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Posts Tagged ‘maxwells equations’

Polarization angles for normal transmission and reflection

Posted by peeterjoot on January 22, 2014

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

Question: Polarization angles for normal transmission and reflection ([1] pr 9.14)

For normal incidence, without assuming that the reflected and transmitted waves have the same polarization as the incident wave, prove that this must be so.

Answer

Working with coordinates as illustrated in fig. 1.1, the incident wave can be assumed to have the form

fig 1.1: Normal incidence coordinates

$\begin{aligned}\tilde{\mathbf{E}}_{\mathrm{I}} = E_{\mathrm{I}} e^{i (k z - \omega t)} \hat{\mathbf{x}}\end{aligned} \hspace{\stretch{1}}(1.0.1a)$

$\begin{aligned}\tilde{\mathbf{B}}_{\mathrm{I}} = \frac{1}{{v}} \hat{\mathbf{z}} \times \tilde{\mathbf{E}}_{\mathrm{I}} = \frac{1}{{v}} E_{\mathrm{I}} e^{i (k z - \omega t)} \hat{\mathbf{y}}.\end{aligned} \hspace{\stretch{1}}(1.0.1b)$

Assuming a polarization $\hat{\mathbf{n}} = \cos\theta \hat{\mathbf{x}} + \sin\theta \hat{\mathbf{y}}$ for the reflected wave, we have

$\begin{aligned}\tilde{\mathbf{E}}_{\mathrm{R}} = E_{\mathrm{R}} e^{i (-k z - \omega t)} (\hat{\mathbf{x}} \cos\theta + \hat{\mathbf{y}} \sin\theta)\end{aligned} \hspace{\stretch{1}}(1.0.2a)$

$\begin{aligned}\tilde{\mathbf{B}}_{\mathrm{R}} = \frac{1}{{v}} (-\hat{\mathbf{z}}) \times \tilde{\mathbf{E}}_{\mathrm{R}} = \frac{1}{{v}} E_{\mathrm{R}} e^{i (-k z - \omega t)} (\hat{\mathbf{x}} \sin\theta - \hat{\mathbf{y}} \cos\theta).\end{aligned} \hspace{\stretch{1}}(1.0.2b)$

And finally assuming a polarization $\hat{\mathbf{n}} = \cos\phi \hat{\mathbf{x}} + \sin\phi \hat{\mathbf{y}}$ for the transmitted wave, we have

$\begin{aligned}\tilde{\mathbf{E}}_{\mathrm{T}} = E_{\mathrm{T}} e^{i (k' z - \omega t)} (\hat{\mathbf{x}} \cos\phi + \hat{\mathbf{y}} \sin\phi)\end{aligned} \hspace{\stretch{1}}(1.0.3a)$

$\begin{aligned}\tilde{\mathbf{B}}_{\mathrm{T}} = \frac{1}{{v}} \hat{\mathbf{z}} \times \tilde{\mathbf{E}}_{\mathrm{T}} = \frac{1}{{v'}} E_{\mathrm{T}} e^{i (k' z - \omega t)} (-\hat{\mathbf{x}} \sin\phi + \hat{\mathbf{y}} \cos\phi).\end{aligned} \hspace{\stretch{1}}(1.0.3b)$

With no components of any of the $\tilde{\mathbf{E}}$ or $\tilde{\mathbf{B}}$ waves in the $\hat{\mathbf{z}}$ directions the boundary value conditions at $z = 0$ require the equality of the $\hat{\mathbf{x}}$ and $\hat{\mathbf{y}}$ components of

$\begin{aligned}\left( \tilde{\mathbf{E}}_{\mathrm{I}} + \tilde{\mathbf{E}}_{\mathrm{R}} \right)_{x,y} = \left( \tilde{\mathbf{E}}_{\mathrm{T}} \right)_{x,y}\end{aligned} \hspace{\stretch{1}}(1.0.4a)$

$\begin{aligned} \left( \frac{1}{\mu} \left( \tilde{\mathbf{B}}_{\mathrm{I}} + \tilde{\mathbf{B}}_{\mathrm{R}} \right) \right)_{x,y} = \left( \frac{1}{\mu'} \tilde{\mathbf{B}}_{\mathrm{T}} \right)_{x,y}.\end{aligned} \hspace{\stretch{1}}(1.0.4b)$

With $\beta = \mu v/\mu' v'$ , those components are

$\begin{aligned}E_{\mathrm{I}} + E_{\mathrm{R}} \cos\theta = E_{\mathrm{T}} \cos\phi \end{aligned} \hspace{\stretch{1}}(1.0.5a)$

$\begin{aligned}E_{\mathrm{R}} \sin\theta = E_{\mathrm{T}} \sin\phi\end{aligned} \hspace{\stretch{1}}(1.0.5b)$

$\begin{aligned}E_{\mathrm{R}} \sin\theta = - \beta E_{\mathrm{T}} \sin\phi\end{aligned} \hspace{\stretch{1}}(1.0.5c)$

$\begin{aligned}E_{\mathrm{I}} - E_{\mathrm{R}} \cos\theta = \beta E_{\mathrm{T}} \cos\phi\end{aligned} \hspace{\stretch{1}}(1.0.5d)$

Equality of eq. 1.0.5b, and eq. 1.0.5c require

$\begin{aligned}- \beta E_{\mathrm{T}} \sin\phi = E_{\mathrm{T}} \sin\phi,\end{aligned} \hspace{\stretch{1}}(1.0.6)$

or $(\theta, \phi) \in \{(0, 0), (0, \pi), (\pi, 0), (\pi, \pi)\}$ . It turns out that all of these solutions correspond to the same physical waves. Let’s look at each in turn

$(\theta, \phi) = (0, 0)$ . The system eq. 1.0.5.5 is reduced to
$\begin{aligned}\begin{aligned}E_{\mathrm{I}} + E_{\mathrm{R}} &= E_{\mathrm{T}} \\ E_{\mathrm{I}} - E_{\mathrm{R}} &= \beta E_{\mathrm{T}},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.7)$

with solution

$\begin{aligned}\begin{aligned}\frac{E_{\mathrm{T}}}{E_{\mathrm{I}}} &= \frac{2}{1 + \beta} \\ \frac{E_{\mathrm{R}}}{E_{\mathrm{I}}} &= \frac{1 - \beta}{1 + \beta}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.8)$
$(\theta, \phi) = (\pi, \pi)$ . The system eq. 1.0.5.5 is reduced to
$\begin{aligned}\begin{aligned}E_{\mathrm{I}} - E_{\mathrm{R}} &= -E_{\mathrm{T}} \\ E_{\mathrm{I}} + E_{\mathrm{R}} &= -\beta E_{\mathrm{T}},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.9)$

with solution

$\begin{aligned}\begin{aligned}\frac{E_{\mathrm{T}}}{E_{\mathrm{I}}} &= -\frac{2}{1 + \beta} \\ \frac{E_{\mathrm{R}}}{E_{\mathrm{I}}} &= -\frac{1 - \beta}{1 + \beta}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.10)$

Effectively the sign for the magnitude of the transmitted and reflected phasors is toggled, but the polarization vectors are also negated, with $\hat{\mathbf{n}} = -\hat{\mathbf{x}}$ , and $\hat{\mathbf{n}}' = -\hat{\mathbf{x}}$ . The resulting $\tilde{\mathbf{E}}_{\mathrm{R}}$ and $\tilde{\mathbf{E}}_{\mathrm{T}}$ are unchanged relative to those of the $(0,0)$ solution above.
$(\theta, \phi) = (0, \pi)$ . The system eq. 1.0.5.5 is reduced to
$\begin{aligned}\begin{aligned}E_{\mathrm{I}} + E_{\mathrm{R}} &= -E_{\mathrm{T}} \\ E_{\mathrm{I}} - E_{\mathrm{R}} &= -\beta E_{\mathrm{T}},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.11)$

with solution

$\begin{aligned}\begin{aligned}\frac{E_{\mathrm{T}}}{E_{\mathrm{I}}} &= -\frac{2}{1 + \beta} \\ \frac{E_{\mathrm{R}}}{E_{\mathrm{I}}} &= \frac{1 - \beta}{1 + \beta}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.12)$

Effectively the sign for the magnitude of the transmitted phasor is toggled. The polarization vectors in this case are $\hat{\mathbf{n}} = \hat{\mathbf{x}}$ , and $\hat{\mathbf{n}}' = -\hat{\mathbf{x}}$ , so the transmitted phasor magnitude change of sign does not change $\tilde{\mathbf{E}}_{\mathrm{T}}$ relative to that of the $(0,0)$ solution above.
$(\theta, \phi) = (\pi, 0)$ . The system eq. 1.0.5.5 is reduced to
$\begin{aligned}\begin{aligned}E_{\mathrm{I}} - E_{\mathrm{R}} &= E_{\mathrm{T}} \\ E_{\mathrm{I}} + E_{\mathrm{R}} &= \beta E_{\mathrm{T}},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.13)$

with solution

$\begin{aligned}\begin{aligned}\frac{E_{\mathrm{T}}}{E_{\mathrm{I}}} &= \frac{2}{1 + \beta} \\ \frac{E_{\mathrm{R}}}{E_{\mathrm{I}}} &= -\frac{1 - \beta}{1 + \beta}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.14)$

This time, the sign for the magnitude of the reflected phasor is toggled. The polarization vectors in this case are $\hat{\mathbf{n}} = -\hat{\mathbf{x}}$ , and $\hat{\mathbf{n}}' = \hat{\mathbf{x}}$ . In this final variation the reflected phasor magnitude change of sign does not change $\tilde{\mathbf{E}}_{\mathrm{R}}$ relative to that of the $(0,0)$ solution.

We see that there is only one solution for the polarization angle of the transmitted and reflected waves relative to the incident wave. Although we fixed the incident polarization with $\mathbf{E}$ along $\hat{\mathbf{x}}$ , the polarization of the incident wave is maintained regardless of TE or TM labeling in this example, since our system is symmetric with respect to rotation.

References

[1] D.J. Griffith. Introduction to Electrodynamics. Prentice-Hall, 1981.

Posted in Math and Physics Learning. | Tagged: boundary value conditions, electromagnetism, maxwells equations, Polarization, reflection, transmission | 1 Comment »

PHY450H1S. Relativistic Electrodynamics Tutorial 4 (TA: Simon Freedman). Waveguides: confined EM waves.

Posted by peeterjoot on March 14, 2011

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

Motivation

While this isn’t part of the course, the topic of waveguides is one of so many applications that it is worth a mention, and that will be done in this tutorial.

We will setup our system with a waveguide (conducting surface that confines the radiation) oriented in the $\hat{\mathbf{z}}$ direction. The shape can be arbitrary

PICTURE: cross section of wacky shape.

At the surface of a conductor.

At the surface of the conductor (I presume this means the interior surface where there is no charge or current enclosed) we have

$\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E} &= - \frac{1}{{c}} \frac{\partial {\mathbf{B}}}{\partial {t}} \\ \boldsymbol{\nabla} \times \mathbf{B} &= \frac{1}{{c}} \frac{\partial {\mathbf{E}}}{\partial {t}} \\ \boldsymbol{\nabla} \cdot \mathbf{B} &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{E} &= 0\end{aligned} \hspace{\stretch{1}}(1.1)$

If we are talking about the exterior surface, do we need to make any other assumptions (perfect conductors, or constant potentials)?

Wave equations.

For electric and magnetic fields in vacuum, we can show easily that these, like the potentials, separately satisfy the wave equation

Taking curls of the Maxwell curl equations above we have

$\begin{aligned}\boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{E}) &= - \frac{1}{{c^2}} \frac{\partial^2 {\mathbf{E}}}{\partial {{t}}^2} \\ \boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{B}) &= - \frac{1}{{c^2}} \frac{\partial^2 {\mathbf{B}}}{\partial {{t}}^2},\end{aligned} \hspace{\stretch{1}}(1.5)$

but we have for vector $\mathbf{M}$

$\begin{aligned}\boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{M})=\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{M}) - \Delta \mathbf{M},\end{aligned} \hspace{\stretch{1}}(1.7)$

which gives us a pair of wave equations

$\begin{aligned}\square \mathbf{E} &= 0 \\ \square \mathbf{B} &= 0.\end{aligned} \hspace{\stretch{1}}(1.8)$

We still have the original constraints of Maxwell’s equations to deal with, but we are free now to pick the complex exponentials as fundamental solutions, as our starting point

$\begin{aligned}\mathbf{E} &= \mathbf{E}_0 e^{i k^a x_a} = \mathbf{E}_0 e^{ i (k^0 x_0 - \mathbf{k} \cdot \mathbf{x}) } \\ \mathbf{B} &= \mathbf{B}_0 e^{i k^a x_a} = \mathbf{B}_0 e^{ i (k^0 x_0 - \mathbf{k} \cdot \mathbf{x}) },\end{aligned} \hspace{\stretch{1}}(1.10)$

With $k_0 = \omega/c$ and $x_0 = c t$ this is

$\begin{aligned}\mathbf{E} &= \mathbf{E}_0 e^{ i (\omega t - \mathbf{k} \cdot \mathbf{x}) } \\ \mathbf{B} &= \mathbf{B}_0 e^{ i (\omega t - \mathbf{k} \cdot \mathbf{x}) }.\end{aligned} \hspace{\stretch{1}}(1.12)$

For the vacuum case, with monochromatic light, we treated the amplitudes as constants. Let’s see what happens if we relax this assumption, and allow for spatial dependence (but no time dependence) of $\mathbf{E}_0$ and $\mathbf{B}_0$ . For the LHS of the electric field curl equation we have

$\begin{aligned}0 &= \boldsymbol{\nabla} \times \mathbf{E}_0 e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \times \mathbf{E}_0 - \mathbf{E}_0 \times \boldsymbol{\nabla}) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \times \mathbf{E}_0 - \mathbf{E}_0 \times \mathbf{e}^\alpha i k_a \partial_\alpha x^a) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \times \mathbf{E}_0 + \mathbf{E}_0 \times \mathbf{e}^\alpha i k^a {\delta_\alpha}^a ) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \times \mathbf{E}_0 + i \mathbf{E}_0 \times \mathbf{k} ) e^{i k_a x^a}.\end{aligned}$

Similarly for the divergence we have

$\begin{aligned}0 &= \boldsymbol{\nabla} \cdot \mathbf{E}_0 e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \cdot \mathbf{E}_0 + \mathbf{E}_0 \cdot \boldsymbol{\nabla}) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \cdot \mathbf{E}_0 + \mathbf{E}_0 \cdot \mathbf{e}^\alpha i k_a \partial_\alpha x^a) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \cdot \mathbf{E}_0 - \mathbf{E}_0 \cdot \mathbf{e}^\alpha i k^a {\delta_\alpha}^a ) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \cdot \mathbf{E}_0 - i \mathbf{k} \cdot \mathbf{E}_0 ) e^{i k_a x^a}.\end{aligned}$

This provides constraints on the amplitudes

$\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E}_0 - i \mathbf{k} \times \mathbf{E}_0 &= -i \frac{\omega}{c} \mathbf{B}_0 \\ \boldsymbol{\nabla} \times \mathbf{B}_0 - i \mathbf{k} \times \mathbf{B}_0 &= i \frac{\omega}{c} \mathbf{E}_0 \\ \boldsymbol{\nabla} \cdot \mathbf{E}_0 - i \mathbf{k} \cdot \mathbf{E}_0 &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{B}_0 - i \mathbf{k} \cdot \mathbf{B}_0 &= 0\end{aligned} \hspace{\stretch{1}}(1.14)$

Applying the wave equation operator to our phasor we get

$\begin{aligned}0 &=\left(\frac{1}{{c^2}} \partial_{tt} - \boldsymbol{\nabla}^2 \right) \mathbf{E}_0 e^{i (\omega t - \mathbf{k} \cdot \mathbf{x})} \\ &=\left(-\frac{\omega^2}{c^2} - \boldsymbol{\nabla}^2 + \mathbf{k}^2 \right) \mathbf{E}_0 e^{i (\omega t - \mathbf{k} \cdot \mathbf{x})}\end{aligned}$

So the momentum space equivalents of the wave equations are

$\begin{aligned}\left( \boldsymbol{\nabla}^2 +\frac{\omega^2}{c^2} - \mathbf{k}^2 \right) \mathbf{E}_0 &= 0 \\ \left( \boldsymbol{\nabla}^2 +\frac{\omega^2}{c^2} - \mathbf{k}^2 \right) \mathbf{B}_0 &= 0.\end{aligned} \hspace{\stretch{1}}(1.18)$

Observe that if $c^2 \mathbf{k}^2 = \omega^2$ , then these amplitudes are harmonic functions (solutions to the Laplacian equation). However, it doesn’t appear that we require such a light like relation for the four vector $k^a = (\omega/c, \mathbf{k})$ .

Back to the tutorial notes.

In class we went straight to an assumed solution of the form

$\begin{aligned}\mathbf{E} &= \mathbf{E}_0(x, y) e^{ i(\omega t - k z) } \\ \mathbf{B} &= \mathbf{B}_0(x, y) e^{ i(\omega t - k z) },\end{aligned} \hspace{\stretch{1}}(2.20)$

where $\mathbf{k} = k \hat{\mathbf{z}}$ . Our Laplacian was also written as the sum of components in the propagation and perpendicular directions

$\begin{aligned}\boldsymbol{\nabla}^2 = \frac{\partial^2 {{}}}{\partial {{x_\perp}}^2} + \frac{\partial^2 {{}}}{\partial {{z}}^2}.\end{aligned} \hspace{\stretch{1}}(2.22)$

With no $z$ dependence in the amplitudes we have

$\begin{aligned}\left( \frac{\partial^2 {{}}}{\partial {{x_\perp}}^2} +\frac{\omega^2}{c^2} - \mathbf{k}^2 \right) \mathbf{E}_0 &= 0 \\ \left( \frac{\partial^2 {{}}}{\partial {{x_\perp}}^2} +\frac{\omega^2}{c^2} - \mathbf{k}^2 \right) \mathbf{B}_0 &= 0.\end{aligned} \hspace{\stretch{1}}(2.23)$

Separation into components.

It was left as an exercise to separate out our Maxwell equations, so that our field components $\mathbf{E}_0 = \mathbf{E}_\perp + \mathbf{E}_z$ and $\mathbf{B}_0 = \mathbf{B}_\perp + \mathbf{B}_z$ in the propagation direction, and components in the perpendicular direction are separated

$\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E}_0 &=(\boldsymbol{\nabla}_\perp + \hat{\mathbf{z}}\partial_z) \times \mathbf{E}_0 \\ &=\boldsymbol{\nabla}_\perp \times \mathbf{E}_0 \\ &=\boldsymbol{\nabla}_\perp \times (\mathbf{E}_\perp + \mathbf{E}_z) \\ &=\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp +\boldsymbol{\nabla}_\perp \times \mathbf{E}_z \\ &=( \hat{\mathbf{x}} \partial_x +\hat{\mathbf{y}} \partial_y ) \times ( \hat{\mathbf{x}} E_x +\hat{\mathbf{y}} E_y ) +\boldsymbol{\nabla}_\perp \times \mathbf{E}_z \\ &=\hat{\mathbf{z}} (\partial_x E_y - \partial_z E_z) +\boldsymbol{\nabla}_\perp \times \mathbf{E}_z.\end{aligned}$

We can do something similar for $\mathbf{B}_0$ . This allows for a split of 1.14 into $\hat{\mathbf{z}}$ and perpendicular components

$\begin{aligned}\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp &= -i \frac{\omega}{c} \mathbf{B}_z \\ \boldsymbol{\nabla}_\perp \times \mathbf{B}_\perp &= i \frac{\omega}{c} \mathbf{E}_z \\ \boldsymbol{\nabla}_\perp \times \mathbf{E}_z - i \mathbf{k} \times \mathbf{E}_\perp &= -i \frac{\omega}{c} \mathbf{B}_\perp \\ \boldsymbol{\nabla}_\perp \times \mathbf{B}_z - i \mathbf{k} \times \mathbf{B}_\perp &= i \frac{\omega}{c} \mathbf{E}_\perp \\ \boldsymbol{\nabla}_\perp \cdot \mathbf{E}_\perp &= i k E_z - \partial_z E_z \\ \boldsymbol{\nabla}_\perp \cdot \mathbf{B}_\perp &= i k B_z - \partial_z B_z.\end{aligned} \hspace{\stretch{1}}(3.25)$

So we see that once we have a solution for $\mathbf{E}_z$ and $\mathbf{B}_z$ (by solving the wave equation above for those components), the components for the fields in terms of those components can be found. Alternately, if one solves for the perpendicular components of the fields, these propagation components are available immediately with only differentiation.

In the case where the perpendicular components are taken as given

$\begin{aligned}\mathbf{B}_z &= i \frac{ c }{\omega} \boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp \\ \mathbf{E}_z &= -i \frac{ c }{\omega} \boldsymbol{\nabla}_\perp \times \mathbf{B}_\perp,\end{aligned} \hspace{\stretch{1}}(3.31)$

we can express the remaining ones strictly in terms of the perpendicular fields

$\begin{aligned}\frac{\omega}{c} \mathbf{B}_\perp &= \frac{c}{\omega} \boldsymbol{\nabla}_\perp \times (\boldsymbol{\nabla}_\perp \times \mathbf{B}_\perp) + \mathbf{k} \times \mathbf{E}_\perp \\ \frac{\omega}{c} \mathbf{E}_\perp &= \frac{c}{\omega} \boldsymbol{\nabla}_\perp \times (\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp) - \mathbf{k} \times \mathbf{B}_\perp \\ \boldsymbol{\nabla}_\perp \cdot \mathbf{E}_\perp &= -i \frac{c}{\omega} (i k - \partial_z) \hat{\mathbf{z}} \cdot (\boldsymbol{\nabla}_\perp \times \mathbf{B}_\perp) \\ \boldsymbol{\nabla}_\perp \cdot \mathbf{B}_\perp &= i \frac{c}{\omega} (i k - \partial_z) \hat{\mathbf{z}} \cdot (\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp).\end{aligned} \hspace{\stretch{1}}(3.33)$

Is it at all helpful to expand the double cross products?

$\begin{aligned}\frac{\omega^2}{c^2} \mathbf{B}_\perp &= \boldsymbol{\nabla}_\perp (\boldsymbol{\nabla}_\perp \cdot \mathbf{B}_\perp) -{\boldsymbol{\nabla}_\perp}^2 \mathbf{B}_\perp + \frac{\omega}{c} \mathbf{k} \times \mathbf{E}_\perp \\ &= i \frac{c}{\omega}(i k - \partial_z)\boldsymbol{\nabla}_\perp \hat{\mathbf{z}} \cdot (\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp)-{\boldsymbol{\nabla}_\perp}^2 \mathbf{B}_\perp + \frac{\omega}{c} \mathbf{k} \times \mathbf{E}_\perp \end{aligned}$

This gives us

$\begin{aligned}\left( {\boldsymbol{\nabla}_\perp}^2 + \frac{\omega^2}{c^2} \right) \mathbf{B}_\perp &= - \frac{c}{\omega} (k + i\partial_z) \boldsymbol{\nabla}_\perp \hat{\mathbf{z}} \cdot (\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp) + \frac{\omega}{c} \mathbf{k} \times \mathbf{E}_\perp \\ \left( {\boldsymbol{\nabla}_\perp}^2 + \frac{\omega^2}{c^2} \right) \mathbf{E}_\perp &= -\frac{c}{\omega} (k + i\partial_z) \boldsymbol{\nabla}_\perp \hat{\mathbf{z}} \cdot (\boldsymbol{\nabla}_\perp \times \mathbf{B}_\perp) - \frac{\omega}{c} \mathbf{k} \times \mathbf{B}_\perp,\end{aligned} \hspace{\stretch{1}}(3.37)$

but that doesn’t seem particularly useful for completely solving the system? It appears fairly messy to try to solve for $\mathbf{E}_\perp$ and $\mathbf{B}_\perp$ given the propagation direction fields. I wonder if there is a simplification available that I am missing?

Solving the momentum space wave equations.

Back to the class notes. We proceeded to solve for $\mathbf{E}_z$ and $\mathbf{B}_z$ from the wave equations by separation of variables. We wish to solve equations of the form

$\begin{aligned}\left( \frac{\partial^2 {{}}}{\partial {{x}}^2} + \frac{\partial^2 {{}}}{\partial {{y}}^2} + \frac{\omega^2}{c^2} - \mathbf{k}^2 \right) \phi(x,y) = 0\end{aligned} \hspace{\stretch{1}}(4.39)$

Write $\phi(x,y) = X(x) Y(y)$ , so that we have

$\begin{aligned}\frac{X''}{X} + \frac{Y''}{Y} = \mathbf{k}^2 - \frac{\omega^2}{c^2}\end{aligned} \hspace{\stretch{1}}(4.40)$

One solution is sinusoidal

$\begin{aligned}\frac{X''}{X} &= -k_1^2 \\ \frac{Y''}{Y} &= -k_2^2 \\ -k_1^2 - k_2^2&= \mathbf{k}^2 - \frac{\omega^2}{c^2}.\end{aligned} \hspace{\stretch{1}}(4.41)$

The example in the tutorial now switched to a rectangular waveguide, still oriented with the propagation direction down the z-axis, but with lengths $a$ and $b$ along the $x$ and $y$ axis respectively.

Writing $k_1 = 2\pi m/a$ , and $k_2 = 2 \pi n/ b$ , we have

$\begin{aligned}\phi(x, y) = \sum_{m n} a_{m n} \exp\left( \frac{2 \pi i m}{a} x \right)\exp\left( \frac{2 \pi i n}{b} y \right)\end{aligned} \hspace{\stretch{1}}(4.44)$

We were also provided with some definitions

\begin{definition}TE (Transverse Electric)

$\mathbf{E}_3 = 0$ .
\end{definition}
\begin{definition}
TM (Transverse Magnetic)

$\mathbf{B}_3 = 0$ .
\end{definition}
\begin{definition}
TM (Transverse Electromagnetic)

$\mathbf{E}_3 = \mathbf{B}_3 = 0$ .
\end{definition}

\begin{claim}TEM do not existing in a hollow waveguide.
\end{claim}

Why: I had in my notes

$\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E} = 0 & \implies \frac{\partial {E_2}}{\partial {x^1}} -\frac{\partial {E_1}}{\partial {x^2}} = 0 \\ \boldsymbol{\nabla} \cdot \mathbf{E} = 0 & \implies \frac{\partial {E_1}}{\partial {x^1}} +\frac{\partial {E_2}}{\partial {x^2}} = 0\end{aligned}$

and then

$\begin{aligned}\boldsymbol{\nabla}^2 \phi &= 0 \\ \phi &= \text{const}\end{aligned}$

In retrospect I fail to see how these are connected? What happened to the $\partial_t \mathbf{B}$ term in the curl equation above?

It was argued that we have $\mathbf{E}_\parallel = \mathbf{B}_\perp = 0$ on the boundary.

So for the TE case, where $\mathbf{E}_3 = 0$ , we have from the separation of variables argument

$\begin{aligned}\hat{\mathbf{z}} \cdot \mathbf{B}_0(x, y) =\sum_{m n} a_{m n} \cos\left( \frac{2 \pi i m}{a} x \right)\cos\left( \frac{2 \pi i n}{b} y \right).\end{aligned} \hspace{\stretch{1}}(4.45)$

No sines because

$\begin{aligned}B_1 \propto \frac{\partial {B_3}}{\partial {x_a}} \rightarrow \cos(k_1 x^1).\end{aligned} \hspace{\stretch{1}}(4.46)$

The quantity

$\begin{aligned}a_{m n}\cos\left( \frac{2 \pi i m}{a} x \right)\cos\left( \frac{2 \pi i n}{b} y \right).\end{aligned} \hspace{\stretch{1}}(4.47)$

is called the $TE_{m n}$ mode. Note that since $B = \text{const}$ an ampere loop requires $\mathbf{B} = 0$ since there is no current.

Writing

$\begin{aligned}k &= \frac{\omega}{c} \sqrt{ 1 - \left(\frac{\omega_{m n}}{\omega}\right)^2 } \\ \omega_{m n} &= 2 \pi c \sqrt{ \left(\frac{m}{a} \right)^2 + \left(\frac{n}{b} \right)^2 }\end{aligned} \hspace{\stretch{1}}(4.48)$

Since $\omega < \omega_{m n}$ we have $k$ purely imaginary, and the term

$\begin{aligned}e^{-i k z} = e^{- {\left\lvert{k}\right\rvert} z}\end{aligned} \hspace{\stretch{1}}(4.50)$

represents the die off.

$\omega_{10}$ is the smallest.

Note that the convention is that the $m$ in $TE_{m n}$ is the bigger of the two indexes, so $\omega > \omega_{10}$ .

The phase velocity

$\begin{aligned}V_\phi = \frac{\omega}{k} = \frac{c}{\sqrt{ 1 - \left(\frac{\omega_{m n}}{\omega}\right)^2 }} \ge c\end{aligned} \hspace{\stretch{1}}(4.51)$

However, energy is transmitted with the group velocity, the ratio of the Poynting vector and energy density

$\begin{aligned}\frac{\left\langle{\mathbf{S}}\right\rangle}{\left\langle{{U}}\right\rangle} = V_g = \frac{\partial {\omega}}{\partial {k}} = 1/\frac{\partial {k}}{\partial {\omega}}\end{aligned} \hspace{\stretch{1}}(4.52)$

(This can be shown).

Since

$\begin{aligned}\left(\frac{\partial {k}}{\partial {\omega}}\right)^{-1} = \left(\frac{\partial {}}{\partial {\omega}}\sqrt{ (\omega/c)^2 - (\omega_{m n}/c)^2 }\right)^{-1} = c \sqrt{ 1 - (\omega_{m n}/\omega)^2 } \le c\end{aligned} \hspace{\stretch{1}}(4.53)$

We see that the energy is transmitted at less than the speed of light as expected.

Final remarks.

I’d started converting my handwritten scrawl for this tutorial into an attempt at working through these ideas with enough detail that they self contained, but gave up part way. This appears to me to be too big of a sub-discipline to give it justice in one hours class. As is, it is enough to at least get an concept of some of the ideas involved. I think were I to learn this for real, I’d need a good text as a reference (or the time to attempt to blunder through the ideas in much much more detail).

Posted in Math and Physics Learning. | Tagged: electromagnetic waves, group velocity, maxwells equations, phase velocity, phasor, PHY450, separation of variables, transverse electric, transverse electromagnetic, transverse fields, transverse magnetic, waveguide | Leave a Comment »

PHY450H1S. Relativistic Electrodynamics Lecture 11 (Taught by Prof. Erich Poppitz). Unpacking Lorentz force equation. Lorentz transformations of the strength tensor, Lorentz field invariants, Bianchi identity, and first half of Maxwell’s.

Posted by peeterjoot on February 24, 2011

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Reading.

Covering chapter 3 material from the text [1].

Covering lecture notes pp. 74-83: Lorentz transformation of the strength tensor (82) [Tuesday, Feb. 8] [extra reading for the mathematically minded: gauge field, strength tensor, and gauge transformations in differential form language, not to be covered in class (83)]

Covering lecture notes pp. 84-102: Lorentz invariants of the electromagnetic field (84-86); Bianchi identity and the first half of Maxwell’s equations (87-90)

Chewing on the four vector form of the Lorentz force equation.

After much effort, we arrived at

$\begin{aligned}\frac{d{{(m c u_l) }}}{ds} = \frac{e}{c} \left( \partial_l A_i - \partial_i A_l \right) u^i\end{aligned} \hspace{\stretch{1}}(2.1)$

$\begin{aligned}\frac{d{{ p_l }}}{ds} = \frac{e}{c} F_{l i} u^i\end{aligned} \hspace{\stretch{1}}(2.2)$

Elements of the strength tensor

\paragraph{Claim}: there are only 6 independent elements of this matrix (tensor)

$\begin{aligned}\begin{bmatrix}0 & . & . & . \\ & 0 & . & . \\ & & 0 & . \\ & & & 0 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.3)$

This is a no-brainer, for we just have to mechanically plug in the elements of the field strength tensor

Recall

$\begin{aligned}A^i &= (\phi, \mathbf{A}) \\ A_i &= (\phi, -\mathbf{A})\end{aligned} \hspace{\stretch{1}}(2.4)$

$\begin{aligned}F_{0\alpha} &= \partial_0 A_\alpha - \partial_\alpha A_0 \\ &= -\partial_0 (\mathbf{A})_\alpha - \partial_\alpha \phi \\ \end{aligned}$

$\begin{aligned}F_{0\alpha} = E_\alpha\end{aligned} \hspace{\stretch{1}}(2.6)$

For the purely spatial index combinations we have

$\begin{aligned}F_{\alpha\beta} &= \partial_\alpha A_\beta - \partial_\beta A_\alpha \\ &= -\partial_\alpha (\mathbf{A})_\beta + \partial_\beta (\mathbf{A})_\alpha \\ \end{aligned}$

Written out explicitly, these are

$\begin{aligned}F_{1 2} &= \partial_2 (\mathbf{A})_1 -\partial_1 (\mathbf{A})_2 \\ F_{2 3} &= \partial_3 (\mathbf{A})_2 -\partial_2 (\mathbf{A})_3 \\ F_{3 1} &= \partial_1 (\mathbf{A})_3 -\partial_3 (\mathbf{A})_1 .\end{aligned} \hspace{\stretch{1}}(2.7)$

We can compare this to the elements of $\mathbf{B}$

$\begin{aligned}\mathbf{B} = \begin{vmatrix}\hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \\ \partial_1 & \partial_2 & \partial_3 \\ A_x & A_y & A_z\end{vmatrix}\end{aligned} \hspace{\stretch{1}}(2.10)$

We see that

$\begin{aligned}(\mathbf{B})_z &= \partial_1 A_y - \partial_2 A_x \\ (\mathbf{B})_x &= \partial_2 A_z - \partial_3 A_y \\ (\mathbf{B})_y &= \partial_3 A_x - \partial_1 A_z\end{aligned} \hspace{\stretch{1}}(2.11)$

So we have

$\begin{aligned}F_{1 2} &= - (\mathbf{B})_3 \\ F_{2 3} &= - (\mathbf{B})_1 \\ F_{3 1} &= - (\mathbf{B})_2.\end{aligned} \hspace{\stretch{1}}(2.14)$

These can be summarized as simply

$\begin{aligned}F_{\alpha\beta} = - \epsilon_{\alpha\beta\gamma} B_\gamma.\end{aligned} \hspace{\stretch{1}}(2.17)$

This provides all the info needed to fill in the matrix above

$\begin{aligned}{\left\lVert{ F_{i j} }\right\rVert} = \begin{bmatrix}0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0.\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.18)$

Index raising of rank 2 tensor

To raise indexes we compute

$\begin{aligned}F^{i j} = g^{i l} g^{j k} F_{l k}.\end{aligned} \hspace{\stretch{1}}(2.19)$

Justifying the raising operation.

To justify this consider raising one index at a time by applying the metric tensor to our definition of $F_{l k}$ . That is

$\begin{aligned}g^{a l} F_{l k} &=g^{a l} (\partial_l A_k - \partial_k A_l) \\ &=\partial^a A_k - \partial_k A^a.\end{aligned}$

Now apply the metric tensor once more

$\begin{aligned}g^{b k} g^{a l} F_{l k} &=g^{b k} (\partial^a A_k - \partial_k A^a) \\ &=\partial^a A^b - \partial^b A^a.\end{aligned}$

This is, by definition $F^{a b}$ . Since a rank 2 tensor has been defined as an object that transforms like the product of two pairs of coordinates, it makes sense that this particular tensor raises in the same fashion as would a product of two vector coordinates (in this case, it happens to be an antisymmetric product of two vectors, and one of which is an operator, but we have the same idea).

Consider the components of the raised $F_{i j}$ tensor.

$\begin{aligned}F^{0\alpha} &= -F_{0\alpha} \\ F^{\alpha\beta} &= F_{\alpha\beta}.\end{aligned} \hspace{\stretch{1}}(2.20)$

$\begin{aligned}{\left\lVert{ F^{i j} }\right\rVert} = \begin{bmatrix}0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.22)$

Back to chewing on the Lorentz force equation.

$\begin{aligned}m c \frac{d{{ u_i }}}{ds} = \frac{e}{c} F_{i j} u^j\end{aligned} \hspace{\stretch{1}}(2.23)$

$\begin{aligned}u^i &= \gamma \left( 1, \frac{\mathbf{v}}{c} \right) \\ u_i &= \gamma \left( 1, -\frac{\mathbf{v}}{c} \right)\end{aligned} \hspace{\stretch{1}}(2.24)$

For the spatial components of the Lorentz force equation we have

$\begin{aligned}m c \frac{d{{ u_\alpha }}}{ds} &= \frac{e}{c} F_{\alpha j} u^j \\ &= \frac{e}{c} F_{\alpha 0} u^0+ \frac{e}{c} F_{\alpha \beta} u^\beta \\ &= \frac{e}{c} (-E_{\alpha}) \gamma+ \frac{e}{c} (- \epsilon_{\alpha\beta\gamma} B_\gamma ) \frac{v^\beta}{c} \gamma \end{aligned}$

But

$\begin{aligned}m c \frac{d{{ u_\alpha }}}{ds} &= -m \frac{d{{(\gamma \mathbf{v}_\alpha)}}}{ds} \\ &= -m \frac{d(\gamma \mathbf{v}_\alpha)}{c \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} dt} \\ &= -\gamma \frac{d(m \gamma \mathbf{v}_\alpha)}{c dt}.\end{aligned}$

Canceling the common $-\gamma/c$ terms, and switching to vector notation, we are left with

$\begin{aligned}\frac{d( m \gamma \mathbf{v}_\alpha)}{dt} = e \left( E_\alpha + \frac{1}{{c}} (\mathbf{v} \times \mathbf{B})_\alpha \right).\end{aligned} \hspace{\stretch{1}}(2.26)$

Now for the energy term. We have

$\begin{aligned}m c \frac{d{{u_0}}}{ds} &= \frac{e}{c} F_{0\alpha} u^\alpha \\ &= \frac{e}{c} E_{\alpha} \gamma \frac{v^\alpha}{c} \\ \frac{d{{ m c \gamma }}}{ds} &=\end{aligned}$

Putting the final two lines into vector form we have

$\begin{aligned}\frac{d{{ (m c^2 \gamma)}}}{dt} = e \mathbf{E} \cdot \mathbf{v},\end{aligned} \hspace{\stretch{1}}(2.27)$

$\begin{aligned}\frac{d{{ \mathcal{E} }}}{dt} = e \mathbf{E} \cdot \mathbf{v}\end{aligned} \hspace{\stretch{1}}(2.28)$

Transformation of rank two tensors in matrix and index form.

Transformation of the metric tensor, and some identities.

With

$\begin{aligned}\hat{G} = {\left\lVert{ g_{i j} }\right\rVert} = {\left\lVert{ g^{i j} }\right\rVert}\end{aligned} \hspace{\stretch{1}}(3.29)$

\paragraph{We claim:}
The rank two tensor $\hat{G}$ transforms in the following sort of sandwich operation, and this leaves it invariant

$\begin{aligned}\hat{G} \rightarrow \hat{O} \hat{G} \hat{O}^\text{T} = \hat{G}.\end{aligned} \hspace{\stretch{1}}(3.30)$

To demonstrate this let’s consider a transformed vector in coordinate form as follows

$\begin{aligned}{x'}^i &= O^{i j} x_j = {O^i}_j x^j \\ {x'}_i &= O_{i j} x^j = {O_i}^j x_j.\end{aligned} \hspace{\stretch{1}}(3.31)$

We can thus write the equation in matrix form with

$\begin{aligned}X &= {\left\lVert{x^i}\right\rVert} \\ X' &= {\left\lVert{{x'}^i}\right\rVert} \\ \hat{O} &= {\left\lVert{{O^i}_j}\right\rVert} \\ X' &= \hat{O} X\end{aligned} \hspace{\stretch{1}}(3.33)$

Our invariant for the vector square, which is required to remain unchanged is

$\begin{aligned}{x'}^i {x'}_i &= (O^{i j} x_j)(O_{i k} x^k) \\ &= x^k (O^{i j} O_{i k}) x_j.\end{aligned}$

This shows that we have a delta function relationship for the Lorentz transform matrix, when we sum over the first index

$\begin{aligned}O^{a i} O_{a j} = {\delta^i}_j.\end{aligned} \hspace{\stretch{1}}(3.37)$

It appears we can put 3.37 into matrix form as

$\begin{aligned}\hat{G} \hat{O}^\text{T} \hat{G} \hat{O} = I\end{aligned} \hspace{\stretch{1}}(3.38)$

Now, if one considers that the transpose of a rotation is an inverse rotation, and the transpose of a boost leaves it unchanged, the transpose of a general Lorentz transformation, a composition of an arbitrary sequence of boosts and rotations, must also be a Lorentz transformation, and must then also leave the norm unchanged. For the transpose of our Lorentz transformation $\hat{O}$ lets write

$\begin{aligned}\hat{P} = \hat{O}^\text{T}\end{aligned} \hspace{\stretch{1}}(3.39)$

For the action of this on our position vector let’s write

$\begin{aligned}{x''}^i &= P^{i j} x_j = O^{j i} x_j \\ {x''}_i &= P_{i j} x^j = O_{j i} x^j\end{aligned} \hspace{\stretch{1}}(3.40)$

so that our norm is

$\begin{aligned}{x''}^a {x''}_a &= (O_{k a} x^k)(O^{j a} x_j) \\ &= x^k (O_{k a} O^{j a} ) x_j \\ &= x^j x_j \\ \end{aligned}$

We must then also have an identity when summing over the second index

$\begin{aligned}{\delta_{k}}^j = O_{k a} O^{j a} \end{aligned} \hspace{\stretch{1}}(3.42)$

Armed with these facts on the products of $O_{i j}$ and $O^{i j}$ we can now consider the transformation of the metric tensor.

The rule (definition) supplied to us for the transformation of an arbitrary rank two tensor, is that this transforms as its indexes transform individually. Very much as if it was the product of two coordinate vectors and we transform those coordinates separately. Doing so for the metric tensor we have

$\begin{aligned}g^{i j} &\rightarrow {O^i}_k g^{k m} {O^j}_m \\ &= ({O^i}_k g^{k m}) {O^j}_m \\ &= O^{i m} {O^j}_m \\ &= O^{i m} (O_{a m} g^{a j}) \\ &= (O^{i m} O_{a m}) g^{a j}\end{aligned}$

However, by 3.42, we have $O_{a m} O^{i m} = {\delta_a}^i$ , and we prove that

$\begin{aligned}g^{i j} \rightarrow g^{i j}.\end{aligned} \hspace{\stretch{1}}(3.43)$

Finally, we wish to put the above transformation in matrix form, look more carefully at the very first line

$\begin{aligned}g^{i j}&\rightarrow {O^i}_k g^{k m} {O^j}_m \\ \end{aligned}$

which is

$\begin{aligned}\hat{G} \rightarrow \hat{O} \hat{G} \hat{O}^\text{T} = \hat{G}\end{aligned} \hspace{\stretch{1}}(3.44)$

We see that this particular form of transformation, a sandwich between $\hat{O}$ and $\hat{O}^\text{T}$ , leaves the metric tensor invariant.

Lorentz transformation of the electrodynamic tensor

Having identified a composition of Lorentz transformation matrices, when acting on the metric tensor, leaves it invariant, it is a reasonable question to ask how this form of transformation acts on our electrodynamic tensor $F^{i j}$ ?

\paragraph{Claim:} A transformation of the following form is required to maintain the norm of the Lorentz force equation

$\begin{aligned}\hat{F} \rightarrow \hat{O} \hat{F} \hat{O}^\text{T} ,\end{aligned} \hspace{\stretch{1}}(3.45)$

where $\hat{F} = {\left\lVert{F^{i j}}\right\rVert}$ . Observe that our Lorentz force equation can be written exclusively in upper index quantities as

$\begin{aligned}m c \frac{d{{u^i}}}{ds} = \frac{e}{c} F^{i j} g_{j l} u^l\end{aligned} \hspace{\stretch{1}}(3.46)$

Because we have a vector on one side of the equation, and it transforms by multiplication with by a Lorentz matrix in SO(1,3)

$\begin{aligned}\frac{du^i}{ds} \rightarrow \hat{O} \frac{du^i}{ds} \end{aligned} \hspace{\stretch{1}}(3.47)$

The LHS of the Lorentz force equation provides us with one invariant

$\begin{aligned}(m c)^2 \frac{d{{u^i}}}{ds} \frac{d{{u_i}}}{ds}\end{aligned} \hspace{\stretch{1}}(3.48)$

so the RHS must also provide one

$\begin{aligned}\frac{e^2}{c^2} F^{i j} g_{j l} u^lF_{i k} g^{k m} u_m=\frac{e^2}{c^2} F^{i j} u_jF_{i k} u^k.\end{aligned} \hspace{\stretch{1}}(3.49)$

Let’s look at the RHS in matrix form. Writing

$\begin{aligned}U = {\left\lVert{u^i}\right\rVert},\end{aligned} \hspace{\stretch{1}}(3.50)$

we can rewrite the Lorentz force equation as

$\begin{aligned}m c \dot{U} = \frac{e}{c} \hat{F} \hat{G} U.\end{aligned} \hspace{\stretch{1}}(3.51)$

In this matrix formalism our invariant 3.49 is

$\begin{aligned}\frac{e^2}{c^2} (\hat{F} \hat{G} U)^\text{T} G \hat{F} \hat{G} U=\frac{e^2}{c^2} U^\text{T} \hat{G} \hat{F}^\text{T} G \hat{F} \hat{G} U.\end{aligned} \hspace{\stretch{1}}(3.52)$

If we compare this to the transformed Lorentz force equation we have

$\begin{aligned}m c \hat{O} \dot{U} = \frac{e}{c} \hat{F'} \hat{G} \hat{O} U.\end{aligned} \hspace{\stretch{1}}(3.53)$

Our invariant for the transformed equation is

$\begin{aligned}\frac{e^2}{c^2} (\hat{F'} \hat{G} \hat{O} U)^\text{T} G \hat{F'} \hat{G} \hat{O} U&=\frac{e^2}{c^2} U^\text{T} \hat{O}^\text{T} \hat{G} \hat{F'}^\text{T} G \hat{F'} \hat{G} \hat{O} U \\ \end{aligned}$

Thus the transformed electrodynamic tensor $\hat{F}'$ must satisfy the identity

$\begin{aligned}\hat{O}^\text{T} \hat{G} \hat{F'}^\text{T} G \hat{F'} \hat{G} \hat{O} = \hat{G} \hat{F}^\text{T} G \hat{F} \hat{G} \end{aligned} \hspace{\stretch{1}}(3.54)$

With the substitution $\hat{F}' = \hat{O} \hat{F} \hat{O}^\text{T}$ the LHS is

$\begin{aligned}\hat{O}^\text{T} \hat{G} \hat{F'}^\text{T} \hat{G} \hat{F'} \hat{G} \hat{O} &= \hat{O}^\text{T} \hat{G} ( \hat{O} \hat{F} \hat{O}^\text{T})^\T \hat{G} (\hat{O} \hat{F} \hat{O}^\text{T}) \hat{G} \hat{O} \\ &= (\hat{O}^\text{T} \hat{G} \hat{O}) \hat{F}^\text{T} (\hat{O}^\text{T} \hat{G} \hat{O}) \hat{F} (\hat{O}^\text{T} \hat{G} \hat{O}) \\ \end{aligned}$

We’ve argued that $\hat{P} = \hat{O}^\text{T}$ is also a Lorentz transformation, thus

$\begin{aligned}\hat{O}^\text{T} \hat{G} \hat{O}&=\hat{P} \hat{G} \hat{O}^\text{T} \\ &=\hat{G}\end{aligned}$

This is enough to make both sides of 3.54 match, verifying that this transformation does provide the invariant properties desired.

Direct computation of the Lorentz transformation of the electrodynamic tensor.

We can construct the transformed field tensor more directly, by simply transforming the coordinates of the four gradient and the four potential directly. That is

$\begin{aligned}F^{i j} = \partial^i A^j - \partial^j A^i&\rightarrow {O^i}_a {O^j}_b \left( \partial^a A^b - \partial^b A^a \right) \\ &={O^i}_a F^{a b} {O^j}_b \end{aligned}$

By inspection we can see that this can be represented in matrix form as

$\begin{aligned}\hat{F} \rightarrow \hat{O} \hat{F} \hat{O}^\text{T}\end{aligned} \hspace{\stretch{1}}(3.55)$

Four vector invariants

For three vectors $\mathbf{A}$ and $\mathbf{B}$ invariants are

$\begin{aligned}\mathbf{A} \cdot \mathbf{B} = A^\alpha B_\alpha\end{aligned} \hspace{\stretch{1}}(4.56)$

For four vectors $A^i$ and $B^i$ invariants are

$\begin{aligned}A^i B_i = A^i g_{i j} B^j \end{aligned} \hspace{\stretch{1}}(4.57)$

For $F_{i j}$ what are the invariants? One invariant is

$\begin{aligned}g^{i j} F_{i j} = 0,\end{aligned} \hspace{\stretch{1}}(4.58)$

but this isn’t interesting since it is uniformly zero (product of symmetric and antisymmetric).

The two invariants are

$\begin{aligned}F_{i j}F^{i j}\end{aligned} \hspace{\stretch{1}}(4.59)$

and

$\begin{aligned}\epsilon^{i j k l} F_{i j}F_{k l}\end{aligned} \hspace{\stretch{1}}(4.60)$

where

$\begin{aligned}\epsilon^{i j k l} =\left\{\begin{array}{l l}0 & \quad \mbox{if any two indexes coincide} \\ 1 & \quad \mbox{for even permutations of$ i j k l=0123$ } \\ -1 & \quad \mbox{for odd permutations of $i j k l=0123$ } \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(4.61)$

We can show (homework) that

$\begin{aligned}F_{i j}F^{i j} \propto \mathbf{E}^2 - \mathbf{B}^2\end{aligned} \hspace{\stretch{1}}(4.62)$

$\begin{aligned}\epsilon^{i j k l} F_{i j}F_{k l} \propto \mathbf{E} \cdot \mathbf{B}\end{aligned} \hspace{\stretch{1}}(4.63)$

This first invariant serves as the action density for the Maxwell field equations.

There’s some useful properties of these invariants. One is that if the fields are perpendicular in one frame, then will be in any other.

From the first, note that if ${\left\lvert{\mathbf{E}}\right\rvert} > {\left\lvert{\mathbf{B}}\right\rvert}$ , the invariant is positive, and must be positive in all frames, or if ${\left\lvert{\mathbf{E}}\right\rvert} {\left\lvert{\mathbf{B}}\right\rvert}$ in one frame, we can transform to a frame with only $\mathbf{E}'$ component, solve that, and then transform back. Similarly if ${\left\lvert{\mathbf{E}}\right\rvert} < {\left\lvert{\mathbf{B}}\right\rvert}$ in one frame, we can transform to a frame with only $\mathbf{B}'$ component, solve that, and then transform back.

The first half of Maxwell’s equations.

\paragraph{Claim: } The source free portions of Maxwell’s equations are a consequence of the definition of the field tensor alone.

Given

$\begin{aligned}F_{i j} = \partial_i A_j - \partial_j A_i,\end{aligned} \hspace{\stretch{1}}(5.64)$

where

$\begin{aligned}\partial_i = \frac{\partial {}}{\partial {x^i}}\end{aligned} \hspace{\stretch{1}}(5.65)$

This alone implies half of Maxwell’s equations. To show this we consider

$\begin{aligned}e^{m k i j} \partial_k F_{i j} = 0.\end{aligned} \hspace{\stretch{1}}(5.66)$

This is the Bianchi identity. To demonstrate this identity, we’ll have to swap indexes, employ derivative commutation, and then swap indexes once more

$\begin{aligned}e^{m k i j} \partial_k F_{i j} &= e^{m k i j} \partial_k (\partial_i A_j - \partial_j A_i) \\ &= 2 e^{m k i j} \partial_k \partial_i A_j \\ &= 2 e^{m k i j} \frac{1}{{2}} \left( \partial_k \partial_i A_j + \partial_i \partial_k A_j \right) \\ &= e^{m k i j} \partial_k \partial_i A_j e^{m i k j} \partial_k \partial_i A_j \\ &= (e^{m k i j} - e^{m k i j}) \partial_k \partial_i A_j \\ &= 0 \qquad \square\end{aligned}$

This is the 4D analogue of

$\begin{aligned}\boldsymbol{\nabla} \times (\boldsymbol{\nabla} f) = 0\end{aligned} \hspace{\stretch{1}}(5.67)$

i.e.

$\begin{aligned}e^{\alpha\beta\gamma} \partial_\beta \partial_\gamma f = 0\end{aligned} \hspace{\stretch{1}}(5.68)$

Let’s do this explicitly, starting with

$\begin{aligned}{\left\lVert{ F_{i j} }\right\rVert} = \begin{bmatrix}0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0.\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(5.69)$

For the $m= 0$ case we have

$\begin{aligned}\epsilon^{0 k i j} \partial_k F_{i j}&=\epsilon^{\alpha \beta \gamma} \partial_\alpha F_{\beta \gamma} \\ &= \epsilon^{\alpha \beta \gamma} \partial_\alpha (-\epsilon_{\beta \gamma \delta} B_\delta) \\ &= -\epsilon^{\alpha \beta \gamma} \epsilon_{\delta \beta \gamma }\partial_\alpha B_\delta \\ &= - 2 {\delta^\alpha}_\delta \partial_\alpha B_\delta \\ &= - 2 \partial_\alpha B_\alpha \end{aligned}$

We must then have

$\begin{aligned}\partial_\alpha B_\alpha = 0.\end{aligned} \hspace{\stretch{1}}(5.70)$

This is just Gauss’s law for magnetism

$\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{B} = 0.\end{aligned} \hspace{\stretch{1}}(5.71)$

Let’s do the spatial portion, for which we have three equations, one for each $\alpha$ of

$\begin{aligned}e^{\alpha j k l} \partial_j F_{k l}&=e^{\alpha 0 \beta \gamma} \partial_0 F_{\beta \gamma}+e^{\alpha 0 \gamma \beta} \partial_0 F_{\gamma \beta}+e^{\alpha \beta 0 \gamma} \partial_\beta F_{0 \gamma}+e^{\alpha \beta \gamma 0} \partial_\beta F_{\gamma 0}+e^{\alpha \gamma 0 \beta} \partial_\gamma F_{0 \beta}+e^{\alpha \gamma \beta 0} \partial_\gamma F_{\beta 0} \\ &=2 \left( e^{\alpha 0 \beta \gamma} \partial_0 F_{\beta \gamma}+e^{\alpha \beta 0 \gamma} \partial_\beta F_{0 \gamma}+e^{\alpha \gamma 0 \beta} \partial_\gamma F_{0 \beta}\right) \\ &=2 e^{0 \alpha \beta \gamma} \left(-\partial_0 F_{\beta \gamma}+\partial_\beta F_{0 \gamma}- \partial_\gamma F_{0 \beta}\right)\end{aligned}$

This implies

$\begin{aligned}0 =-\partial_0 F_{\beta \gamma}+\partial_\beta F_{0 \gamma}- \partial_\gamma F_{0 \beta}\end{aligned} \hspace{\stretch{1}}(5.72)$

Referring back to the previous expansions of 2.6 and 2.17, we have

$\begin{aligned}0 =\partial_0 \epsilon_{\beta\gamma\mu} B_\mu+\partial_\beta E_\gamma- \partial_\gamma E_{\beta},\end{aligned} \hspace{\stretch{1}}(5.73)$

$\begin{aligned}\frac{1}{{c}} \frac{\partial {B_\alpha}}{\partial {t}} + (\boldsymbol{\nabla} \times \mathbf{E})_\alpha = 0.\end{aligned} \hspace{\stretch{1}}(5.74)$

These are just the components of the Maxwell-Faraday equation

$\begin{aligned}0 = \frac{1}{{c}} \frac{\partial {\mathbf{B}}}{\partial {t}} + \boldsymbol{\nabla} \times \mathbf{E}.\end{aligned} \hspace{\stretch{1}}(5.75)$

Appendix. Some additional index gymnastics.

Transposition of mixed index tensor.

Is the transpose of a mixed index object just a substitution of the free indexes? This wasn’t obvious to me that it would be the case, especially since I’d made an error in some index gymnastics that had me temporarily convinced differently. However, working some examples clears the fog. For example let’s take the transpose of 3.37.

$\begin{aligned}{\left\lVert{ {\delta^i}_j }\right\rVert}^\text{T} &= {\left\lVert{ O^{a i} O_{a j} }\right\rVert}^\text{T} \\ &= \left( {\left\lVert{ O^{j i} }\right\rVert} {\left\lVert{ O_{i j} }\right\rVert} \right)^\text{T} \\ &={\left\lVert{ O_{i j} }\right\rVert}^\text{T}{\left\lVert{ O^{j i} }\right\rVert}^\text{T} \\ &={\left\lVert{ O_{j i} }\right\rVert}{\left\lVert{ O^{i j} }\right\rVert} \\ &={\left\lVert{ O_{a i} O^{a j} }\right\rVert} \\ \end{aligned}$

If the transpose of a mixed index tensor just swapped the indexes we would have

$\begin{aligned}{\left\lVert{ {\delta^i}_j }\right\rVert}^\text{T} = {\left\lVert{ O_{a i} O^{a j} }\right\rVert} \end{aligned} \hspace{\stretch{1}}(6.76)$

From this it does appear that all we have to do is switch the indexes and we will write

$\begin{aligned}{\delta^j}_i = O_{a i} O^{a j} \end{aligned} \hspace{\stretch{1}}(6.77)$

We can consider a more general operation

$\begin{aligned}{\left\lVert{{A^i}_j}\right\rVert}^\text{T}&={\left\lVert{ A^{i m} g_{m j} }\right\rVert}^\text{T} \\ &={\left\lVert{ g_{i j} }\right\rVert}^\text{T}{\left\lVert{ A^{i j} }\right\rVert}^\text{T} \\ &={\left\lVert{ g_{i j} }\right\rVert}{\left\lVert{ A^{j i} }\right\rVert} \\ &={\left\lVert{ g_{i m} A^{j m} }\right\rVert} \\ &={\left\lVert{ {A^{j}}_i }\right\rVert}\end{aligned}$

So we see that we do just have to swap indexes.

Transposition of lower index tensor.

We’ve saw above that we had

$\begin{aligned}{\left\lVert{ {A^{i}}_j }\right\rVert}^\text{T} &= {\left\lVert{ {A_{j}}^i }\right\rVert} \\ {\left\lVert{ {A_{i}}^j }\right\rVert}^\text{T} &= {\left\lVert{ {A^{j}}_i }\right\rVert} \end{aligned} \hspace{\stretch{1}}(6.78)$

which followed by careful treatment of the transposition in terms of $A^{i j}$ for which we defined a transpose operation. We assumed as well that

$\begin{aligned}{\left\lVert{ A_{i j} }\right\rVert}^\text{T} = {\left\lVert{ A_{j i} }\right\rVert}.\end{aligned} \hspace{\stretch{1}}(6.80)$

However, this does not have to be assumed, provided that $g^{i j} = g_{i j}$ , and $(AB)^\text{T} = B^\text{T} A^\text{T}$ . We see this by expanding this transposition in products of $A^{i j}$ and $\hat{G}$

$\begin{aligned}{\left\lVert{ A_{i j} }\right\rVert}^\text{T}&= \left( {\left\lVert{g_{i j}}\right\rVert} {\left\lVert{ A^{i j} }\right\rVert} {\left\lVert{g_{i j}}\right\rVert} \right)^\text{T} \\ &= \left( {\left\lVert{g^{i j}}\right\rVert} {\left\lVert{ A^{i j} }\right\rVert} {\left\lVert{g^{i j}}\right\rVert} \right)^\text{T} \\ &= {\left\lVert{g^{i j}}\right\rVert}^\text{T} {\left\lVert{ A^{i j}}\right\rVert}^\text{T} {\left\lVert{g^{i j}}\right\rVert}^\text{T} \\ &= {\left\lVert{g^{i j}}\right\rVert} {\left\lVert{ A^{j i}}\right\rVert} {\left\lVert{g^{i j}}\right\rVert} \\ &= {\left\lVert{g_{i j}}\right\rVert} {\left\lVert{ A^{i j}}\right\rVert} {\left\lVert{g_{i j}}\right\rVert} \\ &= {\left\lVert{ A_{j i}}\right\rVert} \end{aligned}$

It would be worthwhile to go through all of this index manipulation stuff and lay it out in a structured axiomatic form. What is the minimal set of assumptions, and how does all of this generalize to non-diagonal metric tensors (even in Euclidean spaces).

Translating the index expression of identity from Lorentz products to matrix form

A verification that the matrix expression 3.38, matches the index expression 3.37 as claimed is worthwhile. It would be easy to guess something similar like $\hat{O}^\text{T} \hat{G} \hat{O} \hat{G}$ is instead the matrix representation. That was in fact my first erroneous attempt to form the matrix equivalent, but is the transpose of 3.38. Either way you get an identity, but the indexes didn’t match.

Since we have $g^{i j} = g_{i j}$ which do we pick to do this verification? This appears to be dictated by requirements to match lower and upper indexes on the summed over index. This is probably clearest by example, so let’s expand the products on the LHS explicitly

$\begin{aligned}{\left\lVert{ g^{i j} }\right\rVert} {\left\lVert{ {O^{i}}_j }\right\rVert} ^\text{T}{\left\lVert{ g_{i j} }\right\rVert}{\left\lVert{ {O^{i}}_j }\right\rVert} &=\left( {\left\lVert{ {O^{i}}_j }\right\rVert} {\left\lVert{ g^{i j} }\right\rVert} \right) ^\text{T}{\left\lVert{ g_{i j} }\right\rVert}{\left\lVert{ {O^{i}}_j }\right\rVert} \\ &=\left( {\left\lVert{ {O^{i}}_k g^{k j} }\right\rVert} \right) ^\text{T}{\left\lVert{ g_{i m} {O^{m}}_j }\right\rVert} \\ &={\left\lVert{ O^{i j} }\right\rVert} ^\text{T}{\left\lVert{ O_{i j} }\right\rVert} \\ &={\left\lVert{ O^{j i} }\right\rVert} {\left\lVert{ O_{i j} }\right\rVert} \\ &={\left\lVert{ O^{k i} O_{k j} }\right\rVert} \\ \end{aligned}$

This matches the ${\left\lVert{{\delta^i}_j}\right\rVert}$ that we have on the RHS, and all is well.

References

[1] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980.

Posted in Math and Physics Learning. | Tagged: 3), Bianchi identity, cross product, curl, electric field, electrodynamic strength tensor, field invariants, four potential, four vector, Gauss's law, gradient, index lowering, index raising, Lorentz field transformations, Lorentz force, Lorentz invariant, magnetic field, Maxwell Faraday equation, maxwells equations, metric tensor, PHY450, SO(1, tensor algebra | Leave a Comment »

PHY450H1S. Relativistic Electrodynamics Lecture 14 (Taught by Simon Freedman). Wave equation in Coulomb and Lorentz gauges.

Posted by peeterjoot on February 17, 2011

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

Reading.

Covering chapter 4 material from the text [1].

Covering lecture notes pp.103-114: the wave equation in the relativistic Lorentz gauge (114-114) [Tuesday, Feb. 15; Wednesday, Feb.16]…

Covering lecture notes pp. 114-127: reminder on wave equations (114); reminder on Fourier series and integral (115-117); Fourier expansion of the EM potential in Coulomb gauge and equation of motion for the spatial Fourier components (118-119); the general solution of Maxwell’s equations in vacuum (120-121) [Tuesday, Mar. 1]; properties of monochromatic plane EM waves (122-124); energy and energy flux of the EM field and energy conservation from the equations of motion (125-127) [Wednesday, Mar. 2]

Trying to understand “c”

$\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} &= 0 \\ \boldsymbol{\nabla} \times \mathbf{B} &= \frac{1}{{c}} \frac{\partial {\mathbf{E}}}{\partial {t}}\end{aligned} \hspace{\stretch{1}}(2.1)$

Maxwell’s equations in a vacuum were

$\begin{aligned}\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A}) &= \boldsymbol{\nabla}^2 \mathbf{A} -\frac{1}{{c}} \frac{\partial {}}{\partial {t}} \boldsymbol{\nabla} \phi - \frac{1}{{c^2}} \frac{\partial^2 \mathbf{A}}{\partial t^2} \\ \boldsymbol{\nabla} \cdot \mathbf{E} &= - \boldsymbol{\nabla}^2 \phi - \frac{1}{{c}} \frac{\partial {\boldsymbol{\nabla} \cdot \mathbf{A}}}{\partial {t}} \end{aligned} \hspace{\stretch{1}}(2.3)$

There’s a redundancy here since we can change $\phi$ and $\mathbf{A}$ without changing the EOM

$\begin{aligned}(\phi, \mathbf{A}) \rightarrow (\phi', \mathbf{A}')\end{aligned} \hspace{\stretch{1}}(2.5)$

with

$\begin{aligned}\phi &= \phi' + \frac{1}{{c}} \frac{\partial {\chi}}{\partial {t}} \\ \mathbf{A} &= \mathbf{A}' - \boldsymbol{\nabla} \chi\end{aligned} \hspace{\stretch{1}}(2.6)$

$\begin{aligned}\chi(\mathbf{x}, t) = c \int dt \phi(\mathbf{x}, t)\end{aligned} \hspace{\stretch{1}}(2.8)$

which gives

$\begin{aligned}\phi' = 0\end{aligned} \hspace{\stretch{1}}(2.9)$

$\begin{aligned}(\phi, \mathbf{A}) \sim (\phi = 0, \mathbf{A}')\end{aligned} \hspace{\stretch{1}}(2.10)$

Maxwell’s equations are now

$\begin{aligned}\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A}') &= \boldsymbol{\nabla}^2 \mathbf{A}' - \frac{1}{{c^2}} \frac{\partial^2 \mathbf{A}'}{\partial t^2} \\ \frac{\partial {\boldsymbol{\nabla} \cdot \mathbf{A}'}}{\partial {t}} &= 0\end{aligned}$

Can we make $\boldsymbol{\nabla} \cdot \mathbf{A}'' = 0$ , while $\phi'' = 0$ .

$\begin{aligned}\underbrace{\phi}_{=0} &= \underbrace{\phi'}_{=0} + \frac{1}{{c}} \frac{\partial {\chi'}}{\partial {t}} \\ \end{aligned} \hspace{\stretch{1}}(2.11)$

We need

$\begin{aligned}\frac{\partial {\chi'}}{\partial {t}} = 0\end{aligned} \hspace{\stretch{1}}(2.13)$

How about $\mathbf{A}'$

$\begin{aligned}\mathbf{A}' = \mathbf{A}'' - \boldsymbol{\nabla} \chi'\end{aligned} \hspace{\stretch{1}}(2.14)$

We want the divergence of $\mathbf{A}'$ to be zero, which means

$\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{A}' = \underbrace{\boldsymbol{\nabla} \cdot \mathbf{A}''}_{=0} - \boldsymbol{\nabla}^2 \chi'\end{aligned} \hspace{\stretch{1}}(2.15)$

So we want

$\begin{aligned}\boldsymbol{\nabla}^2 \chi' = \boldsymbol{\nabla} \cdot \mathbf{A}'\end{aligned} \hspace{\stretch{1}}(2.16)$

Can we solve this?

Recall that in electrostatics we have

$\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} = 4 \pi \rho\end{aligned} \hspace{\stretch{1}}(2.17)$

and

$\begin{aligned}\mathbf{E} = -\boldsymbol{\nabla} \phi\end{aligned} \hspace{\stretch{1}}(2.18)$

which meant that we had

$\begin{aligned}\boldsymbol{\nabla}^2 \phi = 4 \pi \rho\end{aligned} \hspace{\stretch{1}}(2.19)$

This has the identical form (with $\phi \sim \chi$ , and $4 \pi \rho \sim \boldsymbol{\nabla} \cdot \mathbf{A}'$ ).

While we aren’t trying to actually solve this (just show that it can be solved). One way to look at this problem is that it is just a Laplace equation, and we could utilize a Green’s function solution if desired.

On the Green’s function.

Recall that the Green’s function for the Laplacian was

$\begin{aligned}G(\mathbf{x}, \mathbf{x}') = \frac{1}{{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}}\end{aligned} \hspace{\stretch{1}}(2.20)$

with the property

$\begin{aligned}\boldsymbol{\nabla}^2 G(\mathbf{x}, \mathbf{x}') = \delta(\mathbf{x} - \mathbf{x}')\end{aligned} \hspace{\stretch{1}}(2.21)$

Our LDE to solve by Green’s method is

$\begin{aligned}\boldsymbol{\nabla}^2 \phi = 4 \pi \rho,\end{aligned} \hspace{\stretch{1}}(2.22)$

We let this equation (after switching to primed coordinates) operate on the Green’s function

$\begin{aligned}\int d^3 \mathbf{x}' {\boldsymbol{\nabla}'}^2 \phi(\mathbf{x}') G(\mathbf{x}, \mathbf{x}') =\int d^3 \mathbf{x}' 4 \pi \phi(\mathbf{x}') G(\mathbf{x}, \mathbf{x}').\end{aligned} \hspace{\stretch{1}}(2.23)$

Assuming that the left action of the Green’s function on the test function $\phi(\mathbf{x}')$ is the same as the right action (i.e. $\phi(\mathbf{x}')$ and $G(\mathbf{x}, \mathbf{x}')$ commute), we have for the LHS

$\begin{aligned}\int d^3 \mathbf{x}' {\boldsymbol{\nabla}'}^2 \phi(\mathbf{x}') G(\mathbf{x}, \mathbf{x}') &=\int d^3 \mathbf{x}' {\boldsymbol{\nabla}'}^2 G(\mathbf{x}, \mathbf{x}') \phi(\mathbf{x}') \\ &=\int d^3 \mathbf{x}' \delta(\mathbf{x} - \mathbf{x}') \phi(\mathbf{x}') \\ &=\phi(\mathbf{x}).\end{aligned}$

Substitution of $G(\mathbf{x}, \mathbf{x}') = 1/{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}$ on the RHS then gives us the general solution

$\begin{aligned}\phi(\mathbf{x}) = 4 \pi \int d^3 \mathbf{x}' \frac{\rho(\mathbf{x}') }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}\end{aligned} \hspace{\stretch{1}}(2.24)$

Back to Maxwell’s equations in vacuum.

What are the Maxwell’s vacuum equations now?

With the second gauge substitution we have

$\begin{aligned}\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A}'') &= \boldsymbol{\nabla}^2 \mathbf{A}'' - \frac{1}{{c^2}} \frac{\partial^2 \mathbf{A}''}{\partial t^2} \\ \frac{\partial {\boldsymbol{\nabla} \cdot \mathbf{A}''}}{\partial {t}} &= 0\end{aligned}$

but we can utilize

$\begin{aligned}\boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{A}) = \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A}) - \boldsymbol{\nabla}^2 \mathbf{A},\end{aligned} \hspace{\stretch{1}}(2.25)$

to reduce Maxwell’s equations (after dropping primes) to just

$\begin{aligned}\frac{1}{{c^2}} \frac{\partial^2 \mathbf{A}''}{\partial t^2} - \Delta \mathbf{A} = 0\end{aligned} \hspace{\stretch{1}}(2.26)$

where

$\begin{aligned}\Delta = \boldsymbol{\nabla}^2 = \boldsymbol{\nabla} \cdot \boldsymbol{\nabla} = \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial y^2}\end{aligned} \hspace{\stretch{1}}(2.27)$

Note that for this to be correct we have to also explicitly include the gauge condition used. This particular gauge is called the \underline{Coulomb gauge}.

$\begin{aligned}\phi &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{A}'' &= 0 \end{aligned} \hspace{\stretch{1}}(2.28)$

Claim: EM waves propagate with speed $c$ and are transverse.

\paragraph{Note:} Is the Coulomb gauge Lorentz invariant?
\paragraph{No.} We can boost which will introduce a non-zero $\phi$ .

The gauge that is Lorentz Invariant is the “Lorentz gauge”. This one uses

$\begin{aligned}\partial_i A^i = 0\end{aligned} \hspace{\stretch{1}}(3.30)$

Recall that Maxwell’s equations are

$\begin{aligned}\partial_i F^{ij} = j^j = 0\end{aligned} \hspace{\stretch{1}}(3.31)$

where

$\begin{aligned}\partial_i &= \frac{\partial {}}{\partial {x^i}} \\ \partial^i &= \frac{\partial {}}{\partial {x_i}}\end{aligned} \hspace{\stretch{1}}(3.32)$

Writing out the equations in terms of potentials we have

$\begin{aligned}0 &= \partial_i (\partial^i A^j - \partial^j A^i) \\ &= \partial_i \partial^i A^j - \partial_i \partial^j A^i \\ &= \partial_i \partial^i A^j - \partial^j \partial_i A^i \\ \end{aligned}$

So, if we pick the gauge condition $\partial_i A^i = 0$ , we are left with just

$\begin{aligned}0 = \partial_i \partial^i A^j\end{aligned} \hspace{\stretch{1}}(3.34)$

Can we choose ${A'}^i$ such that $\partial_i A^i = 0$ ?

Our gauge condition is

$\begin{aligned}A^i = {A'}^i + \partial^i \chi\end{aligned} \hspace{\stretch{1}}(3.35)$

Hit it with a derivative for

$\begin{aligned}\partial_i A^i = \partial_i {A'}^i + \partial_i \partial^i \chi\end{aligned} \hspace{\stretch{1}}(3.36)$

If we want $\partial_i A^i = 0$ , then we have

$\begin{aligned}-\partial_i {A'}^i = \partial_i \partial^i \chi = \left( \frac{1}{{c^2}} \frac{\partial^2}{\partial t^2} - \Delta \right) \chi\end{aligned} \hspace{\stretch{1}}(3.37)$

This is the physicist proof. Yes, it can be solved. To really solve this, we’d want to use Green’s functions. I seem to recall the Green’s function is a retarded time version of the Laplacian Green’s function, and we can figure that exact form out by switching to a Fourier frequency domain representation.

Anyways. Returning to Maxwell’s equations we have

$\begin{aligned}0 &= \partial_i \partial^i A^j \\ 0 &= \partial_i A^i ,\end{aligned} \hspace{\stretch{1}}(3.38)$

where the first is Maxwell’s equation, and the second is our gauge condition.

Observe that the gauge condition is now a Lorentz scalar.

$\begin{aligned}\partial^i A_i \rightarrow \partial^j {O_j}^i {O_i}^k A_k\end{aligned} \hspace{\stretch{1}}(3.40)$

But the Lorentz transform matrices multiply out to identity, in the same way that they do for the transformation of a plain old four vector dot product $x^i y_i$ .

What happens with a Massive vector field?

$\begin{aligned}S = \int d^4 x \left( \frac{1}{{4}} F^{ij} F_{ij} + \frac{m^2}{2} A^i A_i \right)\end{aligned} \hspace{\stretch{1}}(4.41)$

An aside on units

“Note that this action is expressed in dimensions where $\hbar = c = 1$ , making the action is unit-less (energy and time are inverse units of each other). The $d^4x$ has units of $m^{-4}$ (since $[x] = \hbar/mc$ ), so $F$ has units of $m^2$ , and then $A$ has units of mass. Therefore $d^4x A A$ has units of $m^{-2}$ and therefore you need something that has units of $m^2$ to make the action unit-less. When you don’t take $c=1$ , then you’ve got to worry about those factors, but I think you’ll see it works out fine.”

For what it’s worth, I can adjust the units of this action to those that we’ve used in class with,

$\begin{aligned}S = \int d^4 x \left( -\frac{1}{{16 \pi c}} F^{ij} F_{ij} - \frac{m^2 c^2}{8 \hbar^2} A^i A_i \right)\end{aligned} \hspace{\stretch{1}}(4.42)$

Back to the problem.

The variation of the field invariant is

$\begin{aligned}\delta (F_{ij} F^{ij})&=2 (\delta F_{ij}) F^{ij}) \\ &=2 (\delta(\partial_i A_j -\partial_j A_i)) F^{ij}) \\ &=2 (\partial_i \delta(A_j) -\partial_j \delta(A_i)) F^{ij}) \\ &=4 F^{ij} \partial_i \delta(A_j) \\ &=4 \partial_i (F^{ij} \delta(A_j)) - 4 (\partial_i F^{ij}) \delta(A_j).\end{aligned}$

Variation of the $A^2$ term gives us

$\begin{aligned}\delta (A^j A_j) = 2 A^j \delta(A_j),\end{aligned} \hspace{\stretch{1}}(4.43)$

so we have

$\begin{aligned}0 &= \delta S \\ &= \int d^4 x \delta(A_j) \left( -\partial_i F^{ij} + m^2 A^j \right)+ \int d^4 x \partial_i (F^{ij} \delta(A_j))\end{aligned}$

The last integral vanishes on the boundary with the assumption that $\delta(A_j) = 0$ on that boundary.

Since this must be true for all variations, this leaves us with

$\begin{aligned}\partial_i F^{ij} = m^2 A^j\end{aligned} \hspace{\stretch{1}}(4.44)$

The RHS can be expanded into wave equation and divergence parts

$\begin{aligned}\partial_i F^{ij}&=\partial_i (\partial^i A^j - \partial^j A^i) \\ &=(\partial_i \partial^i) A^j - \partial^j (\partial_i A^i) \\ \end{aligned}$

With $\square$ for the wave equation operator

$\begin{aligned}\square = \partial_i \partial^i = \frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta,\end{aligned} \hspace{\stretch{1}}(4.45)$

we can manipulate the EOM to pull out an $A_i$ factor

$\begin{aligned}0 &= \left( \square -m^2 \right) A^j - \partial^j (\partial_i A^i) \\ &= \left( \square -m^2 \right) g^{ij} A_i - \partial^j (\partial^i A_i) \\ &= \left( \left( \square -m^2 \right) g^{ij} - \partial^j \partial^i \right) A_i.\end{aligned}$

If we hit this with a derivative we get

$\begin{aligned}0 &= \partial_j \left( \left( \square -m^2 \right) g^{ij} - \partial^j \partial^i \right) A_i \\ &= \left( \left( \square -m^2 \right) \partial^i - \partial_j \partial^j \partial^i \right) A_i \\ &= \left( \left( \square -m^2 \right) \partial^i - \square \partial^i \right) A_i \\ &= \left( \square -m^2 - \square \right) \partial^i A_i \\ &= -m^2 \partial^i A_i \\ \end{aligned}$

Since $m$ is presumed to be non-zero here, this means that the Lorentz gauge is already chosen for us by the equations of motion.

References

[1] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980.

Posted in Math and Physics Learning. | Tagged: Coulomb gauge, gauge transformation, Lorentz Gauge, maxwells equations, PHY450, speed of light, wave equation | Leave a Comment »

Fourier transform solutions and associated energy and momentum for the homogeneous Maxwell equation. (rework once more)

Posted by peeterjoot on December 29, 2009

[Click here for a PDF of this post with nicer formatting]. Note that this PDF file is formatted in a wide-for-screen layout that is probably not good for printing.

These notes build on and replace those formerly posted in Energy and momentum for assumed Fourier transform solutions to the homogeneous Maxwell equation.

Motivation and notation.

In Electrodynamic field energy for vacuum (reworked) [1], building on Energy and momentum for Complex electric and magnetic field phasors [2], a derivation for the energy and momentum density was derived for an assumed Fourier series solution to the homogeneous Maxwell’s equation. Here we move to the continuous case examining Fourier transform solutions and the associated energy and momentum density.

A complex (phasor) representation is implied, so taking real parts when all is said and done is required of the fields. For the energy momentum tensor the Geometric Algebra form, modified for complex fields, is used

$\begin{aligned}T(a) = -\frac{\epsilon_0}{2} \text{Real} \Bigl( {{F}}^{*} a F \Bigr).\end{aligned} \hspace{\stretch{1}}(1.1)$

The assumed four vector potential will be written

$\begin{aligned}A(\mathbf{x}, t) = A^\mu(\mathbf{x}, t) \gamma_\mu = \frac{1}{{(\sqrt{2 \pi})^3}} \int A(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(1.2)$

Subject to the requirement that $A$ is a solution of Maxwell’s equation

$\begin{aligned}\nabla (\nabla \wedge A) = 0.\end{aligned} \hspace{\stretch{1}}(1.3)$

To avoid latex hell, no special notation will be used for the Fourier coefficients,

$\begin{aligned}A(\mathbf{k}, t) = \frac{1}{{(\sqrt{2 \pi})^3}} \int A(\mathbf{x}, t) e^{-i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{x}.\end{aligned} \hspace{\stretch{1}}(1.4)$

When convenient and unambiguous, this $(\mathbf{k},t)$ dependence will be implied.

Having picked a time and space representation for the field, it will be natural to express both the four potential and the gradient as scalar plus spatial vector, instead of using the Dirac basis. For the gradient this is

$\begin{aligned}\nabla &= \gamma^\mu \partial_\mu = (\partial_0 - \boldsymbol{\nabla}) \gamma_0 = \gamma_0 (\partial_0 + \boldsymbol{\nabla}),\end{aligned} \hspace{\stretch{1}}(1.5)$

and for the four potential (or the Fourier transform functions), this is

$\begin{aligned}A &= \gamma_\mu A^\mu = (\phi + \mathbf{A}) \gamma_0 = \gamma_0 (\phi - \mathbf{A}).\end{aligned} \hspace{\stretch{1}}(1.6)$

Setup

The field bivector $F = \nabla \wedge A$ is required for the energy momentum tensor. This is

$\begin{aligned}\nabla \wedge A&= \frac{1}{{2}}\left( \stackrel{ \rightarrow }{\nabla} A - A \stackrel{ \leftarrow }{\nabla} \right) \\ &= \frac{1}{{2}}\left( (\stackrel{ \rightarrow }{\partial}_0 - \stackrel{ \rightarrow }{\boldsymbol{\nabla}}) \gamma_0 \gamma_0 (\phi - \mathbf{A})-(\phi + \mathbf{A}) \gamma_0 \gamma_0 (\stackrel{ \leftarrow }{\partial}_0 + \stackrel{ \leftarrow }{\boldsymbol{\nabla}})\right) \\ &= -\boldsymbol{\nabla} \phi -\partial_0 \mathbf{A} + \frac{1}{{2}}(\stackrel{ \rightarrow }{\boldsymbol{\nabla}} \mathbf{A} - \mathbf{A} \stackrel{ \leftarrow }{\boldsymbol{\nabla}})\end{aligned}$

This last term is a spatial curl and the field is then

$\begin{aligned}F = -\boldsymbol{\nabla} \phi -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A}\end{aligned} \hspace{\stretch{1}}(2.7)$

Applied to the Fourier representation this is

$\begin{aligned}F =\frac{1}{{(\sqrt{2 \pi})^3}} \int\left(- \frac{1}{c} \dot{\mathbf{A}}- i \mathbf{k} \phi+ i \mathbf{k} \wedge \mathbf{A}\right)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(2.8)$

It is only the real parts of this that we are actually interested in, unless physical meaning can be assigned to the complete complex vector field.

Constraints supplied by Maxwell’s equation.

A Fourier transform solution of Maxwell’s vacuum equation $\nabla F = 0$ has been assumed. Having expressed the Faraday bivector in terms of spatial vector quantities, it is more convenient to do this back substitution into after pre-multiplying Maxwell’s equation by $\gamma_0$ , namely

$\begin{aligned}0&= \gamma_0 \nabla F \\ &= (\partial_0 + \boldsymbol{\nabla}) F.\end{aligned} \hspace{\stretch{1}}(3.9)$

Applied to the spatially decomposed field as specified in (2.7), this is

$\begin{aligned}0&=-\partial_0 \boldsymbol{\nabla} \phi-\partial_{00} \mathbf{A}+ \partial_0 \boldsymbol{\nabla} \wedge \mathbf{A}-\boldsymbol{\nabla}^2 \phi- \boldsymbol{\nabla} \partial_0 \mathbf{A}+ \boldsymbol{\nabla} \cdot (\boldsymbol{\nabla} \wedge \mathbf{A} ) \\ &=- \partial_0 \boldsymbol{\nabla} \phi - \boldsymbol{\nabla}^2 \phi- \partial_{00} \mathbf{A}- \boldsymbol{\nabla} \cdot \partial_0 \mathbf{A}+ \boldsymbol{\nabla}^2 \mathbf{A} - \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A} ) \\ \end{aligned}$

All grades of this equation must simultaneously equal zero, and the bivector grades have canceled (assuming commuting space and time partials), leaving two equations of constraint for the system

$\begin{aligned}0 &=\boldsymbol{\nabla}^2 \phi + \boldsymbol{\nabla} \cdot \partial_0 \mathbf{A}\end{aligned} \hspace{\stretch{1}}(3.11)$

$\begin{aligned}0 &=\partial_{00} \mathbf{A} - \boldsymbol{\nabla}^2 \mathbf{A}+ \boldsymbol{\nabla} \partial_0 \phi + \boldsymbol{\nabla} ( \boldsymbol{\nabla} \cdot \mathbf{A} )\end{aligned} \hspace{\stretch{1}}(3.12)$

It is immediately evident that a gauge transformation could be immediately helpful to simplify things. In [3] the gauge choice $\boldsymbol{\nabla} \cdot \mathbf{A} = 0$ is used. From (3.11) this implies that $\boldsymbol{\nabla}^2 \phi = 0$ . Bohm argues that for this current and charge free case this implies $\phi = 0$ , but he also has a periodicity constraint. Without a periodicity constraint it is easy to manufacture non-zero counterexamples. One is a linear function in the space and time coordinates

$\begin{aligned}\phi = p x + q y + r z + s t\end{aligned} \hspace{\stretch{1}}(3.13)$

This is a valid scalar potential provided that the wave equation for the vector potential is also a solution. We can however, force $\phi = 0$ by making the transformation $A^\mu \rightarrow A^\mu + \partial^\mu \psi$ , which in non-covariant notation is

$\begin{aligned}\phi &\rightarrow \phi + \frac{1}{c} \partial_t \psi \\ \mathbf{A} &\rightarrow \phi - \boldsymbol{\nabla} \psi\end{aligned} \hspace{\stretch{1}}(3.14)$

If the transformed field $\phi' = \phi + \partial_t \psi/c$ can be forced to zero, then the complexity of the associated Maxwell equations are reduced. In particular, antidifferentiation of $\phi = -(1/c) \partial_t \psi$ , yields

$\begin{aligned}\psi(\mathbf{x},t) = \psi(\mathbf{x}, 0) - c \int_{\tau=0}^t \phi(\mathbf{x}, \tau) d\tau.\end{aligned} \hspace{\stretch{1}}(3.16)$

Dropping primes, the transformed Maxwell equations now take the form

$\begin{aligned}0 &= \partial_t( \boldsymbol{\nabla} \cdot \mathbf{A} )\end{aligned} \hspace{\stretch{1}}(3.17)$

$\begin{aligned}0 &=\partial_{00} \mathbf{A} - \boldsymbol{\nabla}^2 \mathbf{A} + \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A} ).\end{aligned} \hspace{\stretch{1}}(3.18)$

There are two classes of solutions that stand out for these equations. If the vector potential is constant in time $\mathbf{A}(\mathbf{x},t) = \mathbf{A}(\mathbf{x})$ , Maxwell’s equations are reduced to the single equation

$\begin{aligned}0&= - \boldsymbol{\nabla}^2 \mathbf{A} + \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A} ).\end{aligned} \hspace{\stretch{1}}(3.19)$

Observe that a gradient can be factored out of this equation

$\begin{aligned}- \boldsymbol{\nabla}^2 \mathbf{A} + \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A} )&=\boldsymbol{\nabla} (-\boldsymbol{\nabla} \mathbf{A} + \boldsymbol{\nabla} \cdot \mathbf{A} ) \\ &=-\boldsymbol{\nabla} (\boldsymbol{\nabla} \wedge \mathbf{A}).\end{aligned}$

The solutions are then those $\mathbf{A}$ s that satisfy both

$\begin{aligned}0 &= \partial_t \mathbf{A} \\ 0 &= \boldsymbol{\nabla} (\boldsymbol{\nabla} \wedge \mathbf{A}).\end{aligned} \hspace{\stretch{1}}(3.20)$

In particular any non-time dependent potential $\mathbf{A}$ with constant curl provides a solution to Maxwell’s equations. There may be other solutions to (3.19) too that are more general. Returning to (3.17) a second way to satisfy these equations stands out. Instead of requiring of $\mathbf{A}$ constant curl, constant divergence with respect to the time partial eliminates (3.17). The simplest resulting equations are those for which the divergence is a constant in time and space (such as zero). The solution set are then spanned by the vectors $\mathbf{A}$ for which

$\begin{aligned}\text{constant} &= \boldsymbol{\nabla} \cdot \mathbf{A} \end{aligned} \hspace{\stretch{1}}(3.22)$

$\begin{aligned}0 &= \frac{1}{{c^2}} \partial_{tt} \mathbf{A} - \boldsymbol{\nabla}^2 \mathbf{A}.\end{aligned} \hspace{\stretch{1}}(3.23)$

Any $\mathbf{A}$ that both has constant divergence and satisfies the wave equation will via (2.7) then produce a solution to Maxwell’s equation.

Maxwell equation constraints applied to the assumed Fourier solutions.

Let’s consider Maxwell’s equations in all three forms, (3.11), (3.20), and (3.22) and apply these constraints to the assumed Fourier solution.

In all cases the starting point is a pair of Fourier transform relationships, where the Fourier transforms are the functions to be determined

$\begin{aligned}\phi(\mathbf{x}, t) &= (2 \pi)^{-3/2} \int \phi(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k} \end{aligned} \hspace{\stretch{1}}(4.24)$

$\begin{aligned}\mathbf{A}(\mathbf{x}, t) &= (2 \pi)^{-3/2} \int \mathbf{A}(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k} \end{aligned} \hspace{\stretch{1}}(4.25)$

Case I. Constant time vector potential. Scalar potential eliminated by gauge transformation.

From (4.24) we require

$\begin{aligned}0 = (2 \pi)^{-3/2} \int \partial_t \mathbf{A}(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.26)$

So the Fourier transform also cannot have any time dependence, and we have

$\begin{aligned}\mathbf{A}(\mathbf{x}, t) &= (2 \pi)^{-3/2} \int \mathbf{A}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k} \end{aligned} \hspace{\stretch{1}}(4.27)$

What is the curl of this? Temporarily falling back to coordinates is easiest for this calculation

$\begin{aligned}\boldsymbol{\nabla} \wedge \mathbf{A}(\mathbf{k}) e^{i\mathbf{k} \cdot \mathbf{x}}&=\sigma_m \partial_m \wedge \sigma_n A^n(\mathbf{k}) e^{i \mathbf{x} \cdot \mathbf{x}} \\ &=\sigma_m \wedge \sigma_n A^n(\mathbf{k}) i k^m e^{i \mathbf{x} \cdot \mathbf{x}} \\ &=i\mathbf{k} \wedge \mathbf{A}(\mathbf{k}) e^{i \mathbf{x} \cdot \mathbf{x}} \\ \end{aligned}$

This gives

$\begin{aligned}\boldsymbol{\nabla} \wedge \mathbf{A}(\mathbf{x}, t) &= (2 \pi)^{-3/2} \int i \mathbf{k} \wedge \mathbf{A}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.28)$

We want to equate the divergence of this to zero. Neglecting the integral and constant factor this requires

$\begin{aligned}0 &= \boldsymbol{\nabla} \cdot \left( i \mathbf{k} \wedge \mathbf{A} e^{i\mathbf{k} \cdot \mathbf{x}} \right) \\ &= {\left\langle{{ \sigma_m \partial_m i (\mathbf{k} \wedge \mathbf{A}) e^{i\mathbf{k} \cdot \mathbf{x}} }}\right\rangle}_{1} \\ &= -{\left\langle{{ \sigma_m (\mathbf{k} \wedge \mathbf{A}) k^m e^{i\mathbf{k} \cdot \mathbf{x}} }}\right\rangle}_{1} \\ &= -\mathbf{k} \cdot (\mathbf{k} \wedge \mathbf{A}) e^{i\mathbf{k} \cdot \mathbf{x}} \\ \end{aligned}$

Requiring that the plane spanned by $\mathbf{k}$ and $\mathbf{A}(\mathbf{k})$ be perpendicular to $\mathbf{k}$ implies that $\mathbf{A} \propto \mathbf{k}$ . The solution set is then completely described by functions of the form

$\begin{aligned}\mathbf{A}(\mathbf{x}, t) &= (2 \pi)^{-3/2} \int \mathbf{k} \psi(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k},\end{aligned} \hspace{\stretch{1}}(4.29)$

where $\psi(\mathbf{k})$ is an arbitrary scalar valued function. This is however, an extremely uninteresting solution since the curl is uniformly zero

$\begin{aligned}F &= \boldsymbol{\nabla} \wedge \mathbf{A} \\ &= (2 \pi)^{-3/2} \int (i \mathbf{k}) \wedge \mathbf{k} \psi(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned}$

Since $\mathbf{k} \wedge \mathbf{k} = 0$ , when all is said and done the $\phi = 0$ , $\partial_t \mathbf{A} = 0$ case appears to have no non-trivial (zero) solutions. Moving on, …

Case II. Constant vector potential divergence. Scalar potential eliminated by gauge transformation.

Next in the order of complexity is consideration of the case (3.22). Here we also have $\phi = 0$ , eliminated by gauge transformation, and are looking for solutions with the constraint

$\begin{aligned}\text{constant} &= \boldsymbol{\nabla} \cdot \mathbf{A}(\mathbf{x}, t) \\ &= (2 \pi)^{-3/2} \int i \mathbf{k} \cdot \mathbf{A}(\mathbf{k}, t) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned}$

How can this constraint be enforced? The only obvious way is a requirement for $\mathbf{k} \cdot \mathbf{A}(\mathbf{k}, t)$ to be zero for all $(\mathbf{k},t)$ , meaning that our to be determined Fourier transform coefficients are required to be perpendicular to the wave number vector parameters at all times.

The remainder of Maxwell’s equations, (3.23) impose the addition constraint on the Fourier transform $\mathbf{A}(\mathbf{k},t)$

$\begin{aligned}0 &= (2 \pi)^{-3/2} \int \left( \frac{1}{{c^2}} \partial_{tt} \mathbf{A}(\mathbf{k}, t) - i^2 \mathbf{k}^2 \mathbf{A}(\mathbf{k}, t)\right) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.30)$

For zero equality for all $\mathbf{x}$ it appears that we require the Fourier transforms $\mathbf{A}(\mathbf{k})$ to be harmonic in time

$\begin{aligned}\partial_{tt} \mathbf{A}(\mathbf{k}, t) = - c^2 \mathbf{k}^2 \mathbf{A}(\mathbf{k}, t).\end{aligned} \hspace{\stretch{1}}(4.31)$

This has the familiar exponential solutions

$\begin{aligned}\mathbf{A}(\mathbf{k}, t) = \mathbf{A}_{\pm}(\mathbf{k}) e^{ \pm i c {\left\lvert{\mathbf{k}}\right\rvert} t },\end{aligned} \hspace{\stretch{1}}(4.32)$

also subject to a requirement that $\mathbf{k} \cdot \mathbf{A}(\mathbf{k}) = 0$ . Our field, where the $\mathbf{A}_{\pm}(\mathbf{k})$ are to be determined by initial time conditions, is by (2.7) of the form

$\begin{aligned}F(\mathbf{x}, t)= \text{Real} \frac{i}{(\sqrt{2\pi})^3} \int \Bigl( -{\left\lvert{\mathbf{k}}\right\rvert} \mathbf{A}_{+}(\mathbf{k}) + \mathbf{k} \wedge \mathbf{A}_{+}(\mathbf{k}) \Bigr) \exp(i \mathbf{k} \cdot \mathbf{x} + i c {\left\lvert{\mathbf{k}}\right\rvert} t) d^3 \mathbf{k}+ \text{Real} \frac{i}{(\sqrt{2\pi})^3} \int \Bigl( {\left\lvert{\mathbf{k}}\right\rvert} \mathbf{A}_{-}(\mathbf{k}) + \mathbf{k} \wedge \mathbf{A}_{-}(\mathbf{k}) \Bigr) \exp(i \mathbf{k} \cdot \mathbf{x} - i c {\left\lvert{\mathbf{k}}\right\rvert} t) d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.33)$

Since $0 = \mathbf{k} \cdot \mathbf{A}_{\pm}(\mathbf{k})$ , we have $\mathbf{k} \wedge \mathbf{A}_{\pm}(\mathbf{k}) = \mathbf{k} \mathbf{A}_{\pm}$ . This allows for factoring out of ${\left\lvert{\mathbf{k}}\right\rvert}$ . The structure of the solution is not changed by incorporating the $i (2\pi)^{-3/2} {\left\lvert{\mathbf{k}}\right\rvert}$ factors into $\mathbf{A}_{\pm}$ , leaving the field having the general form

$\begin{aligned}F(\mathbf{x}, t)= \text{Real} \int ( \hat{\mathbf{k}} - 1 ) \mathbf{A}_{+}(\mathbf{k}) \exp(i \mathbf{k} \cdot \mathbf{x} + i c {\left\lvert{\mathbf{k}}\right\rvert} t) d^3 \mathbf{k}+ \text{Real} \int ( \hat{\mathbf{k}} + 1 ) \mathbf{A}_{-}(\mathbf{k}) \exp(i \mathbf{k} \cdot \mathbf{x} - i c {\left\lvert{\mathbf{k}}\right\rvert} t) d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.34)$

The original meaning of $\mathbf{A}_{\pm}$ as Fourier transforms of the vector potential is obscured by the tidy up change to absorb ${\left\lvert{\mathbf{k}}\right\rvert}$ , but the geometry of the solution is clearer this way.

It is also particularly straightforward to confirm that $\gamma_0 \nabla F = 0$ separately for either half of (4.34).

Case III. Non-zero scalar potential. No gauge transformation.

Now lets work from (3.11). In particular, a divergence operation can be factored from (3.11), for

$\begin{aligned}0 = \boldsymbol{\nabla} \cdot (\boldsymbol{\nabla} \phi + \partial_0 \mathbf{A}).\end{aligned} \hspace{\stretch{1}}(4.35)$

Right off the top, there is a requirement for

$\begin{aligned}\text{constant} = \boldsymbol{\nabla} \phi + \partial_0 \mathbf{A}.\end{aligned} \hspace{\stretch{1}}(4.36)$

In terms of the Fourier transforms this is

$\begin{aligned}\text{constant} = \frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(i \mathbf{k} \phi(\mathbf{k}, t) + \frac{1}{c} \partial_t \mathbf{A}(\mathbf{k}, t)\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.37)$

Are there any ways for this to equal a constant for all $\mathbf{x}$ without requiring that constant to be zero? Assuming no for now, and that this constant must be zero, this implies a coupling between the $\phi$ and $\mathbf{A}$ Fourier transforms of the form

$\begin{aligned}\phi(\mathbf{k}, t) = -\frac{1}{{i c \mathbf{k}}} \partial_t \mathbf{A}(\mathbf{k}, t)\end{aligned} \hspace{\stretch{1}}(4.38)$

A secondary implication is that $\partial_t \mathbf{A}(\mathbf{k}, t) \propto \mathbf{k}$ or else $\phi(\mathbf{k}, t)$ is not a scalar. We had a transverse solution by requiring via gauge transformation that $\phi = 0$ , and here we have instead the vector potential in the propagation direction.

A secondary confirmation that this is a required coupling between the scalar and vector potential can be had by evaluating the divergence equation of (4.35)

$\begin{aligned}0 = \frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(- \mathbf{k}^2 \phi(\mathbf{k}, t) + \frac{i\mathbf{k}}{c} \cdot \partial_t \mathbf{A}(\mathbf{k}, t)\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.39)$

Rearranging this also produces (4.38). We want to now substitute this relationship into (3.12).

Starting with just the $\partial_0 \phi - \boldsymbol{\nabla} \cdot \mathbf{A}$ part we have

$\begin{aligned}\partial_0 \phi + \boldsymbol{\nabla} \cdot \mathbf{A}&=\frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(\frac{i}{c^2 \mathbf{k}} \partial_{tt} \mathbf{A}(\mathbf{k}, t) + i \mathbf{k} \cdot \mathbf{A}\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.40)$

Taking the gradient of this brings down a factor of $i\mathbf{k}$ for

$\begin{aligned}\boldsymbol{\nabla} (\partial_0 \phi + \boldsymbol{\nabla} \cdot \mathbf{A})&=-\frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(\frac{1}{c^2} \partial_{tt} \mathbf{A}(\mathbf{k}, t) + \mathbf{k} (\mathbf{k} \cdot \mathbf{A})\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.41)$

(3.12) in its entirety is now

$\begin{aligned}0 &=\frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(- (i\mathbf{k})^2 \mathbf{A}+ \mathbf{k} (\mathbf{k} \cdot \mathbf{A})\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.42)$

This isn’t terribly pleasant looking. Perhaps going the other direction. We could write

$\begin{aligned}\phi = \frac{i}{c \mathbf{k}} \frac{\partial {\mathbf{A}}}{\partial {t}} = \frac{i}{c} \frac{\partial {\psi}}{\partial {t}},\end{aligned} \hspace{\stretch{1}}(4.43)$

so that

$\begin{aligned}\mathbf{A}(\mathbf{k}, t) = \mathbf{k} \psi(\mathbf{k}, t).\end{aligned} \hspace{\stretch{1}}(4.44)$

$\begin{aligned}0 &=\frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(\frac{1}{{c^2}} \mathbf{k} \psi_{tt}- \boldsymbol{\nabla}^2 \mathbf{k} \psi + \boldsymbol{\nabla} \frac{i}{c^2} \psi_{tt}+\boldsymbol{\nabla}( \boldsymbol{\nabla} \cdot (\mathbf{k} \psi) )\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k} \\ \end{aligned}$

Note that the gradients here operate on everything to the right, including and especially the exponential. Each application of the gradient brings down an additional $i\mathbf{k}$ factor, and we have

$\begin{aligned}\frac{1}{{(\sqrt{2 \pi})^3}} \int \mathbf{k} \Bigl(\frac{1}{{c^2}} \psi_{tt}- i^2 \mathbf{k}^2 \psi + \frac{i^2}{c^2} \psi_{tt}+i^2 \mathbf{k}^2 \psi \Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned}$

This is identically zero, so we see that this second equation provides no additional information. That is somewhat surprising since there is not a whole lot of constraints supplied by the first equation. The function $\psi(\mathbf{k}, t)$ can be anything. Understanding of this curiosity comes from computation of the Faraday bivector itself. From (2.7), that is

$\begin{aligned}F = \frac{1}{{(\sqrt{2 \pi})^3}} \int \Bigl(-i \mathbf{k} \frac{i}{c}\psi_t - \frac{1}{c} \mathbf{k} \psi_t + i \mathbf{k} \wedge \mathbf{k} \psi\Bigr)e^{i \mathbf{k} \cdot \mathbf{x} } d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(4.45)$

All terms cancel, so we see that a non-zero $\phi$ leads to $F = 0$ , as was the case when considering (4.24) (a case that also resulted in $\mathbf{A}(\mathbf{k}) \propto \mathbf{k}$ ).

Can this Fourier representation lead to a non-transverse solution to Maxwell’s equation? If so, it is not obvious how.

The energy momentum tensor

The energy momentum tensor is then

$\begin{aligned}T(a) &= -\frac{\epsilon_0}{2 (2 \pi)^3} \text{Real} \iint\left(- \frac{1}{c} {{\dot{\mathbf{A}}}}^{*}(\mathbf{k}',t)+ i \mathbf{k}' {{\phi}}^{*}(\mathbf{k}', t)- i \mathbf{k}' \wedge {\mathbf{A}}^{*}(\mathbf{k}', t)\right)a\left(- \frac{1}{c} \dot{\mathbf{A}}(\mathbf{k}, t)- i \mathbf{k} \phi(\mathbf{k}, t)+ i \mathbf{k} \wedge \mathbf{A}(\mathbf{k}, t)\right)e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x} } d^3 \mathbf{k} d^3 \mathbf{k}'.\end{aligned} \hspace{\stretch{1}}(5.46)$

Observing that $\gamma_0$ commutes with spatial bivectors and anticommutes with spatial vectors, and writing $\sigma_\mu = \gamma_\mu \gamma_0$ , the tensor splits neatly into scalar and spatial vector components

$\begin{aligned}T(\gamma_\mu) \cdot \gamma_0 &= \frac{\epsilon_0}{2 (2 \pi)^3} \text{Real} \iint\left\langle{{\left(\frac{1}{c} {{\dot{\mathbf{A}}}}^{*}(\mathbf{k}',t)- i \mathbf{k}' {{\phi}}^{*}(\mathbf{k}', t)+ i \mathbf{k}' \wedge {\mathbf{A}}^{*}(\mathbf{k}', t)\right)\sigma_\mu\left(\frac{1}{c} \dot{\mathbf{A}}(\mathbf{k}, t)+ i \mathbf{k} \phi(\mathbf{k}, t)+ i \mathbf{k} \wedge \mathbf{A}(\mathbf{k}, t)\right)}}\right\rangle e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x} } d^3 \mathbf{k} d^3 \mathbf{k}' \\ T(\gamma_\mu) \wedge \gamma_0 &= \frac{\epsilon_0}{2 (2 \pi)^3} \text{Real} \iint{\left\langle{{\left(\frac{1}{c} {{\dot{\mathbf{A}}}}^{*}(\mathbf{k}',t)- i \mathbf{k}' {{\phi}}^{*}(\mathbf{k}', t)+ i \mathbf{k}' \wedge {\mathbf{A}}^{*}(\mathbf{k}', t)\right)\sigma_\mu\left(\frac{1}{c} \dot{\mathbf{A}}(\mathbf{k}, t)+ i \mathbf{k} \phi(\mathbf{k}, t)+ i \mathbf{k} \wedge \mathbf{A}(\mathbf{k}, t)\right)}}\right\rangle}_{1}e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x} } d^3 \mathbf{k} d^3 \mathbf{k}'.\end{aligned} \hspace{\stretch{1}}(5.47)$

In particular for $\mu = 0$ , we have

$\begin{aligned}H &\equiv T(\gamma_0) \cdot \gamma_0 = \frac{\epsilon_0}{2 (2 \pi)^3} \text{Real} \iint\left(\left(\frac{1}{c} {{\dot{\mathbf{A}}}}^{*}(\mathbf{k}',t)- i \mathbf{k}' {{\phi}}^{*}(\mathbf{k}', t)\right)\cdot\left(\frac{1}{c} \dot{\mathbf{A}}(\mathbf{k}, t)+ i \mathbf{k} \phi(\mathbf{k}, t)\right)- (\mathbf{k}' \wedge {\mathbf{A}}^{*}(\mathbf{k}', t)) \cdot (\mathbf{k} \wedge \mathbf{A}(\mathbf{k}, t))\right)e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x} } d^3 \mathbf{k} d^3 \mathbf{k}' \\ \mathbf{P} &\equiv T(\gamma_\mu) \wedge \gamma_0 = \frac{\epsilon_0}{2 (2 \pi)^3} \text{Real} \iint\left(i\left(\frac{1}{c} {{\dot{\mathbf{A}}}}^{*}(\mathbf{k}',t)- i \mathbf{k}' {{\phi}}^{*}(\mathbf{k}', t)\right) \cdot\left(\mathbf{k} \wedge \mathbf{A}(\mathbf{k}, t)\right)-i\left(\frac{1}{c} \dot{\mathbf{A}}(\mathbf{k}, t)+ i \mathbf{k} \phi(\mathbf{k}, t)\right)\cdot\left(\mathbf{k}' \wedge {\mathbf{A}}^{*}(\mathbf{k}', t)\right)\right)e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x} } d^3 \mathbf{k} d^3 \mathbf{k}'.\end{aligned} \hspace{\stretch{1}}(5.49)$

Integrating this over all space and identification of the delta function

$\begin{aligned}\delta(\mathbf{k}) \equiv \frac{1}{{(2 \pi)^3}} \int e^{i \mathbf{k} \cdot \mathbf{x}} d^3 \mathbf{x},\end{aligned} \hspace{\stretch{1}}(5.51)$

reduces the tensor to a single integral in the continuous angular wave number space of $\mathbf{k}$ .

$\begin{aligned}\int T(a) d^3 \mathbf{x} &= -\frac{\epsilon_0}{2} \text{Real} \int\left(- \frac{1}{c} {{\dot{\mathbf{A}}}}^{*}+ i \mathbf{k} {{\phi}}^{*}- i \mathbf{k} \wedge {\mathbf{A}}^{*}\right)a\left(- \frac{1}{c} \dot{\mathbf{A}}- i \mathbf{k} \phi+ i \mathbf{k} \wedge \mathbf{A}\right)d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(5.52)$

Or,

$\begin{aligned}\int T(\gamma_\mu) \gamma_0 d^3 \mathbf{x} =\frac{\epsilon_0}{2} \text{Real} \int{\left\langle{{\left(\frac{1}{c} {{\dot{\mathbf{A}}}}^{*}- i \mathbf{k} {{\phi}}^{*}+ i \mathbf{k} \wedge {\mathbf{A}}^{*}\right)\sigma_\mu\left(\frac{1}{c} \dot{\mathbf{A}}+ i \mathbf{k} \phi+ i \mathbf{k} \wedge \mathbf{A}\right)}}\right\rangle}_{{0,1}}d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(5.53)$

Multiplying out (5.53) yields for $\int H$

$\begin{aligned}\int H d^3 \mathbf{x} &=\frac{\epsilon_0}{2} \int d^3 \mathbf{k} \left(\frac{1}{{c^2}} {\left\lvert{\dot{\mathbf{A}}}\right\rvert}^2 + \mathbf{k}^2 ({\left\lvert{\phi}\right\rvert}^2 + {\left\lvert{\mathbf{A}}\right\rvert}^2 )- {\left\lvert{\mathbf{k} \cdot \mathbf{A}}\right\rvert}^2+ 2 \frac{\mathbf{k}}{c} \cdot \text{Real}( i {{\phi}}^{*} \dot{\mathbf{A}} )\right)\end{aligned} \hspace{\stretch{1}}(5.54)$

Recall that the only non-trivial solution we found for the assumed Fourier transform representation of $F$ was for $\phi = 0$ , $\mathbf{k} \cdot \mathbf{A}(\mathbf{k}, t) = 0$ . Thus we have for the energy density integrated over all space, just

Observe that we have the structure of a Harmonic oscillator for the energy of the radiation system. What is the canonical momentum for this system? Will it correspond to the Poynting vector, integrated over all space?

Let’s reduce the vector component of (5.53), after first imposing the $\phi=0$ , and $\mathbf{k} \cdot \mathbf{A} = 0$ conditions used to above for our harmonic oscillator form energy relationship. This is

$\begin{aligned}\int \mathbf{P} d^3 \mathbf{x} &=\frac{\epsilon_0}{2 c} \text{Real} \int d^3 \mathbf{k} \left( i {\mathbf{A}}^{*}_t \cdot (\mathbf{k} \wedge \mathbf{A})+ i (\mathbf{k} \wedge {\mathbf{A}}^{*}) \cdot \mathbf{A}_t\right) \\ &=\frac{\epsilon_0}{2 c} \text{Real} \int d^3 \mathbf{k} \left( -i ({\mathbf{A}}^{*}_t \cdot \mathbf{A}) \mathbf{k}+ i \mathbf{k} ({\mathbf{A}}^{*} \cdot \mathbf{A}_t)\right)\end{aligned}$

This is just

$\begin{aligned}\int \mathbf{P} d^3 \mathbf{x} &=\frac{\epsilon_0}{c} \text{Real} i \int \mathbf{k} ({\mathbf{A}}^{*} \cdot \mathbf{A}_t) d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(5.56)$

Recall that the Fourier transforms for the transverse propagation case had the form $\mathbf{A}(\mathbf{k}, t) = \mathbf{A}_{\pm}(\mathbf{k}) e^{\pm i c {\left\lvert{\mathbf{k}}\right\rvert} t}$ , where the minus generated the advanced wave, and the plus the receding wave. With substitution of the vector potential for the advanced wave into the energy and momentum results of (5.55) and (5.56) respectively, we have

$\begin{aligned}\int H d^3 \mathbf{x} &= \epsilon_0 \int \mathbf{k}^2 {\left\lvert{\mathbf{A}(\mathbf{k})}\right\rvert}^2 d^3 \mathbf{k} \\ \int \mathbf{P} d^3 \mathbf{x} &= \epsilon_0 \int \hat{\mathbf{k}} \mathbf{k}^2 {\left\lvert{\mathbf{A}(\mathbf{k})}\right\rvert}^2 d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(5.57)$

After a somewhat circuitous route, this has the relativistic symmetry that is expected. In particular the for the complete $\mu=0$ tensor we have after integration over all space

$\begin{aligned}\int T(\gamma_0) \gamma_0 d^3 \mathbf{x} = \epsilon_0 \int (1 + \hat{\mathbf{k}}) \mathbf{k}^2 {\left\lvert{\mathbf{A}(\mathbf{k})}\right\rvert}^2 d^3 \mathbf{k}.\end{aligned} \hspace{\stretch{1}}(5.59)$

The receding wave solution would give the same result, but directed as $1 - \hat{\mathbf{k}}$ instead.

Observe that we also have the four divergence conservation statement that is expected

$\begin{aligned}\frac{\partial {}}{\partial {t}} \int H d^3 \mathbf{x} + \boldsymbol{\nabla} \cdot \int c \mathbf{P} d^3 \mathbf{x} &= 0.\end{aligned} \hspace{\stretch{1}}(5.60)$

This follows trivially since both the derivatives are zero. If the integration region was to be more specific instead of a $0 + 0 = 0$ relationship, we’d have the power flux ${\partial {H}}/{\partial {t}}$ equal in magnitude to the momentum change through a bounding surface. For a more general surface the time and spatial dependencies shouldn’t necessarily vanish, but we should still have this radiation energy momentum conservation.

References

[1] Peeter Joot. Electrodynamic field energy for vacuum. [online]. http://sites.google.com/site/peeterjoot/math2009/fourierMaxVac.pdf.

[2] Peeter Joot. {Energy and momentum for Complex electric and magnetic field phasors.} [online]. http://sites.google.com/site/peeterjoot/math2009/complexFieldEnergy.pdf.

[3] D. Bohm. Quantum Theory. Courier Dover Publications, 1989.

Posted in Math and Physics Learning. | Tagged: bivector, clifford algebra, energy-momentum, fourier transform, geometric algebra, maxwells equations, spacetime gradient | Leave a Comment »

Energy and momentum for assumed Fourier transform solutions to the homogeneous Maxwell equation.

Posted by peeterjoot on December 22, 2009

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

Motivation and notation.

In Electrodynamic field energy for vacuum (reworked) [1], building on Energy and momentum for Complex electric and magnetic field phasors [2] a derivation for the energy and momentum density was derived for an assumed Fourier series solution to the homogeneous Maxwell’s equation. Here we move to the continuous case examining Fourier transform solutions and the associated energy and momentum density.