Jan. 23, Slit Experiments and Coherence Patterns
Snarky Puzzle
The coherent source is often modeled as a plane wave. Write the quaternion valued plane wave like so:
$\phi = A_0 e^{(\omega t - k \cdot R + \psi)I}$
where
A0 is the amplitude
omega is an angular frequency
k is a wave vector
psi is a phase shiftSay you know the angular frequency and wave vector exactly. Fix the angular frequency and wave vector, and make the phase zero. [Arg, poorly formed question:]What values of time t and position R will have no signals? [Better: What values of time t and position R will the wave be completely real? What values to be completely imaginary?]
Background
Everyday light is incoherent. Like our everyday lives, light is jumbled in time and space, never certain what is coming next. I doubt an equation for incoherent light appears in most physics books. Instead students are fed a consistent diet of coherent light, which is easy enough to make in the lab with one slit or a laser.

Real if:
$\omega t - k \cdot R = n \pi$

Imaginary if:
$\omega t - k \cdot R = n \pi + \pi/2$

Discussion

From the write-everything-and-its-brother-as-a-quaternion view, I associate real with time-ish quantities, while a pure imaginary is space-ish. This source is predictable. It is the predictability that shines through on interference tests.

Jan. 31, Playing New Gauge Symmetry Games

Snarky Puzzle
Three odd gauge fields used in this blog used a quaternion product.
\begin{align*} \left( e_i \left( \frac{\partial}{\partial t}, \vec{\nabla} \right) e_i \right)^* \times (f, \vec{0}) &= \left( -\frac{\partial f}{\partial t}, +\frac{\partial f}{\partial x_i}, -\frac{\partial f}{\partial x_j}, -\frac{\partial f}{\partial x_k} \right) \\ \\ \left( e_{i+1 \% 3} \left( \frac{\partial}{\partial t}, \vec{\nabla} \right) e_{i+1 \% 3} \right) \times (f, \vec{0}) &= \left( -\frac{\partial f}{\partial t}, +\frac{\partial f}{\partial x_i}, -\frac{\partial f}{\partial x_j}, \frac{\partial f}{\partial x_k} \right) \\ \\ \left( e_{i+1\%3} \left( \frac{\partial}{\partial t}, \vec{\nabla} \right) e_{i+1\%3} \right)^* \times (f, \vec{0}) &= \left( -\frac{\partial f}{\partial t}, -\frac{\partial f}{\partial x_i}, +\frac{\partial f}{\partial x_j}, -\frac{\partial f}{\partial x_k} \right) \end{align*}
Redo them with the Klein 4-group product rule for multiplication.

Background
I was exploring how much I could do with the Klein 4-group. They look pretty flexible.

\begin{align*} \left(e_i \boxtimes \left(\frac{\partial}{\partial t},\vec{\nabla} \right) \right) ^* \boxtimes e_i \boxtimes (f, \vec{0}) &= \left(-\frac{\partial f}{\partial t},+\frac{\partial f}{\partial x_i},-\frac{\partial f}{\partial x_j},-\frac{\partial f}{\partial x_k} \right) \\ \\ \left(\left(e_{i+1\%3} \boxtimes \left(\frac{\partial}{\partial t},\vec{\nabla} \right) \right)^*\boxtimes e_{i+1\%3} \right)^* \boxtimes (f, \vec{0}) &= \left(-\frac{\partial f}{\partial t},+\frac{\partial f}{\partial x_i},-\frac{\partial f}{\partial x_j},+\frac{\partial f}{\partial x_k} \right) \\ \\ \left(e_{i+1\%3} \boxtimes \left(\frac{\partial}{\partial t},\vec{\nabla} \right) \right)^* \boxtimes e_{i+1\%3} \boxtimes (f, \vec{0}) &= \left(-\frac{\partial f}{\partial t},-\frac{\partial f}{\partial x_i},+\frac{\partial f}{\partial x_j},-\frac{\partial f}{\partial x_k} \right) \end{align*}

Discussion
Figure out one, the rest follow. There was not much to the exercise, but exercise is good.

Feb. 21, Diving into Euler-Lagrange: 1D (1 of 2)

Snarky Puzzle
A few blogs ago in the comments section, a little finite group theory happened, leading to the calculation that the quotient group of the quaternion group Q8 modulo the multiplication group {1, -1} Z2 was the Klein 4-group:
$Q_8/Z_2 = V$
[Note: V = Z2xZ2]
Don't feel bad if that doesn't mean much to you. Instead reread the comments so that it will make some sense - I know I tripped over any detail I could before stumbling to the answer. Now calculate:
$Q_8/Z_4 = ?$
What is Z4? It is a finite group with - you guessed it - 4 elements. There are a bunch of ways to represent the group. One applies to the complex numbers: e(i n pi/2) for n between 0 and 3. The result should be a non-Abelian because a quaternion divided by a particular complex number will be non-Abelian. Which group? Don't ask me, I have not done the calculation yet, but it may be interesting.

Background
I was wondering how much I could do with quaternion group Q8 and quotient groups. Not a lot because there are only so many subgroups of Q8. One can only form a quotient group using a subgroup. I think of it as dividing and leaving no remainder. I skipped over the trivial group since it is trivial:

$Q_8 /\{1\} = Q_8$

The next step is with Z2={-1,1}, the path to the Klein 4-group V. There are then thee ways to write the complex numbers using quaternions. Those are all subgroups, and thus the question at hand. The final "subgroup" is the group itself:

$Q_8 /Q_8 = \{1\}$

I bet all groups have those two quotient groups based on the trivial group, unless someone has designed some odd groups.

Answer[Correction: My first effort at solving this question failed. I will keep it as it might be helpful to others who have calculated fewer that a dozen such quotient groups in their years. I was not exhaustive and reductionist enough, plus I did not employ a decent check on the resulting proposal.]

Here is the starting expression in detail:

$\{1, -1, i, -i, j, -j, k, -k\} ~/~\{1, -1, i, -i\}$

This is a game of being systematically exhaustive, yet dropping any redundancy. It always starts easy...

$1/\{1, -1, i, -i\} = \{1, -1, i, -i\}$

In the spirit of dropping any redundancy:

$-1/\{1, -1, i, -i\} = \{1, -1, i, -i\}$

Less obvious, but still true:

$\\ i/\{1, -1, i, -i\} = \{1, -1, i, -i\} \\ -i/\{1, -1, i, -i\} = \{1, -1, i, -i\}$

Time to move onto something new with the j's and k's. No matter which one of the four one chooses, the result will be like the following:

$j/\{1, -1, i, -i\} = \{j, -j, k, -k\}$

This is the result so far:

$\{1, -1, i, -i, j, -j, k, -k\}~/~\{1, -1, i, -i\} = \{\{1, -1, i, -i\},\{j, -j, k, -k\}\}$

Is this reasonable? Quotient groups are like rearrangements of the starting group. This looks like the original group divided into two groups of four. Bingo.

But which group is this? A good first thing to do is calculate the multiplication table:

$\begin{matrix} \times&+j&-j&+k & -k \\ +1 |& +j & -j & +k & -k \\ -1 |& -j & +j & -k & +k \\ +i | & +k & -k & -j & +j \\ -i |& -k & +k & +j & -j \end{matrix}$

Let's compare that to the multiplication table for Z4:

$\begin{matrix} \times & +1 & -1 & +i & -i \\ +1 |& +1 & -1 & +i & -i \\ -1 |& -1 & +1 & -i & +i \\ +i | & +i & -i & -1 & +1 \\ -i |& -i & +i & +1 & -1 \end{matrix}$

The number 1 steps in for j, and i for k, but all the signs stay the same. This is Z4.

$Q_8 / Z_4 = Z_4$

[Correction: A check
One check on division is to see if multiplying the divisor by the result ends up at the numerator, and awkward way of saying if 12/3=4 then 3*4=12. Sixteen is bigger than eight, so I did not reduce things enough.

I started on the right foot:

$\{1, -1, i, -i, j, -j, k, -k\} ~/~\{1, -1, i, -i\}$

which led to:

$\{1, -1, i, -i, j, -j, k, -k\}~/~\{1, -1, i, -i\} = \{\{1, -1, i, -i\},\{j, -j, k, -k\}\}$

What I failed to do was to be exhaustive at this early stage, calculating in addition:

$\{1, -1, i, -i, j, -j, k, -k\}~/~\{1, -1, j, -j\} = \{\{1, -1, j, -j\},\{i, -i, k, -k\}\}$

$\{1, -1, i, -i, j, -j, k, -k\}~/~\{1, -1, k, -k\} = \{\{1, -1, k, -k\},\{i, -i, j, -j\}\}$

Clearly these three are related. What group is "enough' to do the job? There are only 2 elements in each of these cosets (a set of sets), so it must be Z2! I still have to work with my brain to accept that :-)

$Q_8 ~/~Z_4 = Z_2$

This does pass the check: 4 elements times 2 equals 8.]

[Right answer, wrong reason. A couple of things I "knew" and didn't mention. First is that Z4 contains the center of the group Q8. For that reason, the left coset will equal the right coset, and I believe it also means the resulting coset is a group. Which group? Form the multiplication table for the cosets. That is the part I did not do. Back to this calculation which is still valid:
$\{1, -1, i, -i, j, -j, k, -k\}~/~\{1, -1, i, -i\} = \{\{1, -1, i, -i\},\{j, -j, k, -k\}\}$

The question is what to do with such a thing. See if it is a group. Call the first coset A, the second one B.

\begin{align*} 1 \times \{1, -1, i, -i\} &= \{1, -1, i, -i\} \\ -1 \times \{1, -1, i, -i\} &= \{1, -1, i, -i\} \\ i \times \{1, -1, i, -i\} &= \{i, -i, -1, 1\} \\ -i \times \{1, -1, i, -i\} &= \{-i, i, 1, -1\} \\ A \times A &= A\\ \end{align*}

\begin{align*} 1 \times \{j, -j, k, -k\} &= \{j, -j, k, -k\} \\ -1 \times \{j, -j, k, -k\} &= \{-j, j, -k, k\} \\ i \times \{j, -j, k, -k\} &= \{k, -k, -j, j \} \\ -i \times \{j, -j, k, -k\} &= \{-k, k, j, -j \} \\ A \times B &= B \end{align*}

\begin{align*} B \times A &= B \end{align*}

\begin{align*} j \times \{j, -j, k, -k\} &= \{-1, 1, i, -i\} \\ -j \times \{j, -j, k, -k\} &= \{1, -1, -i, i\} \\ k \times \{j, -j, k, -k\} &= \{-i, i, -1, 1\} \\ -k \times \{j, -j, k, -k\} &= \{i, -i, 1, -1\} \\ B \times B = A \end{align*}

This is the group with 2 elements only, good old Z2.]

Discussion
When I posed the question, I was hoping something like SU(2) might pop out. Nope. Anyone know if there is a nice way to link Z4 and U(1)?

Feb. 27, Diving Into Euler-Lagrange: 2D, 3D, And 4D [parameterizations Of 1D Functionals] (2 Of 2)

Snarky Puzzle
Lots of theory, little practice. I will steal a problem from David:
Given a fixed length, closed boundary (like a fence), what shape boundary will maximize the enclosed area?

Background
Solving a calculus of variations problem should look different from anything in first year calculus. When I think about an area, I want to take a length L and multiply it by a width W like I've been doing since third grade. The length of a path would be 2(L+W).

Too bad, I graduated from third grade a while ago. The calculus of variations has to have derivatives and integrals and look darn different. There need to be two "darn different" equations, one for the area, one for the perimeter.

This is a one parameter equation working in a plain, so the variables are:

$x(t), y(t), \frac{d x}{dt}, \frac{d y}{d t}$

Area A
Recall how cross products provide a way to characterize an area? The position cross the velocity for a closed loop integrate over the parameter gets the area

$A = \frac{1}{2} \int{dt \left( x \frac{d y}{dt} - y \frac{d x}{dt}\right) }$

Perimeter P
This is the Pythagorean theorem on speed.

$P = \int {dt \left( \sqrt{ \left( \frac{d x}{dt} \right)^2 + \left( \frac{d y}{dt} \right)^2 } \right)}$

The question asks to find an extremum of a perimeter for a given area, a sum of these two:

$f =\frac{1}{2} \left( x \frac{d y}{dt} - y \frac{d x}{dt} \right) + \alpha \sqrt{ \left( \frac{d x}{dt} \right)^2 + \left( \frac{d y}{dt} \right)^2 }$

The alpha is there because the ratio of one of these to the other shouldn't matter for the extremum. This equation has both position and the derivatives of position, so can be fed into the Euler-Lagrange equation to find an extremum:

$\\ \frac{\partial f\left[x, \frac{d x}{d t} \right] }{\partial x} = \frac{d}{d t} \left( \frac{\partial f\left[x, \frac{d x}{d t} \right] }{\partial ~\frac{d x}{d t}} \right) \\ \\ \frac{\partial f\left[y, \frac{d y}{d t} \right] }{\partial y} = \frac{d}{d t} \left( \frac{\partial f\left[y, \frac{d y}{d t} \right] }{\partial ~\frac{d y}{d t}} \right)$

The answers should be really similar except for a few minus signs.

$\\ \frac{1}{2} \frac{d y }{d t} = \frac{d}{dt }\left( -\frac{1}{2}y + \frac{\alpha \frac{d x}{d t}}{ \sqrt{ \left( \frac{d x}{dt} \right)^2 + \left( \frac{d y}{dt} \right)^2 }} \right) \\ \\ -\frac{1}{2} \frac{d x }{d t} = \frac{d}{dt }\left( \frac{1}{2}x + \frac{\alpha \frac{d y}{d t}}{ \sqrt{ \left( \frac{d x}{dt} \right)^2 + \left( \frac{d y}{dt} \right)^2 }} \right)$

Integrate with respect to time t which is easy in this case due to the fundamental theorem of calculus:

$\\ y = \frac{\alpha \frac{d x}{d t}}{ \sqrt{ \left( \frac{d x}{dt} \right)^2 + \left( \frac{d y}{dt} \right)^2 }} + C_1 \\ \\ x = - \frac{\alpha \frac{d y}{d t}}{ \sqrt{ \left( \frac{d x}{dt} \right)^2 + \left( \frac{d y}{dt} \right)^2 }} + C_2$

This still looks complicated. I don't want two answers, I only want one. How to combine them? Look at the big scary term. If the constants are tossed over to the other side, squaring and adding up the result gets them to fade away. Nice.

$(y - C_1)^2 + (x - C_2)^2 = \alpha^2$

The answer is a circle somewhere int the plane.

[Correction: Thor Russell suggested an answer when this snarky puzzle was first asked: "Isn't it just a circle". The problem is that it looks like the kind of answer one finds in a standard calculus problem: it is all position, not position and derivatives of position. Not only should the question look different, but so should the answer. That is a simple check to apply.

With this change in perspective, it is not the goal to eliminate all traces of the velocities. Put the constants on the other side as before. This time multiply the equation y= by the y velocity, and x= by the x velocity, then add them. The messy term will drop, but the question has both positions and velocities:

$(y-C_1) \frac{d y}{d t} + (x-C_2) \frac{d x}{d t} = 0$

Is this really a circle? Does it solve the initial equations? Consider this effort provisional, I need to do more work with it tonight...]

Discussion
In the world of the calculus of variations, that was an easy problem. It does show how to do it in detail. The big picture is the shift from familiar x and y to functions and the derivatives of functions. One needs a tension between two terms. In this case it was the perimeter and the area expressed in terms of position and velocity.

Think about the jump up to the Maxwell equations. There the tension is between the interaction term and the field equations. That is a 4D extremum problem. No wonder EM is full of wonders.

Doug

Next Monday/Tuesday: BFF with HTFs (Hyperbolic Trig Functions)