How Can Gravity and Electromagnetism Be Unified Through a Rank 1 Field Theory?

[sweetser]

Managing the Even representation

Hello Lut:

You were only suppose to get the cross terms :-) I scanned in my hand-drawn derivations of the field equations here:

http://picasaweb.google.com/dougsweetser/MaxwellFieldEquations/photo#5171869024067390082

The step that may be tripping up your program is multiplying the even quaternion representation that is (0, b + e)(0, b - e) should give the scalar b² - e², not the negative of this. My bet is B² - E² - b² + e² would work in your software as is.

Doug

 

[Mentz114]

Doug:

Yes, I changed a sign and now I get

B^2 - E^2 - b^2 + e^2 =

8\partial^xA^2\partial^yA^1<br /> +8\partial^xA^3\partial^zA^1<br /> +8\partial^yA^3\partial^zA^2<br /> -8\partial^tA^1\partial^xA^0<br /> -8\partial^tA^2\partial^yA^0<br /> -8\partial^tA^3\partial^zA^0
(apologies for mixed notation, but I'm sure you get the drift)
In agreement ( up to a sign) with your doodle. I won't repeat the algebra to get the field equations.

I suppose the only point of this is that one can start with the traceless field tensors and go from there to get the same formulation as your funny quaternions. Much easier for the lay ( non-quat.) person to grasp.

Lut

PS : before I closed down the calculator I did this -

g_{ik}F_{(g)}^{mk}g_{mn}F_{(em)}^{ni}. It comes to zero, zip, nothing.

So the field tensors are orthogonal in a way.

 

[Mentz114]

Another way to lose g

Doug:

there is a neat way to lose g. The symmetric field tensor has a dual, in analogy with EM theory, defined thus -

\tilde{F_g}_{mn} = s_{mnij}F_g^{mn}

where s is the totally symmetric pseudo-tensor equal to |e|. The dual has no diagonal terms and

\tilde{F_g}_{mn}\tilde{F_g}^{mn} = -8(b^2 - e^2)

So, putting the square of the dual in the Lagrangian density does the job.

The symbol s also allows the symmetric curl of a vector to be defined as s_{ijk}\partial^{j}E^{k}

Lut

 

[Lawrence B. Crowell]

Doug:


[later] I seem to be getting

\Box^2A^{\mu} = 0

Too tired to continue right now. I've got a feeling I've made a meal of something simple.

Which is the Lorentz gauge. If you put the g in there and let this be something "dynamical" then this inserts group dependent structure into the theory. The elliptic complex determines the bundle curvatures \Omega^{2}(ad(g)) which give fields "mod-g," for here g means group. The gauge condition is set to constrain the appropriate variables.

Lawrence B. Crowell

 

[sweetser]

Orthogonal tensors

Hello Lut:

I concur with the postscript statement:

g_{ik}F_{(g)}^{mk}g_{mn}F_{(em)}^{ni} It comes to zero, zip, nothing. So the field tensors are orthogonal in a way.

When I thought about GEM in terms of tensors, the symmetric one was the average amount of change in the 4-potential in 4 spatial dimensions, while the EM tensor is the deviation from the average amount of change. My recent shift to quaternions puts a stress on the underlying math tools, the Hamilton versus Even representations. At this time I don't understand all the links between these two perspectives, but it is nice having multiple perspectives. Thanks for adding to my views via this calculation.

Doug

 

[sweetser]

Hello Lut:

I am not clear on post #514. F_{(g)}^{\mu\nu} has nothing down its diagonal - where the terms of a gauge live - so its dual also has this property. That appears to be starting too far down the page as it were.

Here is what I am doing with quaternions, written as your tensors:

\[ \left[ \begin{array}{cccc}<br /> g_t &amp; e_x &amp; e_y &amp; e_z\\\<br /> e_x &amp; g_x &amp; b_z &amp; b_y \\\<br /> e_y &amp; b_z &amp; g_y &amp; b_x \\\<br /> e_z &amp; b_y &amp; b_x &amp; g_z \end{array} \right]\] - <br /> \[ \left[ \begin{array}{cccc}<br /> g_y &amp; -e_x &amp; -e_y &amp; -e_z\\\<br /> -e_x &amp; g_x &amp; b_z &amp; -b_y \\\<br /> -e_y &amp; -b_z &amp; g_y &amp; -b_x \\\<br /> -e_z &amp; -b_y &amp; -b_x &amp; g_z \end{array} \right]\]

The second tensor is the conjugate of a symmetric tensor. It shows operationally why there is a significant difference between F_{(g)} and F_{(em)}, where both define the conjugate as flipping the signs of the later three parts, but F_{(em)} can do so by taking the transpose of the matrix representation. That does not work for F_{(g)}. Nice.

Doug

 

[Mentz114]

Hi Doug,

in post #514, I'm defining the dual of F_g using a pseudo-tensor, in analogy with the EM case. The pseudo-tensor s is something I just invented, because we go from EM/anti-symmetric -> gravity/symmetric. This dual is traceless and gives the right energy.

It's marginally less dodgy than inventing a new quaternion algebra ( maybe).

Lut

 

[sweetser]

Inventing math tools

Hello Lut:

I thought you might be inventing things. This is a good sign. It dovetails with Lawrence's complaint about my work not fitting into the proud tradition of differential geometry.

I think I referred to the wrong post. Post #514 is an extension of #511. It was #511 where you wrote out F_(g), saying one should just ignore the gauge. It would be great if you could write the software to make it so. I don't count setting to zero a good answer, or just ignoring it is good practice. I want to catch the math responsible for gauge symmetry. That is what is most interesting.

Doug

 

[Mentz114]

Hi Doug:

I understand your unwillingness to ignore, or set to zero the trace of the grav. field tensor. We can play at finding mathematical ways to dispose of it, but this might not throw light on the physical justification. I'm not sure why you call it 'the gauge'.

Anyhow, I'll try and justify my use of the 'symmetric dual' for the field tensor.

In EM theory, one gets the dual using the Levi-Civita antisymmetric pseuso-tensor like this,

\tilde{F}_{pq} = \epsilon_{mnpq}F^{mn}.....(1)

furthermore

F_{mn}F^{mn} = -\tilde{F_{mn}}\tilde{F^{mn}} = -2(E^2 - B^2)...(2)

So, to carry on the analogy between the EM theory, with anti-symmetric F and curl, to the gravitation case with a symmetric F and symmetric curl, we make a dual using a symmetric equivalent of the L-C pseudotensor, which we call 's' ( for 'symmetric').

\tilde{F}_{pq} = s_{mnpq}F^{mn}...(3)

from which it follows

F_{mn}F^{mn} - 2g^2 = -\tilde{F_{mn}}\tilde{F^{mn}} = -2(e^2 - b^2)...(4)

because \tilde{F}_{pq} is traceless.

While I was typing the above, I realized that your field Lagrangian can now be very succinctly written as

(s_{\mu\nu\rho\sigma}\partial^{\mu}A^{\nu})^2 - (\epsilon_{\mu\nu\rho\sigma}\partial^{\mu}A^{\nu})^2......(5)

this has no g terms as required, with the two pseudo-tensors doing the work of the odd and even quaternions.

I still think GEM is wrong but the Lagrangian is tidier !

Lut

PS the software worked faultlessly and did all the donkey work.

 

[sweetser]

Using all modes of emission

Hello Lut:

Good work! I particularly like equation (5) and will make a mental note of it. Many group theory pros would object, saying the tensor is reducible, and reducible tensors cannot be used to represent fundamental forces. This issue has been raised indirectly in that thread you recommended reading in the first reply:

The "trouble" with gauge theories is that you are including unphysical degrees of freedom (such as the longitudinally polarized photon) and you have to make sure that they don't appear in the final calculations, and this complicates matters when you have to calculate things. However, despite the headaches, this can be done.

What I have argue since post #1 is the longitudinal mode of emission is a spin 2 graviton traveling at the speed of light doing the work of gravity. It is the scalar mode of emission that causes more serious problems since for a spin 1 field it implies negative probabilities.

People familiar with EM quantum field theory know these problems, and impose a constraint on quantizing EM such that the scalar and longitudinal modes are always virtual. Grad students grinding through these calculations often complain: it looks like a hack. Yet the hack needs to be there for a spin 1 field, and eventually they shut up and accept it.

The hack looks like a golden opportunity to put two modes of emission to work for gravity. Those modes will not harm EM theory in any way. A spin 2 scalar field would not have the indefinite metric problem, no negative probability problem.

No one else has picked up on the omission in Feynman's analysis of the current coupling term. I think that represents the greatest barrier to a rank 1 field theory.

Someone who bought a DVD from me thought I was within spitting distance of a unified field theory. I like that characterization slightly crude as it is.

Doug

 

[sweetser]

Everything is antisymmetric (for now)

Hello:

I thought I'd share a small epiphany I had since it sheds a bit of light on the conflict I have had with Lawrence. The best field theory we have is the Maxwell field equations for EM. The second rank field strength tensor is:

F_{\mu \nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu} \quad eq 1

It gives me confidence to know this is the foundation of my GEM proposal. Yet it also was the basis for the weak, the strong force, and Einstein's work on EM. For EM, the weak, and the strong forces, what changes is what group gets plugged into the machinery, U(1), SU(2), and SU(3) for EM, the weak, and strong forces respectively.

Gravity is a bit different, but not so different. I don't know the details, but I recall reading how Einstein looked to EM for guidance on how to proceed. Take a peak at the Riemann curvature tensor:

R^{\rho}{}_{\sigma \mu \nu}=\partial_{\mu} \Gamma^{\rho}{}_{\sigma \nu}-\partial_{\nu} \Gamma^{\rho}{}_{\sigma \mu}+\Gamma^{\rho}{}_{\mu \lambda}\Gamma^{\lambda}{}_{\nu \sigma}-\Gamma^{\rho}{}_{\nu \lambda}\Gamma^{\lambda}{}_{\mu \sigma} \quad eq 2

We have a subtraction thing going on again. This is a much more complicated animal, but we only need two contractions (the Ricci tensor minus the Ricci scalar), for the Einstein field equations.

Now I can see the resistance to tossing in a "+" for the minus in eq 1. That equation is good enough for EM, the weak, and the strong force, which covers 3/4 known forces of Nature. We have GR which has withstood every subtle test, while failing all large scale ones (velocity profiles of galaxies, the big bang, and our current acceleration). I can be at peace with that resistance. I don't think the objection to a symmetric field strength tensor will stand the test of time, but it is rational to not embrace it.

Doug

 

[Lawrence B. Crowell]

Gravitation differs from the gauge fields as an interaction determined by an exterior symmetry, rather than an inner symmetry modeled as a vector space on a principal bundle. The bundle structure is then a fibration of the symmetries of the space (spacetime) in an atlas of local tangent spaces. This is all spelled out in the Coleman-Mandula theorem, which gives a mathematical reason for the structure of interactions which can possibly exist. It does not tell us their explicit form, but as a "no-go theorem" of sorts it tells us what is not permitted. What you say above about "gravity being different" is frankly a bit on the silly side of things.

Lawrence B. Crowell

 

[sweetser]

No Lie algebra for a symmetric tensor

Hello Lawrence:

Thank you for clarifying the difference between an inner symmetry and an exterior symmetry. I do lack the experience to speak the jargon of what is known. I thought I had admitted I was speaking imprecisely.

So I go read up about the "Coleman-Mandula theorem, which gives a mathematical reason for the structure of interactions which can possibly exist [under the assumptions used in the theorem]." I hope you do not object to the clause, which is appears banal. Plucking out a one-line summary:

http://en.wikipedia.org/wiki/Coleman-Mandula_theorem]In[/PLAIN] other words, every quantum field theory satisfying certain technical assumptions about its S-matrix that has non-trivial interactions can only have a symmetry Lie algebra which is always a direct product of the Poincare group and an internal group if there is a mass gap: no mixing between these two is possible.

A Lie algebra has the property that it anti-commutes. If one makes a strategic decision to walk in a new direction with a field strength tensor \partial_{\mu} A_{\nu} + \partial_{\nu} A_{\mu}[/tex] which is a different combination of exactly the same players in F_{\mu \nu} (when the symmetric field tensor trace is zero). One of the first observations is that any generating algebra for the group would have to commute, ergo any conclusions of the Coleman-Mandula theorem do not apply.<br /> <br /> Sorry Lut, this is the same non-debate Lawrence and I have had before. I had to find out where the imposed domination of anti-commutivity was written.<br /> <br /> Doug<br /> <br /> Big fuzzy picture: If gravity is the force justifying why everything loves and attracts everything else, its algebra must be commuting.

 

[Mentz114]

Hi Doug:

You needn't apologise, it is always diverting to read your and LBC's posts, as long as they are short and to the point.

The gauge group of teleparallel gravity, the translation group, has a commuting Lie algebra.

I don't believe all mathematical theorems necessarily apply in physics, except in so far as they prevent inconsistencies. For example, von Neumann's 'proof' that QM cannot be modeled by 'hidden variables' is disproved by counter-example. Mathematical theorems always have lots of tight conditions that are often violated in practice.

Lut

 

[Lawrence B. Crowell]

Hi Doug:

You needn't apologise, it is always diverting to read your and LBC's posts, as long as they are short and to the point.

The gauge group of teleparallel gravity, the translation group, has a commuting Lie algebra.

I don't believe all mathematical theorems necessarily apply in physics, except in so far as they prevent inconsistencies. For example, von Neumann's 'proof' that QM cannot be modeled by 'hidden variables' is disproved by counter-example. Mathematical theorems always have lots of tight conditions that are often violated in practice.

Lut

I am not trying to say that the Coleman-Mandula theorem is proven in physics, where after all in science nothing is ever proven as such. It is just that the CM theorem is a fairly strong aspect of theoretical physics.

Teleparallel connections are commutative for special cases of nongeodesic motion. Christopher Columbus sailed to the new World on a nearly teleparallel path.

Lawrence B. Crowell

 

[sweetser]

Maxwell and GEM derivation on YouTube

Hello:

I will be going to the New England APS/AAPT Meeting in New London, Connecticut on Saturday to give the talk "Using Quaternions for Lagrangians in EM and GEM." At just under 15 minutes, you will be able to see the Maxwell equations derived from its Lagrangian. Rarely will you see this much detail about this central part of physics. GEM is a small variation on the theme, where you can see the 12 of 24 terms flip signs, generating a relativistic cousin of Newton's field equations for gravity. Enjoy.

Doug

http://youtube.com/watch?v=P9TUqUXGgpE

 

[Mentz114]

Teleparallel connections are commutative for special cases of nongeodesic motion. Christopher Columbus sailed to the new World on a nearly teleparallel path.

I can see that the connections are symmetric for geodesics (force-free) motion. But are the generators of the translation group identified with the connections ? The connections are derivatives of the gauge field whose generators commute. I'm busking this, maybe you can set me straight.

Lut

 

[sweetser]

Connections in Standard EM

Hello Lut:

On my latest YouTube talk, I got asked a good question: what would the Maxwell and GEM derivations look like if I included all the connections? I know that because the EM field strength tensor uses an exterior derivative, I could just they that is why they drop out. I wish to see these drop in detail. I will make the same presumptions as happen in GR: the connection is metric compatible and torsion-free, so I can use Christoffel symbols of the second kind. These are not tensors and have 3 indexes. What I hope to learn is how exactly to play with Christoffel symbols in EM, so that they all happen to drop out. Seeing this work for the B field gives us a good place to start:

\nabla \times A = \frac{\partial A^u}{\partial x^v} - \Gamma_{\mu}^{v u} A^{\mu} - \frac{\partial A^v}{\partial x^u} + \Gamma_{\mu}^{u v} A^{\mu}

= \frac{\partial A^u}{\partial x^v} - \frac{\partial A^v}{\partial x^u}\quad eq 1

Because the Christoffel is symmetric in the indexes, the two gammas cancel. I figured a similar thing should happen with the E field, but I must be missing a sign:

E = -\frac{\partial A^u}{\partial t} + \Gamma_{\mu}^{0 u} A^{\mu} -\frac{\partial \phi}{\partial x^u} + \Gamma_{\mu}^{u 0} A^{\mu}\quad eq 2

This time the gammas add. Oops. Where did I misstep?

Thanks,
Doug

 

[Mentz114]

Losing connections

Hi Doug:
I'll try and repeat the calculations above later. The covariant derivatives cancel out in the field tensor.

F_{\mu\nu} = \nabla_{\mu}A_{\nu} - \nabla_{\nu}A_{\mu}

where

\nabla_{\mu}A_{\nu} = \partial_{\mu}A_{\nu} - \Gamma^{k}_{\mu\nu}A_k

and because of the symmetry in the lower indexes the Gamma factors cancel.

As I've pointed out many times, this does not happen with the symmetric field tensor, so you end up with terms containing A^2 in the lagrangian, as well as functions of the metric.

I just don't see how you can mix up a metric theory and a vector potential. Your field equations are four in number and your unknowns are 10 for the metric ( 40 connections) and 4 for the potential. How do you solve that ?

Lut

 

[sweetser]

Mistake in 529

Hello Lut:

It looks like my mistake in Eq. 2 of post #529 was on the first gamma, it should have been negative. For the E field, I need:

E = \nabla_{0}A_{u} - \nabla_{u}A_{0} = -\frac{\partial A_u}{\partial t} - \Gamma^{k}_{0 u}A_k - (\frac{\partial \phi}{\partial x^u} - \Gamma^{k}_{u 0}A_k)


So the minus in the first term happens from the lower script for the A_u which does not change the sign of the gamma. The minus on the second term does flip the switch on the second gamma, making it positive. Not difficult, but I need to get rock solid on how signs work in standard EM first, then out to GEM. Gammas are always negative unless one tosses a minus sign in front.

Doug

 

[Mentz114]

Hi Doug:

with non-zero Christoffel symbols, raising and lowering indices is a nightmare -

A_{\mu} = g_{\mu\nu}A^{\nu} = A_{0}g^{\mu 0}+A_{1}g^{\mu 1}+A_{2}g^{\mu 2}+A_{3}g^{\mu 3}

and this

\Gamma^{ bc}_{a} = g_{ka}g^{bm}g^{cn}\Gamma^{k}_{ mn}

has 64 terms.

Note that most of the GEM calculations I did are with the Minkowski metric.

Lut

 

[sweetser]

3-vector versus 3-vector

Hello:

I am writing code to handle the Even representation of quaternions on the command line so I can make animations using them. In this context, quaternions are always 4 numbers that get piped from one program to another. It is not possible to know if a quaternion being fed into the next program in a chain of these programs was an Even or Hamilton quaternion.

When I got to the nuts and bolts of the programming, I noticed that this issue only matters for binary operations, and then only for that part of the binary operation that has the 2 3-vectors forming a product. In the Hamilton representation, the dot product gets a minus sign, the even dot product stays positive. In the Hamilton representation, the curl is the curl with half the terms positive, half minus. In the Even representation, all the same terms are all positive.

A few issues are missed by using tensors. One is the issue of knowing there is an inverse for both the Hamilton and Even representations of quaternions. I used to justify this on an issue of quantum field theory which is kind of obscure. A more apparent reason might have to do with the observation of the critical importance of group theory in physics. The definition of a group relies on there being both an identity and an inverse for every member of the group. Those are both there for these two representations of quaternions. Tensors don't have to have inverses.

Doug

 

[sweetser]

Complete analysis of vector current coupling

Hello:

I am preparing a talk for an APS meeting in St. Louis. One of the points I intend to make is that Feynman's analysis of the phase of the current coupling term is incomplete. The image really rocks, but I am unsure how to upload images here, so I'll have to get by with LaTeX.

The current coupling term is also known as the source term: -J^{\mu} A_{\mu}. Take the Fourier transform of the potential to get J'/k². Since the indices are the same, this term evaluates to the following scalar:

-J^{\mu} J&#039;_{\mu}/k^2 = (-\rho \rho&#039; ~+~ Ax Ax&#039; ~+~ Ay Ay&#039;~+~ Az Az&#039;)/k^2 \quad eq~1

What Feynman does is consider a specific case of a current moving along the z axis. While it makes the algebra simpler, it made my slide longer, and not as general. Skip all the z-axis specific stuff, and just multiply these two together using the Hamilton representation of quaternions:

-J J&#039;/k^2 = (-\rho \rho&#039; ~+~ Ax Ax&#039; ~+~ Ay Ay&#039;~+~ Az Az&#039;,

-\rho Ax&#039; ~-~ Ax \rho&#039; ~+~ Ay Az&#039;~-~ Az Ay&#039;,
-\rho Ay&#039; ~-~ Ay \rho&#039; ~+~ Az Ax&#039;~-~ Ax Az&#039;,
-\rho Az&#039; ~-~ Az \rho&#039; ~+~ Ax Ay&#039;~-~ Ay Ax&#039;)/k^2 \quad eq~2

The phase is the final 3 lines of eq 2. What Feynman calculated was part of the third line, Ax Ay&#039;~-~ Ay Ax&#039;. These two terms do not add together, so it will take 2 pi to get back to the starting position, the sign of spin 1 symmetry. That is excellent, since these terms are the transverse modes of EM that need spin 1 symmetry so like charges repel.

Now look at the first two terms of the third line, -\rho Az&#039; ~-~ Az \rho&#039;[/itex]. These do work together, and will require only pi rotations to get back to the starting position. This is a property of spin 2 symmetry. Spin 2 particles are needed for systems where like charges attract. A complete analysis of the phase looks good to me.<br /> <br /> Doug

 

[Lawrence B. Crowell]

I can see that the connections are symmetric for geodesics (force-free) motion. But are the generators of the translation group identified with the connections ? The connections are derivatives of the gauge field whose generators commute. I'm busking this, maybe you can set me straight.

Lut

I guess I missed this one. Non-geodesic motion involves some other force on the motion of a particle. Teleparallel connections have a number of meaning as I understand. One is that the geodesic motion is given by a horizontal bundle and the "force" by a vertical bundle. This is Finsler geometry. Another definition, related maybe, is where the motion is parallel or Euclidean-like (latitudinal lines on a globe) and where the force or vertical bundle terms act to sustain this flow. I think this can result in torsional gravity.

Lawrence B. Crowell

 

[CarlB]

New article out that perhaps reads on this subject:

A Spatially-VSL Gravity Model with 1-PN Limit of GRT
Jan Broekaert
In the static field configuration, a spatially-Variable Speed of Light (VSL) scalar gravity model with Lorentz-Poincaré interpretation was shown to reproduce the phenomenology implied by the Schwarzschild metric. In the present development, we effectively cover configurations with source kinematics due to an induced sweep velocity field w. The scalar-vector model now provides a Hamiltonian description for particles and photons in full accordance with the first Post-Newtonian (1-PN) approximation of General Relativity Theory (GRT). This result requires the validity of Poincaré’s Principle of Relativity, i.e. the unobservability of ‘preferred’ frame movement. Poincaré’s principle fixes the amplitude of the sweep velocity field of the moving source, or equivalently the ‘vector potential’ ξ of GRT (e.g.; S. Weinberg, Gravitation and cosmology, [1972]), and provides the correct 1-PN limit of GRT. The implementation of this principle requires acceleration transformations derived from gravitationally modified Lorentz transformations. A comparison with the acceleration transformation in GRT is done. The present scope of the model is limited to weak-field gravitation without retardation and with gravitating test particles. In onclusion the model’s merits in terms of a simpler space, time and gravitation ontology—in terms of a Lorentz-Poincaré-type interpretation—are explained (e.g. for ‘frame dragging’, ‘harmonic coordinate condition’).
http://www.springerlink.com/content/b3758t680gj383j8/

Back on the first of the year, Foundations of Physics (the above journal) picked up a new editor, Gerardus 't Hooft, who writes:

During my first couple of months in this office, it became clear that fundamental questions in physics and philosophy also attract the interest of many laymen physicists.

We receive numerous submissions from people who venture to attack the most basic premises of theories such as Special Relativity, but instead only succeed in displaying a lack of professional insight in how a physical theory is constructed. I suspect that some of these people may have been working somewhere in an attic, deprived from daylight for decades, determined only to reemerge with a Theory of Everything in their hands. Even though they may be very sincere, we have to disappoint such authors. New insights are gained only by intense interactions with professionals all over the globe, and by solidly familiarizing oneself with their findings, and we must make a selection from only those papers whose authors have a solid understanding of the topics they are discussing.

Fortunately they also submit their work, and their clever inventiveness continues to surprise us.
...
I hope to receive your submissions. Acceptation of a paper may not necessarily mean that all referees agree with everything, but rather that the issues put forward by the author were considered to be of sufficient interest to our readership, and the exposition was clear enough that our readers, whom we assume to be competent enough, can judge for themselves.
...
http://www.springer.com/physics/journal/10701

 

[sweetser]

Variable Speed of Light theory

Hello Carl:

The paper cost $32 to download, ouch. I know I have a bias against spatially-Variable Speed of Light theories. On my dinning room wall is a large artwork, The Speed of Light According to Rene Magritte. I think about c as time's relationship to space. Photons travel at the speed of light because time is space. It is hard to see how that statement can vary.

Any reasonable proposal for gravity must agree to the first parameterized Post-Newtonian accuracy, as these folks do. I am a bit surprised they didn't look at the second level of accuracy.

The "scalar-vector model" might have passing similarities to GEM, but I am skeptical.

The attitude of the editor parallels some of the motivations of the Independent Research section of Physics Forums. 't Hooft has a site, How to become a good theoretical physicist, I always recommend to people working alone in the attic.

Should I ever get satisfied with my draft paper, I would consider sending it there. A number of issues need to be addressed first. I need to shift the action from tensor notation to quaternions because the issue of gauge symmetry must be clearly addressed. Second, I still have to get the energy expression in order.

I am emailing from the April APS Meeting in Saint Louis. I heard a talk by someone who is trying to take EM up a level to use the tools similar to GR, basically the opposite direction I am traveling. That talk gave me a good overview of the various approaches to EM as a geometry phenomena. I asked the speaker to send me his slides.

The talks about dark matter and dark energy make me sad. They certainly don't sound like solid physics. In a question and answer section, I confirmed with the speaker that she knew of no large scale systems where Newtonian gravity produces the observed results. That is what I expect since I think an unusual effect of gravity, labeled the relativistic rocket effect, Vc dm/dR Vhat, is never put into their numerical integration systems. I will recall these talks when trying to approach my fears of doing the calculations (they look hard, and I know my limitations too well).

Doug

 

[Mentz114]

Doug:
Jan Broekaert has several papers in the arXiv. Do a search in GR-QC for his name in authors. I'm having a look at one later, for diversion. Wiil report if it's interesting.

Carl:
I enjoyed the quote from t'Hooft.

Lut

[edit : this appears to be the very paper. Save $32 now !]

arXiv:gr-qc/0405015

 

[Lawrence B. Crowell]

New article out that perhaps reads on this subject:

A Spatially-VSL Gravity Model with 1-PN Limit of GRT
Jan Broekaert
In the static field configuration, a spatially-Variable Speed of Light (VSL) scalar gravity model with Lorentz-Poincaré interpretation was shown to reproduce the phenomenology implied by the Schwarzschild metric.

These types of theories are curious, but as Doug pointed out c is just an invariant measure by which space and time are equivalent. The value of c in its particular comes from other units in nature associated with electromagnetism, e, \epsilon_0,~\mu and so forth. If suspect that a variability in c or some energy dependence might come about near the Planck scale, but then again the Planck scale is defined according to c.

Lawrence B. Crowell

 

[CarlB]

These types of theories are curious, but as Doug pointed out c is just an invariant measure by which space and time are equivalent.

This is true only under the assumption that there is no preferred coordinate system. In that sense, the equivalence is logically circular. Any theory of gravity that is built on a flat underlying space has to have a variable speed of light. And in such a theory, the speed of light is well defined and variable.

My favorite such theory is the gauge theory of gravity by the Cambridge geometry group. The math it uses is based on the Clifford algebra of Dirac's gamma matrices (which they call "geometric algebra" following David Hestenes). Their theory is identical to GR (to all orders) outside of the event horizons but is built on a flat (well, Minkowski) underlying space or coordinates. I started a website supporting their theory here that includes links to the papers:
http://www.gaugegravity.com/
I'm planning on redoing the website to make it more informational soon. There are a lot of articles that use these ideas to model electrons and black holes, and I need to link those in.

When you have a flat gravity theory (and from my point of view, Sweetser's theory is flat), the first thing to do is to write down what the theory does for a singular mass, a black hole. This defines a preferred coordinate system, and that system defines the speed of light in the theory.

Since gauge gravity is identical to GR, the gravity you get from it is equivalent to GR written in a particular choice of coordinates. The stationary black hole becomes Painleve coordinates for the Schwarzschild solution. I created a java applet simulation that shows the relation of these to the usual Schwarzschild coordinates here:
http://www.gaugegravity.com/

Now the original name of the applet was "sweet" because I wanted to include Sweetser's gravity as well as Newton, and Einstein's in Schwarzschild and Painleve coordinates. However, I haven't figured out how to actually write down his equations, in the form of the acceleration felt by a test body, and so I haven't modeled it yet. I quit asking, but if he comes up with an equation, I'll type it up into java and add it to the simulation (hint hint).

 

[lba7]

oscillation can respond ?

(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