# Geometric product in electromagnetism

1. Dec 22, 2009

### Longstreet

Hi. I've been learning how to use geometric algebra and I've been stumbling when I apply it to E&M. I am hoping someone here can point out what I am doing wrong.

The problem comes when trying to represent the field tensor in terms of the 4-potential. Here is the standard form:

$$F^{\mu\nu} = \partial^\mu A^{\nu} - \partial^\nu A^{\mu}$$

$$\partial_\mu A^{\mu} = 0$$

Here is the form presented in "Geometric Algebra for Physicists":

$$F = E + IB = \nabla A$$

Here it uses the space time vectors $$\nabla = \gamma^\mu \partial_\mu , A = \gamma_\nu A^\nu: \mu, \nu= 0,1,2,3$$, the space time bi-vectors $$E = E^i \gamma_i \gamma_0 ; B = B^i \gamma_i \gamma_0: i = 1,2,3$$, and spacetime pseudoscalar $$I = \gamma_0 \gamma_1 \gamma_2 \gamma_3$$

When I apply it to the 4-potential I get:

$$\nabla A = \gamma^\mu \gamma_\nu \partial_\mu A^\nu = \gamma^0 \gamma_0 \partial_0 A^0 + ... + \gamma^3 \gamma_3 \partial_3 A^3$$

Now, at this point I make use of some identities: $$\gamma^0 = \gamma_0; \gamma_0 \gamma_0 = 1; \gamma^i = -\gamma_i; \gamma_i \gamma_i = -1: i = 1,2,3; \gamma_\mu \gamma_\nu = -\gamma_\nu \gamma_\mu$$

When I do this to collect terms I should get a scalar, which represents the divergence of A = 0, and a set of bi-vectors which represents the field. But, instead I get numerous wronge signs.

$$(\partial_0 A^0 + \partial_1 A^1 + \partial_2 A^2 + \partial_3 A^3) + (\partial_0 A^1 + \partial_1 A^0)\gamma_0 \gamma_1 + (\partial_0 A^2 + \partial_2 A^0)\gamma_0 \gamma_2 + ...$$

The scalar is correct, but the bivectors should be, for example: $$(\partial_0 A^1 - \partial_1 A^0)\gamma_0 \gamma_1$$

Thank you for anyone that can help point out my misunderstanding.

2. Dec 23, 2009

### tiny-tim

Hi Longstreet!

(have a nabla: ∇ )

Shouldn't F = Λ A ?

3. Dec 23, 2009

### Longstreet

Thank you for your reply. However, $$\nabla A = \nabla \cdot A + \nabla \wedge A = \nabla \wedge A$$, from the lorenz condition $$\nabla \cdot A = 0$$.

I think I realized where the negative comes in. EG:

$$(\partial_0 A^1 + \partial_1 A^0)\gamma_0 \gamma_1 = (\partial^0 A^1 - \partial^1 A^0)\gamma_0 \gamma_1$$

Because of the fact that $$x^\mu \gamma_\mu = x_\mu \gamma^\mu$$. However, now I am a bit confused as to what difference it makes taking the derivative $$\partial_\mu$$ versus $$\partial^\mu$$ (besides a potential negative sign), and how to know when to use which one. I guess one interpretation is a unit like effect, as in bi-vectors have a area-like quantity associated with them.

Last edited: Dec 23, 2009
4. Dec 27, 2009

### Peeter

You can use either:

$$\nabla = \gamma^\mu \partial_\mu = \gamma_\mu \partial^\mu$$

(the sign change for raising $\partial_k = - \partial^k$ is cancelled by the corresponding sign change lowering $\gamma^k = - \gamma_k$)

Use whichever is most convienent. For example, expanding the four-Laplacian in coordinates, use of both makes sense:

$$\nabla^2 = \gamma^\mu \partial_\mu \cdot \gamma_\nu \partial^\nu = \partial_\mu \partial^\mu$$