Unifying Gravity and EM

In summary, the conversation discusses a proposal for a unified field theory that combines gravity and electromagnetism into a single rank 1 field. The Lagrange density for this proposal is provided, along with a discussion of how the equations are generated and the physical implications of the theory. The proposal is consistent with both weak and strong field tests of gravity and there are no known physical experiments that contradict it.
  • #596
SUSY and dark matter

Hello Vanick:

Supersymmetric (SUSY) proposals have been with us for over 30 years. During that time, every time a new more powerful collider has come on line, part of the justification was the search for the SUSY particles. The same is claimed for the Large Hadron Collider: it might find the first evidence for supersymmetric. Based on the previous decades of effort, I will not bet on it, although there is 8 billion dollars on the line.

As a hypothesis, supersymmetry has two failings in my eyes. The first is that one of the motivating factors behind the proposal is to explain aspects of the standard model. One big mystery is why there should be three groups, U(1), SU(2) and SU(3) - and why those groups in particular. In my reading of this article on grand unified theories, using a group like SO(10) does not answer that question. Oops.

Animating physics with quaternions does give a visual reason why Nature is constrained to the symmetries of U(1), SU(2), SU(3) and Diff(M) for EM, the weak force, the strong force, and gravity respectively. If you go though the math of an expanding and contracting unit sphere with quaternions, these are the groups that come into play. If an observer is at the origin and sees an event, that event must be a member of these groups. It is less that SUSY is wrong than I have a concrete counter proposal. I too have limitations in my math skills, and have no idea how to pitch a model that is smaller than the standard model to particle physicists. The smaller nature of the quaternion proposal might have an advantage: it could justify containment, the observation that there are no free gluons anywhere in the Universe. Formalizing that line of logic is beyond my reach. A search on YouTube for "standard model" should have my blue & yellow colored animation.

That is the backdrop. Now to your specific question about SUSY and dark matter. It is not enough for a light SUSY particle to only interact with gravity, to have the properties of a dark matter particle. That particle would have to be distributed in space around disk galaxies with a distribution that would lead to a momentum profile where the velocity was constant, but the visible mass falls off exponentially. How Nature using only gravity could get the distribution of dark matter such that it is consistent with what we see is a core issue that is unexplained. The distribution of dark matter must be different for clusters of galaxies. Getting non-interacting matter except for gravity into the right place looks like a tough problem.

Doug
 
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  • #597
Disk galaxy momentum profile - the drawing

Hello:

To use Newton's law of gravity, masses and distances must be known. For a disk, one cannot treat it as an effective point mass. Instead, the galaxy needs to be chopped up into little bits, and add up each contribution. My plan is to do this discretely, but it may be possible to do things continuously. I will have to see.

Here is a picture of how I plan to slice up a galaxy:

galaxy-600.jpg


There are quite a few R's:

  • Rmax - the maximum radial distance
  • Rpi - the radius to the passive mass, i steps from the origin
  • Raj - the radius to the active mass, j steps from the origin
  • Rij - the distance between the active and passive masses.
  • Rr - the portion of Rij in the direction of the radius
  • Rv - the portion of Rij perpendicular to the radius, parallel to V
The passive mass is the one that appears on both sides of Newton's force law, and always gets cancelled. With the relativistic rocket effect, that cancelation might not happen (since the mass distribution is an exponential, and we are taking the derivative of an exponential which returns the exponential times the derivative of the exponent, it might drop out in this special case). There are three forces that point in these directions: Fij, Fr, and Fv. For the pure Newtonian calculation, only Fr will be needed. The rocket effect uses Fv.

I have partially revealed how I will be slicing up the galaxy: in i steps with the passive gravitational mass and in j steps for the active gravitational mass. The active mass also needs to revolve around the origin, which will be done in k steps. The differential active mass is dMa, and the differential passive mass is dmp.

The force term will be the sum of these differential terms:

[tex]F_g = \sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n (dFr + dFv) \quad eq~1[/tex]

For the Newtonian case, the Fv is not included. Can all these terms be written in terms of i, j, k, and Rmax? It is not simple, but a weekend of doodling created this picture of the terms involved:

draw_it-600.jpg


The first three terms are about the progression along i, j, and k:

[tex]Rpi = \frac{i}{n} ~ Rmax[/tex]

[tex]Raj = \frac{j}{n} ~ Rmax[/tex]

[tex]B = 2 \pi \frac{k}{n} \quad eq ~ 2-4[/tex]

Although we don't use it everyday, Rij is calculated using the law of cosines, a variation on the Pythagorean theorem with a cosine to account for a non-right angle triangle:

[tex]Rij = \frac{Rmax}{n} \sqrt{i^2 + j^2 - 2 i j ~Cos(2 \pi \frac{k}{n})} \quad eq ~5[/tex]

Break down Rij into that pointing along Rv - a simple application of the definition of a sine, and Rr, which is the venerable Pythagorean theorem at work:

[tex]Rv = Rij ~Sin (2 \pi \frac{k}{n})[/tex]

[tex]Rr = Rij \sqrt{1 - \frac{Raj^2}{Rij^2} ~Sin^2 (2 \pi \frac{k}{n})}\quad eq ~6, ~7[/tex]

There are corresponding equations for the forces that are proportional to these:

[tex]dFv = dFij ~\frac{Rai}{Rij} ~Sin (2 \pi \frac{k}{n})[/tex]

[tex]dFr = dFij \sqrt{1 - \frac{Raj^2}{Rij^2} ~Sin^2 (2 \pi \frac{k}{n})} \quad eq ~8,~9[/tex]

With this much algebra going on, it is good to think of quality controls. The distance Rij should only equal 2 Rmax for one set of values of i, j, and k. The differential forces dFij, dFr, and dFv should satisfy the Pythagorean theorem.

The mass for disk galaxies is often given in terms of mass per unit area. As such, determine what a differential area is. The area of the complete galaxy is [itex]pi Rmax^2[/itex]. This is being sliced into n pieces, so the area of a sector is [itex]pi Rmax^2/n[/itex]. As we step from i-1 to i, how much area is there?

[tex]dAi = \pi ~\frac{Rmax}{n} ~ (\frac{i^2}{n^2} ~-~ \frac{(i ~-~ 1)^2}{n^2}) = \pi ~ \frac{Rmax}{n^3} ~ (2 i ~-~ 1) \quad eq ~10[/tex]

One problem with this approach is that it samples the galaxy near the core more densely than the outer regions, where i is greater. That may be acceptable since the mass distribution is exponential, so most of the mass comes from the center.

I have skimmed from a paper that the number of solar masses/parsec2 is 37 exp (R'/2.23'). Combine the mass/area with the differential area to get the differential masses:

[tex]dmp = 37 \pi ~\frac{Rmax}{n^3} ~ (2 i ~-~ 1) ~ exp (Rpi'/2.23')[/tex]

[tex]dMa = 37 \pi ~\frac{Rmax}{n^3} ~ (2 j ~-~ 1) ~ exp (Raj'/2.23')\quad eq ~11, ~12[/tex]

These are the players needed to calculate the force: the distance Rij and the two masses, dmp and dMa. Yoda has said, simple it is not, the way of relativistic rocket astrophysics.

Doug
 
  • #598
Symmetry and torque forces

Hello:

Symmetry is useful to think about because it can pinpoint what can be ignored. Feynman emphasized looking for things that add up, knowing we can ignore things that cancel. This is what I drew on my board this morning:

opposing_forces-600.jpg


The two Fv's point in opposite directions. Oops, those will never add up and amount to anything. Therefore, none of my calculations should involve Fv, which appeared in the previous post #598, eq 1 and 8.

So is this galaxy momentum profile calculation over? No, because there is a change in momentum that is being omitted from the standard calculation. Recall that energy is force times distance. One only does work in the direction of motion, or the cosine of the angle between two 3-vectors. In the drawing, that would be Fr dotted to Rr. That is what goes into a classical calculation. Fr dotted with Rv is zero. My objection is that our analysis is not complete. Something which has the same units as work, but is not a scalar like energy, is the sine of a force and a distance vector, or [itex]Fij ~ Rv ~Sin(\theta)[/itex]. That will not be zero. It is the 3-momentum times c. This is the torque force term that pairs with the relativistic rocket effect.

Doug
 
  • #599
Dark energy and the relativistic rocket effect

Hello:

All forces must have a cause and an effect. For a disk galaxy, there is the Newtonian force, [itex]Fij Rr Cos(\theta)[/itex], whose effect is the centripital motion. At the current time, this is all that is used to calculate the momentum profiles of disk galaxies. This is not enough, something is missing. The leading hypothesis is dark matter. I am proposing that the omission is the gravitational force in a different direction, [itex]Fij Rr Sin(\theta)[/itex], causes the new relativistic rocket effect, which helps determine where masses are located in space. This would eliminate the need for dark matter. It is important that I get through all the details of this calculation to test the hypothesis.

The proposal at this early stage has consequences for cosmology. As one moves to larger scales, the effect of gravity becomes more about where mass is distributed than about making things move faster or slower. At the big bang - the farthest distance we can go in spacetime, the cosmic background radiation data indicates that gravity was exclusively about the relativistic rocket effect, where the velocity is constant, at least to five significant digits. No matter was changing its velocity relative to other matter at the time of recombination. As the Univserse has aged, it may be that the balance has shifted a bit towards the mA term from the relativistic rocket effect. If so, then the net effect of gravity would be the same, but the Universe would appear to be accelerating more instead of all traveling uniformily. This would eliminate the need for dark energy.

If I could only quantify such a claim...

Doug
 
  • #600
Bicycle wheels and disc galaxies

Hello:

In this post we will think about momentum for a disk bicycle wheel and a disk galaxy. The similarities between the two systems may give support to the relativistic rocket effect discussed here.

A disk has all of its mass in a plane. Both of the disks in question spin. The bicycle wheel is a rigid body, while the galaxy is not. All parts of the wheel rotate at the same rotational velocity, but travel at a different tangential velocity which depends on how far out from the center a point is. For disk galaxies, the stars at the center of a galaxy travel slowly, yet quickly reach a maximum velocity outside the core. From there on, the stars, and then the helium gas, rotate at the same tangential velocity.

If one applies a force to the axis of a spinning bicycle wheel, the wheel will move a distance. This is an easy problem. It is just like applying a force to a brick, which moves, so the energy is the distance times the force.

Now apply exactly the same amount of force to the spinning bicycle disk, but somewhere along the rim. The problem gets much messier! The axle might move a bit - so you can do the force times distance calculation for that energy - but there will also be a wobble. I don't know how to deal mathematically with the energy that goes into wobbling, sorry. I doubt many people are confident about the subject since it involves torques and all that jazz. My big picture view is that if I apply the same amount of energy to moving the bicycle wheel along the axis as to somewhere on the disk, then the disk must distribute that energy between moving the center of mass and the energy of wobbling. Energy must be conserved.

Now think about a disk galaxy. The mass for disk galaxies has two components, a spherical one, and the disk, whose visible mass drops off exponentially with increasing distance from the center. That means the vast majority of mass can be viewed as being central: due to spherical symmetry and the exponential decay of the disk.

Newton's law of gravity has a cause, the inverse square attraction of gravity, and an effect, the centripetal force that keeps things moving in toward the center:

[tex]-\frac{G M m}{R^2} ~\hat{R} = m~\frac{V \cdot V}{R} ~\hat{R} \quad eq ~ 1[/tex]

What I would like to point out is both cause and effect point in exactly the same direction, as they must, along the radius, [itex]\hat{R}[/itex]. This is why modeling of galaxies by Newton's law works near the core, getting the right maximal terminal velocity. For a bicycle wheel, the closer to the axis you apply the force, the less you need to worry about all the messy torque stuff.

On physical grounds, I cannot accept it as reasonable that gravity does not exert a torque force. Gravity would somehow have to favor doing work along the radial direction. One might argue that perhaps all the torques cancel out nicely. Please recall the bicycle wheel. That problem is hard, there is no cancellation going on for the rigid body. There is no reason to expect the disk galaxy to be so clean. Yet in all my readings, I have never seen anyone discuss gravity working in any direction other than [itex]\hat{R}[/itex]. I have a paper from 1963 where Alar Toomre did the calculation of the momentum profile for highly flattened galaxies the right way (using Bessel integrals, too technical for me, but one of the papers that started this area of study). It is clear he was only dealing with force in the radial direction.

Gravity from a system with its mass distributed in a disc must have torque forces that need to be taken into account. That would make the cause term point in a new direction, perpendicular to the radius, or [itex]\hat{V}[/itex]. The effect of gravity must point in this same direction. Dimensional analysis leads to this proposal:

[tex]-\frac{G M m}{R^2} ~\hat{V} = c~ V~ \frac{d m}{d R} ~\hat{V} \quad eq ~ 2[/tex]

Since the torque force points in the V direction, a V can only appear once, and naturally points along [itex]\hat{V}[/itex]. As discussed in earlier posts, this term is a direct result of viewing a force as a change in momentum, and using the product rule, which has a constant V and a changing m.

I call myself a member of the "ultra-conservative fringe". Folks as bright as Alar Toomre frighten me. I don't think it is reasonable to say someone with such skills is wrong. On rare occasions, it may be OK to document something they did not account for, the error by omission. All the complexity of a torque force for gravity of a disc galaxy have been omitted. Based on energy conservation, I think that omission must be corrected.

It is reasonable to ignore this proposal since I have yet to go numerical. It might be the effect is trivial. Yet this problem works in the trivial end of the force spectrum. The accelerations are on the order of 10-10 m/s2! They are measuring how helium II gas moves way out from the core. One needs to be complete. I cannot defend the models used today.

Doug

Note added in proof: Notice in equations 1 and 2, the cause term has in the denominator R2, which is equal to the dot product of R with itself, and thus has no direction. Newton tacked on the radial direction part, as has every one of his students since. There is no "natural" directionality to the universal law of gravitation. The terms on the effect side both get there directions from single vectors, the dividing by R of the directionless V dotted to V for the centripetal force, and the V in the relativistic rocket effect. Nice.
 
  • #601


sweetser said:
Hello:

I am taking the General Relativity course 8.06s at MIT's Professional Institute. It is great fun that I get to think and chat about physics for four days. It has been work the 4 vacation days and $2k for the class.

At lunch, one fellow said his biggest concern was with force. He pressed us to define it. I decided to just play it quiet, so he said it was either energy integrated over space or momentum integrated over time. Most discussion emphasize then energy integrated over time, while spending less effort on the impulse force. There is no reason for why one should be more important than the other except for accidents in the history of teaching physics.

Hi Doug,

I don't reconise the gentleman's definitions. Isn't force dP/dt or dE/dx ?

Four days seems rather short for the average professional to learn GR.

Regards,
Lut
 
  • #602
Three definitions of a 4-force

Hello Lut:

I have used 3 definitions for force. One is the derivative of the 4-momentum with respect to the interval tau. I saw that in a discussion of the Lorentz force of EM in Jackson, the end of chapter 11. This is the dP/dt and dE/dx one. The second definition is based on varying a Lagrange density with respect to velocity. Again that was used to derive the Lorentz force, in Landau and Lifgarbagez "Classical Theory of Fields". The third definition is the integral definition discussed in post #585, p37. All three give the same answer for force, love that consistency.

The class was a survey of issues in gravity and cosmology, not a professional introduction to GR. We got a bit deeper than Scientific American, using a few equations. This thread is more "equation dense" than the course was. A fun aspect was being able to spend the days thinking of physics. I miss that spirit, and have been adrift since then.

Doug
 
  • #603
Of course, any expression with the right dimensions can be interpreted as force. My view was a bit telescopic (?).

I found gravity as a coordinate acceleration independent of mass, by letting inertial mass depend on a field. It doesn't quite work but it was fun doing it.

Lut
 
  • #604


sweetser said:
Hello:

This would eliminate the need for dark energy.

Doug:

Feng and Gallo have been working on the luninocity density relation for a solution to the dark matter issue. You may find their paper useful.

"Galactic Rotation Described with Thin-Disk Gravitational Model"
http://arxiv.org/abs/0803.0556

DT
 
  • #605
Hey Sweets, I was reading this and noticed something which seemed familiar from how I was considering the problem.

Would you object to saying your interpretation of Gravity is more akin to a "stretching" of Space in the direction of the body in question, and EM could be considered a "stretch and a twist" in the direction of the charge orientation?

I didn't come into physics from the mathematics side, but from the thinking about explanations for processes which can be converted into mathematics side.

It could just be that I'm reading what you're describing mathematically wrong, but when I model it in my head, it looks like the explanation I've been developing from the other "direction", if you will.
 
  • #606
Gravity versus EM

Hello Max:

I do have a word description about these two forces that is based on looking at the equations and solutions for quite some time. Here goes...

Stuff in Nature is pretty well isolated from each other after a separation lasting 13 billion years. In an ideal isolationist Universe, nothing would interact with anything else. The problem is that there is other stuff in the Universe. So the math wonk question is what is the least amount of interaction stuff can do with each other?

The answer is a 4D wave equation, with very little waving going on. If there was zero waving, the Minkowski metric would rule all. It does not, although it is darn close. By darn close, I mean the variations happen in parts per million for some systems, parts per hundreds of millions for others.

If we think about gravity versus EM, that has to do with the field strength tensor of gravity versus the field strength tensor of EM. The one for gravity is the average amount of change in a potential. The one for EM is the deviation from the average amount of change in a potential. In baseball terms, gravity is about the batting average, while EM is the deviation from the average - is the batter steady or does he have hot and cold streaks. The average is a different measure than the deviation, yet both are measured in terms of hits because they both are about hits.

If the average is zero, there is no deviation. In Nature, there are no charged particles that have a mass of zero.

As far as your specific effort, I would make the same critique I make of professionals: you cannot say something about space without saying something about time. To my ear, this common habit of taking about space sounds awful. With this correction in mind, is gravity a stretching of spacetime in the direction of the body in question? Even said this way, the "direction of the body" sounds like it wants to only deal with space. It is hard to work consistently with spacetime! Our language has a time separate from space bias built in! If I gave you an assignment, it would be to work on spotting those biases and avoid them.

Good luck in your studies, and good luck to the Red Sox,
Doug
 
  • #607
Well, I say space, but I mean space-time.

The two are very intertwined in my head these days. Relativistic overdose, if you will.

Stretching of space when I say it, implies acceleration through a three space direction, and deceleration through the time direction.

Both bodies are moving through time, so I have a nasty habit of assuming that to be implied. Been working on it.
 
  • #608
Talk last weekend and this weekend

Hello:

Last weekend, I went to the Fall APS/AAPT meeting at the University of Massachusetts, Boston. They have a lovely campus, right next to the JFK library on the water. It was a sunny and warm day thanks to global warming.

The first speaker was also the second, third, and fourth speaker. I hate when people game the system and claim they are working with four different people so they can ramble on about their crap. And it was crap. He claimed that rooms in the Egyptian pyramids were useful for neutrino detection. At one point, he was taking about how he felt he could tell the intent of moose. I almost lost it at that point, but I showed my self-control.

There were two talks that were serious. One concerned the wave function. Should we consider it to be real only if the result of an experiment is obtained? That is the sort of thing one could argue over in a bar, so was worth the time. The second talk solidly refuted the claim of people who imagined a pattern in radioactive decay.

My talk went well, I fit it to the time slot. The longer version (a half hour) is up on YouTube. There were two clarification questions, no probing ones.

I decided to give the http://http://www.nd.edu/~astro/MWRM18/ a try. It will be held next weekend at Notre Dame University, South Bend Indiana. I will be the absolute last speaker of the conference at 2:40-3:00 pm on Saturday for the last session, "New Ideas/Novel Approaches" i.e. fringe work. Since my flight leaves at 6PM from O'Hare which is a two hour drive away, I will have to fly out of the meeting and burn some rubber.

I was feeling kind of down about my role in these meetings. What I am doing to lift my spirits back up is to review the logic presented in these talks. It is rock solid. I may do a few posts based on the talk since it clarified a few points. The logic is getting better, a sign that I am "within spitting distance" of the unified field theory Nature uses.

Doug
 
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  • #609
Quaternions in Lagrangians

Hello:

Preparing the talk for Midwest Relativity Meeting was quite productive. It forced an interesting shift: I actually need to use quaternions for the action, tensors will not suffice. It is ironic that it has taken so long for me to insist on using quaternion on technical grounds since I have owned the domain quaternions.com since 1997.

Tensor have a few great properties. For example, one does not choose a coordinate system for a tensor expression. The same tensor expression applies if one uses Cartesian, spherical, cylindrical, or some sort of oddball coordinate system that applies to a problem at hand. This property is shared by quaternions.

Tensors can be added to other tensors, or multiplied by another scalar. This property is shared by quaternions.

Tensors can be in arbitrary dimensions. Quaternions are limited to 4D. Folks working with strings will find that utterly inadequate. I have no interest in making folks who get the units of spacetime wrong happy. Every measurement of events in spacetime, every measurement of energy and momentum, ever made in physics has been 4D. That covers an absurd volume of data, so I will stick with it.

One cannot take two tensors and form a product. It is this limitation which has forced me to write my Lagrangian only in terms of quaternions. If one wants to make a Lagrangian that has connections to EM, the weak, and the strong forces, then one needs to build a bridge to group theory. Group theory involves having an identity, multiplication, and inverses. That is an option for quaternions, but not so for tensors. Sure, one can tack on a group symmetry operator, but that feels phoney. I recall seeing the symmetry stapled onto terms of the action. Yukko.

Better for group theory is to start with a division algebra to see if various ways to write it are equivalent to the group. In the case of U(1), the canonical representation is a complex number normalized to itself, forming a unit circle in the complex plane, [itex]\frac{z}{|z|}[/itex]. Take either n roots, or n powers, whichever you prefer, and a unit circle in the complex plane can be formed.

The exact same approach can be used for quaternions, [itex]\frac{q}{|q|}[/itex]. The group U(1) is Abelian, members commute with each other. Although quaternions do not commute in general, they do commute if all the quaternions point in the same direction, which would be the case for n roots or powers of a normalized quaternion. Nice.

The weak force has the symmetry SU(2), also known as the unitary quaternions. Sure enough, there is a way to write quaternions as unit quaternions. The Lie algebra has 3 generators. A common approach is to take the exponential of a quaternion where the scalar has been subtracted away, [itex]exp(q-q*)[/itex]. It is natural to combine the U(1) and SU(2) together so as to capture the four degrees of freedom in a quaternion with [itex]\frac{q}{|q|} exp(q-q^*)[/itex]. So wherever there was the rank 1 tensor [itex]A^{\mu}[/itex], drop in its place the quaternion [itex]\frac{q}{|q|} exp(q-q^*)[/itex] so the resulting expression has electroweak symmetry.

There is only one more symmetry in the standard model to get, that for the strong force, SU(3). Its Lie algebra has 8 parts, which is what is found in two quaternions. Multiplying two U(1)xSU(2) symmetries just creates another member of the group. That is a basic property of groups. To make the multiplication table different, one needs to take the conjugate of one quaternion, then multiply it by the other, [itex](\frac{q}{|q|} exp(q-q^*))^*\frac{p}{|p|} exp(p-p^*)[/itex]. Drop this one in for [itex]A^{\mu}[/itex] and you have the symmetries required by three of the four known forces of Nature. This connection to the known forces makes it worthwhile to insist on quaternions.

My sense is that those formally trained will reject moving to quaternions out of hand. I have seen conversations stop because I use the q word, so that has been the source of my reservations. That is too bad, because a quaternion is in fact a tensor whose diminsions happened to be constrained to 4 but has additional abilities, namely multiplication and division, useful to reach out to group theory.

Doug
 
  • #610
I quite like the restriction to 4D myself, never found it too satisfying to try and extend into arbitrary additional dimensions.
 
  • #611
Mathematical fields

Hello Max:

We share a common suspicion about the focus of efforts in the most prestigious centers of study on the planet. Lee Smolin and Peter Woit are the biggest voices in the professional community. My focus is on constructing a viable alternative, so I like to stick to the math. I need to be precise.

So what is a tensor, really? One way to view it is with group theory: a tensor forms a group under the operation of addition. The identity element is zero, the inverse of any tensor is -1 times that tensor. Add the two together, and the result is the identity, zero.

From a math perspective, in order to do calculus, one needs a mathematical field. This means one needs the set of numbers to behave like a group with the addition operator, but the numbers also must behave as a group under multiplication - with some elements excluded, usually just zero. Real and complex numbers fit those requirements for calculus. Quaternions also fit that definition, both the Hamilton representation and the even representation I have discussed here so long as zero and the Eivenvalues are excluded.

The identity element for addition is zero. The identity element for multiplication is one. This is no accident, it is a central observation. Once one has zero and one, the rest of the numbers we know about can be constructed according to work by Peano. If we only have addition, we only have zero built into the group theory of the number system. By insisting that the numbers used in physics must be mathematical fields, then zero and one, and by logical extension, all of number theory, is part of the system.

Addition is easy to generalize for arbitrary dimensions. Multiplication can also be generalized, but division is the hurdle. Frobenius worked out a proof years ago that associative division algebras are limited to one, two, and four dimensions, up to an isomorphism. If my thesis is correct, that we must necessarily shift any expression in physics written as a tensor as a quaternion to get the advantages of mathematical field theory, then we will look back on much effort in the twentieth century as ways to construct isomorphism of quaternions. Physicists are not way off the mark, only slightly off center.

Here is a fun benefit of using a normalized quaternion to get U(1) symmetry: it explains why we cannot build a device to measure the potential directly. Because the potential must always be normalized, its size is always the same, namely equal to one.

Doug
 
  • #612
Quaternions are not only Cl(0, 2)

Hello:

Clifford algebra = Geometric Algebra.

David Hestenes has promoted GA as a label over Clifford algebra because Clifford was one guy who worked on some aspects of this approach, while Geometric Algebra suggests what the math does. I only figured that out at the end of the 8th International Conference on Clifford Algebras and their applications to Mathematical Physics. David was a nice guy, and was fun to talk to so long as you didn't disagree with him on a technical point. As a visionary, he is quick to take out his blade.

According to standard GA,

quaternions = Cl(0,2)

meaning it has a scalar - the 0 - and a bivector - the 2. A bivector is a generalization of a cross product to other dimensions. Look in a mirror, and left hand becomes right hand. It was clear from a brief discussion with David that this bit of dogma was not worth discussing because there were people who were confused on the topic. Anyone who didn't understand that quaternions were Cl(0, 2) needed education.

It is one of these size arguments: GA is bigger than quaternions, ergo study GA. The size of your algebra matters. NOT.

What I think matters most in our efforts to describe Nature is completeness. Quaternions certainly have the property that under a mirror reflection M:

[tex]M: M(t, x, y, z) -> (t, -x, -y, -z)\quad eq~1[/tex]

The problem with this approach is that we need to also consider a time reversal operation (TR = -M):

[tex]TR: TR(t, x, y, z) -> (-t, x, y, z)\quad eq~2[/tex]

The handedness stays the same, which indicates a quaternion should not be viewed as a scalar and a bivector. The artificial constraint that the behavior of quaternions should fit within the GA model doesn't fit.

It was Grassman and Clifford who people credit with the great discovery of the geometric product:

[tex]a b = a \dot b + a \wedge b\quad eq~3[/tex]

Having my groundings with quaternions, this result is incomplete. For quaternions, the dot product will have 4 pairs of number, while the cross product has 6, making 10 out of the possible 16 combinations. Grassman and Clifford came after Hamilton, so they partially recreated quaternions. Arg! Let me write the quaternion a as (a, A) and b as (b, B) where the lowercase letter inside a () is one number, while a capital letter inside is a 3-vector. Compare the products of Hamilton to Clifford:

[tex](a, A)(b, -B) = (ab + A \dot B, Ab - aB - A \times B) \quad eq~4[/tex]
[tex](a, A)(b, B) = (ab + A \dot B, A \times B) \quad eq~5[/tex]

Aside from silly sign issues that arose from the "pro-positive sign" position of Grassman, I won't work with geometric products because there is no accounting for Ab and aB. It is completely legal to adjust the scale of a 3 vector. That completely reasonable operation is missing from the geometric product. It is such a glaring omission that I don't put an effort in bridging the gap between the GA school and my own work. Bad accounting is indefensible.

The even representation of quaternions - or hypercomplex numbers - also doesn't work within the Clifford algebra constraint.

So it goes for Independent Research.

Doug
 
  • #613
Creatively seeing sine

Hello:

On occasion I pitch my quaternion animations to people who have written something on quaternion. The usual reply is no reply. Sometime we exchange a few emails, then go our separate ways.

This time I found out the fellow was visually impared. My animations are meaningless. What I could do is provide a description. Here is what a 4D cube looks like:

4D cube said:
We are all familiar with a 3D cube. Mathematically that is the eight spatial vertices formed from all the combinations of plus or minus one for x, y, and z. In 4D, there are sixteen vertices, the additional ones being plus or minus time. So when time is minus one, we have a 3D cube. At plus one for time we have a 3D cube. In between, when say time is minus one third or plus two sevenths, all we have are the eight vertices, none of the connecting points between the vertices. We see the vertices travel through time from one 3D cube to another.

This approach can be generalized:

general approach said:
For fun, you might want to imagine what a polynomial looks like, or a sine function. Take your mind's image of these, and run a pencil along the image to create the animation. Note you are allowed to hold the pencil at any angle, which can create very different looking animations.

Let us think about the sine function, an endless range of mole hills. If the pencil is vertical, we will see one dot oscillating forever. If the pencil is horizontal, then we see nothing until we meet the bottom of the sine functions at -1. Then we see single dots that split in two, run apart quickly at 0, and slow down and join another, then disappear.

I think the single oscillating dot is how classical physics uses a sine function. We can watch the time-ordered oscillation of one particle for as long as we want. The array of creation, then destruction has the look of quantum field theory. To make it a complete field theory, then we would have to pile one sine atop another, making a graph that looks something like chainmail, so we see the creation and annihilation forever.

Doug
 

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