- #456
sweetser
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Generating the gauge-invariant GEM field equations
Hello:
In this post, I will apply similar, but not quite identical, approaches to generating the field equations for the fields E, B, e, and b.
If one hopes to model particles that travel at the speed of light, that requires that the field theory be invariant under a gauge transformation. This is one of those constraints on finding a solution I have heard, I have accepted, and I don't understand as well I should.
In deriving the Maxwell field equations using quaternion operators, the gauge invariance was achieved by noting all the derivatives that make up a gauge are in the first term of [itex]-A \nabla[/itex]. The first term was subtracted away using the time honored quaternion trick subtracting the conjugate, q - q* = vector(q). The identical method was applied to the symmetric fields e and b. If were were to just add these two results together, there would be no link between the two sets of equations.
This time we will not use the trick of subtracting the conjugate, yet no terms with g appear in the final field equations because the g2 terms in one cancel the g2 terms of the other in the scalar. There is a g in the 3-vector, but that is not used for generating the field equations.
Start by taking the derivatives of 4-potentials in two ways. For quaternions written in the Hamilton basis, change the order of the differential operator with the potential, which flips the sign of B. For quaternions written in the Even basis, change which term gets conjugated. Calculate:
[tex]\frac{1}{2}(-(A \nabla)(\nabla A) ~+~ (\nabla^* A2)(\nabla A2^*))[/tex]
[tex]=((-\frac{\partial \phi}{\partial t} ~+~ c \nabla . A, -\frac{\partial Ax}{\partial t} ~-~ \nabla \phi ~+~ \nabla X A)(\frac{\partial \phi}{\partial t} ~-~ c \nabla . A , \frac{\partial Ax}{\partial t} ~+~ \nabla \phi ~+~ \nabla X A) [/tex]
[tex]~+~ (\frac{\partial \phi}{\partial t} - c \nabla . A,\frac{\partial Ax}{\partial t} ~-~ \nabla \phi ~-~ \nabla .X2. A)(\frac{\partial \phi}{\partial t} ~-~ c \nabla . A,-\frac{\partial Ax}{\partial t} ~+~ \nabla \phi ~-~ \nabla .X2. A))[/tex]
[tex]= ((-g, E ~+~ B)(g, -E ~+~ B) ~+~ (g, e ~+~ b)(g, -e ~+~ b))[/tex]
[tex](-g^2 ~+~ E^2 ~-~ B^2 ~+~ g^2 ~-~ e^2 ~+~ b^2, 2 E X B ~-~ e .X2. e ~+~ b .X2. b ~+~ 2 gE ~+~ 2 gb) \quad eq 1[/tex]
The g field is not in the scalar due to a cancelation, but is in the 3-vector. The field equations are generated from the scalar, not the 3-vector, so any choice for the gauge g will not effect the field equations I am about to derive.
The current coupling term is complicated by the need to have spin 1 and spin 2 symmetry in the phase. This was worked out earlier in the thread, and here is the solution:
[tex]-\frac{1}{4}(J A ~+~ (J A)^* ~-~ J^{*1} A^{*2 *3} ~-~ (J^{*1} A^{*2 *3})^*) = ~-\rho \phi ~+~ Jx Ax ~+~ Jy Ay ~+~ Jz Az \quad eq 2[/tex]
Write out the Lagrangian by its components, including the current coupling terms:
[tex]\mathcal{L}_{BEbe} ~=~ -c^2 \frac{\partial \Ay}{\partial z} \frac{\partial Az}{\partial y} ~-~ c^2 \frac{\partial Ax}{\partial y} \frac{\partial Ay}{\partial x} ~-~ c^2 \frac{\partial Ax}{\partial z} \frac{\partial Az}{\partial x}[/tex]
[tex]-c \frac{\partial \phi }{\partial x} \frac{\partial Ax}{\partial t} ~-~ c \frac{\partial \phi }{\partial y} \frac{\partial Ay}{\partial t} ~-~ c \frac{\partial \phi }{\partial z} \frac{\partial Az}{\partial t}[/tex]
[tex]-\rho \phi ~+~ Jx Ax ~+~ Jy Ay ~+~ Jz Az \quad eq 3[/tex]
There are fewer terms than in either the Maxwell Lagrangian or the symmetric field Lagrangian because terms between the two cancel. The fields in the field equations will need to do the same. Calculate the first field equation by taking the derivative of [itex]\mathcal{L}_{BEbe}[/itex] with respect to the 4 derivatives of phi.
[tex]\frac{\partial}{\partial x^{\mu}}\left( \frac{\partial \mathcal{L}_{BEbe}}{ \left( \frac{\partial \phi}{\partial x^{\mu} \right)}} \right) ~=~ -c \frac{\partial ^2 Az}{\partial t\partial z} ~-~ c\frac{\partial ^2 Ay}{\partial t\partial y} ~-~ c\frac{\partial ^2 Ax}{\partial t\partial x}\right) ~-~ \rho[/tex]
[tex]=~ \frac{1}{2}\nabla . (E ~-~ e) ~-~ \rho ~=~ 0 \quad eq 4[/tex]
Nice, the E and e terms work together to isolate the A derivatives. And yet, you can spot Gauss' law for EM where like charges repel. There is a Gauss-like law for like charges that attract. Repeat for the derivative with respect to Ax:
[tex]\frac{\partial}{\partial x^{\mu}} \left( \frac{\partial \mathcal{L}_{BEbe}}{ \left( \frac{\partial Ax}{\partial x^{\mu} \right)}} \right) ~=~ -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}\right) ~+~ Jx ~=~ \frac{1}{2}(-\nabla X B ~-~ \nabla .X2. b ~+~ \frac{\partial(Ex ~+~ ex)}{\partial t}) ~+~ Jx ~=~ 0 \quad eq 5[/tex]
You should be able to spot Ampere's law.
What has been done
There is now a formulation of the GEM proposal that uses quaternions exclusively. The standard quaternion algebra inherited from the nineteenth century needed to be extended in two ways. First the idea of a conjugate (or anti-involutive automorphism in fancy jargon, or thingie that flips the sign of all but one part in simple words) had to be expanded to *1, *2, and *3 to get the phase symmetry right for the current coupling term.
The second advance in quaternion algebra needed is the Even representation of quaternion multiplication. Here the eigenvectors of the representation must be excluded to make the representation an algebraic field. I have consistently said I am more comfortable when I found out someone else has done this before. The multiplication table is known as the Klein four-group. I will have to see if others have noticed what happens when the eigenvectors are excluded.
With these two innovations, the field equations for E, B, e and b have been generated. These field equations are gauge invariant because g may be whatever one chooses since it is canceled out in the process. This is vital since the graviton and proton both travel at the speed of light.
What needs to be done
I need to develop a non-gauge invariant set of field equations for massive particles, where the gauge symmetry is broken by the mass charge. I have something technical to keep me off the streets.
Doug
Hello:
In this post, I will apply similar, but not quite identical, approaches to generating the field equations for the fields E, B, e, and b.
If one hopes to model particles that travel at the speed of light, that requires that the field theory be invariant under a gauge transformation. This is one of those constraints on finding a solution I have heard, I have accepted, and I don't understand as well I should.
In deriving the Maxwell field equations using quaternion operators, the gauge invariance was achieved by noting all the derivatives that make up a gauge are in the first term of [itex]-A \nabla[/itex]. The first term was subtracted away using the time honored quaternion trick subtracting the conjugate, q - q* = vector(q). The identical method was applied to the symmetric fields e and b. If were were to just add these two results together, there would be no link between the two sets of equations.
This time we will not use the trick of subtracting the conjugate, yet no terms with g appear in the final field equations because the g2 terms in one cancel the g2 terms of the other in the scalar. There is a g in the 3-vector, but that is not used for generating the field equations.
Start by taking the derivatives of 4-potentials in two ways. For quaternions written in the Hamilton basis, change the order of the differential operator with the potential, which flips the sign of B. For quaternions written in the Even basis, change which term gets conjugated. Calculate:
[tex]\frac{1}{2}(-(A \nabla)(\nabla A) ~+~ (\nabla^* A2)(\nabla A2^*))[/tex]
[tex]=((-\frac{\partial \phi}{\partial t} ~+~ c \nabla . A, -\frac{\partial Ax}{\partial t} ~-~ \nabla \phi ~+~ \nabla X A)(\frac{\partial \phi}{\partial t} ~-~ c \nabla . A , \frac{\partial Ax}{\partial t} ~+~ \nabla \phi ~+~ \nabla X A) [/tex]
[tex]~+~ (\frac{\partial \phi}{\partial t} - c \nabla . A,\frac{\partial Ax}{\partial t} ~-~ \nabla \phi ~-~ \nabla .X2. A)(\frac{\partial \phi}{\partial t} ~-~ c \nabla . A,-\frac{\partial Ax}{\partial t} ~+~ \nabla \phi ~-~ \nabla .X2. A))[/tex]
[tex]= ((-g, E ~+~ B)(g, -E ~+~ B) ~+~ (g, e ~+~ b)(g, -e ~+~ b))[/tex]
[tex](-g^2 ~+~ E^2 ~-~ B^2 ~+~ g^2 ~-~ e^2 ~+~ b^2, 2 E X B ~-~ e .X2. e ~+~ b .X2. b ~+~ 2 gE ~+~ 2 gb) \quad eq 1[/tex]
The g field is not in the scalar due to a cancelation, but is in the 3-vector. The field equations are generated from the scalar, not the 3-vector, so any choice for the gauge g will not effect the field equations I am about to derive.
The current coupling term is complicated by the need to have spin 1 and spin 2 symmetry in the phase. This was worked out earlier in the thread, and here is the solution:
[tex]-\frac{1}{4}(J A ~+~ (J A)^* ~-~ J^{*1} A^{*2 *3} ~-~ (J^{*1} A^{*2 *3})^*) = ~-\rho \phi ~+~ Jx Ax ~+~ Jy Ay ~+~ Jz Az \quad eq 2[/tex]
Write out the Lagrangian by its components, including the current coupling terms:
[tex]\mathcal{L}_{BEbe} ~=~ -c^2 \frac{\partial \Ay}{\partial z} \frac{\partial Az}{\partial y} ~-~ c^2 \frac{\partial Ax}{\partial y} \frac{\partial Ay}{\partial x} ~-~ c^2 \frac{\partial Ax}{\partial z} \frac{\partial Az}{\partial x}[/tex]
[tex]-c \frac{\partial \phi }{\partial x} \frac{\partial Ax}{\partial t} ~-~ c \frac{\partial \phi }{\partial y} \frac{\partial Ay}{\partial t} ~-~ c \frac{\partial \phi }{\partial z} \frac{\partial Az}{\partial t}[/tex]
[tex]-\rho \phi ~+~ Jx Ax ~+~ Jy Ay ~+~ Jz Az \quad eq 3[/tex]
There are fewer terms than in either the Maxwell Lagrangian or the symmetric field Lagrangian because terms between the two cancel. The fields in the field equations will need to do the same. Calculate the first field equation by taking the derivative of [itex]\mathcal{L}_{BEbe}[/itex] with respect to the 4 derivatives of phi.
[tex]\frac{\partial}{\partial x^{\mu}}\left( \frac{\partial \mathcal{L}_{BEbe}}{ \left( \frac{\partial \phi}{\partial x^{\mu} \right)}} \right) ~=~ -c \frac{\partial ^2 Az}{\partial t\partial z} ~-~ c\frac{\partial ^2 Ay}{\partial t\partial y} ~-~ c\frac{\partial ^2 Ax}{\partial t\partial x}\right) ~-~ \rho[/tex]
[tex]=~ \frac{1}{2}\nabla . (E ~-~ e) ~-~ \rho ~=~ 0 \quad eq 4[/tex]
Nice, the E and e terms work together to isolate the A derivatives. And yet, you can spot Gauss' law for EM where like charges repel. There is a Gauss-like law for like charges that attract. Repeat for the derivative with respect to Ax:
[tex]\frac{\partial}{\partial x^{\mu}} \left( \frac{\partial \mathcal{L}_{BEbe}}{ \left( \frac{\partial Ax}{\partial x^{\mu} \right)}} \right) ~=~ -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}\right) ~+~ Jx ~=~ \frac{1}{2}(-\nabla X B ~-~ \nabla .X2. b ~+~ \frac{\partial(Ex ~+~ ex)}{\partial t}) ~+~ Jx ~=~ 0 \quad eq 5[/tex]
You should be able to spot Ampere's law.
What has been done
There is now a formulation of the GEM proposal that uses quaternions exclusively. The standard quaternion algebra inherited from the nineteenth century needed to be extended in two ways. First the idea of a conjugate (or anti-involutive automorphism in fancy jargon, or thingie that flips the sign of all but one part in simple words) had to be expanded to *1, *2, and *3 to get the phase symmetry right for the current coupling term.
The second advance in quaternion algebra needed is the Even representation of quaternion multiplication. Here the eigenvectors of the representation must be excluded to make the representation an algebraic field. I have consistently said I am more comfortable when I found out someone else has done this before. The multiplication table is known as the Klein four-group. I will have to see if others have noticed what happens when the eigenvectors are excluded.
With these two innovations, the field equations for E, B, e and b have been generated. These field equations are gauge invariant because g may be whatever one chooses since it is canceled out in the process. This is vital since the graviton and proton both travel at the speed of light.
What needs to be done
I need to develop a non-gauge invariant set of field equations for massive particles, where the gauge symmetry is broken by the mass charge. I have something technical to keep me off the streets.
Doug