Derive the Maxwell and GEM field equations

24 Apr 2009

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Fri, 04/24/2009 - 16:51

dougsweetser

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Joined: 03/13/2009

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Write the Maxwell field strength tensor using quaternion operators only.

$\begin{align*}\frac{1}{4}(\nabla A - (\nabla A)^*)(A \nabla - (A \nabla)^*) &= (0, \nabla_0 A + \nabla_u \phi + \nabla \times A)(0,\nabla_0 A + \nabla_u \phi - \nabla \times A)\\ &= (0, -E + B)(0, -E - B)\\ &= (B^2 - E^2, 2 E \times B) \end{align*} \quad eq ~1$

Notice on the first line the zeros in the scalars. This is how with a quanternion formalism, the Lagrange density can be made gauge invariant. Choose whatever gauge you want, and it will be subtracted away. The scalar is Lorentz invariant. The 3-vector is known as the Poynting vector. The Lagrange density only uses the scalar. Write this out in terms of its components, using numbers for subscripts so this will be valid no matter what the choice of coordinates.

$\begin{align*}\mathcal{L}_{\textrm{EM}} &= -\frac{1}{2}(J A + (J A)^*) + \frac{1}{4} ((B^2 - E^2, 2 E \times B) + (B^2 - E^2, 2 E \times B)^*)\\ &= -\rho \phi + J_1 A_1 + J_2 A_2 + J_3 A_3 \\ & +\frac{1}{2}(-(\nabla_1 \phi)^2 - (\nabla_2 \phi)^2 - (\nabla_3 \phi)^2) - (\nabla_0 A_1)^2 - (\nabla_0 A_2)^2 - (\nabla_0 A_3)^2 \\ & + (\nabla_3 A_2)^2 + (\nabla_2 A_3)^2 + (\nabla_1 A_3)^2 + (\nabla_3 A_1)^2 + (\nabla_2 A_1)^2 + (\nabla_1 A_2)^2) \\ & - (\nabla_3 A_2)(\nabla_2 A_3) - (\nabla_1 A_3)(\nabla_3 A_1) - (\nabla_1 A_2)(\nabla_2 A_1) - (\nabla_1 \phi)(\nabla_0 A_1) - (\nabla_2 \phi)(\nabla_0 A_2) - (\nabla_3 \phi)(\nabla_0 A_3) \end{align*} \quad eq ~2$

Calculate the field equations using the Euler-Lagrange equations. This is much simpler than it appears. Instead of taking a derivative with respect to x, one uses J or on of the fields, like $\nabla \phi$ . The derivative of x^2 with respect to x is just 2x, the most complicated derivative here. Like a suduko puzzle, all that is required is keeping track of what goes where.

$\nabla_{\mu}} ( \frac{\partial \mathcal{L}_\textrm{EM}}{\partial(\nabla_{\mu} \phi)}) = - \nabla^2_1 \phi - \nabla^2_2 \phi - \nabla^2_3 \phi - \nabla_0 \nabla_1 A_1 - \nabla_0 \nabla_2 A_2 - \nabla_0 \nabla_3 A_3 = \nabla \cdot E = \rho \quad eq ~3$

$\nabla_{\mu}} ( \frac{\partial \mathcal{L}_\textrm{EM}}{\partial(\nabla_{\mu} A_1)}) = - \nabla^2_0 A_1 + \nabla^2_3 A_1 + \nabla^2_2 A_1 - \nabla_1 \nabla_3 A_3 - \nabla_1 \nabla_2 A_2 - \nabla_0 \nabla_1 \phi = \nabla_0 E_1 - (\nabla \times B)_1 = -J_1 \quad eq ~4$

$\nabla_{\mu}} ( \frac{\partial \mathcal{L}_\textrm{EM}}{\partial(\nabla_{\mu} A_2)}) = - \nabla^2_0 A_2 + \nabla^2_3 A_2 + \nabla^2_2 A_2 - \nabla_2 \nabla_3 A_3 - \nabla_1 \nabla_2 A_1 - \nabla_0 \nabla_2 \phi = \nabla_0 E_2 - (\nabla \times B)_2 = -J_2 \quad eq ~5$

$\nabla_{\mu}} ( \frac{\partial \mathcal{L}_\textrm{EM}}{\partial(\nabla_{\mu} A_3)}) = - \nabla^2_0 A_3 + \nabla^2_3 A_3 + \nabla^2_2 A_3 - \nabla_2 \nabla_3 A_2 - \nabla_1 \nabla_3 A_1 - \nabla_0 \nabla_3 \phi = \nabla_0 E_3 - (\nabla \times B)_3 = -J_3 \quad eq ~6$

Collect these together, and we have the Maxwell source equations:

$\begin{align*}\rho &= \nabla \cdot E \\ J &= \nabla \times B - \nabla_0 E \end{align*} \quad eq 7$

These equations hold in flat or curved spacetime no matter what the coordinate choice.

Gravity

It has been shown experimentally that gravity can be described using a dynamic metric equation. A metric is symmetric. As the symmetric metric changes depending on where one is in the gravitational field, the change must also be symmetric. Hamilton's rules of quaternion multiplication cannot work in this context since they have an antisymmetric component. As such, we need a different division algebra that is symmetric using 4 numbers. There is something that fits the task called hypercomplex numbers. I dislike the name since there is nothing hyper or complex about these numbers. So long as one allows every one of the standard products f two basis vectors of a quaternion be positive, then we have consistent rules for multiplication. To ensure that one always necessarily has an inverse, it turns out that by excluding the set of Eigenvalues of the real 4x4 matrix representation, one can make a division algebra. It is customary to assume that a division algebra is constricted to the case zero and only zero must be excluded. That is a point set with one member. The set is larger by a factor of 5 - zero and the 4 Eivenvalues.

"Even multiplication" will be represented by the box times symbol, $\boxtimes$ . The symmetric analog to the curl will be written with the o times symbol, $\otimes$ . Here is the even product of two quaternions, where A is the quaternion, and (a, A) is the scalar and 3-vector:

$A \boxtimes B = (a b + A \cdot B, a B + A b + A \otimes B)$

Now I will repeat the derivation of the Maxwell equations precisely in its structure, using even multiplication. A few more conjugates are also required at the start, but then all the old math machinery does its job.

$\begin{align*}\frac{1}{4}(\nabla^* \boxtimes A - (\nabla^* \boxtimes A)^*) \boxtimes (A^* \boxtimes \nabla - (A^* \boxtimes \nabla)^*) &= (0, \nabla_0 A - \nabla_u \phi - \nabla \otimes A) \boxtimes (0, -\nabla_0 A + \nabla_u \phi - \nabla \otimes A)\\ &= (0, e + b)(0, -e + b)\\ &= (b^2 - e^2, b \otimes b - e \otimes e) \end{align*} \quad eq ~1-even$

where

$e \equiv +\nabla_0 A - \nabla_u \phi$
$b \equiv -\nabla \otimes A$

Notice again on the first line the zeros in the scalars, indicating the Lagrange density will be independent of the gauge choice, essential for the graviton to travel at the speed of light. The meaning of the 3-vector is not known to me at this time.

Proceed as before to form the Lagrangian

$\begin{align*}\mathcal{L}_{\textrm{eb}} &= -\frac{1}{2}(J \boxtimes A^* + (J \boxtimes A^*)^*) + \frac{1}{4} ((b^2 - e^2, b \otimes b - e \otimes e) + (b^2 - e^2, b \otimes b - e \otimes e)^*)\\ &= -\rho \phi + J_1 A_1 + J_2 A_2 + J_3 A_3 \\ & +\frac{1}{2}((\nabla_1 \phi)^2 + (\nabla_2 \phi)^2 + (\nabla_3 \phi)^2) + (\nabla_0 A_1)^2 + (\nabla_0 A_2)^2 + (\nabla_0 A_3)^2 \\ & - (\nabla_3 A_2)^2 - (\nabla_2 A_3)^2 - (\nabla_1 A_3)^2 - (\nabla_3 A_1)^2 - (\nabla_2 A_1)^2 - (\nabla_1 A_2)^2) \\ & - (\nabla_3 A_2)(\nabla_2 A_3) - (\nabla_1 A_3)(\nabla_3 A_1) - (\nabla_1 A_2)(\nabla_2 A_1) - (\nabla_1 \phi)(\nabla_0 A_1) - (\nabla_2 \phi)(\nabla_0 A_2) - (\nabla_3 \phi)(\nabla_0 A_3) \end{align*} \quad eq ~2-even$

Going from equation 2 to 2-even, the current coupling term and cross terms (bottom line) remained the same. All the signs flipped for the squared terms.

$\nabla_{\mu}} ( \frac{\partial \mathcal{L}_\textrm{eb}}{\partial(\nabla_{\mu} \phi)}) = \nabla^2_1 \phi + \nabla^2_2 \phi + \nabla^2_3 \phi - \nabla_0 \nabla_1 A_1 - \nabla_0 \nabla_2 A_2 - \nabla_0 \nabla_3 A_3 = -\nabla \cdot e = \rho \quad eq ~3-even$

$\nabla_{\mu}} ( \frac{\partial \mathcal{L}_\textrm{eb}}{\partial(\nabla_{\mu} A_1)}) = \nabla^2_0 A_1 - \nabla^2_3 A_1 - \nabla^2_2 A_1 - \nabla_1 \nabla_3 A_3 - \nabla_1 \nabla_2 A_2 - \nabla_0 \nabla_1 \phi = \nabla_0 e_1 + (\nabla \otimes b)_1 = -J_1 \quad eq ~4-even$

$\nabla_{\mu}} ( \frac{\partial \mathcal{L}_\textrm{eb}}{\partial(\nabla_{\mu} A_2)}) = \nabla^2_0 A_2 - \nabla^2_3 A_2 - \nabla^2_2 A_2 - \nabla_2 \nabla_3 A_3 - \nabla_1 \nabla_2 A_1 - \nabla_0 \nabla_2 \phi = \nabla_0 e_2 + (\nabla \otimes b)_2 = -J_2 \quad eq ~5-even$

$\nabla_{\mu}} ( \frac{\partial \mathcal{L}_\textrm{eb}}{\partial(\nabla_{\mu} A_3)}) = \nabla^2_0 A_3 - \nabla^2_3 A_3 - \nabla^2_2 A_3 - \nabla_2 \nabla_3 A_2 - \nabla_1 \nabla_3 A_1 - \nabla_0 \nabla_3 \phi = \nabla_0 e_3 + (\nabla \otimes b)_3 = -J_3 \quad eq ~6-even$

Collect these together, and we have the even source equations:

$\begin{align*}\rho &= -\nabla \cdot e \\ J &= -\nabla \otimes b - \nabla_0 e \end{align*} \quad eq 7-even$

Please feel free to take a darn good close look at the signs used for the e and b fields - I am not 100% sure I was consistent.

In the coming weeks, I hope to write the unified field equations here, which is an exercise starting from here:

$\frac{1}{2}(-(\nabla A)(A \nabla) + (\nabla^* \boxtimes A) \boxtimes (A^* \boxtimes \nabla))$

must go prepare for BarCampBoston...