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## Unifying Gravity and EM

Hello Lut:

The way the g fields cancel which I write the fields using the Hamilton and Even representations of quaternions, does that mean I now have gauge symmetry for GEM? And would you remind me of why gauge symmetry is so important to have?

The morning realization: to be inverted, the Even representation must exclude its own Eigenvectors and values. Not sure of the consequences of that...

Thanks,
Doug

 To Sweetser, I got a couple of messages from P-forum about this. I have not followed your stuff in some time. But I will illustrate what a gauge theory is. Say you have a manifold M with a bundle over it which has some vector space or algebraic system of roots, that are eigenvector of the weights. For the manifold of dimension = d and the bundle of dimension = p we can define a section s over the base manifold. The principal bundle group is then a set of transformations between bundle sections $$s'~=~gs.$$ Now consider the action of a differential operator d on this transformed section $$ds'~=~d(gs)~=~(dg)s'~+~gds$$ which can be seen in elementary terms with $ds~=~As$ as $$ds'~=~A's'~=~\Big((dg)g^{-1}~+~gAg^{-1}\Big)s'$$ which clearly gives the transformation of the gauge connection A. This equation is not homogeneous with respect to the transformation. Yet for the gauge fields $F~=~dA~+~A\wedge A$ the fields transform as $F'~=~gFg^{-1}$, which is a homegenous covariant transformation. Now the field term can be seen as due to the action of the two-form $$\Omega^2~=~(d~+~A)\wedge(d~+~A)$$ on the bundle section. Now this differential form is antisymmetric in its tensor components for the fields $F_{ab}~=~-F_{ba}$. This can be extended to spacetime curvature as well, where the bundle is a fibre bundle with a hyperbolic group structure and so forth. This antisymmetric structure is grounded in some basic differential geometry and topology, where these systems of forms can define cohomology rings and groups. Now what you have is a theory which has symmetric field components. As a result you are doing something else. This is not to say that what you are doing is utterly wrong, but honestly it is outside the standard canon of field theory and the underlying system of differential geometry it is formulated around. As such what you are proposing is something other than gauge theory, or is an extension of gauge theory into some other arena. Cheers, Lawrence B. Crowell

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Hi Doug,

after a great struggle I realised what you are attempting with the quaternions. You've found a representation that allows you to separate the fields ( symmetric and anti-symmetric curls ) in a way that you find correct, satisfactory. Right now I can't comment on the meaning or validity of this.

I understand that the differential operators correspond to ordinary partial differentiation.

Getting to gauge invariance: the parts of your Lagrangian density that contain the potential are -

$$-A^{\mu}J_{\mu} + \frac{1}{4c}\partial^{\mu}A_{\nu}\partial^{\nu}A_{\mu}$$

One can split the second term into symmetric and antisymmetric parts without changing the energy.

Provided those operators are not covariant derivatives, this part is globally gauge invariant, because adding a constant to A will not affect the energy. Local gauge invariance which applies when we add a value to A which depends on x, is assured by specifying a gauge condition, which ensures that the Lagrangian doesn't change when the potential is altered.
My complaint has always been that when the derivatives are covariant, products of metric terms with the potential itself appear in the symmetric terms. Now if you add a constant to A, the energy changes, and gauge invariance is lost. Unless more constraints are added to make those terms disappear.

Everything I know about this subject comes from the EM field, and I'm not even sure this can be applied to the symmetric fields. For instance, not requiring a fixed value for A ensures energy conservation, as does the gauge condition.

But as Lawrence B. Crowell tells us, you are going outside standard differental geom. definitions ( at least, I think that is part of what he's saying).

 As such what you are proposing is something other than gauge theory, or is an extension of gauge theory into some other arena.
This is not bad news. If you keep track of the energy you won't go wrong.

But I don't think your calculations above affect my ( simplistic ?) view about gauge invariance, because there is only a problem when the derivatives are covariant and introduce terms in A itself.

I've got to go now, but I'll be revisiting soon.

 My comments are meant to show that if you have a theory of fields with $$F_{ab}~=~\{Q_a,~{\bar Q}_b\}$$ then this symmetric term is an anticommutator of grassmannian fields. In the case I write above this defines gauge connections for N > 2 supersymmetry, for $A_i~=~{\sigma_i}^{ab}F_{ab}$. Yet these fields are physically Fermionic. Gauge fields are vector (or chiral) and don't obey this sort of rule. To have symmetric gauge fields F_{ab} = F_{ba}, or components which obey this, runs to my mind in a host of mathematical questions. This appears to be something other than supermanifold theory. There a gauge field will be contained in a super-multiplet with $$\Phi~=~A~+~{\bar\xi}\psi~+~\xi{\bar\psi}~+~F$$ for $\psi$ the fermionic super pair (gaugino) of the gauge potential, $\xi$ the Grassmannian spinor super fields. The gaugino will obey anti-commutator rules. Yet the gauge fields are still bosonic. This GEM theory appears to be something else, and there are to my mind a host of open questions. The biggest question is whether this even defines an appropriate geometry. A crucial fact is "boundary of a boundary is zero," which is from d^2 = 0 why gauge fields are anti-symmetric. In the case of Grassmannians which obey anti-commutators we have $Q^2~=~0$, which is equivalent to the Pauli exclusion principle. the d^2 = 0 principle is also why the Riemannian curvature tensor has anti-symmetric indices. Once you twist this around with symmetric terms then the whole geometric meaning of things is lost to me --- outside of a super-manifold theory. So without a blizzard of tensor analysis and differential terms and the like it seems important to give some indication of what this means geometrically. Without some sense of what this means geometrically, such as the symmetries of a manifold, and how this connects up to conservation laws (eg d^2 = 0 is what gives Bianchi identities etc) then I have a very difficult time knowing how to evaluate this. Lawrence B. Crowell
 Blog Entries: 3 Recognitions: Gold Member Lawrence, thanks for taking the time to post all this. I don't have the math to understand it all, unhappily. I think I can see how energy conservation is linked to the geometry. The GEM lagrangian has no explicit time dependence, which ought to be enough to ensure energy conservation. I can't remember now how gauge invariance came up. There's a lot for Doug to think about still. M
 Gauge theory is a bed rock of the physics of fields and forces. In fact the subject is mathematically rich. If you are familiar with Uhlenbeck, Atiyah, Donaldson, Freed, etc, the structure of gauge theories, in particular their moduli spaces, have proven surprising results on the categorization of four dimensional manifolds. The work I have been doing for the last two years involves these results. My sense is that GEM is geometrically mysterious. All of the stuff we do with physics is ultimately based on on fundamentals of differential geometry. For instance in ordinary three dimensions a function F under the action of a differential operator d is $$dF~=~\frac{\partial F}{\partial x^i} dx^i,$$ which is the gradient. For a one form $\omega~=~\omega_idx^i$ the action of d is $$d\omega~=~\frac{\partial\omega_i}{\partial x_i}dx^j\wedge dx^i,$$ and since $dx^i\wedge dx^j~=~-dx^j\wedge dx^i$ clearly i can't equal j. This in vector language is the curl. You can actually prove all the div, grad & curl stuff this way if you also throw in something called Poincare duality. All the anti-symmetric tensors and the rest in gauge theory are grounded on this, and this has incredible generalizations for manifolds and bundles. GEM appears to throw all of this to the wind, which means there has to be some sort of other foundation to this. As yet I have not seen it. Lawrence B. Crowell

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Hello Lawrence:

Thanks for all your technical responses. I wish to make clear that there is something brand new in the GEM proposal as of posts 424 and 425. It is fair to say I am working on the extension of gauge theory into quaternion representations, but one that does tie in directly to some of what we know of gauge theory.

There are also some clear breaks. Let me point one out. The Bianchi identities will never be relevant to this work. The Bianchi identities come out of the Riemann curvature tensor. While I do work with the connection if and only if it appears as part of a covariant derivative, I never work with the connection outside of a covariant derivative as happens with the Riemann curvature tensor. This makes communicating with someone with your level of training darn near impossible because you find some reason to reintroduce the curvature tensor or its stand ins.

[digression]
Why should the Riemann curvature tensor create problems? After all, it is part of a huge intellectual effort, done by some of the smartest math and physics people ever (Lawrence listed a few of the many authors working on this subject). It is both courageous and daft to claim Riemann curvature tensor is irrelevant to how Nature works. Here is the logic most readers of this thread should be able to follow.

The 4-potential $A^{\nu}$ transforms like a tensor.
The 4-derivative $\partial^{\mu}$ transforms like a tensor.
The 4-derivative of a 4-potential $\partial^{\mu} A^{\nu}$ does not transform like a tensor.
To correct this problem, we need to add in the connection. It turns out there are lots of options with the choice of the connection. I choose to work with the same one used in GR, which is to say the connection is torsion-free and metric compatible, so the connection is the Christoffel symbol of the second kind. The Christoffel symbol has three first derivatives of the metric. Thus we get the definition of the covariant derivative:

$$\nabla^{\mu} A^{\nu} = \partial^{\mu} A^{\nu} - \Gamma_{\sigma}^{\mu \nu} A^{\sigma}$$

Einstein learned this stuff (he didn't invent it) from is math tutors. He focused on the Gamma which contains the first derivatives of the metric. He then asked his math buddies for the math object that transforms like a tensor but has second order derivatives of the metric. He was thinking about good old $F=Ma$, how Nature accomplishes things with second order derivatives. That was his inspired guess, and he needed some direction to implement it. The answer he was given was the Riemann curvature tensor. That was a great answer that has led to lots of great results. The way math wonks sell it, they say it is the only answer without getting into really obscure math.

Yet Einstein missed the obvious way to get second order differential equations of the metric. Take a closer look at what Newton did. Newton was saying $m \frac{dR}{dt}$ is not interesting, but acting again on this with another time derivative operator, $\frac{d}{dt}$, then you get $F=m \frac{d^2 R}{dt^2}$. Follow that program exactly. Einstein is saying $\nabla^{\mu} A^{\nu}$ with one derivative of the metric is not interesting, so take another covariant derivative: $\nabla_{\nu} \nabla^{\mu} A^{\nu}$. There will be the divergence of the Christoffel symbol. Drop the Rosen metric into the Christoffel, take its divergence, and see something even Poisson would recognize. Why I cannot lead a well-trained mathematical horse to do that calculation is beyond me, but that's the way it is.

GEM is about potentials in bed with geometry. The lights are on, the camera is rolling. Someday I'll sell a lot of film, but until then, I will post to the Independent Research forum.
[/digression]

I have an even simpler view of gauge theory, the one that applies directly to EM. I look at it in terms of this specific Lorenz gauge transformation:

$$(\phi, A) \rightarrow (\phi', A')=(\phi, A)+(\frac{\partial f}{\partial t}, - \frac{\partial f}{\partial x}, - \frac{\partial f}{\partial y}, - \frac{\partial f}{\partial z})$$

The function f needs to have a few derivatives. Drop this into the anti-symmetric field strength tensor, form the field equations and all the dependence on f drops out. Drop A' into any symmetric tensor, and any field equations that come out will depend on f. This is a point you made long ago, and looking back, I don't think I addressed it. Why not? Your point is technically correct. That's the way tensors are.

In post 425, I used no tensors. Instead I used two representations of quaternions. The first is the well known Hamilton representation of the quaternion division algebra. It is an asymmetric tensor, with the symmetric part being made of the scalar, and the antisymmetric part being the 3-vector. The other one I am calling the Even representation. This is symmetric and is a division algebra so long as one does not use the Eigenvalues and Eigenvectors of the 4x4 matrix representation. We can point to gulfs in our communication because I am using these tools, and you are skilled with differential geometry.

When I added all 5 of the fields in my proposal together, E, B, e, b, and g, this was the result:

$$\frac{1}{2}\left(-A \nabla + \nabla^* A2\right)$$

$$= -g + E + B + g + e + b$$

$$= E + B + e + b$$

where
$$g = \frac{\partial \phi }{\partial t}-\frac{\partial Ax}{\partial x}-\frac{\partial Ay}{\partial y}-\frac{\partial Az}{\partial z}$$
$$E = (-\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x},-\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y}, -\frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z})$$
$$B = (c \left(\frac{\partial Ay}{\partial z}-\frac{\partial Az}{\partial y}\right), c \left(\frac{\partial Az}{\partial x}-\frac{\partial Ax}{\partial z}\right), c \left(\frac{\partial Ax}{\partial y}-\frac{\partial Ay}{\partial x}\right) )$$
$$e = (\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x},\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y}, \frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z})$$
$$b = (-c \left(\frac{\partial Ay}{\partial z}-\frac{\partial Az}{\partial y}\right), -c \left(\frac{\partial Az}{\partial x}-\frac{\partial Ax}{\partial z}\right), -c \left(\frac{\partial Ax}{\partial y}-\frac{\partial Ay}{\partial x}\right) )$$

Notice how the field g, no matter what it is, drops out. That means you are completely free to work with whatever g field you choose.
Consider a function h such that $\nabla h = g$. Then:

$$A \rightarrow A' = A + h$$

Hit this with a derivative, out comes the field g, and the field g drops. That is a gauge transformation.

 GEM appears to throw all of this to the wind, which means there has to be some sort of other foundation to this. As yet I have not seen it.
I noticed no references to the Hamilton or even representation of quaternions in your replies. That is where the foundation is. We do not have to feel bad if we fail to communicate. You have demonstrated an impressive knowledge of both the history and current efforts to do productive research using the power tools of differential geometry. I respect that. New math is for the next generation. I have no expectation that I will ever be comfortable with it all, but that is the way research goes.

Here is my specific plan of action. I have shown how to write the 5 GEM fields in a compact way using the Hamilton and Even representations of quaternions. I have shown how to write the coupling term using the Hamilton and Even representation in a way that shows the coupling represents both spin 1 and spin 2 fields. I have to form the action from these fields. Then I have to generate the field equations from the action. Finally Lut, I will calculate the stress energy tensor to address your energy conservation question which is a good one. All this work must be checked with Mathematica. There is much to do, like prepare for three talks, make new quaternion animations, do outreach, work full time on something else, walk the dogs, and play nice with the wife.

Doug

 Recognitions: Gold Member Hello: I found a clear technical error with post 424 titled "Generalizing Current Coupling Spin 1 & Spin 2". The two current coupling terms look like so: $$-J J' -J2 J2'^*$$ These both have the same sign. When dropped into the Lagrangian, they will have like charges repel. That is good for EM, but failure for gravity. Now that the issue is defined, let's create a solution. Post 424 was trying to generalize a specific case for motion along the z axis. There I needed to use the first or second conjugates, which I had defined somewhere in this vast discussion as: $$(i q i)^* = (-t, x, -y, -z) === q^{*1}$$ $$(j q j)^* = (-t, -x, y, -z) === q^{*2}$$ We can continue this theme and define the third conjugate like so: $$(k q k)^* = (-t, -x, -y, z) === q^{*3}$$ As an indication of generalization, all three conjugates are needed: $$+J2^{*1}^{*2} J2'^{*3}=(\rho, -Jx, -Jy, Jz)(-\rho', -Jx', -Jy', Jz')=$$ $$=(- \rho \rho' + Jx Jx' + Jy Jy' + Jz Jz',$$ $$-\rho Jx' + Jx \rho' - Jz Jy' - Jy Jz',$$ $$-\rho Jy' + Jy \rho' - Jz Jx' - Jx Jz',$$ $$-\rho Jz' + Jz \rho' - Jy Jx' - Jx Jy')$$ This current coupling term has three defining characteristics: 1. The scalar is invariant under a Lorentz transformation 2. The phase has both spin 1 and spin 2 symmetry, necessary for a field theory where like charges repel for EM and attract for gravity respectively 3. Is a positive coupling between the two currents, so when put in the action will apply to like charges that attract. The analysis of the -J J' coupling is unaltered. The generalzized coupling term using the Hamilton and Even representations is: $$-J J' + J2^{*1}^{*2} J2'^{*3}$$ That looks better. Dodged another bullet tonight. Doug
 Recognitions: Gold Member Hello: I said I had to "form the action from these fields" using quaternions only, no tensors. This was not an easy exercise. Once again, the simplest term gave me problems, specifically the g field. I spent much time thinking about what I was trying to do with these fields before I could proceed effectively. In this post I will give an overview of the math to follow in later posts. The five fields defined in post 432 - the standard electromagnetic E and B fields, their symmetric counterparts e and b, and a diagonal g - together transform like a second rank tensor. That tensor gets contracted to make a Lorentz invariant scalar that goes into the action. I was not able to build an exact analogy with a rank 2 tensor contraction. That does not mean there isn't one, just that I couldn't find it. There are no convenient indexes to work with when using quaternions, so plucking out the right terms did not work. We need to create a Lorentz invariant scalar. I recalled reading in Jackson's "Classical Electrodynamics" that $E^2 - B^2$ is Lorentz invariant. To use a little high school algebra, we know: $$E^2 - B^2 = (E + B)(E - B)$$ There are nice, compact quaternion expressions for -g + E + B and -g + E - B. When will g be zero? If the system has only massless particles. Using that assumption, I am able to calculate $E^2 - B^2$ using quaternions. The next step is to apply the Euler-Lagrange equation to calculate the field equations from the Lagrangian. I suspect the majority of readers have not done this sort of calculation. It looks scary, but the details are quite simple. I am enough decades away from my calculus classes that I can recall only a few derivatives, simple ones like $\frac{d x y}{d x} = y$ and $\frac{1}{2} \frac{d x^2}{d x} = x$. These are the only two derivatives that are required to get to the Maxwell and GEM field equations. What is needed is to have the courage to write everything down, even though it looks complicated. Instead of the simple x, one uses the derivative of a function, $x \rightarrow \frac{\partial \phi}{\partial t}$, along with 15 other partial derivatives. When I do this in subsequent posts, I will point out the simplicity of what is going on in taking the Lagrangian and getting the field equations. Does anyone know if physicists make much of the invariant $E^2 - B^2$? As far as I can remember, it was mentioned once in Jackson, and he did not say much about it at all. Yet in my quaternion calculation, this invariant leads directly to the Maxwell equations. If true, the difference of the squares of these two fields should be on the center stage of EM theory. That has been a surprise benefit of my recent struggles. Doug
 Blog Entries: 3 Recognitions: Gold Member Hi Doug: Do you mean apart from the term $$F^{\mu\nu}F_{\mu\nu} = -2(E^2 - B^2)$$ appearing in the lagrangian of the EM field ? It seems to be center stage.
 Recognitions: Gold Member Hello Lut: Good to hear you are familiar with that result. Perhaps it is somewhere in Jackson, but I had not seen it before in my physics browsing. I gain confidence when I see other people have done something I tried to do before. It indicates I am barking up the right tree. I will go through all the details because that is my idea of a fun technical time. It also sets the stage for variations needed to see where GEM is going. Doug

 Quote by sweetser Hello Lawrence: I noticed no references to the Hamilton or even representation of quaternions in your replies. That is where the foundation is. We do not have to feel bad if we fail to communicate. You have demonstrated an impressive knowledge of both the history and current efforts to do productive research using the power tools of differential geometry. I respect that. New math is for the next generation. I have no expectation that I will ever be comfortable with it all, but that is the way research goes. Doug
I didn't bring up quaternions because they didn't seem relevant.

In what you wrote here it looks a bit like standard Kaluza Klein theory. I am not sure if this differs significantly from rather standard fair. At lest with this the approach appears within the standard construction of gauge theories.

Lawrence B. Crowell

 Recognitions: Gold Member Hello: In this post, I will show in detail how to generate the Maxwell equations using quaternions exclusively. This is achieved by getting to exactly the same terms that appear in the standard tensor approach for the electromagnetic field strength contraction. After the contraction, every step is the same. I will do them anyway, since it may be instructive to the viewers of this thread. Simple quaternion expressions generate combinations of three fields, the standard definition of E and B, along with one I call g: $$g = (\frac{\partial \phi}{\partial t}- \frac{\partial Ax}{\partial x}- \frac{\partial Ay}{\partial y}- \frac{\partial Az}{\partial z})$$ Notice how this one has all the parts of a typical gauge field in standard EM. There are two simple ways to generate g, B, and E using quaternions: $$1. \nabla A = (\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})(\phi, Ax, Ay, Az)$$ $$(\frac{\partial \phi }{\partial t}-c \frac{\partial Ax}{\partial x}-c \frac{\partial Ay}{\partial y}-c \frac{\partial Az}{\partial z},\frac{\partial Ax}{\partial t}+c \frac{\partial \phi }{\partial x}+c \frac{\partial Ay}{\partial z}-c \frac{\partial Az}{\partial y},\frac{\partial Ay}{\partial t}+c \frac{\partial \phi }{\partial y}+c \frac{\partial Az}{\partial x}-c \frac{\partial Ax}{\partial z},\frac{\partial Az}{\partial t}+c \frac{\partial \phi }{\partial z}+c \frac{\partial Ax}{\partial y}-c \frac{\partial Ay}{\partial x})$$ $$= g - E + B$$ $$2. - A \nabla = -(\phi, Ax, Ay, Az)(\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})$$ $$(-\frac{\partial \phi }{\partial t}+c \frac{\partial Ax}{\partial x}+c \frac{\partial Ay}{\partial y}+c \frac{\partial Az}{\partial z},-\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x}+c \frac{\partial Ay}{\partial z}-c \frac{\partial Az}{\partial y},-\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y}+c \frac{\partial Az}{\partial x}-c \frac{\partial Ax}{\partial z},-\frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z}+c \frac{\partial Ax}{\partial y}-c \frac{\partial Ay}{\partial x})$$ $$= -g + E + B$$ By changing the order of the derivative with the potential, we only flip the sign of the curl. Let's get rid of that pesky g by subtracting away the conjugate: $$1. \nabla A - (\nabla A)^* = (0, -E + B)$$ $$2. - A \nabla - (A \nabla)^* = (0, E + B)$$ Take the product of these two: $$(\nabla A - (\nabla A)^*)(- A \nabla - (A \nabla)^*) = (0, -E + B)(0, E + B) = (E^2 - B^2, 2 ExB)$$ It is the scalar that will be used to get the Maxwell field equations. It is the same one that appears in standard EM field theory as Lut pointed out in post 435 if I can get my signs right. The 3-vector is the Poynting vector which plays a role in energy conservation laws. Let's work only with the scalar, and toss in a factor of minus a half, and include a charge coupling term. Write out all of the components: $$-\rho \phi + Jx Ax + Jy Ay + Jz Az$$ $$+\frac{1}{2} c^2 \left(\frac{\partial Ax}{\partial z}\right)^2+\frac{1}{2} c^2 \left(\frac{\partial Ay}{\partial z}\right)^2-\frac{1}{2} c^2 \left(\frac{\partial \phi }{\partial z}\right)^2+\frac{1}{2} c^2 \left(\frac{\partial Ax}{\partial y}\right)^2$$ $$-c^2 \frac{\partial Ay}{\partial z} \frac{\partial Az}{\partial y}+\frac{1}{2} c^2 \left(\frac{\partial Az}{\partial y}\right)^2-\frac{1}{2} c^2 \left(\frac{\partial \phi }{\partial y}\right)^2-c^2 \frac{\partial Ax}{\partial y} \frac{\partial Ay}{\partial x}$$ $$+\frac{1}{2} c^2 \left(\frac{\partial Ay}{\partial x}\right)^2-c^2 \frac{\partial Ax}{\partial z} \frac{\partial Az}{\partial x}+\frac{1}{2} c^2 \left(\frac{\partial Az}{\partial x}\right)^2-\frac{1}{2} c^2 \left(\frac{\partial \phi }{\partial x}\right)^2-c \frac{\partial \phi }{\partial x} \frac{\partial Ax}{\partial t}$$ $$-\frac{1}{2} \left(\frac{\partial Ax}{\partial t}\right)^2-c \frac{\partial \phi }{\partial y} \frac{\partial Ay}{\partial t}-\frac{1}{2} \left(\frac{\partial Ay}{\partial t}\right)^2-c \frac{\partial \phi }{\partial z} \frac{\partial \text{Az}}{\partial t}$$ $$-\frac{1}{2} \left(\frac{\partial \text{Az}}{\partial t}\right)^2$$ From here on out, this is the standard way to derive the Maxwell equations. This is the part that looks scary, but its bark is worse that its bite. The goal is to take the derivative of the above Lagrangian with respect to the four potentials, $\phi, Ax, Jy, Az$, and all 16 derivatives of the four potentials. The first step, taking the derivative with respect to the 4-potentials, gives us back the current, $-\rho, Jx, Jy$, and $Jz$. The derivatives of the Lagrangian with respect to the changing potential is more complicated, but once the rule is learned, it is simple to apply, over and over again. Here are the first 4 of 16: $$\frac{\partial }{\partial x_{\mu }}\left(\frac{\partial \mathcal{L}}{\frac{\partial \phi }{\partial x_{\mu }}}\right)$$ The mu brings in the derivative of phi with respect to t, x, y, z. The derivative out in front will be generating second order derivatives. Let's do this for one of these, say, d phi/dx: $$\frac{\partial }{\partial x}\left(\frac{\partial \mathcal{L}}{\frac{\partial \phi }{\partial x}}\right)=-c \frac{\partial ^2\phi }{\partial x^2}-\frac{\partial ^2 Ax}{\partial t\partial x} = \frac{\partial E}{\partial x}$$ Repeat this for y and z, and you will have Gauss' law, the divergence of E equals rho, $\rho = \nabla . E$. Now we need to take the derivative of the Lagrangian with respect to Ax, running through t, x, y, and z. $$c^2 \frac{\partial ^2 Ax}{\partial z^2}+c^2 \frac{\partial ^2 Ax}{\partial y^2}-c^2 \frac{\partial ^2 Az}{\partial x\partial z}-c^2 \frac{\partial ^2 Ay}{\partial x\partial y}-c \frac{\partial ^2\phi }{\partial t\partial x}-\frac{\partial ^2 Ax}{\partial t^2}$$ The last two terms are a time derivative of Ex. If you were to calculate the curl of B, then you would recognize the first four terms. The way I spot it, is the two "pure" second order derivatives, which don't have an x, are positive, meaning this is minus the curl of Bx. We also have a +J, so tossing all the terms on the other side generates: $$J = \nabla X B - \frac{\partial E}{\partial t}$$ This is Ampere's law. What has been done In this post I showed how to get to the standard EM Lagrangian using the first term of a particular quaternion product. One interesting subtle issue is that the field g was explicitly removed, and so the proposal is gauge invariant - no matter the choice for the terms in g, it does not change a thing because all terms in the g field were subtracted away. In the standard approach, we observe the field equations are gauge invariant. With quaternions, we see the step where the field disappears. There is no difference in the end result, but it is fun to think about. What will be done In the coming posts, I will repeat the exercise done in this post for the symmetric fields, e and b. The resulting equations are gauge invariant in exactly the same sense as the Maxwell equations were gauge invariant in this exercise, because g will be subtracted away. In the third installment, I will not subtract g from either E+B or e+b. The field g goes in, but I hope to show that the g field does not come out. The field equations are the same no matter what g one chooses. It is quite remarkable that g politely disappears from the stage. Doug

 Quote by sweetser Hello: Simple quaternion expressions generate combinations of three fields, the standard definition of E and B, along with one I call g: $$g = (\frac{\partial \phi}{\partial t},- \frac{\partial Ax}{\partial x},- \frac{\partial Ay}{\partial y},- \frac{\partial Az}{\partial z})$$ Notice how this one has all the parts of a typical gauge field in standard EM. There are two simple ways to generate g, B, and E using quaternions: $$1. \nabla A = (\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})(\phi, Ax, Ay, Az)$$ $$(\frac{\partial \phi }{\partial t}-c \frac{\partial Ax}{\partial x}-c \frac{\partial Ay}{\partial y}-c \frac{\partial Az}{\partial z},\frac{\partial Ax}{\partial t}+c \frac{\partial \phi }{\partial x}+c \frac{\partial Ay}{\partial z}-c \frac{\partial Az}{\partial y},\frac{\partial Ay}{\partial t}+c \frac{\partial \phi }{\partial y}+c \frac{\partial Az}{\partial x}-c \frac{\partial Ax}{\partial z},\frac{\partial Az}{\partial t}+c \frac{\partial \phi }{\partial z}+c \frac{\partial Ax}{\partial y}-c \frac{\partial Ay}{\partial x})$$ $$= g - E + B$$ Doug
I am a bit unsure about the nature of g. You have a "nabla" operator on the vector potential, or four potential, which gives a vector quantity. Maybe this nabla is supposed to be a "d" operator?

If you want to work in quaternions things should be in a Clifford basis on Cl(3,1) with the generators

$$\gamma_1~=~\sigma_2\otimes\sigma_1,~\gamma_2~=~\sigma_2\otimes\sigma_2, ~\gamma_3~=~\sigma_2\otimes\sigma_3,~\gamma_4~=~i\sigma_1\otimes{\bf 1}$$

The electromagnetic potential are then on the Clifford frame if

$${\underline A}~=~(e_a^{\mu})\gamma_\mu A^a$$

for $e_a^{\mu}$ a tetrad or vierbein. The Maxwell equations will then arise from $d\wedge{\underline A}~=~{\underline F}$ and the gauge condition will come from $\nabla_a A^a~=~C$, for C a constant, set to zero in the Lorentz gauge. This can come from setting $(e_a^{\mu})\gamma_\mu~=~D_a$ which is orthogonal to the vector potential A_a on the bundle section with $D_aA^a~=~0$ --- again for the Lorentz gauge.

Lawrence B. Crowell

 Recognitions: Gold Member Hello Lawrence: I wrote out g incorrectly. I have corrected post #438. Nabla is a quaternion, the 4-potential is a quaternion, the Nabla acting on the 4-potential makes a quaternion, and the three fields of g, E, and B together form a quaternion. The g is the first term of the quaternion 4-derivative of the 4-potential. Doug
 Blog Entries: 3 Recognitions: Gold Member Doug, Very nice. I look forward to the next installment. Could you number your equations, please ? Also using ExB for $$E\times B$$ caused me some confusion until I repeated the calculation. Lut
 Recognitions: Gold Member Hello: In this post, I will apply the techniques used to generate the Maxwell equations with quaternions to the symmetric analogs of the E and B fields. The three symmetric fields involved in this analysis are: $$g = (\frac{\partial \phi}{\partial t}- \frac{\partial Ax}{\partial x}- \frac{\partial Ay}{\partial y}- \frac{\partial Az}{\partial z}) \quad eq 1$$ $$e = (\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x},\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y}, \frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z}) \quad eq 2$$ $$b = (c \left(-\frac{\partial Ay}{\partial z}-\frac{\partial Az}{\partial y}\right), c \left(-\frac{\partial Az}{\partial x}-\frac{\partial Ax}{\partial z}\right), c \left(-\frac{\partial Ax}{\partial y}-\frac{\partial Ay}{\partial x}\right) ) \quad eq 3$$ Although I first spotted these fields by peering into a symmetric rank 2 tensor, for this exercise I will use quaternions. There are two reasons to do so. First, we see exactly why the field equations are invariant under a gauge transformation - such a field will be subtracted away. The second point is more subtle. To do quantum field theory, one needs to take the field equations and invert them to make a propagator. If we consistently use a division algebra, then we necessarily will be able to do that step. With gauge theories such as Maxwell and general relativity, an arbitrary gauge must be chosen before the equation can be inverted. I am trying to get a theory that one is free to choose the gauge, yet remains invertible to get the propagator. The Hamilton representation of quaternion multiplication will not suffice for generating the symmetric b field, where all the terms that go into a curl have a minus sign. For this reason, I have introduced the Even representation of quaternion multiplication which excludes the Eigen vectors and Eigen values in order to be a division algebra. This representation will not be a Clifford algebra, since $e_0^2 = e_1^2 = e_2^2 = e_3^2 = +1$. Two simple expressions can generate g, e, and b: $$\nabla A2^* = (\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})(\phi, Ax, Ay, Az)$$ $$=(\frac{\partial \phi }{\partial t}-c \frac{\partial Ax}{\partial x}-c \frac{\partial Ay}{\partial y}-c \frac{\partial Az}{\partial z},$$ $$-\frac{\partial Ax}{\partial t}+c \frac{\partial \phi }{\partial x}-c\frac{\partial Ay}{\partial z}-c \frac{\partial Az}{\partial y},$$ $$-\frac{\partial Ay}{\partial t}+c \frac{\partial \phi }{\partial y}-c \frac{\partial Az}{\partial x}-c \frac{\partial Ax}{\partial z},$$ $$-\frac{\partial Az}{\partial t}+c \frac{\partial \phi }{\partial z}-c \frac{\partial Ax}{\partial y}-c \frac{\partial Ay}{\partial x})$$ $$= g - e + b \quad eq 4$$ $$A2 \nabla^* = (\phi, Ax, Ay, Az)(\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})$$ $$=(\frac{\partial \phi }{\partial t}-c \frac{\partial Ax}{\partial x}-c \frac{\partial Ay}{\partial y}-c \frac{\partial Az}{\partial z},$$ $$\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x}-c \frac{\partial Ay}{\partial z}-c \frac{\partial Az}{\partial y},$$ $$\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y}-c \frac{\partial Az}{\partial x}-c \frac{\partial Ax}{\partial z},$$ $$\frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z}-c \frac{\partial Ax}{\partial y}-c \frac{\partial Ay}{\partial x})$$ $$= g + e + b \quad eq 5$$ To demonstrate this is a subtle proposal, I cut and paste this calculation from my derivation of the Maxwell equations, and made the appropriate sign and letter changes. There are not many, and hopefully I did it right. changing who gets conjugated flips the sign of e, but not b. Let's get rid of that pesky g by subtracting away the conjugate: $$\nabla A2^* - (\nabla A2^*)^* = (0, -e + b) \quad eq 6$$ $$- A2 \nabla^* - (A2 \nabla^*)^* = (0, e + b) \quad eq 7$$ Take the product of these two: $$(\nabla A2^* - (\nabla A2^*)^*)(- A2 \nabla^* - (A2 \nabla^*)^*) = (0, -e + b)(0, e + b) = (-e^2 + b^2, -e.X2.e + b.X2.b) \quad eq 8$$ It is the scalar that will be used to get the symmetric field Maxwell equations. No Poynting vector this time, I don't know what that means. Write out all the components, the same ones as before, but with a different collection of signs: $$-\rho \phi + Jx Ax + Jy Ay + Jz Az$$ $$+\frac{1}{2} c^2 \left(-\frac{\partial Ax}{\partial z}\right)^2-\frac{1}{2} c^2 \left(\frac{\partial Ay}{\partial z}\right)^2+\frac{1}{2} c^2 \left(\frac{\partial \phi }{\partial z}\right)^2-\frac{1}{2} c^2 \left(\frac{\partial Ax}{\partial y}\right)^2$$ $$-c^2 \frac{\partial Ay}{\partial z} \frac{\partial Az}{\partial y}-\frac{1}{2} c^2 \left(\frac{\partial Az}{\partial y}\right)^2+\frac{1}{2} c^2 \left(\frac{\partial \phi }{\partial y}\right)^2-c^2 \frac{\partial Ax}{\partial y} \frac{\partial Ay}{\partial x}$$ $$-\frac{1}{2} c^2 \left(\frac{\partial Ay}{\partial x}\right)^2-c^2 \frac{\partial Ax}{\partial z} \frac{\partial Az}{\partial x}-\frac{1}{2} c^2 \left(\frac{\partial Az}{\partial x}\right)^2+\frac{1}{2} c^2 \left(\frac{\partial \phi }{\partial x}\right)^2-c \frac{\partial \phi }{\partial x} \frac{\partial Ax}{\partial t}$$ $$+\frac{1}{2} \left(\frac{\partial Ax}{\partial t}\right)^2+c \frac{\partial \phi }{\partial y} \frac{\partial Ay}{\partial t}+\frac{1}{2} \left(\frac{\partial Ay}{\partial t}\right)^2-c \frac{\partial \phi }{\partial z} \frac{\partial \text{Az}}{\partial t}$$ $$+\frac{1}{2} \left(\frac{\partial \text{Az}}{\partial t}\right)^2 \quad eq 9$$ Fear not, the derivative of this Lagrangian with respect to the potential again generates the current density. Now take the derivative with respect to the for derivatives of phi: $$-\rho +\left(c^2 \frac{\partial ^2\phi }{\partial x^2}+c^2 \frac{\partial ^2\phi }{\partial y^2}+c^2 \frac{\partial ^2\phi }{\partial z^2}-c\frac{\partial ^2 Ax}{\partial t\partial x}-c\frac{\partial ^2 Ay}{\partial t\partial y}-c\frac{\partial ^2 Az}{\partial t\partial z}\right)==0 \quad eq 10$$ Plug in the definition of e and rearrange: $$-\nabla . e = \rho \quad eq 11$$ This looks similar to Gauss' law, but there is an important difference. In the static case, the time derivatives of A make no contribution, and we have: $$\nabla^2 \phi = \rho \quad eq 12$$ For Gauss' law of EM, under the same static conditions: $$-\nabla^2 \phi = \rho \quad eq 13$$ The reason like charges repel in EM has to do with the above minus sign, while for gravity to attract, the Laplacian of phi must have the same sign as the current density. Take the derivative with respect to the 4 derivatives of Ax: $$Jx-c^2 \frac{\partial ^2 Ax}{\partial z^2}-c^2 \frac{\partial ^2 Ax}{\partial y^2}-c^2 \frac{\partial ^2 Az}{\partial x\partial z}-c^2 \frac{\partial ^2 Ay}{\partial x\partial y}-c \frac{\partial ^2\phi }{\partial t\partial x}+\frac{\partial ^2\text{Ax}}{\partial t^2}==0 \quad eq 14$$ The time derivative of e is the last two terms. The symmetric curl is negative, but two symmetric curls is positive, so this is minus the symmetric curl of b. Plug in and rearrange: $$\nabla . X2 . b - \frac{\partial e}{\partial t} = J \quad eq 15$$ Looks similar to Ampere's law. The symmetric Gauss and Ampere's law should not be shocking, since the symmetric fields e and b are built from the same terms as E and B. There are important differences, such as the symmetric Gauss' law has like charges that attract as happens for gravity, whereas the EM Gauss' law has like charges repel. The two Ampere's laws have two different sorts of curls (since curl is one of the most confusing concepts I have come across, I will not venture about the meaning of two different curls). Again we can choose whatever we want for $g = (\frac{\partial \phi}{\partial t}- \frac{\partial Ax}{\partial x}- \frac{\partial Ay}{\partial y}- \frac{\partial Az}{\partial z})$ will not change these equations at all. Some of my readers will say that is a good thing :-) Doug