# Geometric product in electromagnetism

• Longstreet
In summary, the problem seems to be that when trying to represent the field tensor in terms of the 4-potential, the form presented in "Geometric Algebra for Physicists" does not work. The form presented in "Geometric Algebra for Physicists" 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^
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.

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

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

Hi Longstreet!

(have a nabla: ∇ )

Shouldn't F = Λ A ?

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:
You can use either:

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

(the sign change for raising $\partial_k = - \partial^k$ is canceled 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:

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

## What is a geometric product in electromagnetism?

A geometric product in electromagnetism is a mathematical operation that combines the concepts of vectors and tensors to describe the electromagnetic field in three-dimensional space. It is a powerful tool for analyzing and solving problems related to electric and magnetic fields.

## How is a geometric product different from other mathematical operations?

Unlike traditional mathematical operations such as addition and multiplication, a geometric product takes into account both the direction and magnitude of vectors and tensors. This allows for a more accurate representation of the electromagnetic field and its interactions.

## What are some real-world applications of the geometric product in electromagnetism?

The geometric product is commonly used in the design and analysis of electrical circuits, antennas, and other electromagnetic systems. It is also used in fields such as optics, telecommunications, and robotics to model and predict the behavior of electromagnetic waves.

## Are there any limitations to using the geometric product in electromagnetism?

While the geometric product is a powerful tool, it is not always the most efficient method for solving all electromagnetic problems. In certain cases, simpler mathematical operations may be more appropriate. Additionally, the geometric product is limited to three-dimensional space and may not accurately describe the behavior of electromagnetic fields in higher dimensions.

There are many resources available for those interested in learning more about the geometric product in electromagnetism. Textbooks, online courses, and research articles are all great places to start. It is also helpful to have a strong foundation in vector calculus and tensor analysis before delving into the geometric product.

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