Unifying Gravity and EM

by sweetser
Tags: gravity, unifying
 P: 4 (translation from french observation) I think a new theory for source gravity. Maybe the origin of gravity is an oscillation of electrostatic wave. The oscillation of each particle is very very small compared an electron charge. Each particle send a wave, and each wave is accumulate to another wave. The charge of a particle is 0 but the wave is something around 0, one moment + and later - etc. So, each particle send + and - wave around it. It's possible to synchronize all wave because there is a difference between forces if you imagine two sinusoide waves face to face, there is two waves very close and two waves far, the force is with 1/d² so the difference can create a synchronism. This can explain deviation of light and redshift. The formula of perihelion precession of Mercury is the same with electrostatic forces integrate in the formula. I have create a site for explain (in french for the moment) my ideas: 3w eisog dot com Maybe this idea is not new, tell me ... Ludovic
 P: 190 The breaking of Lorentz symmetry has been proposed by some. I am not sure how I stand on this for certain --- leaning against the idea I suppose. One can work with all sort of pseudotensors and the like with this, though I am not sure if that is what Sweetser is doing. Gauge gravity, or in a general form with the B-F or Plebanski Lagrangian, recovers Einstein's gravitation at least for weak $\alpha^2~\sim~G$, though corrections on this for stronger G and higher energy (also I suspect cosmological scales) probably lead to consequences beyond a naive breaking of Lorentz symmetry. Lawrence B. Crowell
 P: 361 Hello Lawrence: The GEM proposal does not appear to break Lorentz symmetry. This issue of tensors and GEM remains a concern. We know that $\nabla_{\mu} A_{\nu}$ transforms like a tensor because that is why the machinery of a covariant derivative was built. We also know that the field strength tensor of EM transforms like a tensor, $\nabla_{\mu} A_{\nu} ~-~ \nabla_{\nu} A_{\mu}$. Correct me if I am wrong, but I believe that the difference of two tensors of the same order is also a tensor. If that is the case, then: $$\nabla_{\mu} A_{\nu} - \frac{1}{2}(\nabla_{\mu} A_{\nu} ~-~ \nabla_{\nu} A_{\mu})$$ $$= \frac{1}{2}(\nabla_{\mu} A_{\nu} ~+~ \nabla_{\nu} A_{\mu})\quad eq ~1$$ Is it fair to call this object with the plus sign a tensor? Sure would hope so. Yes all three have different properties, but they all look like they belong in the family of tensors, not pseudotensors. Doug
HW Helper
P: 1,204
 Quote by Lawrence B. Crowell Gauge gravity, or in a general form with the B-F or Plebanski Lagrangian, recovers Einstein's gravitation at least for weak $\alpha^2~\sim~G$, though corrections on this for stronger G and higher energy (also I suspect cosmological scales) probably lead to consequences beyond a naive breaking of Lorentz symmetry.
This is a different gauge gravity than that of the Cambridge geometry group. I guess "gauge" is a cool word to use to describe a theory. The Cambridge version is identical to GR to all orders, at least locally. Where it differs is in that you can't do weird topological things like wormholes. Here's how David Hestenes puts it:

The specialization of GA [i.e. Geometric Algebra] to Minkowski spacetime is called Spacetime Algebra (STA). As explained below, STA clarifies the geometric significance of the Dirac Algebra and thereby extends its range of application to the whole of physics. In particular, STA provides the essential mathematical framework for the new Gauge Theory Gravity (GTG).

The foundations of GTG are fully expounded in a seminal paper, so we can concentrate
on highlights of its unique features. GTG is a gauge theory on Minkowski spacetime, but locally it is equivalent to General Relativity (GR), so it can be regarded as an alternative formulation of GR.8 However, GTG reformulates (or one might say, replaces) Einstein’s vague principles of equivalence and general relativity with sharp gauge principles that have clear physical consequences (Section IV). These gauge principles are more than mere rephrasing of Einstein’s ideas. They lead to intrinsic mathematical methods that simplify modeling and calculation in GR and clarify physical meaning of terms at every stage. In particular, they provide clean separation between gauge transformations and coordinate transformations, thus resolving a point of longstanding confusion in GR. Moreover, GTG simplifies and clarifies the analysis of singularities, for example, in assignment of time direction to a black hole horizon. Finally, since tensors and spinors are fully integrated in STA, GTG unifies classical GR and relativistic quantum mechanics with a common system of gauge principles. Besides facilitating the application of quantum mechanics to astrophysics, this opens up new possibilities for a grand unification of gravitation and electroweak theories, as explained in Section V.
http://modelingnts.la.asu.edu/pdf/procGTG-RQM.pdf

So far, GR has only been verified to 1-PN. I don't think that it will make 3-PN. What I like about Sweetser's gravity is that it seems natural to do as the consequences of a particle theory. The problem I see in doing this is that to end up with a variable speed of light type gravity theory requires that the force of gravity be carried by superluminal particles.

In other words, just as Maxwell's equations end up being a force mediated by a speed of light particle, a theory which produces a variable speed of light needs to be mediated by something that moves at some higher speed. Comments from Doug?
 P: 361 Hello Carl: Your email inspired me to go make another effort to wrap my head around geometric algebra. David Hestenes will be giving a keynote address at the conference in Brazil. Both Clifford algebras and geometric algebra fans view there tools as more general that quaternions, since quaternions are just Cl(0, 2) or a 0,2-multivector. There is not much difference between these three approaches, but there is some. I think of the basis vectors differently, due to looking at this with my physics glasses on. A quaternion is described as a 0,2 multivector, which means a scalar for the first part, and three bivectors, e2/\e3, e3/\e1, and e2/\e1 for the 3-vector. There are two issues the physicists in me objects. First, the main message of special relativity is to bring the scalar in with the 3-vector on equal footing. I look at quaternons at being 4 scalars with 4 basis vectors: (a0 e0, a1 e1, a2 e2, a3 e3). The GA folks like the wedge because it explains the minus signs. I find that claim hollow since is merely shift where the minus comes in to the wedge itself. In my view of quaternions, the four parts look like equals. The second objection comes from my view of gravity. In GEM, gravity can be about the potential, the basis vectors, or a combination of both. When it is about the basis vectors, then I want to be able to shift e0, e1, e2, and e3. For the exponential metric, e0 = exp(-2 GM/c^2 R) and e0 e1 = e0 e2 = e0 e3 = 1. I don't see how I could make similar statements with the GA formalism. In the cited paper, Hestene explains how to do the Dirac algebra using geometric algebra. The details are different from how I do that with quaternions as triple products (all sixteen possibilities for e_u Q e_v make up the action of the gamma matrices). My guess is the GA crowd will not be happy with the even representation of quaternions. I will get to find out, since that will lead off my 25 minute talk in Brazil. The Cambridge geometry group looks like it is trying to recreate general relativity, with some more polite features. The GEM work is in line with a long tradition of directly confronting the dominant idea of the day, saying it is good up to 1-PN, but not 2-PN. One of those binary pulsar guys was at the APS meeting. I asked him if it is correct to take the equations developed for super-weak perihelion precession and apply it to a strong, dynamic system. If he writes back, I'll commmunicate to this thread. I stay away from "faster than light". Using quaternions for everything is radical enough! Doug
P: 190
 Quote by CarlB This is a different gauge gravity than that of the Cambridge geometry group. I guess "gauge" is a cool word to use to describe a theory. The Cambridge version is identical to GR to all orders, at least locally. Where it differs is in that you can't do weird topological things like wormholes. Here's how David Hestenes puts it: The specialization of GA [i.e. Geometric Algebra] to Minkowski spacetime is called Spacetime Algebra (STA). As explained below, STA clarifies the geometric significance of the Dirac Algebra and thereby extends its range of application to the whole of physics. In particular, STA provides the essential mathematical framework for the new Gauge Theory Gravity (GTG).
Spinor algebra can reproduce general relativity in a pretty straightforward manner. The Dirac matrices $\gamma_\mu$ define the metric as

$$\gamma_\mu\gamma_\nu~=~\frac{1}{4}g_{\mu\nu}$$

and so if the representation of the spinor basis is local or chart dependent one can pretty easily reproduce GR with the Dirac operator. A gauge invariant form of the Dirac operator would then be

$$\partial_\mu(\gamma^\mu\psi)~=~\gamma^\mu D_\mu\psi~=~\gamma^\mu(\partial_\mu~+~A_\mu)\psi,$$

for $A_\mu$ a gauge term for GR.

For higher Clifford basis elements or with vierbiens extended GR theories can be derived. I suspect these are related in some ways to the string theory for GR which obtains for scales larger than the string length, but has deviations for scales approaching the string length. Of course there are some conformal invariance abuses with the string theoretic approach to GR as a bimetric form of theory. Yet string theory as a form of math-method does have some suggestive elements to it.

It is my sense that GR will survive to very high PN orders. I think deviations from GR obtain on two complementary scales: one scale near the Planck scale or say $\le~10^3\sqrt{G\hbar/c^3}$, and the other is on the cosmological scale where time translation invariance is "deformed" and there are inequivalent vacua states on scales across the cosmological event horizon. The verdict on this will likely come from gravity wave detection, such as LIGO, and the connection between these measurements with astrophysical events.

Lawrence B. Crowell
 Sci Advisor HW Helper P: 1,204 Doug, I'm swamped right now with correspondence and I don't have internet connectivity at home so let me postpone writing more about what you've said here until later. But here's a link on another exponential gravitation theory. Can you talk about how this differs from your own? [Uh, read down to the end of the article] http://www.insidegnss.com/node/451 Carl
P: 190
 Quote by lba7 (translation from french observation) This can explain deviation of light and redshift. The formula of perihelion precession of Mercury is the same with electrostatic forces integrate in the formula. I have create a site for explain (in french for the moment) my ideas: 3w eisog dot com Maybe this idea is not new, tell me ... Ludovic
What is possible is that the spin quantum Hall effect might result in deviations from the equivalence principle. A circularly polarized photon as the superposition of two polarization states might be split by the effective index of refraction due to a gravity field. This would mean that gravitation has a Berry phase or a topological index associated with the spin of particles. This would be a possible deviation from classical general relativity.

An Einstein lens might provide the way to detect this. A radio telescope that detects photons from a lensed source will for a very narrow band pass filter measure quantum effect for entangled photons in this cosmic beam splitter. In principle the Wheeler Delayed Choice experiment could be performed this way. Also there might be for the two arms of this cosmic beam splitter a spin dependency in how the photons are split.

Lawrence B. Crowell
 P: 361 Hello: I can finally write about the New England American Physical Society Meeting on Saturday, April 5. Lut can skip this post since it involves the personal stories. The three posts that follow in quick succession will be technical calculations. My hopes were not high for this local meeting at the Coast Guard Academy. It was held at a military facility, with an official entry gate. People were marching around all dressed to the military nines at 8 am. One positive aspect of the military is they get to wear sharp hats to work, hats that indicate one's station. The usual cast of characters where at this APS meeting. There were a dozen people in the room, half of whom I was familiar with. The session started off with a familiar fringe guy. As far as I can tell, he claims that everything is in motion. As a skeptic, I attempt to see what notions I need to stretch to see some grain of truth in his near incomprehensible riffs. He claimed that everything was in motion. What I would say is everything moves in spacetime, mostly as time, as things continue to persist. He does make fun of himself a little which helps, otherwise his talks are tragic. Larry Gold gave his "Let's doubt global warming". It reminds me of Fox News in the US which claims to be unbiased, then makes every effort to selectively sample data needed to support its position. My talks was titled: "Using Quaternions for Lagrangians in EM and GEM Unified Field Theory." For an audience this tiny and quirky, I could not expect them to know about Lagrangians, or how to use them to derive field equations. I had thought about doing all the equations in LaTeX, but decided to use a fountain pen, good old black ink written by hand, to indicated that despite the complications, this sort of calculation can be done by a person. These sheets of paper packed full of partial differential equations are prized possessions. It looks so hard core. Two of these sheets - a derivation of the Maxwell equations from the Lagrangian, and a derivation of the gravity part of GEM - were at the core of the talk (the rest being a setup for the heavy duty math). The talk had one technical glitch. My connecting wire is flaky so everything was yellow. Otherwise, the speech went according to plan. My talks often do not elicit questions, but this time someone started asking about the symmetric curl. It was like he was channeling Lawrence. He was wondering why I had all these coordinate dependent equations. After the session broke up, I asked him some more questions, trying to see what he found unsettling. It certainly is my intension to write coordinate-independent equations that are valid in flat or curved spacetime. When I write a quaternion as "q", it could be in any coordinate system. When I write $\nabla q$, that is a covariant derivative, the same whether spacetime is flat or curved, independent of coordinates. When I went to write out the derivative by its component parts, I invariably wrote things in terms of Cartesian coordinates, the rectilinear t, x, y, and z. The progression of symbols for derivatives goes something like this: $$\frac{d}{dt} \rightarrow \frac{\partial}{\partial t} \rightarrow \nabla_0 \quad eq~1$$ Only the last symbol is independent of coordinates, valid in flat or curved spacetime. I had never seen the Maxwell equations derived in a coordinate-free manifestly covariant way, so it was natural for me to just use partial derivatives with respect to t, x, y, and z. Based on this criticism, I will be changing how I write component expressions. I will only use subscripts 0-3, not t-z. This way, equations that I think of as being manifestly covariant will have the notation required to be manifestly covariant. For the record, I will repeat the quaternion operator derivations for the Maxwell equations, the gravity part of GEM, and the GEM unified field equations using the covariant notation in the next three posts. Doug Stills: http://picasaweb.google.com/dougswee...FieldEquations The talk: http://www.youtube.com/watch?v=Kd4nNb7nGOc
 P: 361 Hello: In this post I will derive the Maxwell field equations using quaternion operators in a manifestly covariant notation. Notice that the quaternion differential operator acting on a 4-potential creates a scalar gauge field, the -E and B fields: $$\nabla A = (\nabla_0 \phi ~-~ \nabla \cdot A, \nabla_0 A ~+~ \nabla_u \phi ~+~ \nabla \times A) = (g, -E ~+~ B)\quad eq ~1$$ The starting point for the derivation of the Maxwell 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 covariant differential operator with respect to the 4-potential, which flips the sign of the curl (B), but not the time derivative of A or gradient of phi which make up E. The scalar gauge field can be subtracted away: $$-\frac{1}{8}(\nabla A ~-~ (\nabla A)^*)(A \nabla ~-~ (A \nabla)^*)$$ $$=\frac{1}{8}(0, -\nabla_0 A ~-~ \nabla_u \phi ~-~ \nabla \times A)(0, \nabla_0 A ~+~ \nabla_u \phi ~-~ \nabla \times A)$$ $$=\frac{1}{2}(0, E ~-~ B)(0, -E ~-~ B) = \frac{1}{2}(E^2 ~-~ B^2, -2 E \times B)\quad eq ~2$$ 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 know the 3-vector part of this expression is the Poynting vector, a conserved quantity in EM. Write out the Lagrange density - the scalar part of eq 2 - in terms of its components, including the vector coupling, $-\frac{1}{2 c}(J A + (J A)^*)$: $$\mathcal{L}_{EB} = \frac{1}{2}(-(\nabla_1 \phi)^2 ~-~(\nabla_2 \phi)^2 ~-~(\nabla_3 \phi)^2 ~-~ (\nabla_0 A_1)^2 ~-~ (\nabla_0 A_2)^2 ~-~ (\nabla_0 A_3)^2$$ $$~+~ (\nabla_3 A_2)^2 ~+~ (\nabla_2 A_3)^2 ~+~ (\nabla_1 A_3)^2 ~+~ (\nabla_3 A_1)^2 ~+~ (\nabla_2 A_1)^2 ~+~ (\nabla_1 A_2)^2)$$ $$~-~ (\nabla_3 A_2)(\nabla_2 A_3) ~-~ (\nabla_1 A_3)(\nabla_3 A_1) ~-~ (\nabla_1 A_2)(\nabla_2 A_1) ~-~ (\nabla_1 \phi)(\nabla_0 A_1) ~-~ (\nabla_2 \phi)(\nabla_0 A_2) ~-~ (\nabla_3 \phi)(\nabla_0 A_3)$$ $$-\rho \phi ~+~ J_1 A_1 ~+~ J_2 A_2 ~+~ J_3 A_3 \quad eq ~3$$ Apply the Euler-Lagrange equation to this Lagrangian, treating the 4-potential as the variable: $$\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} \phi)}) = -\nabla_1^2 \phi ~-~ \nabla_2^2 \phi ~-~ \nabla_3^2 \phi ~-~ \nabla_0 \nabla_1 A_1 ~-~ \nabla_0 \nabla_2 A_2 ~-~ \nabla_0 \nabla_3 A_3 ~-~ \rho$$ $$= \nabla \cdot E - \rho = 0 \quad eq ~4$$ $$\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} A_1)}) = -\nabla_0^2 A_1 ~+~ \nabla_3^2 A_1 ~+~ \nabla_2^2 A_1 ~-~ \nabla_1 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_2 ~-~ \nabla_0 \nabla_1 \phi ~+~ J_1$$ $$= \nabla_0 E_1 - (\nabla \times B)_1 + J_1 = 0 \quad eq ~5$$ $$\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} A_2)}) = -\nabla_0^2 A_2 ~+~ \nabla_3^2 A_2 ~+~ \nabla_1^2 A_2 ~-~ \nabla_2 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_1 ~-~ \nabla_0 \nabla_2 \phi ~+~ J_2$$ $$= \nabla_0 E_2 - (\nabla \times B)_2 + J_2 = 0 \quad eq ~6$$ $$\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} A_3)}) = -\nabla_0^2 A_3 ~+~ \nabla_2^2 A_3 ~+~ \nabla_1^2 A_3 ~-~ \nabla_2 \nabla_3 A_2 ~-~ \nabla_1 \nabla_3 A_1 ~-~ \nabla_0 \nabla_3 \phi ~+~ J_3$$ $$= \nabla_0 E_3 - (\nabla \times B)_3 + J_3 = 0 \quad eq ~7$$ This work can be summarized with the Maxwell source equations: $$\nabla \cdot E = \rho \quad eq 8$$ $$\nabla \times B ~-~ \nabla_0 E = J$$ This has been a manifestly covariant derivation of the Maxwell equations. 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: http://picasaweb.google.com/dougswee...FieldEquations The talk: http://www.youtube.com/watch?v=Kd4nNb7nGOc
 P: 361 Hello: In this post I will derive the Maxwell field equations using quaternion operators in a manifestly covariant notation. Notice that the quaternion differential operator acting on a 4-potential creates a scalar gauge field, the -E and B fields: $$\nabla A = (\nabla_0 \phi ~-~ \nabla \cdot A, \nabla_0 A ~+~ \nabla_u \phi ~+~ \nabla \times A) = (g, -E ~+~ B)\quad eq ~1$$ The starting point for the derivation of the Maxwell 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 covariant differential operator with respect to the 4-potential, which flips the sign of the curl (B), but not the time derivative of A or gradient of phi which make up E. The scalar gauge field can be subtracted away: $$-\frac{1}{8}(\nabla A ~-~ (\nabla A)^*)(A \nabla ~-~ (A \nabla)^*)$$ $$=\frac{1}{8}(0, -\nabla_0 A ~-~ \nabla_u \phi ~-~ \nabla \times A)(0, \nabla_0 A ~+~ \nabla_u \phi ~-~ \nabla \times A)$$ $$=\frac{1}{2}(0, E ~-~ B)(0, -E ~-~ B) = \frac{1}{2}(E^2 ~-~ B^2, -2 E \times B)\quad eq ~2$$ 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 know the 3-vector part of this expression is the Poynting vector, a conserved quantity in EM. Write out the Lagrange density - the scalar part of eq 2 - in terms of its components, including the vector coupling, $-\frac{1}{2 c}(J A + (J A)^*)$: $$\mathcal{L}_{EB} = \frac{1}{2}(-(\nabla_1 \phi)^2 ~-~(\nabla_2 \phi)^2 ~-~(\nabla_3 \phi)^2 ~-~ (\nabla_0 A_1)^2 ~-~ (\nabla_0 A_2)^2 ~-~ (\nabla_0 A_3)^2$$ $$~+~ (\nabla_3 A_2)^2 ~+~ (\nabla_2 A_3)^2 ~+~ (\nabla_1 A_3)^2 ~+~ (\nabla_3 A_1)^2 ~+~ (\nabla_2 A_1)^2 ~+~ (\nabla_1 A_2)^2)$$ $$~-~ (\nabla_3 A_2)(\nabla_2 A_3) ~-~ (\nabla_1 A_3)(\nabla_3 A_1) ~-~ (\nabla_1 A_2)(\nabla_2 A_1) ~-~ (\nabla_1 \phi)(\nabla_0 A_1) ~-~ (\nabla_2 \phi)(\nabla_0 A_2) ~-~ (\nabla_3 \phi)(\nabla_0 A_3)$$ $$-\rho \phi ~+~ J_1 A_1 ~+~ J_2 A_2 ~+~ J_3 A_3 \quad eq ~3$$ Apply the Euler-Lagrange equation to this Lagrangian, treating the 4-potential as the variable: $$\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} \phi)}) = -\nabla_1^2 \phi ~-~ \nabla_2^2 \phi ~-~ \nabla_3^2 \phi ~-~ \nabla_0 \nabla_1 A_1 ~-~ \nabla_0 \nabla_2 A_2 ~-~ \nabla_0 \nabla_3 A_3 ~-~ \rho$$ $$= \nabla \cdot E - \rho = 0 \quad eq ~4$$ $$\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} A_1)}) = -\nabla_0^2 A_1 ~+~ \nabla_3^2 A_1 ~+~ \nabla_2^2 A_1 ~-~ \nabla_1 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_2 ~-~ \nabla_0 \nabla_1 \phi ~+~ J_1$$ $$= \nabla_0 E_1 - (\nabla \times B)_1 + J_1 = 0 \quad eq ~5$$ $$\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} A_2)}) = -\nabla_0^2 A_2 ~+~ \nabla_3^2 A_2 ~+~ \nabla_1^2 A_2 ~-~ \nabla_2 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_1 ~-~ \nabla_0 \nabla_2 \phi ~+~ J_2$$ $$= \nabla_0 E_2 - (\nabla \times B)_2 + J_2 = 0 \quad eq ~6$$ $$\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} A_3)}) = -\nabla_0^2 A_3 ~+~ \nabla_2^2 A_3 ~+~ \nabla_1^2 A_3 ~-~ \nabla_2 \nabla_3 A_2 ~-~ \nabla_1 \nabla_3 A_1 ~-~ \nabla_0 \nabla_3 \phi ~+~ J_3$$ $$= \nabla_0 E_3 - (\nabla \times B)_3 + J_3 = 0 \quad eq ~7$$ This work can be summarized with the Maxwell source equations: $$\nabla \cdot E = \rho \quad eq ~8$$ $$\nabla \times B ~-~ \nabla_0 E = J \quad eq ~9$$ This has been a manifestly covariant derivation of the Maxwell equations. 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: http://picasaweb.google.com/dougswee...FieldEquations The talk: http://www.youtube.com/watch?v=Kd4nNb7nGOc
 P: 361 Hello: 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. 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: http://picasaweb.google.com/dougswee...FieldEquations The talk: http://www.youtube.com/watch?v=Kd4nNb7nGOc (this derivation was not part of the talk)
 P: 361 Hello: 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. 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: http://picasaweb.google.com/dougswee...FieldEquations The talk: http://www.youtube.com/watch?v=Kd4nNb7nGOc
 P: 361 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
P: 361
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:
 Quote by Feynman ...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
 P: 361 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 slide from a talk. 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
 P: 361 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
 P: 193 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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