of the

  • original orbit, based on smaller and smaller angles for those slices, the relevant regular

  • polygon defining the tips of the vectors in our velocity diagram will approach a circle.

  • Isn’t that really neat?

  • Hopefully, at this point youre looking at this circle with a special eccentric point,

  • and your just itching to see it give rise to an ellipse the way we saw earlier. But,

  • it’s a little weird, right? Were looking at a diagram in velocity space, how exactly

  • will this give us the shape of the orbit? What follows is tricky, but very clever.

  • Step back and consider what we know: We don’t know the specific shape of the orbit, only

  • the shape the velocity vectors trace. But more specifically than that, we know that

  • once the planet has turned an angle theta off the horizontal with respect to the sun,

  • this corresponds to walking theta degrees around our circle in the velocity diagram,

  • since the acceleration vectors rotate just as much as the radius vector.

  • This tells us the tangency direction for each point on the orbit; whichever vector from

  • our velocity diagram touches that point theta degrees around, that’s the velocity vector

  • of our orbiting object, and hence the tangency direction of the curve.

  • In fact, let me just start drawing all those velocity vectors as lines, since all well

  • need to use is the information they cary about the slope of the orbit curve; the specific

  • magnitude of each velocity will not be as important.

  • Notice, it's not that the angle of the velocity vector at this point is an angle theta off

  • the vertical. No no no. The angle I’m referencing in the velocity diagram is with respect to

  • the circle’s center, which is almost certainly a little different from where the velocity

  • vectors are rooted. So the question is, what special curve satisfies

  • the property that the tangency direction for a point theta radians off the horizontal is

  • given by this vector from a special eccentric point of a circle to a point theta degrees

  • around the circle from the vertical?

  • Well, here’s the trick. First, rotate this whole circle setup 90 degrees.

  • Then take each of those individual velocity directions and rotate them 90 degrees back

  • the other way, so that theyre oriented as they were before, it’s just that each

  • is rooted in a different spot. Aha! Weve spotted our ellipse! But we still

  • have just a little thinking ahead of us to really understand how this emergent ellipse

  • is related to the astronomical orbit. Importantly, I didn’t just rotate these

  • lines about any point, I rotated each about its center, which means we can leverage the

  • geometric proof we saw several minutes ago. And this is the moment where you kind of have

  • to furrow your brow and think backwait, what was going on in that proof again?”

  • One of the key points was that when you have two lines, one from the center of the circle,

  • one from the eccentric point, both to a common spot on the circle’s circumference, the

  • perpendicular bisector to the eccentric line will be tangent to the ellipse. What’s more,

  • the point of tangency is where it intersects with the radial line from the center.

  • What that means is that the point of our little ellipse which is theta degrees off the horizontal,

  • with respect to the circle’s center, has a tangent slope perpendicular to this eccentric

  • line. And because of the whole 90 degree rotation, this means it’s parallel to the velocity

  • vector we need it to be. So this little emergent curve inside the velocity

  • diagram has exactly the tangency property we need our orbit to have!

  • And hence, the shape of the orbit must be an ellipse. QED.

  • Alright, pat yourself on the back, because there’s no small amount of cleverness required

  • to follow this. First there was this peculiar way of constructing

  • an ellipse, requiring some geometry savviness to prove.

  • Then there’s the pretty clever step of even thinking to ask the question about what shape

  • the velocity vectors trace out when you move all their tails to the same spot.

  • And showing that this is a circle requires mixing together the inverse square law with

  • Kepler’s second law in another sly move. But the cleverness doesn’t end there! Showing

  • how this velocity diagram with vectors rooted at a point off the circle’s center implies

  • an elliptical orbit brings in this neat 90 degree rotation trick.

  • I just love this. Watching Feynman do physics, even elementary physics, is like watching

  • Bobby Fischer play chess.

  • (Henry): Thanks again to Grant, and you should definitely go check out his videos on 3blue1brown

You may be aware that I’m a huge fan of the YouTube channel 3blue1brown, run by Grant

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ファインマンの失われた講義 (ft. 3Blue1Brown) (Feynman's Lost Lecture (ft. 3Blue1Brown))

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    林宜悉 に公開 2021 年 01 月 14 日
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