[NOTES/EM-11003]-Electromagnetic waves in vacuum

1. Solution in vacuum

Assuming that there is no charge or current density present, we set $\rho=0$ $\vec{j}=0$. In this case the free Maxwell's equations take the form: \begin{align*} &\bar{\nabla}\cdot\bar{E}=0& \bar{\nabla}\cdot\bar{B}=0& \\ \bar{\nabla}\times\bar{E}=&-~\frac{\partial\vec{B}}{\partial t}, &\bar{\nabla}\times\bar{B}=&\mu_0\epsilon_0~\frac{\partial\vec{E}}{\partial t}.& \end{align*} Taking curl of the third equation, we get \begin{align*} \bar{\nabla}\times(\bar{\nabla}\times\bar{E}) = &-~\frac{\partial}{\partial t}(\vec{\nabla}\times\vec{B}) =-\mu_0\epsilon_0~\frac{\partial^2\vec{E}}{\partial t^2}\\ \bar{\nabla}(\bar{\nabla}\cdot\bar{E})-\nabla^2\vec{E} = & - \mu_0\epsilon_0~\frac{\partial^2\bar{E}}{\partial t^2} \end{align*} since \(\nabla\cdot\vec{E}=0\), we get the wave equation. \begin{align*} \nabla^2\vec{E}-&\mu_0\epsilon_0~\frac{\partial^2\vec{E}}{\partial t^2} = 0 \end{align*} Proceeding in a similar manner, we can now derive wave equation for the magnetic field. \begin{align*} \bar{\nabla}\times(\bar{\nabla}\times\bar{B}) =& \mu_0\epsilon_0\vec{\nabla}\times\left(\frac{\partial\vec{E}} {\partial t^2}\right)\\ \nabla(\bar{\nabla}\cdot\bar{B})-\nabla^2\bar{B}=&-\mu_0\epsilon_0 \frac{\partial^2\vec{B}}{\partial t^2} \end{align*} or $$ \nabla^2\vec{B}-\mu_0\epsilon_0~\frac{\partial^2\vec{B}} {\partial t^2}=0 $$ Therefore, even if $\rho=0$, $\vec{j}=0$ we have equations $$ \square~\bar{E} = 0 \qquad\qquad \square~\vec{B}=0 $$ which have propagating wave solutions.

2.Propagating wave solutions

Assuming plane wave form for the electric field $$ \vec{E} = \vec{E}_0~\exp(i\vec{k}\cdot\vec{r}-i\omega t) $$ we get the relation $$ \vec{k}^2-(\mu_0\epsilon_0)\omega^2 = 0 $$ Here \(k, \omega,c\) are the wave vector, the frequency and the velocity of the electromagnetic waves, where the velocity is given by $$c=\frac{1}{\sqrt{\mu_0\epsilon_0}}.$$ and $\lambda=2\pi/k\qquad\qquad \nu=\omega/2\pi$\\ The magnetic field $\vec{B}$ and the potentials have similar solutions. \noindent In general \begin{align*} \phi = & Re[\phi_0\exp(-\vec{k}\cdot\vec{r}-i\omega t)]\\ \phi = & A \cos(\vec{k}\cdot\vec{r}-i\omega t)+B\sin(\vec{k}\cdot\vec{r}-i\omega t) \end{align*} where $\phi_0=$ is a complex constant. The solutions for the fields can be written down as \begin{eqnarray} \vec{E}&=&\vec{E}_0\exp(i\vec{k}\cdot\vec{r}-i\omega t)\\ \vec{B}&=&\vec{B}_0\exp(i\vec{k}\cdot\vec{r}-i\omega t) \end{eqnarray} and \(\vec{E}_0, \vec{B}\) are complex constants. The Maxwell's equations \begin{equation} {\nabla}\cdot\vec{E}=0, \quad \text{and}\quad \bar{\nabla}\cdot\vec{B}=0, \end{equation} imply that $\vec{E}$ and $\vec{B}$ are $\perp$ to the direction of propagation, i.e. \begin{equation} \vec{k}\cdot\bar{E}=0 \qquad \vec{k}\cdot\vec{B}_0=0. \end{equation} \noindent The Maxwell's equation \(\bar{\nabla}\times\bar{E} = - \frac{\partial\vec{B}}{\partial t} \) implies \begin{align*} \vec{k}\times\vec{E}_0 = & \omega \vec{B}_0 \end{align*} This equation is equivalent to two results that $\vec{B}$ is $\perp~\vec{E}$ and \(|\vec{k}||\vec{E}_0| = \omega |\vec{B}_0|\Rightarrow B_0= ~\frac{1}{c}|\vec{E}_0|\) To summarize we have the results that

• $\vec{E}, \vec{B}$ are mutually $\perp$ and also $\perp$ to the direction of propagation.
• $\vec{k},\vec{E},\vec{B}$, form a set of right handed orthonormal vectors.

References

1. Sec 20-1  Waves in free space , plane wavesR. P. Feynman, Robert B. Leighton and Mathew Sands Lectures on Physics, vol-II, B.I. Publications (1964)

2. Sec 9.2.1  The wave equation for \(\vec{E}\) and \(\vec{B}\)Sec 9.2.2  Monochromatic plane waves. David Griffiths, Introduction to Electrodynamics, 3rd EEE edn, Prentice Hall of India Pvt Ltd New Delhi, (2002).