# Unifying Gravity and EM

Gold Member
Covariant GEM field equations

Hello:

<Preamble>
This post is very similar to the last two. This time I have few of the set of the last two, but unify gravity and EM in the process. Nice.
</Preamble>

In this post I will derive gravity and the Maxweill field equations using quaternion operators in a manifestly covariant notation.

The Hamilton representation will be used for the Maxwell field equations. The even representation of quaternions - where all signs are the same, and the Eivenvalues ore excluded so the quaternion can be inverted - will be used in the first phase of this derivation. What the even representation does is flip the relative signs of the E and B signs, so now the e filed has two signs, and the b field has one, like so:

$$E = -\nabla_0 A ~-~ \nabla_u \phi$$
$$e = \nabla_0 A2 ~-~ \nabla_u \phi$$

$$B_w = \nabla_u A_v ~-~\nabla_v A_u = \nabla \times A$$
$$b_w = -\nabla_u A2_v ~-~\nabla_v A2_u = \nabla \Join A2 \quad eq ~0$$

The fifth play is the term that makes up the gauge field:

$$g = \nabla_0 \phi ~-~ \nabla \cdot A$$

None of these transform like tensors, but together they do from $\nabla A$.

Let's generate all 5 fields:

$$\nabla A = (\nabla_0 \phi ~-~ \nabla \cdot A2, \nabla_0 A2 ~+~ \nabla_u \phi ~+~ \nabla \times A) = (g, -E ~+~ B)$$
$$\nabla^* A2 = (\nabla_0 \phi ~-~ \nabla \cdot A2, -\nabla_0 A2 ~+~ \nabla_u \phi ~-~ \nabla \Join A2) = (g, -e ~+~ b)\quad eq ~1$$

The starting point for the derivation of the unified GEM equations is the Lagrangian which can be viewed as the difference between the scalars of E,b squared and B,e squared. This can be achieved by using the first set of quaternion operators used in the previous posts:

$$\frac{1}{8}((\nabla A)(A \nabla) ~-~(\nabla^* A2)(\nabla A2^*))$$

$$=\frac{1}{8}(\nabla_0 \phi ~-~ \nabla \cdot A, \nabla_0 A ~+~ \nabla_u \phi ~+~ \nabla \times A)(\nabla_0 \phi ~-~ \nabla \cdot A, \nabla_0 A ~+~ \nabla_u \phi ~-~ \nabla \times A)$$
$$-(\nabla_0 \phi ~-~ \nabla \cdot A, \nabla_0 A ~-~ \nabla_u \phi ~-~ \nabla \Join A)(\nabla_0 \phi ~-~ \nabla \cdot A, -\nabla_0 A ~+~ \nabla_u \phi ~-~ \nabla \Join A)$$

$$=\frac{1}{2}(g, -E ~+~ B)(g, -E ~-~ B) ~-~ (g, e ~+~ b)(g, -e ~+~ b) = \frac{1}{2}(B^2 ~-~ E^2 ~-~ b^2 ~+~ e^2, 2 E \times B ~-~ b \Join b ~+~ e \Join e)\quad eq ~2$$

Any expression derived from this one will be invariant under a scalar gauge transformation which uses the terms $\nabla_0 \phi, \nabla \cdot A$ because they have miraculously been cancelled!

Write out the Lagrange density - the scalar part of eq 2 - in terms of its components, including the vector coupling whose phase has spin 1 and spin 2 symmetry[/itex]:

$$\mathcal{L}_{EBeb} = -~(\nabla_3 A_2)(\nabla_2 A_3) ~-~ (\nabla_1 A_3)(\nabla_3 A_1) ~-~ (\nabla_1 A_2)(\nabla_2 A_1) ~-~ (\nabla_1 \phi)(\nabla_0 A_1) ~-~ (\nabla_2 \phi)(\nabla_0 A_2) ~-~ (\nabla_3 \phi)(\nabla_0 A_3)$$
$$-\rho \phi ~+~ J_1 A_1 ~+~ J_2 A_2 ~+~ J_3 A_3 \quad eq ~3$$

[sidebar: The assertion that the current coupling term has both spin 1 and spin 2 symmetry for all terms in the phase requires a small calculation. I decided to leave that as a problem. My guess is that until someone starts confirming these by hand, I'll be the only one to see that even these cool details work out.[/sidebar]

Apply the Euler-Lagrange equation to this Lagrangian, treating the 4-potential as the variable:

$$\nabla_{\mu}(\frac{\partial \mathcal{L}_{EBeb}}{\partial (\nabla_{\mu} \phi)}) = - \nabla_0 \nabla_1 A_1 ~-~ \nabla_0 \nabla_2 A_2 ~-~ \nabla_0 \nabla_3 A_3 ~-~ \rho$$
$$= -\frac{1}{2}\nabla \cdot (E ~-~ e) - \rho = 0 \quad eq ~4$$

$$\nabla_{\mu}(\frac{\partial \mathcal{L}_{EBeb}}{\partial (\nabla_{\mu} A_1)}) = -~ \nabla_1 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_2 ~-~ \nabla_0 \nabla_1 \phi ~+~ J_1$$
$$= \frac{1}{2}(\nabla_0 (E_1 ~+~ e_1) - (\nabla \times B)_1 - (\nabla \Join b)_1) + J_1 = 0 \quad eq ~5$$

$$\nabla_{\mu}(\frac{\partial \mathcal{L}_{EBeb}}{\partial (\nabla_{\mu} A_2)}) = - \nabla_2 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_1 ~-~ \nabla_0 \nabla_2 \phi ~+~ J_2$$
$$= \frac{1}{2}(\nabla_0 (E_2 ~+~ e_2) - (\nabla \times B)_2 - (\nabla \Join b)_2) + J_2 = 0 \quad eq ~6$$

$$\nabla_{\mu}(\frac{\partial \mathcal{L}_{EBeb}}{\partial (\nabla_{\mu} A_3)}) = - \nabla_2 \nabla_3 A_2 ~-~ \nabla_1 \nabla_3 A_1 ~-~ \nabla_0 \nabla_3 \phi ~+~ J_3$$
$$= \frac{1}{2}(\nabla_0 (E_3 ~+~ e_3) - (\nabla \times B)_3 - (\nabla \Join b)_3) + J_3 = 0 \quad eq ~7$$

This work can be summarized with the GEM gravity source equations:

$$\frac{1}{2} \nabla \cdot (E ~-~ e) = \rho \quad eq ~8$$

$$\frac{1}{2}(\nabla \times B ~+~ \nabla \Join b - \nabla_0 (E ~+~ e)) = J \quad eq ~9$$

This has been a manifestly covariant derivation. This derivation remains the same in flat or curved spacetime, in Cartesian or spherical coordinates.

I have spent the day trying get all the signs and factors right. If there are any questions, send me a private note, and I will recheck it. If needed, we can put up a new post.

Doug

Stills:

The talk:
(this derivation was not part of the talk)

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Gold Member
Covariant gravity field equations

Hello:

<Preamble>Looks like two of my posts were identical, sorry. The software here doesn't appear to like this much LaTeX.

This post is very similar to the last. I am starting out the draft by cutting and pasting the text, as all the same terms are the same. used in the same locations of the same equations. What changes are a few signs. The E field is made of two parts, $-\nabla A_0$ and $-\nabla \phi$. These happen to have the same sign. In this post, the case where these have opposite signs will be explored. A similar thing will be done for the magnetic field, where the two terms $\nabla_u A_v$ and $\nabla_v A_u$ have the same sign.

Those readers concerned about how these objects transform can rest easy. We know that $\nabla A$ transforms like a rank 2 tensor - it was the justification behind developing the covariant tensor $\nabla[/tex] in the first place. We also know that the EM field strength tensor, [itex]\frac{1}{2}(\nabla_u A_v - \nabla_v A_u)$, transforms like a tensor. The difference between these two tensors which is $\frac{1}{2}(\nabla_u A_v + \nabla_v A_u)$, also transforms like a tensor since the difference of two tensors remains a tensor.
</Preamble>

In this post I will derive the gravity part of the GEM proposal using quaternion operators in a manifestly covariant notation.

The even representation of quaternions - where all signs are the same, and the Eivenvalues ore excluded so the quaternion can be inverted - will be used in the first phase of this derivation. What the even representation does is flip the relative signs of the E and B signs, so now the e filed has two signs, and the b field has one, like so:

$$E = -\nabla_0 A ~-~ \nabla_u \phi$$
$$e = \nabla_0 A2 ~-~ \nabla_u \phi$$

$$B_w = \nabla_u A_v ~-~\nabla_v A_u = \nabla \times A$$
$$b_w = -\nabla_u A2_v ~-~\nabla_v A2_u = \nabla \Join A2 \quad eq ~0$$

None of these transform like tensors, but together they do from $\nabla A$.

Notice that the quaternion differential operator acting on a 4-potential creates a scalar gauge field, and two fields the -e and b fields:

$$\nabla^* A2 = (\nabla_0 \phi ~-~ \nabla \cdot A2, \nabla_0 A2 ~-~ \nabla_u \phi ~-~ \nabla \Join A2) = (g, e ~+~ b)\quad eq ~1$$

The starting point for the derivation of the GEM gravity equations is the Lagrangian which can be viewed as the difference between the scalars of b squared and e squared. This can be achieved by changing the order of the conjugation operator with respect to the 4-potential, which flips the sign of the time derivative of A and gradient of phi which make up e, but not the symmetric curl b. The scalar gauge field can be subtracted away:

$$\frac{1}{8}(\nabla^* A2 ~-~ (\nabla^* A2)^*)(\nabla A2^* ~-~ (\nabla A2^*)^*)$$

$$=\frac{1}{8}(0, \nabla_0 A ~-~ \nabla_u \phi ~-~ \nabla \Join A)(0, -\nabla_0 A ~+~ \nabla_u \phi ~-~ \nabla \Join A)$$

$$=\frac{1}{2}(0, e ~+~ b)(0, e ~-~ b) = \frac{1}{2}(b^2 ~-~ e^2, b \Join b ~-~ e \Join e)\quad eq ~2$$

Any expression derived from this one will be invariant under a scalar gauge transformation which uses the terms $\nabla_0 \phi, \nabla \cdot A$ because they have been explicitly subtracted away at this early stage. It is also of interest to think about the properties of the 3-vector, since in the EM case it was the Poynting vector.

Write out the Lagrange density - the scalar part of eq 2 - in terms of its components, including the vector coupling, $-\frac{1}{2 c}(J A2* + (J A2^*)^*)$:

$$\mathcal{L}_{eb} = \frac{1}{2}((\nabla_1 \phi)^2 ~+~(\nabla_2 \phi)^2 ~+~(\nabla_3 \phi)^2 ~+~ (\nabla_0 A_1)^2 ~+~ (\nabla_0 A_2)^2 ~+~ (\nabla_0 A_3)^2$$
$$~-~ (\nabla_3 A_2)^2 ~-~ (\nabla_2 A_3)^2 ~-~ (\nabla_1 A_3)^2 ~-~ (\nabla_3 A_1)^2 ~-~ (\nabla_2 A_1)^2 ~-~ (\nabla_1 A_2)^2)$$
$$~-~ (\nabla_3 A_2)(\nabla_2 A_3) ~-~ (\nabla_1 A_3)(\nabla_3 A_1) ~-~ (\nabla_1 A_2)(\nabla_2 A_1) ~-~ (\nabla_1 \phi)(\nabla_0 A_1) ~-~ (\nabla_2 \phi)(\nabla_0 A_2) ~-~ (\nabla_3 \phi)(\nabla_0 A_3)$$
$$-\rho \phi ~+~ J_1 A_1 ~+~ J_2 A_2 ~+~ J_3 A_3 \quad eq ~3$$

Apply the Euler-Lagrange equation to this Lagrangian, treating the 4-potential as the variable:

$$\nabla_{\mu}(\frac{\partial \mathcal{L}_{eb}}{\partial (\nabla_{\mu} \phi)}) = \nabla_1^2 \phi ~+~ \nabla_2^2 \phi ~+~ \nabla_3^2 \phi ~-~ \nabla_0 \nabla_1 A_1 ~-~ \nabla_0 \nabla_2 A_2 ~-~ \nabla_0 \nabla_3 A_3 ~-~ \rho$$
$$= -\nabla \cdot e - \rho = 0 \quad eq ~4$$

$$\nabla_{\mu}(\frac{\partial \mathcal{L}_{eb}}{\partial (\nabla_{\mu} A_1)}) = \nabla_0^2 A_1 ~-~ \nabla_3^2 A_1 ~-~ \nabla_2^2 A_1 ~-~ \nabla_1 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_2 ~-~ \nabla_0 \nabla_1 \phi ~+~ J_1$$
$$= \nabla_0 e_1 - (\nabla \Join b)_1 + J_1 = 0 \quad eq ~5$$

$$\nabla_{\mu}(\frac{\partial \mathcal{L}_{eb}}{\partial (\nabla_{\mu} A_2)}) = \nabla_0^2 A_2 ~-~ \nabla_3^2 A_2 ~-~ \nabla_1^2 A_2 ~-~ \nabla_2 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_1 ~-~ \nabla_0 \nabla_2 \phi ~+~ J_2$$
$$= \nabla_0 e_2 - (\nabla \Join b)_2 + J_2 = 0 \quad eq ~6$$

$$\nabla_{\mu}(\frac{\partial \mathcal{L}_{eb}}{\partial (\nabla_{\mu} A_3)}) = \nabla_0^2 A_3 ~-~ \nabla_2^2 A_3 ~-~ \nabla_1^2 A_3 ~-~ \nabla_2 \nabla_3 A_2 ~-~ \nabla_1 \nabla_3 A_1 ~-~ \nabla_0 \nabla_3 \phi ~+~ J_3$$
$$= \nabla_0 e_3 - (\nabla \Join b)_3 + J_3 = 0 \quad eq ~7$$

This work can be summarized with the GEM gravity source equations:

$$-\nabla \cdot e = \rho \quad eq ~8$$
$$\nabla \Join b - \nabla_0 e = J \quad eq ~9$$

This has been a manifestly covariant derivation. This derivation remains the same in flat or curved spacetime, in Cartesian or spherical coordinates. Once you get used to keeping track of all these terms, it is kind of fun, honest!

Doug

Stills:

The talk:

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Gold Member
Hatch's work on an exponential force equation

Hello Carl:

Ronald Hatch sent me his paper which has an exponential in the force equation. For me as a member of the ultra-conservative fringe, it was not a happy read. He did not define a Lagrangian, so the proposal felt ad hoc. The way to get the force equation is to vary the Lagrangian with respect to velocity.

He thinks there is an absolute ether which strikes me as wrong.

He wants to give up the equivalence principle. His logic for this struck me as muddled. The discussion of mass was particular confusing. It looked like he had the classic self-taught blindspot vis-a-vis E=mc2, that the real relation is between the invariant m2 c4 and the square of the covariant 4-momentum, E2 - P2 c2. One of his equations looked like a trivial rearrangement of terms, a rearrangement he took seriously.

The precession of the perihelion of Mercury is a tough calculation. He wanted to add a second order effect in as if it was a first order effect.

By a strict application of the rules for the Independent Research area of Physics Forums, I don't think his work would be accepted. I forwarded my concerns to him directly, but he said I needed to think about things more carefully. The paper is published in Physics Essays.

Doug

Gold Member
Seeing spin 1 and spin 2 symmetry

Hello:

I exchanged a dozen emails with Steve Carlip over the coupling current $J^{\mu} A_{\mu}$. The discussion started this way: "I have thought about Feynman's analysis of the spin of the vector current coupling." This post represents the back story, what I got from looking at chapter 3, pages 29-39 for about three to four weeks.

The first thing I did was work on my speed of going through the algebra in section 3.2, "Amplitudes and polarizations in electrodynamics, our typical field theory". You can only motor if you are confident about all the steps, just like in video games. The logic runs like this:

2. Take the Fourier transformation of the 4-potential to get a current.
3. Simplify the current-current interaction along one axis, z.
4. Write out the contraction in terms of its components.
5. Use charge conservation to eliminate one term.
6. One term is the standard Coulomb interaction, the others are the relativistic corrections.

Then Feynman wants to know what that correction term is. This is were it gets a little odd. He talks about plane polarized light, and how looking at that you can see the angular momentum projections. I admit, I never quite saw those. What I did instead was try to strip away all the physics-speak, and just find the kernel of math underlying the operation. Looking back, that is what took the time: reducing the physics to a simple math expression. The Rosetta stone was a line on page 39:
Feynman said:
...we know that $(x ~\pm ~ i y)(x ~\pm~ i y)$ are evidently of spin 2 and projection $\pm 2$; these products are $(xx ~-~ yy ~\pm~ 2 i xy)$, which have the same structure as our terms (3.4.1)
This is the pure math way to spot a system with spin 2: start with the product of two complex numbers, and check that the imaginary part has a 2ixy. This will require a change in x of pi radians to get back to the start point since there is an multiplier of 2. To speak like Feynman, I should talk up the projection operators, but I like to keep the math kernel free of that jargon.

Looking back on an earlier calculation, I was able to transcribe an earlier bit of algebra into a similar set of complex numbers:

$$(x ~\pm ~ i y)(x' ~\pm~ i y')^* = (xx' ~+~ yy' ~+~ (yx' ~-~ xy') i \quad eq~1$$

This is a system which has spin 1 projection operators, to use physics-speak. In math terms, the xy' does not help out the yx', there are no factors of 2 or 1/2, so this would take 2 pi radians to get the imaginary part of this back to where it was.

In terms of the math, the difference between a spin 1 system (eq 1) and a spin 2 system (see the quote) is no more complicated than looking at the imaginary part of a complex product.

Doug

Gold Member
Feynman, Steve Carlip, and I

Hello:

I have the exchange of emails between Prof. Steve Carlip and myself up on the screen. It is not easy for me to read. Steve is a professional, I am not. On occasion, I babble. Babbling is a form of exploring, a process used to learn how to speak a language like a native. Studies have shown that deaf children do so learning sign language, and that baby birds do so before they can sing exactly like adults do. Recognizing this process, I have great patience for others that babble physics. Steve probably is that way in the right context, but in this email exchange, I got the book tossed at my head.

The discussion began with a http://picasaweb.google.com/dougsweetser/SpinAndPhaseOfCurrentCoupling/photo#5196141258835741330 [Broken]. I went by the book, section 3.2 of Feynman's lectures, for three of four steps:

1. Start with the coupling term, $J^{\mu} A_{\mu}$.

2. Take the Fourier transformation, $J^{\mu} A_{\mu} = -\frac{1}{K^2} J^{\mu} J'_{\mu} \quad eq~3.2.2$

3. Write out 2 in terms of the components:
$$J^{\mu} A_{\mu} = -\frac{1}{K^2}(\rho \rho' - J_1 J'_1 - J_2 J'_2 - J_3 J'_3 ) \quad eq~3.2.5$$

[Note on imprecise notation: in the slide I use x, y, and z which imply a coordinate choice. I should have used numbers for subscripts. I also didn't toss in the minus sign as Feynman does for the Fourier transformation step.]

Up to this point, I have exactly walked down the path Feynman wrote about. When it came to writing up the slides, I initially put in a further step Feynman used: he imagines picking a coordinate system such that all the current goes along one direction (the 3 axis). Everyone is accustom to this step. Yet it bothered me. I would have to rewrite the derivation if someone chose the 2 axis instead. A more general position would involve choosing no axis, yet spotting the symmetries of the spin in the phase anyway. I decided to work with that as a goal.

4. Multiply out the two currents as quaternions:

-J J' = (-rho rho' + J1 J'1 + J2 J'2 + J3 J'3,

-rho J'1 - J1 rho' - J2 J'3 + J3 J'2,

-rho J'2 - J2 rho' - J3 J'1 + J1 J'3,

-rho J'3 - J3 rho' - J1 J'2 + J2 J'1)

The terms in italics are equation 3.2.5, the underlined terms are in an expression about spin 2 symmetry on page 39, and the terms in bold are in 3.2.10 in a discussion of spin 1 symmetry.

Steve had no idea what I was doing, none. Communication was broken by step 4. I was at work, trying to do my job, or appear to be doing my job, and quickly come up with a response to someone with far more intellectual precision. It did not work out so well. The worse thing I did was about equation 3.2.10 concerning circularly polarized light. Feynman writes out the currents for two circularly polarized light whose imaginary parts cancel. I got a sign wrong, so they didn't cancel, and I got to look stupid.

Both Steve and Feynman talked about projection operators, the relevant machinery from quantum field theory. I did not talk about projection operators at all. I am not going to do so now since I would probably just babble about them. I understand why the well-trained would say that if I don't discuss projection operators intelligently, then this has nothing to do with the spin of particles. Nothing. For me, projection operators are a patina on the underlying algebra (patina def.: the sheen on the surface of an old object, caused by age and much handling).

Steve left the discussion convinced I didn't even understand the basics of spin in physics. He won the word game, but the algebra in step 4 still stands consistent with what Feynman wrote. At the end of the day, algebra trumps words.

Doug

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Gold Member
Three new bits of math

Hello:

Preparing for my talk in Brazil, I had an interesting insight. The thesis behind "Doing Physics with Quaternions" at quaternions.com is that physics describes patterns of events in spacetime using quaternions up to an isomorphism. Most of physics works great without quaternions because the quaternion expression would not provide new information.

My epiphany was this: new quaternion math equals new physics. Here in this thread, the new math is the Even representation for quaternions (a reinvention of Clyde Daven's hypercomplex numbers). The current coupling J2 A2 has the spin 2 symmetry in the phase, and the field strength tensor $\nabla A2$ contains the symmetric curl needed for the symmetric field b.

I have also mentioned here the work in animating quaternions which lead to an understanding why the groups U(1), SU(2), SU(3) and Diff(M) must be all that makes up the symmetry forces of Nature. With the visual perspective, the forces are more tightly linked algebraically. The standard model is written as U(1)xSU(2)xSU(3), which would have 1+3+8=12 generators for its Lie algebra. One mystery of the standard approach is why the electro and weak forces should team up to form the electroweak force. Another mystery is why should confinement exist for the strong force SU(3)? These phenomena are suggesting something more like [U(1)xSU(2)]xSU(3) than three equal players. That is what happens with the quaternion representation of the symmetry of forces:

$$A^* B = (\frac{A}{|A|} exp (A - A^*))^* (\frac{B}{|B|} exp (B - B^*))\quad eq~1$$

which is (U(1) SU(2))*(U(1) SU(2)). The Lie algebra only has 8 generators. This smaller model has a chance to provide a cause for the confinement of quarks.

The third quaternion math innovation I do not talk about much because I have yet to see how it impacts a calculation, although it helps with a big riddle, a "why" question, in physics. Many who work with quaternion derivatives accept the idea of a left handed versus right handed derivative. This comes from the limit definition, putting the differential on the left or right. This definition is ineffective since one cannot show that a function as simple as f(q)=q2 is analytic in q. For me that indicates the definition has no utility. What I did was steal a move from L'Hospital's rule and use a dual limit process. Let the pesky 3-vector with its three imaginary basis vectors go to zero first, leaving only the real number which commutes with all. Effectively this is a directional derivative along the real axis. Things work out great for proofs using this definition (if one is good at doing proofs, which I am not). All events are ordered by the real scalar. If this definition is applied to events in spacetime, the scalar is time, and thus all the events are order in time like a movie.

What happens when the limit processes are reversed, and the pesky 3-vector goes to zero after the scalar gets frozen? All one can do in this case is to determine the norm of the derivative. Although not well known, there is a branch of math that studies norms of derivatives. I think this is the domain of quantum mechanics. We cannot order things in time, but we can tell on average how much change is going to happen after making many measurements.

Now to make the slides...
Doug

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but when doing the product of quaternions if Q is a quaternion then where you put $$Q.Q=Q^{2}$$ it should read $$Q.Q^{*}$$ , for example for Minkowsky metric

$$dQ=dt-idx-jdy-kdz$$ then $$ds^{2}=(dQ).(dQ^{*})=dt^{2}-dx^{2}- dy^{2}-dz^{2}$$

and as i pointed in other part of the forum, you have the problem of non-commutativity so

$$ij(dx.dy)=-ji (dy.dx)$$,

anyway the idea seems very interesting

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Gold Member
The Even representation

Hello mhill:

In the standard Hamilton representation, for a quaternion dQ = (c dt, dx, dy, dz) we have the product:

$$dQ dQ = (c^2 dt^2 ~-~ dx^2 ~-~ dy^2 ~-~ dz^2, 2 c ~dt~ dx, 2 c~ dt~ dy, 2 c ~dt~ dz)\quad eq~1$$

When one uses tensors, the Einstein summation convention ignores the three other terms. That is a mistake in my opinion. One thing I have noticed about the GEM proposal: the 3-vector (2 dt dx, 2 c dt dy, 2 c dt dz) is an invariant in the presence of a gravitational source. That is really cool because special relativity is about the invariant scalar, and for GEM, gravity is about the invariance of the 3-vector.

If we look at the product of two quaternions that are nearby each other, a dQ and dQ', we get a similar result with a cross product:

$$dQ dQ' = (c^2~ dt~ dt' ~-~ dx~ dx' ~-~ dy~ dy' ~-~ dz ~dz',$$
$$c~ dt ~dx' ~+~ c ~dx~ dt' ~+~ dy ~dz' ~-~ dz~ dy',$$
$$c ~dt ~dy' ~+~ c ~dy ~dt' ~+~ dz ~dx' ~-~ dx ~dz',$$
$$c ~dt ~dz' ~+~ c ~dz ~dt' ~+~ dx ~dy' ~-~ dy ~dx')\quad eq~2$$

Hamilton could have done that one, so this is not new math. For the Even representation of quaternions, the rules are easy to remember: everything is sunny and positive in California, and so it goes for the Even representation:

$$i^2 = j^2 = k^2 = ijk = 1\quad eq~3$$

$$ij = ji = k \quad eq~4$$

$$ik = ki = j \quad eq~5$$

$$jk = kj = i \quad eq~6$$

There is not a minus sign to write. This will be a division algebra if and only if the eigenvalues of the real 4x4 matrix representation are excluded from the set of quaternions.

There will be math wonks who insist that the name "quaternions" is reserved for non-commutative 4D division algebras. Historically, that has been the use. Doing new math can cause conflict, so be it. The non-commutative aspect is a result of an arbitrary choice in how to represent a 4D division algebra. I have chosen a different representation. It still is a division algebra, but one where multiplication commutes. Clyde Daven did this first and called them hypercomplex numbers. That makes them sound like a separate animal, and I don't think it is. This is representation theory in practice, applied to quaternions.

To indicate the Even representation is being used, I toss in a "2" every now and then. Repeat eq. 1 with the even representation:

$$dQ2^* dQ2 = (c^2 ~dt^2 ~-~ dx^2 ~-~ dy^2 ~-~ dz^2, - 2~ dy ~dz, - 2 ~dx ~dz, - 2 ~dx ~dy)\quad eq~7$$

The scalar is the same, but the 3-vector has changed. Instead of rescaling the (dx, dy, dz) by a 2 dt factor, this time it is all about the symmetric curl. The dt factor does not make it into the 3-vector.

Repeat the dQ2 dQ2' in the Even representation:

$$dQ2^* dQ2' = (c^2 ~dt ~dt' ~-~ dx ~dx' ~-~ dy ~dy' ~-~ dz ~dz',$$
$$c ~dt ~dx' ~-~ c ~dx ~dt' ~-~ dy ~dz' ~-~ dz ~dy',$$
$$c ~dt ~dy' ~-~ c ~dy ~dt' ~-~ dz ~dx' ~-~ dx ~dz',$$
$$c ~dt ~dz' ~-~ c ~dz ~dt' ~-~ dx ~dy' ~-~ dy ~dx')\quad eq~8$$

If dQ2' where to get acted on by the conjugate operator instead of dQ2, then the only terms to flip signs involve dt dR.

What makes the Even representation interesting is it gives a more complete way of viewing a quaternion product. In the Hamilton representation, the rescaling (dt dR) was unaltered by changing the order of multiplication, while the cross product flipped signs. In the Even representation, the symmetric curl is unaltered by changing the order of conjugation, while the rescaling flips signs. The scalar remains steady no matter what. That sounds like a more complete survey of the product of two sets of four numbers.

Doug

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Gold Member
Pitching the program

Hello:

I spend time and effort pitching this research project to both the upper elite and the technical masses. I took off a Wednesday from work to see a talk by Nobel Laureate Sheldon Glashow, who got one in the 70s for the electroweak theory. I told him I had an animation for U(1)xSU(2) on my MacBook Pro. He was too busy to look right then, but I could send him an email, which I did.

Later that evening, Michio Kaku was doing book promo work. He agreed to sign his latest book, "Physics of the Impossible", which is doing well on the New York Times Best Seller list. I got in line late, and the book was sold out by the time I got there. I bought a different book, and had him sign "Maxwell is the best!" I also dropped off a version of my proposal where I derive the Maxwell equations first - to established I am much better than your average crank - and with a variation get equations for a metric approach to gravity. He thanked me as he went on to sign another book.

Max Tegmark gave a talk on the Physics of Super Heroes, along with a screening of Superman. He is a big survey astronomer by day, so it was fun to cruise through the known Universe with his software. He knew me from a previous outreach program he did, and had traded a few stories back then about quaternions. I told him of the quaternion animation project, how it could be found on YouTube with a search for "Quaternions Standard Model". I gave him a business card with the search instructions after my 20 second pitch.

Low odds on getting a reply from these busy folks, but I need to try, so I do.

There are many more folks who read a high traffic site such as slashdot.com They had an article on Lectures On the Frontiers of Physics Online. Buch of big names in physics have videos up there:
Neil Turok's 'What Banged?,' John Ellis with 'The Large Hadron Collider,' Nima Arkani-Hamed with 'Fundamental Physics in 2010,' Paul Steinhardt with 'Impossible Crystals,' Edward Witten with 'The Quest for Supersymmetry,' Seth Lloyd with 'Programming the Universe,' Anton Zeilinger with 'From Einstein to Quantum Information,' Raymond Laflamme with 'Harnessing the Quantum World,'
You might be able to see the site in a day or two here, http://perimeterinstitute.ca/index.html [Broken]

So I posted a note there which reflects my current thoughts:

Title: Maxwell Trumps General Relativity

General Relativity rocks. It is elegant in its minimialism. All efforts to add a little extra have failed, usually by allowing a dipole gravity wave mode of emission which has been ruled out by binary pulsar data.

The only field theory that is manifestly better than GR is the Maxwell field equations. Every time we have added to it in the name of symmetry, the theory has done more. James did it himself by tacking on the Ampere current. Einstein looked to get rid of a duplicate law, and so special relativity was born. With the huge supply of new particles coming out of atom smashers, the gauge symmetry in EM (U(1)) was expanded to SU(2) for the weak force, and SU(3) for the strong.

None of those smart cats listed in the initial post will be talking about the Maxwell equations. Too bad, the history of physics is clear: expand Maxwell, you win.

Max depends on the field strength tensor d_u A_v - d_v A_u. There is a subtraction in there, a great thing (called an exterior derivative). But in the name of symmetry, we need to work with the rest of it, d_u A_v + d_v A_u. Do that right, and you get a unified field theory that Einstein failed to find by looking for workable extensions of GR. Extend Max, not GR.

If anyone here wants to see the nuts and bolts of deriving the Maxwell equations using the Euler-Lagrange equations, search for "GEM action" on YouTube. A small variation - two minus signs - on the Maxwell equations leads to equations for gravity. Yes, I show that there is a metric solution (the Rosen metric if you are up on your GR jargon, a bunch of exponentials if not). Yes I know there is an issue of spin 1 and spin 2 which can be addressed if you get what the phase of current coupling really is.

Doug

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Gold Member
Brazil Talk Video

Hello:

I am about to power down the laptop and get it ready for a trip to a foreign land. I will be giving this Thursday, May 29, in Campinas, Brazil at 5:30 pm. Stop by if you are in town. Since I figure about zero of you can do that, I just uploaded a video of the talk to YouTube. The video is 41 minutes long, but I only have a 25 minute slot, so I will have to be more efficient. As always, critiques are appreciated.

I really like the central thesis: that only by doing new math can I do new physics. Will report back on if I can get others excited by this work.

Doug

Bon voyage, Doug.

Lut

I have a brand new theory on the mechanics of gravity I believe the whole science world should review for possible validity.

Gold Member
Hello Sunsphere:

The rules for starting your own thread are here. The rules may appear harsh, but there are many people who make grand claims and the rules are there to filter out ideas that are not precise enough to have a discussion about.

Doug
Brazil is tropical, what can I say?

Gold Member
Report from ICCA8 in Brazil

Hello:

I am back in the US after the trip to the 8th International Conference on Clifford Algebras and the Applications to Mathematical Physics in Campinas Brazil. The exotic location was irrelevant. If the meeting were held in Indiana, it would have made little difference. I went from the plane to the hotel to one building on the campus, and back to the hotel. Sure the vegetation was odd, but when technical talks go from 8:30 AM to 6:30 PM, there was not time or structure to get any exposure to Brazilian culture. Campinas means "grass fields" in Portuguese, a place not far outside San Paulo. thirty years ago it had 100,000 people. Now there are over a million, attracted by high tech companies. When a farm town blows up quickly, the effort is focused on infastructure and not on art and public works. In the opening remarks, one of the organizers said one of the benefits of the meeting was that there was nothing to see in Campinas so we could focus on work.

There were about 70 participants. This was an international conference, with accents in French, Italian, German, British, Polish, and South American (few people from the Far East). Each day had a dozen talks. The morning and early afternoon plenary talks were an hour long. The last afternoon session had half hour talks for the "not invited" who wanted to present their work. This was a PC crowd (I may have had the only Keynote presentation on a Mac). Much of the meeting was video taped, but how they hope to make that available is not known. My own session was not recorded since I was shuffled to the alternate room of a parallel session.

There were many lessons I learned. One of the basic ones is that geometric algebra and Clifford algebra are one and the same (the similarities were clear, but not the formal identity). The story is that David Hestenes has made a big push to bring Clifford algebras onto the center stage of physics. He did not think that the name "Clifford" had information content, preferring to use the name Clifford himself used, geometric algebra, to discuss this area of work.

David attended the meeting. It was clear he was the "star", but he was a decent and approachable guy. It is also clear that despite his efforts, geometric algebra has remained a minor contributor to physics. I could relate to his position (excluding the star part). Quaternions are an even smaller footnote in physics. David said he discovered things using geometric algebra that he needed to translate into tensor lingo to publish. I also find myself translating quaternions insights back to tensors to communicate with the broader physics community.

During a coffee break, I asked him what he thought about quaternions. He said they are just a part of the bigger Clifford algebra (specifically CL(0, 2)). I expected that one, the typical male thing - mine is bigger. I will write a separate post about my critique about such a perspective. He invited me, like I saw him do with others, to grow up and embrace the real math tool of Nature, geometric algebra.

David's talk was great. He brought up an idea proposed by Louis de Broglie, that each atom has an internal clock with a frequency of $B = m c^2/h$. The math is simple but has been largely ignored by the physics community. The frequency is in the zeptoseconds range (if they skipped this in your schooling, it is because it is new, representing a factor of $10^{-21}$). David proposed a way to detect this absurdly fast clock as a resonance in a crystal. The message is darn deep: TIME IS MASS (an ironic aside: I recall Archimedes Plutonium, an Internet crackpot who has passed away, that would always write this IN CAPS).

David did not give a mechanism for the clock. My own work might provide that. In GEM, one has a relativistic 4D wave equation. That has a 1/R2 potential solution, leading to a 1/R3 force law. Dipoles have this sort of potential and force law. Since charged massive particles will both attract and repel their brothers, they are born as dipoles. I have been wondering what those dipoles do. Now I have a candidate: it could be the basis of the de Broglie internal clock and connected to mass. Neat.

Over the course of the next few weeks, I may discuss other talks that have already influenced my research. As for my own talk, I was scheduled to be the next to last talk on the next to last day. I had pitched the talk to a number of people. Most of the people I gave the personal pitch showed up, about 20 folks in the room. This may have been the biggest crowd with good technical credentials I have presented to (Hestenes was in the group).

The thesis was that quaternions, as invented by Hamilton and Rodrigues, cannot make new contributions to physics, but new math with quaternions can lead to new physics. I went through three examples.

First, I have a new definition of a quaternion derivative that splits into a directional derivative along the real line for classical physics, and a normed derivative for quantum mechanics. Getting the derivative of a quaternion derivative on a spacetime manifold might resolve the Bohr/Einstein debate as to why quantum mechanics is different from classical physics.

Second, the analytic animations represents a logical extension of analytic geometry. We can be precise, exact, with the treatment of time's relationship to space. For this crowd, I developed some new animations about cross products which I will discuss in another post.

Third, the Maxwell equations have a better track record at being extendable and connected to quantum mechanics than general relativity. I have figured out how to write the Maxwell action using quaternions. When I wanted to write a symmetric tensor contraction using quaternions, I faced a wall. The way to tunnel through the wall was to find a 4D division algebra where elements commute so long as the Eigen values are excluded. That leads to the GEM action, a way to unify gravity and EM. I showed the solution to the field equations that was physically relevant. There was a slide for the spin of the current coupling term.

The crowd reaction was standard: no one asked a question (other than one clarification). Most other talks had two or three questions. No one came up to ask about the content of the talks during later coffee breaks or over lunch. People were interested in the fact that this was a hobby of mine.

This trip was well worth the time and money, I am glad I went.

Doug

Hi Doug,
thanks for ther report. It's very interesting. The thing is that if we're geometrizing something, then to handle curved spaces you need tensors It's very handy that all the observables are scalar contractions of tensors.

I will have to bone up on geometric algebra, which may be has a different way of doing curved space.

Lut

Gold Member
Curved spacetime quaternions

Hello Lut:

It would take less effort to "bone up" on how to handle curved spacetime with quaternions. In flat spacetime, an event at:

$$event = (t e_0, x_1 e_1, x_2 e_2, x_3 e_3) \quad eq~1$$

where:

$$|e_0| = |e_1| = |e_2| = |e_3| = 1 \quad eq~2$$

$$e_0^2 = +1$$
$$e_1^2 = -1$$
$$e_2^2 = -1$$
$$e_3^2 = -1\quad eq~3-6$$

In spacetime curved by a static, spherically symmetric mass charge, equation 1 remains the same, while the other ones are altered:

$$|e_0| = \frac{1}{|e_1|} = \frac{1}{|e_2|} = \frac{1}{|e_3|} = exp(-G M/c^2 R) \quad eq~7$$

$$e_0^2 = +exp(-2 G M/c^2 R)$$
$$e_1^2 = -exp(+2 G M/c^2 R)$$
$$e_2^2 = -exp(+2 G M/c^2 R)$$
$$e_3^2 = -exp(+2 G M/c^2 R)\quad eq~8-11$$

This might not be too hard to implement, now that I have written it down clearly (the key step to programming). q_metric -mass 10e-6 1 2 3 4 would return the event (1, 2, 3, 4) in a space curved by a 10e-6 mass.

The problem with my software as written is handling this very tiny numbers.

Doug

Hi Doug,
yes, one way of losing tensors is to use differential geometry ( if I haven't got my nomenclatures wrong). I think that's what you did above, by introducing basis vectors in your your quaternion space. Makes sense.

What technology are you using for computing ? I have floating point down to 10^-300.

Lut

Gold Member
Precision in C

Hello Lut:

I am using C. I print to the command line which by default prints to 8 digits. I bet I can adjust printf to do more. I am not certain how to set the internal precision accurately of a standard C program.

I have always had the basis vectors, but had not put them to use in a calculation.
Doug

Hi Doug,

sounds like you need a better C compiler.

You seem to have all the apparatus for space-time in your algebra. Letting gravity appear in the basis and not the coordinates is a good practice.

When you conjugate this

$$event = (t e_0, x_1 e_1, x_2 e_2, x_3 e_3) \quad eq~1$$

do you change sign and invert the basis vectors ? I'm thinking about a conjugate space, and what scalars you can make by multiplying a vector times a conjugate ?

Lut

Gold Member
Conjugates

Hello Lut:

That is a great question! My initial reaction was the signs just flip. The thought is that conjugation is about the work of mirrors. In flat spacetime, three sign flips is all one needs to do a mirror reflection. If the mirror is curved, I am not sure if flipping the sign is enough. I'll have to think about it some more.

I use the gcc compiler. I am sure there is a way to make the precision high, but at this time I don't know how to do it. What is your programming trick/tools to get such high precision?

Doug

Hi Doug,
I was free thinking there. I'll have to ponder it too.

I use Delphi Win32 and I've got the following floating type amongst others

Extended 3.6 x 10^-4951 .. 1.1 x 10^4932 digits 19-20 uses bytes 10

10^-4951 !

Lut

Gold Member
Precision in gcc

Hello Lut:

I figured out how to crank up the precision. The default single precision float has about 7 significant digits. At double precision, there are 16 significant digits. All I have to do is set the -DFmode flag and set all programs to print 16 digits. There is the third level, -XFmode which will do 24 digits, but that may just be wasteful of resources, using 12 bytes. There is a Tetra Floating number for those that need 16 bytes.

Thanks for getting me to look this up.
Doug

Gold Member
Quaternions versus Clifford algebras

Hello:

Quaternions as events in spacetime physics are not the Clifford algebra CL(0, 2). Let me first explain what that bit of jargon means.

Clifford algebras were an attempt to generalize the math started by Hamilton, passing through Grassman. These algebras are independent of coordinates, and can be written in arbitrary dimensions. As algebras, it is not important that they are invertible, although that can happen.

The Clifford algebra CL(0, 2) is a multivector, meaning it has two parts. First there is the scalar, the 0 part, a pure number. The bivector is the second part. They use a wedge product, $\wedge$, a generalized cross product to form a 3-term bivector.

$$q(t, x_1, x_2, x_3) = t ~+~ x_1 e_2 \wedge e_3 ~+~ x_2 e_3 \wedge e_1 ~+~ x_3 e_1 \wedge e_2 \quad eq~1$$

Why use a wedge product? The reason is that wedge products are axial vectors, those that switch handedness in a mirror. The other sort is a polar vector. If one looks at an event in a mirror, this makes physical sense.

It is not worth the time to argue with math types who think they can define whatever they want, however they see fit. I have the perspective of a mathematical physicists, where any math definition can be cow-roped by physical meaning (I am a crude mathematical physics who uses rodeo analogies).

The unspoken assumption behind CL(0, 2) is that one should use a mirror on this event. That is like taking (t, x1, x2, x3) to (t, -x1, -x2, -x3). That sort of transformation does have the handedness needed for an axial vector. Yet it is easy enough to think of counter examples. What about a time reflection, where an event (t, x1, x2, x3) goes to (-t, x1, x2, x3)? This transformation does not change handedness, it is represented by a polar vector. Compare the two functions

mirror reflection: q -> q' = q*
time reflection: q -> q' = -q*

Since these are so close to each other as functions, I don't think one should take precedence over the other, as asserting one must use a bivector for the spatial part of a quaternion does.

The fundamental currency of the Universe is a bare event which stands alone in a vacuum, not in front of a mirror waving a left hand. General relativity has a message about basis vectors. Should we have a toy Universe with two events, the interval between these two events is found by taking the dot product of the difference between the two events:

$$(c dt, dx_1, dx_2, dx_3).(c dt, dx_1, dx_2, dx_3) = c^2 dt^2 ~+~ dx_1^2 (e_2 \wedge e_3).(e_2 \wedge e_3) ~+~ dx_2^2 (e_3 \wedge e_1).(e_3 \wedge e_1) ~+~ dx_3^2 (e_1 \wedge e_2).(e_1 \wedge e_2)$$

$$= c^2 dt^2 ~-~ dx_2^2 ~-~ dx_2^2 ~-~ dx_2^2 \quad eq~2$$

So far, so good. Now repeat the measurement in a toy Universe where nothing has been altered for the generators of the two events, but one has added a mass. A metric theory of gravity does not alter events, but does change the measure of events. In this setup, the differences are the same. The only thing that could change are the sizes of the basis vectors themselves. The problem here is that there is no basis vector associated with the dt2 term.

The deep message of special relativity is to treat time as we do space. In the geometric algebra approach, time is a scalar, whereas the space parts use two basis vectors as a bivector. Contrast this with the way I define a quaternion:

$$q(t, x_1, x_2, x_3) = t e_0 + x_1 e_1 + x_2 e_2 + x_3 e_3 \quad eq~3$$

This is not a Clifford algebra because the basis vector e_0 commutes with the others. In the definition of a Clifford algebra, all the basis elements anti-commute. Physicists would call this a 4-vector, because one can add them or multiply it by a scalar, and it transforms like a 4-vector. It is this last phrase, on how it transforms under a Lorentz boost, that holds the magic. The goal of a great definition is to remove the magic and let the math speak for itself. Square the difference between two events using the quaternion definition as written in equation 3:

$$(c dt, dx_1, dx_2, dx_3)^2 = (c^2 dt^2~ e_0^2 ~+~ dx_1^2 ~e_1^2 ~+~ dx_2^2 ~e_2^2 ~+~ dx_3^2 ~e_3^2, 2 c ~dt ~dx_1~ e_0 ~e_1, 2 c ~dt ~dx_2 ~e_0 ~e_2, 2 c ~dt ~dx_3 ~e_0 ~e_3) \quad eq~4$$

To be consistent with special relativity, we make the following map:

$$e_0^2 = +1$$
$$e_1^2 = -1$$
$$e_2^2 = -1$$
$$e_3^2 = -1 \quad eq~5-8$$

To be consistent with general relativity to first order tests of weak gravity fields for non-rotating, spherically symmetric sources, all we need is:

$$e_0 = exp(-G M/c^2 R)$$
$$1/e_0 = e_1 = e_2 = e_3 \quad eq~9-10$$

so according to the GEM proposal,

$$e_0^2 = exp(-2 G M/c^2 R)$$
$$e_1^2 = -exp(2 G M/c^2 R)$$
$$e_2^2 = -exp(2 G M/c^2 R)$$
$$e_3^2 = -exp(2 G M/c^2 R) \quad eq~11-15$$

I forget who said it, but one can only hope to find an important law in physics if one also finds an invariance principle. When there is a gravitational source, the terms in the 3-vector part of the square (2 dt dx e_0 e_1, 2 dt dy e_0 e_2, 2 dt dz e_0 e_3) remain invariant under the presence of such a gravitational field.

In summary, I see two flaws to the assertion that quaternions are the Clifford algebra CL(0, 2). First, while some quaternions are axial vectors, others are polar vectors. Second, the scalar needs to have a basis vector associated with it to work with metric theories of gravity. As far as I can tell, everyone working with geometric algebra in physics works with the assumption that the quaternions are faithfully represented by CL(0, 2).

Doug

Gold Member
The geometric product

Hello:

Today I will discuss the geometric product, a tool used extensively for people applying Clifford algebras to problems in physics and robotics. Here is the definition:

$$ab = a.b ~+~ a \wedge b \quad eq~1$$

There are a number of great features to this definition. It will work for whatever dimension you have in a Clifford algebra. The product is conformal, meaning angles do not change. The dot product is the magnitude of a time b times the cosine of the angle between a and b, while the wedge is the magnitude of a time b times the sine of the angle between a an b. How clean!

The road to Clifford's geometric product is convoluted. It started with Hamilton writing out the 4D product one day, carving it into a bridge, and the very next day working to get rid of one dimension with "pure quaternions" that had a zero scalar. The scalar that looked completely alien to Hamilton for the square of a quaternion, $t^2 ~-~ x^2 ~-~ y^2 ~-~ z^2$, looks familiar to us as the interval of special relativity. By setting t=0, one gets something the Egyptians would have recognized instead of Einstein, so long as one ignores a pesky minus sign.

Yet that minus sign did bother people. The nice thing about a.b that appears in the geometric product is that all the signs are the same and positive (fears of minus signs have wasted more time historically than almost any other math issue). It was Grassman that focused on the wedge product, the new math to come out of the quaternions. He built up an entire system of algebra using basis vectors and wedge products. Clifford put the dot product together with the wedge product to make the geometric product. All is good...

...unless you are a stickler for technical details. Hamilton developed quaternions that are defined for one scalar dimension, and three spatial vectors. One starts with 4 numbers. Look for the product of those four numbers with another set of 4 numbers, and the complete story necessarily has 16 parts. The geometric algebra only has 10. Oops. What it is missing is a scalar times a 3 vector, done twice. This is the math omission needed to make the product conformal - there is no vector stretching.

The geometric product, written as a quaternion operation, looks like so:

$$ab = (a^* b ~+~ b^* a)/2 ~+~ (a b ~-~ b a)/2 \quad eq~2$$

Recall the history. Quaternions were used first to form a product that always had an inverse. Grassman isolates on the wedge, then Clifford glues things back together. The result to my eye does not look pretty. The geometric product makes sense if you fear the unknown Lorentz invariant interval, but that is not where we are today. The conformal nature of the geometric product looks to me like a rigged outcome because the 6 problem terms are dropped.

I understand why people have put a huge investment in the geometric product, yet it does not look like a safe bet to me.

Doug

Hi Doug,

most illuminating. What exactly does this

The geometric product makes sense if you fear the unknown Lorentz invariant interval,
say ?

Lut