Electromagnetic Waves

In a vacuum where there are no currents and no charges, so that

$${\bf J} = 0, \qquad \rho_{c} = 0,$$

Maxwell's equations become

$$\nabla \cdot {\bf E} = 0,$$ (6.12)

$$\nabla \times {\bf E} = -{\partial {\bf B}\over \partial t},$$ (6.13)

$$\nabla \cdot {\bf B} = 0,$$ (6.14)

$$\nabla \times {\bf B} = \mu_{0}\epsilon_{0}{\partial {\bf E}\over \partial t}.$$ (6.15)

Consider equations (6.13) and (6.15). If we take, $\partial /\partial t$ of (6.13) we obtain

$${\partial^{2}{\bf B}\over \partial t^{2}} = - \nabla \times {\partial {\bf E}\over \partial t}.$$

Now we use the curl of (6.15) to rewrite the right hand side in terms of ${\bf B}$ alone. Hence,

$${\partial^{2}{\bf B}\over \partial t^{2}} = -{1\over \mu_... ...t (\nabla \cdot {\bf B}\right ) - \nabla^{2}{\bf B}\right ].$$

Finally, using (6.14) we obtain

$${\partial^{2}{\bf B}\over \partial t^{2}} = {1\over \mu_{0}\epsilon_{0}}\nabla^{2}{\bf B},$$ (6.16)

where

$$c^{2} = {1\over \mu_{0}\epsilon_{0}},$$

is the speed of light squared. Thus, the magnetic field and the electric field support electromagnetic waves that propagate at the speed of light. Note, that this represents very rapid time variations so that the usual MHD approximation does not apply.

In Cartesian coordinates, the wave equation has constant coefficients and solutions can be represented in the form

$${\bf B} = {\bf B}_{0}e^{i({\bf k}\cdot {\bf r} - \omega t)},$$

$${\bf E} = {\bf E}_{0}e^{i({\bf k}\cdot {\bf r} - \omega t)},$$

where

$${\bf k} = k{\bf i} + l {\bf j} + m {\bf k}, \qquad {\bf k}\cdot {\bf r} = kx + ly + mz,$$

and the dispersion relation is

$$\omega^{2} = \left (k^{2} + l^{2} + m^{2}\right )c^{2},$$

or

$$\omega = \pm \sqrt{k^{2} + l^{2} + m^{2}}c.$$ (6.17)

Electromagnetic waves, such as gamma rays, x-rays, UV, visible light, infra-red waves and radio waves, have wavelengths ranging from $3\times 10^{-16}m$ to $3\times 10^{4}m$, all propagate at the speed of light. (6.12) and (6.14) imply that

$${\bf k}\cdot {\bf E} = 0 \hbox{ and } {\bf k}\cdot {\bf B} = 0,$$

and so the electromagnetic waves are transverse since they propagate in a direction that is at right angles to the direction of the electric and magnetic field vectors. In addition, (6.13) gives

$${\bf k}\times {\bf E} = \omega {\bf B},$$

so that

$${\bf E}\cdot {\bf B} = 0$$

and the electric and magnetic fields associated with the waves are perpendicular to each other. Finally, (6.15) implies

$${\bf k}\times {\bf E} = \vert{\bf k}\vert\vert{\bf E}\vert\hat{{\bf n}},$$

where $\hat{{\bf n}}$ is a unit vector in the direction of the magnetic field,

$$\omega {\bf B} = \omega \vert{\bf B}\vert\hat{{\bf n}},$$

and so

$${\omega \over \vert{\bf k}\vert} = {E\over B} = \pm c.$$

Example 6. .2Suppose that ${\bf E} = E_{y}(x,t){\bf j}$, ${\bf B} = B_{z}(x,t){\bf k}$ then (6.12) and (6.14) are satisfied identically and (6.13) and (6.15) imply

$$-{\partial B_{z}\over \partial t} = {\partial E_{y}\over \partial x},$$ (6.18)

$$\mu_{0}\epsilon_{0}{\partial E_{y}\over \partial t} = - {\partial B_{z}\over \partial x},$$ (6.19)

and so

$${\partial^{2}B_{z}\over \partial t^{2}} = c^{2}{\partial^{2}B_{z}\over \partial x^{2}}.$$ (6.20)

One possible solution is

$$B_{z} = B_{0}\cos (kx - \omega t),\qquad E_{y} = E_{0}\cos (kx - \omega t), \qquad \omega = kc,$$

and

$$\omega B_{0} = k E_{0}.$$