# Quaternions and Maxwell

1. Dec 6, 2003

### arcnets

This thread is the sequel to my other thread, 'Quaternions and SR'.
My goal is to write some physical equations with quaternions.
Because I think quaternions represent the 4-dimensionality and metric used in special relativity. See:
A quaternion is a generalized complex number:
$$A = a_t + ia_x + ja_y + ka_z$$
with the fundamental equation
$$i^2 = j^2 = k^2 = ijk = -1.$$
Quaternions are not commutative, for instance
$$ij = -ji = k.$$
Let's define
$$A_3 = ia_x + ja_y + ka_z$$
and the dot and cross products as usual for 3-vectors, then the product of two quaternions is
$$AB = a_tb_t + a_tB_3 + A_3b_t - A_3 \cdot B_3 + A_3 \times B_3.$$
Thus,
$$\frac{1}{2}(AB - BA) = A_3 \times B_3$$
and
$$\frac{1}{2}(AB + BA) = a_tb_t + a_tB_3 + A_3b_t - A_3 \cdot B_3.$$
Let's define the commutator
$$\left[A,B\right] = \frac{1}{2}(AB - BA)$$
and the anticommutator
$$\left<A,B\right> = \frac{1}{2}(AB + BA).$$
Now for physics. Let's define the differential operator
$$\nabla = \frac{\partial}{\partial t} + i \frac{\partial}{\partial x} + j \frac{\partial}{\partial y} + k \frac{\partial}{\partial z}.$$
Then Maxwell's equations can be written
1. Coulomb's law: $$\left<\nabla, E\right> = \nabla_tE - 4\pi J_0$$
2. Ampere's law: $$\left[\nabla, B\right] = \nabla_tE + 4\pi J_3$$
3. Faraday's law: $$\left[\nabla, E\right] = -\nabla_tB$$
4. No magnetic monopoles: $$\left<\nabla, B\right> = \nabla_tB.$$
Now if we use a vector potental written as a quaternion A, which satisfies Lorentz's condition
$$\nabla_t a_t - \nabla_3 \cdot A_3 = 0$$
and let
$$E = -\frac{1}{2}\left<\nabla,A\right>$$
$$B = \frac{1}{2}\left[\nabla,A\right]$$
then Maxwell's equations reduce nicely to two wave equations:
$$4 \pi J = \frac{1}{2}\left<\nabla^2,A\right>$$
$$\nabla_t B = \frac{1}{2}\left[\nabla^2,A\right].$$
That's my result so far. Any comments?

Last edited: Dec 6, 2003
2. Dec 6, 2003

### Hurkyl

Staff Emeritus
I know what you mean, but taking things literally, $\nabla A$ is a quaternion function, but $A \nabla$ is a differential operator. Looking at individual pieces, then for instance, $x (\partial / \partial x) \neq (\partial / \partial x) x$

3. Dec 6, 2003

### arcnets

Hurkyl,
yes they look like differential operators, but that's no problem. If you analyze my formulae, you'll see:
$$B\nabla = \nabla_t(B-E) - 4 \pi J_3$$
and
$$E\nabla = \nabla_t(B+E) - 4 \pi J_0.$$
Similarly for $A\nabla$ and $A\nabla^2$.

4. Dec 7, 2003

### arcnets

"No problem"? - I must have been blind!
Hurkyl, I think you pointed out the weak spot.
Of course, the expressions for dot and cross products don't work if the vectors themselves don't commute.
Maybe that can be repaired with extra definitions... but probably it won't be so elegant...