Unifying Gravity and EM

[sweetser]

Light Bending Around the Sun

Hello:

The Bad Astronomy discussion is going OK. If I write up a note over there that I like, I'll port it over here. The following deserves "double posting" because it is a simple calculation with consequences. I was asked to derive the light bending around the Sun effect which is different between GR and GEM. Here goes...

The calculation comes directly out of a paper: “Post-post-Newtonian deflections of light by the Sun” by Reuben Epstein and Irwin Shapiro, Physical Review D, 22:2947, 1980. They write a generalized Schwarzschild metric like so:

dτ² = A(R) dt² – B(R) dR.dR

where

A(R) = 1 – 2 G M/c² R + 2 β (G M/c² R)² + O(3)

and

B(R) = 1 + 2 γ G M/c² R + 3/2 ε (G M/c² R)² + O(3)

There are three Greek letters in play here: β, γ, and ε. This is a research paper, so naturally they write "Calculating the deflection angle in the usual manner, we find for the ppN contribution

Δθ_ppN = π (2 + 2 γ - β + 3/4 ε) (G M/c² R)²"

For the Schwarzschild metric, all the Greeks are 1, so these constants add up to 3.75. For GEM, the constants add up to 4. Now collect the numbers needed. I fetch mine online, http://physics.nist.gov/cuu/Constants/index.html, http://en.wikipedia.org/wiki/Sun

G = 6.674 28 x 10^-11 m³ kg^-1 s^-2
M = 1.989 x 10³⁰ kg
R = 6.955 x 10⁸ m
c = 299 792 458 m s^-1
π = 3.14159
206265 arcseconds/degree

Go to Google, and type it in:
206265 * pi * 3.75 * 6.67428^2 * 10^-22 * 1.989^2 * 10^60 / 299792458^4 / 6.955^2 /10^16
and the answer is 10.96 microarcseconds.
For GEM, the answer is 11.69.
The difference is 0.73 microarcseconds.

That's how its done. GEM predicts more bending that GR, as earlier GR predicted more bending than Newtonian theory. Tradition!

doug

 

[CarlB]

I noticed that the GPB is indicating frame dragging effects that are centered about 25 or 35% higher than GR would indicate, but with large enough error bars that GR is consistent. Does GEM get higher frame drag than GR?

 

[sweetser]

Hello Carl:

GEM always predicts a bit more bending since the coefficients on the Taylor series are a bit larger. I have NOT tried to do a frame dragging calculation. Do you have a reference that spells it all out in detail?

doug

 

[sweetser]

Note: by "a bit more", this applies only to second order PPN effects, not first order PPN where the two are exactly the same. I don't have a metric for a rotating reference frame, so the short answer is I don't know at this time.

 

[Don J]

Just by pure curiosity I want to know if GEM can describe "spin orbit coupling"

Apparently from Wikipedia,
As the master theory of classical physics general relativity has one known flaw: it cannot describe "spin orbit coupling"
So they have developed
http://en.wikipedia.org/wiki/Einstein–Cartan_theory

 

[sweetser]

Spin Orbit Coupling

Hello Don:

Great question! If I was a stellar student of GR, I would know about this issue. Now I know something based on the wikipedia article.

All the players in the Einstein field equations are symmetric: the Ricci tensor and metric tensor are both symmetric. That creates a problem. Things like spin and angular momentum need an antisymmetric tensor. This is a fundamental problem for GR since there is no place to put such energy that exists in the real world.

The problem is averted in GR by including torsion in the connection. This is tough stuff to follow for me. One no longer uses the Christoffel symbol of the second kind which again is a symmetric tool. Bring in torsion to the connection, and there is a place to handle angular momentum conservation in an antisymmetric tensor.

With GEM, spin orbit coupling is handled easily. It is in the EM part of GEM. The Lagrangian has the rank 2 antisymmetric field strength tensor. To quote from http://en.wikipedia.org/wiki/Spin_tensor:

Examples of materials with a nonzero spin density are molecular fluids, the electromagnetic field and turbulent fluids.

Spin orbit coupling will not be a problem for GEM. Unification does has some advantages!

doug

 

[sweetser]

Pitch to Peter Coles

Hello:

Here is another one of my pitches, but this time I included a t-shirt, one that has an oil painting I did called "Turquoise Einstein" along with the GEM field equations. I've invested $16.99, so I'll have to see if the investment pays off.

doug


Hello Peter:

I enjoyed your book, "Cosmology: A Very Short Introduction", which was assigned reading for a class I planned to take at MIT's Professional Institute on "Relativity, Gravity, and Cosmology" taught by Paul Joss. Unfortunately he had a significant medical issue which led to a cancellation of the class. Near the end of your book, you speculated that one might be able to put a central equation for understanding the Universe on a t-shirt. You should have by now, or within a few days, received from cafepress.com my effort toward that elusive goal. It should be popular due to the image of "Turquoise Einstein", even if people ignore the math.

I'll spend a few paragraphs describing my idea for how to unify gravity with the rest of physics. In my own experience, I have found it is usually quite easy to spot if someone is taking a logically coherent approach to a problem. If I can be coherent in my reasoning, perhaps like a laser I can project an idea over the Atlantic and share some of the profound excitement I feel.

Newton's law is still the most useful approach to gravity. It is what engineers use for most of their calculations. The exception are those rocket scientists that bring on board atomic clocks. Then the flaw in Newton's approach - that gravity works instantaneously - becomes apparent.

Over time, I have come to appreciate the sophistication of general relativity. I particularly like the Hilbert action, which only needs the Ricci scalar to do its magic. Yet more than eighty years of trying to quantize the approach have come up short.

Our very best theory today for gravity is a rank 2 field equation. Our most useful theory is rank 0. Goldielocks would wonder, what about a rank 1 field theory for gravity? This turns out to be a subtle theory in 4D.

Surely we have a nice, solid reason why a rank 1 field theory is dead on arrival. Someone told me to look up Clifford Will's Living Review article on tests of GR, that I would find the reason there. In the third section, a pure 4-vector theory was not mentioned. Go to his green book, p. 19, a vector theory by Kustaanheimo, which was not published (p. 360).

I found two papers in the 1950's, one by Gupta, the other by Thirring, that said it was not possible to make a rank 1 field theory where like charges attract. This was an assertion, it was not proved. If you were to copy the approach to the Maxwell equations keeping all the signs the same, then yes, like charges repel. If one flips the sign of the charge coupling term, both the force equation which results from varying the action relative to the velocity, and the field equations which result from varying the action with respect to the potential, have like charges attract. A well-placed sign change can work wonders.

Earlier in the Spring, I found another fun problem concerning the spin of the mediating particle for gravity. The graviton has to be spin 2 to bend light. In my action, I have a symmetric, rank 2 field strength tensor, a natural place for a spin 2 particle to call home. If you read chapter 3 of Feynman's "Lectures on Gravitation", he shows how to spot a spin 1 particle in the current coupling term. At a critical point, he multiplies to different currents together, J' J* (p. 34). Looking at the resulting phase, it will take 2 pi to come back around, as we expect of a spin 1 particle. What I realized was if a slightly different current product was used, J' (iJi)*, then the resulting phase will return home in pi, the signal that a spin 2 particle can live in the current coupling term (p. 39).

The immediate reaction to the equation on the t-shirt is to dismiss it as the Maxwell equations written in the Lorenz gauge (at least I care enough about details to know the difference between the Danish Ludwig Lorenz and the more famous Dutch Hendrik A. Lorentz). If I had meant the d'Alembertian operator, I would only need the box, not a box squared. The box squared is suppose to symbolize two covariant 4-derivatives, one taken after the other. A covariant derivative has a normal derivative and a connection, which is the derivative of a metric. Take two in a row, and there will be a derivative of a connection, which is a second order differential of a metric. The shirt has a second order differential equation that can determine the metric, so the metric does not have to be provided as part of the background mathematical structure.

I have this speculation that no one of your station in life will find this body of work of interest until they do a simple calculation: calculate the Christoffel of the Rosen metric. I prefer to call that metric the exponential metric for obvious reasons: static exponentials live on the diagonal. Only one of the three metric derivatives matter, and the contravariant exponential walks into its covariant cousin leaving a charge/distance that would look familiar to Poisson himself. Math magic like that means something.

I'll stop here, although there is much more. I've done my responsibility. I have no doubts you are overwhelmed with the standard pace of life. This is a low-odds, high rewards game. At least you have a t-shirt out of it. Feel free to ask any questions.

doug

 

[sweetser]

The Big Idea

Hello

As it was presented, GEM looks like the fish none of the other fishermen saw. Since this cast of casters is made up of all the brightest kids from all the brightest classrooms on the planet, that doesn't sound likely.

Work in 11 dimensions - oh my! - that will impress the unknowning crowd. Say it is so tiny no one will ever see the stuff claim is amazing! Might as well use three bangs since it will forever be beyond reach.

Let me tell you how I dreamed up this old dream. I was preparing to go to the Second Meeting on Quaternionic Structures in Mathematics and Physics, in Rome, September 1999. As usual, I had to pay my way, but me and my traveling partner, Prof. Guido Sandri from BU added a week on to spend time looking at the art that has been gathering at the birthplace of western civilization. I was making up my transparencies for my talk which was how to write the Maxwell equations using only real-valued quaternions. It turns out to be easy to write the Maxwell equations with complex-valued quaternions, which are not a division algebra, and therefore of no more interest than any other arbitrary Clifford algebra. Making things work with real quaternions, that was a trick. James Clerk Maxwell himself speculated that someday someone would be able to do the trick. Knowing I accomplished something Maxwell himself wanted will always be one of my more ridiculous achievements. I should be noted that a year earlier, Peter Jack, another non-professional figured out all the hoops that have to be passed through.

The homogeneous equations:
$$(\frac{\partial}{\partial t}, \nabla)((\frac{\partial}{\partial t}, \nabla)(, A) - (\phi, A)(\frac{\partial}{\partial t}, \nabla)) + ((\frac{\partial}{\partial t}, \nabla)(\phi, A) - (\phi, A)(\frac{\partial}{\partial t}, \nabla))(\frac{\partial}{\partial t}, \nabla)$$
$$+(\frac{\partial}{\partial t}, \nabla)((\frac{\partial}{\partial t}, \nabla)^*(\phi, A)^* + (\phi, A)^*(\frac{\partial}{\partial t}, \nabla)^*) - ((\frac{\partial}{\partial t}, \nabla)^*(\phi, A)^* + (\phi, A)^*(\frac{\partial}{\partial t}, \nabla)^*)(\frac{\partial}{\partial t}, \nabla) = (0, 0)$$

The source equations:

$$(\frac{\partial}{\partial t}, \nabla)((\frac{\partial}{\partial t}, \nabla)(\phi, A) - (\phi, A)(\frac{\partial}{\partial t}, \nabla)) - ((\frac{\partial}{\partial t}, \nabla)(\phi, A) - (\phi, A)(\frac{\partial}{\partial t}, \nabla))(\frac{\partial}{\partial t}, \nabla)$$
$$- (\frac{\partial}{\partial t}, \nabla)((\frac{\partial}{\partial t}, \nabla)^*(\phi, A)^* + (\phi, A)^*(\frac{\partial}{\partial t}, \nabla)^*) + ((\frac{\partial}{\partial t}, \nabla)^*(\phi, A)^* + (\phi, A)^*(\frac{\partial}{\partial t}, \nabla)^*)(\frac{\partial}{\partial t}, \nabla) = 4 \pi (\rho, J)$$

Impressively ugly. I can justify why it took me six months to find this particular combination of terms, and why Maxwell did not find them.

Now imagine me with a transparency and a sharpie, trying to come up with something I could be proud to travel several thousand miles to present to a half dozen people. This was my one result, and it was so bulky particularly in comparison to the Maxwell equations themselves. The technical struggle was to toss away just the right stuff.

But does Nature toss away anything? I felt the answer had to be a flat "no". At this point, I was focusing on EM. Why bother doing all this work to throw things away? It appeared to me that the EM equations liked things that were antisymmetric, and things that were symmetric were getting disposed of (which now makes sense, understanding the antisymmetric field strength tensor are the heart of EM). I recalled a quote at the start of one of the chapters of Misner, Thorne, and Wheeler dealt with symmetry. Those quotes are often the only part I understand, so I have read a decent fraction of them. I pulled the black phone book off the shelf, and started to hunt for the quote. Here it is, Chapter 17:

The physical world is represented as a four-dimensional continuum. If in this I adopt a Riemannian metric, and look for the simplest laws which such a metric can satisfy, I arrive at the relativistic gravitational theory of empty space. If I adopt in this space a vector field, or the antisymmetrical tensor field derived from it, and if I look for the simplest laws which such a field can satisfy, I arrive at the Maxwell equations for free space...at any given moment, out of all conceivable constructions, a single one has always proved itself absolutely superior to all the rest...

After reading that, I thought it was possible that if I did not toss away information, the field equations might be able to do the work of both gravity and EM. I traveled to Rome and had a grand time, even if no one in the group of six appeared that interested in the talk.

I had my unified field equations:


$$Jq^u - Jm^u = \square^2 A^u$$

This is when I had the meeting with Prof. Alan Guth (24, #355). He told me I needed to figure out the action, derive the field equations from the action, find solutions consistent with current tests and different for more refined ones. That several year march took me away from quaternions. There also was a strategic decision. I knew my intended audience is trained and comfortable with tensors. I rewrote it all to use tensors, and did not mention quaternions. I found it amusing that quaternions could play such a subtle role.

My experience is that theoretical physicists are frenetically busy. They are only comfortable getting involved in a discussion close to their area of expertise. Since no one works on rank 1 field theories for gravity as documented by the lack of coverage in the literature, there is no one to target.

The discussion here has convinced me that I have to return this proposal to its quaternion roots. The technical reasons have to do with discoveries made only in this calendar year. The first was to address Steve Carlip's complaint about a spin 2 particle in the current coupling term, which was dealt with in post p 22, #319 and #320. One cannot multiply one 4-tensor by another 4-tensor and get a third 4-tensor unless all the machinery for automorphic multiplication is there, which is set if one works with quaternions.

The second reason has to do with the weak and the strong forces. To this day, I still have not seen anyone write something like: "...and this force equation does a similar thing to Coulomb's law but for the weak force". There are excellent reasons for this, but still, it makes the weak and the strong forces feel unapproachable. My one handle on them is their symmetries. In particular, the weak force has the symmetry of SU(2), known as the unit quaternions (I am not making that up). If I formulate the GEM proposal in terms of quaternions, then where I can place a unit quaternion will be the symmetric house where the weak force can do its work. That is so direct and simple, odds are good that it is true.

The sexy math idea is that the properties of quaternions - when done right - dictate every fundamental aspect of Nature, bar none. There are four forces of Nature because when you consider the symmetries of two quaternions interacting:


$$(\frac{J}{|J|} exp(J - J^*))^* (\frac{J'}{|J'|} exp(J' - J'^*)) = 1 + \delta$$

this one expression has the symmetries of U(1), SU(2), SU(3) and Diff(M), EM, weak force, strong force and gravity respectively. The video is on YouTube because the new math cannot be shown in a PDF.

doug

 

[Mentz114]

Hi Doug, I hope you are going well.

I've been catching up in the arXiv and I wonder if you might be interested in these ( take a break from GEM ! ).

[40] arXiv:0708.1507 [ps, pdf, other]
Title: Quaternionic and Poisson-Lie structures in 3d gravity: the cosmological constant as deformation parameter
Authors: C Meusburger, B J Schroers
Comments: 34 pages
Subjects: General Relativity and Quantum Cosmology (gr-qc)

arXiv:0708.1154 (cross-list from hep-th) [ps, pdf, other]
Title: Antisymmetric-Tensor and Electromagnetic effects in an alpha'-non-perturbative Four-Dimensional String Cosmology
Authors: Jean Alexandre, Nick E. Mavromatos, Dylan Tanner
Subjects: High Energy Physics - Theory (hep-th)

 

[sweetser]

Hello Mentz:

The body is not doing so well. Looks like I have forgotten how to eat. I have diabetes, and one of the possible complications is to mess up neurons that signal things, such as "hey, let's empty the belly". I am working on living with gastroparesis. Life is always a challenge.

I did download both papers, and honestly understood about nothing. I was pleased to see a Lagrange density early one in one paper, but was unable to parse it.

GEM is in a good state. Personally, I am not aware of a "killer" problem with the proposal as it is today. There have been times when there were killer problems. Initially, the current for gravity had the same sign as the one for EM, and I got slapped for writing that one. Just this Spring I had the spin 2 current coupling scare that was avoided by the use of the first conjugate (iqi)* at just the right time.

doug

 

[sweetser]

Constructing a link between GEM and a quantum SHO

Hello:

In this post I will start from the GEM unified field equation, and see what needs to be done to get to a quantum simple harmonic oscillator. This will establish a link between the GEM field equations and quantum mechanics.

A 4D homogeneous wave equation is the GEM vacuum equation:

$$(\frac{1}{c} \frac{\partial^2 phi}{\partial t^2} - c \nabla^2 \phi, \frac{1}{c} \frac{\partial^2 phi}{\partial t^2} - c \nabla^2 A) = 0$$

The 4D inhomogeneous equation is the source GEM equation:

$$(\frac{1}{c} \frac{\partial^2 phi}{\partial t^2} - c \nabla^2 \phi, \frac{1}{c} \frac{\partial^2 phi}{\partial t^2} - c \nabla^2 A) = (\rho_q - \rho_m, J_q - J_m)$$

We can see that the scalar part of this equation can deal with Gauss' law of EM and Newton's law of gravity in a way consistent with special relativity since this equation in manifestly covariant under a Lorentz transformation (a fancy way of saying we know exactly how everything changes). The waves travel at the speed c. It is critical to note that this equation uses covariant derivatives, the sort with a connection that can work fine on a curved manifold. It is unfortunate that some skilled people see a d'Alemberian operator on a flat Minkowski background (it is there), but do not also see this expression can have a dynamic metric and work in a curved spacetime. I have shown earlier how the Rosen metric solves this 4D inhomogeneous equation for a static point charge.

The next step is to add in a term to convert the wave equation into a simple harmonic oscillator. One needs to subtract the norm of a factor that depends on R times the 4-potential. I'll write out the expression in its smallest parts, so you can notice the similarities:

$$(\frac{1}{c} \frac{\partial}{\partial t}, c \nabla)^*(\frac{\partial}{\partial t}, \nabla)(\phi, A) - (\epsilon, R)^*(\epsilon, R)(\phi, A)= (\rho_q - \rho_m, J_q - J_m)$$

Multiply things out.

$$(\frac{1}{c} \frac{\partial^2 \phi}{\partial t^2} + c \nabla^2 \phi - (\epsilon^2 + R^2) \phi, \frac{1}{c} \frac{\partial^2 A}{\partial t^2} + c \nabla^2 A - (\epsilon^2 + R^2) A) = (\rho_q - \rho_m, J_q - J_m)$$

To make this simpler, imagine that the potential in no way depends on time. A few terms drop:

$$(c \nabla^2 \phi - (\epsilon^2 + R^2) \phi, c \nabla^2 A - (\epsilon^2 + R^2) A) = (\rho_q - \rho_m, J_q - J_m)$$

What we now have looks almost like a quantum simple harmonic oscillator, as written here: quantum SHO. The one difference is that the $\epsilon$ is positive in that expression, and it is a squared negative here. Since that term contains energy, it may end up depending how positive and negative energy are defined.

The quantum simple harmonic oscillator is a particular class of the time-independent Schrödinger equation. I'll put up a quaternion derivation of the Schrödinger equation soon.

doug

 

[sweetser]

Video on Spin 2 Symmetry Available

Hello:

I gave a talk this morning at an APS meeting titled "Seeing Spin 2 Symmetry in GEM, a Unified Field Theory", and already it is up on YouTube



There were seven people in the room, and due to the nature of the meeting, they were all folks who work with atoms, molecules, and lasers, not gravity. I am using Keynote '08 for my presentations these days, and they have made it simpler to record and upload to YouTube. I did that at 11am on Friday night, and now three people have viewed the 15 minute talk, one of them giving it 5 stars. I like the global reach of YouTube! The number of folks that flock to a Saturday 8am meeting in Connecticut was limited exclusively to the other people giving talks and the moderator.

The video contains another presentation of the content in posts #319 and #320, but I got some new insights in preparing for the talk. Let me explain again what the problem is. There is a current coupling term in my vector unified field theory, $J^{\mu}A_{\mu}$. In Feynman's lecture book on gravity, chapter 3, he shows why a vector coupling term like $J^{\mu}A_{\mu}$ has spin 1 symmetry, a great thing for EM because spin 1 particles dictate that like charges repel. That creates a HUGE problem for a unified field theory because for gravity, like charges attract. I think this is the key calculation why bright folks don't play with something like GEM.

The calculation that shows the current coupling term has spin 1 symmetry is correct as far as its assumptions go. The hidden assumption has to do with the orientation of the current and the 4-potential. You are free to work with a 4-potential pointing in any darn direction you want relative to the current J. The freedom to choose a different directions for the orientation of the potential means there might be different symmetries to the current coupling term. Look at it this way: $J^{\mu}A_{\mu}$ evaluates to a scalar, and we cannot detect $A_{\mu}$ directly (only how it changes), so we must consider different possible angles between the currents. In short, Feynman shows that $J^* J'$ has spin 1 symmetry, and I show that $(i J i)^* J'$ has both spin 1 and spin 2 symmetry. It is a great little calculation, and helps me understand why folks as bright as Feynman never stopped to take a serious look at a rank 1 field theory for gravity. The particle physics is wrong, until you recognized the freedom to orient the coupled currents differently to each other.

doug

 

[Graviman]

I feel humbled corresponding here. Your YouTube video is about the only leading edge physics on the internet that makes any sense! I tried to order your DVDs, but there seems to be something wrong with the PayPal setup.

I have studied Maxwell, Relativity, and Quantum with the Open University. I am intending to teach myself Differential Geometry, and will probably try to understand Dirac too. What else do i need to read to do your GEM principle justice?

Thanks in advance,

Mart

 

[Mentz114]

Hi Doug,

re-the APS talk. I don't know the details of the Feynman calculation, and I'd like to see how the Fourier transforming of the potential leads to the Grassman (?) product between the currents. From that point on, it looks reasonable. But if there's a degree of freedom to rotate one vector independently without affecting the coupling scalar - is it physical ?

I'd like to read what Feynman has to say on gravity - do you have a fuller reference ?

Regards,
Lut

 

[sweetser]

Learning GEM

Hello Mart:

The humbling thing here is how cool Nature is. I work by making every possible mistake along the way, and learning from those mistakes, putting policies in place so I don't repeat a mistake.

Take for example relativity. People say they have studied relativity, they can solve problems, they "get it". All true, but how do you make sure you always work in a way that will be consistent with special relativity and the effects of gravity? [note to careful readers: I did not say general relativity since GEM is in technical conflict with GR, but both approaches must be able to explain gravity as the result of a dynamic metric]. My answer to the question is to always use quaternions which are just 4-vectors that one can also multiply or divide. Some folks might think that silly for problems in classical physics, but I say solving such problems in 4D is great practice: you accept the results, can believe them in a classical context, so when the same problem appears (same diff. eq.), the same approach will solve the relativistic problem. This comment is relevant to GEM, since the central equation for GEM is a 4D simple harmonic oscillator.

To learn the GEM proposal specifically, download and print these PDFs:

http://theworld.com/~sweetser/quaternions/ps/day1.pdf
http://theworld.com/~sweetser/quaternions/ps/day2.pdf
http://theworld.com/~sweetser/quaternions/ps/day3.pdf

Then you need to push a pencil. I do lots of calculations, always with a specific "start", always having 8 or fewer steps. It is important for you to be able to do those yourself, and repeat it until you can do it quickly (that is a sign that the brain does really get it). Redraw the little graphics that accompany the slides. The little graphics represent some of the information contained in the algebra, but the drawing will work a different part of the brain than the algebra, so you will get a wider grip on what is going on with this obtuse stuff. I toss around the phrases symmetric and antisymmetric tensors a bit, but I also have a small doodle that gives me a visualization on the issue. It has black and red squares, could be art, but is physics.

If you understand all the calculations in those notes from 2003, that is the foundation of this proposal. The YouTube lecture on spin 2 is an important technical issue, but could not be a complete explanation of the proposal, which probably would run on the order of five hours (the Maxwell equations are not simple, and the necessary bridges to gravity takes time to develop).

Understanding the GEM proposal take a real block of time and effort. If at any point, you get stuck, I want to know about it.

doug

 

[sweetser]

Feynman's Lecture Notes

Hello Lut:

Here is the reference:

https://www.amazon.com/dp/0813340381/?tag=pfamazon01-20
Feynman Lectures on Gravitation

I liked Brian Hatfield's introduction. I have not read the entire book, instead focusing on chapter 3, particularly pages 34-38. I used to be able to "look inside" on Amazon at those pages, but it is not working for me now.

You bring up a most excellent point, one I have been thinking about. EM is a gauge invariant theory, GEM is not. With EM, you can do a transformation like so:

$$A \rightarrow A' = A - \nabla \phi$$

This new A' points in a different direction. That is a "legal move" in EM, not GEM, an allowable gauge transformation. I am making this a bit more specific, by looking at a rotation of A:

$$A \rightarrow A' = A - \nabla \phi = (i A i)^*$$

The exercise is to compare the symmetry of the phase of A (spin 1) to A' (spin 1 and spin 2).

For GEM, I am not free to rotate A. Instead, I am free write A like so:

$$A = 1/2(A + (i A i)^*) + 1/2(A - (i A i)^*)$$

The 1/2(A + (i A i)*) has spin 1 symmetry, while 1/2(A - (i A i)*) has spin 2 symmetry in the phase. I have the freedom to represent the 4-potential in this way.

doug

 

[Mentz114]

Doug,

that looks promising.
The book is not available at the moment. I'm lost on the notation. Is your i from the vector basis i,j,k ? What is the '*' operator doing ?

Lut

 

[Mentz114]

I've been looking at the Lagrangian and your splitting of the potential. I've alway been worried about the hidden (implied) coupling between EM and grav currents.

If you write

$$A_g = \frac{1}{2}(A + L(\theta)A)$$
and
$$A_{em} = \frac{1}{2}(A - L(\theta)A)$$

Where L is a 3D spatial rotation, you can split the interaction term
into two
$$A_g^{\mu}J_{\mu} + A_{em}^{\mu}I_{\mu}$$, where J and I are the currents.
Now the coupling is explicit, and depends on theta. For me, this takes away one major problem with GEM.

The meaning of theta is a bit obscure, but as theta changes, energy moves between the spin 2 and the spin 1 fields.

[later] I've just realized that this changes the Lagrangian too much. Back to the scratch-pad.

 

[sweetser]

Spliting the Potential

Hello Lut:

The "i" is from the 4-basis vector (e, i, j, k). The way I think about curved spacetime is none of these has the norm of one, although ei=ej=ek=1 (e gets smaller, i,j, and k get larger).

The * is the conjugate operator, so (e, i, j, k)* = (e, -i, -j, -k). On my flight, I plan on playing with two sorts of splits of the potential:

$$A_g = \frac{1}{2}(A + i A i) = (0, 0, A_y, A_z)$$
$$A_{EM} = \frac{1}{2}(A - i A i) = (\phi, A_x, 0, 0)$$

One thing I like about this is that it has two degrees of freedom for gravity and two degrees of freedom for EM. This is done for the current coupling term, but not the field strength tensor, because that splits the spin 1 and spin 2 fields based on the symmetry of the second rank tensors.

The second possibility was my initial suggestion:

$$A_g = \frac{1}{2}(A + (i A i)^*) = (\phi, 0, A_y, A_z)$$
$$A_{EM} = \frac{1}{2}(A - (i A i)^*) = (0, A_x, 0, 0)$$

Gravity has too much of the potential, EM to little, so I am skeptical of this one working out.

I had been ignoring the coupling term because it looked too simple to do anything with. Now there may be a way to separate the coupling of like charges that attract from the coupling for like charges that repel. Need to do some work with pencil and paper... And do my day job, and pack.

doug

 

[sweetser]

One Current Density, Two Coupling Terms

Hello:

I realize there is a logical inconsistency in my proposal, and therefore there is a technical problem that must be addressed or the GEM proposal is wrong.

I have discussed how the 4D wave has two transverse modes of emission for EM and two modes for gravity. Yet I have written many times about 2 separate 4-currents, $J_q$ and $J_m$. Together, those 4-currents have eight degrees of freedom which is too many and thus is wrong. I must make due with one 4-current density.

There are three constraints on the current coupling term.

1. The coupling term must be spilt into two: one for gravity and another for EM since they have different properties in regards to the behavior of like charges.

2. The two coupling terms must have opposite signs, so for EM like charges repel and for gravity like charges attract.

3. The EM coupling term must have a phase with spin 1 symmetry, while the gravity coupling term has spin 2 symmetry.

Nothing like a fun riddle!

To make things easier, consider motion along the z axis. The 4-potential can be split along the the transverse part of the potential, Ax and Ay, from the scalar and longitudinal parts, phi and Az:

$$A = \frac{1}{2} (A - k A k) + \frac{1}{2} (A + k A k)$$

$$A = (0, A_x, A_y, 0) + (\phi, 0, 0, A_z)$$

If I wanted to generalize this, I could refer to the parallel versus the transverse parts of the potential, $A_{tr}$ and $A_{||}$. Constraint one solved.

The end result of these games must be the Lorentz invariant contraction of the current and the potential as it appears in the GEM action:

$$-\rho \phi + J_x A_x + J_y A_y + J_z A_z$$

Let's get the EM part first. I will be treating the current and potential 4-vectors as quaternions equipped with the usual ijk kind of algebra, but keep only the scalar part since that is all the action needs. Just multiply the current and potential together for the transverse part:

$$scalar (J \frac{1}{2} (A - k A k)) = - J_x A_x - J_y A_y$$

Close, but not quite. One needs to include a minus sign, which is great, since it satisfies half of the second constraint. To be explicit:

$$scalar (- J \frac{1}{2} (A - k A k)) = J_x A_x + J_y A_y$$

Where there is a scalar, there is a vector, necessary to look at the symmetry of the phase. Fourier transform the 4-potential to a current:

$$vector (- J \frac{1}{2} (A - k A k)) = J_x J'_y - J_y J'_y$$

This satisfies half of constraint 3, not a surprise since that is standard EM analysis.

Now on to the gravity part. If we do exactly the same as before, we get exactly the same result:

$$scalar (J \frac{1}{2} (A + k A k)) = \rho \phi - J_z A_z$$

These have the wrong signs, and so we would have to toss in a minus sign and fail to meet constraint number 2. This would be good in the sense that I would finally be able to forget the GEM proposal. Yet I am too crafty for such a defeat! This is one scalar product, but not the only one. Let's take what I call the first conjugate, $(i q i)^* = (-t, x, -y, -z)$. This will flip the sign of phi and Az:

$$scalar (J \frac{1}{2} (i (A + k A k)) i)^*$$
$$=scalar((\rho, J_x, J_y, J_z)(-\phi, 0 , 0, -J_z))$$
$$=-\rho \phi + J_z A_z$$

Now we have the correct signs for the scalar with a positive J A contraction. That completes the requirements for constraint number 2. Now look at the vector after the potential has been Fourier transformed into a current:

$$vector (J \frac{1}{2} (i (A + k A k)) i)^* =$$
$$=vector((\rho, J_x, J_y, J_z)(-\rho', 0, 0, -J'_z))$$
$$=(0, 0, 0, -\rho J'_z - J_z \rho')$$

The phase has spin 2 symmetry! These two will subtract together, so will take only pi radians to return.

The rain in London was not heavy enough to prevent my writing this note while waiting for the tickets to Wicked. I hope the math stands up to anyone's analysis.

doug

ps. I did get front row tickets, nice.

 

[Mentz114]

Hi Doug,

I look forward to seeing the new Lagangian. I've been looking at current conservation and have found the energy-momentum tensor of the EM fileld is (from Stephani )-

$$T^{mn} = g_{ak}F^{am}F^{nk}-\frac{1}{4}g^{mn}F^{pq}F^{ba}g_{ap}g_{bq}$$
The condition for energy conservation is -

$$(1) T^{mn}_n = -\frac{1}{c}F^{mn}j_n$$

The thing on the left is the divergence of T, as you'll recognise. Having found a nice new tensor problem I set it up in the calculator and tried to verify equation 1. The interesting thing is that I get expressions like
$$T^0 = B_x(\partial_zE_y-\partial_yE_z)$$
$$+B_y(\partial_xE_z-\partial_zE_x)$$
$$+B_z(\partial_yE_x-\partial_xE_y)$$
$$+E_z(\partial_xB_y-\partial_yB_x)$$
$$+E_x(\partial_yB_z-\partial_zB_y)$$
$$+E_y(\partial_zB_x-\partial_xB_z)$$

which goes to

$$T^0 = -B_x\dot{B_x}-B_y\dot{B_y}-B_z\dot{B_z}+E_x(\dot{E_x}+j_x)+E_y(\dot{E_y}+j_y)+E_z(\dot{E_z}+j_z)$$

In the source free case, this is zero because B and E are orthogonal and J = 0. Yippee. I'm not so sure what emerges if there is a current.

The other terms of $$T^n$$ look like this -

$$T^1 = B_x\partial^i\dot{B_i}+E_x\partial^i\dot{E_i}-B_y\dot{E_z}-B_z\dot{E_y}+E_y\dot{B_z}+E_z\dot{B_y}$$

The first two terms look like divs of curls, but I'm not sure about the other four. They could be parts of a curl of a curl and so zero.

The reason I've shown all this detail is because it shows that the energy conservation conditions are actually Maxwell's equations.

This means that this analogy, and the use of the E-M tensor could show coupling of the current and potential as the equations we get by working out the divergence of the E-M tensor of the gravitational force tensor.

Watch this space.

 

[Mentz114]

Doug:
Some time later : I have realized that the terms I get for T_0 are not zero. Something wrong somewhere, back to the drawing board.

[Later]
For the source free case, with an EM field traveling in the z direction, we can say

$$T^0 = -B_y\dot{B_y}+E_x\dot{E_x}$$

which comes to

$$2kB_0E_0 - \omega (B_0^2+E_0^2)$$ multiplied by a phase,

which may well be 0.

 

[Mentz114]

Hi Doug,
I've been re-reading your last post (#371) and disregarding my dislike of using the same potential and current for gravity and EM, it looks OK.

Back to the energy conservation. I did some pencil work and discovered that the divergence of the EM stress tensor actually hinges on this $$F^{\mu\nu}_{,\nu}$$, which is easily seen if you apply a differentiation operator to $$T^{\mu\nu}$$ above.

If $$F^{\mu\nu}_{,\nu} = J^{\mu}$$ then $$T^{\mu\nu}_{,\nu}$$ goes to zero even if there's a current. This is a good thing in my opinion. So $$F^{\mu\nu}_{,\nu}$$ is an interesting thing to calculate. For the EM anti-symmetric F the result looks like a 4-current. With $$A^{\mu} = ( \phi, \vec{A} )$$

$$J^0 = -\nabla^2\phi + \partial_{t} \vec{\nabla}.\vec{A}$$

$$J^1 = \partial_{t}E_x + \partial_{y}B_z + \partial_{z}B_y$$

Repeating the above with the symmetric field tensor

$$F^{\mu\nu} = \partial^{\mu}A^{\nu} + \partial^{\nu}A^{\mu}$$

I get,

$$F^{\mu\nu}_{,\nu} = \partial^{\mu}(\partial_{k}A^{k}) + \partial^{k}\partial_{k}A^{\mu} = J^{\mu}$$

[edit] I found the Tex 'Box' operator so is this -
$$F^{\mu\nu}_{,\nu} = \partial^{\mu}(\Box A) + \Box^2A = J^{\mu}$$


which looks like your field equations. Or is that an extra term ?

So, provided the conditions above are met, we have energy conservation in the field tensor. I'm not sure if this only works in Minkowski space where covariant differentiation commutes. The gravity is all in the potential, and the field equations tell us how to get the potential from the current. It looks too linear to be a theory of gravity without the coupling term, which maybe will tell us how the current changes the potential, which restores the non-linearity.

I might have a go at constructing a full stress tensor from the gravity part of the Lagrangian, then check the divergence for non-linearity.

 

[sweetser]

Non-linearity

Hello Lut:

It is quite common to think that non-linearity is a requirement for a full theory of gravity. What is going on in my opinion is one of those linked circle of statements one sees in math, where one thing implies another thing implies yet another. They all hang together tough. One can even change the order of the links, and the logic is consistent. If you are familiar with that idea in general, here is how it applies for GR:

1. The field equations are rank 2
2. The gravity field can be a source of gravity
3. The non-interacting vacuum field equations are non-linear
4. No efforts to quantize a GR type of theory of gravity have succeeded

For GEM, there is a different collection:

1. The field equations are rank 1
2. The gravity field cannot be a source of gravity just like in EM
3. The non-interacting vacuum field equations are linear (interactions make the equations non-linear, and must be approximated using thngs like Feynman diagrams)
4. The completely relativistic approach to quantizing EM (known as Gupta/Bleuler) can be applied to GEM, but the scalar and longitudinal modes of emission are spin 2

I asked Clifford Will once if there was direct experimental data about non-linearity, and I think he said no (but worry that my memory read in what it wanted to hear, or at least filtered out whatever technical issues he raised. Damn brain). I know there are people who claim there is data for the non-linearity. What I recall Will saying is that so far, all of our best, most consistent theories for gravity have to date all been non-linear. Time for a change! It is just the non-interacting, vacuum equations that are linear.

doug

 

[CarlB]

For GEM, there is a different collection:

1. The field equations are rank 1
2. The gravity field cannot be a source of gravity just like in EM
3. The non-interacting vacuum field equations are linear (interactions make the equations non-linear, and must be approximated using thngs like Feynman diagrams)
4. The completely relativistic approach to quantizing EM (known as Gupta/Bleuler) can be applied to GEM, but the scalar and longitudinal modes of emission are spin 2

I agree completely with the first three. I think that they make complete sense from a parrticle theory point of view. Number 4 is unfortunately over my head.

By the way, have you given the question of the equations of motion any more thought?

Carl

 

[Mentz114]

Doug:

How did you derive your field equations ? Do you have any remark about the formula in my last post ?

I think that can be written ( now I've found 'Box' )
$$F^{\mu\nu}_{,\nu} = \partial^{\mu}(\Box A) + \Box^2A^{\mu} = J^{\mu}$$


Lut

 

[sweetser]

I'll give a try at answering both of these questions over the weekend.

doug

 

[sweetser]

Getting the Field Equations, the Standard Way

Hello Lut:

The field equations were the first things I derived when I wrote the GEM Lagrangian. They were the test that I had the right Lagrangian. What I did was write the Lagrangian in terms of the components:

I used this definition of the field equations:

which if you write out all the components looks like so:

Although there are lots of terms, it is kind of simple, since everything ends up being positive. The field equations are generated by taking the derivative of the Lagrangian with respect to the sixteen possible combinations of taking the 4-derivative of the 4-potential. The result is this:

This looks like

$$\Box^2A^{\mu} = J^{\mu}$$

Notice that I am using the complete asymmetric tensor, not the symmetric term in isolation. I'll have to look at what the scalar is from the contraction of the symmetric tensor. That should be a fun calculation! In other words, I have not calculated the field equations just for gravity.

doug

 

[Mentz114]

Field equations

Hi Doug,

thanks for posting all that. I can see you've used Euler-Lagrange with the
$$A^{\mu}$$'s as dynamical variables. I reckon the result adds up.

I've been finding out some interesting wrinkles about the EM Lagrangian from Itzykson&Zuber ( Warning; they write $$\Box$$ for our $$\Box^2$$ )

Starting with

$$L = \frac{1}{4}F^2+j\cdot A$$

they get field equations

$$\Box^{2}A^{\mu} - \partial^{\mu}(\partial_{\nu}A^{\nu}) = j^{\mu}$$

and by applying the Lorentz condition $$\partial_{\nu}A^{\nu} = 0$$ the field equations become

$$\Box^{2} A^{\mu} = j^{\mu}$$

Now, under the Lorentz condition, only gauge transformations of the type

$$A^{\mu} \rightarrow A^{\mu} + \partial^{\mu}\phi$$ where $$\Box\phi = 0$$ are allowed.

The point of this is to show how they lose the extra term from the field equations by restricting allowed gauge transformations.

Back to GEM. We have,

for EM field : $$\partial_{\nu} (\partial^{\nu}A^{\mu}-\partial^{\mu}A^{\nu}) = j_{EM}^{\mu}$$

for Grav field : $$\partial_{\nu} (\partial^{\nu}A^{\mu}+\partial^{\mu}A^{\nu}) = j_{G}^{\mu}$$

which when summed gives your original field equations. There is no need to apply the Lorentz condition - it has canceled out.

What can it mean ? I'm off for food, more later.

Lut

 

[sweetser]

Hello Lut:

I haven't really integrated the difference in notation yet, but this looks cool. Although I started out with 2 4-currents, now there is only one. So a bit of simplification is:

For EM:
$$\partial_{\nu} (\partial^{\nu}A^{\mu}-\partial^{\mu}A^{\nu}) = j_{tr}^{\mu}$$
For Gravity:
$$\partial_{\nu} (\partial^{\nu}A^{\mu}+\partial^{\mu}A^{\nu}) = j_{||}^{\mu}$$
For GEM:
$$\partial_{\nu} \partial^{\nu}A^{\mu} = j^{\mu}$$

So far, I have missed the detail concerning the $\partial_{\nu} \partial^{\mu}A^{\nu}$ factor.

doug

 

[sweetser]

The G field equations

Hello Lut:

Great question about the G field equations. The Lagrangian needed for this looks like so:

$$\mathcal{L}_G = \frac{1}{c} J^{\mu} (A_{\mu} - I_{||} A_{\mu} I_{||}) -\frac{1}{4c^{2}}(\nabla_{\mu}A_{\nu}+\nabla_{\nu}A_{\mu})(\nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu})$$

Make sure you write out the coupling term in a way that splits out the transverse from the parallel parts. The parallel part is mediated by a spin 2 field where like charges attract.

Write this out in its component parts:

$$\mathcal{L}_G = -\frac{1}{c} \rho A_0 + \frac{1}{c}J_{||} A_{||}$$

$$+ \frac{1}{2c^2}( -(\frac{\partial \theta}{\partial t})^2 + (\frac{\partial A_x}{\partial x})^2 + (\frac{\partial A_y}{\partial y})^2 + (\frac{\partial A_z}{\partial z})^2$$

$$- (\frac{\partial A_x}{\partial t})^2 + 2 \frac{\partial A_x}{\partial t} \frac{\partial \theta}{\partial x} - (\frac{\partial \theta}{\partial x})^2$$

$$- (\frac{\partial A_y}{\partial t})^2 + 2 \frac{\partial A_y}{\partial t} \frac{\partial \theta}{\partial y} - (\frac{\partial \theta}{\partial y})^2$$

$$- (\frac{\partial A_z}{\partial t})^2 + 2 \frac{\partial A_z}{\partial t} \frac{\partial \theta}{\partial z} + (\frac{\partial \theta}{\partial z})^2$$

$$+ (\frac{\partial A_y}{\partial x})^2 + 2 \frac{\partial A_x}{\partial y} \frac{\partial A_y}{\partial x} + (\frac{\partial A_y}{\partial x})^2$$

$$+ (\frac{\partial A_z}{\partial x})^2 + 2 \frac{\partial A_x}{\partial z} \frac{\partial A_z}{\partial x} + (\frac{\partial A_x}{\partial z})^2$$

$$+ (\frac{\partial A_y}{\partial z})^2 + 2 \frac{\partial A_y}{\partial z} \frac{\partial A_z}{\partial y} + (\frac{\partial A_y}{\partial z})^2)$$

Wow, that's a lot of partial derivatives! Notice the simple pattern: either a term is squared with a 1/2 in front of it, or the term is mix, and has a coefficient of 1. When we start taking the derivatives of these things, the derivative of $d \frac{1}{2}x^2/dx = x$ and $d (x y)/dx = y$. Now it is time to apply the Euler-Lagrange equation. First do the current, which is the derivative of the Lagrangian with respect to the potential:

$$\frac{\mathcal{L}_G}{\partial A^{\mu}} = -\frac{1}{c} \rho + \frac{1}{c}J_{||}}$$

The tougher part is taking the derivative with respect to all the fields. This part takes some practice, but it is cool if you can get it to all work out. Do this one guy at a time:

$$\frac{\partial}{\partial x^{\mu}}\frac{\mathcal{L}_G}{\partial \frac{\theta}{x^{\mu}}} = -\frac{\partial^2 \theta}{\partial t^2} + \frac{\partial^2 A}{\partial t \partial x} - \frac{\partial^2 \theta}{\partial x^2} + \frac{\partial^2 A}{\partial t \partial y} - \frac{\partial^2 \theta}{\partial y^2} + \frac{\partial^2 A}{\partial t \partial z} - \frac{\partial^2 \theta}{\partial z^2}$$

There are a few things to notice from this, the first of four field equations. The current density rho has the same negative sign as the second derivative terms of theta. That is good, because it is consistent with like charges attracting each other. We can rewrite pairs of terms in this kind of way:

$$+ \frac{\partial^2 A}{\partial t \partial x} - \frac{\partial^2 \theta}{\partial x^2} = + \frac{\partial}{\partial x}(\frac{\partial A}{\partial t} - \frac{\partial \theta}{\partial x})$$

Notice the difference in the sign between these two terms. For the E field, they are both minus signs. For gravity, they have different signs. Time to get the next field equation:

$$\frac{\partial}{\partial x^{\mu}}\frac{\mathcal{L}_G}{\partial \frac{A_x}{x^{\mu}}} = -\frac{\partial^2 A_x}{\partial t^2} + \frac{\partial^2 \theta}{\partial t \partial x} + \frac{\partial^2 A_x}{\partial x^2} - \frac{\partial^2 A_x}{\partial y^2} - \frac{\partial^2 A_y}{\partial x \partial y} - \frac{\partial^2 A_x}{\partial z^2} - \frac{\partial^2 A_z}{\partial x \partial z}$$

Here again we can rewrite a pair of these terms like so:

$$- \frac{\partial^2 A_x}{\partial y^2} - \frac{\partial^2 A_y}{\partial x \partial y} = - \frac{\partial}{\partial y}(\frac{\partial A_x}{\partial y} + \frac{\partial A_y}{\partial x})$$

This is "curl-like" in the way the different derivatives swap places, but it certainly is not the curl since both have the same sign. There is also the term $+ \frac{\partial^2 A_x}{\partial x^2}$ which is probably the "Lorentz gauge canceling" term.

I did this all first with pencil and eraser. It might be a good exercise in 19th century math to see if you can spot mistakes (it is hard to proof in LaTeX format). Kind of amazing to work with this many partial differential equations, and have it make sense. It does look consistent with Lut's statements based on tensor analysis.

doug

 

[Mentz114]

Hi Doug:

on first reading this, I recognise those not curl-like terms. I've calculated currents and field-tensor contractions for potential A^m and A^m+phi^m to get the extra terms, and check if they vanish like divergences. I'm struggling to make sense of the extra terms, but I'll post them later so you can have a go at solving the puzzle, or even checking the calculation ( Mathematica ?).

Later ...

Lut

 

[Mentz114]

Hi Doug,

I have to say that I don't think a current can be represented with 2 degrees of freedom, so I have some doubts about the one-current proposal. I've been thinking about stress tensors instead of currents.

I'm intrigued by the cancellation of the extra terms from the field equations when EM and gravity are included. This could mean something or just be an mathematical accident.

Another thing is that if the potential is augmented thus -

$$A^{\mu} \rightarrow A^{\mu} + \phi^{\mu}$$ where $$\phi^{\mu} = \partial^{\mu}\Phi$$

the field equations $$\Box^{2}A^{\mu} = j^{\mu}$$ have an extra current

$$2\partial_{\nu} \partial^{\nu}(\partial_{\mu}\phi^{\mu})$$

which means we can add any potential as long as the above is zero.

But the important thing is to check the Lagrangian when the gauge transformation is done. After much effort I've ended up with 32 terms which I've emailed to you. It would be good to reduce them, and even to check the calculation, if Mathematica is up to it !

My software takes about 4 minutes to run the script, but with no simplification. Some of the extra terms will be divergences and will go to zero when integrated. The remainder are extra conditions to ensure energy conservation. If they cannot be met physically, the action is not gauge invariant. In EM this is bad thing, but whether it has any significance in the GEM context, I don't know yet.

Check your email...

Lut

 

[sweetser]

Not gauge invariant in the usual sense

Hello Lut:

Looks like we have a few technical disagreements based on your last post. Let's look at them seperately.

> I have to say that I don't think a current can be represented with 2 degrees of freedom, so I have some doubts about the one-current proposal.

Let me say it this way: a 4-current has 4 degrees of freedom, and a current-current interaction has two types of symmetry, both spin 1 and spin 2 symmetry. The two sorts of symmetry seen in the contraction of a current with a potential comes from the ability to arbitrarily point the potential relative to the current density it represents. Part of the potential will be parallel to the current it represents, part of it will be transverse. If you split the current into these two parts, that algebraic act by itself changes nothing. The phase of the parallel (||) and transverse (tr) potentials are:

$$J A_{||}->J FT[A_{||}] = J J' = \rho \vec{J'} - \vec{J} \rho'$$

$$J A_{tr}->J FT[A_{tr}] = J J' = \rho \vec{J'} - \vec{J} \rho'$$

The phase of these two contractions are the same, spin 1 symmetry.

What is needed is another twist of the potential for the transverse current:

$$J (I_{tr} A_{||} I_{tr})^*->J FT[(I_{tr} A_{||} I_{tr})^*] = -J J' = - \rho \vec{J'} - \vec{J} \rho'$$

Notice those minus sign! When I did this work initially, I had to work with 2 4-currents, and put in the sign difference by hand. Now I get to work with 1 4-current with 4-degrees of freedom, and by twisting the orientation of the 4-potential, get to find phases with spin 1 and spin 2 symmetry and two signs.

Onward...

This gauge transformation:

$$A^{\mu} \rightarrow A^{\mu} + \phi^{\mu} where \phi^{\mu} = \partial^{\mu}\Phi$$

is not allowed in the GEM proposal. The reason is there are two fields, E for EM, and e for gravity, and this transformation will mess up one or the other. The fields are defined like so:

$$E = - \frac{\partial A}{\partial t} - \nabla \phi$$

$$e = + \frac{\partial A}{\partial t} - \nabla \phi$$

If I were to work only with E, there is a gauge transformation of the sort you refer to. If I were to work only with e, there is also such a gauge transformation. But these two gauge transformations are not the same.

The title here refers to an unusual way to think about gauge symmetry. Gauge means "to measure". You are exploring the usual use of gauge symmetry, the ability to add in an arbitrary derivative of a scalar field. When I look at the GEM action, and see the covariant derivative, I think of gauge symmetry as the ability to measure the action as the result of a potential in flat spacetime, or as the curvature of the spacetime manifold with a dull potential, or any combination in between.

I will work on the gauge symmetry, to make certain the kind of gauge transformation you suggest is not allowed for the full GEM proposal.

doug

 

[Mentz114]

Hi Doug :

I thought that

$$e = + \frac{\partial A}{\partial t} + \nabla \phi$$ ?

I assume that $$A^{\mu} = (\phi, \vec{A})$$.

I can see your point, I think.

The only reason gauge transformations are of interest is to see if energy is conserved, and I think it is because the gauge current can be eliminated with the field condition.

Regarding the the current business. You can split the 4 degrees of freedom into 2 orthogonal vectors but you have an electrical and a matter current to describe, and only 2 df each. Or do I misunderstand ?

Lut