GEM Fields & Currents

Math Tools

To write these terms compactly using only quaternions, two representations of quaternions are needed.  The first is the standard Hamilton quaternion where his choice of the represtation leads to non-commuting multiplication.

q[t_, x_, y_, z_] :=  (({{t, -x, -y, -z}, {x, t, -z, y}, {y, z, t, -x}, {z, -y, x, t}}))

To express the symmetric quaternions, a new representation of the 4D quaternion algebra is required where multiplication commutes.  Even quaternion representation tools:

q2[t_, x_, y_, z_] :=(({{t, x, y, z}, {x, t, z, y}, {y, z,  t, x}, {z, y, x, t}}))
Conjugate2[q1_]:=q2[q1[[1,1]],-q1[[2,1]],-q1[[3,1]],-q1[[4,1]]]

Multiplication for the representation commutes:

(q2[a,b,c,d].q2[e,f,g,h]-q2[e,f,g,h].q2[a,b,c,d]).{1,0,0,0}

There is always an inverse so long as zero and any quaternion where no sum of two terms equals the other two.

Factor[Inverse[q2[t,x,y,z]].{1,0,0,0}]

If there are an odd number of minus signs, the excluded group is slightly different: any quaternion that is the sum of the other three must be excluded:

Factor[Inverse[q2[t,-x,-y,-z]].{1,0,0,0}]

The 4-crrunt and 4-potential for the Hamilton and Even representations will be used.

J=q[ρ[t,x,y,z],Jx[t,x,y,z],Jy[t,x,y,z],Jz[t,x,y,z]];
A=q[ϕ[t,x,y,z],Ax[t,x,y,z],Ay[t,x,y,z],Az[t,x,y,z]];
J2= q2[ρ[t,x,y,z],Jx[t,x,y,z],Jy[t,x,y,z],Jz[t,x,y,z]];
A2= q2[ϕ[t,x,y,z],Ax[t,x,y,z],Ay[t,x,y,z],Az[t,x,y,z]];

Tools to take derivatives of Hamilton quaternion representation.

dq[A_]:=q[D[A[[1,1]],t]-D[A[[2,1]],x]-D[A[[3,1]],y]-D[A[[4,1]],z],D[A[[2,1]],t]+c D[A[[1,1]],x]+c D[A[[3,1]],z]-c D[A[[4,1]],y],D[A[[3,1]],t]+c D[A[[1,1]],y]+c D[A[[4,1]],x]-c D[A[[2,1]],z],D[A[[4,1]],t]+c D[A[[1,1]],z]+c D[A[[2,1]],y]-c D[A[[3,1]],x]]

qd[A_]:=q[D[A[[1,1]],t]-D[A[[2,1]],x]-D[A[[3,1]],y]-D[A[[4,1]],z],D[A[[2,1]],t]+c D[A[[1,1]],x]-c D[A[[3,1]],z]+c D[A[[4,1]],y],D[A[[3,1]],t]+c D[A[[1,1]],y]-c D[A[[4,1]],x]+c D[A[[2,1]],z],D[A[[4,1]],t]+c D[A[[1,1]],z]-c D[A[[2,1]],y]+c D[A[[3,1]],x]]

Here are the tools to take derivatives of the Even representation.  Theses three derivatives differ by which term gets conjugated:
dq2 - none
dcq2 - the differential
dqc2 - the potential

dq2[A_]:= q2[D[A[[1,1]],t]+D[A[[2,1]],x]+D[A[[3,1]],y]+D[A[[4,1]],z],D[A[[2,1]],t]+c D[A[[1,1]],x]+c D[A[[3,1]],z]+c D[A[[4,1]],y],D[A[[3,1]],t]+c D[A[[1,1]],y]+c D[A[[4,1]],x]+c D[A[[2,1]],z],D[A[[4,1]],t]+c D[A[[1,1]],z]+c D[A[[2,1]],y]+c D[A[[3,1]],x]]

dcq2[A_]:= q2[D[A[[1,1]],t]-D[A[[2,1]],x]-D[A[[3,1]],y]-D[A[[4,1]],z],D[A[[2,1]],t]-c D[A[[1,1]],x]-c D[A[[3,1]],z]-c D[A[[4,1]],y],D[A[[3,1]],t]-c D[A[[1,1]],y]-c D[A[[4,1]],x]-c D[A[[2,1]],z],D[A[[4,1]],t]-c D[A[[1,1]],z]-c D[A[[2,1]],y]-c D[A[[3,1]],x]]

dqc2[A_]:= q2[D[A[[1,1]],t]-D[A[[2,1]],x]-D[A[[3,1]],y]-D[A[[4,1]],z],-D[A[[2,1]],t]+c D[A[[1,1]],x]-c D[A[[3,1]],z]-c D[A[[4,1]],y],-D[A[[3,1]],t]+c D[A[[1,1]],y]-c D[A[[4,1]],x]-c D[A[[2,1]],z],-D[A[[4,1]],t]+c D[A[[1,1]],z]-c D[A[[2,1]],y]-c D[A[[3,1]],x]]

By default, Mathematica does not write the derivatives of variables in a human readible way.  After consulting with their support department, I have figured out how to change its ways.  Unless you are interested, these Mathematica notation specific issues can be skipped.

Needs["Notation`"]

Currents and potentials.

Notation[ρ ⟸ ρ[t,x,y,z]]
Notation[Jx ⟸ Jx[t,x,y,z]]
Notation[Jy ⟸ Jy[t,x,y,z]]
Notation[Jz ⟸ Jz[t,x,y,z]]
Notation[ρ' ⟸ ρ'[t,x,y,z]]
Notation[Jx' ⟸ Jx'[t,x,y,z]]
Notation[Jy' ⟸ Jy'[t,x,y,z]]
Notation[Jz' ⟸ Jz'[t,x,y,z]]
Notation[ϕ ⟸ ϕ[t,x,y,z]]
Notation[Ax ⟸ Ax[t,x,y,z]]
Notation[Ay ⟸ Ay[t,x,y,z]]
Notation[Az ⟸ Az[t,x,y,z]]

First derivatives

Notation[((∂f_)/(∂t)) ⟸ Derivative[1, 0, 0, 0][f_][t, x, y, z]]
Notation[((∂f_)/(∂x)) ⟸ Derivative[0, 1, 0, 0][f_][t, x, y, z]]
Notation[((∂f_)/(∂y)) ⟸ Derivative[0, 0, 1, 0][f_][t, x, y, z]]
Notation[((∂f_)/(∂z)) ⟸ Derivative[0, 0, 0, 1][f_][t, x, y, z]]

nth derivatives

Notation[(((∂^(n_))f_)/(∂(t^(n_)))) ⟸ Derivative[n_,0,0, 0][f_][t, x, y, z]]
Notation[(((∂^(n_))f_)/(∂(x^(n_)))) ⟸ Derivative[0,n_,0,0][f_][t, x, y, z]]
Notation[(((∂^(n_))f_)/(∂(y^(n_)))) ⟸ Derivative[0,0,n_,0][f_][t, x, y, z]]
Notation[(((∂^(n_))f_)/(∂(z^(n_)))) ⟸ Derivative[0,0,0,n_][f_][t, x, y, z]]

Mixed derivatives.

Notation[(((∂^2)f_)/(∂t∂x)) ⟸ Derivative[1, 1, 0, 0][f_][t, x, y, z]]
Notation[(((∂^2)f_)/(∂t∂y)) ⟸ Derivative[1, 0, 1, 0][f_][t, x, y, z]]
Notation[(((∂^2)f_)/(∂t∂z)) ⟸ Derivative[1, 0, 0, 1][f_][t, x, y, z]]
Notation[(((∂^2)f_)/(∂x∂y)) ⟸ Derivative[0, 1, 1, 0][f_][t, x, y, z]]
Notation[(((∂^2)f_)/(∂x∂z)) ⟸ Derivative[0, 1, 0, 1][f_][t, x, y, z]]
Notation[(((∂^2)f_)/(∂y∂z)) ⟸ Derivative[0, 0, 1, 1][f_][t, x, y, z]]

Notation[(((∂^3)f_)/(∂(t^2)∂x)) ⟸ Derivative[2, 1, 0, 0][f_][t, x, y, z]]
Notation[(((∂^3)f_)/(∂(t^2)∂y)) ⟸ Derivative[2, 0, 1, 0][f_][t, x, y, z]]
Notation[(((∂^3)f_)/(∂(t^2)∂z)) ⟸ Derivative[2, 0, 0, 1][f_][t, x, y, z]]
Notation[(((∂^3)f_)/(∂(x^2)∂y)) ⟸ Derivative[0, 2, 1, 0][f_][t, x, y, z]]
Notation[(((∂^3)f_)/(∂(x^2)∂z)) ⟸ Derivative[0, 2, 0, 1][f_][t, x, y, z]]
Notation[(((∂^3)f_)/(∂(y^2)∂z)) ⟸ Derivative[0, 0, 2, 1][f_][t, x, y, z]]

Notation[(((∂^3)f_)/(∂t∂(x^2))) ⟸ Derivative[1, 2, 0, 0][f_][t, x, y, z]]
Notation[(((∂^3)f_)/(∂t∂(y^2))) ⟸ Derivative[1, 0, 2, 0][f_][t, x, y, z]]
Notation[(((∂^3)f_)/(∂t∂(z^2))) ⟸ Derivative[1, 0, 0, 2][f_][t, x, y, z]]
Notation[(((∂^3)f_)/(∂x∂(y^2))) ⟸ Derivative[0, 1, 2, 0][f_][t, x, y, z]]
Notation[(((∂^3)f_)/(∂x∂(z^2))) ⟸ Derivative[0, 1, 0, 2][f_][t, x, y, z]]
Notation[(((∂^3)f_)/(∂y∂(z^2))) ⟸ Derivative[0, 0, 1, 2][f_][t, x, y, z]]

Notation[(((∂^4)f_)/(∂(t^3)∂x)) ⟸ Derivative[3, 1, 0, 0][f_][t, x, y, z]]
Notation[(((∂^4)f_)/(∂(t^3)∂y)) ⟸ Derivative[3, 0, 1, 0][f_][t, x, y, z]]
Notation[(((∂^4)f_)/(∂(t^3)∂z)) ⟸ Derivative[3, 0, 0, 1][f_][t, x, y, z]]
Notation[(((∂^4)f_)/(∂(x^3)∂y)) ⟸ Derivative[0, 3, 1, 0][f_][t, x, y, z]]
Notation[(((∂^4)f_)/(∂(x^3)∂z)) ⟸ Derivative[0, 3, 0, 1][f_][t, x, y, z]]
Notation[(((∂^4)f_)/(∂(y^3)∂z)) ⟸ Derivative[0, 0, 3, 1][f_][t, x, y, z]]

Notation[(((∂^4)f_)/(∂t∂(x^3))) ⟸ Derivative[1, 3, 0, 0][f_][t, x, y, z]]
Notation[(((∂^4)f_)/(∂t∂(y^3))) ⟸ Derivative[1, 0, 3, 0][f_][t, x, y, z]]
Notation[(((∂^4)f_)/(∂t∂(z^3))) ⟸ Derivative[1, 0, 0, 3][f_][t, x, y, z]]
Notation[(((∂^4)f_)/(∂x∂(y^3))) ⟸ Derivative[0, 1, 3, 0][f_][t, x, y, z]]
Notation[(((∂^4)f_)/(∂x∂(z^3))) ⟸ Derivative[0, 1, 0, 3][f_][t, x, y, z]]
Notation[(((∂^4)f_)/(∂y∂(z^3))) ⟸ Derivative[0, 0, 1, 3][f_][t, x, y, z]]

Notation[(((∂^4)f_)/(∂(t^2)∂(x^2))) ⟸ Derivative[2, 2, 0, 0][f_][t, x, y, z]]
Notation[(((∂^4)f_)/(∂(t^2)∂(y^2))) ⟸ Derivative[2, 0, 2, 0][f_][t, x, y, z]]
Notation[(((∂^4)f_)/(∂(t^2)∂(z^2))) ⟸ Derivative[2, 0, 0, 2][f_][t, x, y, z]]
Notation[(((∂^4)f_)/(∂(x^2)∂(y^2))) ⟸ Derivative[0, 2, 2, 0][f_][t, x, y, z]]
Notation[(((∂^4)f_)/(∂(x^2)∂(z^2))) ⟸ Derivative[0, 2, 0, 2][f_][t, x, y, z]]
Notation[(((∂^4)f_)/(∂(y^2)∂(z^2))) ⟸ Derivative[0, 0, 2, 2][f_][t, x, y, z]]

These are all the tools needed to beguin the analysis.

Currents

In the tensor approach to EM, the current and potential are contracted. To look at the issue of spin, one usually take the Fourier transform of the potential.  We do this for both the Hamilton and Even representation of quaterrnions:

(-J.A).{1,0,0,0}/.{ϕ->ρ',Ax->Jx',Ay->Jy',Az->Jz'}
(-J2.Conjugate2[A2]).{1,0,0,0}/.{ϕ->ρ',Ax->Jx',Ay->Jy',Az->Jz'}

The scalars are identical as they must be to be identical to the Lorentz invariant contraction found in EM.  What requires closer study is the signs of terms in the phase.  There are pairs of terms which have the same sign.  if the J current points in the directin of the J' current, these will add together and only take π radians to return, a sign of spin 2 symmetry.  There are other pairs where the signs are different.  They will require 2π radians to return, a sign of spin 1 symmetry.  The Hamilton product has spin 1 for the cross terms, spin 2 for the ρ source terms.  The even product does the opposite: spin 2 for the cross terms and spin 1 for the ρ source terms.  Sounds like a complete story to me.

Look at the sum:

Simplify[(1/2)(-J.A-J2.Conjugate2[A2]).{1,0,0,0}]/.{ϕ->ρ',Ax->Jx',Ay->Jy',Az->Jz'}

Recall the phase does not play a direct roll in electromagnetic theory.  Instead it is used to confirm what particles particoipate in a current-current interaction.

Fields

There are 5 fields in the GEM poposal.  Two are familiar: the electric field E and the magnetic field B.  For gravity, there are two symmetric analogues to these fields I call e and b, which have the same differential terms but different signs.  The final field is composed of terms down the diagonal, which arise in both expressions.

The field strength tensor is a combination of E and B

-g+E+B=- A∇

Simplify[-qd[A]].{1,0,0,0}

This expression also has a scalar which I argue is zero for massive particles, but mass breaks gauge symmetry for all charged particles.

E=-(1/2)( ∇A+ A∇)

(Subscript[E, quat])=Simplify[((1/2)(-dq[A]-qd[A]))].{1,0,0,0}

B=(1/2)(∇A- A∇)

(Subscript[B, quat])=Simplify[(1/2)(dq[A]-qd[A])].{1,0,0,0}

The difference between the E and B field is the two ways to order the differential operator have the same sign for E, but different signs for B.

This is the symmetric field strength tensor, where the e field flips the sign of the time derivative of A and the symmetric b all have negative signs.

g+e+b=(SuperStar[∇])A2

Simplify[dcq2[A2].{1,0,0,0}]

Isolate the scalar:

g=(1/2)((SuperStar[∇])A2+(SuperStar[((SuperStar[∇])A2)]))

Simplify[(1/2)(dcq2[A2] + Conjugate2[dcq2[A2]]).{1,0,0,0}]

The signs of the scalar here are opposite the ones from the E field, so the could cancel.  The job for this scalar is to break EM gauge symmetry. If the cancellation is not perfect because like electrical charges repel while like mass charges attract, then this would be the source for mass.

The symmetric e and b fields arise from the difference of conjugating either the differential or the potential.

e=(1/2)((SuperStar[∇])A2-∇(SuperStar[A2]))

(Subscript[e2, quat])=Simplify[(1/2)(dcq2[A2]-dqc2[A2]).{1,0,0,0}]

b=(1/2)((SuperStar[∇])A2-(SuperStar[(∇(SuperStar[A2]))]))

(Subscript[b2, quat])=Simplify[(1/2)(dcq2[A2]-Conjugate2[dqc2[A2]]).{1,0,0,0}]

Sum the 5 fields

Simplify[(1/2)(-qd[A]+dcq2[A2])].{1,0,0,0}

No term lives in the scalar, and in the vector is the gradient of the scalar ϕ and one of the curl terms.