Higgs? We ain't got no Higgs
By Doug Sweetser | April 18th 2011 11:06 PM | 37 comments | Print | E-mail | Track Comments

Trying to be a semi-pro amateur physicist (yes I accept special relativity is right!). I _had_ my own effort to unify gravity with other forces in...

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The title is a playful variation on a line from a Bogart movie, The Treasure of the Sierra Madre.  That was an adaptation from a novel of the same title:
"All right," Curtin shouted back. "If you are the [LHC] police, where are your [Higgs]? Let's see them."
"[Higgs], to god-damned hell with [Higgs]! We have no [Higgs]. In fact, we don't need [Higgs]. I don't have to show you any stinking [Higgs], you god-damned cabrón and ching' tu madre! Come out from that shit-hole of yours. I have to speak to you."
This blog will explain why we need to see the Higgs at the LHC. My studies offer an alternative. The big idea is that if a way to account for how gravity changes a symmetric metric, and build that into a cleverly constructed Lagrangian, then there will be no need for the Higgs.[click or skip this reading of the blog]

In the lingo of the World Series of Poker, I am all in for my bet against the Higgs particle being detected. The claim is on the front of the t-shirt. I have an inventory of $800 worth of t-shirts I sell globally. The IRS knows about them via a Schedule C. If the Higgs is found, I will be sending a check to Prof. Arkani-Hamed. This is not a bet, it is a payout. Nami has a position at the Institute for Advanced Study at Princeton, one of the highest perches in academia. You can have a bet with a peer, but he does not work the AM shift in the basement as I do. In an article, he said he was so confident the Higgs would be found, he was willing to wager a home mortgage payment. Time runs out on this payout in 2017. The Lagrangian for the standard model had a problem: all the gauge bosons had to be massless. That works for EM, it works for the strong force, but it does not work for the weak force. In 1964, three generations of researchers came up with a fix: Higgs, Engler and Brout, and Guralnik-Hagen-Kibble. There are only 5 people in the photo. I’ll let you guess who is missing, a modest man who has decided to dodge the lime light. Peter Higgs got the opportunity to rewrite his paper after a rejection. He put in a closing line about there being a boson to detect. This may be why his name is reference more than the others. If anyone has a blog devoted to the history of this discovery by the gang-of-6, send me an email and I will add it to the post. The idea is not to alter the Lagrangian. If one breaks symmetry in the Lagrangian, that can lead to charges not being conserved. Electric charge conservation has been tested to an absurd degree. I am not familiar with the level of experimental constraints on weak and strong charge conservation. Instead one postulates that the vacuum solution state is spontaneously broken. A false vacuum energy is not where any particle wants to be. It is the Mexican Sombero (available for$15 from http://mariachiconnection.com). The difference between the height of the hat and low part dictates the mass. The Higgs mechanism may also have consequences for the mass of fermions via a Yukawa coupling, but I don’t understand that stuff.

The Higgs is not the only game in town for providing mass for the gauge bosons. The only two I had heard about before were technicolor and higher dimensions. I will let others with in depth knowledge comment or link to those subjects if they wish.

The search for the Goddamn Particle has gone on for decades. Leon Lederman wanted to use that title for his book as a reflection of the frustration. An editor convinced him that the shorter title, God Particle, would generate more sales. It certainly was effective. Peter H. was not pleased because he appreciates that religious people could be offended. The slogan is not too modest either. Will the LHC be consistent with all previous searches and not see the Higgs boson? Us folks on the sideline only have to wait until the end of 2012 to know. There should be enough data collected by then to give the Higgs boson a thumbs down. If there is a signal on the positive side, the machine will be closed down and given an upgrade, focusing the efforts on finding relevant new information about the Higgs.

Let’s reflect on the gauge invariance of the EM Lagrangian. That sounds so fancy only experts can understand it. With quaternions, the subject is approachable. Take the 4-derivative of a 4-potential:
$\nabla \times A = (\frac{\partial}{\partial t}, c \vec{\nabla})\times(\phi, \vec{A})$
$= (\frac{\partial \phi}{\partial t} - c \nabla \cdot A, \frac{\partial \vec{A}}{\partial t} + c \vec{\nabla} \phi + c \vec{\nabla} \times \vec{A})$
$=(g, -\vec{E} + \vec{B}) \quad eq.\,1$
Note: the letter g is defined as a gauge field.
This is the simplest derivative of a quaternion potential one could construct. It contains two of the most important fields in physics, the electric and magnetic fields in a quaternion 3-vector that has the Lorentz transformation properties of an antisymmetric rank 2 tensor. An accident? I don’t think so.

What is the quaternion scalar g? It is not a Lorentz invariant scalar which requires a plus sign to be the Lorenz gauge.

[sidebar: Notice the different spelling between the invariant scalar name and the gauge? The Danish physicist Ludvig Lorenz was not the Dutch physicist Hendrik Lorentz. A silly American wrote an influential textbook with the error years ago. Most don’t know about this issue. Some do know, and go with the cultural flow. Out here on the ultra conservative fringe, we don’t care about popularity. We only hope to become more right, having ample examples of earlier errors. Lessons learned should be used consistently.]

The quaternion scalar can be found along the diagonal of the rank 2 asymmetric tensor Du Av. The E and B fields make up the off-diagonal terms. To make the quaternion 4-derivative of a 4-potential invariant under a gauge transformation, subtract these terms away:
$\nabla \times A - (\nabla \times A)^* = (0, -\vec{E} + \vec{B}) \quad eq.\,2$
Choose whatever combo-meal of the time derivative of phi or the divergence of A you wish. It doesn’t matter because they are subtracted away. The path from this quaternion 3-vector expression to the quaternion scalar that is a Lorentz scalar for the Lagrangian will be left for the snarky puzzle.

We have experimental data that says the changes caused by gravity induce changes in a symmetric metric. Symmetric changes are symmetric. Quaternions cannot do the job. I am exploring using an algebra that is even less popular than quaternions to do the job. The rules of multiplication are identical except that all the minus signs get erased. That much is simple. A long running discussion with David Halliday on my “Time in Bed with Space” blog has refined what I can say about these numbers. For this blog, let’s call them California quaternions to reflect the all-positive nature of their products.

Unlike quaternions, the California quaternions commute for multiplication as well as addition. There are more ways to make zero in California, so this is not a division algebra. The multiplication part is a remake of the Klein 4-group, extending them by including addition. At this time I do not fully appreciate the consequences of the California quaternions not being closed under addition and multiplication. A test to exclude those not wanted would be easy to implement and surprisingly similar to what must be done for Hamilton’s quaternions. Too bad California quaternions are not closed because then they would be a mathematical field like the real and complex numbers (but not Hamilton’s quaternions which do not commute).

Take the same 4-derivative and the same 4-potential, but use the California quaternion rules of multiplication:
$\nabla \boxtimes A^* = (\frac{\partial}{\partial t}, c \vec{\nabla})\boxtimes(\phi, -\vec{A})$
$= (\frac{\partial \phi}{\partial t} - c\nabla \cdot A, -\frac{\partial \vec{A}}{\partial t} + c \vec{\nabla} \phi + c \vec{\nabla} \otimes \vec{A})$
$\equiv (g, -\vec{e} + \vec{b}) \quad eq.\,3$
Notice that the quaternion scalar g is identical to the one seen before. Wipe it out the same way:
$\nabla \boxtimes A^* - (\nabla \boxtimes A^*)^* = (0,-\vec{e} + \vec{b}) \quad eq.\,4$
The quaternion 3-vectors small e and b are gauge invariant. I cannot prove to the skeptical reader that these fields have anything to do with gravity as that would take too long (I don’t know how to write in the margins of this web site). The same trick to go from a quaternion 3-vector to a Lorentz invariant scalar is used.

Is it conceivable that the small e and b fields are related to how gravity works? A sanity check is to look at the symmetric curl in b which only has positive signs, so looks like it could account for symmetric changes in a symmetric metric. The most practical approach to gravity is Newton’s pure scalar theory. A more refined and sublime theory is Einstein’s pure rank 2 theory of general relativity. Between rank 0 and rank 2 is a pure rank 1 hypothesis. That is where my work lives. I like having Newton on the ground floor and Einstein in the penthouse. On my level is Maxwell’s theory for EM which has been extended with two additional gauge groups to cover the weak and the strong forces. In a chess analogy, it feels like all my pieces cover the center of the board.

I will proceed with the hypothesis that the e and b fields are related to gravity. To have a mass for all four fields in EM and gravity - E, B, e and b - the fields must be gauge dependent. That has already been done in equations 1 and 3. What goes into the Lagrangian is a Lorentz scalar. That is formulated by forming the product of two of these field strengths. What will show up as far as the gauge is concerned is g2. To make the term going into the Lagrangian invariant under a gauge transformation, subtract one g2 from the other:

\begin{align*} \mathcal{L}_{GEM} &= Scalar(J \boxtimes A^* - J \times A\\ &+ \frac{1}{4}((\nabla^* \boxtimes A)^* \boxtimes (A^* \boxtimes \nabla) - (\nabla \times A) \times (A \times \nabla)))\\ &= b^2 - e^2 + B^2 - E^2\quad eq.\,5 \end{align*}

Notice that no current density terms show up in the Lagrangian. One cannot unify gravity where like charges attract and EM where like charges repel, and expect to have a unified charge that knows how to both attract and repel. What is more appealing is this Lagrangian will apply to a vacuum state, and thus resonate with all the work done on vacuum states for the Higgs. To be honest, this is not a vacuum state. For that to be the case, the quaternion scalar and 3-vector would both have to be zero. While the quaternion scalar is zero, the quaternion 3-vector is not. Much work would be required to putting that information to good use.

Separately the field strengths all depend on gauge. Because the gauges exactly cancel, the Lagrangian overall is invariant under a gauge transformation, good news for photons, gluons, and now gravitons. If you want to see the field equations that result by applying Euler-Lagrange, they are on the front of the t-shirt.

Most people who enter the World Series of Poker don’t win. A fellow I played poker with for serveral years, Rob Varkonyi, did win $2 million in 2002. Time will tell if my all in bet against the Higgs wins the$10 billion jackpot the work at the LHC represents.

Doug

Snarky puzzle.
Factor B2 - E2. That should remind you of problems back in high school, maybe junior high, maybe first grade if you are Gauss II. Earlier in this post, it was shown how to make a gauge invariant combo of -E + B. When the order of operators flip, only terms with a cross product or curl change signs. Do that here. Crash the two gauge invariant quaternion 3-vectors to make a quaternion and Lorentz invariant scalar. Poynt to the meaning of the resulting quaternion 3-vector.
Next Late Monday: The Sexy Standard Model Symmetries

Doug,

There are many Higgsless models nowadays yet virtually noone is seriously paying attention, except maybe for Technicolor. Same applies to BSM models. Perhaps things will turn around by the end of 2012 (hopefully not the doomsday...) but, for now, theoretical HEP appears to have fallen in a state of utter confusion.

Ervin Goldfain

Ervin:

Glad to see my sense of the state of the art is reasonable. I camp out at the edge of what I know, so live a life filled with confusion. I won't say I have a great alternative unless I find a reason for the -1/3 e, +2/3 e charge seen with quarks.

Thanks for the encouragement. Mostly I need to do more work. Luck tends to follow new calculations.
Doug
Doug,

Maybe some non-conventional efforts centered around nonlinear dynamics and fractal topology in BSM models might be of interest to you:

http://vixra.org/pdf/1104.0038v2.pdf

http://vixra.org/pdf/1011.0061v3.pdf

Take care,

Ervin

Doug,

A quick follow-up note: in the upcoming couple of weeks I plan on posting a derivation of the charge quantization condition using fractal description of Euclidean space-time in the asymptotic limit. The derivation is not based on magnetic or Yang-Mills monopoles.

Regards,

Ervin

So does this make the physics of that surfer dude guy more likely?

The surfer dude, Garrett Lisi, is my competition. He works with something known as the Monster group E8. As the name suggests, the Monster group is scary because it is so big. Lisi has professional training, but an unconventional home economics. His work has gotten both global and peer attention. Lisi knows his Monster, and how to make many connections to the details of the standard model.

Here is my own connection to Lisi. I noticed his work was funded by an unusual organization. I decided to pitch his funding group with my work. They were such an unusual group, someone actually called me back. We had a fun discussion, lasted a few hours. The guy on the other end of the line asked what he could do to help me. My work needs professional validation. I thought some sort of open line of communication with Lisi would be a good step forward. The funder was not willing to just hand over Garrett's cell phone number. I like unusual solutions, so I said I would tailor a pitch to Lisi and put it up on YouTube. That video now has over 30k of downloads.

One of the people who saw the video was Garrett Lisi. He also said he spent some time clicking around my web sites. He concluded that I had done nothing new. That was at least part of the email send to the funder. Garrett and I did not trade emails directly.

I could go about quantifying that the claim I have done nothing new is inaccurate. That would be a waste of my time. I think what it really means is that I did not effectively communicate with the man. If one fails to communicate the technical content, then there is no there there. Seeing nothing is equivalent to seeing nothing new.

Lisi showed emotional maturity. He noted that my quest would probably be lonely. That is a common plight for fringe physicists. I think the web has allowed me to avoid that situation. I know I have fans on every continent due to this work (very sparsely distributed). I am working with a grad student in physics, and hope to see him at the 14th Easter Gravity Meeting at Princeton if those people would organize it already. I also have material goods like the t-shirts. I know there are about 50 members of the MIT community with the No Stinkin' Higgs t-shirts wandering around the campus. And now Science2.0 is beginning to feel like a community to me.

Given the complexity of the moster, and the simplicity I work with, it is not surprising that neither of us think the other's research program will stand the test of time. All we can do is wish each other luck so the better hypothesis can win.

Doug
" I know there are about 50 members of the MIT community with the No Stinkin' Higgs t-shirts wandering around the campus"

You would think (hope) the Course VIII guys would not call it the "Higgs" given Hagen and Guralnik were MIT alums (see below).

http://www.technologyreview.com/article/24610/

Doug:

You say
Unlike quaternions, the California quaternions commute for multiplication as well as addition. There are more ways to make zero in California, so this is not a division algebra. The multiplication part is a remake of the Klein 4-group, extending them by including addition. At this time I do not fully appreciate the consequences of the California quaternions not being closed under addition and multiplication. A test to exclude those not wanted would be easy to implement and surprisingly similar to what must be done for Hamilton’s quaternions. Too bad California quaternions are not closed because then they would be a mathematical field like the real and complex numbers (but not Hamilton’s quaternions which do not commute).
Forgetting that I had suggested (and I thought you had agreed—heh) that you call the entire system, with both multiplication operators, the "California quaternions", while calling the potion with just the commuting form of multiplication the "California numbers" (what's in a name, anyway  :)  ); I did think I had helped you see that the distinction between these numbers and those of a division algebra (AKA division ring—when approached from another perspective) is not about "closure" or about "There [being] more ways to make zero".
1. There are exactly as many "ways to make zero" in these numbers as there are in the quaternions.
2. Since for any system with both operations of addition and multiplication (with addition forming an abelian group of the system, and multiplication distributing over addition in the usual way, etc.) in which 0≠1 (zero, the additive identity, is not the same as one, the multiplicative identity), the minimal set of elements without multiplicative inverses is the set consisting of the single element 0 (the additive identity); the distinction between a division algebra and these numbers is the number of elements without multiplicative inverses.
3. So long as these numbers contain all numbers that can be formed by application of the allowed operations (addition, multiplication, and whatever else you wish to throw in, like conjugations) to any and all other such numbers, then your numbers are closed (under such operations).  Otherwise, they are not.  So the only place where any question of closure comes into play is if you wish to try and exclude all non-zero elements that do not have multiplicative inverses (in an attempt to obtain a division algebra, perhaps).  (Note:  To then obtain closure, one needs to then eliminate all elements that can be combined to form such non-invertible elements, and all those that can be combined to form any of the newly eliminated set, ad nauseam.)
Does that help clarify?

I'm sorry if I wasn't clear then, or if I'm not clear now.  However, please let me know if I'm still not helping you understand the issue.

David
David:

This does help.
I have not embraced the union of California numbers with Hamilton's quaternions to make California quaternions because then I would have to have another sign, other than the \times X and the \boxtimes. to symbolize they are in use. I don't think unification here would add clarity. It is easier to point at those parts that use the Californian number rules of multiplication, and those that use Hamilton's rules of multiplication.

When I use the California numbers, I do mean that the eigenvalues of the matrix representation that are zero should be excluded. If I don't do that, then I should say I am using the Klein 4-group for multiplication plus addition, since much work has been done with the Klein 4-group.

Conjugation is required for the California numbers. That is the only way to make a Lorentz invariant in the scalar selection step for the Lagrangian.
So the only place where any question of closure comes into play is if you wish to try and exclude all non-zero elements that do not have multiplicative inverses (in an attempt to obtain a division algebra, perhaps).  (Note:  To then obtain closure, one needs to then eliminate all elements that can be combined to form such non-invertible elements, and all those that can be combined to form any of the newly eliminated set, ad nauseam.)
This sounds like it would leave nothing left, termites that eat all the numbers.  Just be be pedantic.
1 has an inverse.
i has an inverse
1 + i does not have an inverse
If I understand correctly, I would need to eliminate 1 and i to get closure. That doesn't sound good. I think I have to accept that I don't have closure for the California numbers (addition&Klein 4-group multiplication & conjugates - eingenvalues = 0). It is all the consequences of these not being closed I don't understand. One consequence is clear: by these rules, I cannot call California numbers a division algebra.

Doug

Doug:

Yes, you are right about the "termites that eat all the numbers."  :D  (However, perhaps not all the numbers need be eliminated, but you will be left with subsets of your numbers that will be division algebras [fields, actually, by the nature of the operators].)

Now, you say that "When I use the California numbers, I do mean that the eigenvalues of the matrix representation that are zero should be excluded."  Of course, I believe you mean to say that you exclude all non-zero non-invertible (singular) numbers (all the non-zero numbers that have zero eigenvalues in the matrix representation).  Correct?

So, yes, if you only do this, you don't have closure, because you will, at times, generate such excluded numbers using the allowed operations.  However, I'm wondering why you think you must eliminate these non-zero non-invertible (singular) numbers?  You have said that you don't use division in your work with these numbers.  Right?

You then proceed to assert that "If I don't do that [excluding non-zero singular elements], then I should say I am using the Klein 4-group for multiplication plus addition, since much work has been done with the Klein 4-group."  This most certainly doesn't follow, since:
1. The Klein 4-group consists only of the four elements you label as e, i, j, k.
2. The Klein 4-group is a group, so it has only a single binary operation.
3. The single binary operation of the Klein 4-group can be labelled "addition" or "multiplication", since it is commutative.
4. However, this single binary operation cannot be labelled as both, of course.  ;)
5. The operation of addition you are using has nothing truly to do with the Klein 4-group.  (It simply treats the elements of the Klein 4-group as basis vectors of a vector space over the real number field.)
6. The operation of multiplication you are using does depend upon the binary operation of the Klein 4-group, but also upon operations that are completely outside the Klein 4-group.
No, your California numbers (or what you called "California quaternions" within the article) are an algebra based upon the real numbers and the Klein 4-group.  (More or less technically, it is an algebra over the real number field, where the operation of multiplication treats the basis elements {e,i,j,k} as elements of the Klein 4-group, where we map the binary Klein 4-group operator into the multiplication operator of the algebra.)

So, again, why do you feel you must exclude the non-zero singular elements of this algebra?

David

P.S.  This algebra is a Ring, when one doesn't treat it so much as an algebra over a field (the field being the real number field), but as a closed entity in and of itself (containing the real number field as a proper subfield).  The distinction is
1. Algebra:  We focus on the way this system is constructed.
2. Ring:  We focus on the system as an entity of its own, and focus on what occurs within it.
Personally, it appears to me that your focus would rather be #2. But that's just my opinion.  :)
Doug:

You say:
I have not embraced the union of California numbers with Hamilton's quaternions to make California quaternions because then I would have to have another sign, other than the \times X and the \boxtimes. to symbolize they are in use. I don't think unification here would add clarity. It is easier to point at those parts that use the Californian number rules of multiplication, and those that use Hamilton's rules of multiplication.
I most certainly agree that "It is easier to point at those parts that use the Californian number rules of multiplication, and those that use Hamilton's rules of multiplication."  However:
• What you have within your article (such as the GEM Lagrangian) do utilize "the union of California numbers with Hamilton's quaternions", where you have used your "\times X and the \boxtimes. to symbolize" the respective multiplication operations, when "they are in use".
So why do you think you would need "to have another sign"?  (Unfortunately, I do see that sometimes you use "\times X" ("×"), and other times you use the \Otimes ⊗ operator.  [OOPS:  I think what I was seeing was the accidental use of the \Otimes ⊗ operator where it was meant to be the "\boxtimes".  :(  ])

David
David:
Now, you say that "When I use the California numbers, I do mean that the eigenvalues of the matrix representation that are zero should be excluded."  Of course, I believe you mean to say that you exclude all non-zero non-invertible (singular) numbers (all the non-zero numbers that have zero eigenvalues in the matrix representation).  Correct?
Correct.
However, I'm wondering why you think you must eliminate these non-zero non-invertible (singular) numbers?  You have said that you don't use division in your work with these numbers.  Right?
When I do math, I think of physics. When doing physics, I think of math. Since this is math, I am thinking of physics. The only reason I started working with these tricky-for-me-to-characterize numbers thinking about field equations for gravity. Many of these non-zero non-invertible (singular) numbers live on the light cone (like (1, 1, 0, 0). The path of light is changed by gravity. One place where I differ from GR is that GR predicts the energy of a bunch of photons would cause a bending of spacetime. My work predicts the zero rest mass of photons would not bend spacetime. We have more than enough data showing that path of light is changed by gravity. I don't think we have any data that says light can be a source for bending spacetime. If you accept the premises underlying GR, then to be consistent, the energy of light must bend spacetime.
This most certainly doesn't follow, since:
Darn it, I thought I was quoting you on that one. You said somewhere about extending the Klein 4-group by including another operation, addition.

Speaking correctly in math is obviously tough for me. It is like speaking in Chinese without hearing any of the inflections. Oops:
(More or less technically, it is an algebra over the real number field, where the operation of multiplication treats the basis elements {e,i,j,k} as elements of the Klein 4-group, where we map the binary Klein 4-group operator into the multiplication operator of the algebra.)
"I am using the Klein 4-group for multiplication" to my untrained ear sounds like it carries this intent.

There is another reason in physics to want to do reverse multiplication. If one has field equations, then inverting those field equations generates the propagator. I fantasize about having field equations that are invariant under a gauge transformation that are invertible without picking a gauge because multiplication is invertible.

I can live without it, but I plan ahead.
Doug
Doug:

You are essentially correct that "The path of light is changed by gravity."  However, light still travels along null geodesics ("on the light cone", at each point along its path).  It's only the curving/warping of spacetime (space and time together) that causes our perception of "The path of light is changed by gravity."

Now, you say:
One place where I differ from GR is that GR predicts the energy of a bunch of photons would cause a bending of spacetime. My work predicts the zero rest mass of photons would not bend spacetime. We have more than enough data showing that path of light is changed by gravity. I don't think we have any data that says light can be a source for bending spacetime. If you accept the premises underlying GR, then to be consistent, the energy of light must bend spacetime.
I have no qualms with you wishing to "differ from GR" (with the ever present caveat, of course, that the ultimate test is comparison against observation/experiment).  You are correct that "We have more than enough data showing that path of light is changed by gravity."  You are also correct that we have precious little, if "any data that says light can be a source for bending spacetime."  (The coupling is so small that we would have to generate an extremely large EM field [in human terms] in order to generate any appreciable "bending" of spacetime.  That's an awfully huge number of photons to have in one place at one time.  ;)  )

You are also correct that "If you [or we] accept the premises underlying GR, then to be consistent, the energy of light must bend spacetime."  Of course, if we accept the premises of Quantum Chromodynamics (QCD), and the experimental evidence, then the primary contribution to the mass of the protons and neutrons are from massless gluons, whipping around inside.  ;)  (There's certainly the virtual, massive particles that "pop in and out" of existence through this quark field, of course.  Oh, and this is to say nothing of the EM contribution to the mass difference between neutrons and protons.)  Seems like an interesting connection, at least to me.  ;)

David

P.S.  The problem, of course, is that we do not have a good candidate for a unified view of all this, at this time.  (I know, that's what you, and so many others are working on.  ;)  )
Doug:

By the way, since, as you say, "Many of these non-zero non-invertible (singular) numbers live on the light cone (like (1, 1, 0, 0)[)]", to me, that's all the more reason not to exclude such elements:  The light cone elements are very important to EM, and the interactions mediated by such.

David
Doug:

To my parenthetical comment,
(More or less technically, it is an algebra over the real number field, where the operation of multiplication treats the basis elements {e,i,j,k} as elements of the Klein 4-group, where we map the binary Klein 4-group operator into the multiplication operator of the algebra.)
you respond with
"I am using the Klein 4-group for multiplication" to my untrained ear sounds like it carries this intent.
I suppose I can see how that can be so.  :)  However, I specifically used "map the binary Klein 4-group operator into the multiplication operator of the algebra" (emphasis added) to distinguish what I was saying, "More or less technically" (in other words, not being as verbose as I would need to be in order to be completely technically correct), from any statement like "using the Klein 4-group for multiplication" (again, emphasis added).

However, you are correct that you are "using the Klein 4-group" in the definition of the "multiplication", for your algebra/ring.

In any case, what you are doing with your algebra/ring is a far cry from simply "using the Klein 4-group for multiplication plus addition".

By the way, it would seem to me that someone must surely have looked at this algebra/ring before.  So it would seem to be reasonable that it would already have a name.  However, until someone steps up to declare a preexisting name for this algebra/ring, I see no harm in giving it your own name.

David
Doug:

A (more or less) technically correct name for this algebra would be:  A Klein 4-group based algebra over the real field (or some such).  However, I like California numbers, or California ring, or California algebra much better.  (Personally, I think California ring has a nice ring to it.  ;)  Besides, as I mentioned above, I would think you would want to emphasize it as an entity unto itself [ring], rather than emphasizing its construction [algebra].)

David
Doug:

There is another reason in physics to want to do reverse multiplication. If one has field equations, then inverting those field equations generates the propagator. I fantasize about having field equations that are invariant under a gauge transformation that are invertible without picking a gauge because multiplication is invertible.
While it would certainly be niceTM to always have divisibility, unless you have a division ring/algebra, simply excluding the non-zero singular elements isn't going to buy you anything—not one single iota.  Whether you "exclude" such elements or not, you will still—formally—take your inverses, but you will still have the potential for particular values to not exist (or be formally infinite, if you don't exclude such).

So I simply don't see that excluding such elements buys you anything.  I think you'll either have to look for a division ring/algebra that will do your bidding (or dirty work), or simply "live without it".  (I most certainly don't recommend merely excluding such elements, regardless of whatever false sense of security you may glean thereby.)

David
David:

You sure got me thinking today (OK, I was suppose to be working, but this is more fun :-) I do a little quote and response:
I have no qualms with you wishing to "differ from GR" (with the ever present caveat, of course, that the ultimate test is comparison against observation/experiment).
One prediction I have is that the Higgs boson will not be found. That one is sure to get done, and we will read about it here. I am null hypothesis man, so I am predicting there is no dark matter, no dark energy, and no source for inflation. We will continue to collect data that says we have problems describing the motion of large systems, that the galaxies are going faster today than we expect them to, and something wild happened at the big bang. Even though I lack the skill and resources to apply the math found in the relativistic rocket effect blog, I am willing to go all in. To clarify, if they announce here that dark matter has been found, and it fits in with our understanding of how particles work so well the community accepts it as true, then I will retire my GEM unification. A similar bet is in place for dark energy and inflation which are separate problems that are under active investigations. Another prediction is second order parameterized post-Newtonian weak gravity effects, like light bending around the Sun. The last time I ask Clifford Will about it (2006?) he said there were no plans in the works. That is 4 nulls and a not-being done for the experimental side.That is 4 nulls and a not-being done for the experimental side.
So why do you think you would need "to have another sign"?  (Unfortunately, I do see that sometimes you use "\times X" ("×"), and other times you use the \Otimes ⊗ operator.  [OOPS:  I think what I was seeing was the accidental use of the \Otimes ⊗ operator where it was meant to be the "\boxtimes".  :(  ])
I use the \otimes for 2 3-vectors that have a symmetric curl between them. I use a \boxtimes between 2 4-vectors that get the full California number treatment. Perhaps I have not hung out with enough odd algebras, but I am unfamiliar with working with two multiplication operators in the same expression.
P.S.  The problem, of course, is that we do not have a good candidate for a unified view of all this, at this time.  (I know, that's what you, and so many others are working on.  ;)  )
I believe this is classified as a difficult problem, so I will blame some of my floundering on that. Another huge complicating factor is I am trying to make things smaller instead of larger, two quaternions instead of more dimensions, super symmetry, E8, and others.
By the way, since, as you say, "Many of these non-zero non-invertible (singular) numbers live on the light cone (like (1, 1, 0, 0)[)]", to me, that's all the more reason not to exclude such elements:  The light cone elements are very important to EM, and the interactions mediated by such.
The light cone is handled well by the Maxwell equations and quaternions. Light is absorbed and emitted by particles with mass constantly. Particles with mass only get to be light via anti-particle annihilation.
By the way, it would seem to me that someone must surely have looked at this algebra/ring before.
My thought too. I found hypercomplex numbers had 4 numbers and did commute. Yet the ones being used by anyone alive I could find were by Clyde Davenport, also working out of his own place. He uses i2=j2=-k2. I was not a fan of hypercomplex because I didn't see anything hyper or complex about them. I used the phrase "California representation of hypercomplex numbers", but that is too bulky. Someone playing around with the Klein 4-group must have wanderer around this direction. Until then, I'll use California numbers as a label. I haven't taken the modern abstract algebra course yet to appreciate the difference between a ring (how it is constructed) and a ring (the thing as a separate entity).

Doug

Doug:

I'm not quite sure what you mean by your "prediction":
... Another prediction is second order parameterized post-Newtonian weak gravity effects, like light bending around the Sun.
First, the Parameterized Post-Newtonian (PPN) formalism for weak gravity effects is already at second order (missing only terms of third order and beyond).  Second, the second order PPN does include light bending around the Sun type effects.  Third, there have been tests that target the appropriate PPN parameter, gamma.  (See, for instance, Clifford M. Will's living review entitled "The Confrontation between General Relativity and Experiment".)

David
David:

Here is my understanding of how to translate from metrics to PPN. There is zeroth order PPN, or Newton's theory. As a metric equation it would be:

d tau2 = (1 - 2 GM/c2 R) dt2 - dR2/c2

The first order PPN reaches these terms:

d tau2 = (1 - 2 GM/c2 R + 2 (GM/c2 R)2) dt2 - (1 + 2 GM/c2 R) dR2/c

where I only show the values that show up for the Schwarzshild metric of GR. I believe there are 10 values determined by ones proposal for gravity. In GR, gamma and beta are equal to unity, the rest are zero. The same is true for my proposal. What has not been tested is second order PPN terms. For GR, the next terms in a Taylor series expansion are:

GR: d tau2 = (1 - 2 GM/c2 R + 2 (GM/c2 R)- 3/2 (GM/c2 R)3) dt2 - (1 + 2 GM/c2 R+ 3/2 (GM/c2 R)2) dR2/c

My proposal is different at this level.  It is more symmetric, often a good thing:

GEM: d tau2 = (1 - 2 GM/c2 R + 2 (GM/c2 R)- 4/3 (GM/c2 R)3) dt2 - (1 + 2 GM/c2 R+ 2 (GM/c2 R)2) dR2/c

This paper is germane to the topic at hand: R, Epstein and I. I. Shapiro. Post-post-Newtonian deflection of light by the Sun. 22:12 2947--2949, 1980.

Doug
Doug:

Your use of the term "order" is not entirely standard, but understandable.  The "standard" usage has the Minkowski metric as zeroth order (no dependence upon the distribution of matter).  Newton's, then, is at first order, while the standard PPN is at second order.

So, what you are talking about is at the next higher order than the standard PPN.  Therefore, you are talking third order.

A simple difference in terminology, but one that can get in the way of good communication.  (Incidentally, I would need to verify the third order terms for GR.)

By the way, when you say your "proposal" for the metric, am I to suppose you are referring to "the Rosen or exponential metric"?  (Wouldn't that actually make it Rosen's "proposal"?)

David
David:

Looking at the article, they use Post-Post Newtonian (ppN) contributions. I have also seen deflection to second order. The most common phrase is parameterized post-Newtonian formalism (PPN). I think I made up second order PPN. It is not really a parameterization - there are not a whole bunch of new Greek letters to fill in.

If I said the Rosen metric, would that imply the Rosen bimetric theory with a fix background? The metrics happen to be the same, but I don't want to imply I use his field equations to get there. The GEM field equations have an exponential metric solution that is identical to that from Rosen's work.

Doug
Doug:

When you say "Looking at the article", are you referring to "R, Epstein and I. I. Shapiro. Post-post-Newtonian deflection of light by the Sun. 22:12 2947--2949, 1980"?

By the way, you may also like to check out the post-post-Newtonian effect formulated within Will, C.M., Theory and experiment in gravitational physics, (Cambridge University Press, Cambridge, U.K.; New York, U.S.A., 1993), 2nd edition.

Of course, such effects are almost certainly very near the experimental noise of present measurements.

If I said the Rosen metric, would that imply the Rosen bimetric theory with a fix background? The metrics happen to be the same, but I don't want to imply I use his field equations to get there. The GEM field equations have an exponential metric solution that is identical to that from Rosen's work.
Well, saying you are using Rosen's metric does not, necessarily, imply that you are using or ascribing to Rosen's bimetric theory (with its fixed background).*  However, in your case, even though the Rosen metric is a solution (almost everywhere) to your vacuum equations, you have yet to show that it is the only solution, or even the "best" solution.

David

*  Actually, you do have a fixed background metric, in your choice of quaternion basis, since, as you have actually shown (by the various conflicting transformational propertied of many of your quaternion equations), your quaternions do not transform strictly according to the emergent metric of which you are referring.

Additionally, since your GEM system (or even strictly your "gravitational" portion) depends upon the California ring/numbers, which have been shown to have preferential directions that don't adhere to the transformational properties of your metric, we once again see the earmarks of a background.
Doug:

... Perhaps I have not hung out with enough odd algebras, but I am unfamiliar with working with two multiplication operators in the same expression.
I can't say I know of many "odd algebras" that have two (or more) multiplication operators (except Clifford/Geometric Algebras*).  However, as long as you can make the concepts clear, and devise a workable notation, why not?

Admittedly, on the other hand, I prefer a simpler situation wherein my "numbers" don't have to have any more than the two usual binary operations (addition and multiplication).

David

*  Strictly speaking, Geometric Algebra has but a single, non-commutative, multiplication operation.  However, one can, in general, split a non-commutative multiplication, within an algebra, into symmetric (commutative) and antisymmetric (anti-commutative) parts.  In the case of Geometric Algebra, this is then used to make the connections with Clifford Algebra.
Doug:

While E8 is wonderful and beautiful, and I think it would be wonderful if it were fundamentally connected with Nature and the Universe; it also appears to be far too big for the number of fundamental particles we seem to observe (E8 would require no less than 240 fundamental particles, while we have no more than 96 Fermions, and 12 Bosons, plus however many total particles may be under the Higgs "umbrella", and the "graviton"?).

Unfortunately, perhaps, E8 appears to be far more than necessary.

David
David:

My fun driving home speculation...In classical physics, dt >>> (dx+dy+dz)/c. Under that condition, there will always be an inverse. Is it OK to say I have a mathematical field under that constraint? The California numbers commute, and none of its eigenvalues will be zero if changes in time dominate changes in space.
Doug

Doug:

You speculate:
My fun driving home speculation...In classical physics, dt >>> (dx+dy+dz)/c. Under that condition, there will always be an inverse. Is it OK to say I have a mathematical field under that constraint? The California numbers commute, and none of its eigenvalues will be zero if changes in time dominate changes in space.
Ahhh.  So I'm not the only one that mulls over scientific and/or mathematical speculation during my commute.  :)

Unfortunately, even "under that constraint" ("dt >>> (dx+dy+dz)/c"), your California numbers do not form "a mathematical field".  The problem is that simply applying this constraint to your California numbers violates closure.  OOPS, no "mathematical field".

On the other hand, you can say that under such conditions, computations with your California numbers (that is, within your California ring) will be invertible, for the reasons you have stated.  Again, no reason to exclude non-zero singular values, they are simply not encountered (at least much less likely to be) under such "classical" conditions.

David
Excellent. My way sound like cheating, yours like it is within the rules.
Doug:

I've taken another look at your set of eigenvalues to avoid (rewriting them in terms of dt, dx, dy, and dz, then dividing by dt, written using velocity components, and letting c ≠ 1):
c = -vx + vy + vz
c = vx - vy + vz
c = vx + vy - vz
c = -vx - vy - vz
The set of such eigenvalues to avoid (constraints) on the light cone
vx2 + vy2 + vz2 = c2
is a set of order zero (meaning that if one picks a random value from the set of points that satisfy any of the four constraints [eigenvalues to avoid], the likelihood that you will be on the light cone is zero).

The fact is that each one of your set of eigenvalues to avoid (constraints) is a plane, in the three space of velocities.  Each of these planes does cut through the sphere of the light cone, and extends far beyond (points faster than the speed of light).  The intersections with the light cone are circles, on that sphere.  The points to be concerned about are the set of points with velocities less than (or equal to) the speed of light.

As it turns out, the minimal speed to have to worry about, for each of the eigenvalues to avoid (constraints) is c/sqrt(3).  (Actually, there is an enclosed volume, around the origin, that is tetrahedral like.  I think it is a tetrahedron, but I haven't analyzed it that closely.)

David
Doug:

The tetrahedral like volume is exactly a tetrahedron.  If you take a cube, centered on the origin, with edges of length 2c, aligned with the coordinate axes; take the edges of the inscribed tetrahedron to be along diagonals of the square faces of the cube.

There are, of course, two ways to inscribe a tetrahedron inside a cube in this way.  In this case, choose the orientation that puts one vertex at vx = vy = vz = c.

David
Doug:

I actually have a nice 3D model of this tetrahedron and sphere inscribed within a 2c×2c×2c cube:

Unfortunately, a single 2D perspective projection doesn't do it justice.  However, I hope it will be sufficient.  (The flat planes, clipped at the boundaries of the cube, in this image, are the four "constraints" [eigenvalues to avoid], talked about above.  They are the surfaces of the non-zero singular values of the California numbers/ring.)

However, this brings up a troubling issue with your use of these California numbers (or the California ring):  They do not respect the spacetime symmetries we have observed, namely, in this instance, isotropy.  These "special" values (singular, non-zero values) exhibit preferential (special) directions!

I most definitely don't see that being of any help to you.  :(

David
David:

The California products always have a conjugate. For example, there is the scalar of the current coupling term, where if I take the Fourier transform of the potential, becomes a current-current interaction. That looks like so:
$scalar(J \boxtimes J'^*) = \rho \rho' - J_x J'_x - J_y J'_y - J_z J'_z$

This is my central use of California numbers, and is isotrophic, no? Looks invariant under a Lorentz transformation to me.

Doug
Doug:

Of course I'm not saying that you (or anyone) cannot create equations using the California ring (these California numbers) to create expressions that have almost whatever invariance properties you (or anyone) may choose.  I'm just pointing out that they exhibit characteristics that are antithetical to such purposes.  (As I said:  "I most definitely don't see that being of any help to you.")

David
Doug:

The fact that you have an expression, using the California ring/numbers that produces an equation that exhibits Lorentz Invariance, even though the underlying system (the California ring) does not adhere to such, simply reminds me of those that use ict, in the time slot of Euclidean 4-vectors, with a Euclidean metric (the identity), in order to do Special Relativity while making it look like they are doing Euclidean Geometry.  (Or, perhaps, they are trying to make Special Relativity look more like Euclidean Geometry.)

David
David:

Quaternion squares like Lorentz invariance in their scalar, but with an additional 3-vector (I'll use the times to indicate I am using quaternion rules of multiplication):
$(dt, \vec{dR}/c) \times (dt, \vec{dR}/c)=(dt^2-dR^2/c^2, 2 \,dt \,\vec{dR}/c)$

If one wants to calculate the norm, a mirror is required:
$(dt, \vec{dR}/c)^* \times (dt, \vec{dR}/c)=(dt^2+dR^2/c^2, \vec{0})$

Notice I include the zero 3-vector. Why? If one needs to feed the norm into an animation, the location in 3-space must be defined. The animation software won't work with those zeroes omitted.

Now go to California numbers. The story flips. A norm with a non-zero 3-vector is generated from the square:
$(dt, d \vec{R}/c) \boxtimes (dt, d \vec{R}/c) = (dt^2 + dR^2/c^2, 2 \,dt \,d\vec{R}/c+d\vec{R}\otimes d\vec{R}/c^2)$
The Lorentz invariant needs a mirror:
$(dt, d \vec{R}/c)^* \boxtimes (dt, d \vec{R}/c) = (dt^2 - dR^2/c^2, -\, d\vec{R}\otimes d\vec{R}/c^2)$
Because both a Lorentz invariant interval and the norm are both vital for calculations in physics, I avoid taking about an underlying system. I need both when working with quaternions. I need both with California numbers. I need to use a spatial mirror sometimes, sometimes not.

The biggest problem in tensor calculus accounting is the way it is set up to ignore these well-defined 3-vectors, focusing excessively on the scalar part. It is this detail that makes the approach different than the ict crowd.

Doug
Doug:

Of course I know that the details "make[ your] approach different than the ict crowd."  My point was and is that using a system that doesn't match the physics, while it can still be made to "work", doesn't help, and can hurt.

By the way, "tensor calculus accounting" is not "set up to ignore these well-defined 3-vectors, focusing excessively on the scalar part."  Tensor calculus within 3D Euclidean space absolutely focusses on such (scalar and 3-vector) things.  (This is the usual vector calculus most learn early on.)  While, on the other hand, higher dimensional tensor calculus (such as the 4D tensor calculus of General Relativity) focusses well beyond "the scalar part", going all the way from scalars, 4-vectors, 2-forms (like the 2-form of the Electromagnetic field, containing both electrical and magnetic fields), and beyond.

By the way, I have never seen cases where the 4D Euclidean norm is "vital for calculations in physics".  Furthermore, so long as you are going to have curved spacetime metrics, even the Rosen exponential metric, the retention of either the 4D Euclidean norm, or the Minkowski 4D invariant ([pseudo-]norm) becomes a background metric within your work.  (It's at least as troublesome as the background Minkowski metric within Rosen's theory.)

David
David:
By the way, "tensor calculus accounting" is not "set up to ignore these well-defined 3-vectors, focusing excessively on the scalar part."  Tensor calculus within 3D Euclidean space absolutely focusses on such (scalar and 3-vector) things.  (This is the usual vector calculus most learn early on.)
The start is good: here is the scalar and 3-vector together. The critique was focused on the next step, which is often the Einstein summation convention. It is that summation convention that implicitly ignores the 3-vectors written in my comment.
By the way, I have never seen cases where the 4D Euclidean norm is "vital for calculations in physics".
Certainly not in gravity work, I can see that. However, in quantum mechanics, I would differ. One is always calculating a* b. In the quaternion representation, that is the norm.

Let's see if I can explain my dream of life without metrics, but with dynamic basis vectors. I standing in the basement, my wife is a floor above me, the proverbial higher power. We both have lasers and atomic clocks to prove we know what the speed of light is, and unity (this must be a thought experiment). A third party, my daughter Elle, decides to compare the two (OK, this is just the Pound-Rebka experiment done at home, nothing more or less). She is a sharp 2 year old, so can detect the difference between the rulers and clocks of daddy in the basement and mommy as higher power.

Let us use the 4 basis vectors (e, i, j, k) for mom, and (e', i', j', k') for dad. Chose to let mom's basis be the start of all things, and all have exactly the magnitude of one. Dad is in the basement. As far as dad's own instruments are concerned, the magnitude of the basis vectors are also one. It is Elle who can spot the difference between the two. What she does is assign the following values to daddy's basis vectors:

|e'| = exp(-G M/c2 R)
|i'| = |j'| = |k'| = 1/|e'|

I cannot see this, I am in the basement. Mommy cannot see this, she is a higher power in her own world. Elle can spot the differences in several ways. For a pair of events in mommy's devices, she would calculate the following:
$(dt, \vec{dR}/c) \times (dt, \vec{dR}/c)=(dt^2-dR^2/c^2, 2 \,dt \,\vec{dR}/c)$

$(dt, \vec{dR}/c)^* \times (dt, \vec{dR}/c)=(dt^2+dR^2/c^2, \vec{0})$
$(dt, d \vec{R}/c) \boxtimes (dt, d \vec{R}/c) = (dt^2 + dR^2/c^2, 2 \,dt \,d\vec{R}/c+d\vec{R}\otimes d\vec{R}/c^2)$
$(dt, d \vec{R}/c)^* \boxtimes (dt, d \vec{R}/c) = (dt^2 - dR^2/c^2, -\, d\vec{R}\otimes d\vec{R}/c^2)$

This is the same data, dt, dR/c, plugged into 4 calculations. Notice that while there are only 2 distinct scalars, all four 3-vectors are different from each other. The Einstein summation convention will get the first scalar and ignore these other possibilities.

Now she does the same 4 calculations with the dynamic basis vectors for daddy's data:
$(e^{-G M/c^2 R} dt, e^{G M/c^2 R}d \vec{R}/c) \times (e^{-G M/c^2 R} dt, e^{G M/c^2 R}d \vec{R}/c) = (e^{-2G M/c^2 R} dt^2 - e^{2 G M/c^2 R}dR^2/c^2, 2\, d t \,d\vec{R}/c)$

$(e^{-G M/c^2 R} dt, e^{G M/c^2 R}d \vec{R}/c)^* \times (e^{-G M/c^2 R} dt, e^{G M/c^2 R}d \vec{R}/c) = (e^{-2G M/c^2 R} dt^2 + e^{2 G M/c^2 R}dR^2/c^2, \vec{0})$

$(e^{-G M/c^2 R} dt, e^{G M/c^2 R}d \vec{R}/c) \boxtimes (e^{-G M/c^2 R} dt, e^{G M/c^2 R}d \vec{R}/c) = (e^{-2G M/c^2 R} dt^2 + e^{2 G M/c^2 R}dR^2/c^2, 2 dt \, d\vec{R}/c + e^{2 G M/c^2 R} d\vec{R} \otimes d \vec{R}/c^2)$

$(e^{-G M/c^2 R} dt, e^{G M/c^2 R}d \vec{R}/c)^* \boxtimes (e^{-G M/c^2 R} dt, e^{G M/c^2 R}d \vec{R}/c) = (e^{-2G M/c^2 R} dt^2 - e^{2 G M/c^2 R}dR^2/c^2, - e^{2 G M/c^2 R} d\vec{R} \otimes d \vec{R}/c^2)$

Elle can tell that her parents are in curved spacetime because the scalars are different in a way consistent with current weak field tests of gravity. The quaternion 3-vectors are the same in flat or curved spacetime. The California 3-vector symmetric curls are different in curved spacetime by a knowable amount.

Since I have the rules of multiplication for quaternions and California numbers, dynamic values for the basis vectors looks sufficient to account for the observations in curved spacetime. This approach is different from that used by folks schooled in differential geometry because the changes it the time component times the changes in the 3-vector component go into the 3-vector slot. To my eye, that is logically more consistent than a metric which stuffs that value back in the the scalar.

Doug