During the past 24 hours, I heard, I swear!, 6 commentators blame biofuels for one of the world’s pains or another. What bothers me about this is that the complaints are, by and large, misguided in one way or another. So I’m starting a blog on the subject, “Ethanol Fuel”, or “Moses Lake”. It’s gotten to the point where the people in the industry joke with each other by pointing out yet another news article blaming biofuels for one thing or another.

So I’m starting another blog. I’m going to provide informed commentary on the energy situation, the food supply, land use, regulation, the economics of biofuels and fuels in general, the engineering, the politics of the subject, etc. And perhaps posting will leave me feeling that I’ve made a little effort towards stopping our country from being led by ignorance into policies that will take years to retract. I don’t think these belong on Mass which is devoted more to physics.

Producing fuel and electricity are engineering problems, the best decisions will be made by engineers who specialize in those areas, not by the general public. Nevertheless, the general public has the ultimate political power and their decisions, rational or otherwise, could effect the future of this country and this planet. And so I will write on the issues with the objective of education.

And it’s not like the US (or any other country) is immune to engineering decisions made on the basis of emotion. As a youth, I saw the nuclear power industry in the US destroyed by ignorance and fear masquerading as wisdom. Now, with the benefit of 30 years of hindsight, and another wave of ignorance and fear (this time related to global warming rather than radiation) support for nuclear power is increasing.

Louise Riofrio recently pointed out that the inflation is in a bit of trouble due to the fact that it predicts a different curve than the one seen for the angular correlation of anisotropies of the Cosmic Microwave Background (CMB). An easily understood review of the CMB is given by The Cosmic Microwave Background for Pedestrians: A Review for Particle and Nuclear Physicists, astro-ph/0803.0834. The data excludes the curve expected by inflation at well above the 99% level:

Previously, Louise had explained the anomaly in a manner that I was too obtuse to understand, for example:

Views of the Cosmic Microwave Background may also indicate a spherical Universe. By measuring distances between acoustic peaks, scientists hope to complete a triangle and determine curvature. When a changing speed of light is accounted for, the angles do not add up to 180 degrees and the triangle is not flat. Most telling, the scale of density fluctuations is nearly zero for angles greater than 60 degrees. Like a ship disappearing over Earth’s horizon, the lack of large-angle fluctuations is smoking-gun evidence that the Universe is curved. Both lines of CMB data indicate that the curvature has radius R = ct.

And, in the comments of this post:

Mendo, I dearly wish for the Space/Time to answer your questions properly. Both relate to density fluctuations grown large by expansion. These can be modelled as Fourier series. The old inflationary paradigm says that fluctuations should be the samne at all scales, leading to a prediction corved ruled out by WMAP and COBE.

Theory predicts that the largest possible fluctuation has half-wavelength of 180 degrees. On the chart, these fluctuations cancel each other out above 60 degrees.

I didn’t find these very convincing, or even understandable. So I sat down to work out the CMB fluctuations under her model myself, from first principles, and sure enough, it does work! Quite nicely! Here’s the calculations:

The Riofrio Variable Speed of Light

Given the speed of light c, and age of the universe t, the observable universe has size R = ct. The usual is to think of this as R(t) = c t, that is, the observable universe increases as the universe ages. Riofrio’s insight was to put the time dependence onto the speed of light instead: c(t) = R/t. There’s a bunch of highbrow general relativity calculations to support this which you can find on her blog, and do not concern us here.

The reason I find this version of cosmology attractive is that it means that spacetime is stable. This is compatible with the flat space version of GR that was found by the Cambridge Geometry Group, (which uses Clifford algebra) and about which I need to write a longer post, and to improve my website on the subject. I want to redo it in a faster loading, simpler, version like my density matrix website with lots of literature references.

A flat space version of GR avoids the science fiction problems of the standard GR, and it also makes it natural that current space appears to be very close to flat. The usual gravity theories have difficulty explaining how it comes to be that the universe hasn’t collapsed into a little ball or whatever. Similarly, Riofrio’s cosmology is clean and simple. Space is big, flat, and matter doesn’t mess it up much.

There’s a great deal of modern commentary to the effect that variable speed of light theories are pointless because the speed of light can always be assumed to be constant, so light itself defines length. They see the speed of light as simply being the conversion ratio between distance and time. But there is nothing unnatural in having a conversion ratio be non constant. And most significantly, c(t) = R/t is a very simple equation. This is far far far more restrictive than what one can do while keeping the speed of light constant but allowing the metric of spacetime to vary. So Louise’s idea has fewer arbitrary constants or assumptions.

And Louise’s cosmology is a lot easier to make calculations with. Most particularly, in the flat space version of Louise’s cosmology, angles are trivial to compute. This makes it particularly easy to compute the angular dependency of the correlations of the CMB anisotropies.

World’s Easiest Cosmology Calculation

Suppose we travel at speed v(t) = kc(t) = kR/t, where k is a constant, from time t = a to t=b, in the direction x. In Louise’s model, the calculation is easy to make:


The distance travelled depends only on the ratio of the beginning and ending time, and is logarithmic.

The CMB light comes to us from the time of “recombination,” by which word the cosmologists mean the time at which ionized matter combined to form hydrogen, helium, and “metals”. Before that time, the universe was largely opaque to light because it was scattered by all the free charges. After that time, the universe was largely transparent to light. This happened at around 380,000 years after the big bang, about 14,000,000,000 years ago.

Since the universe has been transparent since the time of recombination, we can see all the way back to that time, and the CMB is what we see. Furthermore, when we look in different directions, what we see is the recombination as it happened approximately 13,999,620,000 years ago [LOL]. That means that the recombinations we see at different directions now were once very widely separated. How far apart were they? Recombination light travels at speed c(t) so k=1. Using the above integral, the distance the recombination light has travelled is:

When we are looking at recombination light coming from two different directions with an angle between them of , this gives the distance between them as as can be seen from trigonometry:

Acoustic Waves in Plasma

In order for the two recombination regions to be correlated, they must have been in causal connection. When you read the cosmology literature, you find that before recombination, the universe was filled with ionized gas, a plasma. In this plasma, matter and energy moves around with “acoustic waves”. The speed of these waves is approximately as is explained in section 1.6 of The CMB for Pedestrians.

These acoustic waves are just like sound waves that we’re all familiar with. They tend to move stuff away from where it’s crowded, towards where it’s not crowded. Now suppose the universe was anisotropic at some early time s << 380,000 years. And suppose that there was a crowded spot over here, and a less crowded spot over there. If the distance between these two spots is short enough that the acoustic wave could get between them, then they can be correlated. If it’s longer than the distance sound can go between s and 380,000 years, then the two regions will be uncorrelated.

If the two spots are very close together, sound can pass back and forth between them multiple times. This will tend to even out there differences so we will see a positive correlation. But that correlation will turn negative if they are far enough apart that sound can only make one trip. This is all explained in section 1.6 of the above reference where they talk about multiple poles. The pole we’re concerned with here is the biggest one, the l=1 pole.

Inflation

Before the theory of inflation, it was assumed that the CMB would not be well correlated. When it was discovered that the CMB is very highly correlated in all directions, inflation was brought in as an explanation for the correlation. The inflation assumption was that distant parts of the universe were once in causal contact and this was when the regions became correlated. Later, as the universe expanded, the correlations remained. The inflation was supposed to have happened at some very very early time in the big bang, seconds, (which is conveniently before our physics theories would have to fall on their swords).

So far so good. But in requiring inflation to happen so early, they were able to calculate that the maximum angle in which the CMB would be correlated would be about 1 degree. You can move this number around by assuming various degrees of curvature in the universe. But the beauty of Riofrio’s theory on a flat space is that one does not need to do this. But to get the fact that the CMB is quite nearly isotropic, we do need to add another epoch.

Corrleation Angles and 2nd Recombination

So there are two problems in the CMB correlations. The first is that the data are close to isotropic. The second is that the very small anisotropies disappear beyond an angle of 60 degrees. To explain this data, we will suppose that the universe had three epochs. The first is a wide period where everything pre plasma happens. We’ll call it the “snuark epoch”. In this scheme, the overall high isotropies are caused by the very long duration snuark epoch, while the small anisotropies arise from the plasma epoch. The speeds of causality (i.e. how matter moves around in bulk) in the three eras are:

(1) The snuark epoch, from the Planck time, seconds to s years. Snuark acoustic waves move at speed the speed of light, or some high fraction of it, i.e. divided by the square root of 3.

(2) The plasma epoch, from s = 3 minutes to 380,000 years. Plasma acoustic waves carry information at speed .

(3) The dark epoch, from 380,000 years to present around 14 billion years. Light carries information at speed .

The ratio of the time durations of the snuark epoch will be sufficiently long, compared to the ratio of times since recombination, that this period would let its acoustic waves die down resulting in the very isotropic overall distribution of CMB radiation. Since this period is so isotropic, we can’t easily verify its duration. However, for the plasma epoch, we can verify s from the approximate maximum angle with nonzero correlation, around 60 degrees.

The Photon Epoch

When the temperature of the universe drops low enough that the snuarks recombine, the effective speed of causality drops, and this sudden decrease in the speed of sound allows the remaining failures of correlation to appear to us now in the above graph. According to the simplest cosmology trig calculation ever, the distance between two barely correlated spots in the recombination plasma is:

We make this equal to the distance that causality can travel during the plasma era, that is, and get:
or
years = 0.0048 years or about 1.75 days, which is kind of late, Wikipedia gives the photon epoch as lasting from 3 minutes to 380,000 years, so let’s reverse the calculation; we’ll use 3 minutes and compute the maximum correlation angle.

According to the Wikipedia timeline of the big bang, the photon epoch lasted from about 3 minutes to 380,000 years. Putting s at 3 minutes gives ln(380,000 years / 3 minutes ) = 25. Multiplying by sqrt(1/3) gives 14.4, and fitting that into the world’s simplest cosmology trig equation gives the maximum correlation angle as:
degrees
so theta should be about 86 degrees.

So really, for a back of the envelope calculation, this isn’t too far off. In fact, looking at the chart at the top of this post, 86 degrees looks about right. Victory!

The Varying Speed of Light

Given a flat background space, a natural interpretation of a varying speed of light is to suppose that the rate at which time passes is changing. Our clocks measure time based, in the final analysis, on the de Broglie frequencies of matter waves. This was discussed in the previous post, David Hestenes Electron Model, which suggested that the source of that frequency is the rate at which the left and right handed portions of particles oscillate back and forth.

In order to arrange for the speed of light to follow the pattern c(t) = R/t, we could assume that there is a particle which is present throughout the universe which stimulates the oscillation. As time goes on, the density of these particles steadily increases and this causes the oscillations to increase, and therefore our clocks to speed up, and therefore our measurement of the speed of light (relative to the scale of the universe as a whole) to increase.

On the quantum mechanical level, this sort of thing, a stimulated oscillation between left and right handed states, can be modeled by the same methods used to calculate the stimulated emission of radiation in lasers. I’ll eventually write a post showing how this can be used to calculate the Painleve gravitational field, at least to first order post Newtonian.

John R Ramsden, as a comment on Motl’s blog, points us to an article by David Hestenes, Reading the Electron Clock, 0802.3227. There are two good reasons for looking at this paper, and the experiment that inspired it. First, we can use it as an excuse to discuss Hestenes’ electron theory, and what it looks like in the density operator language. Second, in a later post, we can discuss de Broglie’s matter waves.

Zitterbewegung

As the Wikipedia Zitterbewegung article states, Zitterbewegung [German for trembling motion] is an “interference between positive and negative energy states produces what appears to be a fluctuation (at the speed of light) of the position of an electron around the median, with a circular frequency of , or approximately .” It was first noticed by Schoedinger, I think. It comes about when you compute the position operator, as a function of time, for an electron with momentum. The electron’s momentum p and mass m define a velocity v = p/m, and the position operator x ends up being given partly by vt = pt/m, but there is another term, an oscillatory term. [In quantum mechanics, the Hamiltonian H gives the energy, and setting H = mc^2, so the equation in the Wikipedia article looks like . Also, I'm mixing metaphors a bit between relativistic and non relativistic definitions of mass. Read the original articles for the correct derivation.] So long as you restrict to solutions of the Dirac equation which have only positive or negative energy, there is no Zitterbewegung, and this is how the frequency is usually ignored.

Hestenes is a long time advocate of The Zitterbewegung Interpretation of Quantum Mechanics. In his interpretation, Zitterbewegung (sometimes called zbw or zitter) is not due to interference between positive and negative frequencies but instead arises from the complex phase factor present in all quantum mechanics. Hestenes looks at the zbw as being associated with spin. The consequences are wide. He writes:

The essential feature of the zbw idea is the association of the spin with a local circulatory motion characterized by the phase factor. Since the complex phase factor is the main feature which the Dirac wave function shares with its nonrelativistic limit, it follows that the Schroedinger equation for an electron inherits a zbw interpretation from the Dirac theory. It follows that such familiar consequences of the Schroedinger theory as barrier penetration can be interpreted as manifestations of the zbw.

From my point of view, zitterbewegung corresponds to the frequency with which the electron switches back and forth from its left handed to right hand form. At its deepest level, it arises from the Feynman checkerboard model previously discussed and the requirement that the left and right handed fields are massless and travel at speed c. (So the electron must oscillate back and forth in order to be stationary.) Hestenes also sees zitterbewegung as arising from spin, but in a manner that is a little different from mine. And he also begins with geometric (Clifford) algebra so our models have deep similarities. To see where the differences come about, it’s useful to analyze Hestenes’ mathematics from my (density matrix) point of view.

Hestenes’ model in Density Matrix form

Where Hestenes and I differ is in that while he uses the state vector formalism, for which the geometrization is kind of complicated and a little arbitrary, I use the pure density matrix formalism, which has only one unique and very simple geometrization. For example, in spin-1/2 qubits, the natural density matrix (operator) form for a spin up quantum state is (1+z)/2 where {x,y,z} are the Pauli spin matrices. The quantum state for spin in the -x direction is (1-x)/2, etc.

To review the traditional relationship between pure density operators and state vectors [i.e. where the density operator is defined from the state vector rather than vice versa], let’s discuss the quantum state for spin up. We can write a state vector representation as |a), where the curved bracket is used to prevent WordPress from misinterpreting. This state vector is not unique, but may be multiplied by any complex phase to obtain an equivalent representative, as in exp(+ik) |+z). This is the “ket.” For each such there is also a “bra” which takes the negative phase: exp(-ik) (+z|. The density matrix form is given by the product:
(1+z)/2 = exp(+ik) |+z) exp(-ik) (+z| = |+z)(+z|.

Now the density matrix was very naturally geometrized (written in Clifford algebra with geometric interpretation) as (1+z)/2, but this does not define a unique state vector form. And so, of course, Hestenes’ geometrization of the state vector is also not unique. He must provide information that corresponds to that arbitrary phase. Since he is building a geometric theory, he must provide that information in geometric form.

I’ve been discussing this with the Pauli algebra (or the “Pauli Spin Matrices” if you absolutely have to perform the ugly and unnecessary act of choosing an arbitrary representation for the Pauli algebra), but the differences are general to all Clifford algebras and also apply to the Dirac algebra (i.e. Dirac’s gamma matrices), and I will switch back and forth between the algebras.

In Hestenes’ paper, the arbitrary complex phase is given by equation (34), . The origin of this arbitrary phase is better described in an earlier Hestenes paper, The Kinematic Origin of Complex Wave Functions, where it is defined on page 3 as follows:

The next step in explicating the geometric structure of the Dirac theory is to reformulate
it in terms of the spacetime algebra. One easy way to do that is to choose a fixed unit
spinor u satisfying the eigenvalue equations


where i is the usual unit imaginary of the complex field. Equation ( 8 ) is especially signiciant because it relates i to the bivector , and thus reveals that i has an implicit geometric meaning in the Dirac theory. This meaning can be made explicit by eliminating i from the theory in favor of the geometric quantity . It is easy to prove that each Dirac spinor can be written in the form

where is a unique even element of the real Dirac algebra.

The real Dirac algebra is just the real Clifford algebra on x, y, z, and t. The even elements are the elements of the even subalgebra, that is, the subalgebra whose basis is {1, xy, xz, xt, yz, yt, zt, xyzt}.

Density matrices and the even subalgebra

Why is he talking about the “even subalgebra here?” It seems very arbitrary until you understand how the density operator language treats quantum states. A quantum state is given by a primitive idempotent of the Dirac algebra (which is a Clifford algebra). A complete set of basis elements means four quantum states that form a basis for any quantum state (just as two quantum states, spin up and spin down, form a complete basis set for a Pauli spin-1/2 state). In the density matrix language, these four states could be Q, R, S, and T which are 4×4 matrices that annihilate each other and sum to 1.

One finds complete basis sets for a Clifford algebra by first defining a “complete set of commuting roots of unity.” This means as large a set of operators (that square to 1) that we can put together, provided that they all commute and we don’t include trivial products. For the Dirac algebra, a complete set of commuting roots of unity is {ixy, ixyzt}. In fact, these are the usual set chosen for the usual spinors used in the Dirac equation. Note that they are both elements of the even Dirac algebra.

With this choice of the commuting roots of unity, the four states making up a complete basis set (electron, positron, spin up and spin down) is:
Q = (1+ixy)(1+ixyzt)/4,
R = (1+ixy)(1-ixyzt)/4,
S = (1-ixy)(1+ixyzt)/4,
T = (1-ixy)(1-ixyzt)/4.

More generally, we might want to look at spin in some other direction say w = ax+by+cz where . So “w” is another element of the Dirac algebra, a spatial vector. To get the basis set oriented in that direction, we need to replace the “xy” with something that has to do with u. Working in the +++- signature so that zz = +1, we can write xy = (xyz)(z). Thus for a general direction u, the complete basis set is:
Q = (1+ixyz w)(1+ixyzt)/4,
R = (1+ixyz w)(1-ixyzt)/4,
S = (1-ixyz w)(1+ixyzt)/4,
T = (1-ixyz w)(1-ixyzt)/4.
Note that the above is in the even subalgebra, the same place which Hestenes seems to arbitrarily pick out for . And it should be clear that considering all the possible directions, the above sorts of things cover the even subalgebra. For example, we can write 1=Q+R+S+T, and ixyzt = Q-R+S-T, and ixyzu = Q+R-S-T, and more difficultly, ut = Q-R-S+T. Thus we get the scalar 1, the pseudoscalar ixyzt, an arbitrary spatial bivector ixyzu, and an arbitrary space-time bivector ut is covered by the basis set. This exhausts the even subalgebra’s basis set, {{1}, {ixyzt}, {ixy, iyz, izx}, {xt, yt, zt}} so any even element of the subalgebra can be written as a sum over multiples of the pure density operators given above (which do not mix particle with antiparticle).

The arbitrary spinor u as “fictitious vacuum”

The other part of Hestenes’ Dirac wave function prescription is the fixed spinor u. It’s quite complicated, and the confusion associated with this has generated some interesting literature, see the excellent paper by Baylis, quant-ph/0202060. Hestenes goes through a lot of gymnastics to explain this arbitrary item, but in the context of density matrix formalism and Julian Schwinger’s measurement algebra, its meaning is quite clear. It is a choice of vacuum.

There are two Schwinger papers that define the fictitious vacuum. The first is The Algebra of Microscopic Measurements which is quite short and easy to understand. The vacuum is defined on the second page of the second article, The Geometry of Quantum States. The reader who finds the following discussion impossible to fathom might read Schwinger’s version. The difference is that his is written in a field theory language while I’m writing here in a wave function vernacular.

Spinors Defined using Ficitious Vacuum

Let’s work entirely in the density matrix formalism. We choose a (more or less) arbitrary pure state “V” and call it the quantum vacuum. Given a (pure) density matrix state Q, we define corresponding bras and kets as follows:
|q) = Q V,
(q| = V Q,
providing QV and VQ are nonzero. (If they are zero, then we go back and choose another vacuum. Schwinger covers this case more elegantly, by defining a new state outside of the universe of states, that does not annihilate anything.) Since Q and V are pure density matrices, they satisfy QQ = Q, VV = V, and tr(Q) = tr(V) = 1. We now show that |q) and (q| act just like kets and bras.

As a first step, let’s compute (q|q):
(q|q) =(V Q) (Q V) = VQV.
Now the state V is fully defined by the fact that it has some set of quantum numbers defined by a complete set of operators. As a pure density matrix, it has these quantum numbers double sided, that is, they apply to either side. But the operator VQV has V on both sides and so it will also be an eigenstate for these operators and it will have the same eigenvalues. Therefore, it is a multiple of the eigenstate V. To figure out the multiple, write v = tr(VQV). Then VQV = vV. The reader can see that v must be real because this is just the transition probability between the states V and Q, that is, tr (VQV) = (q|v) (v|q). This tells us how to scale our spinor definition. And it also tells us that in computing scalar products, our answers will be in multiples of the vacuum V.

Now, let M be an arbitrary quantum operator. We wish to compute its expectation value for the state Q. In the density matrix language, this is given by:
m = (M) = tr( Q M ) = tr (Q Q M ) = tr (Q M Q) which must be equal to our (q| M |q) so tr (QMQ) = m, and as before, QMQ = mQ. Using this fact:
(q| M |q) = (VQ) M (QV) = (V Q M Q V)
= (V (QMQ) V) = (V mQ V) = m (VQV) = mv V.
The “v” is just the normalization for our choice of vacuum. That is, (q|q) = v V so we need to divide by v. The result is that (q| M |q) = m, and our definitions of (q| and |q) give the same expectation values as usual.

So the arbitrary “u” that Hestenes’ model of the electron uses corresponds to the arbitrary vacuum choice that Schwinger introduces when he converts his measurement algebra into state vector form. It corresponds to a way of picking an arbitrary geometric phase when converting from density matrix form to a spinor form.

Lubos Motl brings to our attention a paper by Ted Jacobson and Aron C. Wall on black hole theremodynamics and Lorentz invariance, hep-ph/0804.2720 and claims that theories that violate Lorentz invariance are ruled out because they will also violate the second law of thermodynamics, the law that requires that entropy never decreases. Lubos concludes, “At any rate, this is another example showing that the “anything goes” approach does not apply to quantum gravity and if someone rapes some basic principles such as the Lorentz symmetry or any other law that is implied by string theory, she will likely end up not only with an uninteresting, ugly, and umotivated theory but with an inconsistent theory.” I disagree with this.

First, the abstract of the article:

Recent developments point to a breakdown in the generalized second law of thermodynamics for theories with Lorentz symmetry violation. It appears possible to construct a perpetual motion machine of the second kind in such theories, using a black hole to catalyze the conversion of heat to work. Here we describe the arguments leading to that conclusion. We suggest the implication that Lorentz symmetry should be viewed as an emergent property of the macroscopic world, required by the second law of black hole thermodynamics.

From the abstract, we see that Lubos has put the cart in front of the horse. Rather than proving that Lorentz symmetry has to be exact “all the way down”, the authors instead say that Lorentz symmetry does not have to be present at the foundations of elementary particles because it will automatically emerge macroscopically as a result of requiring that the second law of thermodynamics apply to black holes. And I agree wholeheartedly with this.

When physics students first learn the facts of Lorentz invariance, there is a tendency to think of the results as being so bizarre, so beyond our usual experience, that the only possible conclusion one could come to is that Lorentz invariance itself is a deep property of nature itself. Not an emergent property of matter, but something that must be satisfied by everything at all times. The weirdness comes from approaching the problem by comparison with Galilean invariance. So much for physicists.

For engineers, Lorentz invariance is less of a surprise. Engineers are concerned with elastic deformations in things like steel and concrete. One writes down the equations of motion. In solving them, one splits wave motion into longitudinal and transverse (what the earthquake people call P and S waves), and one finds that these two waves have different speeds, with longitudinal waves travelling around 50 to 80% faster than transverse waves. This reduces the more complicated elastic equations of motion into two branches, each a massless Klein-Gordon equation, but with different wave speeds. So from an engineering point of view, Lorentz symmetry follows naturally from assuming that space is an elastic solid, but only one branch of its elastic equations of motion are present. The other is suppressed by quantum effects, presumably becoming significant at much higher (Planck) energies.

The other point that Lubos makes is that there is no major group of physicists working on a theory that microscopically violates Lorentz invariance. This is exactly the sort of physics argument that causes mathematicians the most laughter; a combination of “truth put to the vote” and “absence of evidence is evidence of absence.” So let’s describe a theory of the electron that microscopically violates Lorentz invariance.

Feynman Checkerboard Model

Wikipedia has a nice 1 paragraph article on the Feynman checkerboard model. It is a discrete model of spacetime in 1+1 dimensions. If you look through the photographs in Genius: The Life and Science of Richard Feynman, you will find a photocopy of Feynman’s notebook page on the subject.

The modern tendency is to see elementary particles as determined only by symmetry principles. For this point of view, there is no need for a microscopic explanation. Feynman was evidently looking for a microscopic explanation for the electron propagator. In his checkerboard model, the electron always travels at speed c, and at each unit of time makes a choice of reversing direction or continuing. One sums over paths. Reversals of direction take a phase contribution of . In the limit as the size of the checkerboard square size goes to zero, one ends up with the right propagator for the electron. An illustration showing two paths:

The wikipedia article says that the Feynman checkerboard model has not been generalized to 3+1 dimensions but this is not the case. The generalization is trivial. Peter Plavchan, now a post doc at Caltech, wrote a paper on the Feynman checkerboard model. This is far more complete than the Wikipedia article and includes, in addition to the 3+1 generationalization, the relationship to the 1-d Ising model and the full Dirac equation as well.

And the 3+1 dimension generalization? What sort of weirdness do you suppose it has? In order to allow the particle to be able to travel at speed c in the (1,1,1) direction as a series of steps equally balanced between +x, +y, and +z directions, the speed in those directions must be . The natural 3+1 generalization of Feynman’s checkerboard model is a counterexample to Lubos’ claim that no such theories exist. As to why it’s not worked on much, well, with attitudes like Lubos’ what else can one expect? As Connes said, physicists are bosons, they all tend to think alike.

Preons

Feynman’s checkerboard model splits the electron into two parts which move in opposite directions at speed c. This is a little prescient in that elementary particle theory treats the electron as composed of left and right handed fields, each of which is massless and travels at speed c. For a stationary electron in 1+1 = z+t dimensions, these two massless fields are quite similar to the checkerboard model. We suppose that the electron is aligned with its spin in the +z direction. We associate the right handed electron with movement in the +z direction and the left handed electron with movement in the -z direction.

Generalizing this to 3+1 dimensions means splitting the left and right handed electrons into three components. Let’s suppose that spin is in the (1,1,1) direction so that the right handed electron moves in this direction while the left handed electron goes in (-1,-1,-1). Then the right handed electron will be modeled with three preons going in the +x, +y, and +z directions while the left handed electron will be made from -x, -y, and -z travelling objects. Following the tradition, we label the right handed directions +x, +y, and +z as R, G, and B, while their left handed relations are /R, /G, and /B.

Let’s take a look at just the right handed electron (though our arguments will not depend on handedness). According to the 3+1 Feynman checkerboard model of the electron, there will be three state we can find the right handed electron in, R, G, and B. Planchan considered a slightly simpler thing; he considered the right and left travelling electron and its becoming a right or left travelling. There were therefore 4 propagators, left to left, left to right, right to left, and right to right. He called these propagators . In our case we have three initial and three final states so we will have nine propagators, which we can similarly label with pairs of indices.

Path Integrals and idempotency

One of the properties of path integrals is that they are idempotent, an obvious fact but it isn’t stressed much in most student’s educations so we will give a short (somewhat simplified in that we’ll deal with only one dimension) proof. Let K(x,x’) be the path integral for some wave equation. Then we can use K as follows (see equation 12 of Plavchan’s checkerboard paper or read the wikipedia article on path integrals):

Applying this twice, we have:

and therefore

This is an idempotency relation, that is, we have K = KK where “multiplication” is defined as convolution. But the above was done in position space. If we take the Fourier transform, the convolution is turned into a more convenient multiplication, and so if G is the Fourier transforms of the path integral K, we have:

In other words, considered as matrices, G is idempotent, . And matrices are easier to calculate with so that’s what we’ll assume from here on. And what happened to ? It’s gone. We’re looking not at the wave function, but instead at the properties of the path integral that transforms the wave function. This is the field equivalent of working with the density matrix instead of the state vector.

Generations

Path integrals reduce to complex numbers so given R, G, and B as our indices, we are concerned with 3×3 matrices of complex numbers:

So we look for 3×3 matrices that are idempotent. In doing this, we have a choice of how to represent this matrix. The simplest choice is to diagonalize them so that we are looking at the 3×3 idempotent diagonal matrices. There are 8 such matrices; they consist of all the diagonal matrices with 0s and 1s on their diagonal. How many particles do these correspond to?

Now one of these 8 matrices is the 0 matrix. That makes a heck of a useless path integral. What we really want are the path integrals that preserve particle type. As with density matrix theory, these are the matrices with trace = 1. There are three of these; their diagonals are (1,0,0), (0,1,0), and (0,0,1). All the other idempotents are sums of pairs of these (or the unit matrix, which is the sum of all three). Interpreting these as particles, the sums of a pair is a two particle state and the unit matrix is a three particle state.

Thus, when looking in the 3×3 complex idempotent matrices, we find three single particle solutions. With reference to the electron, we interpret these as the electron, muon, and tau.

Koide’s Lepton Mass Formulas

To make the electron not have any orientation other than spin, we need to treat the three directions R, G, and B equivalently. We need the path integrals from R to G be the same as the path integral frmo G to B and B to R. Similarly for the R to R, G to G, and B to B path integrals. This implies that we need to consider circulant matrices. Going back to the matrix, and putting , etc., we make the matrix idempotent and solve for the primitive idempotents:

It would take us a little too far afoot to derive Koide’s mass formula for the lepton, how I rewrote it, and how I used it to predict the masses of the neutrinos. The details are in my paper, The Lepton Masses (2006). This paper ended up with four citations in the published, peer reviewed physics literature, but the citations were for the formula only. They did not cite the theory behind it, which I find kind of annoying, but not surprising. Everyone wants to keep doing the thing that got them tenure, or, alternatively, to get tenure by not sticking their neck out very far on anything different. This post has mostly been an attempt to explain the theory behind the Koide formulas from another point of view, one that is more complete and requires less understanding of Clifford algebra.

The Fermion Cube

Obviously I’ve done a lot more work on this since two years ago. After dealing with the leptons, the next obvious place to apply this sort of thing is to the quarks. In fact, the theory was applied to the structure of the leptons and quarks before it was applied to the generation structure. The paper was The Geometry of Fermions (2004), which used the assumption that the elementary particles would be primitive idempotents of a Clifford algebra (in the form of the Geometric Algebra originated by David Hestenes) to count the number of hidden dimensions in spacetime. Of course this paper is 4 years old and has a numberr of wrong guesses in it, especially the method of ending up with generations.

Avoiding the Clifford algebra complications, the claim of The Geometry of Fermions was that the best way to organize the quarks and leptons is on the basis of their weak hypercharge , and weak isospin quantum numbers, the quarks and leptons of a single generation graph as follows:

The quantum states of the above are the handed quarks and leptons. They come in columns with an up and down quark (or antiquark) surrounded by an electron and a neutrino (or anti particles). With color, the multiplicity of the quark states is 3 while the electron and neutrinos have mulitplicity 1. With respect to the 3+1 Feynman checkerboard model, the implication is that the quarks in the middle are made from the same stuff as the leptons on the end, but as mixtures.

For each column, there are two different preons, one related to the electron, the other related to the neutrino. Since the electron is negatively charged, we’ll draw these with a circle with a line through them and leave the neutrinos drawn as a circle. Each of the quarks and lepton is built from one of each color preon; these correspond to the Feynman steps in the +x, +y, and +z direction and are colored red, green, and blue:

In the above, electric charge becomes more negative up the chart with the electron having a charge of -1, the anti up quark -2/3, the down quark at -1/3, and the neutrino with 0. The electric charge is given by the number of “electron” preons contained in the particle. Each of the quarks is composed of two preons of one type and one of the other. The color of the quark can be defined by the color of the odd man out, so in the above drawing, the quarks are listed in red, green, blue order. The color force is attributed to the mismatch when one mixes preons of different types. Evidently, the requirement that the preon color force is even stronger than the strong force.

Weak Hypercharge and Weak Isospin

In addition to electric charge (Q) and color, quarks also have the weak hypercharge quantum number and weak isospin . There is a relationship between electric charge and these quantum numbers, so we really only need to define one of these two. We will choose weak hypercharge; it’s the quantum number associated with the U(1) part of U(1)xSU(2)xSU(3) while weak isospin follows SU(2).

The possible values for weak hypercharge are (-1, -2/3, -1/2, -1/3, -1/6, 0, +1/6, +1/3, +1/2, +2/3, +1). As in the above demonstration of Koide’s mass formula, we seek a simple idempotency equation that will have these values as its roots (and nothing else). Not much of a hope, eh?

The various elements of the matrix correspond to the “b” color switching to the “a” color. Following the lead of the color force, we assume that there has to be equal parts R, G and B in the overall state. This implies that the transitions from one color to another have to be related to one another.

We solved the lepton case by symmetry; we required that the states treat the three colors equally. This need not apply to the quarks. However, we still need to balance the transitions. Accordingly, we will analyze not the individual transitions, but instead collections of transitions that permute the colors.

There are six permutations on three elements. We will label these as I, J, K, R, G, and B, where I is the identity, R, G, and B are the permutations that leave R, G, and B alone, and J and K are the remaning two even permutations:

These permutations form a non commutative group under composition:

When we looked at the circulant matrices for the lepton case, we made the assumption that R=G=B=0 so that only I, J, and K contributed (and were, respectively, A, B, and C in those matrices). The requirement of idempotency, when translated from the 3×3 matrices into I, J, and K, was



These are 3 coupled quadratic equations. The above can also be attributed to the group structure on the even permutations of three elements. That is, one has I = II = JK = KJ, and this becomes the first line. Similarly, J = KK = IJ = JI and K = JJ = IK = KI.

We can generalize the above relations to the full permutation group. The result is six coupled quadratic equations in I, J, K, R, G, and B:

My audience is wAAAAy too lazy to solve six coupled quadratic equation, LOL, so I will refer you to my temporary paper where the above are given as equation (112), with slightly different notation, as the column . The spectrum of solutions for I turn out to be (0, 1/6, 1/3, 1/2, 2/3, 1). This is exactly what we need except that we also need the negatives.

We can transform the above six coupled quadratic equations into a set of six that will have the spectrum (-1,-2/3,-1/2,-1/3,-1/6,0) by replacing I, J, K, R, G, and B with their negatives. This will negate the left hand side of the equations but leave the right hand side unchanged. So we can get the negative signs by adding a parameter w to the left hand side that can be either +1 or -1. Since negating a wave function leaves the meaning of the wave function unchanged, these sorts of transformations have to be included in the theory. They amount to putting a minus sign into the definition of the permutation group definition (so that in addition to the R permutation, we also include a permutation -R).

Generations and Weak Hypercharge

The generation structure that gave Koide’s lepton formulas uses 3×3 matrices in a manner that appears, at first glance, to be incompatible with the method of getting the weak hypercharge quantum numbers out of 3×3 matrices. The method obtained to get the negative weak hypercharge numbers gives a clue on how to unify the two models.

Let the six permutations, I, J, K, R, G, and B, all have arbitrary phases associated with them. What could those phases be if we required that the objects still satisfy the same 6 coupled equations? In other words, given a solution to these six equations, what phase changes can I make to I, J, K, R, G, and B that is still a solution. This amounts to a symmetry of the group of permutation path intergrals. For the negative hypercharge numbers, the symmetry was that we could negate all six numbers.

With very little effort, the reader will find that the arbitrary phases are restricted in that they must all be cubed roots of unity. In fact, there are three solutions corresponding to the three generations found for the leptons, but tihs generalizes the solution to the case for the quarks. This topic is discussed further in a previous post, quarks, leptons and generations.

Further work

I’m signed up to give a talk at the northwest APS meeting in Portland, Oregon in mid May. I did not choose the above as the topic as it is too long to give in a 10 minute talk. Instead, I’ll be discussing another way of looking at this, which is to think of it as a generalization of the orthonormality of bound states. This was also briefly discussed in a previous postQuantum Bound States and the Hydrogen Atom, but I’ll be adding stuff having to do with the baryons and mesons.

In a sci.physics.foundations post, Jay Yablon has brought to light an obscure article by Hans C. Ohanian on the nature of the intrinsic spin of quantum objects and kindly loaded it onto the web: What is Spin? Am J. Phys. 54 (6) June 1986. The abstract is:

According to the prevailing belief, the spin of the electron or some other particle is a mysterious internal angular momentum for which no concrete physical picture is available, and for which there is no classical analog. However, on the basis of an old calculation by Belinfante [Physica 6 887 (1939)], it can be shown that the spin may be regarded as an angular momentum generated by a circulating flow of energy in the wave field of the electron. Likewise, the magnetic moment may be regarded as generated by a circulating flow of charge in the wave field. This provides an intuitivelyl appealing picture and establishes that neither the spin nor the magnetic moment are “internal” — they are not associated with the internal structure of the electron, but rather with the structure of the field. Furthermore, a comparison between calculations of angular momentum in the Dirac and electromagnetic fields shows that the spin of the electrons is entirely analogous to the angular momentum carried by a classical circularly polarized wave.

If you’re interested in the foundations of physics, the above is well worth reading. My efforts on quantum mechanics has been to look at things at a qubit level, where one reduces the number of degrees of freedom down to an absolute minimum. The calculations in the above are of momentum density, and energy density and the like. It’s nice to see them done explicitly.

The decorative knots to be discussed here are those which are tied with one or more cords that may be repeated through several plies. These sorts of knots can be represented by self-intersecting loops on the plane, set up so that no more than two loops intersect at any single point. One generates a tying diagram from such by picking which of the two paths are uppermost at each intersection point. While this could be done more arbitrarily, for the knots discussed here the paths will be selected so that each path alternates over and under as in:

My eventual objective here is to tie a knot with approximate dodecahedral or icosahedral symmetry. Let’s begin with a line drawing that has the right symmetry. Flattened out to the plane, the dodecahedron looks like the following planar graph:

But this is not in the form we need; it is not in the form of a collection of loops. The basic problem is that, as a graph, there are three edges meeting at each vertex.

There are ways of transforming diagrams so that the symmetry is retained. The reader is probably familiar Euler’s formula for planar graphs: V + F - E = 2 where the letters stand for the numbers of vertices, faces, and edges. If a triple (V,F,E) satisfies this relation, then so does (F,V,E). That is, Euclid’s formula allows the existence of a graph with the number of vertices and faces swapped. And in fact, just this can be done. At the center of each face, define a new vertex and draw edges from these vertices to other vertices if the old figure had an edge between the corresponding faces. The result will be that each vertex will be transformed to a face, each face will be transformed to a vertex, and the edges will be spun around. This duality transformation converts a dodecahedron into an icosahedron:

where the outer edges of the icosahedron need to be a bit broken. Unfortunately, this is also not in the form of a knot. The vertices of the icosahedron have 5 edges meeting; we need instead 4.

There is a transformation that will take an arbitrary graph and turn it into a knot. We need to replace each edge with two edges. The trivial way to do this is as follows:

But this isn’t much of a knot; it falls apart. Instead, we may make progress by replacing each edge with two edges with a crossing. The result is that each edge is doubled so E’ <= 2E, each vertex becomes a face and the old faces are kept so that F’ <= F+V, and each edge becomes a vertex so V’ <= E. Computing, we have V’ + F’ - E’ = (E) + (F+V) - (2E) = V + F - E = 2. Drawing over the black dodecahedron with colored lines, we find that there are six loops:

If a graph is in solid form, such as the edges of an icosahedron, and is regular enough, the above transformation can be obtained by cutting the vertices back to the centers of the edges. That is, the above modified icosahedron graph is also the edge graph of an icosahedron modified by a dodecahedron, the icosidodecahedron.

The tying diagram for the Monkey’s fist is built from 3 loops. One ties this with a single line by diverting the line from one loop to another after each loop has been traced an appropriate number of times (the usual is 3). The result is a knot with cubic symmetry. One could use the same method to tie a knot with icoshedral symmetry. There would be 40 crossings, so it’s a fairly complicated knot.

Another way of modifying a knot is to draw a line parallel to an already existing line. When the existing line is in the form of a loop, this can be thought of as splitting a loop into two. This can be repeated. Suppose we take all the loops of a knot and split them into four or five loops each. The result will be that each intersection (or vertex or crossing) will a square array of crossings. These regions will become cross hatches. Note that a square field of cross hatches has no contribution to Euler’s formula as each square has one face, four 1/4 vertices, and four half edges so the change in V+F-E is zero. In this sense, rectangles meeting four at a corner are flat.

Since Euclid’s formula gives V+F-E = 2, there has to be regions of the knot that contribute nonzero to this formula. For icosahedral symmetry in a knot, the natural way to arrange for this is to have 12 pentagons and 20 triangles. That is, the contribution to vertices from each pentagonal face will be 5/4 as the vertex is shared by 4 faces; the contribution to edges is 5/2. Similarly, the number of vertices from each triangle is 3/4 and the edge contribution is 3/2. The contribution from 12 pentagons is (12) + (12×5/4)-(12×5/2) = -3, while the contribution from 20 triangles is (20) + (20×3/4) - (20×3/2) = 5. Thus, as expected, a knot with overall approximate icosahedral symmetry will need 12 pentagons and 20 triangles.

Another basic topological operation that can be done on a knot, cutting and gluing. In order to preserve the flatness and/or approximate symmetry of the knot, this can be done by shifting the cut ends by one. if the cut and paste is made across quadrilaterals, then the cut will conveniently preserve the shape of the cut pieces, that is, the cut and paste will transform quadrilaterals to quadrilaterals:

Now if we split a loop, the result will be two loops separated by a series of quadrilaterals. Therefore, combining these two transforms is a way of changing the connectivity of a knot while keeping the number and types of shapes the same.

So. Here’s the challenge. Begin with the icosidodecahedron. It requires 6 loops. Duplicate each loop it now needs 12. Consider cuts between loops. Is there a way of choosing these so that the resulting knot becomes a single loop knot? If there is more than one way, what is the most elegant? And can any of them be tied as a Turk’s Head, that is, in a simple repetitive manner on a cylinder?

Kea recently brought up the subject of Zeno of Elea and his now long lost book of 40 paradoxes dealing with the continuum. His nominal 2500th birthday should be celebrated relatively soon. Let me paraphrase an example paradox is the following:

If one assumes that space and time are continuous, then an arrow shot from a bow, before reaching its target, must first travel half the distance. And then travel half the remaining distance. And so on. And therefore, there are an infinite number of distances to be travelled and the arrow could never reach the target. But arrows do reach targets. Therefore, space and time are not continuous.

Surprisingly, there is an echo of this thought in quantum mecahnics. The echo is so close to the original paradox that it is known as the Quantum Zeno’s Effect or sometimes “Paradox” depending on the writer. The subject is discussed in many arXiv articles.

In quantum mechanics, when one measures a system, the formalism requires that the system collapse to the result of the measurement. If one examines this carefully, one finds that if one measures a system at a sufficiently high rate, the effect of the repeated measurements is to prevent the quantum system from changing. In effect, if one examines the position of the arrow too frequently, the arrow cannot move. It’s worthwhile looking at the simple mathematics that causes this effect.

In the usual quantum mechanics, one represents a quantum system by a wave function or state vector. One obtains probabilities from this object by squaring its absolute magnitude. Given the state of the system at time t, a linear differential equation defines its state at later times. For instance, with Schroedinger’s wave equation, one has:

Suppose we have a very short time period . How much will the wave function change? As for any differential equation, we have:

Now suppose that we are looking at a quantum state that decays. For example, a uranium atom. At time t=0, we measure the system and determine that it has not decayed. According to the laws of quantum mechanics, this puts the uranium atom into a pure “not decayed” state. If this seems a funny way of talking about a thing, some physicists think so too.l You might try reading about Schroedinger’s Cat which will likely just confuse you further but may be more entertaining than one of those confusing foreign movies.

According to quantum mechanics, it is possible to define a complete set of basis states that fully define the wave functions that one could get if one made a measurement on the system. Now one of those basis states is the undecayed state and the rest of them consist of the various decayed states. Now at time t=0, we’re supposing the state to be the undecayed state (because we measured or checked it, and that is what it was). After a short time, the state vector moves to be a mixture of this undecayed state and a decayed state. But since the motion is linear (i.e. Schroedinger’s wave equation is linear), the amount of the decayed state that we have entered must be linear for small time.

To compute the probability of decay, we take the state we wish to measure and compute the dot product of it with the possible state we might get for a measurement. This is the “amplitude” and is a complex number. Again this amplitude must be linear in . To get the probability we compute the squared magnitude of the amplitude. Therefore, the probability of decaying has to be approximately proportional to the square of .

Exponential Decay

Now this is a very odd thing. When you have a radioactive atom, we assume that the rate at which it decays does not depend on time and so it should follow an exponential distribution. In exponential decay, the probability of a decay over a very short time interval is proportional to . It is not proportional to . This is what we intuitively expect, but it is not what happens.

Since the square of small numbers gives even smaller numbers, the effect of this nonlinear decay rate is that decay is suppressed near a quantum measurement. If we measure the system again, then again the decay rate will be suppressed. And if we repeatedly do this, then the decay rate will be reduced accordingly. Furthermore, in the limit as the time delay between measurements goes to zero, decay is completely suppressed; its probability goes to zero. This is the Quantum Zeno Effect.

As one can learn by reading the literature, one can show that the exponential distribution does obtain at longer times. Even more interestingly, the deviation from exponential decay rates reappears for extremely long times as well as short times, and it is possible for the decay rate to be increased (the “Quantum Anti-Zeno Effect”), for reference, see papers like quant-ph/9708024.

Density Matrices

From the density matrix point of view, part of the reason we see a paradox in the QZE is due to our natural inclination to see a quantum state as it were a little photograph of the condition of the system at a given time. We see a sequence of states as if it were a movie showing how the system changes with time. Similarly, when we split a system into things being acted upon and things that are acting, of course it becomes confusing.

Consequently, adding a measurement to the system, for example by using a laser pulse, is a modification of the process. Leaving the laser pulse out of the equations (and thinking of them as a “measurement” that uses no matter or energy) is unphysical and it is not surprising that the dynamics are unexpected. In fact, one can derive the decay rate change of the QZE without any need for a collapse hypothesis by using density matrix theory that includes the laser pulse used to make the measurement. This is well known in the literature and will show up if you search for “density matrix” along with “quantum zeno”. For example, see quant-ph/9611020:

The QZE and this experiment have not only aroused considerable interest in the literature [8, 9], but the very relevance of the above experimental results for the QZE has given rise to controversies. In particular the projection postulate and its applicability in this experiment have been cast into doubt, and it was pointed out that the experiment could be understood without recourse to the QZE by simply including the probe laser in the dynamics, e.g. in the Bloch equations or in the Hamiltonian [9]. Since the Bloch equations describe the density matrix of the complete ensemble, including the probe pulse as an interaction in them gives, however, no direct insight on how such a pulse acts on a single system.

The tying diagram Ashley gives for ABOK #2217 has a 4-fold axis of symmetry:

Tying a knot according to a diagram like this is quite time consuming. One must redraw the diagram by photocopying to the size needed. And in tying the knot, one pins the rope to the diagram. This is a pain because the rope moves around, the pins come out, etc. And the pins can damage the appearance of the rope.

In this post I give an alternative method of tying this knot, and several others like it, that is easier to set up, is much faster for each knot, and uses cheaper materials. Rather than an expensive cork board, we will use a 2×2 and build the knot as if it were a sort of Turk’s Head knot, on a cylindrical of square form.

First, to show that this method of tying the knot gives an identical knot, observe that the following diagram gives the same pattern of numbers as the Ashley diagram:

These drawings are how one converts a path in a 2-manifold into an over/under knot. This knot has 40 crossings (or 41, as the bitter end is hidden). Following the path of the rope, every other crossing will be an “under.” Beginning with the start point, label the under crossings in sequence. Along the way, you will also visit all the “over” crossings. If there is already a number there, then great, at that point in making the knot you will be laying rope on top of rope. But if there is no number yet there, then the “under” rope is not yet placed. This means that when you do get around to that crossing, you will have to dig under the “over” rope that is already there. Mark these crossings with a circle.

The result is an Ashley tying diagram. To tie the knot with this diagram, pin it out on a cork board. Lay rope along the diagram. Whenever you meet a crossing marked with a circle, there should already be a rope there and you need to go under it. This is quite a pain.

Turk’s heads are cylindrical knots. Most Boy Scouts know how to tie small Turk’s Heads in the hand. Larger Turk’s heads are tied on a cylinder. Suppose one wishes to tie a Turk’s head with 8 bights. One finds 16 nails and a wood cylinder of appropriate size. One marks two rings on the cylinder and puts the nails on these rings in equally spaced intervals. One chooses an offset for the left to right and right to left movement. Starting at the left side at nail 0, you go to a right nail and then back to a left nail with the offsets one has chosen. To tie the Turk’s Head with a single line, one must return to nail #1, 3, 5, or 7 as these are relatively prime to 8. The offset on the right side is typically chosen to be close to half the total offset for left to right and back. The crossing diagram is trivial in that the rings on the cylinder alternate between left over rigth and right over left. So there is no need to deal with a tying diagram. In addition, one runs the cord freely from side to side and so there is no need to pin the cord.

The above cylindrical diagram for ABOK #2217 means that it possible to tie this beautiful knot using the Turk’s Head’s method.

Materials: 50 feet of 5/32″ cord. 1 1/2″ diameter wood ball. 6 inches of wood 2×2. 16 screws or nails, 5/8″ long. 15 feet of thin wire or thread to be used as a guide. [Metric: 10 meters of 4mm cord. 40mm wood ball. 20cm of wood 5cmx5cm. 16 screws or nails, 15mm long. 5 meters of thin wire or thread to be used as a guide.]

No need to be exact, just get the nails or screws into the square wood piece approximately as follows:

Next we tie the guide wire to the wood. We’re going to label the four rings of screws, from the left in the picture, “outer left”, “inner left”, “inner right”, and “outer right.” Tie one end of the guide to an outer left screw. We’re going to wind it diagonally around the wood, back and forth. The rule for winding is fairly simple, but it is different when you are going from left to right versus when you’re going from right to left.

Starting on the right, from an outer screw go two sides to an outer screw; And from an inner screw go one side to an inner screw. Starting on the left, from an outer screw go two sides to an inner screw; And from an inner screw go two sides to an outer screw:

OL -> OR + 2,
IL -> IR + 1,
IR -> OL + 2,
OR -> IL + 2.

Eventually, after visiting all 16 screws, you should get back to where you started and your form will look something like this:

As with any Turk’s Head, different rings will be either left over right, or right over left. You need to mark this on the form. I used a black magic market to mark the places where left would go over right. (The unmarked crossings will have right over left.) The result looks something like this:

Now tie the cord to a screw on the left side of your wood. Wind the cord along the guide wire, bringing it under itself where needed to be compatible with the marked crossings. This doesn’t take very long at all. Early on your knot will look like this:

And eventually, following the guide wire and going over or under as needed, you return to where you started:

Lift a few bights off of the nails and remove the knot from the form:

The knot is now done with 1-ply, but it is too loose to be pretty. So pull out some of the excess cord, bringing the knot to a more spherical shape:

Before you get it too small, put the wooden ball inside. Pull more cord out of the knot, leaving it loosely around the wood ball. Then tighten it up a little, but still leaving it pretty loose:

Now we double the knot. Note that the beginning and ending points of the knot are next to each other. That means you can lead the working end back into the knot, following alongside the other end. To do this you will have to go over and under where the other lead does this. After a few minutes your knot should progess like this:



In doing this, you may find it useful to have a large crochet needle.

When you get back to the origin, take one last extra dive so that the two ends of the rope come out separated by one crossing. This makes it look more symmetric (and is shown in the Ashley drawing as crossing number 41):

For the dimensions suggested, you should have enough room to add a 3rd ply. Our photos are for a smaller ball with room only for two plies. So add another ply according as you think appropriate.

Finally, it’s time to tighten up the knot. When you’re done, the knot should be quite hard as the rope will be pulled tight. To tighten up the knot, work out the excess line the same way you’ve done it before. Uh, did I forget to mention how this is done? Ok. You move a loop of excess cord through the knot starting at one end and working the loop through to the other end. By the time you get from one end to the other, you will have tightened up each individual part of the knot. Here’s a drawing showing a loop partway worked through the knot:

I suppose I should also mention that it is important to eliminate twists in the knot as you pull excess material through it by moving a loop. Twists could become kinks if you leave them in. And try to avoid tearing up the material while you’re doing this. The finished 2-ply knot:

For those of you who aren’t familiar with blue collar knots, “ABOK” means “The Ashley Book of Knots,” an ecyclopedic book on knots written in 1944 that has since become the reference for knot identification. I think it was my maternal grandfather that gave me my, somewhat rare, now the worse from love and use, 1st edition copy back in the 1960s; but what with the natural self-naturedness of a boy I cannot recall for sure. #2217 refers to a particularly handsome knot in the chapter “The Monkey’s Fist and Other Knot Coverings.” To justify “handsome” requires a sample photo from cbrew6 on Knot Heads World Wide:

This knot is “tied on the table,” which means that one uses a diagram to draw it. From a topological point of view, a table diagram in this case is a mapping of the surface of the sphere to the plane. A line drawing shows the path that the cord takes. The path is a loop, that is, it ends at the same point at which it starts. The path is restricted to never cross itself twice at a single point. At each crossing, some sort of notation indicates which line is to be on top, but for planar knots like the above, it is arranged so that the cord will weave over, under, over, under … And it is a trivial fact of practical folks topology that one can always assign such a pattern.

The above knot is “5-ply,” by which is meant that the knot was tied once and then duplicated four times by following the initial path over again four times. This is made convenient by the fact that the end point for the knot is the same as the initial point. The difficulty in designing knots like these is to arrange this, that is, one wishes the path to be a single loop.

ABOK #2217 has a 4-fold symmetry axis. In the above example, the axis runs vertically, roughly between the two purple cords extending through the top. This allows us to show how it was designed, perhaps, and to redraw its tying diagram into a form that is easier to generalize. The 4-fold axis is around the center of the tie diagram:

In the above knot, the bights are made of 3 or 4 cords. This is ideal for a covering knot; a larger number would tend to let the encased object peak through:

A convenient way of designing a knot is to assume that it has some symmetry and to draw a region that can be stitched together several times. In stitching together two regions, we will simply number the free ends from each region and connect them in order.

Since the knot is going to have at least one 4-bight, we might as well assume that the knot has a 4-fold axis through a 4-bight. The regions we’ll stick together will need to be approximately diamond shaped, with a height around half their width:

The diamond shaped regions need to be approximately flat. This means that they should be constructed from 4-bights, that is, squares. And this construction works for knot #2217. The diamond shaped pattern is:

The diamond shaped region has 7 ends on the left and 7 ends on the right. Numbering these in order, we find that the diamond pattern defines a permutation which is a 7-cycle, (1475263):

Furthermore, the mapping takes rising lines to rising lines except for the top and bottom two rows. This ensures that there aren’t any sudden direction changes in the middle regions.

To get the 2217 knot, we use the above diagram four times. In fact, a single cycle knot would result if we reproduced it N times providing N is not a multiple of 7. More generally, for this sort of scheme to work, we need for the section of the knot to be an M-cycle if it has 2M free ends, and we need to reproduce it a number of times that is relatively prime to M. This is the equivalent of a well known result for the Turk’s heads. Ashley managed to write out a table classifying Turk’s head knots by number of bights and number of leads for up to 40 leads and 24 bights (in our example, there are 4 bights and 7 leads) but didn’t quite manage to figure out the relatively prime detail. As close as he could come was “A good practical way to plan Turk’s Heads is to take a prime number for the larger dimension (5, 7, 11, 13, 17, 19, 23, 29, 37, 41, etc.) and to use any smaller number, either odd or even, for the other dimension.” This quote is from ABOK # 1314. An example of several Turk’s heads, along with an ABOK #2217, see: Bell rope with ABOK #2217. For those who aren’t knotees, I should mention that white cord is the most difficult to tie as it shows errors most easily.

So, drawn in Turk’s head form, here’s my simplified diagram for the ABOK #2217:

This means we now also have a decent covering knot with a 3-fold axis of symmetry instead of 4. Such a knot has 30 crossings where #1314 has 40. Does the ABOK list such a knot? No, we may have found a new covering knot, assuming that it’s not already in the literature. But in any case we had fun and the method should generalize to produce other beautiful knots.

P.S. For any sort of plane (or spherical surface) polygonal region, Euler characteristic tells us that V-E+F = 2 where V = the number of vertices, E = number of edges, and F = the number of faces. For planar knots, E = 2F so this reduces to V = F+2. For this knot, V=10B, F = 10B+2, where “B” is the number of bights. For ABOK #2217, B=4.

For an arbitrary convex polygon with all vertices having four faces meet (as applies to a loop) and assembled from 4A triangles and B squares, V = 3A+B, E= 6A + 2B, F = 4A+B so we have (3A+B)-(6A+2B)+(4A+B) = 2 or A=2. So there will be 8 triangles and B squares. This suggests that the most symmetric covering knot will be one that distributes the 8 triangles as evenly as possible around the sphere. For knot ABOK #2217, the 8 triangles are distributed with a 4A2 symmetry. A more symmetric symmetry would be cubic, with the triangles on the corners of the cube. That is approximately obtained when the #2217 is tied with three bights instead of four. Of the 8 triangles, two are on the north and south poles. The rest alternate equally spaced above and below the equator. This is as close as you can get to cubic symmetry:

I may tie a 3-bight knot and put a picture up here. And sorry for the rough diagrams, WordPress has just changed their software and it defaults to medium resolution on my efficient little PNG files.

P.S. The “Turk’s Head” method of tying ABOK #2217 and variations works quite nicely. I’ll put up a series of photos showing how to do this soon. And the 3-bight version is a pretty knot: