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## Wave equation form of Maxwell’s equations

Posted by peeterjoot on July 6, 2009

[Click here for a PDF of this post with nicer formatting]

# Motivation.

In ([1]), on plane waves, he writes “we find easily…” to show that the wave equation for each of the components of $\mathbf{E}$, and $\mathbf{B}$ in the absence of current and charge satisfy the wave equation. Do this calculation.

# Vacuum case.

Avoiding the non-vacuum medium temporarily, Maxwell’s vacuum equations (in SI units) are

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} = 0 \end{aligned} \quad\quad\quad(1)
\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{B} = 0 \end{aligned} \quad\quad\quad(2)
\begin{aligned}\boldsymbol{\nabla} \times \mathbf{B} = \frac{1}{c^2} \frac{\partial \mathbf{E}}{\partial t} \end{aligned} \quad\quad\quad(3)
\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} \end{aligned} \quad\quad\quad(4)

The last two curl equations can be decoupled by once more calculating the curl. Illustrating by example

\begin{aligned}\boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{E}) = -\frac{\partial }{\partial t} \boldsymbol{\nabla} \times \mathbf{B} = -\frac{1}{c^2} \frac{\partial^2 \mathbf{E}}{\partial t^2} \end{aligned} \quad\quad\quad(5)

Digging out vector identities and utilizing the zero divergence we have

\begin{aligned}\boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{E}) = \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{E}) - \boldsymbol{\nabla}^2 \mathbf{E} = -\boldsymbol{\nabla}^2 \mathbf{E} \end{aligned} \quad\quad\quad(6)

Putting (5), and (6) together provides a wave equation for the electric field vector

\begin{aligned}\frac{1}{c^2} \frac{\partial^2 \mathbf{E}}{\partial t^2} - \boldsymbol{\nabla}^2 \mathbf{E} = 0 \end{aligned} \quad\quad\quad(7)

Operating with curl on the remaining Maxwell equation similarly produces a wave equation for the magnetic field vector

\begin{aligned}\frac{1}{c^2} \frac{\partial^2 \mathbf{B}}{\partial t^2} - \boldsymbol{\nabla}^2 \mathbf{B} = 0 \end{aligned} \quad\quad\quad(8)

This is really six wave equations, one for each of the field coordinates.

# With Geometric Algebra.

Arriving at (7), and (8) is much easier using the GA formalism of ([2]).

Pre or post multiplication of the gradient with the observer frame time basis unit vector $\gamma_0$ has a conjugate like
action

\begin{aligned}\nabla \gamma_0&=\gamma^0 \gamma_0 \partial_0 + \gamma^k \gamma_0 \partial_k \\ &=\partial_0 - \boldsymbol{\nabla} \\ \end{aligned}

(where as usual our spatial basis is $\sigma_k = \gamma_k \gamma_0$).

Similarly
\begin{aligned}\gamma_0 \nabla &=\partial_0 + \boldsymbol{\nabla} \\ \end{aligned}

For the vacuum Maxwell’s equation is just
\begin{aligned}\nabla F = \nabla (\mathbf{E} + I c \mathbf{B}) = 0 \end{aligned}

With nothing more than an algebraic operation we have

\begin{aligned}0 &= \nabla \gamma_0 \gamma_0 \nabla F \\ &=( \partial_0 - \boldsymbol{\nabla} ) ( \partial_0 + \boldsymbol{\nabla} ) (\mathbf{E} + I c \mathbf{B}) \\ &=\left( \frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \boldsymbol{\nabla}^2 \right) (\mathbf{E} + I c \mathbf{B}) \\ \end{aligned}

This equality is true independently for each of the components of $\mathbf{E}$ and $\mathbf{B}$, so we have as before

These wave equations are still subject to the constraints of the original Maxwell equations.

\begin{aligned}0 &= \gamma_0 \nabla F \\ &= (\partial_0 + \boldsymbol{\nabla}) (\mathbf{E} + I c \mathbf{B}) \\ &= \boldsymbol{\nabla} \cdot \mathbf{E} + (\partial_0 \mathbf{E} - c \boldsymbol{\nabla} \times \mathbf{B})+ I ( c \partial_0 \mathbf{B} + \boldsymbol{\nabla} \times \mathbf{E} )+ I c \boldsymbol{\nabla} \cdot \mathbf{B} \\ \end{aligned}

# Tensor approach?

In both the traditional vector and the GA form one can derive the wave equation relations of (7), (8). One can obviously summarize these in tensor form as

\begin{aligned}\partial_\mu\partial^\mu F^{\alpha\beta} = 0 \end{aligned} \quad\quad\quad(9)

working backwards from the vector or GA result. In this notation, the coupling constraint would be that the field variables $F^{\alpha\beta}$ are subject to the Maxwell divergence equation (name?)

\begin{aligned}\partial_\mu F^{\mu\nu} = 0 \end{aligned} \quad\quad\quad(10)

and also the dual tensor relation

\begin{aligned}\epsilon^{\sigma\mu\alpha\beta} \partial_\mu F_{\alpha\beta} = 0 \end{aligned} \quad\quad\quad(11)

I cannot seem to figure out how to derive (9) starting from these tensor relations?

This probably has something to do with the fact that we require both the divergence and the dual relations (10), (11) expressed together to do this.

# Electromagnetic waves in media.

Jackson lists the Macroscopic Maxwell equations in (6.70) as

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{B} &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{D} &= 4 \pi \rho \\ \boldsymbol{\nabla} \times \mathbf{E} + \frac{1}{c}\frac{\partial \mathbf{B}}{\partial {t}} &= 0 \\ \boldsymbol{\nabla} \times \mathbf{H} - \frac{1}{c}\frac{\partial \mathbf{D}}{\partial {t}} &= \frac{4 \pi}{c} \mathbf{J} \\ \end{aligned}

(for this note this means unfortunately a switch from SI to CGS midstream)

For vacuum ($\rho =0$, and $\mathbf{J} =0$), plus $\mathbf{B} = \mu \mathbf{H}$, and $\mathbf{D} = \epsilon \mathbf{E}$, we can assemble these into his (7.1) equations

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{B} &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{E} &= 0 \\ \boldsymbol{\nabla} \times \mathbf{E} + \frac{1}{c}\frac{\partial \mathbf{B}}{\partial {t}} &= 0 \\ \boldsymbol{\nabla} \times \mathbf{B} - \frac{\epsilon \mu}{c}\frac{\partial \mathbf{E}}{\partial {t}} &= 0 \\ \end{aligned}

In this macroscopic form, it isn’t obvious how to assemble the equations into a nice tidy GA form. A compromise is

\begin{aligned}\boldsymbol{\nabla} \mathbf{E} + \partial_0 (I\mathbf{B}) &= 0 \\ \boldsymbol{\nabla} (I \mathbf{B}) + \epsilon \mu \partial_0 \mathbf{E} &= 0 \end{aligned} \quad\quad\quad(12)

Although not as pretty, we can at least derive the wave equations from these. For example for $\mathbf{E}$, we apply one additional spatial gradient

\begin{aligned}0 &= \boldsymbol{\nabla}^2 \mathbf{E} + \partial_0 (\boldsymbol{\nabla} I \mathbf{B}) \\ &= \boldsymbol{\nabla}^2 \mathbf{E} + \partial_0 ( -\epsilon \mu \partial_0 \mathbf{E} ) \\ \end{aligned}

For $\mathbf{B}$ we get the same, and have two wave equations

\begin{aligned}\frac{\mu \epsilon}{c^2} \frac{\partial^2 \mathbf{E}}{\partial t^2} - \boldsymbol{\nabla}^2 \mathbf{E} &= 0 \\ \frac{\mu \epsilon}{c^2} \frac{\partial^2 \mathbf{B}}{\partial t^2} - \boldsymbol{\nabla}^2 \mathbf{B} &= 0 \end{aligned} \quad\quad\quad(14)

The wave velocity is thus not $c$, but instead the reduced speed of $c/{\sqrt{\mu\epsilon}}$.

The fact that it is possible to assemble wave equations of this form means that there must also be a simpler form than (12). The reduced velocity is the clue, and that can be used to refactor the constants

\begin{aligned}\boldsymbol{\nabla} \mathbf{E} + \sqrt{\mu\epsilon}\partial_0 \left(\frac{I\mathbf{B}}{\sqrt{\mu\epsilon}}\right) &= 0 \\ \boldsymbol{\nabla} \left(\frac{I \mathbf{B}}{\sqrt{\mu\epsilon}}\right) + \sqrt{\mu\epsilon} \partial_0 \mathbf{E} &= 0 \end{aligned}

These can now be added

\begin{aligned}\left(\boldsymbol{\nabla} + \sqrt{\mu\epsilon} \partial_0\right)\left(\mathbf{E} + \frac{I\mathbf{B}}{\sqrt{\mu\epsilon}}\right) = 0 \end{aligned}

This allows for the one liner derivation of (14) by premultiplying by the conjugate
operator $-\boldsymbol{\nabla} + \sqrt{\mu\epsilon} \partial_0$

\begin{aligned}0 &=\left(-\boldsymbol{\nabla} + \sqrt{\mu\epsilon} \partial_0\right)\left(\boldsymbol{\nabla} + \sqrt{\mu\epsilon} \partial_0\right)\left(\mathbf{E} + \frac{I\mathbf{B}}{\sqrt{\mu\epsilon}}\right) \\ &=\left(-\boldsymbol{\nabla}^2 + \frac{\mu\epsilon}{c^2} \partial_{tt}\right)\left(\mathbf{E} + \frac{I\mathbf{B}}{\sqrt{\mu\epsilon}}\right) \end{aligned}

Using the same hint, and doing some rearrangement, we can write Jackson’s equations (6.70) as

\begin{aligned}\left(\boldsymbol{\nabla} + \sqrt{\mu\epsilon} \partial_0\right)\left(\mathbf{E} + \frac{I\mathbf{B}}{\sqrt{\mu\epsilon}}\right) = \frac{4\pi}{\epsilon}\left( \rho - \frac{\sqrt{\mu\epsilon}}{c}\mathbf{J} \right) \end{aligned} \quad\quad\quad(17)

# References

[1] JD Jackson. Classical Electrodynamics Wiley. 2nd edition, 1975.

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

1. ### Willie Wongsaid

About the Tensor approach: what you need to do is take the divergence of equation (11). But in the form you have written you will end up with a full anti-symmetrisation of two derivatives of F, which will be unpleasant to deal with (you’d have to re-write everything in components to see the answer). A better way is to re-write equation (11) using its dual: that d_aF_bc + d_bF_ca + d_cF_ab = 0. Here you use the fact that a form is zero if and only if its Hodge dual is zero. Then just take the divergence of this equation with d^a. The first term gives the wave equation, the second and third terms vanish because F is divergence free. cheers.

• ### peeterjootsaid

I’ll try taking the divergence to see how it works (even if it is messy). I know of the forms approach, but not well enough to use it (but can probably translate your statement since duality I know in the context of geometric algebra as multiplication by the pseudoscalar)

2. ### Sunnysaid

You sick intellectual fck……You need help and at the very least require PhD. to level set you !

It’s been along time, Sunny

• ### peeterjootsaid

We all have our vices. Have people learned to keep the bix boxes near you locked yet?