Mastering Geometry is a Lost Art

I am trying to learn Quantum Field Theory the hard way: Alone and from text books. But there is something harder than the abstract math of advanced quantum physics:

You can aim at comprehending ancient texts on physics.

If you are an accomplished physicist, chemist or engineer – try to understand Sadi Carnot’s reasoning that was later called the effective discovery of the Second Law of Thermodynamics.

At Carnotcycle’s excellent blog on classical thermodynamics you can delve into thinking about well-known modern concepts in a new – or better: in an old – way. I found this article on the dawn of entropy a difficult ready, even though we can recognize some familiar symbols and concepts such as circular processes, and despite or because of the fact I was at the time of reading this article a heavy consumer of engineering thermodynamics textbooks. You have to translate now unused notions such as heat received and the expansive power into their modern counterparts. It is like reading a text in a foreign language by deciphering every single word instead of having developed a feeling for a language.

Stephen Hawking once published an anthology of the original works of the scientific giants of the past millennium: Corpernicus, Galieo, Kepler, Newton and Einstein: On the Shoulders of Giants. So just in case you googled for Hawkins – don’t expect your typical Hawking pop-sci bestseller with lost of artistic illustrations. This book is humbling. I found the so-called geometrical proofs most difficult and unfamiliar to follow. Actually, it is my difficulties in (not) taming that Pesky Triangle that motivated me to reflect on geometrical proofs.

I am used to proofs stacked upon proofs until you get to the real thing. In analysis lectures you get used to starting by proving that 1+1=2 (literally) until you learn about derivatives and slopes. However, Newton and his processor giants talk geometry all the way! I have learned a different language. Einstein is most familiar in the way he tackles problems though his physics is on principle the most non-intuitive.

This amazon.com review is titled Now We Know why Geometry is Called the Queen of the Sciences and the reviewer perfectly nails it:

It is simply astounding how much mileage Copernicus, Galileo, Kepler, Newton, and Einstein got out of ordinary Euclidean geometry. In fact, it could be argued that Newton (along with Leibnitz) were forced to invent the calculus, otherwise they too presumably would have remained content to stick to Euclidean geometry.

Science writer Margaret Wertheim gives an account of a 20th century giant trying to recapture Isaac Newton’s original discovery of the law of gravitation in her book Physics on the Fringe (The main topic of the book are outsider physicists’ theories, I have blogged about the book at length here.).

This giant was Richard Feynman.

Today the gravitational force, gravitational potential and related acceleration objects in the gravitational fields are presented by means of calculus: The potential is equivalent to a rubber membrane model – the steeper the membrane, the higher the force. (However, this is not a geometrical proof – this is an illustration of underlying calculus.)

Gravity Wells Potential Plus Kinetic Energy - Circle-Ellipse-Parabola-Hyperbola

Model of the gravitational potential. An object trapped in these wells moves along similar trajectories as bodies in a gravitational field. Depending on initial conditions (initial position and velocity) you end up with elliptical, parabolic or hyperbolic orbits. (Wikimedia, Invent2HelpAll)

(Today) you start from the equation of motion for a object under the action of a force that weakens with the inverse square of the distance between two massive objects, and out pops Kepler’s law about elliptical orbits. It takes some pages of derivation, and you need to recognize conic sections in formulas – but nothing too difficult for an undergraduate student of science.

Newton actually had to invent calculus together with tinkering with the law of gravitation. In order to convince his peers he needed to use the geometrical language and the mental framework common back then. He uses all kinds of intricate theorems about triangles and intersecting lines (;-)) in order to say what we say today using the concise shortcuts of derivatives and differentials.

Wertheim states:

Feynman wasn’t doing this to advance the state of physics. He was doing it to experience the pleasure of building a law of the universe from scratch.

Feynman said to his students:

“For your entertainment and interest I want you to ride in a buggy for its elegance instead of a fancy automobile.”

But he underestimated the daunting nature of this task:

In the preparatory notes Feynman made for his lecture, he wrote: “Simple things have simple demonstrations.” Then, tellingly, he crossed out the second “simple” and replaced it with “elementary.” For it turns out there is nothing simple about Newton’s proof. Although it uses only rudimentary mathematical tools, it is a masterpiece of intricacy. So arcane is Newton’s proof that Feynman could not understand it.

Given the headache that even Corpernicus’ original proofs in the Shoulders of Giants gave me I can attest to:

… in the age of calculus, physicists no longer learn much Euclidean geometry, which, like stonemasonry, has become something of a dying art.

Richard Feynman has finally made up his own version of a geometrical proof to fully master Newton’s ideas, and Feynman’s version covered hundred typewritten pages, according to Wertheim.

Everybody who indulges gleefully in wooden technical prose and takes pride in plowing through mathematical ideas can relate to this:

For a man who would soon be granted the highest honor in science, it was a DIY triumph whose only value was the pride and joy that derive from being able to say, “I did it!”

Richard Feynman gave a lecture on the motion of the planets in 1964, that has later been called his Lost Lecture. In this lecture he presented his version of the geometrical proof which was simpler than Newton’s.

The proof presented in the lecture have been turned in a series of videos by Youtube user Gary Rubinstein. Feynman’s original lecture was 40 minutes long and confusing, according to Rubinstein – who turned it into 8 chunks of videos, 10 minutes each.

The rest of the post is concerned with what I believe that social media experts call curating. I am just trying to give an overview of the episodes of this video lecture. So my summaries do most likely not make a lot of sense if you don’t watch the videos. But even if you don’t watch the videos you might get an impression of what a geometrical proof actually is.

In Part I (embedded also below) Kepler’s laws are briefly introduced. The characteristic properties of an ellipse are shown – in the way used by gardeners to creating an elliptical with a cord and a pencil. An ellipse can also be created within a circle by starting from a random point, connecting it to the circumference and creating the perpendicular bisector:

Part II starts with emphasizing that the bisector is actually a tangent to the ellipse (this will become an important ingredient in the proof later). Then Rubinstein switches to physics and shows how a planet effectively ‘falls into the sun’ according to Newton, that is a deviation due to gravity is superimposed to its otherwise straight-lined motion.

Part III shows in detail why the triangles swept out by the radius vector need to stay the same. The way Newton defined the size of the force in terms of parallelogram attached to the otherwise undisturbed path (no inverse square law yet mentioned!) gives rise to constant areas of the triangles – no matter what the size of the force is!

In Part IV the inverse square law in introduced – the changing force is associated with one side of the parallelogram denoting the deviation from motion without force. Feynman has now introduced the velocity as distance over time which is equal to size of the tangential line segments over the areas of the triangles. He created a separate ‘velocity polygon’ of segments denoting velocities. Both polygons – for distances and for velocities – look elliptical at first glance, though the velocity polygon seems more circular (We will learn later that it has to be a circle).

In Part V Rubinstein expounds that the geometrical equivalent of the change in velocity being proportional to 1 over radius squared times time elapsed with time elapsed being equivalent to the size of the triangles (I silently translate back to dv = dt times acceleration). Now Feynman said that he was confused by Newton’s proof of the resulting polygon being an ellipse – and he proposed a different proof:
Newton started from what Rubinstein calls the sun ‘pulsing’ at the same intervals, that is: replacing the smooth path by a polygon, resulting in triangles of equal size swept out by the radius vector but in a changing velocity.  Feynman divided the spatial trajectory into parts to which triangles of varying area e are attached. These triangles are made up of radius vectors all at the same angles to each other. On trying to relate these triangles to each other by scaling them he needs to consider that the area of a triangle scales with the square of its height. This also holds for non-similar triangles having one angle in common.

Part VI: Since ‘Feynman’s triangles’ have one angle in common, their respective areas scale with the squares of the heights of their equivalent isosceles triangles, thus basically the distance of the planet to the sun. The force is proportional to one over distance squared, and time is proportional to distance squared (as per the scaling law for these triangles). Thus the change in velocity – being the product of both – is constant! This is what Rubinstein calls Feynman’s big insight. But not only are the changes in velocity constant, but also the angles between adjacent line segments denoting those changes. Thus the changes in velocities make up for a regular polygon (which seems to turn into a circle in the limiting case).

Part VII: The point used to build up the velocity polygon by attaching the velocity line segments to it is not the center of the polygon. If you draw connections from the center to the endpoints the angle corresponds to the angle the planet has travelled in space. The animations of the continuous motion of the planet in space – travelling along its elliptical orbit is put side-by-side with the corresponding velocity diagram. Then Feynman relates the two diagrams, actually merges them, in order to track down the position of the planet using the clues given by the velocity diagram.

In Part VIII (embedded also below) Rubinstein finally shows why the planet traverses an elliptical orbit. The way the position of the planet has finally found in Part VII is equivalent to the insights into the properties of an ellipse found at the beginning of this tutorial. The planet needs be on the ‘ray’, the direction determined by the velocity diagram. But it also needs to be on the perpendicular bisector of the velocity segment – as force cause a change in velocity perpendicular to the previous velocity segment and the velocity needs to correspond to a tangent to the path.

Revisiting the Enigma of the Intersecting Lines and That Pesky Triangle

Chances are I made a fool of myself when trying to solve an intriguing math/physics puzzle described in this post.

I wanted to create a German version but found it needs a revision. I will just give you my stream of consciousness as I cannot make it worse anyway.

The puzzle is presented as a ‘physics puzzle’ but I think its enigmatic nature is described better if stated in purely mathematical terms:

Consider three lines in a flat plane, not parallel to each other and not intersecting in a single point. Their mutual intersection points are the corners of a triangle.

Assuming that the probability to find an arbitrary point on either side of each line is 50% – what is the probability to find a point within the triangle?

I had proposed a solution 1/7. My earlier line of reasoning was this:

The three lines divide the full area into 7 parts – the triangle in the center and 6 sections adjacent to the triangle: Each of these parts is located either ‘to the left’ or ‘to the right’ of each line, called the ‘+’ and ‘-’ parts.

Center of mass, physics puzzle.

Proposed solution in post published in Feb. 2013: The body is divided into 7 parts; the center of mass being located in either with equal probability. (Image Credits: Mine)

There are 8 possible combinations of + and – signs, but note that the inverse of the symbols assigned to the triangle is missing: (-+-)  Digression: It would be there if we painted these lines on a ball instead of a flat plane – then each line would close on itself in a circle and there would be 8 equivalent triangles. The combination missing here would correspond to the triangle opposite to the distinguished and singular triangle in our flat plane.

I had assumed that the 7 areas are equivalent based on ‘symmetry’ – each area being positioned on either side of one of three planes – and assuming that the condition given (50%) is not physical anyway. A physical probability would vary with distance from the line – imagine something like a Gaussian symmetrical distribution function centered around each line. Than the triangle would approximately correspond to the area of highest probability (where the peaks of the three Gaussians overlap most).

Do you spot the flaw?

Intersecting lines, two halves

The lower half contains 4 different parts (1x triangle, 2x open trapezoid, 1x open wedge), and the lower half contains two open wedges and one open trapezoid. Probabilities should add up to 50% in each half though.

We can do two cross-checks:

1) the sum of the probabilities of all parts should add up to 1 – OK as 7 x 1/7 is 1. But :

2) the sum of probabilities of all pieces on either side of a line should add up to 0,5! This was the assumption after all.

Probabilities don’t add up correctly if I assign the same probability to each of the 7 pieces – it is 4/7 for the lower half and 3/7 for the upper half.

So I need to amend my theory and rethink the probability assigned to different kinds of areas (I guess mathematicians have a better term for ‘kinds of areas’ – more like ‘topologically equivalent’ or something.).

We spot three distinct shapes:

  • A triangle formed by the three lines.
  • Three ‘open wedges’ formed by two lines – e.g. part (- – -) in the lower half.
  • Three ‘open trapezoids’ formed by three lines, e.g. part (+++) in the upper half.

I am assuming now that probabilities assigned to all wedges are the same and those assigned to trapezoids are the same. I am aware of the fact that this will not work out if we consider a limiting case: Assume the angle between two of the three lines gets smaller and smaller – this will result in one very small wedge (between the red and the blue line) and two ‘wedges’ which are nearly equivalent to a quarter of the total area:

Intersecting lines, narrow wedges

In the limiting case of the blue and red lines coalescing we would end up with four quarters, and you would find an arbitrary point with a probability of 25% in either quarter.

In the video Quantum Boffin has asked for the probability of the triangle – which can be any triangle, of any arbitrary shape and size, and he states that there is a definitive answer. Therefore I think also the details of size and shape of the other areas does not matter, and the 50% assumption is somewhat unphysical.

As there are three distinct types of shapes – I need three equations to calculate them all.

Notation in the following: p… probability. T…triangle, W…wedge, Z…trapezoid. p(T) denotes the probability to find a point in the triangle.

The sum of all probalities to meet the point in either of the 7 pieces must be 1, and we have 3 wedges and 3 trapezoids:
i) p(T) + 3p(W) +3p(Z) = 1

We need 50% on either side of a line.

There is one Z and 2W on one side…
ii) p(Z) + 2p(W) = 0,5

…and T and 2W on the other side:
iii) p(T) + 2p(Z) + p(W) = 0,5

Now the sum ii) and iii) is just i) that these equations are not independent. We need one more information to solve for p(T), p(W), and p(Z)!

And here is my great educated guess: You have to make another assumption and this has to be based on a limiting case. How else could we make an assumption for an arbitrary shape?

I played with different ones, such as letting iv) p(W) = 0,25 motivated by the limiting case of a nearly right angle. Interestingly, you obtain a self-consistent solution. Just plugging in and solving you get: p(T)=0,25 and p(Z)=0. Cross-checking you see immediately that this is consistent with the assumptions – probabilities sum of to 50%: You either have two Ws or one W and the T in one half of the plane.

Assigning 0 to the trapezoid does not seem physical though. We can do better.
So what about assigning equal probabilities to Z and W? iv) p(Z) = p(W)?

I don’t need to do the algebra to see that p(T) has to be zero as you would have 3 equivalent pieces on each side, but the triangle can only be located one one side.

This assumption is in line with the limiting case of a really infinite plane. The triangle has finite size compared to 6 other infinite areas.

I change my proposal to: The probability to find an arbitrary point in the triangle is zero – given the probability to find it on either side of each line is 50% and given that the area is infinite.

Again I’d like to stress that I consider this a math puzzle as the 50% assumption does not make sense without considering a spatial variation of probability (probability density, actually).

Addition as per November 21:

Based on the ingenious proposal by Jacques Pienaar in the comments, I am adding a sketch highlighting his idea.

Theoretically, the center of mass would correspond to the intersection of the 3 “perfect” solid lines. Now allow for some “measurement error” and add an additional line denoting the deviations. I depicted the “left” and “right” lines as dashed and dotted, respectively.

Now take a break, get a coffee, and look at the position of the true center of mass with respect to the triangles made up by the dashed and dotted lines:

Intersecting-Lines-Proposal-Jacques-PienaarSince we have 3 colors and either a dashed or a dotted lines, there are 8 distinct triangles. I tried to make the angles and distances as random as possible, so I think Jacques’ proof does not depend on the details of the configuration or the probability distribution function (yet beware the limiting cases such as parallel lines). The intersection of the solid lines is within 2 of 8 triangles – hence a probability of 2/8 = 1/4.

I was intrigued by an odd coincidence as I had played with o,25, too (see above), but based on the assumption that of a probability of 0,25 for the wedges/corners – which by cranking the algebra or just cross-checking the 50% criterion results in p(Triangle)=0,25, too, and in p(Trapezoid)=0.

Looking hard at this new figure introduced by Jacques I see something closely related, but unfortunately a new puzzle as well: The true center of mass is in exactly two of eight trapezoids built from dashed or dotted lines. So I am tempted to state:

p(Trapezoid)=0,25.

But it is difficult to make a statement on the corners or wedges as any intersecting two lines cut the plane in 4 parts and any point is found in one of them. I was tempted to pick p(W) = 0 though, but this would result in p(Triangle)= -0,25.

So this was probably not the last update or the last post related to the enigma of the intersecting lines.

Physics / Math Puzzle: Where Is the Center of Mass?

On randomly searching for physics puzzles I have come across QuantumBoffin‘s site.

The puzzle is about how to determine the center of mass of a body using the plumb line method, given there is some uncertainty due to experimental errors:

You have found three plumb lines not intersecting in a single point; thus the three intersections of two lines each form a triangle. The question is:

What is the probability that the center of mass is actually located within the triangle?

The following assumption should be used:

The probability of finding the center of mass on either side of each line is 50%

If you want to entertain yourself with trying to solve the puzzle on your own, do not scroll down. I did not find a published solution, so the following is just my proposal. Thus there is a chance that I make a fool of myself. But I enjoy those deceptively simple science puzzles, and there is always a good chance so-called intuition might lead you astray.

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I have tackled the problem as follows:

If there is a single line, it divides the body into two pieces, the center of mass (COM) is found with 50% probability in either piece.

If there are two lines, they would intersect in a point, and the body would be divided into four quadrants. The COM is located in one of these quadrants with a probability of 25%. I’d like to stress that the reason is that the body is divided into 4 equivalent pieces, each of them is located either to the left or to the right of each line. In this case the result is the same as 0,5 times 0,5.

However, if there are three lines, the probability is not simply equal to 0,53 = 1/8 (as confirmed by QuantumBoffin)

I propose: The solution is 1/7.[*]

The three lines divide the piece into 7 (not 8!) parts – the triangle in the center and 6 sections adjacent to the triangle: Each of these parts is located either ‘to the left’ or ‘to the right’ of each line, I call these the ‘+’ and ‘-‘ parts (halves) of the body:

Center of mass, physics puzzle.

Proposed solution: The body is divided into 7 parts; the center of mass being located in either with equal probability. (Image Credits: Mine)

The seven parts are equivalent. This may sound a bit awkward as you might expect the probability to find the COM on the periphery lower than in the middle. But this is due to the assumption that should be made – the assumption (50%) should be replaced by a probability distribution. So in a sense, this is more a math puzzle than a physics puzzle.

[*] Edit as per Nov. 2013:  I think I have found a flaw in my argument as the sum of probabilities for areas on either side of a line would not add up to 0,5 if I assign the same probability to each of the 7 areas depicted below (It is: 4/7 on one side versus 3/7 on the other side). I am now searching for a system of equations that let me determining the probabilities for the three different types of areas: the triangle, the corner areas (wedges) and the open trapezoids.
You can find my updated proposal here. I keep the rest of the post unchanged as I consider my random musings about curved spaces (on the bottom) correct.

I did not use the number of 0,5 in a calculation, but as a justification of the equivalence of the 7 pieces.

Now there is a remaining – mathematical – puzzle that still baffles me: There are 8 (23) possibilities of combining + and -: Which is missing and why?

-+- is the missing combination.

The picture exhibits some nice symmetry, but note that the definition of + and – as such is arbitrary. So we could have ended up with other missing combinations.

  • +-+ in the center. The missing combination is the ‘negation’ of the combination reflecting the triangle.
  • +++ and — opposite to each other (opposite = one attached to a corner of the triangle and the other to the side of the triangle not connected to this corner)
  • Other pairs / opponents:
    –+ and ++-
    -++ and +–

Imagine the negative of the triangle now: To the left of the red line, to the right of the blue and above the green line. Obviously there is no intersection between these, and this is the only combination that does not result in any intersection at all. I suppose there is a better way to state that in stricter mathematical terms. And if we put the figure on a sphere (Non-Euclidian geometry), there would be an intersection – actually the figure would collapse onto 4 equivalent triangles (a ‘bulged tetrahedron’).

In case of two lines only, all combinations can be ‘realized’ in flat space, therefore the solution is simply equal to 1 over the total number of combinations (4).

This stuff is addictive – finally I understand why Feynman was fascinated by flexagons.