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 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.
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:
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.
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.
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.
10 responses to “Lorentz Violation and Feynman’s Checkerboard Model”
Gee, good to see this all collected in one post. Thanks. I didn’t see the point in protesting over Lubos’ post, since this is one of the more common mudslinging topics for the people hanging over the sides of the spiffy stringy bandwagon.
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“Reversals of direction take a phase contribution of -iemc^2/hbar .”
I recently discovered that (Gm^2)^1/3 / ke^2 = 128.5383303^2
As expected theoretically, the newly obtained value of 1/128.5 is significantly larger than the 1/137 observed for a fully screened electron.
“Ours is a clean measurement of the electromagnetic effect,” Koltick says. In higher-energy experiments at other accelerators, the effect is swamped by additional factors, including the strong force, which holds neutrons and protons together in an atomic nucleus and binds quarks into protons and neutrons.
Didn’t you say that you’re still around after Fermi falsified the Lorentz violation up to 100 times the Planck scale? This article of yours surely doesn’t seem to say so – you wrote the same irrational rants against the crucial nature of Lorentz symmetry as all other incompetent physics commentators on the blogosophere did.
We both agree that Fermi showed that photon speeds do not depend on frequency, at least to first order. If’ my theory had predicted that the speed of photons depend on frequency then I agree that Fermi would be evidence against it. But if you’ll look carefully you’ll find I’ve never supported that idea.
On the other hand, Louise and I do believe that photons travel at speeds that depend on time. This hasn’t been tested by Fermi; it’s a statement about cosmology.
Where we disagree is that you believe that if light travels at speed c, then Lorentz invariance has to be perfect. This is a gross logic error; ask any competent mathematician. You’re basically an idiot with a mouth and some math skills.
I think that the electron does not become divided. The two waves come from the Dirac sea.The particles of the Dirac sea move all of the routes, not the electron.
is E1==E2 always
some detail on my website