- #551

sweetser

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**Covariant GEM field equations**

Hello:

<Preamble>

This post is very similar to the last two. This time I have few of the set of the last two, but unify gravity and EM in the process. Nice.

</Preamble>

In this post I will derive gravity and the Maxweill field equations using quaternion operators in a manifestly covariant notation.

The Hamilton representation will be used for the Maxwell field equations. The even representation of quaternions - where all signs are the same, and the Eivenvalues ore excluded so the quaternion can be inverted - will be used in the first phase of this derivation. What the even representation does is flip the relative signs of the E and B signs, so now the e filed has two signs, and the b field has one, like so:

[tex]E = -\nabla_0 A ~-~ \nabla_u \phi[/tex]

[tex]e = \nabla_0 A2 ~-~ \nabla_u \phi[/tex]

[tex]B_w = \nabla_u A_v ~-~\nabla_v A_u = \nabla \times A[/tex]

[tex]b_w = -\nabla_u A2_v ~-~\nabla_v A2_u = \nabla \Join A2 \quad eq ~0[/tex]

The fifth play is the term that makes up the gauge field:

[tex]g = \nabla_0 \phi ~-~ \nabla \cdot A[/tex]

None of these transform like tensors, but together they do from [itex]\nabla A[/itex].

Let's generate all 5 fields:

[tex]\nabla A = (\nabla_0 \phi ~-~ \nabla \cdot A2, \nabla_0 A2 ~+~ \nabla_u \phi ~+~ \nabla \times A) = (g, -E ~+~ B)[/tex]

[tex]\nabla^* A2 = (\nabla_0 \phi ~-~ \nabla \cdot A2, -\nabla_0 A2 ~+~ \nabla_u \phi ~-~ \nabla \Join A2) = (g, -e ~+~ b)\quad eq ~1[/tex]

The starting point for the derivation of the unified GEM equations is the Lagrangian which can be viewed as the difference between the scalars of E,b squared and B,e squared. This can be achieved by using the first set of quaternion operators used in the previous posts:

[tex]\frac{1}{8}((\nabla A)(A \nabla) ~-~(\nabla^* A2)(\nabla A2^*))[/tex]

[tex]=\frac{1}{8}(\nabla_0 \phi ~-~ \nabla \cdot A, \nabla_0 A ~+~ \nabla_u \phi ~+~ \nabla \times A)(\nabla_0 \phi ~-~ \nabla \cdot A, \nabla_0 A ~+~ \nabla_u \phi ~-~ \nabla \times A)[/tex]

[tex]-(\nabla_0 \phi ~-~ \nabla \cdot A, \nabla_0 A ~-~ \nabla_u \phi ~-~ \nabla \Join A)(\nabla_0 \phi ~-~ \nabla \cdot A, -\nabla_0 A ~+~ \nabla_u \phi ~-~ \nabla \Join A)[/tex]

[tex]=\frac{1}{2}(g, -E ~+~ B)(g, -E ~-~ B) ~-~ (g, e ~+~ b)(g, -e ~+~ b) = \frac{1}{2}(B^2 ~-~ E^2 ~-~ b^2 ~+~ e^2, 2 E \times B ~-~ b \Join b ~+~ e \Join e)\quad eq ~2[/tex]

Any expression derived from this one will be invariant under a scalar gauge transformation which uses the terms [itex]\nabla_0 \phi, \nabla \cdot A[/itex] because they have miraculously been cancelled!

Write out the Lagrange density - the scalar part of eq 2 - in terms of its components, including the vector coupling whose phase has spin 1 and spin 2 symmetry[/itex]:

[tex]\mathcal{L}_{EBeb} = -~(\nabla_3 A_2)(\nabla_2 A_3) ~-~ (\nabla_1 A_3)(\nabla_3 A_1) ~-~ (\nabla_1 A_2)(\nabla_2 A_1) ~-~ (\nabla_1 \phi)(\nabla_0 A_1) ~-~ (\nabla_2 \phi)(\nabla_0 A_2) ~-~ (\nabla_3 \phi)(\nabla_0 A_3)[/tex]

[tex]-\rho \phi ~+~ J_1 A_1 ~+~ J_2 A_2 ~+~ J_3 A_3 \quad eq ~3[/tex]

[sidebar: The assertion that the current coupling term has both spin 1 and spin 2 symmetry for all terms in the phase requires a small calculation. I decided to leave that as a problem. My guess is that until someone starts confirming these by hand, I'll be the only one to see that even these cool details work out.[/sidebar]

Apply the Euler-Lagrange equation to this Lagrangian, treating the 4-potential as the variable:

[tex]\nabla_{\mu}(\frac{\partial \mathcal{L}_{EBeb}}{\partial (\nabla_{\mu} \phi)}) = - \nabla_0 \nabla_1 A_1 ~-~ \nabla_0 \nabla_2 A_2 ~-~ \nabla_0 \nabla_3 A_3 ~-~ \rho[/tex]

[tex]= -\frac{1}{2}\nabla \cdot (E ~-~ e) - \rho = 0 \quad eq ~4[/tex]

[tex]\nabla_{\mu}(\frac{\partial \mathcal{L}_{EBeb}}{\partial (\nabla_{\mu} A_1)}) = -~ \nabla_1 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_2 ~-~ \nabla_0 \nabla_1 \phi ~+~ J_1[/tex]

[tex]= \frac{1}{2}(\nabla_0 (E_1 ~+~ e_1) - (\nabla \times B)_1 - (\nabla \Join b)_1) + J_1 = 0 \quad eq ~5[/tex]

[tex]\nabla_{\mu}(\frac{\partial \mathcal{L}_{EBeb}}{\partial (\nabla_{\mu} A_2)}) = - \nabla_2 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_1 ~-~ \nabla_0 \nabla_2 \phi ~+~ J_2[/tex]

[tex]= \frac{1}{2}(\nabla_0 (E_2 ~+~ e_2) - (\nabla \times B)_2 - (\nabla \Join b)_2) + J_2 = 0 \quad eq ~6[/tex]

[tex]\nabla_{\mu}(\frac{\partial \mathcal{L}_{EBeb}}{\partial (\nabla_{\mu} A_3)}) = - \nabla_2 \nabla_3 A_2 ~-~ \nabla_1 \nabla_3 A_1 ~-~ \nabla_0 \nabla_3 \phi ~+~ J_3[/tex]

[tex]= \frac{1}{2}(\nabla_0 (E_3 ~+~ e_3) - (\nabla \times B)_3 - (\nabla \Join b)_3) + J_3 = 0 \quad eq ~7[/tex]

This work can be summarized with the GEM gravity source equations:

[tex]\frac{1}{2} \nabla \cdot (E ~-~ e) = \rho \quad eq ~8[/tex]

[tex]\frac{1}{2}(\nabla \times B ~+~ \nabla \Join b - \nabla_0 (E ~+~ e)) = J \quad eq ~9[/tex]

This has been a manifestly covariant derivation. This derivation remains the same in flat or curved spacetime, in Cartesian or spherical coordinates.

I have spent the day trying get all the signs and factors right. If there are any questions, send me a private note, and I will recheck it. If needed, we can put up a new post.

Doug

Stills:

http://picasaweb.google.com/dougsweetser/MaxwellFieldEquations [Broken]

The talk:

(this derivation was not part of the talk)

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