For the unpopular cutting edge, there is no book with the answers at the back. This post will need to suffice for the snarky puzzles at the end of my previous posts. I am a playful snark, not a caustic one. I used to teach retarded citizens how to add. [clarification: while paying for 1 year of math grad school in Bloomington, Indiana, I volunteered an hour a week for a year at a local center.] I am that patient with myself, and will be so with you. Physics discussions can devolve into that 6 year-old boy fighting kind of groove (hi Sascha!). Since this is research, we do have a chance to learn something new together.
[click or skip a dramatic reading of this post]

The puzzles are meant to be simple. The answer has four parts. First is the quote of the problem, easy to skip is you recall the question. Second is the back story: why I originally did the calculation. Third is the one or two line mathematical answer. Fourth is a discussion of the implication of the solution.

March 14 - Quaternion Multiplication for a Third Grader

The Snarky Puzzle:
"Quaternions came from Hamilton after his really good work had been done; and though beautifully ingenious, have been an unmixed evil to those who have touched them in any way, including Maxwell." Lord K.

Quaternions have been a one-trick pony, good for doing 3D rotations since the 1850’s. We presume you have received a good enough education to write it down as:
R -> R’ = (cos(a), I Sin(a)) R (cos(a), I Sin(a))*
where I is an arbitrary 3-vector. Show how if hyperbolic sine and hyperbolic cosine functions are swapped in, evil results: while two terms are correct, two terms are missing, and two other terms that don’t belong soil the calculation any finely educated simpleton would try first. Your job: unmix the evil. For this not simply connected group, find another two combinations of these hyperbolic functions, R, and conjugate operators that correct the flaws. That is what “not simply connected” means, that things must be added together, or did they skip the obvious in your group theory training? Should you tire of the strain, the answer is on a preprint server. Note, that is not the preprint server, since those living on the ultra-conservative fringe have to build their own servers out of fairy dust and Drupal. Very old school here, pre-school, grad school, fringe school, whatever.

The Back Story:

Take another look at that quote: Lord Kelvin is bitch slapping James Clerk Maxwell. Given how much respect I have for Max, the historical level of contempt for work on quaternions is extreme.

Ivan, from Parahyangan University, Bandung, Indonesia, was majoring in Physics. For his final project, he decided to do rotations using quaternions. This was not the most original idea since that is what Rodrigues did with them in the middle of the 1800s. It was only an undergraduate thesis after all. The bigger problem was his professor who pointed out this had little to do with physics. Professors!

A new idea was proposed: work with the Maxwell equations using quaternions only. I suspect my detailed work on derivation of the four Maxwell equations from the Lagrange density would be on the first page of a web search. That pesky professor wanted still more, like showing what happened to the field strength tensor under a Lorentz boost. Ivan asked for help on this one.

Being a practical guy, I said there are 2 approaches. One says that since the quaternion terms are exactly like the antisymmetric rank 2 tensor, nothing new means nothing new. A second idea was to use a paper by De Leo on the subject of quaternion boosts written in 1996. He thanked me for my time, and did get his degree. Way to go Ivan.

I had to live with guilt. Yes, I had read the De Leo paper. I kinda got it, but not really. When I really get it, then I can turn it into code, either C or Perl, depending on convenience. That did not happen here. I knew I didn’t know it. I felt bad about my reply, even if Ivan didn’t.

This is what motivated me to look slowly and carefully again at regular old 3D rotations as done with quaternions for a century and a half. The plan was to switch to hyperbolic cosines and sines, repairing as needed.

$b'=h b h^*+\frac{1}{2}\left((h h b)^*-\left(h^*h^*b\right)^*\right) \quad eq. 5$

$=(cosh(2\alpha ) c t - sinh(2\alpha ) x, cosh(2\alpha ) x - sinh(2\alpha ) c t, y, z)}$

$=(\gamma c t - \gamma \beta x, \gamma x - \gamma \beta c t, y, z)}$

where b is a spacetime quaternion being boosted, and h is a quaternion with hyperbolic cosines and hyperbolic sines.

The Discussion:
Quaternions are in a lose-lose situation. So long as this result remains obscure, physicists are right to avoid quaternions as boosts are too critical for work in special relativity. Should the result gain some traction, it will not generate interest since boosts are so early 1900s.

Here is the new win. A compact Lie group can be used for 3D rotations. Using quaternions that is the product of 3 quaternions. The Lie group for the Lorentz group is not compact. Why not? Using quaternions, the reason is reflected in using both multiplication and addition. Of course the Lorentz group is not compact, because multiple parts are added together. It is as simple as that.

March 21, 2011 - Lunchbox for a Theoretical Physicist

The Snarky puzzle:
This is Gauss’s Law:
These are two gauges:
1. The Coulomb gauge
2. The Lorenz gauge
Apply these gauges to Gauss’s law. Prove that the resulting equations are not the same. Do not try and impress your friends with this obvious result.
This is the GEM version of Gauss’s Law:
Apply the gauges from above if you can.
“Toto, I have a feeling we're not in [Normandy] any more” Dorothy, misquoted to reference Pierre-Simone Laplace’s home state.
“Only bad witches are ugly” Good looking Good Witch of the North
First and last lines spoken here: http://www.youtube.com/watch?v=EPWenQxryr4

The Back Story:

I was made to feel like an incompetent dweeb by a professor in a rushed discussion of spin angular momentum projection operators for a current coupling term in a Lagrangian. I had never had a discussion with anyone about a spin angular momentum projection operator before. I bet few people have. He has won awards for his work, and is a classically frantic fellow. The message was go study Feynman, don’t bother me again, ever. The message was received.

After getting a beat down, it is easy to feel paranoid. I was writing up my results to submit to a real publication. As a gambler, I thought the odds of getting published were not worth the ante cost, the first chips on the table, chump change. If I cannot read and understand the titles and abstracts of a single paper in “Classical and Quantum Gravity”, why should I submit a paper to that very journal? It was out of respect for the process of science. As I am doing the draft, I am worried about the professor - how would he trash the paper? We are not talking about helpful suggestions, a note of encouragement. I imagined a shredder snark, one that might get some blood.

I thought he would ask me about how my unified standard model proposal would look under a gauge transformation. If I didn’t know, I would have to sit in the corner and sulk. That didn’t sound like fun, fear of failure never is. I lived with that feeling for about a month. Now I know the answer.

Gauss’s law under the Coulomb gauge
$\rho = -\nabla^2 \phi$
Gauss’s law under the Lorenz gauge
$\rho = \frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2}- \nabla^2 \phi$
GEM Gauss’s law under the Coulomb or Lorenz gauge
$0=\frac{G \hbar}{c^3} \nabla^2 \phi$

The Discussion:

Crank a Lagrange density through the Euler-Lagrange equations and out comes the field equations. All the discussions of gauge symmetries I have seen appear at the Lagrangian level, not for the fields. If I am just ignorant on this issue, please politely educate me. What I saw going on was people thinking carefully about their gauge choice so the resulting field equations could be solved for the problem at hand.

The GEM unified standard model field equations look every bit as invariant under a gauge choice as does the EM Lagrangian, but not the EM field equations. That feels like I rung a deep bell. It is only a feeling as I don’t know what fancy sounding jargon to cloak the observation.

March 18 - Time in Bed with Space

The Snarky Puzzle:

Consider an event in spacetime (t, x, y, z) in Cartesian coordinates so you do not have to think too hard. First show how this member of the Lorentz group:
diag{-1, 1, 1, 1}
would take the 4-vector into another 4-vector where only time had flipped its sign, no matter if time was positive or negative. Take that new vector and apply the same member of the Lorentz group to it. Repeat 1729 times. Admit you took the short cut of induction and did not complete the letter of the assignment. Nature lacks the power of induction, so does the long form.
Define a 4-vector R. I now bestow upon you the power to multiply 2 4-vectors together, a power you have always had Dororthy, via the rules of quaternion multiplication. Find R such that:
(t, x, y, z) R = (-t, x, y, z)
There can be no doubt such an R exists since quaternions are a mathematical field. The results should look so complicated the weak will leave the room. Let them go. Show how if the scalar is 10 orders of magnitude larger than the 3-vector as happens in classical physics, then to a wonderful approximation, R = (-1, 0, 0, 0). Maybe the weak return, since -1 is easy enough for them to remember. Notice that R is not a member of the Lorentz group. If the 3-vector is tiny, ignore the 3-vector, and flip the sign on time with a -1. That is an approximation whose flaw becomes obvious to systems using 1023 atoms. The arrow of spacetime is obvious to any child, so it should be to any physicist too.
[correction: the t, x, y, and z's should be deltas, dt, dx/c, dy/c, and dz/c. Events in classical physics have tiny relativistic velocities. For example, walking speed translates into relativistic velocity of 5x10-9, indicating the change in space is tiny compared to changes in time.]
The Back Story:

I was lucky to have one of the world’s best teachers on the subject of special relativity, Edwin F. Taylor. One of his classes was on the problem of time. He recommended the book by Price on the subject. Given the fundamental nature of the asymmetry of time reversal, this was an obvious thing to try early in my quaternion adventures (1998?).

$R=(-t^2+x^2/c^2+y^2/c^2+z^2/c^2, 2 \, t \, x, 2 \, t \, y, 2 \, t \, z) / (t^2+x^2+y^2+z^2)$
If t >>> x, y, z:
$R\approx (-1, \vec{0})$

The Discussion:

Any issue that has been around since Boltzman’s day must be tricky. This could be a removing the thorn from the paw of a lion story. Other professionals think the issue has been resolved, ignoring the modern analysis by Price. I have too much experience with frenetic physicists too busy to do a quaternion calculation, so I don’t worry. Spacetime has 3 arrows, more than enough to solve the arrow of time problem.

April 4, 2011 - Relativistic Rocket Science for Astrophysics

The Snarky Puzzle:

Here again is the relativistic rocket science technical speculation:
With the bold stroke of a pen, eliminate Newton. Isolate the small m and its differential to one side of the equation. Integrate both sides. If you are rusty, ask someone like alpha for help:
Integrate 1/m
Integrate 1/R2
Take the exponential of both sides. Only the second one is not exceedingly trivial:
exponential of the integral of 1/R2 series expansion for large R
Notice the series solution for R as R goes to infinity delivers what was promised.

The Back story:

I found a way to go from a force law for gravity to the exponential metric. It was a long, odd path. The first draft was shown to be wrong using Mathematica. I found a reasonable alternative that got the green light from my impartial, strict judge. At one step, the product rule came into play. I chose mA for sentimental reasons. My experience with shotgun DNA sequencing said I must also work the other road. Being complete is good. That led to a derivation of the relativistic rocket science term, not some inspired guess.

$\sqrt{\frac{G}{\hbar c}} m = \left(1 - \frac{G M}{V c R} + …\right)$
$\frac{G}{c^4} m \frac{d V}{d t} \hat{R}\rightarrow \frac{G}{c^3} V \frac{d m}{d R} \hat{V}$