# Peeter Joot's (OLD) Blog.

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## Sumarizing: Transverse electric and magnetic fields

Posted by peeterjoot on August 1, 2009

There’s potentially a lot of new ideas in the previous transverse field post (some for me even with previous exposure to the Geometric Algebra formalism). There was no real attempt to teach GA here, but for completeness the GA form of Maxwell’s equation was developed from the traditional divergence and curl formulation of Maxwell’s equations. That was mainly due to use of CGS units which differ since this makes Maxwell’s equation take a different form from the usual (see [1]).

This time a less exploratory summary of the previous results above is assembled.

In these CGS units our field $F$, and Maxwell’s equation (in absence of charge and current), take the form

\begin{aligned}F &= \mathbf{E} + \frac{I\mathbf{B}}{\sqrt{\mu\epsilon}} \\ 0 &= \left(\boldsymbol{\nabla} + \frac{\sqrt{\mu\epsilon}}{c}\partial_t\right) F \end{aligned} \quad\quad\quad(30)

The electric and magnetic fields can be picked off by selecting the grade one (vector) components

\begin{aligned}\mathbf{E} &= {\left\langle{{F}}\right\rangle}_{1} \\ \mathbf{B} &= {\left\langle{{-I F}}\right\rangle}_{1} \end{aligned} \quad\quad\quad(32)

With an explicit sinusoidal and $z$-axis time dependence for the field

\begin{aligned}F(x,y,z,t) &= F(x,y) e^{\pm i k z - i \omega t} \end{aligned} \quad\quad\quad(34)

and a split of the gradient into transverse and $z$-axis components $\boldsymbol{\nabla} = \boldsymbol{\nabla}_t + \hat{\mathbf{z}} \partial_z$, Maxwell’s equation takes the form

\begin{aligned}\left(\boldsymbol{\nabla}_t \pm i k \hat{\mathbf{z}} - \sqrt{\mu\epsilon}\frac{i\omega}{c}\right) F(x,y) = 0 \end{aligned} \quad\quad\quad(35)

Writing for short $F = F(x,y)$, we can split the field into transverse and $z$-axis components with the commutator and anticommutator products respectively. For the $z$-axis components we have

\begin{aligned}F_z \hat{\mathbf{z}} \equiv E_z + I B_z = \frac{1}{{2}} (F \hat{\mathbf{z}} + \hat{\mathbf{z}} F) \end{aligned} \quad\quad\quad(36)

The projections onto the $z$-axis and and transverse directions are respectively

\begin{aligned}F_z &= \mathbf{E}_z + I \mathbf{B}_z = \frac{1}{{2}} (F + \hat{\mathbf{z}} F \hat{\mathbf{z}}) \\ F_t &= \mathbf{E}_t + I \mathbf{B}_t = \frac{1}{{2}} (F - \hat{\mathbf{z}} F \hat{\mathbf{z}} ) \end{aligned} \quad\quad\quad(37)

With an application of the transverse gradient to the $z$-axis field we easily found the relation between the two
field components

\begin{aligned}\boldsymbol{\nabla}_t F_z &= i \left( \pm k \hat{\mathbf{z}} - \sqrt{\mu\epsilon}\frac{\omega}{c}\right) F_t \end{aligned} \quad\quad\quad(39)

A left division of the exponential factor gives the total transverse field

\begin{aligned}F_t &= \frac{1}{{i \left( \pm k \hat{\mathbf{z}} - \sqrt{\mu\epsilon}\frac{\omega}{c}\right) }} \boldsymbol{\nabla}_t F_z \end{aligned} \quad\quad\quad(40)

Multiplication of both the numerator and denominator by the conjugate normalizes this

\begin{aligned}F_t &= \frac{i}{k^2 - \mu\epsilon\frac{\omega^2}{c^2}} \left( \pm k \hat{\mathbf{z}} + \sqrt{\mu\epsilon}\frac{\omega}{c}\right) \boldsymbol{\nabla}_t F_z \end{aligned} \quad\quad\quad(41)

From this the transverse electric and magnetic fields may be picked off using the projective grade selection operations of (32), and are

\begin{aligned}\mathbf{E}_t &= \frac{i}{\mu\epsilon\frac{\omega^2}{c^2} -k^2} \left( \pm k \boldsymbol{\nabla}_t E_z - \frac{\omega}{c} \hat{\mathbf{z}} \times \boldsymbol{\nabla}_t B_z \right) \\ \mathbf{B}_t &= \frac{i}{\mu\epsilon\frac{\omega^2}{c^2} -k^2} \left( {\mu\epsilon}\frac{\omega}{c} \hat{\mathbf{z}} \times \boldsymbol{\nabla}_t E_z \pm k \boldsymbol{\nabla}_t B_z \right) \end{aligned} \quad\quad\quad(42)

# References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.