RETRACTION: Deriving The Unified GEM Field Equations (5/5)
By Doug Sweetser | October 11th 2011 12:37 AM | 12 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...

View Doug's Profile
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 diligent reader of this mini series on fields which comes to a close today will know what is coming: a Lagrange density will be defined, the Euler-Lagrange equation will be applied to create the field equations, and something will be said about gauges. This is the simplest of those seen so far, having only twelve terms in the Lagrange density instead of eighteen or twenty two. There is nothing grand about the process. No extra dimensions peak in from peek-a-boo branes. The simple math swings out the strange, making the GEM field equations glitter.

There is one word in the title, unified, that will cause controversy. There already is an equation for gravity and EM known as the Hilbert-Maxwell Lagrange density:
$\mathcal{L}_{\rm{H-M}} = \mathcal{L}_{\rm{matter}} + R +J^{\mu} A_{\mu} + \frac{1}{4}(\nabla^{\mu} A^{\nu} - \nabla^{\nu} A^{\mu})(\nabla_{\mu} A_{\nu} - \nabla_{\nu} A_{\mu})$

This is a combo meal of our two best theories for gravity and EM. There is no give-and-take between terms. Addition is not enough to claim unification.
Click or skip this reading of the blog.
It is easy enough to follow in the footsteps of failure by adding the Lagrange densities discussed so far in this mini series [IMPORTANT CORRECTION: they hypercomplex Lagrange density current coupling term must be distinct from the EM coupling term to have a force where like charges attract]:
\begin{align*} \mathcal{L}_{\rm{EM\;source}} &= scalar(J \times A + \frac{1}{4}(\nabla \times A - (\nabla \times A)^*) \times (A \times \nabla - (A \times \nabla)^* ) \\ \mathcal{L}_{\rm{hcG}} &= scalar(J \boxtimes A - \frac{1}{4}(\nabla^* \boxtimes A - (\nabla^* \boxtimes A)^*)^* \boxtimes (A^* \boxtimes \nabla - (A^* \boxtimes \nabla)^* ) \\ \mathcal{L}_{\rm{empty\; U}} &= \mathcal{L}_{\rm{EM\;source}} + \mathcal{L}_{\rm{hcG}} \end{align*}

A few comments on nomenclature for anyone jumping in late. The \times symbol X is for a quaternion product. The \boxtimes is for the hypercomplex or California product defined in the forth blog. It is like a quaternion product, but no minus signs are used on the road to generating four equations for universal attraction. The three players in these field equations are a 4-current density J, the 4-potentials A, and covariant 4-derivatives (upside down triangles).

The Lagrange density for the EM source equation can be used to derive two of the four Maxwell equations: Gauss's and Ampere's laws, as done in the second post. The hypercomplex gravity Lagrange density was used to derive an updated form of Newton's field theory for gravity. Controversy in the comments section remains over whether there is enough there there to provide a differential equation for the metric. For the Maxwell equations, a metric must be supplied as part of the background mathematical structure. I have claimed there is a potential/metric theory duality for the hypercomplex field equations. Work in a flat spacetime should you choose with the static equation:
$\rho = \partial_{\mu} \partial^{\mu} \phi$

Can this equation be solved if the potential is a constant? No, those are the differentials one learned in high school.
For a spacetime that is ever so slightly curved, which is the state of nearly all of spacetime in the Universe, treat one of the derivatives as a covariant derivative:
$\rho = \partial_{\mu} \nabla^{\mu} \phi$

The upside down triangle has a connection, which if metric compatible and torsion free is called the Christoffel symbol of the second kind. Those are technical conditions used in GR. The Christoffel symbol has first order derivatives of the metric, so this equation has second order derivatives of the metric. It is this equation along with a similar hypercomplex Ampere's law equation that has the exponential metric solution. I devoted a blog to this subject, and had a friend confirm the calculation independently.

Finally there is the form of the equation derived in the forth blog:
$\rho = \nabla_{\mu} \nabla^{\mu} \phi$

This form may apply to exceptionally strong, yet static fields. I have not found a metric solution to this equation.

This blog assumes the four hypercomplex gravity equations are a valid and interesting competing alternative with general relativity.

A good unification of gravity and the three other forces of Nature would be... no one knows. Gravity and EM are not enough. There must be a clear road to the weak and strong forces. There is no pattern of unification to follow. The unification of electricity and magnetism can be spotted here:
\begin{align*} \nabla \times A - (\nabla \times A)^* &= (0, \frac{\partial \vec{A}}{\partial t} + \vec{\nabla} \phi + \vec{\nabla} \times \vec{A}) \\ &= (0, -\vec{E} + \vec{B}) \end{align*}

Traditionally, one does not use quaternions, saying instead the two fields are part of the same second rank antisymmetric field strength tensor. Unification of EM and the weak force is done with group theory. General relativity with its lonely Lagrange density consisting of the Ricci scalar R has turned down all proposals by the brightest minds in physics for almost a hundred years. It is time to leave that beautiful bitch at the alter. Some women are not the marrying type. See Lady Gaga's nuptial bed at the end of the video for "Bad Romance".

When climbing at high altitude, it is vital to take a gentle, indirect assent. Let's ask a different question related to mass and the other forces of Nature: what does the Higgs mechanism accomplish?
Right out of the box, the standard model has particles with zero mass for all, including the bosons. That works fine for photons and gluons, but not the three bosons of the weak force. The two W's and Z's need to gain some mass in a way that does not break the symmetries found in the Lagrange density. The Higgs mechanism leaves the Lagrange density untouched. Instead, it uses the Mexican hat trick, spontaneous symmetry breaking of the vacuum by the scalar Higgs field. Just like the game "Clue", the LHC knows by now where Professor Plum must have used the candlestick to foil high energy physicists for decades on end: 119 GeV. That is the money bet. Bookies in England won't take a bet for finding no Higgs. That is the bet I want to make, no Goddamn Particle, not now, not in the near future with 5 inverse femtobarns of data from the LHC.

The basic fields have gauge fields, using either California or quaternion products:
\begin{align*} \nabla^* \boxtimes A &= (\frac{\partial \phi}{\partial t} - \nabla \cdot A, \frac{\partial \vec{A}}{\partial t} - \vec{\nabla} \phi - \vec{\nabla} \otimes \vec{A}) = (g, \vec{e} - \vec{b}) \\ A^* \boxtimes \nabla &= (\frac{\partial \phi}{\partial t} - \nabla \cdot A, -\frac{\partial \vec{A}}{\partial t} + \vec{\nabla} \phi - \vec{\nabla} \otimes \vec{A}) = (g, -\vec{e} - \vec{b}) \\ \nabla \times A &= (\frac{\partial \phi}{\partial t} - \nabla \cdot A, \frac{\partial \vec{A}}{\partial t} + \vec{\nabla} \phi + \vec{\nabla} \times \vec{A}) = (g, -\vec{E} + \vec{B}) \\ A \times \nabla &= (\frac{\partial \phi}{\partial t} - \nabla \cdot A, \frac{\partial \vec{A}}{\partial t} + \vec{\nabla} \phi - \vec{\nabla} \times \vec{A}) = (g, -\vec{E} - \vec{B}) \end{align*}

No matter how the 4-derivative and 4-potential are multiplied together, the first term of the product is the same (it transforms like the diagonal of $\nabla^{\mu} A^{\nu}$ for those more familiar with tensor notation, which is not a Lorentz invariant). This is a perfect setup for a perfect cancelation. In blogs 2-4, the gauge field was wiped out first thing. This time, the gauge field is a keeper. Rewrite the Lagrange densities including the gauges [IMPORTANT CORRECTION: The current coupling term is a sum of rho phi and J.A for the hypercomplex gravity proposal.]:
\begin{align*} \mathcal{L}_{hcG\;+\;gauge}&=J \boxtimes A + \frac{1}{2}(\nabla^* \boxtimes A)^*\boxtimes (A^* \boxtimes \nabla) \\ &= J \boxtimes A + \frac{1}{2}(g, -\vec{e} - \vec{b})^* \boxtimes(g, \vec{e} - \vec{b}) \\ &=(\rho \phi + J \cdot A, \rho \vec{A} - \vec{J} \phi - \vec{J} \otimes \vec{A}) \\ &\quad+ \frac{1}{2}(g^2 + e^2 - b^2, -2 g \vec{e} + \vec{e} \otimes \vec{e} - \vec{b} \otimes \vec{b}) \end{align*}
\begin{align*} \mathcal{L}_{EM\;+\;gauge}&=J \times A + \frac{1}{2}(\nabla \times A)\times (A \times \nabla) \\ &= J \times A + \frac{1}{2}(g, -\vec{E} + \vec{B}) \times(g, -\vec{E} - \vec{B}) \\ &=(\rho \phi - J \cdot A, \rho \vec{A} + \vec{J} \phi + \vec{J} \times \vec{A}) \\ &\quad+ \frac{1}{2} (g^2 + B^2 - E^2, -2 g \vec{E} + 2 \vec{E} \times \vec{B}) \end{align*}

The phases are worthy of study. When I have a small army of grad students, I promise to do so. It looks like fertile soil to work because Poynting's vector shows up by a happy accident. For now, I will follow the rules laid down by the Einstein summation convention and ignore the phase terms.

Define the GEM Lagrange density by subtracting one from the other [IMPOERTAND CORRECTION: there is a 3-vector current due to the hypercomplex gravity current coupling term change]:
\begin{align*} \mathcal{L}_{GEM}&= scalar(J \boxtimes A - J \times A \\&\quad+ \frac{1}{4}(\nabla^* \boxtimes A)^*\boxtimes (A^* \boxtimes \nabla) \\&\quad- \frac{1}{4}(\nabla \times A)\times (A \times \nabla) )\\ &=2(Jx+Jy+Jz)+ \frac{1}{4}(e^2 - b^2 + E^2 - B^2) \end{align*}

The gauge field is gone due to a perfect cancelation. There are 4 distinct field strengths, 4-derivatives of the 4-potential A. Each one is gauge dependent. That means they are relevant to particles that have mass and are unable to travel at the speed of light. Yet the overall Lagrange density is invariant under any gauge choice because of the cancelations. All of this happens at the level of the Lagrange density, not the solutions.

As done in previous blogs, the Lagrange density must be written out in the component parts. We have been there, done that, so copy in the results:

We need to toss in a minus sign for each of these to get to the GEM Lagrange density. Only the mixed terms have opposite signs so will cancel. We are left with only squared terms [correction: location of first '(' per comment below] [IMPORTANT CORRECTION: the GEM Lagrangian has a 3-vector current density]:

\begin{align*} \mathcal{L}_{GEM}&=2(Jx+Jy+Jz)\\ \ &\quad\frac{1}{2}(-(\frac{\partial Az}{\partial y})^2 - (\frac{\partial Ay}{\partial z})^2 + (\frac{\partial Ax}{\partial t})^2 + (\frac{\partial \phi}{\partial x})^2 \\ &\quad \quad - (\frac{\partial Ax}{\partial z})^2 - (\frac{\partial Az}{\partial x})^2 + (\frac{\partial Ay}{\partial t})^2 + (\frac{\partial \phi}{\partial y})^2 \\ &\quad \quad -(\frac{\partial Ay}{\partial x})^2 - (\frac{\partial Ax}{\partial y})^2 + (\frac{\partial Az}{\partial t})^2 + (\frac{\partial \phi}{\partial z})^2) \end{align*}

Focus on the terms with a phi:

Apply Euler-Lagrange:

Summarize by rewriting using fields:

Focus on the Ax terms, there are only three:

Apply Euler-Lagrange:

Summarize:

Compare the two laws:

There is a lot to talk about. Actually, the issue is that there is so little to talk about. Front row and center is the zero that is the current coupling term that is not there. Whatever this is, it cannot be about EM, there is no home for charge. Physics doesn't need Maxwell II since the Maxwell equations have stayed solid all these years. The GEM field equations don't encroach on that turf.

If there is no mass density, it cannot be about gravity. Gravity requires big honking sources. Zero is not big.

Bring together a force where like charges attract with other forces where like charges repel creates a conflict. A charge cannot do both. The zero looks good.

The GEM field equations are not grand, that was promised. They do accomplish two feats. The first is to offer an alternative to the Higgs mechanism. A gauge field is included and cancelled away in the same Lagrange density. That is what GEM does. The second bit of mathemagic is a field theory justification for the equivalence principle. The terms that use the California product house gravitational mass. The terms that use the quaternion product house inertial mass. The two always exactly cancel. Nice.

By repeating this exercise with a unit quaternion, the work will apply to a force that has SU(2) symmetry, the weak force being the obvious candidate. One could even plug in two quaternions, A* B, in the hopes of reaching out to SU(3) symmetry, but I am doubting that result since demanding the norm be one means there are not eight degrees of freedom. There is always something to struggle with.

Here is a property that surprises me to this day, and I have no idea what it means, but it sounds deep. There are technical quantum field theory books devoted to gauges. It seams like many important Lagrange densities are invariant under a gauge transformation, examples including the Hilbert and the Maxwell Lagrange densities. All the Lagrange densities that have appeared in this mini series have been invariant under a gauge transformation. The GEM field equations are also invariant under a gauge transformation, unlike the Einstein or Maxwell equations. What does it mean that the field equations are independent of a gauge choice? If being invariant under a gauge transformation is a deep idea for Lagrange densities, that might also be the case for field equations.

Doug

Snarky puzzle: I doubt it is meaningful, but do the sum of the gauge dependent Lagrange densities instead of the difference. Practice, practice, practice. Someday those equations might solve a deep mystery. Or not.

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.
Now that you completely understand what is on these simple garments, bet against the Higgs being found and buy the t-shirt

Next Monday/Tuesday: Snarky Puzzle Answers VI, including showing E2 - B2 is invariant under a Lorentz boost (well known), and e2 - b2, Doug might be wrong, will have to calculate and see.

I'm sorry, but this is just filled with claims that either come out of no-where (your "version" of the equivalence principle !?), or when they can be checked with math they are wrong. To focus these comments, I'll stick to the major things that I can relate to math.

Usually, a four-current is like: charge * four velocity. Above you implicitly assume J^\mu for your version of gravity is the same four-current as J^\mu for electrodynamics. It is by this assumption that you get all sources to cancel to get a "source-less" theory. However the "charges" for your gravity are not the same as the charges for electrodynamics, unless you want to deny the possibility of particles with positive mass that have negative electric charge.

You claim: "Yet the overall Lagrange density is invariant under any gauge choice because of the cancelations."
Electrodynamics has the gauge symmetry ( http://en.wikipedia.org/wiki/Gauge_fixing ) of
A -> A + grad f
phi -> phi - (partial / partial t) f
for any scalar field f.

To save space, for this next part I'll use the notation f_t to mean (partial / partial t) f, and f_xy is (partial / partial x) (partial / partial y) f, etc. Also, it looks like you have the parentheses in the wrong place for the first term in your Lagrangian written out in components, I'll assume that is a typo. Looking at your Lagrangian written in components, we can just read off what an arbitrary gauge choice would do:
L -> L + (1/2)(
- (f_zy^2 + 2 f_z Az_y) - (f_yz^2 + 2 f_y Ay_z) + (f_xt^2 + 2 f_x Ax_t) + (f_tx^2 - 2 f_t phi_x)
- (f_xz^2 + 2 f_x Ax_z) - (f_zx^2 + 2 f_z Az_x) + (f_yt^2 + 2 f_y Ay_t) + (f_ty^2 - 2 f_t phi_y)
- (f_yx^2 + 2 f_y Ay_x) - (f_xy^2 + 2 f_x Az_y) + (f_zt^2 + 2 f_z Az_t) + (f_tz^2 - 2 f_t phi_z)
)
This can be simplified a bit,
L -> L + (
- (f_zy^2 + f_z Az_y + f_y Ay_z) + (f_xt^2 + f_x Ax_t - f_t phi_x)
- (f_xz^2 + f_x Ax_z + f_z Az_x) + (f_yt^2 + f_y Ay_t - f_t phi_y)
- (f_yx^2 + f_y Ay_x + f_x Az_y) + (f_zt^2 + f_z Az_t - f_t phi_z)
)

If the Lagrangian was invariant (like in electrodynamics) we would just get
L -> L
that is clearly not the case here. Due to the lingering A and phi terms, we can see that for this to work independently of A and phi, f must be independent of x,y,z,t. Which means we can only use f=constant. Which if we go back to our "gauge transformation" just gives:
A -> A
phi -> phi
In other words, there is no gauge freedom left. Since (E^2 - B^2) is gauge invariant, this means (e^2 - b^2) is not. (This also affects your previous article where you assumed you could choose the Lorenz gauge to simplify things.)

Also, looking at your equations of motion, the four components of the four-vector A are completely decoupled. As there is no phi_t term in your Lagrangian, phi is not a physical degree of freedom. That leaves three degrees of freedom. Since there is no gauge freedom, these can not be reduced further.

Therefore, unlike electrodynamics with 2 physical polarizations, or gravity with 4 physical polarizations, here there are 3 physical degrees of freedom. You don't have a theory of gravity or of electrodynamics. As currently written, since there aren't even any source terms, you have a free field theory of three decoupled fields.

Hello Henry:

Usually, a four-current is like: charge * four velocity. Above you implicitly assume J^\mu for your version of gravity is the same four-current as J^\mu for electrodynamics. It is by this assumption that you get all sources to cancel to get a "source-less" theory. However the "charges" for your gravity are not the same as the charges for electrodynamics, unless you want to deny the possibility of particles with positive mass that have negative electric charge.

Thanks for this question. The charge for EM in the Maxwell equations can take two signs. In tensor terms, the antisymmetric field strength tensor flips signs if the indexes are swapped. The hypercomplex gravity field equation uses the rest mass as the charge, which only has one sign. In tensor terms, the field strength tensor is symmetric and will not flip signs if the indexes are swapped. Now we have a third set of equations. If the particles in question happen to be electrically neutral, the charge is the rest mass. If there is an electric charge, then it is a simple enough exercise in dimensional analysis to make mass and electric charge have the same units so can be added or subtracted as needed. Such a mass-electric current does not appear in the final equations.

This highlights a social problem with the proposal. We all like to keep inertial mass separate from electric charge. Of course we don't know what either of these things actually are. My guess is that only if one develops a solid math and visual depiction of what inertia and electric charge are, will the idea have a chance. Since I also was brought up with the wall of separation between mass and electric charge, I can accept a rejection on those grounds.

Thanks for working through the common gauge symmetry example for EM. Everybody does it, so it must be good.

As you have pointed out correctly, the 4-potential (phi, A) transforms as a 4-vector. The gauge symmetry everyone adds to this, (-df/dt, Del f) transforms like part of a rank 2 tensor that is not generally covariant (jump to a different inertial reference frame, and other terms will show up, like Del.A and DA/dt). This looks to me like sloppy - but common - accounting. It is wrong to add parts of a rank 2 tensor to a 4-potential and reach any conclusion. [ERROR: the critique in this paragraph is inaccurate, as the field f must be a Lorentz invariant scalar field, not phi, the first component of a 4-vector.]

Here are three gauges one can choose:

Coulomb gauge: $\nabla \cdot A = 0$
Static gauge: $\frac{\partial \phi}{\partial t} = 0$

Lorentz gauge: $\frac{\partial \phi}{\partial t} + \nabla \cdot A = 0$

Write the hypercomplex fields in these three gauges:

\begin{align*} \left( \frac{\partial}{\partial t}, \vec{\nabla} \right)^* \boxtimes \left(\phi, \vec{A} \right) &= \left( \frac{\partial \phi }{\partial t} - \nabla \cdot A, \frac{\partial \vec{A}}{\partial t} - \vec{\nabla} \phi - \vec{\nabla} \otimes \vec{A} \right) \\ & =\left( \frac{\partial \phi }{\partial t}, \frac{\partial \vec{A}}{\partial t} - \vec{\nabla} \phi - \vec{\nabla} \otimes \vec{A} \right) \rm{Coulomb} \\ & =\left( -\nabla \cdot A, \frac{\partial \vec{A}}{\partial t} - \vec{\nabla} \phi - \vec{\nabla} \otimes \vec{A} \right) \rm{Static} \\ & =\left( 2 \frac{\partial \phi }{\partial t}, \frac{\partial \vec{A}}{\partial t} - \vec{\nabla} \phi - \vec{\nabla} \otimes \vec{A} \right) \rm{Lorenz} \end{align*}

Here is the way I define the hypercomplex gravity Lagrange density:

$\mathcal{L}_{\rm{hcG}} &= scalar(J^* \boxtimes A - \frac{1}{4}(\nabla^* \boxtimes A - (\nabla^* \boxtimes A)^*)^* \boxtimes (A^* \boxtimes \nabla - (A^* \boxtimes \nabla)^* )$

I don't see how this Lagrangian can change under those three choices of gauge. That is what the subtraction step does.

Henry,
Mr. Sweetser already admits "Whatever this is, it cannot be about EM, there is no home for charge. ... If there is no mass density, it cannot be about gravity." My biggest complaint would merely be: Calling it "unified" or making strange claims about the "equivalence principle", is pointless if even the author is unable to state what these equations are supposed to be representing in the real world.

I think that is why there weren't any responses to this for about a week. When I first read the Lagrangian, all I could think was "what in the world is he even trying to do?" Then the author went on to say essentially the same thing. There is no substance here to discuss.

Oh, and Henry, it looks like you made a small typo: f_x Az_y should be f_x Ax_y. That of course doesn't change your conclusion on gauges though.

Hello LagrangiansForBreakfast:

I thought I was offering an alternative to the Higgs mechanism. Looks like I failed to communicate that idea. If the Higgs is found at 119 GeV, the GEM proposal dies on Tammaso's post. This proposal can be rejected on experimental grounds within a few months.

On the other hand, if no Higgs is found, then all the good little theoretical physicists get to wipe their blackboards clean and start over. Right, like they are waiting. What will they try? I have no idea, but suspect they have no idea either.

It looks like you are using the method of gauge transformation everyone teaches: let's just drop random shit in until something nice happens. My approach here was to be complete: using multiple products of 4-derivatives with 4-potentials. The gauge terms just show up that way. What I failed to communicate was the way I change the gauge: by altering the gauge in situ (in place). That is not a natural thing to do with tensors, but it is with California or quaternion products. I didn't appreciate that would be hard to explain.

A longer post on the equivalence principle is available. I should have been more precise: The GEM Lagrangian tries to address the weak equivalence principle, which is about the gravitational forces (ones with hypercomplex products) always exactly balancing the inertial forces (ones with quaternion products).
That we cannot agree on simple math frustrates me to no end. The math is unambiguous. However, instead of accepting the answer or pointing out errors, you have this habit of just ignoring the results and stating something in quaternions as if that answers the question. If our only common ground is the components, let's please focus on the components.

"The gauge symmetry everyone adds to this, (-df/dt, Del f) transforms like part of a rank 2 tensor that is not generally covariant ... It is wrong to add parts of a rank 2 tensor to a 4-potential and reach any conclusion."
Stop saying incorrect statements about the math. You are now claiming that we start with the four-vector field A and the gauge transformation gives something that transforms neither like a four-vector nor a tensor. That is just wrong.

To match my sign choice above, the gauge transformation in tensor notation is:
A^\mu --> A^\mu - partial^\mu f
Nothing is being taken from parts of a rank 2 tensor as you claim.
Please take the time to accept your error and learn from it.

You seem to be confusing something about the gauge transformation, so in an effort to explain possibly what you were remembering, let me add some info. If we use the gauge freedom to enforce a particular "gauge condition" (an equation we want the components of A^\mu to satisfy), while A^\mu is still a four-vector and will transform like such, unless the gauge condition itself is Lorentz invariant (such as the Lorenz gauge choice) the transformation will not preserve the gauge condition. Again, this does not in any way support your complaint "It is wrong to add parts of a rank 2 tensor to a 4-potential and reach any conclusion."

"I don't see how this Lagrangian can change under those three choices of gauge."
How can you not see it? Again you go back to quaternions, make some incorrect claims, and use this to justify ignoring the explicit math someone wrote out for you. If your quaternion argument convinces you, then go and see if there is a flaw in the logic and math presented with the components.

"Write the hypercomplex fields in these three gauges: ..."
Your way of "writing" or applying those gauges conditions is assuming you have a gauge symmetry. As shown, you do not have a gauge symmetry. So you haven't actually shown anything useful here.

You've made this error before. You are assuming what you are trying to prove in your attempt to prove it. Despite your claims otherwise, this is not a valid method of proof.

Hello Henry:

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

In the static case:

$\vec{E} = - \vec{\nabla} A^0$

The phi is part of a 4-vector. What you wrote was f, which must be a Lorentz invariant scalar field. My mistake was to equate phi with f. Simple enough. I drop all my complaints about the second rank tensors, and edited my comment above to make that clear.

Now that I have a more accurate account of f, I realize it must be a Lorentz invariant scalar field for the standard example you work out in detail to make sense.

Is there any example of a Lorentz invariant scalar field in all of Nature?

This is too funny:

No fundamental scalar fields have been observed in nature, though the Higgs boson may yet prove the first example.

http://en.wikipedia.org/wiki/Scalar_field_theory

I have a lot riding on the LHC's effort to find the Higgs! If the Higgs is found, then my house of cards collapses. If no Higgs is found, my t-shirt sales will spike (probably not :-) I will be able to return here and say this is an empty technical challenge. I don't worry about what ghosts can do to me, they are not real. I will not worry about a kind of field that only exists on paper.

Technical disagreements rock, no matter how bad they make me or my opponents feel. I got a new reason to suspect bad accounting. That f must be a spin 0 field. This is the way everyone does the gauge transformation for photons, a spin 1 field? Excuse the techno-nerd me? Spin 0 fields can be used for forces where like charges attract. In EM, like charges repel. Does taking the derivative of a spin 0 field convert it to the ways of spin 1? Anyone with lots of knowledge of spin, please speak up.

The frustration is mutual. I have not communicated the difference between the tensor approach where people get to be smart, spotting the symmetry, versus both the quaternion and hypercomplex methods where the players in gauge symmetry show up naturally. Multiplication is not a presumption in gauge symmetry. As the gauges are wiped out in the definition of the Lagrange density, that provides the ability to pick whatever gauge one wants without altering the Lagrange density.

Doug

"Anyone with lots of knowledge of spin, please speak up."
You are misunderstanding so much here, that it is not worth trying to explain all of this. So here is just an overview/starting point. The gauge freedom in E&M does not represent a physical scalar field. In fact, the gauge freedom is not telling us that there are more physical degrees of freedom, but LESS physical degrees of freedom. In classical electrodynamics, we choose the four components of the four-vector potential to be our generalized coordinates in the Lagrangian. This may lead one to think there are four physical degrees of freedom. However if you work through the math, you see that there are actually only two physical degrees of freedom in the field (two linearly independent polarizations).

"As the gauges are wiped out in the definition of the Lagrange density, that provides the ability to pick whatever gauge one wants without altering the Lagrange density."
Please stop doing this. Please stop just presuming your answer in defending your math. You see something reminiscent of a single gauge condition in a component of your quaternion and decided that this therefore can be interpreted as the gauge freedom. There is absolutely no math or real logic behind such an interpretation. And as shown to you with explicit math, this interpretation is wrong, as you do NOT have "the ability to pick whatever gauge one wants without altering the Lagrange density".

The math is clear here. There is no denying the example with the components. If you are really on a quest to learn, please stop denying the simple math and actually learn here.

Hello Henry:

This is a research blog, not a teaching blog. The most important thing in research is to know and explore what is already well known. That is what the first blog in this series derived the Euler-Lagrange equation which I could have just quoted. As much as I admire Jackson, whether the book is dressed in a red or blue cover, he blurts out the Maxwell equations in chapter two, not getting around to deriving them from the Lagrange density until chapter eleven or twelve.

I had seen the gauge symmetry analysis many times. To start on common ground, I will take your written out proof and dress it up.

$\\ 1:\quad A^i \rightarrow A'^i = A^i + \nabla^i f \\ 2:\quad A^0 \rightarrow A'^0 = A^0 - \frac{\partial f}{\partial t}$

where i ranges from 1-3.

These need to be dropped into the GEM Lagrange density, and see if things nicely cancel out. Here is the Lagrange density before the great gauge transformation:

\begin{align*} 3:\quad \mathcal{L}_{GEM} &= \frac{1}{2}((-\frac{\partial Az}{\partial y})^2 - (\frac{\partial Ay}{\partial z})^2 + (\frac{\partial Ax}{\partial t})^2 + (\frac{\partial \phi}{\partial x})^2 \\ &\quad \quad - (\frac{\partial Ax}{\partial z})^2 - (\frac{\partial Az}{\partial x})^2 + (\frac{\partial Ay}{\partial t})^2 + (\frac{\partial \phi}{\partial y})^2 \\ &\quad \quad -(\frac{\partial Ay}{\partial x})^2 - (\frac{\partial Ax}{\partial y})^2 + (\frac{\partial Az}{\partial t})^2 + (\frac{\partial \phi}{\partial z})^2) \end{align*}

Now I go ahead and to the gauge transformation:

Is there a way for the two Lagrange densities in lines 3 and 4 to be equal? Sure, but in an empty way as the scalar function f would have to not be a function of t, x, y, or z.

$\therefore \mathcal{L_{GEM}} \ne \mathcal{L'_{GEM}} \quad QED$

The good student gets a good grade, good for him.

Presume 1 and 2 are OK, and the rest follows. I get that. Inever denied the simple math.

As a researcher, I have to understand that bit of math. Hopefully this work at my lunch hour proves I get your proof.

What researchers do is challenge assumptions. What I explicitly challenged was lines 1 and 2, whether those made sense from a physics standpoint. The scalar function f, what is it? You propose it is nothing. Physics uses virtual photons all the time. It is a bad label because virtual photons get to push real electrons apart. If the field f is a virtual field, then it is a virtual spin 0 field, and thus would be involved with a force where like charges attract. That is logically inconsistent with EM. I am skeptical of any analysis that relies on Nature being inconsistent.

Here is a fine, outstanding transformation:

$R \rightarrow R' = U \times R \times U^*$

Where R is a quaternion, U is a quaternion the times indicates a quaternion product. Everyone is like everyone else.

Contrast that with 1+2: above which has a 4-vector Au, a dual 4-vector space derivative, and a Lorentz invariant scalar field f. My gut says adding together a vector and a dual vector might be found on, but I am not certain of that detail.

One productive element of this exchange is that I see the origins of my own error: the phi in my gauge term is the first term of a 4-vector. Yes, it is different from the standard. The nice thing is this is not some arbitrary ass function f, which is what the world works with. Instead it is constrained to be the same phi that makes itself known in the field. The antisymmetric tensor does precisely the same thing, subtracting away the very same symbols, the ones down the diagonal of $\nabla^{\mu} A^{\nu}$

I know what the gauge freedom does. I believe I provided a darn similar description of why it is needed I think in 4 out of the 5 blogs in this series (didn't double check).

You have demonstrated you don't follow my line of logic as shown by your focus on quaternions. The issue has not a thing to do with quaternions. I get the same results as EM when using the quaternion product. It is the properties of the hypercomplex product that are both different and not acknowledged. If you choose to not do a hypercomplex product in any calculation, then frustration with me must remain. That is your choice, not mine.

...you have this habit of just ignoring the results and stating something in quaternions as if that answers the question. If our only common ground is the components, let's please focus on the components.

This is not the forrest versus the trees plea, it is a vote for the pine needles. Work only with the components. If I would only agree to that one rule, you would win. I provided the reason to work at the forrest level: this is a systematic exploration of all the symmetric ways a 4-potential can change with a 4-derivative.

Doug

"Presume 1 and 2 are OK, and the rest follows. I get that. Inever denied the simple math."

You are denying the simple math. This is just like the argument about the other symmetries of your lagrangian. Someone gives simple logic or explicit math showing your theory doesn't have a symmetry, and you deny it for some reason until you do it yourself in mathematica. If that is what it takes, then please pause, and check for the gauge symmetry in Mathematica.

"Contrast that with 1+2: above which has a 4-vector Au, a dual 4-vector space derivative, and a Lorentz invariant scalar field f. My gut says adding together a vector and a dual vector might be found on, but I am not certain of that detail."
More of this again? We are not adding a four-vector with a dual-vector. Look at what I wrote previously, and note that partial^\mu transforms just like A^\mu. In tensor notation it is partial_\mu (with the lowered index) that would specify the dual vector to the four-vector partial^\mu.

"What I explicitly challenged was lines 1 and 2, whether those made sense from a physics standpoint. The scalar function f, what is it? You propose it is nothing. Physics uses virtual photons all the time. It is a bad label because virtual photons get to push real electrons apart. If the field f is a virtual field, then it is a virtual spin 0 field, and thus would be involved with a force where like charges attract. That is logically inconsistent with EM. I am skeptical of any analysis that relies on Nature being inconsistent"
This is like a word salad of physics terms. It is not worth tackling this until we can agree on some more basic things.

"Here is a fine, outstanding transformation:
R --> R' = U x R x U*
Where R is a quaternion, U is a quaternion the times indicates a quaternion product. Everyone is like everyone else."

We agree that in EM there is a gauge freedom in the fields. If you can find a way to write this freedom in quaternions, it would be the same exact thing just in different notation. You even use the phrase "same up to an ismorphism" to convey the same idea. So if I can prove it wrong using the components, you cannot undo this with quaternions. So let's be clear and honest here, your potential field does not have the same freedom as EM. If you are going to now make up an inequivalent transformation on the spot here, that is a different question.

As your theory does not have rotational or Lorentz symmetry, it is clear your Lagrangian isn't invariant to all of your new proposed "transformations" of the potential either. Stop making new errors, slow down, pause, and learn from your mistakes. Progress in this discussion is painfully slow, as you keep plunging ahead into new errors before fixing the old ones.

I will reply to this in a separate blog, backed up by a Mathematica notebook. Here is the kernel of the observation. What goes on in EM gauge transformations is that E->E' and B->B' where E=E' and B=B' because of Faraday's and the no monopoles field equations. Those vector identities do not exist for hypercomplex products. They better not exist for gravity where there is only one form of charge that always attracts. Gravity and EM are different in how one can demonstrate gauge symmetry. It is easy to do, just pick a gauge and look how it will alter the Lagrange density, no need to transform.
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.

sir, i have simply found a time travel formula. i have went trough it throughly, i can't find a mistake in it, i derived a simple GEM field equation separately, through it only i got this formula, i mean the time travel formula.