RETRACTION: Deriving And Fixing The Force Equations (6/5+1)
By Doug Sweetser | November 1st 2011 01:39 AM | 41 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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RETRACTION: I have decided to retract three blogs (Deriving … 4/5, 5/5, 6/5+1). I was unable to figure out a reasonable statement concerning gauge symmetry. When the blogs were initially written, I focused on the field equations, mainly the Gauss-like law, and ignored the force equations entirely. Finding a solution that works with the the field and force equations were not looked for. A consistent proposal should do all three things (fields, forces, and solutions) with grace. I have concluded it is not possible to achieve these goals with the Lagrangian as written, hence the retraction.
The hypercomplex gravity and unified GEM Lagrange densities was wrong.

Nice clear admission of error, so rare these days. My critics think my fix remains wrong, so you will have to be the judge. Skepticism is a good thing. Here is the fix:
$J^* \boxtimes A \rightarrow J \boxtimes A$

The conjugate operator is the *, basically I am removing a mirror. As much as is practical, I will go through my 30+ blogs and make the correction (should not be many posts). I have stopped the global sales of all t-shirts since the errant conjugate appears on the back. The redesign is complete, but new shirts have not been printed.

click or skip this reading of the blog:

To a fault, I have a hope that my equations will be super similar to those of EM. The current coupling term and field strength tensor of EM are invariant under 3D rotations and boosts. When I started to work with hypercomplex products, I wanted them to have that property too. It was easy to do for the current coupling term - that is why I hung a conjugate on the J. Nice, simple, direct. There are conjugates needed for the field strength.

What I did not calculate was how the hypercomplex field equations transformed under 3D rotations and boosts. Why not? There turns out to be a mountain of detail in such a calculation. I did try to do it, but did not complete the effort. That happens. So what I did since I didn't know the answer was to ask a snarky puzzle question about it, to force myself to answer the question.

Several sharp readers could see what the answer was: the hypercomplex field strength tensor was not invariant under 3D rotations and boosts. It did take me ten days to do my own effort and confirm their observations. All well and good.

Not really though. If one term in a Lagrange density is invariant under a Lorentz transformation and another one isn't, that is a problem. I actually did not spot that internal inconsistency at the time.

Let's go back to two things I did derive. I got Gauss's law for EM:
$\rho = - \nabla \frac{\partial A}{\partial t} - \nabla^2 \phi$
In the static case, this reduces to:
$\rho = - \nabla^2 \phi$
Looks old school.

Here are my hypercomplex gravity field equations:
$\rho = - \nabla \frac{\partial A}{\partial t} + \nabla^2 \phi$
In the static case, this reduces to:
$\rho = + \nabla^2 \phi$
This one appears in Misner, Thorne, and Wheeler as a good old field theory for gravity (not my invention). Not much different between those two forces or those two equations except a sign. Like charges repel for EM, but attract for gravity.

Both of those equations have similar, but not the same solutions. It is a really cool cancellation that solves the vacuum equation, when there is no rho. What is darn tricky is showing that the q/R solution works when there is a rho. It does work, if and only if rho is negative. That's how it goes in EM.

To get a solution for the gravity equations takes more thought. There is a wonderful article on the subject by Kevin Brown, "Why Maxwell Couldn't Explain Gravity". It is well worth the read. Maxwell was so sharp. He wanted to treat gravity as something that had positive, definite energy. Yet the phi needed a minus so that like charges attract. He realized that by the rules of integral calculus, one could staple on a HUGE positive value to the phi to keep it positive overall. Unfortunately, he was not able to justify that move, so he remained puzzled about this subject to the end of his days.

What Brown argued was the Schwarzschild solution to the field equations of general relativity resolved this issue. The first two terms of the Taylor series expansion of the square root of g00 as R goes to infinity could be used to define phi:
$\phi = 1 - \frac{G M}{c^2 R}$

Almost no one includes that big factor of 1. I call this this Ed Miksch big factor of one. He is an older guy, going to APS meetings, talking about negative mass. Personally, I think he is channeling the issue Maxwell knew and solved but could not justify until after his death, with the work of Einstein and Schwarzschild.

So to my eye, I think I have delivered to the Science20 community field equations where like charges attract. That has not changed at all, even with the above admission of error which does not alter the first field equation.

The super tricky thing is that everything absolutely has to fit in together perfectly. That phi up there, that is the one I have to use in whatever force equation I derive. I did say I had never done that before right? Oh, I had done the EM Lorentz force lots of times, but not one for gravity.

People are so much more comfortable with forces than field equations. Forces appear so direct, good old pushing and shoving. There is a direct link between the force and field equations. This was in the Bible according to Purcell, "Electricity and Magnetism, Vol 2", p. 24
"We call the statement in the box [Gauss's Law] because it is equivalent to Coulomb's [force] law and it could server equally well as the basic law of electrostatic interactions, after charges and fields have been defined. Gauss's law and Coulomb's law are not two independent physical laws, but the same law expressed in different ways."
He does add a footnote that Gauss's law is more general that Coulomb's force law.
Here was my logic thinking that, in general, I must have a force equation where like charges attract. Coulomb's law says like electric forces repel. To be consistent with that, Gauss's law must also say that like electric forces repel. Newton's field equation is the opposite of Gauss's field equation, so like charges must attract. That must result in a force law where like charges attract.

That is not a circle of logic, it is a path, from Coulomb, to Gauss, to Newton, to a force law. I could say with utmost confidence that my field equations demand that like charges attract.

So where was my force equation?

I didn't have one.

Oh, I had tried a couple of times, but the calculation never worked out. I figured I was doing something wrong. It was one of the rare times when I made a file and put the calculation away. I knew I would have to do the direct calculation someday. I would revisit it from time to time, but when it flopped I just figured I was not bright enough to do it.

And getting the form of the equation is not enough. The same solution has to work between the force and the field equations. I was slow to pick up on that constraint. It is easier to treat them independently, which was my own fault.

In this derivation series, several people said the force equation, the one I had not figured out how to calculate, showed that like charges repelled. This is what drove me to pull out the folder and try again. Last weekend I was able to see what they said, that the standard derivation of the Lorentz force equation of EM always gave the Lorentz force equation of EM if one used the same current coupling term. Like charges repel for the Lorentz force equation.

The derivation was always ending up in the house of EM were like charges repel. The details that bothered me last weekend as it did in the past was I got all kinds of cross products. The entire thrust of the hypercomplex gravity proposal was symmetric cross products took the place of the standard cross product. That was not happening. Why?

Like the thorough experimenter I am, I decided to challenge an assumption I had made a few years ago, that the current coupling term for gravity should be a Lorentz invariant scalar. I erased the conjugate operator, repeated the calculation, and got my symmetric curls. The form was good. Too bad there was no way to find a solution that worked for both of them. A few more fair kicks to the groin by the critics, and I saw how to make the force equations and the field equations use the same solution. That is all I will present here, not the less than elegant process that got me this far.

Here is how I will proceed. I will derive the Lorentz force for EM nice and neat, so those folks who like the math can follow along. I will include the extension that gets to Coulomb's law. No matter which side of the fence you sit on in regards to my own work, that derivation will be a value or review. I will then repeat the calculation, side by side, so you can spot both Coulomb's law and the one Newton wrote down a good number of years ago. He did not derive the gravitational force this way, what with the calculus of variations a few years off into the future, relying instead on observational data.

The derive series was focused on fields. This is the modern way to do physics. I actually had some trouble switching back to Lagrangians for good old particles. That is what is needed to derive the Lorentz force. This is about pushing around a particle. There are two things one varies: position and velocity. I don't have the time/energy/stamina to do that, show this should look familiar to the standard Euler-Lagrange, but differing in some detail, like using the total time derivative (a critical detail by the way).

We start with a Lagrangian - no longer a density - that has a kinetic energy term and one for the current coupling and... nothing else. This is some serious minimalism. [Update: included the free field terms]\begin{align*} \mathcal{L}_{EM}&=\sum_{particles}(- m c^2/\gamma - q \phi + q V \cdot A) -\frac{1}{2}\int(B^2-E^2)~dx~dy~dz \\ &\approx\sum_{particles}(\frac{1}{2} m v^2 - q \phi + q V \cdot A) -\frac{1}{2}\int(B^2-E^2)~dx~dy~dz \\ \end{align*}

The V is the velocity. There is a real charge q instead of current densities. Here is the relevant equation:
$\frac{\partial \mathcal{L}_{EM}}{\partial R} = \frac{d}{dt}\frac{\mathcal{L}_{EM}}{\frac{\partial R}{\partial t}}$

This has a similar structure to the Euler-Lagrange equation used before because it is a non-relativistic variant. An important difference is that total time derivative.

Who changes with space (R)? The potentials phi and A. Who changes with respect to velocity (V)? The kinetic energy term and the current. Write that out:
$-q \nabla \phi + q \nabla(V \cdot A) = \frac{d}{dt}( m V + q A)$

Give the force an entire side:
$F = -q \nabla \phi + q \nabla(V \cdot A) -q \frac{d A}{dt}$

Total time derivative sidebar: The total time derivative is made up of two parts. One is the partial derivative with respect to time, and the other has to do with moving changes in space. Recall the definition of the E field:
$E = -\frac{\partial A}{\partial t} - \nabla \phi$

Notice the change in space has the same sign as the change in time. So the total time derivative of A can be viewed as this partial time derivative:
$\frac{d A}{d t} = \frac{\partial A}{\partial t} + (V \cdot \nabla) A$

It takes a serious dose of work with vectors to prove that one, but please accept it for this blog. There is another bit of magic sauce/fancy vector identity I will use soon enough:
$\nabla(V \cdot A) - (V \cdot \nabla)A = V \times (\nabla \times A)$

Ouch. I did check that one this weekend, takes about ten minutes, but not easy to summarize, sorry, other than the one that goes on the left is dotted twice.
Sidebar complete.

Plug in the partial derivatives for the total derivative of A:
\begin{align*} F &= q(-\nabla \phi - \frac{\partial A}{\partial t}) + q (\nabla(V \cdot A) - (V \cdot \nabla)A \\ &= q(-\nabla \phi - \frac{\partial A}{\partial t}) + q (V \times (\nabla \times A)) \\ &= q E + q (V \times B) \end{align*}

That is the proof of the Lorentz force law, at least the non-relativistic form.

So that is enough, right? No. This was my own failure up to sometime late on Monday. One has to take the phi one found as a solution to the field equations, plug that equation in here, then show that the Lorentz force law indicates like charges repel (the force is positive).

The solution was Q/R, the derivative of that with respect to R is -Q/R2, so in the static case:
$F = k \frac{q Q}{R^2}$

Good old Coulomb's law, first guessed by Joseph Priestly.

To do the hypercomplex gravity proposal, here is my Lagrangian [updated to include the free field terms]:
\begin{align*} \mathcal{L}_{hcG}&=\sum_{particles}(-m c^2 /\gamma- q \phi - q V \cdot A) +\frac{1}{2}\int(b^2-e^2)~dx~dy~dz \\ &\approx\sum_{particles}(\frac{1}{2} m v^2 - q \phi - q V \cdot A) +\frac{1}{2}\int(b^2-e^2)~dx~dy~dz \\ \end{align*}

There is one vital difference when evaluating the total time derivative. The field used is small e:
$e = \frac{\partial A}{\partial t} - \nabla \phi$

Therefore changes in time have the opposite sign from changes in space for the hypercomplex gravity proposal.

Here is the side-by-side derivation of the Coulomb and Newtonian gravity equations via the hypercomplex Lagrangian:

The ink is drying on this one. Based on previous efforts, it should take about two weeks for me to come up with the Mathematica notebook.
[The image above was updated. Based on a discussion below, the initial version had a simpler proposal for the total derivative term, which unfortunately lead to terms that could not be express in terms of the fields e and b. Since the hypercomplex product has the symmetry of Z2xZ2, it may make sense to treat those terms with 2 spatial components as the space parts as was done in the above calculation.]

I wish to thank my critics who pushed me to make the solution to the field equations consistent with the force equations. The process is rarely pretty in any business, but I feel better about having a force equation after hoping it would appear for quite a few years.

Doug

Snarky puzzle. Show how to derive the Coulomb force equation if you start from Gauss's law. If you get stuck, read chapter 28-5 of David Halliday and Resnick, "Physics", part 2.

Google+ hangout: 11:00-11:45pm Eastern time, Tuesday-Friday. http://gplus.to/sweetser

This could be an efficient way to exchange a few ideas. If you have a question or two, hangout.

The tentative new t-shirt design, not yet for sale:

Next Monday/Tuesday: Gauge symmetries

L = L_matter + q phi - J . A - (1/2) (b^2 - e^2)
then you made it clear L_matter = (1/2) mv^2. Then changed the Lagrangian to
L = L_matter + q phi + J . A - (1/2) (b^2 - e^2)
now you are trying
L = L_matter - q phi - J . A - (1/2) (b^2 - e^2)

ALL of these have a repulsive force between like charges.
I would hope that this would be clear to you by now. If after all this you are still making incorrect claims about your Lagrangians due to the same type of errors, it should be a real warning sign that you are not taking the time to learn from your mistakes. Please please take a break from trying to "fix" your theory, and learn about the bigger picture here.

Del^2 phi = - rho
F = - q Del phi
Look familiar?
Stop, go back, and learn from your errors.

ROFL! Has "- q phi + J . A" been tried yet?
Let's be constructive though: The T-shirts should be sold just with minus signs and no stars printed. Little Velcro stars and bars come with the shirts and can be added or removed according to which minus signs are plus signs any given week.
Indeed it has. You end up back at Maxwell when you use the difference between those two terms.

It is funny, agreed. I am going to leave money on the table. The folks who bought the bad one are stuck with mathematically faulty merchandize. The Chief Financial Officer has said I cannot order new t-shirts until I have a Mathematica notebook that has the forces work included.
It is easy to get the field equations right and their solutions. That is what I delivered long ago. It is tricky to get the field equations right, the force equations right, and one potential solution solution that works for both. I make no apologies for that struggle I had. Usually they are done privately with a person you trust. Not so here, but that is water under the bridge.

What I cannot do is deal with misquotes. The static field equation was unaltered by the correction made to the hypercomplex current coupling term. Only the 3-vector changed The static field equation equation on the derivation blog, and on this one is:
$\nabla^2 = +\rho$

Based on discussion on a Google+ stream plus my own calculations, I did finally see what the relationships between these four equations (force/field equations for EM/gravity) )must be: the field equations must be different (which has always been the case literally for years), and the static force equations must be the same between, as you correctly wrote out, $\inline F=-q \nabla \phi$.

So no, it does not look familiar.
I'm having trouble following this with all the changes.

"What I cannot do is deal with misquotes. The static field equation was unaltered by the correction made to the hypercomplex current coupling term. Only the 3-vector changed "

The interaction term you write here in this blog post is:
L_interaction = - q phi - v . A

Previously, when working out the field equations here
http://www.science20.com/standup_physicist/deriving_hypercomplex_gravity...
you used
L_interaction = + rho phi - J . A

This is exactly what Henry is saying. So I agree with his conclusion.
If there is miscommunication, how about just writing the entire lagrangian to absolutely remove any doubt?

Is your full lagrangian the following? (written out fully for total clarity)
L = (1/2) m v^2 - q phi - q v . A - (1/2) Integral (b^2 - e^2) dx dy dz

If not, please state what it is. It is hard to tell if you are making yet another sign error, or there is currently confusion on what your Lagrangian is. Please clarify. Thanks.

I am in the process of making the math here consistent. On my lunch hour, I was able to fix the LaTeX ones that were generated using the capital Omega on the edit bar. It will show "edit equation" if you hover over it. The software is provided by CodeCogs, well worth a few minutes to learn. Since I had minor edits (removing the conjugate from a J, then switching a few signs), the switch to Plain Editor mode allows the LaTeX to be altered directly without inputing the entire thing again.

A good fraction of the equations are also jpg's from talks I have given in the past. I hope to edit those and correct them this evening.

What has been interesting is the conviction shown by both sides that, golly gee, they were right. Only by deriving the force law again could I feel the power of the source for the other side. It was so darn simple. The only two things that go into the derivation of the Lorentz force law are 1/2mv2 and the current coupling term. So if that is all there is, case closed, no need to lift a pencil. Quite simple if you recall the derivation. Wrong, wrong, wrong. I can see today why they would not budge, their path was too simple.

My path was simple too. So the situation wasn't we are both right (self-contradictory) or one was right (ignore the other team), but that the proposal itself was logically inconsistent. The proposal had to change, which was the focus of the job I tried to do here (lead paragraphs, but apparently it didn't work).

Many people who comment choose not to log in. I always do so because that way I can edit what I have written.

The 3 part Lagrangian to my eye looks incorrect. There is the Lagrange density used to generate the force equation. That is pushing around particles, so none of the things going in are densities. The Lagrange density for the hypercomplex gravity proposal are all densities. One varies position and velocity in the former, and potential and changes in the potential for the latter. Since math is so hard to communicate, I think it is well worth the time to do the LaTeX. Here are the two Lagrangians in play for the gravity proposal:
$\\ \mathcal{L}_{\rm{hcG\;force}}=\frac{1}{2}mv^2-q \phi - q ~v \cdot A \\ \mathcal{L}_{\rm{hcG\;fields}}= \rho \phi + \rho ~v \cdot A -\frac{1}{2}(b^2 - e^2)$

[here was the LaTeX:
\\
\mathcal{L}_{\rm{hcG\;force}}=\frac{1}{2}mv^2-q \phi - q ~v \cdot A \\
\mathcal{L}_{\rm{hcG\;fields}}= \rho \phi + \rho ~v \cdot A -\frac{1}{2}(b^2 - e^2)

Hope that clarifies things,
[Apparently it did not. One could subtract one from the other. At this time, I see no reason why the first should be subtracted from the second or to do the reverse. To my eye, what matters are the relative signs: the two current coupling terms always are counter to the kinetic energy or free field terms. It is important not to gloss over the differences between force and field equations. The force equation fixes the potential, so doesn't use the free field equations, and the fields require fixing position and velocity.]
Doug

First equation should be
L_interaction = - q phi - q v . A

How are you typing math in comments? Niether a < math > tag like at wikipedia, nor a $$tag like at physicsforums, are listed in "allowed tags" below the comment section. [tex] y = x^2$$
y = x^2

He's creating them in LaTeX and pasting in the image

As for that gauss law in quaternions, you might like to like look at the
extension of the residue theorem to quaternions. But what do you do
for the time component?

The multiplication rules are the same as the Klein 4-group.  Everything is positive and commutes:

$\\ ij=ji=k \\ ik=ki= j\\ jk=kj=i \\ i^2=j^2=k^2=+1$

What may be interesting about that is the center of the quaternion group is the Klein 4-group (also the derived subgroup, Fratini subgroup, Jacobson radical, socle, join of elementary abelian subgroups of maximum order, ZJ-subgroup all things I do not understand).

I don't understand math at this level, which is too bad, since I might be able to present it as a "big idea". The way I explain it now, it looks like I plucked out 2 ways to form 4 products out of a hat.

My dream job would involve taking the book "Visaul Complex Analysis" by Tristan Needham and translating it into quaternion animations (an odd dream, I confess). That would cover the residue on page 434.

I consistently view the first term of a quaternion as involving time (a scalar as apposed to a vector, having magnitude without a direction). So energy is "time-ish", a kind of Fourier transformation of space-time.

Doug: "using the capital Omega on the edit bar. It will show "edit equation" if you hover over it"

I don't have an edit bar. Is this an external application you use, and then you embed the results in the your post?

Doug: "Here are the two Lagrangians in play for the gravity proposal"

That doesn't make any sense.
There needs to be a single action describing the entire system. You can't have part of the system described with one action and some other part described with another action. One action should describe the entire system.

For example, the action for electrodynamics with point particles is (I didn't look up the signs, so Henry may complain on this one, but I just want to give all the pieces):
L = Sum_across_particles [ (1/2) m v^2 - q phi + q v . A ] - (1/2) Integral (B^2 - E^2) dx dy dz

With all these pieces, we get Maxwell's equations and the Lorentz force law.
Maybe this is what you didn't realize. ALL the equations for electrodynamics come out this one single Lagrangian.
Often textbooks make it conceptually clearer by grouping it like:
L = L_free_matter + L_interaction + L_free_fields
and then ignore the free matter terms when working out the equation of motion for the field coordinates (becuase the free matter term has no dependence on the fields and therefore doesn't contribute to variation of the action with respect to the field coordinates). Similarly they ignore the free field terms when working out the variation of the action with respect to a particle coordinate, for the converse reason.

If we leave out a piece, then Maxwell's equation, Lorentz's equation, or both will be affected, and it will not be describing electrodynamics anymore.

You seem to be claiming we can do electrodynamics entirely with JUST the lorentz force law, OR, entirely with JUST Maxwell's equations. That is wrong. Both are needed to specify electrodynamics.

Given just F = q (E + v x B), we can figure out how a charge moves due to the fields E and B. However, with only that equation we cannot specify how a charge reacts to another charge. To figure that out, we'd need to know the additional information of how the fields E and B relate to that other charge. The force equation alone does not give this information.

Similarly, if we only have Maxwell's equations, we can't solve for how the particles move.

So again, I ask that you clear up any confusion by clearly specifying the Lagrangian you are using. Please just specify a single Lagrangian, so we don't have to guess on how to combine them. I think that is where the present confusion is coming from.

Membership has its privileges. Creating an account provides access to that most excellent LaTeX equation editor. Another great thing is to be able to edit the post. I know you meant an S up there, I make those kinds of errors too (and writing hear for here, etc). I always login so I can always edit.
So let me make the action for EM as you wrote pretty [corrections: added integrals and dt per LagrangiansForLunch comment]:
$S_{EM}=\int dt \sum_{all\;particles}\left[\frac{1}{2}mv^2 - q \phi + q v \cdot A \right] - \frac{1}{2} \int dt(B^2 - E^2) dx dy dz$

Which you claim is all that is needed to derive both the Lorentz force equation and the Maxwell source equations. You did get the signs right :-)  On a technical level, I don't think you can derive Maxwell equations because the charge density may well need to be smooth and/or continuous - meaning q is not rho. The integral is invariant under a Lorentz transformation, but I think the sum is a low velocity approximation. The Lorentz force derived here was [non-relativistic] because it was only a 3-force.

This is a blog, not a journal, so let's ignore those technical details. With those caveats, my one action for the hypercomplex gravity proposal becomes [similar correction]:
$S_{hcG}=\int dt\sum_{all\;particles}\left[\frac{1}{2}mv^2 - q \phi - q v \cdot A \right] + \frac{1}{2} \int dt(b^2 - e^2) dx dy dz$

Hope that helps,

Doug
Yes, I assume most people knew what you meant, but unfortunately Doug seemed to misinterpret your clarification (or maybe he just did some typos?). I agree with your assessment. Rereading Doug's article here, it does appear he was claiming there are two equal but _separate_ descriptions : the force laws, and the field laws. But thanks to your comment, we all seem to be on the same page now.

Doug,
What he wrote was the Langrangian. The action, as he explained, is S= Integral L dt. So you forgot the integral with respect to time when you wrote out the action.

Thank you for writing out your action. That was indeed where the confusion was coming from.
Notice this action is changed compared to your original derivation of the field equations. In this article you talk only about changing the signs of the JA terms. If you only changed a sign in these terms, then Henry's comments are correct. But it is clear now that you ALSO changed signs for the free-field terms. This latest change then now gives you static case you wanted:
Del^2 phi = rho
F = - q Del phi
Note however that there is still a relative sign change between your free-field terms and the current term, compared to your original derivation. So the field evolution equations DID still change. However you do now have an attraction between like forces since you changed that other sign. I do hope you are done changing signs now.

"On a technical level, I don't think you can derive Maxwell equations because the charge density may well need to be smooth and/or continuous - meaning q is not rho."

I'm sure you've worked with Maxwell's equations for non smooth densities before. Deriving the magnetic field for a line current, or the electric field for a point charge. It works out fine, with the distributions given with delta functions. So from the EM Lagrangian with point charges you would still get:
Del . E = rho
where in this case, since we have a collection of point charges, rho = Sum_over_particles_i q_i delta(r_i). Similarly for the other Maxwell source equation.

"The integral is invariant under a Lorentz transformation, but I think the sum is a low velocity approximation."

Yes he used (1/2) m v^2, but that is the only term that is a low velocity approximation. To make the EM action Lorentz invariant, change the free matter term from "(1/2) m v^2" to "- m c^2 / gamma". Of course you still get the Lorentz force law
dp/ dt = q ( E + v x B )
the only difference between the two cases being that p=mv (the non-relativistic momentum) in the first case, and p=gamma m v (the relativistic momentum) in the second case.

"The Lorentz force derived here was classical because it was only a 3-force."

Since we are discussing clarity, it is usually best to use "classical" to refer to non-quantum theories. General relativity is called a classical theory by physicists. Usually physicists use "relativistic" or "non-relativistic" to make the distinction you are trying to make there. Sometimes "non-relativistic" is also called "Newtonian" for obvious reasons.

Just reread and I see I am kind of using the phrase action and lagrangian interchangeably. They are of course different but closely related. Here, the action
S = Integral L dt
and then the equations of motion for all the generalized coordinates involved are obtained via the action principle. This is how the evolution equations for the system are obtained once a Lagrangian is specified. I realize Doug, Sascha, Henry, etc. already know this. I just figured if I'm asking for clarity, I should give clarity myself and point out I was using some of the language sloppily / incorrectly even if the meaning was (hopefully) clear to the main parties reading.

Doug, what do you hope to accomplish here?
This is not a rhetorical question. I am actually interested in the answer.

Even if we did accept your claimed force law written there, it is already ruled out experimentally for many different reasons. You once claimed "My goal is not to win people over, it is to find flaws in my own efforts." If this truly is what you hope to accomplish here, then because there are so many things are problematic, there is a lot that can be explored and learned here. This could also make for a fascinating blog to read. Here's some topics right off the bat that I've thought or seen other posters mention:
- how to notice there is no local Lorentz symmetry without doing all the complicated math
- what the consequences of violating local Lorentz symmetry are, and what experiments rule this theory out based on this
- how to notice there is no rotational symmetry without doing all the complicated math
- what the consequences of violating rotational symmetry are, and what experiments rule this out
- how to notice gauge freedom has been removed, and what useful things about gauge symmetry we lose
- the differences between coupling gravity to the invariant mass instead of to the stress energy tensor, and what experiment can comment about the general predictions from these
- discussion on what allows us to write gravity as a curvature of spacetime in GR (but not, for instance, allow us replace the interaction of EM in 3+1 spacetime dimensions with a curved spacetime)
- follow-up discussion on how your exponential metric is not equivalent to your force equations in flat spacetime
- and this isn't even touching on all your incorrect claims about reproducing the weak or strong force (you don't currently appear to know what we mean when we say EM is a U(1) gauge theory, or the strong force is an SU(3) gauge theory)
...

But to do this, you need to start approaching this with a true learning attitude. Currently you have two separate things, one is what you "want your theory to say" and then all kinds of squabbling over details of "what the math actually says". You are very slow to recognize mathematical results and consequences of your theory that don't fit with the former.

Since you are an experimentalist, you know that often we have "what we want to check with our experiment" and "what the experiment actually outputs" as two separate things when things go wrong (the results are affected by something we forgot to control, so the experimental is actually checking something else) or when a very happy unexpected discovery occurs. To benefit from the unexpected discovery (in this case, you finding out your theory says something unexpected to you), we need to explore what is going on. Instead you just keep marching along with your theory.

I agree with the others. It would be much more beneficial (and interesting to readers), to stop and learn about the bigger picture. What was the mistake? What nugget of information should be learned and incorporated into our intuition when approaching further problems? Stop just patching your theory or marching ahead. There are many interesting and beneficial things that can be learned if you switch modes from "promote my theory" to "investigate and learn from my theory's mistakes".

Hello Justin:

I added your name to the list of folks who do not think there is value in the hypercomplex gravity hypothesis (this comment is effectively the same thing). There are five names on the list so far, and not one on Science20 thinks I am right (I do still have my grad student buddy).

Even if we did accept your claimed force law written there..

Accept my claim? I have no interest in anyone who accepts my claim. The reason I spend the time with LaTeX or write out a derivation neatly is so you can look at it yourself. Henry has given it a thumbs down. He also did not write my proposal down correctly, so I agree with Henry, write my proposal wrong as he did, and it ain't right. If you repeated the calculation yourself, say so. If you skipped it, I understand how it goes in a busy life.

You have provided a list of possible future blogs. I cannot address such a long list here.

You are very slow to recognize mathematical results

Yup, ten days to accept that my proposal for gravity locally breaks Lorentz invariance versus GR that only does so globally. I was a full month to get the force thing. In my experience doing research, that is not such a bad rate.

I don't have a learning attitude. I have a research attitude. We had some honest confrontations. The hypercomplex gravity Lagrangian took a bullet to the head. Death to J*, long live J. I have fixed both the LaTeX equations and gifs in the 4/5 and 5/5 blogs. And marked they were "IMPORTANT CORRECTIONS".

Such a bad boy I am, I should "investigate and learn from my theory's mistakes". How about admit the errors? Oh, I did that for the force. How about see if I can find a solution to the problem at hand? I did that for the force. How did I do that? The force terms are the result of the 3-vector part of the current, and the Gauss's law is the first term of the current. The fact that I could pull that rabbit out of my new Lagrangian is one of the strongest possible statements in support of the proposition I may be barking near a real tree.

Yes, I know there is an issue with gauge symmetry. It was dealt with in the 2/5-5/5 blogs, but I will revisit it next week. I will do some reading about all the scary things breaking Lorentz locally versus globally will do, but the word "breaking" is a huge exaggeration. The effects of rotation in GR required 40+ years and \$750 million to spot in gravity probe B, an effect less than 1 degree per year. We will have to see if any such tests of local Lorentz symmetry get to this level of precision. Gravity is very, very, very weak, a good thing for this issue. It is also possible that what once was a bug, could become a feature. For example, it could be the arrow space-time needs to make the future different from the past. Plenty of stuff to think about, the sign of a productive research subject.

Wow, Sascha was right. In addition to the q phi sign change, did you add your newest sign change (the free field terms) on your tee-shirt as well?

Well, at least with this extra sign change you added in the comments here, you finally get an attractive force in the static case. Yeah for you. Some consequences that you may not have realized.
1) Note that with this change, your "unified" equation now finally has a source since the source terms don't fully cancel anymore.
2) In your "gravity", Del . b != 0 .... do you want to interpret your gravity as having magnetic monopoles?
3) Since you couple to the invariant mass instead of the stress energy tensor, light will not be affected by a mass source. Pound-Rebka rules out your theory. Gravitational Lensing rules out your theory.
4) If we neglect the spinning of the sun, your force equation is just a simple central force equation on the planets like Newtonian gravity. Therefore the measurements of Mecury's orbit rule our your theory (unless you believe the spinning of the sun somehow perfectly accounts for this, but note that the GR gravitomagnetic effects are negligible here ... so changing your theory to make this match would ruin many other measurements).

Hello Margo:

I added your name to the list of folks who do not think there is value in the hypercomplex gravity hypothesis. You are #6. If they find the Higgs boson at 119 MeV in a few months, I will add my name to the list and fade away into the wayback machine.
Well, at least with this extra sign change you added in the comments here, you finally get an attractive force in the static case. Yeah for you.
It is not about me, it is about some equations. An hour after I posted this blog, Henry wrote to say the attractive force issue was not resolved. This is the first comment made by someone other than me that the math does work out for the force as I had written. That is measurable progress.
1) Note that with this change, your "unified" equation now finally has a source since the source terms don't fully cancel anymore.
Indeed I did notice that, writing the J on the front of the t-shirt. There is no static charge for the GEM proposal, but things do happen in the current. I don't understand what that means. Such is a life in research.
2) In your "gravity", Del . b != 0 .... do you want to interpret your gravity as having magnetic monopoles?
This was discussed earlier, but yes, my interpretation of gravity is that it has magnetic monopoles because it is a universally attractive force. If putting a mass in motion creates two poles, then one could get moving gravitational systems to repel each other. Gravity is different from EM in this way.
3) Since you couple to the invariant mass instead of the stress energy tensor, light will not be affected by a mass source. Pound-Rebka rules out your theory. Gravitational Lensing rules out your theory.
4) If we neglect the spinning of the sun, your force equation is just a simple central force equation on the planets like Newtonian gravity.
Perhaps you did not hear or choose to deny on some sort of technical grounds my assertion that the exponential metric is a solution to the field equations. That dynamic metric solution has both the bending of time and of space needed to account for the bending of light around the Sun.

The number of physicist who were able to spot the flaw in the force term was three (and I give a *10 factor for those who know and do not comment, so about 30 out of a thousand). Most folks don't keep the derivation of the Lorentz force law in RAM since human minds are quite limited in this way. Once spelled out like done in this blog, it is obvious this was a deadly flaw. What is not obvious is that there could be a reasonable solution. I bet none of the fictional 30 would have thought the problem could be fixed. Yeah for GEM.

Things happen because of pressure, diamonds being a classic example. If it gives you a headache, take some aspirin and stop reading. There is plenty of work for me to do, with or without you.

Doug

4) If we neglect the spinning of the sun, your force equation is just a simple central force equation on the planets like Newtonian gravity. Therefore...
I hope people recognize a rigged test. Assume the proposal is exactly like Newton's theory, then it will fail every test where Newton goes head-to-head against GR.

There is not just phi (which Newton is constrained to work with). There is also the 3 -potential A. Light bending around the Sun is a relativistic effect. There is a dA/dt term. For light, changes in time are equal to changes in space. One would have to show that there was a relevant solution to the field equations of A=1+GM/c2 R. In a relativistic context, that could double the bending found from the phi=1-GM/cR term. That is a sketch of a possible solution, nothing more, which is why I didn't make it pretty.

I didn't realize you were using a built in feature to post the math equations.  That was enough to finally convince me to create an account.
An hour after I posted this blog, Henry wrote to say the attractive force issue was not resolved. This is the first comment made by someone other than me that the math does work out for the force as I had written. That is measurable progress.
Maybe you missed it, but as Margo and LagrangiansForBreakfast pointed out, you have added ADDITIONAL sign changes since your article on Monday night.  In the comment you wrote on "11/02/11 | 23:34 PM" you revealed that you now ALSO changed the sign of the free field terms.

So now the static equations are indeed, as LagrangiansForBreakfast stated:
\begin{align*} \nabla^2 \phi &= \rho \\ F &= - q \nabla \phi \end{align*}

To make absolutely sure we are on the same page now, is the matter term for your theory this
$\mathcal{L}_{\mathrm{free\ matter}} = \frac{1}{2}mv^2$
or did you use that just to make the derivation a bit simpler, and your theory actually uses the relativistic form below?
$\mathcal{L}_{\mathrm{free\ matter}} = -mc^2\sqrt{1-(v/c)^2}$

Assuming the former, then is it correct that your theory is (or at least equivalent to) the following?
1) Definition
\begin{align*} e &\equiv \frac{\partial}{\partial t} A - \nabla \phi \\ b &\equiv \nabla \boxtimes A \end{align*}
\begin{align*} \mathcal{L} &= \mathcal{L}_{free\ matter} + \mathcal{L}_{interaction} + \mathcal{L}_{free\ field} \\ \mathcal{L}_{free\ matter} &= \sum_{particles} \frac{1}{2}mv^2 \\ \mathcal{L}_{interaction} &= \sum_{particles} \left[ -q\phi - q~v \cdot A \right] \\ \mathcal{L}_{free\ field} &= + \frac{1}{2} \int (b^2 - e^2)dx~dy~dx \end{align*}
3) Choice of generalized coordinates (what is varied in the action):
the particle positions, and the fields phi and A
4) equations of motion come from the action principle, where the action is
$S = \int \mathcal{L}~dt$

Alright, this fully specifies the theory. So the predictions should follow unambiguously from the math. So let's double check each other. Propagating your sign changes to the field equations I think you now have:
\begin{align*} \nabla \cdot e &= -\rho \\ \nabla \boxtimes b &= J + \frac{\partial}{\partial t}e \end{align*}
For the force equations, it looks like the result is
$F = q (e - v\times\nabla\times A)$
This force is different than what you got.

Looking at your derivation, you wrote:
$\frac{d}{dt}A = \frac{\partial}{\partial t}A -(v\cdot\nabla)A$
That is not correct.  I'll write this out so we can see what is going on.  Written out to make it more clear where the field is evaluated, the coupling is:
$- q \phi(x,y,z,t) - J \cdot A(x,y,z,t)$
where x,y,z is the position of the charge q at time t (the coupling between the field and particle depends on the field just at the location where the particle is).
\begin{align*} \frac{d}{dt}A(x,y,z,t) &= \frac{\partial A}{\partial t} + \frac{\partial A}{\partial x} \frac{dx}{dt} + \frac{\partial A}{\partial y} \frac{dy}{dt} + \frac{\partial A}{\partial z} \frac{dz}{dt} \\ &= \frac{\partial A}{\partial t} + \left(v_x \frac{\partial}{\partial x} + v_y \frac{\partial}{\partial y} + v_z \frac{\partial}{\partial z}\right)A\\ &= \frac{\partial}{\partial t}A + \left(v\cdot\nabla\right)A \end{align*}

You could have kept the rotational symmetry and lorentz symmetry if you had just started with electrodynamics and changed from:
L = L_{free matter} + L_{interaction} + L_{free field}
to
L = L_{free matter} + L_{interaction} - L_{free field}
that gives you an attractive theory.  This was explained to you earlier, but you must have just ignored it since after all your sign flips you essentially did this to make the force attractive but claim "I bet none of the fictional 30 would have thought the [non-attractive force] problem could be fixed."   If you learned the bigger picture here, you'd have seen that nothing magical is happening here.
There is no static charge for the GEM proposal, but things do happen in the current. I don't understand what that means. Such is a life in research.
It is not clear what anything in your GEM proposal is supposed to be.  Yet you claim it does away with the Higgs.  Can't we agree to stick to the actual math consequences of your theory?  You yourself say GEM doesn't have electromagnetism nor gravity.  Sure, you can get some equations from it by cranking the Lagrangian through the mathematical machinery.  But how you leap from these equations to claiming anything about the equivalence principle or the Standard model makes no sense.  Your GEM theory doesn't reproduce any of these things.

Perhaps you did not hear or choose to deny on some sort of technical grounds my assertion that the exponential metric is a solution to the field equations. That dynamic metric solution has both the bending of time and of space needed to account for the bending of light around the Sun.
And here you are making claims that do not follow from the math.  You keep ignoring the very clear reasons why you can't replace your interactions with a curved space-time theory.  Do you at least understand that not all theories and interactions can be described just due to moving on a curved space-time?  Do you understand for example that this cannot be done with electrodynamics?  Your theory doesn't have local rotational symmetry, so it clearly can't be represented as just moving on a Reimannian manifold.  What are you not understanding here?  Please tell us, instead of just ignoring the math and making false claims again and again.

Another way to look at this is calculating the orbit of Mercury using your force laws.  Does it give what you are claiming your curved spacetime gives? No. It clearly does not.

Hello Ed "Henry" Brown:

Thanks for logging in, really :-)  Math is hard to communicate, but the LaTeX makes it much easier to spot issues to address.

Nice summary of my work by the way. I prefer to keep my 4-vectors [clarification: 4-vectors in the math sense of 4 things that can added together or multiplied by a scalar, not how they transform, since this is a current density] together on the same line, and like to keep rho positive, so would write:

$(\rho, \vec{J})=(-\nabla \cdot e, \vec{\nabla} \boxtimes \vec{A} - \frac{\partial e}{\partial t})$

This is a style, not substance issue.

We have such a nice clean difference on a technical issue. It centers around moving from the total time derivative of the 3-potential A to the partial derivatives of time and of space.

I have been teased here about not seeing the proverbial big picture. I need a big picture view for this issue. What is the EM field E about? It has two parts:

$E = -\frac{\partial A}{\partial t} - \nabla \phi \quad eq.~1$

Part of the change in E comes from changes with respect to time, the first term. The rest of the changes come from changes in space. In the derivation of the EM Lorentz force, there is the total time derivative of A. That total time derivative is a combination of changes in time plus changes in space. So for EM:

\begin{align*} \frac{d A_E}{d t} &= \frac{\partial A}{\partial t} + \left[ \frac{\partial A}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial A}{\partial y} \frac{\partial y}{\partial t} + \frac{\partial A}{\partial z} \frac{\partial z}{\partial t} \right] \\ &= \frac{\partial A}{\partial t} + \left[ \left( v_x \frac{\partial }{\partial x} + v_y \frac{\partial }{\partial y} +v_z \frac{\partial }{\partial z} \right)A \right] \\ &= \frac{\partial A}{\partial t} + \left[ (v \cdot \nabla) A \right] \quad eq. ~2 \end{align*}

The content is the same as you had, but there is a stylistic difference. The square brackets were put there to emphasize that those changes are the changes in space contribution to the total time derivative of A.

Let me cut and paste this argument for the hypercomplex "gravity" field small e.

What is the hypercomplex "gravity" field e about? It has two parts:

$e = +\frac{\partial A}{\partial t} - \nabla \phi \quad eq.~3$

Part of the change in e comes from changes with respect to time, the first term. The rest of the changes come from changes in space. In the derivation of the hypercomplex "gravity" Lorentz force, there is the total time derivative of A. That total time derivative is a combination of changes in time minus changes in space. So for the hypercomplex "gravity" proposal:

\begin{align*} \frac{d A_e}{d t} &= \frac{\partial A}{\partial t} - \left[ \frac{\partial A}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial A}{\partial y} \frac{\partial y}{\partial t} + \frac{\partial A}{\partial z} \frac{\partial z}{\partial t} \right]\\ &= \frac{\partial A}{\partial t} - \left[ \left( v_x \frac{\partial }{\partial x} + v_y \frac{\partial }{\partial y} +v_z \frac{\partial }{\partial z} \right) \right] \\ &= \frac{\partial A}{\partial t} - \left[ (v \cdot \nabla) A \right] \quad eq. ~4 \end{align*}

This is the technical difference that remains between us.

I do appreciate that if equation 2 applies to the hypercomplex "gravity" force equation, then one ends up with a force equation that has nothing to do with the field b, not a thing, it is the typical EM field B.

We can part ways over this issue. To me the logic is clear: the total change in the field e has a time contribution minus a space contribution. If you insist that the total change is the same for both EM and hyerpcomplex "gravity", then you get the same force law where like charges repel. I certainly want a force equation that involves both the fields e and b. I do get the ramifications of your math. At least we can point to a specific fork in the road.

Lunch is over, have to go.

Doug

I think my lunchtime calculation is unambiguously wrong. The problem is that one gets the "All three" index terms with a minus sign. Those fellows add up in the force equation, and hang around. Since they cannot be expressed in terms of the e and b fields, dangling terms like $v_x \frac{\partial A_x}{\partial x}$ is a sign of failure.

Go back to the big picture. I am working with products that have the structure of Z2xZ2. What I proposed for the total time derivative was quaternion-like, flipping 3 signs, not two. I much prefer the simple situation in EM, with no sign flipping. Alas, I must be consistent with my self:
That total time derivative is a combination of changes in time minus changes in space.
The question is how to implement such a proposal, in a rational Z2xZ2 kind of way.

The total time derivative has all of 12 terms in it. Play with the terms using Ax:
$\frac{d A_{x\;e}}{d t} = \frac{\partial A_x}{\partial t} + \frac{\partial x}{\partial t} \frac{\partial A_x}{\partial x} -\frac{\partial y}{\partial t}\frac{\partial A_x}{\partial y} - \frac{\partial z}{\partial t} \frac{\partial A_x}{\partial z} \quad eq. ~5$

The changes in space parts require two directions in space, either an x and a y or an x and a z. If there is only one direction (x), then it is positive. Here is the entire expression:
\begin{align*} \frac{d A_{e}}{d t} &= \frac{\partial A_x}{\partial t} + \frac{\partial x}{\partial t} \frac{\partial A_x}{\partial x} -\frac{\partial y}{\partial t}\frac{\partial A_x}{\partial y} - \frac{\partial z}{\partial t} \frac{\partial A_x}{\partial z} \\ &\quad \;\frac{\partial A_y}{\partial t} - \frac{\partial x}{\partial t} \frac{\partial A_y}{\partial x} +\frac{\partial y}{\partial t}\frac{\partial A_y}{\partial y} - \frac{\partial z}{\partial t} \frac{\partial A_y}{\partial z} \\ &\quad \;\frac{\partial A_z}{\partial t} - \frac{\partial x}{\partial t} \frac{\partial A_z}{\partial x} -\frac{\partial y}{\partial t}\frac{\partial A_z}{\partial y} + \frac{\partial z}{\partial t} \frac{\partial A_z}{\partial z} \\ \end{align*} \quad eq. ~6

I don't have a short pithy way to write this ((v.Del)A is not it). My sense here is that the language, useful nomenclature and shortcuts of vector calculus are going to trip me up in Z2xZ2 land, as happened here.

I'll put a "see below" correction comment in the blog.
straight from the FAQ
http://www.science20.com/science_2.0_faq

Articles and comments that fall into any of the following categories will be deleted by the moderators:
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2. Posts promoting pseudoscience:
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B. Baseless speculative nonsense - claims, made without supporting evidence by people with no research training or track record in the relevant field, that suggest that major, fundamental knowledge of biology, chemistry, and/or physics is wrong.

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Mr. Sweetser may use more math that the average crackpot, but he ultimately makes wild claims divorced from the math. Sure, crackpots cause a flurry of responses and comments and may even give a temporary spike in "interest" in readers, but is this really what people want Science2.0 to be?

There should be value in showing how the EM Lorentz force is derived, #7. I thought I was still married to math, hence my admission to infidelity at the start of this post. I found a fix to the like charges attract force problem (and only the like forces attract force problem).

Join the peaceful revolution. Work on strings. Get paid well, travel the globe, work on elegant math. In a year, if there is not a Higgs or single supersymmetric particle to be found, and people are saying we gotta start from the ground up, well, I am giving something specific enough to work on.
Doug,
Can you give feedback on Henry's comment above? He has a very self-contained summary of the mathematical specification of your theory. Are the signs and terms of the Lagrangian correct? I'm hoping there is enough clarity now that we all agree "exactly what the theory is", and all that is left is to do the math so we can figure out "exactly what the theory predicts". If so, I might try verifying the outcome of the action principle -> equations of motion myself this weekend to double check everything.

Also his comments regarding the "magnetic" part of the force look correct. In addition to the sign issue in your derivative of A that he mentions, I also don't follow your derivation in the step from:
Del (v . A) + (v . Del) A
to
v boxtimes (Del boxtimes A)
Is this just a vector identity for your boxtimes operator?

Thanks.

Sure, that is my plan for lunch.

This is a standard vector identity:
$\nabla (v \cdot A) - (v \cdot \nabla) A = v \times (\nabla \times A)$

I don't have a quickie proof, but note that the left-most one is in both dot products. There are all kinds of negative signs around. What if there was not a single minus sign on either side? That would be "a vector identity for your boxtimes operator".
$\nabla (v \cdot A) + (v \cdot \nabla) A = v \boxtimes (\nabla \boxtimes A)$

The same terms sit in the same places at the table, but no minus signs are in the house.

Doug
I checked it by hand, and the standard identity worked (of course). But your proposed identity doesn't. So it is not always possible to convert a vector identity to your boxtimes just by removing minus signs and replacing cross products with boxtimes.

For example, for the x component I get:
[ Del (v . A) + (v . Del) A ]_x = [ v boxtimes (Del boxtimes A) ]_x + 2 (v_x partial_x A_x)
and similarly for the y and z components.

I'll recheck it tonight and post a reply.
Indeed, you are spot on right there. I am brand new to working with Z2xZ2, which does not want to toss anything away. I consider an extra term like:  $v_x \frac{\partial A_x}{\partial x}$ a proposal killer because it is found in neither the e or b fields.
Wow! A lot has happened since I last was here.

Based on CuriousReader's comment, I checked that boxtimes vector identity you suggested, and I got exactly what CuriousReader got. So while everyone now seems to agree on the new field equations and the "electric" part of the force equation, there seem to still be a couple sign issues in the "magnetic" part.

"I prefer to keep my 4-vectors together on the same line ... This is a style, not substance issue."

It is a substance issue if you are claiming that combination of fields gives you a four-vector. Just a friendly reminder that it does not give you a four-vector.

Oh, and I noticed that you skipped Henry's question regarding the free matter term. For clarity can you please tell us if your theory actually uses (1/2) mv^2, or if you are just using this as an approximation for now and the term is actually - mc^2 / gamma. This could be important later when making orbit predictions from your theory, so can you please clarify that?

"I have been teased here about not seeing the proverbial big picture. I need a big picture view for this issue."

The point of the "big picture" is to allow one to quickly asses the situation and guide investigation. Its because of understanding the "big picture" that Henry and I were able to hone in on explicit examples that showed various symmetries were broken for example... because we understood what was breaking the symmetries without having to dig through all the detailed math.

But what you just did by fudging that sign to match what you want is not using "a big picture view for this issue". The big picture can (hopefully) help guide your intuition, but ultimately the math says what it says. You can't appeal to something unrelated to the calculation to claim you can change what it means to take a derivative. If you saw a scientist work out the prediction of a theory, and then change the meaning of the math mid calculation to make it fit his expectations, would you accept that? I hope not. The beauty of math is that once you specify your theory mathematically, it is out of your hands. The math is impartial and says what it says.

I have updated the blog to have the 3 term Lagrangian. It has the -gamma mc2 for the free matter term.

We all agree there is a problem with the magnetic part of the force law. I have put in a proposal in a comment above. The hypercomplex product has Z2xZ2 symmetry, so it may be valid to only consider the terms with two spatial directions as being the space part. If so, the math would work out better.

I run into the lingo issue all the time. I always work with 4-terms, using either quaternion or hypercomplex products. The word that gets me into the most trouble is "scalar". To a physicists ear, that means it is invariant under a Lorentz transformation. I use it to talk about the first term in that collection of 4-terms. How the first term transforms can be figured out, but is not implied.

Now a naive question: is there any difference in the physics that one gets switching the overall sign of the Lagrangian? The Lagrangian gets fed into the Euler-Lagrange machinery. The global sign should not matter, since the equations that result will just flips signs overall. Relative signs I get, this Lagrange term versus that one.
"Now a naive question: is there any difference in the physics that one gets switching the overall sign of the Lagrangian?"

You are correct, there is no difference if you change the overall sign.
Classically you can multiply the entire Lagrangian by any non-zero constant as it doesn't change anything. Also, classically there can be multiple Lagrangians that give the same equations of motion:
http://en.wikibooks.org/wiki/Classical_Mechanics/Lagrange_Theory#Is_the_...

In quantum mechanics the story is a bit different (though you can still change an over all sign with no effect). Different actions which have the same classical equations of motion, can lead to different quantum theories after they are quantized. This is why there is no unique prescription to go from a classical theory -> quantum theory.

Doug,
Since I have an account now, I was considering writing something up about magnetic monopoles or something. What exactly is the difference between a column and a blog? If I click on "my column" in the dashboard it just gives me an "access denied" and redirects to main page. Sorry to bother you with this, but the help page I found looks quite outdated or something.  Figured I'd ask the resident expert.

Well now onto the math and physics...
I'd much appreciate it if you could comment on some things.
Part of the change in e comes from changes with respect to time, the first term. The rest of the changes come from changes in space. In the derivation of the hypercomplex "gravity" Lorentz force, there is the total time derivative of A. That total time derivative is a combination of changes in time minus changes in space.
Besides the change in the free matter term (Note: in the blog you accidentally wrote -gamma mc^2 instead of -mc^2 / gamma ... I keep writing that on accident as well, maybe because gamma mc^2 is the energy), I thought what I wrote above was a decent summary of your theory.  Apparently it is not, because what you are claiming doesn't follow from the math.

'e' and 'b' are just defined values.  Just like 'E' and 'B' in EM, we could completely ignore them and just work everything out in terms of your actual chosen field coordinates (the potential field A).

1] It wasn't even necessary to define 'e' or 'b', so why do you think how we define the 'e' or 'b' field affects how we take a derivative of A?

The math says what it says
http://en.wikipedia.org/wiki/Total_derivative
if you want to change your theory again, so be it.  But you can't change the meaning of a derivative to fudge the math to fit your expectations.

One point of note is this: your derivation is trying to get the equations of motion from the action principle.  The derivation of the Euler-Lagrange equation assumes the actual derivative is used.  So if you really think your redefinition of a derivative during your derivation is justified, note that it no longer satisfies the stationary action condition.  So you are violating even the action principle when you do this.

If you want, you could even take your solutions and plug it back into the action equations to check if it really does give a stationary path.  This might be difficult with Mathematica, but it is probably possible.  I hope you don't turn to Mathematica for this though, since you shouldn't need Mathematica to tell you that changing the definition of what it means to take a derivative means your "result" will no longer be the solution to the equations you started with.

Snarky puzzle. Show how to derive the Coulomb force equation if you start from Gauss's law.
I found some of your other comments very fascinating, but it requires a solid understanding of vector calculus to discuss.  So let me take this opportunity to answer your snarky puzzle as well as review some vector calculus.  You probably already know most of this, but if these discussions have taught me anything, it is that it is always a good idea to build agreement on a common mathematical foundation before digging into details.

Alright, the Coulomb force equation is:
http://en.wikipedia.org/wiki/Coulomb's_law#Table_of_derived_quantities
Force on charged particle 1 from charged particle 2, if both charges are stationary.
$\vec{F}_{21} = \frac{q_1 q_2}{4 \pi \epsilon_0 r^2} \hat{r}_{21}$
Now for Gauss's law, that is just what some people call the Divergence theorem, which is just a vector field identity. For any vector field A in 3 dimensions (well, mathematicians would probably add some requirements to make the field "suitably nice" or something, but I won't be worrying about such details here) the integral of the divergence of the field over some volume can be related to an integral over the surface of that volume.
$\int_V (\nabla \cdot A) ~dV = \int_{\partial V} A \cdot d\vec{a}$
Since this is true for any vector field, it can't tell us anything about the physics by itself. We'd have to assume a vector field of some kind, assume how it relates to charges, and assume how charges react to this field.  In short, we'd have to just outright assume the answer.

For course instead of just the Divergence theorem, you probably meant the more specific Gauss's Law in electrodynamics for the electric field, which is the "integral form" of one of the Maxwell source equation:
\begin{align*} \nabla \cdot E &= \rho/\epsilon_0 \\ \int_V (\nabla \cdot E) ~dV &= \int_V \rho/\epsilon_0 ~dV \\ \int_{\partial V} E \cdot d\vec{a} &= \frac{1}{\epsilon_0} Q_{enclosed} \end{align*}

Here's where the vector calculus review comes in.  An arbitrary vector field can be specified if three pieces of information are given: its divergence, its curl, and some boundary values (to handle a possible spatially constant value that is added everywhere).

When I helped students with electrodynamics, a common mistake was for them to think that this integral equation (Gauss's law) would allow them to calculate the electric field for any distribution of charges just by considering a surface that surrounded the charges.  What's the electric field of a dipole, zero?, no.  Is the field from a moving charge just (q/4 pi e_0 r^2) ? Gauss's law only tells you the field integrated over the entire surface.  Even if you know this over every possible surface (or equivalently, the divergence everywhere), you still can't solve for E.  You still need something to specify the curl (ie. the other Maxwell equations), and some boundary condition (usually that E is zero off at infinity).

If you've ever used Gauss's law to solve for the electric field of some charge distribution, you know the intuitive way of describing this is usually something like: we can only use Gauss's law to easily get the fields from static highly symmetric charge distributions (cylindrical symmetry, symmetry of an infinite charged plane, spherical symmetry).

If I'm allowed to assume that the curl is zero, and the field is zero off at infinity then I can use Guass's law to get the electric field from a static point charge is:
$E = \frac{q}{4\pi \epsilon_0 r^2} \hat{r}$
Even with this additional assumptions, I still cannot get to the Coulomb force law.  Without additional information on how the force depends on the electric field, or maybe information on how the energy of the system depends on the electric field, we can't get to a force with only this information.
Since they cannot be expressed in terms of the e and b fields, dangling terms like V_x partial_x A_x is a sign of failure.
This fascinated me for two reasons.

2] Why does everything need to be in terms of 'e' and 'b'; why isn't your chosen field coordinates for the Lagrangian, A, enough?

Notice the two field evolution equations you got so far only specify the divergence of 'e', and the 'boxtimes curl' of 'b'.  If someone was given ONLY those field equations, it would not be enough to solve for 'e' and 'b' given the sources rho and J.

Let's focus on 'e' first.  Since you defined it in terms of A, we can of course check what the curl of e is.  This cannot be written in terms of e or b.  I'm not sure if the 'boxtimes curl' specifies enough information, but even if it did, even that cannot be written in term of 'e' or 'b'
\begin{align*} \nabla \times e &= \nabla \times (\partial_t A - \nabla \phi) = \partial_t \nabla \times A = \partial_t B_{EM} \\ \nabla \boxtimes e &= \nabla \boxtimes (\partial_t A - \nabla \phi) =\partial_t b - [\hat{x}2\partial_y \partial_z + \hat{y}2\partial_z \partial_x + \hat{z}2\partial_x \partial_y ] \phi \end{align*}

3] Why would you go to the point of trying to redefine what a derivative even means to make your force equation only refer to 'e' and 'b', when your field equations can't even be written fully in terms of 'e' and 'b'?

If one term in a Lagrange density is invariant under a Lorentz transformation and another one isn't, that is a problem. I actually did not spot that internal inconsistency at the time.
I realize this probably wasn't meant to be a mathematically precise point, and probably more of an insight into your model building "intuition".  But then I have to ask

4] If we know the free matter term is Lorentz invariant, then why don't you consider it an "internal inconsistency" failure that the rest of your theory isn't Lorentz invariant?

Have a good weekend,
-Henry
Hello Henry:

I have never written a column, only a blog. Go with the blog (and send me a private message should you do so, since the topic will be of interest to me).
1] It wasn't even necessary to define 'e' or 'b', so why do you think how we define the 'e' or 'b' field affects how we take a derivative of A?
The math says what it says
http://en.wikipedia.org/wiki/Total_derivative
if you want to change your theory again, so be it.  But you can't change the meaning of a derivative to fudge the math to fit your expectations.
That sure sounds like a reasonable line of logic looking only at the math. What I don't want to do is fudge my physics. Of course you are correct in e and b just being defined values. The two in turn involve 2 derivatives, so four over all. They do all live in the same second rank symmetric [questionable if I can call it] tensor. In a different reference frame, the terms can mix with the other.

Even with all this ability to mix things around, there is a sense that electricity is separate from magnetism, the dividing line between the two is not phony it is real (sorry to get vague here). The terms that go together in EM are, as you well know:
$E=-\frac{\partial A}{\partial t} - \nabla \phi$

This comes out of an antisymmetric tensor. It has parts that are changing in time and in space. If the time derivative gets bigger, E gets more negative. If the gradient gets steeper, then E gets more negative.

Now we go to my proposal. It is the same thing, well almost. It has the same derivatives, just one sign different:
$e=\frac{\partial A}{\partial t} - \nabla \phi$

This comes out of a symmetric tensor. If the time derivative gets bigger, small e gets more positive. If the gradient gets steeper, then e gets more negative.

I used the total derivative term correctly for EM, so I do know how to do the right thing. It is also consistent with the description of how the electric field changes. And we get to the Lorentz force law. Everything rocks for EM.

In my proposal, changes in time are not moving in the same direction as changes in space for the thing that is the homologue to the electric field. I think it is reasonable that those changes are seen in going from a total time derivative to partial derivatives of both time and space. Algebraically, I do end up with a force law that is written in terms of the symmetric curl and a difference between the time derivative of A and the gradient of phi.

The only way I see to really answer your question however is to do a manifestly covariant derivative of the force law. That would not have this step.

The Gauss->Coulomb I saw said presume spherical symmetry for a point change, and proceed. Will have to study up on it again.

I was worried that a term like "partial_x A_x" indicates a problem since to my eye it looks like something one uses for gauges.

I will have to puzzle over your games with the e and b field.

Quite a tough weekend, worrying about gauge transformations. Wallowing in the lack of rotational symmetry as it were.

Doug
Since I have an account now, I was considering writing something up about magnetic monopoles or something. What exactly is the difference between a column and a blog? If I click on "my column" in the dashboard it just gives me an "access denied" and redirects to main page.
Your column is the section where readers can find all of your work in one spot, search just your articles for terms, etc. but there is nothing in it until you have at least one thing, like a blog, so it isn't generated yet.  'Access denied' is a rather inelegant screen to ever get, though.
I read through all the comments tonight, and followed back to figure out what some of the stuff Margot was talking about.

She's right, and like she said, it was mentioned before but I didn't notice it : if you just took your usual E&M (and use quaternions if you want), and change ONLY the sign of the free-field terms, then you get the static equations:
F = - q Del phi
Del^ phi = rho

Exactly what you are trying to do here!
However:
1) it still has rotational symmetry
2) it still has Lorentz symmetry
3) it still is invariant to gauge transformations
4) the equations can be done in terms of E and B, instead of A if you want

Your current theory doesn't have any of these things. So why did you reject using your quaternion EM and just changing the sign of the free field terms? All these problems with the "hypercomplex" equations, and it looks like you were almost there if you just tweaked normal EM. Maybe it would be more profitable to go back and look at that?

Bingo, bingo (see new title for the blog).
Hmm... I could have sworn I typed that correctly.
Second try
F = - q Del phi
Del^2 phi = rho

Hello:

I have decided to retract the content of the main blog because I was unable to find a reasonable approach to the issue of gauge symmetry. While I could delete the blog, there may be some value in the struggles as detailed here. A deletion also looks more like a cover-up than an honest effort that did fail on technical grounds, darn rotational symmetry.

Thanks for the comments, they helped me reach this conclusion.

Thank you for not deleting everything. Some of the people here seem to understand the issues much quicker than I. Some of it I'm still trying to figure out. The posts + comments = lots of interesting info about EM and gravity that I never realized was important for the theories to work.