By Doug Sweetser | June 28th 2011 01:43 AM | 32 comments | Print | E-mail | Track Comments

Trying to be a semi-pro amateur physicist (yes I accept special relativity is right!). I _had_ my own effort to unify gravity with other forces in...

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I think of myself as non-violent, a purple guy in the blue state of Massachusetts.
When it comes to proposals in science though, I go for head shots. Nature doesn’t care if we keep repackaging bullshit. A well-fed bull does not run out of bullshit.

To stop a science zombie, a bullet to the brain is all that might stop it. I know from experience that the easiest person to fool is myself. I am on the lookout for bullets for subjects I speculate on.

I am a poor student of my own personal history, so this is only an approximate timeline. In 1997, I got interested in quaternions based on a contest I came up with to state a brief definition of time. In 1999, I was the second person to find a road to the Maxwell equations using combinations of symmetric and antisymmetric quaternion operators. The complexity of that result led to my first expression to unify gravity and EM - ignoring the weak and strong forces as many do but should not. Right before my 20th high school reunion, I found a tenuous road to the exponential metric.
$g_{\mu \nu}=\begin{bmatrix}exp(-2 G M/c^2 R)&0 & 0 & 0 \\ 0 & -exp(2 G M/c^2 R) & 0 & 0 \\ 0 & 0 & -exp(2 G M/c^2 R) & 0 \\ 0 & 0 & 0 & -exp(2 G M/c^2 R) \end{bmatrix}\quad eq ~1$
I loved the exponentials, the calling card of deep insights in physics. The metric is consistent with weak field tests, but differs from general relativity at the next level of precision.

Experiments for the first order bending of light at the edge of the Sun pitted Newton’s scalar potential theory of gravity versus Einstein’s rank 2 metric theory. The predictions were different by 100%: Newton with the low bid of under an arcsecond, 0.875. Einstein doubled down, predicting 1.75 arcseconds for the bending after a paper initially agreeing with Newton. The first results from solar eclipse experiments in 1919 gave Einstein the victory. It took radio astronomy work without the need for eclipses to solidify the case. Newton’s scalar potential theory has but one term to warp time, effectively the g00 term of a metric. Einstein had another, equal contributor that bent space. Pessimist say Newton’s theory got it wrong. Optimists say the prediction is half right.

Second order light bending requires a measurement a million times more sensitive. That experiment has not been done yet. The Schwarzswchild metric predicts an additional 10.8 microarcseconds should be viewable. If the exponential metric rules the heavens, then the experimentalists will see 11.5 microarcseconds more bending, a 6.6% difference. The state of the art measurements are about 100 microarcseconds, a 10,000 fold improvement over 1919, but still needing a hundred fold improvent. The fact that the Sun spins and has a quadrapole moment could obscure the difference.

I spend years on SPR (sci.physics.research) pounding the drum for the exponential metric. It was a solution to a differential equation! It was consistent with current tests! It was different at higher resolution! I thought that triple play would be enough to elicit interest. Triple plays are rare. I had a specific testable hypothesis while work on strings to this day does not. Alas, people are into their own areas of study, or their kissing cousins. Rank one proposals for gravity are the work of idiot savants, without the savant part.

In 2004, I found a bullet with the exponential metric’s name on it. I had my funky derivation of the metric which relied on a perturbation solution of a potential equation, plugged into a force equation, rearranged to look like a metric. I also had field equations “Done the right way”™, derived from the Lagrange density. I had yet to take the exponential metric and see if it solved my field equations.

I recall a particular walk from 1340 Comm Ave to my gym.
I had only a vague idea how to "take the exponential metric and see if it solved my field equations." I didn't sweat those details. I was worried about the results of any calculation. Let's say I did figure out how to plug the exponential metric into the field equations and the result was plain old wrong. That got me sweating before I reached the gym. I had been talking about the exponential metric for years.

I remember the realization I had right in front of Natalie's Pizza: if the exponential metric does not solve my field equations, then I would have to announce that negative result on sci.physics.research and on my web site, quaternions.com. Nice clear line of action, but there was backlash in my brain. I wouldn't really have to do that, would I? Yes, I would. Think about anyone else reading up on my claims in SPR about the exponential metric. If and when I found the flaw, I would be obligated to describe it in order to save others from wasting effort in this direction. By the time I reached the gym, I knew I would have to fire a bullet at the exponential metric and report the result.

For three weeks, I did nothing. I was waiting for my mind to settle down. I also did not know how to do the calculation. There was a disincentive to learn how to do a calculation that would lead to a report of failure of four years of work. This was not about me though, it was about Nature. She does so much simultaneously in silence. I was in the deep end of the pool. I didn’t have any swimming buddies as happens in graduate school. At least I was not in the basement, living on the second floor at the time, with a view of 1325 Comm. Ave., Aerosmith’s place in Boston.

Let me set the stage, kinda knowing the calculation to do, but missing some essential details. The first of my four field equations is a variation on Gauss’s law:
$\rho=-\nabla \cdot \frac{\partial A}{\partial t} + \nabla^2 \phi \quad eq ~ 2$
Only the sign on the Laplace operator changes. Fringe physicists before me have constructed a mountain of Maxwellian-gravity crap. I understand why many a pro will avoid this sort of expression. In the static case, the equation is exactly Newton’s law of gravity in the potential form:
$\rho=\nabla^2 \phi \quad eq ~3$
This is the sort of problem given to high school students starting out in calculus: show the charge/R potential is a solution to this differential equation.

Now imagine the potential is constant. Can you find a solution if the charge density is not zero?

This sounds like a trick question. There is nothing to do if the potential is constant. That is the case with the tools provided in high school. The derivatives they teach are all about the function and how it changes. A completely reasonable place to begin an education.

Mathematicians thought about taking derivatives if the background geometry was changing. In that case, one would use a covariant derivative. A covariant derivative has the normal derivative plus something called the connection.
$\nabla A = \partial A + \Gamma A \quad eq~4$
The partial derivative is the one from high school. The gamma is the connection which accounts for how the geometry changes. Math heavyweights can play all kinds of games with the connection. I used the playbook of general relativity. The connection there is metric compatible and torsion free. I spent 4-5 quality months with Sean Carroll’s “Lecture Notes on General Relativity” to get a sense of what those technical constraints mean. The bottom line is that the connection bears the name Christoffel symbols of the second kind. Sounds like a movie by Steven Spielberg.

We live in nearly perfectly flat spacetime. To be super right, I could view the Laplace operator as two covariant derivatives. I was skeptical of that approach, remembering the lesson of early quantum mechanics where if one used two relativistic derivatives, the equations did not describe a hydrogen atom. For almost perfectly flat spacetime, I thought I should take the normal divergence of a covariant derivative of a potential:
$\nabla^2 \phi \rightarrow \partial (\partial \phi + \Gamma \phi) \quad eq ~5$
The Christoffel take three first order derivatives of a metric. A divergence of a Chrisoffel will thus take a second order derivative of a metric. This is the minimum required to constrain a metric by physical conditions, a second order differential equation. Think of it like Newton’s second law for spacetime geometry.

This is where I was stuck for three weeks: I had never calculated the Christoffel symbol for any metric. I was not sure I was going to be able to do the calculation. Plus I was scared of the result. I recall it was a nice Saturday morning. I was wondering what to do, and then I remembered the calculation. Time to crack open a book. I happened to have good luck understanding “Gravitation and Spacetime” by Ohanian and Ruffini. The index leads to a definition soon enough:
$\Gamma^{\alpha}\,_{\nu \mu}=\frac{1}{2}g^{\alpha \beta}(g_{\beta \mu, \nu} + g_{\nu \beta, \mu} - g_{\mu \nu, \beta}) \quad eq. ~6$
Ouch. I started reading around about Christoffels. I found out that one of these three derivatives can be ignored if the metric is static, nice. Another can be ignored if the metric is diagonal, nice. The derivative of an exponential is one I can still remember decades later:
$(e^a)’ = a’ e^a \quad eq.~ 7$
The inverse of the exponential metric will put a minus sign in front of the exponential, so
$\Gamma = e^{-a} (e^a)’ = a’ \quad eq~8$
Nice. The exponent is charge/R, so the divergence is something like:
$\partial \Gamma = \nabla^2 \frac{G M}{c^2 R} \quad eq~9$
Bingo, bingo. This is something Laplace would recognize.

That is the sketch. Details can kill, like one sign error. I have run the exponential metric through Mathematica using functions written by James Hartle. It is section 4, page 7 in html , a pdf or notebook, if you wish to cross check. Just as important, I told a friend to see if he could independently confirm this observation about the exponential metric in Gauss’s gravity law. He came back and said no, it did not work. I went over the details with him. He had done two covariant derivatives. When he did just one, it worked.

I have a memory of reading in one of my GR books that one should never bother to take the divergence of a Christoffel symbol. The calculation is banished because a Christoffel doesn’t transform like a tensor. Fortunately I am doing something different. I am taking the divergence of a covariant derivative of a 4-potential. The 4-potential is constant while the metric is dynamic, thus reducing the problem to the divergence of a Christoffel. Did I tell you my father went to Harvard and Harvard Law as did his father? I got wait listed, so went to the tech school down the road, my first choice.

What is cool about this observation is that Newton’s old scalar potential law for gravity can be expanded to a 4-potential. Do gravity as a potential theory or as a metric theory, the choice is yours. The exponential metric is not a solution to the general relativity field equations. In 1919, Einstein’s gang of 10 equations beat Newton’s scalar theory. In 2011, a 4-vector Newton could challenge Einstein for a rematch. This time Newton predicts more bending. Go new Newton.

The exponential metric survived the potentially deadly calculation.

Doug

Snarky puzzle. If you go down in a gravity field, time ticks slower, rulers look longer to someone else. That is 2 changes. Newton’s scalar theory accounts for just one, duh. Write a 4-potential that could account for both g00 and gRR. If you don’t want Newton to do all the lifting, combine a metric with a potential. Since mixing potential theory and dynamic metrics is blasphemy, expect to take a bullet for the team.

Next Monday/Tuesday: The Certainty and Uncertainty Principles

So the point the exercise is to find a wrong answer? I think I can do this!

...But grr and gtt are potentials...and they form a 2nd Rank Tensor... (And 4-Vectors are Rank 1)...

Doug Sweetser,
You should go to my homepage to read my 'Beyond Albert Einstein's Relativity: UFT Physics' because I am an elite genius at mathematics and spent more than seven thousand hours on pondering about and creating this UNIFIED FIELD THEORY. My paper goes to about one thousand times the universal physics formulae incorporation depth that General Relativity goes to. Although my paper would only be comprehended by elite geniuses at mathematics, the formulae produced within it should simplify the present haphazard layout of Physics education in schools and universities, and simplify formulae for practical users of physics, such as engineers. My extensive physics paper is held at Australia's national library because I decided to stop my obsessive improvement of physics formulae. One day, after I die impoverished, physicists will realise how brilliant I was. In 1999, I was tested by university psychology researchers and found to be the cleverest person out of millions tested at Abstract Reasoning.

Hello Peter:

Take none of this personally, it is all technical.

The site is http://uftphysics.webs.com/.

I run a number of tests. The first one is to see if there are any equations. Yours has a few, so it passes that test.

If any site claims to have field equations, I search for "action", "Lagrangian", "Lagrange density", or "Hamiltonian". A modern field theory proposal requires a Lagrangian which can reveal critial symmetries in the proposal. A Lagrangian also provides roads to quantizing the work. I spent a few years looking for the one behind my efforts. The site failed that test.

If the work has not passed a peer-review process, I search for "Mathematica", "Math Works", "Matlab", or "Maple". These are symbolic math programs that can be used to confirm the validity of the math shown on the site. Quality control is essential in mathematical work. The site failed that test.

"Maxwell’s Equations do not apply to a multi-particled system of charged particles."

I am confident this claim is wrong. Please study "Classical Electrodynamics" by Jackson.

"Maxwell’s Equations do not use the centre-of-mass of the entire system as the reference point for distances involved in the equations."

I think you want the Lorentz force law. That can be derived by keeping the potential constant, but varying the 4-velocity. Read Landau and Lifshitz "Classical fields.

"Maxwell’s Equations do not include the gravitational components as in Schwarzschild’s equation to predict gravitational bending of the space-time continuum."

The Maxwell equations are about EM, not gravity, so no surprise.

"Maxwell’s Equations do not explain particles with similar charges attracting each other when at very small distance apart."

There is more than enough experimental data to say this claim is silly. Such an effect would surely be noticed by folks building integrated circuits.

"Maxwell’s Equations do not explain why mass, like energy, is variable with charge."

Yup, we do not understand why there are various charge to mass ratios for different particles. This does not mean that the Maxwell field equations are incorrect. Our understanding of Nature is incomplete, a different issue.

There are many equations weh site that make no sense in terms of their symbols. Under "What is time in modern physics", I see a with a circle over it. I have no idea what that is. If you need help formating equations, go here: http://www.codecogs.com/latex/eqneditor.php. That will make a very pretty image that can be included in a web site. It is the technology used here at Science20.com.

"THIS INDICATES THAT A RELATIVISTIC EFFECT ON THE COULOMB FORCE EXISTS."

This was important enough to warrant ALL CAPS. It is also well understood. Look up the Lorentz force law, which for low speeds becomes the Coulomb force law.

"Everything in existence, including photons, sub-atomic particles, massive suns, and humans, consists of wave-particles of the same formulae, and is a manifestation of electromagnetic waves commonly called light."

I don't take this claim written in larger type and put in a box seriously. Photons have spin 1. Humans are made up of many fermions, most with a spin of 1/2. Light and electrons do play nicely with each other, the subject of their study being quantum electrodynamics.

Of the 39 major equations, I recommend a study of "Spacetime Physics" by Taylor and Wheeler. It is doubtful you understand the energy-momentum 4-vector. There is not a M_i and M_io.

$(E, \vec{Pc})^2 = (E^2 - P^2 c^2, 2 E \vec{P} c) = (m^2 c^4, 2 E \vec{P})$

I wish this expression was more widely taught. Both E and P are Lorentz covariant, meaning we understand how they change under a Lorentz boost. The inertial mass m is a Lorentz invariant quantity, so it does not change under a boost.

I don't think your work is technically of any value. I do hope you find a way out of poverty and your other listed ailments. If you want to write a blog at Science20.com, you are free to do so. You will be able to monitor how many people view your posts. The comments might be helpful, but I suspect would be more caustic than mine (I saw a few such appraisals through a web search).

I don't expect you to accept my critiques. After all, you are more clever than I. I was good enough to get into MIT, but so where a thousand of my classmates. All I have done here is to be specific into why I do not accept your claims which is my responsibility as a skeptic.

Doug

Important note: I will delete any further comments about UFT Physics since this blog is not about that body of work.

Doug Sweetener.
Dear Sir,
After seeing your comments about my UFT Physics, I am very thankful and have a lot of respect for you.
Due to progressive brain-damage, at 57, I find it difficult to read much and am significantly less intelligent than I was, though I am very creative at more artistic endeavours. Therefore, I do not intend to pursue the complicated analysis of such physics beyond the effort I sent to Australia's National Library.
A few times, while reading your responses, I believed that you expressed ultra-conservative attitudes that affected your mathematical appreciation of what I was proposing, so you denied real possibilities. I began my intellectual quest on this topic when I was 16, about 41 years ago. We did not have computers, did not have advanced books on relativity in our Brisbane libraries, and did not have most of our present technological advances including probably integrated circuits, so you should realise that my statements are not 'silly' statements. (Young people with modern technological advancements are as silly as older people were at their ages. Maybe young people are much more flippant because they rely too heavily upon other people's misguided opinions expressed on the internet, rather than rely on struggling, for thousands of hours, to create their personal opinions.)
I did not use any 'a with a circle above it', so I believe that you are seeing some, or maybe all, of my formulae differently from how I intended them to be.
When you stated that it is doubtful that I understand the energy-momentum 4-vector, you are completely incorrect. After modifying that equation about a thousand times or more, I did understand that equation very well, but decided that it was too simplistic and inappropriate. Understanding what I created is much more difficult than understanding that. The equation is an aesthetically pleasing equation in appearance, but physics is not poetry. That equation should not be in textbooks because it has been very damaging, probably the most damaging equation, to the advancement of theoretical physics.
As I have stated, I am very thankful to you and respect you for your efforts concerning this topic. My physics paper introduces much interesting conjecture, so I hope that young people consider the possible ramifications of its equations. After spending more than seven thousand hours on very many thousands of possible equations, I am happy when thinking about other non-mathematical topics.
Keep up your excellent work, and enjoy a good life.
Peter Donald Rodgers 30/06/2011

I believe Peter is one of the most dedicated trolls on the internet. I'm sceptical of anyone who mentions their IQ as as support for their ideas. Einstein didn't need to mention how smart he was. His ideas were correct.

Doug, regarding equation 5 and "I thought I should take the normal divergence of a covariant derivative of a potential."

A subtle point: The Laplacian is "the divergence of the gradient." The divergence [of a vector-here the vector is the covariant derivative of a scalar] is formed contracting the indices. But the gradient isn't a vector [It is a one form]! So you must convert it to a vector before taking the covariant derivative and contracting.

(See "The Divergence and Laplacian" Schutz Intro to GR page 137-138 if possible. Calculating the Laplacian in polar coordinates. There the Laplacian is "the divergence of the gradient" )

...I'm not 100% sure you didn't do that or something equivalent to it by the way. You've obviously left out the details to make the blog readable. Maybe you could attach a short mathematical appendix to the blog with calculations worked out in detail for people like myself. Just a suggestion.

The details do get difficult. That is why I insisted on getting Mathematica and an independent nerd to confirm the calculation. I am going to cut and paste from that notebook to see if this works in showing the details underlying the claim in the blog.
<start of quoted text>
Here is the starting equation:
$\rho = \nabla \cdot \nabla \phi \quad$
[Note: I am going to skip the terms involving J. If you can follow this blog and want to see J, see the paper where this was lifted from.] Assume gravity is weak. Although there are two covariant derivatives, if gravity is weak then we do not need to account for the second-order curvature. To use the machinery of curvature, this section will require the use of tensors, indexes, and Christoffel symbols. Write out the covariant derivative [Note: I switched from 1-4 to 0-3 from the cited link because it make more sense to my eye, and I included the A's as I should have due to the suggestion in a comment below]:
$\rho \approx \partial _{\mu } \left(\partial^{\mu } A_0+ \Gamma _0{}^{\mu \sigma }A_{\sigma }\right) =\partial _x \left(\partial^x \phi + \Gamma _{0 }{}^{x, 0-3}A_{0-3}\right)+ \partial _y \left(\partial^y \phi + \Gamma _{0 }{}^{y, 0-3}A_{0-3}\right)+\partial _z \left(\partial^z \phi + \Gamma _{0 }{}^{z, 0-3}A_{0-3}\right) \quad eq. 24$
The approximation arises from the first derivative symbol in each equation which is approximately the standard derivative without the connection. The expressions with x's, y's, and z's are not written in a standard way, but I do this for pedagogical reasons. Everything should be in Greek letters or numbers.  Yet my mind is tied to x-y-z. Instead of the Einstein summation convention, I wrote 1-4 to remind the reader that all four Gammas and potentials will be required to evaluate the derivative. There may be one derivatives, but four possible contributions from the connection.  In practices, for the metric in question there are at most two non-zero terms.

The derivatives of the potentials are all zero given the assumptions, leaving only derivatives of Christoffel symbols:
$\rho \approx \partial _x \left(\Gamma _{0 }{}^{x, 0-3}A_{0-3}\right)+ \partial _y \left(\Gamma _{0 }{}^{y, 0-3}A_{0-3}\right)+\partial _z \left(\Gamma _{0 }{}^{z, 0-3}A_{0-3}\right) \quad eq. 25$
Find a metric that solves these four equations. The metric must reduce to the Minkowski metric as the gravitational mass goes to zero. Test if the exponential metric below provides a solution with these properties:
$g_{\mu \nu }=diag\left(\exp \left(-\frac{2 G M}{c^2R}\right),-\exp \left(\frac{2 G M}{c^2R}\right),-\exp \left(\frac{2 G M}{c^2R}\right),-\exp \left(\frac{2 G M}{c^2R}\right)\right)$

There are 21 non-zero Christoffel symbols. The ones relevant to the field equations follow a simple pattern for n and m going from 1 to 3:$\frac{G M x_n}{c^2 R^3} = \Gamma _0{}^{0n}A_{0-3}=\Gamma _n{}^{m m}A_{0-3}=-\Gamma _m{}^{n m}A_{0-3} \quad eq. 27$
To make this more explicit, I have swapped x->n, y->m:

$\frac{G M x}{c^2 R^3} = \Gamma _0{}^{0x}A_{0-3}=\Gamma _x{}^{y y}A_{0-3}=-\Gamma _y{}^{x y}A_{0-3} \quad eq. 28$

Notice the simple form: the Christoffel symbol equals the first derivative of a charge over distance. The reason this happens is that the Christoffel symbol has a contraction between three derivatives of a metric and the inverse of that metric. The derivative of the exponential metric will have the exponential times the first derivative of charge over distance. The exponential and its inverse politely cancel each other out, leaving only the first derivative of the charge over distance. For the symmetric Gauss's law, one then takes derivatives of the Christoffel symbols  which is precisely the same as the Laplacian of a charge over distance. This is remarkable since the machinery of the Christoffel symbol is so much more complicated than the Laplace operator. The calculation for the symmetric Ampere's law involves more terms. The reader is invited to confirm that they all cancel, which sounds reasonable for a static, spherically symmetric source.

Proving the connections solve the field equations require Mathematica functions found in a notebook by James Hartle from his book, Gravity: An Introduction to Einstein's General Relativity [skipping the details which can only be found in the mathematica notebook form]

For Gauss's law, the only derivative that comes into play is the Gamma{44x}'s because the 4's are the phi's in equation 25 (sorry, some eq's numbers were lost in translation).Take their derivatives:

$\frac{\partial \Gamma _0{}\,^{0 x} A_{0-3}}{\partial x}=-\frac{3 G M x^2}{c^2 \left(x^2+y^2+z^2\right)^{5/2}}+\frac{G M}{c^2 \left(x^2+y^2+z^2\right)^{3/2}} \quad eq. 29$

$\frac{\partial \Gamma _0{}\,^{0 y} A_{0-3}}{\partial y}=-\frac{3 \, G M y^2}{c^2 \left(x^2+y^2+z^2\right)^{5/2}}+\frac{G M}{c^2 \left(x^2+y^2+z^2\right)^{3/2}}$

$\frac{\partial \Gamma _0{}\,^{0 z} A_{0-3}}{\partial z}=-\frac{3 \, G M z^2}{c^2 \left(x^2+y^2+z^2\right)^{5/2}}+\frac{G M}{c^2 \left(x^2+y^2+z^2\right)^{3/2}}$

$\frac{\partial \Gamma _0{}\,^{0 x} A_{0-3}}{\partial x}+\frac{\partial \Gamma _0{}\,^{0 y} A_{0-3}}{\partial y}+\frac{\partial \Gamma _0{}\,^{0 z} A_{0-3}}{\partial z}=0 \quad eq. 30$
Zero? The same thing happens with the 1/R solution for Gauss's law. In fact, despite all the complications of the Christoffel symbols, this solution is precisely the same as the 1/R potential, using the same trick of three terms adding together and canceling.
<End quoted source>

This is the only calculation I have ever done with a covariant derivative. I am not a bastion of confidence on this subject. Hopefully this provides enough detail so you don't think I cut corners.

Funny story... On that Saturday, I was doing this calculation with paper and pencil, lost in that math phase of mind where you lose track of time. I added up the three derivatives and saw the zero. I was crushed. Did a recheck, things looked good, the cancellation looked too perfect. I put the pencil down, a little bewildered. I went around for a day, surprised I had done such a hard calculation and gotten a wrong answer. I wanted to see a charge on the other side, not zero. The next day I remembered the same think happens for a 1/R potential which I looked up in Jackson. The result was better than I had thought. The exponential metric and 1/R potential will be spot on exactly the same mathematically. Nice.

Ubi major , minor cessat !

When it comes to proposals in science though, I go for head shots. Nature doesn’t care if we keep repackaging bullshit. A well-fed bull does not run out of bullshit.
That's some fine prose!
Thank you. The logic is clear.

But shouldn't there be "A"s in equation 29? I understand that you say that the partial derivatives of the "A"s are zero, but, I believe, the "A"s should still be there. You're Laplacian was d(dA+PA) where P=Christoffel symbol. Shouldn't this give: dP A? assuming dA=0)? I was expecting something that involved A and end with the Christoffel symbols.

My understanding is this: The vector potential is different from the ordinary potential. It doesn't involve the gradiant, but the curl. The scalar potential f satisfies del(f)=o and the vector potential satisfies curl{A} =0. I personally would have started there.

Glad to hear the logic is clear, now to work on notation. I went back and shifted from 1-4 to 0-3 in the above comment. While I normally use 0-3, the software of Hartle used the 1-4 convention which is why I will keep it that way in the Mathematica notebook. That is not a requirement for the blog, and the 0 makes my eye happier.
The A's have been added per your suggestion. I hope the notation becomes as clear as the logic. I recorded the LaTeX to a library in LaTeXiT, a good tool for equations on the Mac.

Gauss's law only checks the g00. An Ampere-like equation is required to check the other terms in the metric.
Thanks. I'm unsure about the physical meaning of the equations, but the mathematical ideas were very interesting.

I am unsure what the difference is between the physical and the mathematical. Unsure is a good feeling, it is the sign that research and learning are happening. I'll try to make this more physical.
What every particle strives to do is absolutely nothing. A universal vacuum is a great place because all the calculations are simple: the Universe is 100% the Minkowski metric. Actually, I am a quaternions guy, so I would say you get quaternion products. You can skip the quaternions angle if you like, but they say similar things. Snap your fingers twice in an utterly empty universe and the interval between the finger snaps is:
$(dt, d\vec{R}/c)^2=(dt^2 - dR^2/c^2, 2~ dt ~d\vec{R}/c)$

The Lorentz invariant interval is the first term. Quaternions give the Lorentz covariant 3-vector, not that any one cares about it but me. Anyway, the point is that in a totally empty Universe, that is the interval between two events.

What happens when something else is in the Universe? The deep message of both general relativity and my own approach to gravity is this: not much. The amount that interval changes is joke small.

What is sitting in front of the dt2 and the dR/c2 in empty space is a 1. Ones are popular in physics. Now we cannot have just a 1, we need a 1 and something else. We want something like is a friend to harmonic motion. Think about the Earth and the Sun. The Earth has gone around the Sun 4 billion times, kind of a regular pattern. When an apple falls from a tree, it is beginning an hour and a half journey from the tree, through the center of the Earth, to its antipode, then back again. The apple never completes the journey since all the other particles are trying to do the same thing. If you could hold a cup of neutrinos, then spill them, they would oscillate like that, showing up every hour and a half at the spill site.

If you want to model simple harmonic motion, exponentials are the way to go. In my approach to gravity, they do show up. In GR, in order to derive the Schwarzschild metric, exponential are used in an intermediate stage, but not the end result.

In GR, gravity is exclusively about dynamic metrics. In some ways, my rank 1 field theory is simpler because a rank 2 field theory just has more parts to it. Yet what is surprising is that the rank 1 field theory is not exclusively about dynamic metrics. Instead it is a choice between dynamic metrics and potentials. If you want to play only with 4-potentials, fine, work with a flat metric. Yet you can do everything with a dynamic metric, writing off the potentials. For those who like to compromise, both can be used at once. It makes no difference physically if one uses potentials or metric.

Doug
Have you read any of Einstein's papers? The Special Relatvity (1905) and General Relativity (1916) papers are suprisingly straightforward. They are also excellent examples of what I'm getting at when I ask what the physical meaning of your work is. The structure in both is 1.A physical assumption is expressed mathematically and, from the math, consequences are derived. In the first time dilation etc. is derived and in the second the precession of the perihelion of mercury is derived. The 1916 paper is the archetype of what physics should be (in my opinion). [Actually, the papar is an excellent introduction to GR in its own right. The first half is basically "an introduction to differential geometry and tensors" for the physicists of the day who didn't know what a Christoffel symbol was].

My feeble attempt at a physical explanationof your equations:

It appears to me that the final equations are saying something like "the components of the vector potential act individually like scalar potentials due to electrical point charges, therefore a masspoint, in a static gravitational field should field an influence of three scalar field such that...[insert experimental prediction here if possible]."

[This is nonsense, but that it is the flavor of what i mean by "physical meaning"]

I recall reading that Sean Carrol proved, as his Ph.D thesis, that gravity must be described geometrically. An attempt to create a regular field theory is bound to fail, but I'm not sure about the details. I've always accepted that it must be so. It's hard to imagine rank 1 theory capturing all the same effects that the rank two theory captures. Actually, it isn't just about the rank of the vector because you can have rank 2 cartesian tensors that are not related to gravity at all. It is really the fact that gravity is metrical (and a general 2nd Rank Tensor) that captures everything. The Equivalence Principle etc.

The most direct method to show that your theory is correct would be to show that it reduces (or correscponds) to GR in the same manner that GR reduces to Newtonian Gravity....I think you might have done this already somewhere else.

Glad to see you set the bar high :-) I did make one serious effort at a paper that started with quaternions and hypercomplex numbers. That paper was voted down by a journal editor because he thought those tools too obscure to be of interest to his audience. He recommended a journal that was into alternative math. An editor there pointed out how I did not reference articles from their own journal or similar sources as inspiration. Since I could not make up such a reference, I have to back to finding more results. It is not worth my scarce time to go editor shopping. It is more important to learn how to state my proposal clearly, perhaps someday catching the eye of a real professional.
Here is a second swing of the bat on the physical interpretation...

The Lagrangian for the hypercomplex gravity proposal has a structure similar to that of the Maxwell equations written with quaternions. It thus has mass charges and will have linear charges in a vacuum. Gravity fields do not gravitate in this proposal, a key difference from GR. The hypercomplex Lagrangian uses only completely symmetric multiplication, so it can capture completely symmetric changes that a change in a metric represent, changes that are lost to quaternions due to the antisymmetric cross product. First order weak field gravity predictions are identical for they exponential versus Schwarzschild metric. For second order bending, the two theories differ, with the exponential metric predicting 12% more bending in the microacrsecond range.
I recall reading that Sean Carrol proved, as his Ph.D thesis, that gravity must be described geometrically. An attempt to create a regular field theory is bound to fail, but I'm not sure about the details. I've always accepted that it must be so.
This sounds like a widely held belief. It may be a reason I have to use a different kind of algebra. The only way to work today is with tensors. A move to quaternions for EM, the weak, and the strong force, and hypercomplex numbers for gravity, would require .. I don't know what it will take. Such proofs asume tensors are the underlying accounting system of mathematical physics. I view quaternions are a restricted subset of tensors (they work only in 4D), but the restriction comes with benefits, namely multiplication. That is a hook into group theory, so worth the cost. In the standard model, groups are stapled to the chicken, a phrase I made up to describe how awkwardly they are included in the Lagrangian.
The most direct method to show that your theory is correct would be to show that it reduces (or correscponds) to GR in the same manner that GR reduces to Newtonian Gravity....I think you might have done this already somewhere else.
GR is a fundamentally different proposal than Newton's work. The Taylor series expansion of the g00 term in the Schwarzschild metric has a contribution that is equivalent to Newton's scalar theory. The hypercomplex gravity proposal is different from GR (solutions to one are not solutions to the other). There is overlap in the Taylor series:
1. The Schwarzschild metric:

2. The exponential metric:
The underlined terms are those that get tested in current weak field tests, and they are identical between the two metrics. Here is another way of comparing them:
Both metric start out as the Minkowski metric. Note that exponential metric remains symmetric for time versus space terms for higher order terms, a good thing I hope.

Doug

"GR is a fundamentally different proposal than Newton's work." I meant, in the weak field limit, Einstien's Field Equations reduce to the Poisson Equation and Newton's Law of Gravity follows from the Einstein's Geodesic Law of Gravity. I'm going to read your post again (I'm tired now), but, on a first reading, it looks like you've answered my question. Your work recaptures (almost) the Scw. solution.

It is interesting that this exp metric is so close to the Scwh. metric, yet it isn't a solution to Einstein's equations. I wonder how you guessed it would be a solution to your field equation before calculating. Getting a Field equation that is close to Einsteins, without being the same, is an accomplishment.

But the Scw. Solution isn't all there is to the equation. I wonder how yours accounts for the principle of equivalence. I don't expect you to answer all my questions. I'm sure you're busy. Thanks for providing the details.

I wonder how yours accounts for the principle of equivalence.
I will blog about that issue someday. The short-of-a-blog answer is that having a dynamic metric that applies to all particles no matter what they are made of goes a long way for the classical principle of equivalence. If I am reading the back of my t-shirt correctly, there may be something to say about a quantum field theory justification for the equivalence of gravitational and inertial mass. That calim will have to remain speculative until confirmed by RSP [Really Smart People].
"I thought I should take the normal divergence of a covariant derivative of a potential: df + ChristoffelXf "

I was watching your video when I spotted a potential error. I believe that statement is mistaken. The covariant derivative of a scalar is (a set of) regular partial derivative(s). So there shouldn't be a Christoffel symbol there at all. However, the derivatives (with respect to time, and space coordinates) of the scalar potential form a covariant tensor (the gradient).

So, I believe, you should take the derivatives (and these form a vector- the gradient,in fact) and then take the covariant derivative of that rank one tensor.

I researched this before posting and found this confirmation:

"What is the covariant derivative of a scalar? The cov derivative differs from the partial derivative with respect to the coordinates only because the basis vectors change. But a scalar does not depend on the basis vectors, so its covariant derivative is the same as the partial derivative, which is is gradiant" Schutz. A First Course in GR page 137

I would've said derivativeS with a "s" because the gradient is actually a set of derivatives (one with respect to each component and forming a vector. I don't know why they use that language. They call the gradient "a derivative" when it is really a set of derivatives-in tensor notation at least)

It is possible that the derivation is still correct for some reason, but it would be strange because the Christoffels are not changing.In fact, they are 0.

Schutz describes the Laplacian as the divergence of the gradient (page 138) where the gradient is the standard gradient (a rank 1 covariant tensor) and the divergence is the covariant derivative (here it acts on the gradient) with contracted indices. So it's sort of the opposite of what you did.

http://en.wikipedia.org/wiki/Covariant_derivative
http://mathworld.wolfram.com/TensorLaplacian.html

I have an implied "4-potential" in that potential statement. I never work with scalar potentials, only 4-potentials as happens in EM theory. With that clarification, does that resolve this issue?
Doug
I'm sorry if I interrupt your arguments. I would really attend to your arguments but I would ask to the 'opinion of bloggers.
My problem is the following, is possible that the correlation between mass and energy right now regarded as in fact does not cover the material point ?
The way I observe it : the correlation is well known, E = mc ^ 2, or also E = mcc , now with mass almost zero
E = cc
Now I reduce time almost zero , the smallest fraction of time , will be E = L x L. ( L is simple distance)
At this point I can see that for infinitesimal masses and infinitesimal time the energy E is equivalent to square (LxL) ... A surface . If it is correct what I wrote until now the correlation between mass and energy is not proposing a point-like mass, but a plane or flat body !
Now my question: in physics, mass is a point or a surface ?.
Again... If the ultimate constituent of mass is a surface, the gap between two masses is a point or an area too?
I believe that to take a break is a valid argument ... Giacomo. (Italian citizen) I have another question Heisenberg's Uncertainty Principles . Thank you very much.

the correlation is well known, E = mc ^ 2, or also E = mcc , now with mass almost zero
E = cc"

According to E=mc^2, setting the mass equal to zero results in E=0 (just plug in 0 for mass).

"Now I reduce time almost zero , the smallest fraction of time , will be E = L x L. ( L is simple distance)"

You seem to have made the same mistake here. If you know anything about calculus, you should now that, for infinitismal time displacements, the finite velocities merely reduce to infinitismal velocities. The units remain the same. You appear to think that things go to 1 as they get infinitly small. In fact, they go to 0.

Now, you might be thinking, "but the energy of light isn't 0." That's correct. E=mc^2 only applies to rest masses.
E^2=(mc^2)^2 + (pc)^2. A photon (the mass 0 particle you were interested in) therefore has E=p/c.

"If it is correct..."

It wasn't correct. So your question is meaningless. Sorry.

I'm sorry, but you don't understand physics. If you think you do, I have a test for you. Pick up an elementary calculus based physics textbook and answer one problem from each chapter. If you can do that you have some understanding. If you pass this first test, continue advancing to higher level textbooks. But, from what you've written, I doubt you could get past the basics. The point is, if you can't solve problems you don't understand the subject.

My name is really Giacomo.

I'm not sure how to interpret your response. Why is that relevent? Perhaps, you feel that I was rude, but I really wasn't trying to be. I was merely answering your question and giving honest advice. If I'm correct and you take my advice, then I've probably saved you an enormous amount of time you might have spent thinking about something that is actually nonsense.

Correction to my previous post:

E=pc or p=E/c.
(Not "E=p/c")

Sorry but I think I expressed myself badly. The 'exposure is wrong and the translation is worse than usual. I apologize for writing something so wrong. In math I am a disaster.

As I have done before, I recommend reading "Spacetime Physics" by Taylor and Wheeler. The ideas are deep, but that math is not too hard. Good luck in your studies.
Thank you really, I read a book by Wheeler. I repeat, I'm not trying to show off my mathematical knowledge, I only asked to those participating or following, the possibility that fundamentals of reality, the mass and the void, can be are squares forms! If so it might be easier to understand what connection there is between the deep mathematics of quaternions and the reality. This was my intention.So if the mass is a square or point, introduced the theme. Ubi major minor cessat! Thank you for your patience.

In "Spacetime Physics", they go into something called momenergy, a word that has not caught on. People call momenergy 4-momentum. In the first slot is energy, followed by 3 values of momentum. Both energy and momentum are conserved. They are a 4-vector:$(E,c \vec{P})=(E,c P_x,c P_y,c P_z) = m c^2(\gamma, \gamma \vec{\beta})$

where:
$\vec{\beta}=\vec{v}/c\quad\gamma=\frac{1}{\sqrt{1 - \beta^2}}$

Different observers have different values for the beta.

To my ear, what you have written sounds like physics babbling. You are using words that do appear in physics, but the logic doesn't make any sense to me. I babble before I learn to speak about new technical areas.

One thing I avoid is claiming too much about quaternions. Nearly everything done with quaternions can be done with other math tools, in particular tensors. The 4-momentum looks different to different inertial observers whether one uses tensors or quaternions. Square the 4-momentum as a quaternion:
$(E,c\vec{P})^2=m^2 c^4(1, 2 \gamma^2 \vec{\beta})$

The "1" on the right hand side is 1 for all intertial observers. The same thing happens from a tensor contraction. As far as I can tell, physicists do not work with the gamma2 beta. It does change for different intertial observers in ways we understand. Since it is well-formed, I think it should be put to use. I am sure there is a way to that term with tensors, but I suspect most would say "Why bother?" I don't have a grand reply other than to appeal to completeness. Nature keeps track of everything, so should we with our equations.