PX284 - P1 - introduction to EM waves

starting from maxwell's equations with no sources, and vector identities:

\begin{matrix} \vec{\nabla} \times (\vec{\nabla} \times \vec{ω}) = \vec{\nabla} (\vec{\nabla} \cdot \vec{ω}) - \nabla^{2} \vec{ω} \\ \vec{\nabla} \times (\vec{\nabla} \times \vec{E}) = \vec{\nabla} \times (- \frac{\partial \vec{B}}{\partial t}) \\ \underset{0}{\underset{⏟}{\vec{\nabla} (\vec{\nabla} \cdot \vec{E})}} - \nabla^{2} \vec{E} = - \frac{\partial}{\partial t} (\vec{\nabla} \times \vec{B}) = - μ_{0} ε_{0} \frac{\partial^{2} \vec{E}}{\partial t^{2}} \\ ∴ \nabla^{2} \vec{E} = μ_{0} ε_{0} \frac{\partial^{2} \vec{E}}{\partial t^{2}} \end{matrix}

there are the wave equations for $\vec{E}$ and $\vec{B}$
considering a generic plane wave:

u = u_{0} \exp (i (\vec{k} \cdot \vec{x} - ω t))

it is propagating in the direction, $\hat{k}$ , with wavevector, $\vec{k}$ , and angular frequency, $ω$
for a plane wave:

\begin{matrix} \vec{\nabla} u = + i \vec{k} u \\ \nabla^{2} u = - k^{2} u \\ \frac{\partial u}{\partial t} = - i ω u \\ \frac{\partial^{2} u}{\partial t^{2}} = - ω^{2} u \\ \nabla^{2} u = \frac{1}{c^{2}} \frac{\partial^{2} u}{\partial t^{2}} \end{matrix}

therefore, the phase velocity of the wave is: $c = ω / k$
for EM waves, $c^{2} = 1 / μ_{0} ε_{0}$
plugging in the values, $c$ is found to be the speed of light

\begin{matrix} \vec{\nabla} \cdot \vec{E} = 0 & ⟹ \vec{k} \cdot \vec{E} = 0 & \vec{k} ⊥ \vec{E} \\ \vec{\nabla} \cdot \vec{B} = 0 & ⟹ \vec{k} \cdot \vec{B} = 0 & \vec{k} ⊥ \vec{B} \\ \vec{\nabla} \times \vec{E} = - \frac{\partial \vec{B}}{\partial t} & ⟹ \vec{k} \times \vec{E} = ω \vec{B} & \vec{E} ⊥ \vec{B} \\ \vec{\nabla} \times \vec{B} = μ_{0} ε_{0} \frac{\partial \vec{E}}{\partial t} & ⟹ \vec{k} \times \vec{B} = - μ_{0} ε_{0} \vec{E} \end{matrix}

taking the curl of both sides:

\begin{aligned} \vec{\nabla} \times (\vec{\nabla} \times \vec{B}) & = - μ_{0} ε_{0} ω \vec{\nabla} \times \vec{E} \\ \vec{k} \times (\vec{k} \times \vec{B}) & = μ_{0} ε_{0} ω \frac{\partial \vec{B}}{\partial t} \\ - k^{2} \vec{B} & = μ_{0} ε_{0} ω^{2} \vec{B} \\ k^{2} & = μ_{0} ε_{0} ω^{2} \\ \frac{ω^{2}}{k^{2}} = c^{2} & = \frac{1}{μ_{0} ε_{0}} \end{aligned}

summary

EM waves propagate at the speed of light:

c = \frac{1}{μ_{0} ε_{0}}

this is because light is an EM wave

waves are transverse

$\vec{E} ⊥ \vec{k}, \vec{B} ⊥ \vec{k}$

the fields are perpendicular to each other and they are in phase

$\vec{E} ⊥ B$

for a given wavevector, there are two independent modes - polarizations

$\vec{E}$ and $\vec{B}$ are oriented such that $\vec{k} \times \vec{E} = ω B$
this can form any linear combination of two modes

Info

$\begin{matrix} | \vec{k} \times \vec{E} | = k | \vec{E} | = ω | \vec{B} | \\ | \vec{B} | = \frac{1}{c} | \vec{E} | \\ \vec{B} = R \frac{1}{c} \vec{E} \end{matrix}$

where, $R$ is a matrix that represents a $90 °$ clockwise rotation around the direction of $\vec{k}$