Even Great Mathematicians Guess Wrong

http://rjlipton.wordpress.com/2011/04/13/even-great-mathematicians-guess-wrong/

by fogus • 15 years ago

54 10 15 years ago

rjlipton.wordpress.com

iwwr• 15 years ago

Another fascinating idea is to consider the split-complex numbers (and their split-quaternion cousins) as ways to express hyperbolic geometry (in 2 or 3/4 dimensions respectively).

aksbhat• 15 years ago

     Complex numbers are a powerful tool for studying the two-dimensional plane: each point corresponds to a unique complex number . The beauty of this correspondence is that it allows you to add, subtract, and multiply points in the plane

I wish someone would have told me this five years ago, it is such a neat way of thinking!

cdavid• 15 years ago

if you are interested in new insights about complex numbers and analysis from a geometric POV, this book is wonderful: http://www.amazon.com/Visual-Complex-Analysis-Tristan-Needha... and quite accessible.

happy4crazy2• 15 years ago

My introductory physics class in college included a problem that investigated the effects of "2d" gravity (a generic attractive force proportional to 1/r instead of 1/r^2) on a point mass in the plane.

When the professor mentioned that I could write all the vectors as complex numbers, and do regular old algebra with them, my brain almost melted.

xyzzyz• 15 years ago

Well, his argument is unconvincing -- it only proves that there is no normed division algebra on R^3. Division algebra on R^3 would be interesting and maybe useful even without norm. However, its existence is still impossible. One can prove it for instance by noticing that if it existed, then {1, v, v^2} for any non-scalar v would span whole R^3 (the hardest part of proof), so v^3 would be a linear combination of them, say v^3 = a 1 + b v + c v^2 for some a, b, c \in R. Then v is the zero of polynomial f(x) = x^3 - c x^2 - b^x - a. But deg f = 3, so it has real zero, so for some q, w, p \in R, f(x) = (x-q)(x^2 + w x + p), and 0 = f(v) = (v-q)(v^2 + w v + p 1), and factors on the right side are non-zero, so we have non-trivial zero divisors which is imposible in division algebra.

There is also a theorem due to H. Hopf, stating that the only finite dimensional, commutative division algebras over R are C and R itself, but the only proof I know requires pretty heavy topological machinery.

happy4crazy2• 15 years ago

If you enjoy playing around with quaternions, you may also have fun with geometric algebra.

Geometric Algebra for Physicists is very clear and gently paced. http://www.amazon.com/Geometric-Algebra-Physicists-Chris-Dor...

This one also looks good: http://www.amazon.com/Linear-Geometric-Algebra-Alan-Macdonal...

yummyfajitas• 15 years ago

Another great example (technically a physicist, not a mathematician) is Einstein and his quest for a local theory of quantum mechanics.

In his original paper deriving EPR ( http://en.wikipedia.org/wiki/EPR_paradox ), he believed it was a reducto ad absurdum which invalidated configuration-space based quantum mechanical theories:

We are thus forced to conclude that the quantum mechanical description of physical reality given by wavefunctions is not complete [...] No reasonable definition of reality could be expected to permit this.

http://prola.aps.org/pdf/PR/v47/i10/p777_1

Nevertheless, he thought a local and complete theory of QM was possible and spent many years searching. Bell showed this to be impossible (after Einstein's death). Eventually experiments showed the predictions of the EPR paper to hold, thereby implying that the definition of reality is not "reasonable".

If a scientist never gets things wrong, they aren't doing anything interesting.

ArbitraryLimits2• 15 years ago

I think it's an even better story how he "guessed wrong" with Hamiltonian mechanics, also; the only difference is that he never fixed the mistake and it wasn't really fixed until the early 20th century with the development of optimal control.

See "The Brachistochrone Problem and Modern Control Theory" (http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.26....)

The TL;DR is that classical mechanics can be formulated as finding a time history of position and velocity which minimizes a certain function of them both. Hamiltonian mechanics essentially optimizes the function over position and velocity as if position and velocity were independent, then adds the constraint that velocity be the time derivative of position via a Lagrange multiplier. This Lagrange multiplier is the momentum.

Of course, no one in a physics course actually learns it that way; undergraduates just memorize some equations and graduate students, if they are lucky, learn if from the Hamilton-Jacobi equation (which is how Hamilton developed it in the first place) where the momenta become the spatial gradient of the function to be minimized when evaluated at all positions. (It's exactly the same as the Bellman return function in dynamic programming; hence the name "Hamilton-Jacobi-Bellman equation.")

IMHO, neither of these other developments is as intuitive as the Lagrange multipler one; but because Hamilton's original formulation obscures it, no one in physics learns it that way.

A fun postscript: when you consider the equations of classical mechanics in terms of position and momenta (this combination is what physicists call "phase space") they form a manifold with a special property called "symplectic"; symplectic geometry can be formulated in terms of matrices over the field of quaternions. According to some people, this idea was a big breakthrough, but I don't believe Hamilton himself, despite having formulated both, ever noticed.

starwed• 15 years ago

"IMHO, neither of these other developments is as intuitive as the Lagrange multipler one; but because Hamilton's original formulation obscures it, no one in physics learns it that way."

I'm pretty sure learning a bit about Lagrange multipliers is standard for a grad classical mechanics class. Granted, I've forgotten all I learned back then, b/c it has never come up in anything else I do, but it was definitely part of the course.

ArbitraryLimits2• 15 years ago

I believe it used to be standard to use Lagrange multipliers to handle constraints on the state space in R^n as an alternative to using generalized coordinates. For example, when you handle the case of motion on the surface of a sphere you can either use R^3 with a Lagrange multiplier enforcing the constraint that you stay on the surface of the sphere, or you could use latitude/longitude as coordinates.

The interpretation of the momenta themselves as Lagrange multipliers is completely nonstandard, though. Try googling "hamiltonian mechanics lagrange multipliers"; you'll get either examples like the sphere thing or hits from optimal control tutorials.

Even Great Mathematicians Guess Wrong

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