From Maxwell’s equations to waves

Deriving the wave equation from Maxwell’s equations

Here we show how Maxwell’s equations of electromagnetism, in regions with neither charge nor current, imply a phenomenon obeying the wave equation in three dimensions.¹ Maxwell’s equations in such regions are \[ \begin{aligned} \nabla.\mathbf{E} &= 0,\\ \nabla.\mathbf{B} &= 0,\\ \nabla\times\mathbf{E} &= -\frac{\partial\mathbf{B}}{\partial t},\\ \nabla\times\mathbf{B} &= \mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t}. \end{aligned} \] We can apply the curl to the third equation to get \[ \nabla\times\left(\nabla\times\mathbf{E}\right) = -\nabla\times\left(\frac{\partial\mathbf{B}}{\partial t}\right). \] Applying the identity, \(\nabla\times\left(\nabla\times\mathbf{A}\right) = \nabla\left(\nabla.\mathbf{A}\right) - \nabla^2\mathbf{A}\), we get \[ \nabla\left(\nabla.\mathbf{E}\right) - \nabla^2\mathbf{E} = -\frac{\partial}{\partial t}\left(\nabla\times\mathbf{B}\right). \] Using the first and fourth of Maxwell’s equations, we get \[ \nabla^2\mathbf{E} = \mu_0\epsilon_0\frac{\partial^2\mathbf{E}}{\partial t^2} \] and we have the wave equation, with the speed of the waves given by \[ c = \frac{1}{\sqrt{\mu_0\epsilon_0}}. \] Notice how we have the wave equation for all three components of \(\mathbf{E}\).

¹ With waves preceding electromagnetism in this unit and vector calculus being taught in parallel, it is unlikely that you will be able to fully understand this section initially. However, this appendix is provided so that you can understand how Maxwell’s equations imply the existence of electromagnetic waves, once you have the requisite background knowledge.

We can also apply the curl to the fourth of Maxwell’s equations. Applying the same steps, we get \[ \nabla\times\left(\nabla\times\mathbf{B}\right) = \mu_0\epsilon_0\nabla\times\left(\frac{\partial\mathbf{E}}{\partial t}\right) \] and thence \[ \nabla\left(\nabla.\mathbf{B}\right) - \nabla^2\mathbf{B} = \mu_0\epsilon_0\frac{\partial}{\partial t}\left(\nabla\times\mathbf{E}\right). \] The second and third of Maxwell’s equations then give \[ \nabla^2\mathbf{B} = \mu_0\epsilon_0\frac{\partial^2\mathbf{B}}{\partial t^2} \] and again the wave equation appears.

Plane electromagnetic waves: transverse and perpendicular

Plane, sinusoidal electromagnetic waves, propagating in the \(x\) direction, can be described using the complex representation², \[ \begin{aligned} \mathbf{E}(x, t) &= \mathbf{E}^0e^{i\left(kx - \omega t\right)},\\ \mathbf{B}(x, t) &= \mathbf{B}^0e^{i\left(kx - \omega t\right)}. \end{aligned} \] But now Maxwell’s equations give us more information. Firstly, the first of Maxwell’s equations, \(\nabla.\mathbf{E} = 0\), implies that \(\mathbf{E}^0_x = 0\).

² The superscript here is not \(\mathbf{E}\) raised to a power. I am just using a superscript to avoid clashing with subscripts indicating components.

Exercise (practice with the divergence)

Take the divergence of the electric field provided to show that \(\mathbf{E}^0_x\) must be zero.

Solution

Using the product rule for the divergence, we get \[ \nabla.\mathbf{E} = \left(\nabla.\mathbf{E}^0\right)e^{i\left(kx - \omega t\right)} + \mathbf{E}^0.\nabla e^{i\left(kx - \omega t\right)}. \] The first term is zero, since \(\mathbf{E}^0\) is a constant. We therefore get \[ \nabla.\mathbf{E} = \begin{pmatrix} \mathbf{E}^0_x\\ \mathbf{E}^0_y\\ \mathbf{E}^0_z \end{pmatrix} . \begin{pmatrix} ike^{i\left(kx - \omega t\right)}\\ 0\\ 0 \end{pmatrix} = ik\mathbf{E}^0_xe^{i\left(kx - \omega t\right)}. \] Maxwell’s equations tell us that this must be zero. Since neither the wave number \(k\) nor the exponential can be zero, we are forced to conclude that \(\mathbf{E}^0_x = 0\) and so the electric field has no component parallel to the propagation of the wave.

By the same reasoning, we can use \(\nabla.\mathbf{B} = 0\) to show that \(\mathbf{B}^0_x=0\) and so the magnetic component of the wave is also transverse.

We may also use the third of Maxwell’s equations, i.e. \[ \nabla\times\mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} \] to show that \[ -k\mathbf{E}^0_z = \omega\mathbf{B}^0_y, \qquad k\mathbf{E}^0_y = \omega\mathbf{B}^0_z. \]

Exercise (practice with the curl)

Confirm that \[ \nabla\times\mathbf{E} = ike^{i\left(kx - \omega t\right)} \begin{pmatrix} 0\\ -\mathbf{E}^0_z\\ \mathbf{E}^0_y \end{pmatrix} . \] Compare this with \[ -\frac{\partial\mathbf{B}}{\partial t} = i\omega e^{i\left(kx - \omega t\right)} \begin{pmatrix} 0\\ \mathbf{B}^0_y\\ \mathbf{B}^0_z \end{pmatrix} \] to obtain \[ -k\mathbf{E}^0_z = \omega\mathbf{B}^0_y, \qquad k\mathbf{E}^0_y = \omega\mathbf{B}^0_z. \]

We now note that \[ \frac{k}{\omega}\hat{\mathbf{x}}\times\mathbf{E}^0 = \frac{k}{\omega} \begin{pmatrix} 0\\ -\mathbf{E}^0_z\\ \mathbf{E}^0_y \end{pmatrix} = \begin{pmatrix} 0\\ \mathbf{B}^0_y\\ \mathbf{B}^0_z \end{pmatrix} = \mathbf{B}^0. \] We therefore find that \(\mathbf{B}^0\) is perpendicular to both the \(x\)-axis and \(\mathbf{E}^0\). This in turn implies that \(\mathbf{B}\) is perpendicular to both the \(x\)-axis and \(\mathbf{E}\).

Figure 1: A vertically polarised, sinusoidal electromagnetic wave.