Electromagnetic Waves

Let us demonstrate that Maxwell's equations possess wave-like solutions that can propagate through a vacuum. These solutions are known as electromagnetic waves. Let us start from Maxwell's equations in free space (i.e., with no charges and no currents):

$$\nabla\cdot{\bf E} =0,$$ (2.504)

$$\nabla\cdot{\bf B} =0,$$ (2.505)

$$\nabla\times{\bf E} = - \frac{\partial {\bf B}}{\partial t} ,$$ (2.506)

$$\nabla\times{\bf B} = \epsilon_0\,\mu_0\,\frac{\partial {\bf E}}{\partial t}.$$ (2.507)

[See Equations (2.484)–(2.487).]

There is an easy way to show that the previous equations possess wave-like solutions, and a hard way. The easy way is to assume that the solutions are going to be wave-like beforehand. Specifically, let us search for plane-wave solutions of the form:

$${\bf E}({\bf r}, t) ={\bf E}_0 \cos\,({\bf k}\cdot {\bf r} - \omega\, t),$$ (2.508)

$${\bf B}({\bf r}, t) = {\bf B}_0 \cos\,({\bf k}\cdot {\bf r} - \omega\, t-\phi).$$ (2.509)

Here, ${\bf E}_0$ and ${\bf B}_0$ are constant vectors, ${\bf k}$ is known as the wavevector, and $\omega$ is the angular frequency of oscillation of the wave. The frequency in hertz, $f$, is related to the angular frequency via $\omega = 2\pi\,f$; this frequency is conventionally defined to be positive. The quantity $\phi$ is a phase difference between the electric and magnetic fields. Actually, it is more convenient to write

$${\bf E}({\bf r}, t) = {\bf E}_0 \,{\rm e}^{\,{\rm i}\,({\bf k}\,\cdot\,{\bf r} - \omega \,t)},$$ (2.510)

$${\bf B}({\bf r}, t) = {\bf B}_0 \,{\rm e}^{\,{\rm i}\,({\bf k}\,\cdot\,{\bf r} - \omega \,t)},$$ (2.511)

where, by convention, the physical solution is the real part of the previous equations. The phase difference $\phi$ is absorbed into the constant vector ${\bf B}_0$ by allowing it to become complex. Thus, ${\bf B}_0 \rightarrow {\bf B}_0 \,{\rm e}^{-{\rm i}\,\phi}$. In general, the vector ${\bf E}_0$ is also complex.

Now, assuming (without loss of generality) that ${\bf E}_0$ is real, a wave maximum of the electric field satisfies

$${\bf k}\cdot {\bf r} = \omega\, t + n\,2\pi,$$ (2.512)

where $n$ is an integer. The solution to this equation is a set of equally-spaced parallel planes (one plane for each possible value of $n$), whose normals are parallel to the wavevector ${\bf k}$, and that propagate in the direction of ${\bf k}$ with phase velocity

$$c = \frac{\omega}{k}.$$ (2.513)

The spacing between adjacent planes (i.e., the wavelength) is given by

$$\lambda = \frac{2\pi}{k}.$$ (2.514)

See Figure 2.40.

Figure 2.40: Wavefronts associated with a plane wave.

Consider a general plane-wave vector field

$${\bf A}({\bf r},t)= {\bf A}_0 \,{\rm e}^{\,{\rm i}\,({\bf k}\,\cdot\,{\bf r} - \omega \,t)}.$$ (2.515)

What is the divergence of ${\bf A}$? This is easy to evaluate. We have

$$\nabla\cdot {\bf A} \equiv \frac{\partial A_x}{\partial x}+ \frac{\partial A_y}{\part... ...{\rm i}\,({\bf k}\,\cdot\,{\bf r} - \omega t)}= {\rm i}\, {\bf k}\cdot {\bf A}.$$ (2.516)

(See Section A.20.) How about the curl of ${\bf A}$? This is slightly more difficult. We have

$$(\nabla\times{\bf A})_x \equiv \frac{\partial A_z}{\partial y} -\... ...({\rm i}\,k_y \,A_z - {\rm i}\, k_z \,A_y)= {\rm i}\,({\bf k} \times {\bf A})_x$$ (2.517)

(see Section A.22), which easily generalizes to

$$\nabla\times{\bf A} = {\rm i}\, {\bf k} \times{\bf A}.$$ (2.518)

Hence, it is apparent that vector field operations on a plane-wave vector field are equivalent to replacing the $\nabla$ operator with ${\rm i}\,{\bf k}$. Of course, the $\partial/\partial t$ operator can be replaced by $-{\rm i}\,\omega$.

The first Maxwell equation, (2.504), reduces to

$${\rm i}\, {\bf k} \cdot {\bf E}_0 = 0,$$ (2.519)

using the assumed electric and magnetic fields, (2.510) and (2.511), and Equation (2.516). Thus, the electric field is perpendicular to the direction of propagation of the wave. (See Section A.6.) Likewise, the second Maxwell equation, (2.505), gives

$${\rm i}\,{\bf k} \cdot {\bf B}_0 = 0,$$ (2.520)

implying that the magnetic field is also perpendicular to the direction of propagation. Clearly, the wave-like solutions of Maxwell's equation are a type of transverse wave. The third Maxwell equation, (2.506), yields

$${\rm i}\,{\bf k}\times {\bf E}_0 = {\rm i}\, \omega\, {\bf B}_0,$$ (2.521)

where use has been made of Equation (2.518). Forming the scalar product of this equation with ${\bf E}_0$ gives

$${\bf E}_0 \cdot {\bf B}_0 = \frac{ {\bf E}_0 \cdot {\bf k} \times {\bf E}_0 }{ \omega }= 0.$$ (2.522)

Thus, the electric and magnetic fields are mutually perpendicular. (See Sections A.6 and A.10.) Forming the scalar product of Equation (2.521) with ${\bf B}_0$ yields

$${\bf B}_0 \cdot {\bf k} \times{\bf E}_0 = \omega\,B_0^{\,2} > 0.$$ (2.523)

Thus, the vectors ${\bf E}_0$, ${\bf B}_0$, and ${\bf k}$ are mutually perpendicular, and form a right-handed set. (See Section A.10.) The final Maxwell equation, (2.507), gives

$${\rm i}\, {\bf k}\times{\bf B}_0 = -{\rm i}\,\epsilon_0\,\mu_0 \,\omega\, {\bf E}_0.$$ (2.524)

Combining this equation with Equation (2.521) yields

$${\bf k}\times ({\bf k} \times {\bf E}_0) \equiv ({\bf k} \cdot {\... ...- k^2\,{\bf E}_0 =-k^2\,{\bf E}_0 = - \epsilon_0 \,\mu_0\,\omega^2\, {\bf E}_0,$$ (2.525)

or

$$k^2 = \epsilon_0 \,\mu_0 \,\omega^2,$$ (2.526)

where use has been made of Equation (2.519). (See Section A.11.) However, we know, from Equation (2.513), that the phase velocity, $c$, of the wave is related to the magnitude of the wavevector and the angular wave frequency via $c = \omega/k$. Thus, we obtain

$$c = \frac{1}{\sqrt{\epsilon_0 \,\mu_0}}.$$ (2.527)

We have found transverse plane-wave solutions of the free-space Maxwell equations propagating at some phase velocity $c$, that is given by a combination of $\epsilon_0$ and $\mu_0$, and is, thus, the same for all frequencies and wavelengths. The constants $\epsilon_0$ and $\mu_0$ are easily measurable. The former is related to the force acting between stationary electric charges, and the latter to the force acting between steady electric currents. Both of these constants were fairly well known in Maxwell's time. Maxwell, incidentally, was the first person to look for wave-like solutions of his equations, and, thus, to derive Equation (2.527). The modern values of $\epsilon_0$ and $\mu_0$ are

$$\epsilon_0 = 8.8542\times 10^{-12} \,{\rm C}^2\,{\rm N}^{-1}\,{\rm m}^{-2},$$ (2.528)

$$\mu_0 =4\pi\times 10^{-7}\,{\rm N} \,{\rm A}^{-2}.$$ (2.529)

Let us use these values to find the phase velocity of electromagnetic waves. We obtain

$$c = \frac{1}{\sqrt{\epsilon_0\, \mu_0}} = 2.998\times 10^8\,{\rm m}\,{\rm s}^{-1}.$$ (2.530)

Of course, we immediately recognize this as the speed of light in vacuum. Maxwell also made this connection back in the 1870's. He conjectured that light, whose nature had previously been unknown, was a form of electromagnetic radiation. This was a remarkable prediction. After all, Maxwell's equations were derived from the results of bench-top laboratory experiments involving charges, batteries, coils, and currents, that apparently had nothing whatsoever to do with light.

Maxwell was able to make another remarkable prediction. The wavelength of light was well-known in the late nineteenth century from studies of diffraction through slits, et cetera. Visible light actually occupies a surprisingly narrow wavelength range. The shortest wavelength blue light that is visible to the typical human eye has a wavelength of $\lambda= 0.38$ microns (one micron is $10^{-6}$ meters). The longest wavelength red light that is visible has a wavelength of $\lambda= 0.75$ microns. However, there is nothing in our analysis that suggests that this particular range of wavelengths is special. Electromagnetic waves can have any wavelength. Maxwell concluded that visible light was a small part of a vast spectrum of previously undiscovered types of electromagnetic radiation. Since Maxwell's time, virtually all of the non-visible parts of the electromagnetic spectrum have been observed.

Table 1 gives a brief guide to the electromagnetic spectrum. Electromagnetic waves are of particular importance to us because they are our main source of information regarding the universe around us. Radio waves and microwaves (which are comparatively hard to scatter) have provided much of our knowledge about the center of our own galaxy. This is completely unobservable in visible light, which is strongly scattered by interstellar gas and dust lying in the galactic plane. For the same reason, the spiral arms of our galaxy can only be mapped out using radio waves. Infrared radiation is useful for detecting protostars, which are not yet hot enough to emit visible radiation. Of course, visible radiation is still the mainstay of astronomy. Satellite-based ultraviolet observations have yielded invaluable insights into the structure and distribution of distant galaxies. Finally, X-ray and $\gamma$-ray astronomy usually concentrates on exotic objects, such as pulsars and supernova remnants.

Table 2.1: The electromagnetic spectrum

Radiation type	Wavelength range ($m$)
Gamma Rays	$<10^{-11}$
X-Rays	$10^{-11}$–$10^{-9}$
Ultraviolet	$10^{-9}$–$10^{-7}$
Visible	$10^{-7}$–$10^{-6}$
Infrared	$10^{-6}$–$10^{-4}$
Microwave	$10^{-4}$–$10^{-1}$
TV-FM	$10^{-1}$–$10^1$
Radio	$>10^1$

Equations (2.519), (2.521), and the relation $c = \omega/k$, imply that

$$B_0= \frac{E_0}{c}.$$ (2.531)

Thus, the magnetic field associated with an electromagnetic wave is smaller in magnitude than the electric field by a factor $c$. Consider an electrically charged particle interacting with an electromagnetic wave. The force exerted on the particle is given by the Lorentz force law,

$${\bf f} = q \,({\bf E} +{\bf v}\times{\bf B}).$$ (2.532)

(See Section 2.2.4.) The ratio of the electric and magnetic forces is

$$\frac{f_{\rm magnetic}}{f_{\rm electric}} \simeq \frac{v\,B_0}{E_0} \simeq \frac{v}{c}.$$ (2.533)

So, unless the particle is moving close to the speed of light (i.e., unless the particle is relativistic), the electric force greatly exceeds the magnetic force. Clearly, in most terrestrial situations, electromagnetic waves are an essentially electrical phenomenon (as far as their interaction with matter is concerned). For this reason, electromagnetic waves are usually characterized by their wavevector, ${\bf k}$ (which specifies the direction of propagation and the wavelength), and the plane of polarization (i.e., the plane of oscillation) of the associated electric field. For a given wavevector, ${\bf k}$, the electric field can have any direction in the plane normal to ${\bf k}$. [See Equation (2.519).] However, there are only two independent directions in a plane (i.e., we can only define two linearly independent vectors in a plane). This implies that there are only two independent polarizations of an electromagnetic wave, once its direction of propagation is specified.

But, how do electromagnetic waves propagate through a vacuum? After all, most types of wave require a medium before they can propagate (e.g., sound waves require air). The answer to this question is evident from Equations (2.506) and (2.507). According to these equations, the time variation of the electric component of the wave induces the magnetic component, and the time variation of the magnetic component induces the electric component. In other words, electromagnetic waves are self-sustaining, and, therefore, require no medium through which to propagate.

Let us now search for the wave-like solutions of Maxwell's equations in free-space the hard way. Suppose that we take the curl of the fourth Maxwell equation, (2.507). We obtain

$$\nabla\times\nabla\times{\bf B} \equiv \nabla(\nabla\cdot{\bf B})... ...\bf B} = \epsilon_0\,\mu_0\,\frac{\partial\, \nabla\times {\bf E}}{\partial t}.$$ (2.534)

[See Equation (A.187).] Here, we have made use of the fact that $\nabla\cdot{\bf B} = 0$, according to the second Maxwell equation, (2.505). The third Maxwell equation, (2.506), yields

$$\left(\frac{1}{c^2} \frac{\partial^2}{\partial t^2}-\nabla^2\right) {\bf B} = {\bf0},$$ (2.535)

where use has been made of Equation (2.530). A similar equation can obtained for the electric field by taking the curl of Equation (2.506):

$$\left( \frac{1}{c^2} \frac{\partial^2}{\partial t^2}-\nabla^2\right) {\bf E} = {\bf0},$$ (2.536)

Figure 2.41: An arbitrary wave-pulse.

We have found that electric and magnetic fields both satisfy equations of the form

$$\left(\frac{1}{c^2} \frac{\partial^2}{\partial t^2}-\nabla^2\right) {\bf A} = {\bf0}$$ (2.537)

in free space. As is easily verified, the most general solution to this equation is

$$A_x = F_x ({\bf n}\cdot{\bf r} -c\,t),$$ (2.538)

$$A_y = F_y ({\bf n}\cdot{\bf r} - c\,t),$$ (2.539)

$$A_z = F_z ({\bf n}\cdot{\bf r} - c\,t),$$ (2.540)

where ${\bf n}$ is a unit vector, and $F_x(\phi)$, $F_y(\phi)$, and $F_z(\phi)$ are arbitrary one-dimensional scalar functions. Looking along the direction of ${\bf n}$, so that ${\bf n}\cdot {\bf r} = r$, we find that

$$A_x = F_x (r-c\,t),$$ (2.541)

$$A_y = F_y (r-c\,t),$$ (2.542)

$$A_z = F_z (r-c\,t).$$ (2.543)

The $x$-component of this solution is shown schematically in Figure 2.41. The solution clearly propagates along the $r$-axis, at the speed $c$, without changing shape. If we look along a direction that is perpendicular to ${\bf n}$ then ${\bf n}\cdot{\bf r} = 0$, and there is no propagation. Thus, the components of ${\bf A}$ are arbitrarily-shaped pulses that propagate, without changing shape, along the direction of ${\bf n}$ with speed $c$. These pulses can be related to the sinusoidal plane-wave solutions which we found earlier by Fourier transformation; that is,

$$F_x(r-c\,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \overline{F}_x(k)\,{\rm e}^{\,{\rm i}\,k\,(r-c\,t)}\,dk,$$ (2.544)

where

$$\overline{F}_x(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} F_x(x)\,{\rm e}^{-{\rm i}\,k\,x}\,dx.$$ (2.545)

(See Section 4.2.4.) Thus, any arbitrary-shaped pulse propagating in the direction of ${\bf n}$ with speed $c$ can be broken down into a superposition of sinusoidal oscillations of different wavevectors, $k\,{\bf n}$, propagating in the same direction with the same speed.