# Peeter Joot's Blog.

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

Posted by Peeter Joot on December 22, 2009

# 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)

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}}(2.9)

# The tensor integrated over all space. Energy and momentum?

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}}(3.10)

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}}(3.11)

Observing that $\gamma_0$ commutes with spatial bivectors and anticommutes with spatial vectors, and writing $\sigma_\mu = \gamma_\mu \gamma_0$, one has

\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}}(3.12)

The scalar and spatial vector grade selection operator has been added for convenience and does not change the result since those are necessarily the only grades anyhow. The post multiplication by the observer frame time basis vector $\gamma_0$ serves to separate the energy and momentum like components of the tensor nicely into scalar and vector aspects. In particular for $T(\gamma^0)$, one could write

\begin{aligned}\int T(\gamma^0) d^3 \mathbf{x} = (H + \mathbf{P}) \gamma_0,\end{aligned} \hspace{\stretch{1}}(3.13)

If these are correctly identified with energy and momentum then it also ought to be true that we have the conservation relationship

\begin{aligned}\frac{\partial {H}}{\partial {t}} + \boldsymbol{\nabla} \cdot (c \mathbf{P}) = 0.\end{aligned} \hspace{\stretch{1}}(3.14)

However, multiplying out (3.12) yields for $H$

\begin{aligned}H &= \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}}(3.15)

The vector component takes a bit more work to reduce

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

Canceling and regrouping leaves

\begin{aligned}\mathbf{P}&=\epsilon_0 \int d^3 \mathbf{k} \text{Real} \left(\mathbf{A} \left( \mathbf{k}^2 {{\phi}}^{*} + \mathbf{k} \cdot {{\dot{\mathbf{A}}}}^{*} \right)+ \mathbf{k} \left( -{{\phi}}^{*} (\mathbf{k} \cdot \mathbf{A}) + \frac{i}{c} ({\mathbf{A}}^{*} \cdot \dot{\mathbf{A}})\right)\right).\end{aligned} \hspace{\stretch{1}}(3.16)

This has no explicit $\mathbf{x}$ dependence, so the conservation relation (3.14) is violated unless ${\partial {H}}/{\partial {t}} = 0$. There is no reason to assume that will be the case. In the discrete Fourier series treatment, a gauge transformation allowed for elimination of $\phi$, and this implied $\mathbf{k} \cdot \mathbf{A}_\mathbf{k} = 0$ or $\mathbf{A}_\mathbf{k}$ constant. We will probably have a similar result here, eliminating most of the terms in (3.15) and (3.16). Except for the constant $\mathbf{A}_\mathbf{k}$ solution of the field equations there is no obvious way that such a simplified energy expression will have zero derivative.

A more reasonable conclusion is that this approach is flawed. We ought to be looking at the divergence relation as a starting point, and instead of integrating over all space, instead employing Gauss’s theorem to convert the divergence integral into a surface integral. Without math, the conservation relationship probably ought to be expressed as energy change in a volume is matched by the momentum change through the surface. However, without an integral over all space, we do not get the nice delta function cancellation observed above. How to proceed is not immediately clear. Stepping back to review applications of Gauss’s theorem is probably a good first step.

# 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.