From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Thu, 20 Oct 2022 18:25:31 +0200 Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex' --- .../concept/electromagnetic-wave-equation/index.md | 72 +++++++++++----------- 1 file changed, 36 insertions(+), 36 deletions(-) (limited to 'source/know/concept/electromagnetic-wave-equation') diff --git a/source/know/concept/electromagnetic-wave-equation/index.md b/source/know/concept/electromagnetic-wave-equation/index.md index c4cd9eb..a27fe6f 100644 --- a/source/know/concept/electromagnetic-wave-equation/index.md +++ b/source/know/concept/electromagnetic-wave-equation/index.md @@ -21,8 +21,8 @@ in order to derive the wave equation. ## 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 $\vb{J}_\mathrm{free} = 0$: +but we start with Ampère's circuital law for the "free" fields $$\vb{H}$$ and $$\vb{D}$$, +in the absence of a free current $$\vb{J}_\mathrm{free} = 0$$: $$\begin{aligned} \nabla \cross \vb{H} @@ -39,9 +39,9 @@ $$\begin{aligned} \end{aligned}$$ Which, upon insertion into Ampère's law, -yields an equation relating $\vb{B}$ and $\vb{E}$. +yields an equation relating $$\vb{B}$$ and $$\vb{E}$$. This may seem to contradict Ampère's "total" law, -but keep in mind that $\vb{J}_\mathrm{bound} \neq 0$ here: +but keep in mind that $$\vb{J}_\mathrm{bound} \neq 0$$ here: $$\begin{aligned} \nabla \cross \vb{B} @@ -49,7 +49,7 @@ $$\begin{aligned} \end{aligned}$$ Now we take the curl, rearrange, -and substitute $\nabla \cross \vb{E}$ according to Faraday's law: +and substitute $$\nabla \cross \vb{E}$$ according to Faraday's law: $$\begin{aligned} \nabla \cross (\nabla \cross \vb{B}) @@ -58,7 +58,7 @@ $$\begin{aligned} \end{aligned}$$ Using a vector identity, we rewrite the leftmost expression, -which can then be reduced thanks to Gauss' law for magnetism $\nabla \cdot \vb{B} = 0$: +which can then be reduced thanks to Gauss' law for magnetism $$\nabla \cdot \vb{B} = 0$$: $$\begin{aligned} - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{B}}{t} @@ -66,8 +66,8 @@ $$\begin{aligned} = - \nabla^2 \vb{B} \end{aligned}$$ -This describes $\vb{B}$. -Next, we repeat the process for $\vb{E}$: +This describes $$\vb{B}$$. +Next, we repeat the process for $$\vb{E}$$: taking the curl of Faraday's law yields: $$\begin{aligned} @@ -77,8 +77,8 @@ $$\begin{aligned} \end{aligned}$$ Which can be rewritten using same vector identity as before, -and then reduced by assuming that there is no net charge density $\rho = 0$ -in Gauss' law, such that $\nabla \cdot \vb{E} = 0$: +and then reduced by assuming that there is no net charge density $$\rho = 0$$ +in Gauss' law, such that $$\nabla \cdot \vb{E} = 0$$: $$\begin{aligned} - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t} @@ -87,8 +87,8 @@ $$\begin{aligned} \end{aligned}$$ We thus arrive at the following two (implicitly coupled) -wave equations for $\vb{E}$ and $\vb{B}$, -where we have defined the phase velocity $v \equiv 1 / \sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}$: +wave equations for $$\vb{E}$$ and $$\vb{B}$$, +where we have defined the phase velocity $$v \equiv 1 / \sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}$$: $$\begin{aligned} \boxed{ @@ -103,7 +103,7 @@ $$\begin{aligned} \end{aligned}$$ Traditionally, it is said that the solutions are as follows, -where the wavenumber $|\vb{k}| = \omega / v$: +where the wavenumber $$|\vb{k}| = \omega / v$$: $$\begin{aligned} \vb{E}(\vb{r}, t) @@ -115,9 +115,9 @@ $$\begin{aligned} In fact, thanks to linearity, these **plane waves** can be treated as terms in a Fourier series, meaning that virtually -*any* function $f(\vb{k} \cdot \vb{r} - \omega t)$ is a valid solution. +*any* function $$f(\vb{k} \cdot \vb{r} - \omega t)$$ is a valid solution. -Keep in mind that in reality $\vb{E}$ and $\vb{B}$ are real, +Keep in mind that in reality $$\vb{E}$$ and $$\vb{B}$$ are real, so although it is mathematically convenient to use plane waves, in the end you will need to take the real part. @@ -125,8 +125,8 @@ in the end you will need to take the real part. ## Non-uniform medium A useful generalization is to allow spatial change -in the relative permittivity $\varepsilon_r(\vb{r})$ -and the relative permeability $\mu_r(\vb{r})$. +in the relative permittivity $$\varepsilon_r(\vb{r})$$ +and the relative permeability $$\mu_r(\vb{r})$$. We still assume that the medium is linear and isotropic, so: $$\begin{aligned} @@ -148,7 +148,7 @@ $$\begin{aligned} = \varepsilon_0 \varepsilon_r(\vb{r}) \pdv{\vb{E}}{t} \end{aligned}$$ -We then divide Ampère's law by $\varepsilon_r(\vb{r})$, +We then divide Ampère's law by $$\varepsilon_r(\vb{r})$$, take the curl, and substitute Faraday's law, giving: $$\begin{aligned} @@ -157,7 +157,7 @@ $$\begin{aligned} = - \mu_0 \mu_r \varepsilon_0 \pdvn{2}{\vb{H}}{t} \end{aligned}$$ -Next, we exploit linearity by decomposing $\vb{H}$ and $\vb{E}$ +Next, we exploit linearity by decomposing $$\vb{H}$$ and $$\vb{E}$$ into Fourier series, with terms given by: $$\begin{aligned} @@ -176,9 +176,9 @@ $$\begin{aligned} = \mu_0 \varepsilon_0 \omega^2 \mu_r \vb{H} \exp(- i \omega t) \end{aligned}$$ -Dividing out $\exp(- i \omega t)$, -we arrive at an eigenvalue problem for $\omega^2$, -with $c = 1 / \sqrt{\mu_0 \varepsilon_0}$: +Dividing out $$\exp(- i \omega t)$$, +we arrive at an eigenvalue problem for $$\omega^2$$, +with $$c = 1 / \sqrt{\mu_0 \varepsilon_0}$$: $$\begin{aligned} \boxed{ @@ -187,12 +187,12 @@ $$\begin{aligned} } \end{aligned}$$ -Compared to a uniform medium, $\omega$ is often not arbitrary here: -there are discrete eigenvalues $\omega$, -corresponding to discrete **modes** $\vb{H}(\vb{r})$. +Compared to a uniform medium, $$\omega$$ is often not arbitrary here: +there are discrete eigenvalues $$\omega$$, +corresponding to discrete **modes** $$\vb{H}(\vb{r})$$. -Next, we go through the same process to find an equation for $\vb{E}$. -Starting from Faraday's law, we divide by $\mu_r(\vb{r})$, +Next, we go through the same process to find an equation for $$\vb{E}$$. +Starting from Faraday's law, we divide by $$\mu_r(\vb{r})$$, take the curl, and insert Ampère's law: $$\begin{aligned} @@ -201,7 +201,7 @@ $$\begin{aligned} = - \mu_0 \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t} \end{aligned}$$ -Then, by replacing $\vb{E}(\vb{r}, t)$ with our plane-wave ansatz, +Then, by replacing $$\vb{E}(\vb{r}, t)$$ with our plane-wave ansatz, we remove the time dependence: $$\begin{aligned} @@ -209,8 +209,8 @@ $$\begin{aligned} = - \mu_0 \varepsilon_0 \omega^2 \varepsilon_r \vb{E} \exp(- i \omega t) \end{aligned}$$ -Which, after dividing out $\exp(- i \omega t)$, -yields an analogous eigenvalue problem with $\vb{E}(r)$: +Which, after dividing out $$\exp(- i \omega t)$$, +yields an analogous eigenvalue problem with $$\vb{E}(r)$$: $$\begin{aligned} \boxed{ @@ -220,13 +220,13 @@ $$\begin{aligned} \end{aligned}$$ Usually, it is a reasonable approximation -to say $\mu_r(\vb{r}) = 1$, -in which case the equation for $\vb{H}(\vb{r})$ +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 $\vb{E}(\vb{r})$. +and is thus easier to solve than for $$\vb{E}(\vb{r})$$. Keep in mind, however, that in any case, -the solutions $\vb{H}(\vb{r})$ and/or $\vb{E}(\vb{r})$ +the solutions $$\vb{H}(\vb{r})$$ and/or $$\vb{E}(\vb{r})$$ must satisfy the two Maxwell's equations that were not explicitly used: $$\begin{aligned} @@ -237,8 +237,8 @@ $$\begin{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 $\vb{E}_0$. +the wavevector $$\vb{k}$$ must be perpendicular to +the amplitudes $$\vb{H}_0$$ and $$\vb{E}_0$$. ## References -- cgit v1.2.3