diff options
Diffstat (limited to 'source/know/concept/electromagnetic-wave-equation')
| -rw-r--r-- | source/know/concept/electromagnetic-wave-equation/index.md | 347 |
1 files changed, 196 insertions, 151 deletions
diff --git a/source/know/concept/electromagnetic-wave-equation/index.md b/source/know/concept/electromagnetic-wave-equation/index.md index a27fe6f..559d943 100644 --- a/source/know/concept/electromagnetic-wave-equation/index.md +++ b/source/know/concept/electromagnetic-wave-equation/index.md @@ -1,7 +1,7 @@ --- title: "Electromagnetic wave equation" sort_title: "Electromagnetic wave equation" -date: 2021-09-09 +date: 2024-09-08 # Originally 2021-09-09, major rewrite categories: - Physics - Electromagnetism @@ -9,236 +9,281 @@ categories: layout: "concept" --- -The electromagnetic wave equation describes -the propagation of light through various media. -Since an electromagnetic (light) wave consists of +Light, i.e. **electromagnetic waves**, consist of an [electric field](/know/concept/electric-field/) and a [magnetic field](/know/concept/magnetic-field/), -we need [Maxwell's equations](/know/concept/maxwells-equations/) -in order to derive the wave equation. +one inducing the other and vice versa. +The existence and classical behavior of such waves +can be derived using only [Maxwell's equations](/know/concept/maxwells-equations/), +as we will demonstrate here. - -## Uniform medium - -We will use all of Maxwell's equations, -but we start with Ampère's circuital law for the "free" fields $$\vb{H}$$ and $$\vb{D}$$, -in the absence of a free current $$\vb{J}_\mathrm{free} = 0$$: - -$$\begin{aligned} - \nabla \cross \vb{H} - = \pdv{\vb{D}}{t} -\end{aligned}$$ - -We assume that the medium is isotropic, linear, -and uniform in all of space, such that: +We start from Faraday's law of induction, +where we assume that the system consists of materials +with well-known (linear) relative magnetic permeabilities $$\mu_r(\vb{r})$$, +such that $$\vb{B} = \mu_0 \mu_r \vb{H}$$: $$\begin{aligned} - \vb{D} = \varepsilon_0 \varepsilon_r \vb{E} - \qquad \quad - \vb{H} = \frac{1}{\mu_0 \mu_r} \vb{B} + \nabla \cross \vb{E} + = - \pdv{\vb{B}}{t} + = - \mu_0 \mu_r \pdv{\vb{H}}{t} \end{aligned}$$ -Which, upon insertion into Ampère's law, -yields an equation relating $$\vb{B}$$ and $$\vb{E}$$. -This may seem to contradict Ampère's "total" law, -but keep in mind that $$\vb{J}_\mathrm{bound} \neq 0$$ here: +We move $$\mu_r(\vb{r})$$ to the other side, +take the curl, and insert Ampère's circuital law: $$\begin{aligned} - \nabla \cross \vb{B} - = \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{\vb{E}}{t} + \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg) + &= - \mu_0 \pdv{}{t} \big( \nabla \cross \vb{H} \big) + \\ + &= - \mu_0 \bigg( \pdv{\vb{J}_\mathrm{free}}{t} + \pdvn{2}{\vb{D}}{t} \bigg) \end{aligned}$$ -Now we take the curl, rearrange, -and substitute $$\nabla \cross \vb{E}$$ according to Faraday's law: +For simplicity, we only consider insulating materials, +since light propagation in conductors is a complex beast. +We thus assume that there are no free currents $$\vb{J}_\mathrm{free} = 0$$, leaving: $$\begin{aligned} - \nabla \cross (\nabla \cross \vb{B}) - = \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{}{t}(\nabla \cross \vb{E}) - = - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{B}}{t} + \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg) + &= - \mu_0 \pdvn{2}{\vb{D}}{t} \end{aligned}$$ -Using a vector identity, we rewrite the leftmost expression, -which can then be reduced thanks to Gauss' law for magnetism $$\nabla \cdot \vb{B} = 0$$: +Having $$\vb{E}$$ and $$\vb{D}$$ in the same equation is not ideal, +so we should make a choice: +do we restrict ourselves to linear media +(so $$\vb{D} = \varepsilon_0 \varepsilon_r \vb{E}$$), +or do we allow materials with more complicated responses +(so $$\vb{D} = \varepsilon_0 \vb{E} + \vb{P}$$, with $$\vb{P}$$ unspecified)? +The former is usually sufficient: $$\begin{aligned} - - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{B}}{t} - &= \nabla (\nabla \cdot \vb{B}) - \nabla^2 \vb{B} - = - \nabla^2 \vb{B} + \boxed{ + \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg) + = - \mu_0 \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t} + } \end{aligned}$$ -This describes $$\vb{B}$$. -Next, we repeat the process for $$\vb{E}$$: -taking the curl of Faraday's law yields: +This is the general linear form of the **electromagnetic wave equation**, +where $$\mu_r$$ and $$\varepsilon_r$$ +both depend on $$\vb{r}$$ in order to describe the structure of the system. +We can obtain a similar equation for $$\vb{H}$$, +by starting from Ampère's law under the same assumptions: $$\begin{aligned} - \nabla \cross (\nabla \cross \vb{E}) - = - \pdv{}{t}(\nabla \cross \vb{B}) - = - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t} + \nabla \cross \vb{H} + = \pdv{\vb{D}}{t} + = \varepsilon_0 \varepsilon_r \pdv{\vb{E}}{t} \end{aligned}$$ -Which can be rewritten using same vector identity as before, -and then reduced by assuming that there is no net charge density $$\rho = 0$$ -in Gauss' law, such that $$\nabla \cdot \vb{E} = 0$$: +Taking the curl and substituting Faraday's law on the right yields: $$\begin{aligned} - - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t} - &= \nabla (\nabla \cdot \vb{E}) - \nabla^2 \vb{E} - = - \nabla^2 \vb{E} + \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg) + &= \varepsilon_0 \pdv{}{t} \big( \nabla \cross \vb{E} \big) + = - \varepsilon_0 \pdvn{2}{\vb{B}}{t} \end{aligned}$$ -We thus arrive at the following two (implicitly coupled) -wave equations for $$\vb{E}$$ and $$\vb{B}$$, -where we have defined the phase velocity $$v \equiv 1 / \sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}$$: +And then we insert $$\vb{B} = \mu_0 \mu_r \vb{H}$$ to get the analogous +electromagnetic wave equation for $$\vb{H}$$: $$\begin{aligned} \boxed{ - \pdvn{2}{\vb{E}}{t} - \frac{1}{v^2} \nabla^2 \vb{E} - = 0 - } - \qquad \quad - \boxed{ - \pdvn{2}{\vb{B}}{t} - \frac{1}{v^2} \nabla^2 \vb{B} - = 0 + \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg) + = - \mu_0 \varepsilon_0 \mu_r \pdvn{2}{\vb{H}}{t} } \end{aligned}$$ -Traditionally, it is said that the solutions are as follows, -where the wavenumber $$|\vb{k}| = \omega / v$$: - -$$\begin{aligned} - \vb{E}(\vb{r}, t) - &= \vb{E}_0 \exp(i \vb{k} \cdot \vb{r} - i \omega t) - \\ - \vb{B}(\vb{r}, t) - &= \vb{B}_0 \exp(i \vb{k} \cdot \vb{r} - i \omega t) -\end{aligned}$$ - -In fact, thanks to linearity, these **plane waves** can be treated as -terms in a Fourier series, meaning that virtually -*any* function $$f(\vb{k} \cdot \vb{r} - \omega t)$$ is a valid solution. +This is equivalent to the problem for $$\vb{E}$$, +since they are coupled by Maxwell's equations. +By solving either, subject to Gauss's laws +$$\nabla \cdot (\varepsilon_r \vb{E}) = 0$$ and $$\nabla \cdot (\mu_r \vb{H}) = 0$$, +the behavior of light in a given system can be deduced. +Note that Gauss's laws enforce that the wave's fields are transverse, +i.e. they must be perpendicular to the propagation direction. -Keep in mind that in reality $$\vb{E}$$ and $$\vb{B}$$ are real, -so although it is mathematically convenient to use plane waves, -in the end you will need to take the real part. -## Non-uniform medium +## Homogeneous linear media -A useful generalization is to allow spatial change -in the relative permittivity $$\varepsilon_r(\vb{r})$$ -and the relative permeability $$\mu_r(\vb{r})$$. -We still assume that the medium is linear and isotropic, so: +In the special case where the medium is completely uniform, +$$\mu_r$$ and $$\varepsilon_r$$ no longer depend on $$\vb{r}$$, +so they can be moved to the other side: $$\begin{aligned} - \vb{D} - = \varepsilon_0 \varepsilon_r(\vb{r}) \vb{E} - \qquad \quad - \vb{B} - = \mu_0 \mu_r(\vb{r}) \vb{H} + \nabla \cross \big( \nabla \cross \vb{E} \big) + &= - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t} + \\ + \nabla \cross \big( \nabla \cross \vb{H} \big) + &= - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{H}}{t} \end{aligned}$$ -Inserting these expressions into Faraday's and Ampère's laws -respectively yields: +This can be rewritten using the vector identity +$$\nabla \cross (\nabla \cross \vb{V}) = \nabla (\nabla \cdot \vb{V}) - \nabla^2 \vb{V}$$: $$\begin{aligned} - \nabla \cross \vb{E} - = - \mu_0 \mu_r(\vb{r}) \pdv{\vb{H}}{t} - \qquad \quad - \nabla \cross \vb{H} - = \varepsilon_0 \varepsilon_r(\vb{r}) \pdv{\vb{E}}{t} + \nabla (\nabla \cdot \vb{E}) - \nabla^2 \vb{E} + &= - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t} + \\ + \nabla (\nabla \cdot \vb{H}) - \nabla^2 \vb{H} + &= - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{H}}{t} \end{aligned}$$ -We then divide Ampère's law by $$\varepsilon_r(\vb{r})$$, -take the curl, and substitute Faraday's law, giving: +Which can be reduced using Gauss's laws +$$\nabla \cdot \vb{E} = 0$$ and $$\nabla \cdot \vb{H} = 0$$ +thanks to the fact that $$\varepsilon_r$$ and $$\mu_r$$ are constants in this case. +We therefore arrive at: $$\begin{aligned} - \nabla \cross \Big( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \Big) - = \varepsilon_0 \pdv{}{t}(\nabla \cross \vb{E}) - = - \mu_0 \mu_r \varepsilon_0 \pdvn{2}{\vb{H}}{t} + \boxed{ + \nabla^2 \vb{E} - \frac{n^2}{c^2} \pdvn{2}{\vb{E}}{t} + = 0 + } \end{aligned}$$ -Next, we exploit linearity by decomposing $$\vb{H}$$ and $$\vb{E}$$ -into Fourier series, with terms given by: - $$\begin{aligned} - \vb{H}(\vb{r}, t) - = \vb{H}(\vb{r}) \exp(- i \omega t) - \qquad \quad - \vb{E}(\vb{r}, t) - = \vb{E}(\vb{r}) \exp(- i \omega t) + \boxed{ + \nabla^2 \vb{H} - \frac{n^2}{c^2} \pdvn{2}{\vb{H}}{t} + = 0 + } \end{aligned}$$ -By inserting this ansatz into the equation, -we can remove the explicit time dependence: +Where $$c = 1 / \sqrt{\mu_0 \varepsilon_0}$$ is the speed of light in a vacuum, +and $$n = \sqrt{\mu_0 \varepsilon_0}$$ is the refractive index of the medium. +Note that most authors write the magnetic equation with $$\vb{B}$$ instead of $$\vb{H}$$; +both are correct thanks to linearity. + +In a vacuum, where $$n = 1$$, these equations are sometimes written as +$$\square \vb{E} = 0$$ and $$\square \vb{H} = 0$$, +where $$\square$$ is the **d'Alembert operator**, defined as follows: $$\begin{aligned} - \nabla \cross \Big( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \Big) \exp(- i \omega t) - = \mu_0 \varepsilon_0 \omega^2 \mu_r \vb{H} \exp(- i \omega t) + \boxed{ + \square + \equiv \nabla^2 - \frac{1}{c^2} \pdvn{2}{}{t} + } \end{aligned}$$ -Dividing out $$\exp(- i \omega t)$$, -we arrive at an eigenvalue problem for $$\omega^2$$, -with $$c = 1 / \sqrt{\mu_0 \varepsilon_0}$$: +Note that some authors define it with the opposite sign. +In any case, the d'Alembert operator is important for special relativity. + +The solution to the homogeneous electromagnetic wave equation +are traditionally said to be the so-called **plane waves** given by: $$\begin{aligned} - \boxed{ - \nabla \cross \Big( \frac{1}{\varepsilon_r(\vb{r})} \nabla \cross \vb{H}(\vb{r}) \Big) - = \Big( \frac{\omega}{c} \Big)^2 \mu_r(\vb{r}) \vb{H}(\vb{r}) - } + \vb{E}(\vb{r}, t) + &= \vb{E}_0 e^{i \vb{k} \cdot \vb{r} - i \omega t} + \\ + \vb{B}(\vb{r}, t) + &= \vb{B}_0 e^{i \vb{k} \cdot \vb{r} - i \omega t} \end{aligned}$$ -Compared to a uniform medium, $$\omega$$ is often not arbitrary here: -there are discrete eigenvalues $$\omega$$, -corresponding to discrete **modes** $$\vb{H}(\vb{r})$$. +Where the wavevector $$\vb{k}$$ is arbitrary, +and the angular frequency $$\omega = c |\vb{k}| / n$$. +We also often talk about the wavelength, which is $$\lambda = 2 \pi / |\vb{k}|$$. +The appearance of $$\vb{k}$$ in the exponent +tells us that these waves are propagating through space, +as you would expect. + +In fact, because the wave equations are linear, +any superposition of plane waves, +i.e. any function of the form $$f(\vb{k} \cdot \vb{r} - \omega t)$$, +is in fact a valid solution. +Just remember that $$\vb{E}$$ and $$\vb{H}$$ are real-valued, +so it may be necessary to take the real part at the end of a calculation. -Next, we go through the same process to find an equation for $$\vb{E}$$. -Starting from Faraday's law, we divide by $$\mu_r(\vb{r})$$, -take the curl, and insert Ampère's law: + + +## Inhomogeneous linear media + +But suppose the medium is not uniform, i.e. it contains structures +described by $$\varepsilon_r(\vb{r})$$ and $$\mu_r(\vb{r})$$. +If the structures are much larger than the light's wavelength, +the homogeneous equation is still a very good approximation +away from any material boundaries; +anywhere else, however, they will break down. +Recall the general equations from before we assumed homogeneity: $$\begin{aligned} - \nabla \cross \Big( \frac{1}{\mu_r} \nabla \cross \vb{E} \Big) - = - \mu_0 \pdv{}{t}(\nabla \cross \vb{H}) - = - \mu_0 \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t} + \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg) + &= - \frac{\varepsilon_r}{c^2} \pdvn{2}{\vb{E}}{t} + \\ + \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg) + &= - \frac{\mu_r}{c^2} \pdvn{2}{\vb{H}}{t} \end{aligned}$$ -Then, by replacing $$\vb{E}(\vb{r}, t)$$ with our plane-wave ansatz, -we remove the time dependence: +In theory, this is everything we need, +but in most cases a better approach is possible: +the trick is that we only rarely need to explicitly calculate +the $$t$$-dependence of $$\vb{E}$$ or $$\vb{H}$$. +Instead, we can first solve an easier time-independent version +of this problem, and then approximate the dynamics +with [coupled mode theory](/know/concept/coupled-mode-theory/) later. + +To eliminate $$t$$, we make an ansatz for $$\vb{E}$$ and $$\vb{H}$$, shown below. +No generality is lost by doing this; +this is effectively a kind of [Fourier transform](/know/concept/fourier-transform/): $$\begin{aligned} - \nabla \cross \Big( \frac{1}{\mu_r} \nabla \cross \vb{E} \Big) \exp(- i \omega t) - = - \mu_0 \varepsilon_0 \omega^2 \varepsilon_r \vb{E} \exp(- i \omega t) + \vb{E}(\vb{r}, t) + &= \vb{E}(\vb{r}) e^{- i \omega t} + \\ + \vb{H}(\vb{r}, t) + &= \vb{H}(\vb{r}) e^{- i \omega t} \end{aligned}$$ -Which, after dividing out $$\exp(- i \omega t)$$, -yields an analogous eigenvalue problem with $$\vb{E}(r)$$: +Inserting this ansatz and dividing out $$e^{-i \omega t}$$ +yields the time-independent forms: $$\begin{aligned} \boxed{ - \nabla \cross \Big( \frac{1}{\mu_r(\vb{r})} \nabla \cross \vb{E}(\vb{r}) \Big) - = \Big( \frac{\omega}{c} \Big)^2 \varepsilon_r(\vb{r}) \vb{E}(\vb{r}) + \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg) + = \Big( \frac{\omega}{c} \Big)^2 \varepsilon_r \vb{E} } \end{aligned}$$ -Usually, it is a reasonable approximation -to say $$\mu_r(\vb{r}) = 1$$, -in which case the equation for $$\vb{H}(\vb{r})$$ -becomes a Hermitian eigenvalue problem, -and is thus easier to solve than for $$\vb{E}(\vb{r})$$. +$$\begin{aligned} + \boxed{ + \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg) + = \Big( \frac{\omega}{c} \Big)^2 \mu_r \vb{H} + } +\end{aligned}$$ -Keep in mind, however, that in any case, -the solutions $$\vb{H}(\vb{r})$$ and/or $$\vb{E}(\vb{r})$$ -must satisfy the two Maxwell's equations that were not explicitly used: +These are eigenvalue problems for $$\omega^2$$, +which can be solved subject to Gauss's laws and suitable boundary conditions. +The resulting allowed values of $$\omega$$ may consist of +continuous ranges and/or discrete resonances, +analogous to *scattering* and *bound* quantum states, respectively. +It can be shown that the operators on both sides of each equation +are Hermitian, meaning these are well-behaved problems +yielding real eigenvalues and orthogonal eigenfields. + +Both equations are still equivalent: +we only need to solve one. But which one? +In practice, one is usually easier than the other, +due to the common approximation that $$\mu_r \approx 1$$ for many dielectric materials, +in which case the equations reduce to: $$\begin{aligned} - \nabla \cdot (\varepsilon_r \vb{E}) = 0 - \qquad \quad - \nabla \cdot (\mu_r \vb{H}) = 0 + \nabla \cross \big( \nabla \cross \vb{E} \big) + &= \Big( \frac{\omega}{c} \Big)^2 \varepsilon_r \vb{E} + \\ + \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg) + &= \Big( \frac{\omega}{c} \Big)^2 \vb{H} \end{aligned}$$ -This is equivalent to demanding that the resulting waves are *transverse*, -or in other words, -the wavevector $$\vb{k}$$ must be perpendicular to -the amplitudes $$\vb{H}_0$$ and $$\vb{E}_0$$. +Now the equation for $$\vb{H}$$ is starting to look simpler, +because it only has an operator on *one* side. +We could "fix" the equation for $$\vb{E}$$ by dividing it by $$\varepsilon_r$$, +but the resulting operator would no longer be Hermitian, +and hence not well-behaved. +To get an idea of how to handle $$\varepsilon_r$$ in the $$\vb{E}$$-equation, +notice its similarity to the weight function $$w$$ +in [Sturm-Liouville theory](/know/concept/sturm-liouville-theory/). + +Gauss's magnetic law $$\nabla \cdot \vb{H} = 0$$ +is also significantly easier for numerical calculations +than its electric counterpart $$\nabla \cdot (\varepsilon_r \vb{E}) = 0$$, +so we usually prefer to solve the equation for $$\vb{H}$$. + ## References |