We will study magnetic and electric fields with a time dependence in a simple situation.

To solve the equations we will require trial solutions as they can be too complicated to solve analytically. We then apply physical constraints such as $\omega=ck$

Solution to the wave equation in 1D

💼 Case: suppose that $\phi(x,t)$ is some time and location dependent field where the dependence of $\phi$ on space and time is determined by the 1D wave equation

• 1D wave equation

  $$ \frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2}-\frac{\partial^2\phi}{\partial x^2}=0 $$

• The solution to this equations are of the form

  $$ \phi(x,t)=f(x-vt)+g(x+vt) $$

  🗒️ Note: the first term is the right moving solution and the second a left moving solution.

💃 Example: if we take a plane moving ansatz

$$ \phi(x,t)=Ae^{i(kx-\omega t)}+Be^{i(kx+\omega t)} $$

and substitute it into the wave equation we find that the ansatz should be subject to the dispersion relation $\omega=vk$. The phase of these waves is $\mathcal K=kx-\omega t$ and hence parts of the field with the same phase are those for whom $\mathcal K= 2\pi n$ where $n\in \Z$.

The phase velocity is $v_p=\frac \omega k$

The group velocity is $v_g=\frac{\partial \omega} {\partial k}$

• if $\phi$ is a real scalar field the physical part of the solution is

  $$ \mathrm{Re}(\phi)=|A|\cos(kx-\omega t+\arg[A]) $$

  For example if $\arg(a)=0$

  $$ \mathrm{Re}(\phi)=A\cos\left (kx-\omega t \right ) $$

  or if $\arg(A)=\pm \frac{\pi}{2}$ then

  $$ \mathrm{Re} (\phi)=A\cos(kx-\omega t\pm \frac{\pi}{2})=\mp A\sin (kx-\omega t) $$

  🗒️ Note: this is for the right moving part but the equivalent is also true for the left moving part

• 3D wave equation where the Cartesian basis vectors are denoted $\rm \hat x,\hat y,\hat z$ we use $\hat k$ to denote the direction vector

  $$ \frac 1 {v^2}\frac{\partial^2 \phi}{\partial t^2}-\nabla^2\phi=0 $$

• the solutions are given by

  $$ \phi=Ae^{i(\vec k\cdot \vec r-\omega t)}+Be^{i(\vec k\cdot \vec r+\omega t)} $$

  🗒️ Note that the wave-number $k$ from the $1$D case has been promoted to the wave vector $\vec k$

• For simplicity we will write

  $$ \phi=Ae^{i(\vec k \cdot \vec r-\omega t)}=Ae^{i(k_xx+k_yy+k_zz-\omega t)} $$

• We can calculate the spatial derivative

  $$ \frac{\partial \phi}{\partial x}=ik_x Ae^{i(\vec k\cdot \vec r-\omega t)}=ik_x\phi $$

• if we calculate the other spatial derivatives we find

  $$ \vec \nabla \phi=i\vec k\phi $$

  the Laplacian of the solution is

  $$ \nabla ^2 \phi=-|\vec k|^2 \phi $$

• Furthermore if we calculate the second time derivative of the solution we would obtain

  $$ \frac{\partial^2 \phi}{\partial t^2}=-\omega^2 \phi $$

• We can sub in these derivatives into the 3D wave equation to get

  $$ -\frac{\omega ^2}{v^2}+|\vec k|^2 =0 $$

  which readily rearranges to

  $$ \omega =v|\vec k| $$

• points of constant phase which move at speed $v$

  $$ \begin{aligned} &\text{for 1D:} \qquad &kx-\omega t=2\pi n \\ &\text{for 3D:} \qquad &\vec k \cdot \vec r-\omega t=2\pi n \end{aligned} $$

  which at a given time describes a plane of constant phase.

💎 Conclusion: the solution to the plane wave equation is a plane wave. Planes of constant phase travel in the direction of $\vec k$ with speed $v$

Maxwell’s equation in free space & the electric wave equation

In free space there are no charges or currents $\rho=0\,,\,\vec j=0$ Hence Maxwells equations in free space are

$$ \begin{matrix} \vec \nabla \cdot \vec E=0 & \vec \nabla \cdot \vec B=0 \\ \dot {\vec B}=-\vec \nabla \times \vec E & \dot{\vec E} =c^2 \vec \nabla \times \vec B \end{matrix} $$