Electromagnetic Waves

Abstract

The Maxwell equations are shown to lead, in a straightforward way, to separate wave equations for the electric and the magnetic field. The characteristics of monochromatic plane electromagnetic (e/m) waves are studied. The propagation of e/m waves in a conducting medium is examined and the concept of plasma frequency is introduced. The production of e/m radiation by various kinds of sources is described.

Appendices

Questions

• 1. Give the general mathematical expression for a harmonic plane wave $\xi \kern0.2em \left(\overrightarrow{r},t\right)$ traveling (a) in the +y direction; (b) in the –z direction.

• 2. A plane e/m wave is traveling in the –x direction. The magnetic field at some point of space is instantaneously oriented in the –z direction. What is the corresponding instantaneous orientation of the electric field at that point?

• 3. Show that the standing waves (10.60) satisfy the wave eq. (10.61).

• 4. Two media, a conducting and a non-conducting one, share the same constant values of ε and μ. Compare the propagation speeds of plane e/m waves of a certain frequency ω in these media.

• 5. A conducting medium has constant values of ε, μ, σ. Compare the propagation speeds υ₁ and υ₂ in this medium, of two plane e/m waves of frequencies ω₁ and ω₂ where ω₁ < ω₂ .

• 6. Two plane waves of low frequencies ω₁ and ω₂ = 2ω₁ fall on the surface of a conductor. Which wave will penetrate deeper into the conductor?

• 7. Consider two metals Μ₁ and Μ₂. The skin depths in these metals for the visible part of the spectrum of e/m radiation are such that Δ₁ < <Δ₂. Which of the two metals would you use to make a mirror?

• 8. Examine which of the following two systems emits e/m radiation: (a) an electron in uniform circular motion; (b) a steady circular current Ι.

• 9. At the center of a very large sphere there is a small electric dipole oscillating along a vertical axis passing through the center of the sphere. At which points of the spherical surface is the intensity of the emitted e/m radiation a maximum and at which points is it a minimum? Repeat the problem, this time considering an oscillating magnetic dipole at the center of the sphere.

• 10. Where is it preferable to listen to the radio, inside a metal room or inside a glass room? Explain.

• 11. On a sunny summer day we lock ourselves inside a dark metal room. Do we need to put on our sunscreen?

• 12. A sailor in a submarine (which is submerged at a very low depth) and an astronaut in a spacecraft are both interested in the radio broadcasting of a football game. The game is being broadcasted at two radio stations: SPORT FM (107.4 MHz) and SPORT AM (560 kHz). Which station should each of these persons choose? (Hint: The plasma frequency of the sea is much higher than radio frequencies while the plasma frequency of the ionosphere is in the region of FM radio waves. How does the penetrating ability of an e/m wave vary with the frequency ω of the wave for ω < <ω_p? What happens when ω > ω_p?).

• 13. Maxwell is assigned the post of commander of a warship. The ship possesses (a) a radar, (b) a sonar, (c) an AM radio transmitter, and (d) a microwave emitter. His multitask mission is (1) to detect enemy submarines; (2) to check for enemy warships; (3) to send a radio signal to a friendly submarine diving at a depth of 5 m; (4) to send a signal to a spacecraft in orbit around the Earth. Which apparatus must he use for each task?

• 14. How do you explain the fact that in nighttime we can hear remote AM radio stations that we are not able to hear in daytime? Why do we receive these stations in the shortwave frequency band?

• 15. Imagine our world if the plasma frequency of the ionosphere happened to be as high as that of the sea. How many hours would you study Physics during daytime and how many hours during nighttime? (Hint: The plasma frequency of the sea is much above the visible spectrum.)

Problems

• 1. Consider a plane wave of the form ξ (x,t) = f (x–υt). Show that, if the function ξ is periodic in t, then it is necessarily periodic in x as well, and vice versa.

• Solution: Assume that ξ is periodic in t with period Τ:

ξ(x, t) = ξ(x, t + Τ) ⇒ f(x – υ t) = f[x – υ(t + Τ)] = f[(x – υT) – υt] = f[(x – λ) – υ t]

where we have set λ = υΤ. Hence, ξ (x, t) = ξ (x–λ, t), which means that ξ is periodic in x, with period λ. Conversely, by assuming that ξ is periodic in x with period λ, it is easy to show that it will also be periodic in t, with period Τ = λ/υ . If ξ represents a harmonic plane wave, what is the physical significance of λ and Τ?

• 2. According to the principle of relativity , the laws of Physics must have the same form in all inertial frames of reference. In particular, the Maxwell equations must be of the same form (i.e., must “look the same”) for all inertial observers. Moreover, as found by experiment, the speed of light in empty space has the same value c in every inertial reference frame (indeed, from the Maxwell equations it follows that the value of c depends only on the constants ε₀ and μ₀). Consider now a charge q in uniform rectilinear motion relative to an inertial observer O. Show that, by the principle of relativity, this charge cannot emit e/m radiation.

• Solution: Since q moves with constant velocity relative to the inertial observer O, it itself defines the origin of an inertial frame of reference. With respect to an observer O′ of this frame, q is at rest. Thus the only thing recorded by O′ is a static electric field with no presence of any e/m wave. Let us now assume that, according to O, the charge q emits e/m radiation. This means that O records the presence of an e/m wave in her own frame, which wave travels at speed c. But, since the propagation speed of an e/m wave (not just of light!) is the same in all inertial frames, it follows that, relative to O′, the wave observed by O also propagates at speed c. This, however, contradicts the fact that O′ does not perceive any e/m wave whatsoever! We conclude that neither O can perceive an emission of e/m radiation from q.

In the case of an accelerating charge q the above rationale breaks down since q no longer defines the origin of an inertial frame of reference (indeed, there is no inertial frame relative to which q is at rest at all moments). Therefore the principle of relativity does not apply in this case.

• 3. Consider two harmonic e/m waves of frequencies ω₁ and ω₂ , propagating in the +x direction in empty space. The electric fields in the waves are

$$ {\overrightarrow{E}}_1\left(x,t\right)={E}_0\;\cos \kern0.2em \left({k}_1x-{\omega}_1\kern0.1em t\right)\kern0.22em {\hat{u}}_y,\kern0.36em {\overrightarrow{E}}_2\left(x,t\right)={E}_0\;\cos \kern0.2em \left({k}_2x-{\omega}_2\kern0.1em t\right)\kern0.22em {\hat{u}}_y;\kern0.6em \frac{\omega_1}{k_1}=\frac{\omega_2}{k_2}=c $$

(a) Find the corresponding magnetic fields in the waves. (b) Find the expression for the e/m wave that results from the superposition of these waves. Is this a standing or a traveling wave?

• Solution: In both waves the fields $\overrightarrow{E}$ and $\overrightarrow{B}$ oscillate in phase and their amplitudes are related by Β₀ = Ε₀ /c . Furthermore, the cross product $\overrightarrow{E}\times \overrightarrow{B}$ is in the direction of propagation, i.e., in the +x direction. Therefore we must have:

$$ {\overrightarrow{B}}_1\left(x,t\right)=\frac{E_0}{c}\kern0.24em \cos \kern0.2em \left({k}_1x-{\omega}_1\kern0.1em t\right)\kern0.22em {\hat{u}}_z,\kern0.48em {\overrightarrow{B}}_2\left(x,t\right)=\frac{E_0}{c}\kern0.24em \cos \kern0.2em \left({k}_2x-{\omega}_2\kern0.1em t\right)\kern0.22em {\hat{u}}_z $$

By using the trigonometric identity

$$ \cos\;\alpha +\cos\;\beta =2\ \cos\;\left[\left(\alpha +\beta \right)/2\right]\ \cos\;\left[\left(\alpha \hbox{--} \beta \right)/2\right] $$

and by noticing that ω₁ ± ω₂ = c (k₁ ± k₂), show that

$$ {\displaystyle \begin{array}{l}\kern0.2em \overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2=2{E}_0\;\cos\;\left[\frac{k_1-{k}_2}{2}\kern0.24em \left(x-c\kern0.2em t\right)\right]\kern0.32em \cos\;\left[\frac{k_1+{k}_2}{2}\kern0.24em \left(x-c\kern0.2em t\right)\right]\kern0.32em {\hat{u}}_y\\ {}\overrightarrow{B}={\overrightarrow{B}}_1+{\overrightarrow{B}}_2=\frac{2{E}_0}{c}\kern0.24em \cos\;\left[\frac{k_1-{k}_2}{2}\kern0.24em \left(x-c\kern0.2em t\right)\right]\kern0.32em \cos\;\left[\frac{k_1+{k}_2}{2}\kern0.24em \left(x-c\kern0.2em t\right)\right]\kern0.32em {\hat{u}}_z\end{array}} $$

The wave is a traveling one since it is of the form F(x–ct) (compare with the example of Sect. 10.8). We can view it as a harmonic wave of frequency (ω₁ + ω₂) /2, the amplitude of which constitutes a separate harmonic wave of frequency |ω₁–ω₂| /2 . This is the simplest example of amplitude modulation, a method on which AM radio broadcasting is based.

• 4. A submarine submerged at depth 20 m can barely receive an e/m signal of very long wavelength λ (in the air). To what depth must the submarine shift in order for it to receive a signal of wavelength λ/4 ? (Assume that, approximately, c=λf in the air, where f is the frequency of the original signal)

• Solution: According to Eq. (10.71), the skin depth for a wave of (angular) frequency ω = 2π f = 2π c/λ is

$$ \kern0.24em \varDelta \simeq \sqrt{\frac{2}{\omega \kern0.1em \mu \kern0.1em \sigma }}=\sqrt{\frac{\lambda }{\pi c\mu \kern0.1em \sigma }} $$

Hence,

$$ {\varDelta}_2/{\varDelta}_1=\sqrt{\lambda_2/{\lambda}_1} $$

By substituting Δ₁ = 20 m and λ₂ = λ₁ /4, we get: Δ₂ = 10 m.

• 5. We will study the characteristics of a general (non-monochromatic) plane e/m wave in empty space. Such a wave can be expressed as a linear combination of monochromatic waves of various frequencies ω, which waves are traveling in the same direction, say, +x. If ω varies continuously, this linear combination has the general form

$$ \overrightarrow{E}\kern0.2em \left(x,t\right)=\int {\overrightarrow{E}}_0\kern0.2em (k)\kern0.22em {e}^{i\kern0.22em \left(k\kern0.1em x-\omega \kern0.22em t\right)}\;d\kern0.1em k,\kern0.72em \overrightarrow{\mathrm{B}}\kern0.2em \left(x,t\right)=\int {\overrightarrow{\mathrm{B}}}_0\kern0.2em (k)\kern0.22em {e}^{i\kern0.22em \left(k\kern0.1em x-\omega \kern0.22em t\right)}\;d\kern0.1em k $$

where ω/k = c ⇔ ω = ck, with c = (ε₀ μ₀)^–1/2 (note that the wave vector is written $\overrightarrow{k}=\mid \overrightarrow{k}\mid {\hat{u}}_x=k\;{\hat{u}}_x$). As we know (Sect. 10.5), the monochromatic waves

$$ {\overrightarrow{E}}_0(k)\;{e}^{i\kern0.22em \left(k\kern0.1em x-\omega \kern0.22em t\right)},\kern0.84em {\overrightarrow{B}}_0(k)\;{e}^{i\kern0.22em \left(k\kern0.1em x-\omega \kern0.22em t\right)} $$

are solutions of the Maxwell equations in empty space, provided that

$$ {\displaystyle \begin{array}{l}\overrightarrow{k}\cdot {\overrightarrow{E}}_0(k)=k\;{\hat{u}}_x\cdot {\overrightarrow{E}}_0(k)=0,\kern0.72em \overrightarrow{k}\cdot {\overrightarrow{B}}_0(k)=k\;{\hat{u}}_x\cdot {\overrightarrow{B}}_0(k)=0\\ {}\overrightarrow{k}\times {\overrightarrow{E}}_0(k)=k\;{\hat{u}}_x\times {\overrightarrow{E}}_0(k)=\omega\;{\overrightarrow{B}}_0(k),\kern0.6em \overrightarrow{k}\times {\overrightarrow{B}}_0(k)=k\;{\hat{u}}_x\times {\overrightarrow{B}}_0(k)=-\frac{\omega }{c^2}\kern0.22em {\overrightarrow{E}}_0(k)\end{array}} $$

• (a) Show that $\overrightarrow{E}\left(x,t\right)$ and $\overrightarrow{B}\left(x,t\right)$ are plane waves traveling in the +x direction with speed c = ω/k . [Hint: Such waves are expressed as functions of the form F(x–ct).]

• (b) Show that $\overrightarrow{E}\left(x,t\right)$ and $\overrightarrow{B}\left(x,t\right)$ satisfy the Maxwell equations in empty space. (Remember that ε₀ μ₀ = 1/c² .)

• (c) Show that ${\hat{u}}_x\cdot \overrightarrow{E}=0,\kern0.48em {\hat{u}}_x\cdot \overrightarrow{B}=0$. That is, the $\overrightarrow{E}\left(x,t\right)$ and $\overrightarrow{B}\left(x,t\right)$ are normal to the direction of propagation of the wave (transverse wave ).

• (d) Show that ${\hat{u}}_x\times \overrightarrow{E}=c\kern0.1em \overrightarrow{B},\kern0.48em {\hat{u}}_x\times \overrightarrow{B}=-\left(1/c\right)\;\overrightarrow{E}$. That is, the $\left(\overrightarrow{E},\overrightarrow{B},{\hat{u}}_x\right)$ form a right-handed rectangular system.

• (e) Show that the instantaneous values of $\overrightarrow{E}$ and $\overrightarrow{B}$ are related by E = cB.

• Solution:

• (a) $\overrightarrow{E}\kern0.2em \left(x,t\right)=\int {\overrightarrow{E}}_0\kern0.2em (k)\kern0.22em {e}^{i\;k\kern0.1em \left(x-c\kern0.1em t\right)}\;d\kern0.1em k\equiv \overrightarrow{F}\left(x-c\kern0.2em t\right)$; similarly for $\overrightarrow{B}$

• (b) The Maxwell equations in empty space are

$$ {\displaystyle \begin{array}{l}\overrightarrow{\nabla}\cdot \overrightarrow{E}=0\kern1.92em \overrightarrow{\nabla}\times \overrightarrow{E}=-\frac{\partial \kern0.1em \overrightarrow{B}}{\partial \kern0.2em t}\kern0.6em \\ {}\overrightarrow{\nabla}\cdot \overrightarrow{B}=0\kern1.9em \overrightarrow{\nabla}\times \overrightarrow{B}={\varepsilon}_0\kern0.1em {\mu}_0\kern0.24em \frac{\partial \kern0.1em \overrightarrow{E}}{\partial \kern0.2em t}=\frac{1}{c^2}\kern0.24em \frac{\partial \kern0.1em \overrightarrow{E}}{\partial \kern0.2em t}\end{array}} $$

By using Eqs. (10.41) and by taking into account that

$$ \overrightarrow{\nabla}\kern0.22em {e}^{i\kern0.2em \left(k\kern0.1em x-\omega \kern0.22em t\right)}=i\;k\;{e}^{i\kern0.2em \left(k\kern0.1em x-\omega \kern0.22em t\right)}\;{\hat{u}}_x $$

we have:

$$ \overrightarrow{\nabla}\cdot \overrightarrow{E}=\int {\overrightarrow{E}}_0\kern0.2em (k)\cdot \overrightarrow{\nabla}\kern0.22em {e}^{i\kern0.22em \left(k\kern0.1em x-\omega \kern0.22em t\right)}\;d\kern0.1em k=\int i\;k\;{\hat{u}}_x\cdot {\overrightarrow{E}}_0(k)\;{e}^{i\kern0.22em \left(k\kern0.1em x-\omega \kern0.22em t\right)}\;d\kern0.1em k\kern0.34em =0;\kern0.34em \mathrm{similarly},\overrightarrow{\nabla}\cdot \overrightarrow{B}=0 $$

$$ {\displaystyle \begin{array}{l}\overrightarrow{\nabla}\times \overrightarrow{E}+\frac{\partial \kern0.1em \overrightarrow{B}}{\partial \kern0.2em t}=\int \left[\overrightarrow{\nabla}\kern0.1em {e}^{i\kern0.22em \left(k\kern0.1em x-\omega \kern0.22em t\right)}\right]\times {\overrightarrow{E}}_0\kern0.2em (k)\;d\kern0.1em k\kern0.36em -\kern0.34em \int i\;\omega\;{\overrightarrow{B}}_0(k)\;{e}^{i\kern0.22em \left(k\kern0.1em x-\omega \kern0.22em t\right)}\;d\kern0.1em k\\ {}\kern3.179999em =i\int d\kern0.1em k\kern0.32em {e}^{i\kern0.22em \left(k\kern0.1em x-\omega \kern0.22em t\right)}\kern0.22em \left[k\;{\hat{u}}_x\times {\overrightarrow{E}}_0\kern0.2em (k)-\omega\;{\overrightarrow{B}}_0(k)\right]\kern0.36em =0\\ {}\overrightarrow{\nabla}\times \overrightarrow{B}-\frac{1}{c^2}\kern0.24em \frac{\partial \kern0.1em \overrightarrow{E}}{\partial \kern0.2em t}=i\int d\kern0.1em k\kern0.32em {e}^{i\kern0.22em \left(k\kern0.1em x-\omega \kern0.22em t\right)}\kern0.22em \left[k\;{\hat{u}}_x\times {\overrightarrow{B}}_0\kern0.2em (k)+\frac{\omega }{c^2}\;{\overrightarrow{E}}_0(k)\right]\kern0.36em =0\end{array}} $$

• (c) ${\hat{u}}_x\cdot \overrightarrow{E}=\int {\hat{u}}_x\cdot {\overrightarrow{E}}_0\kern0.2em (k)\kern0.22em {e}^{i\kern0.22em \left(k\kern0.1em x-\omega \kern0.22em t\right)}\;d\kern0.1em k=0,\kern0.84em {\hat{u}}_x\cdot \overrightarrow{B}=\int {\hat{u}}_x\cdot {\overrightarrow{B}}_0\kern0.2em (k)\kern0.22em {e}^{i\kern0.22em \left(k\kern0.1em x-\omega \kern0.22em t\right)}\;d\kern0.1em k=0$

• (d) ${\hat{u}}_x\times \overrightarrow{E}=\int {\hat{u}}_x\times {\overrightarrow{E}}_0\kern0.2em (k)\kern0.22em {e}^{i\kern0.22em \left(k\kern0.1em x-\omega \kern0.22em t\right)}\;d\kern0.1em k=\int \frac{\omega }{k}\;{\overrightarrow{B}}_0(k)\;{e}^{i\kern0.22em \left(k\kern0.1em x-\omega \kern0.22em t\right)}\;d\kern0.1em k=c\kern0.1em \overrightarrow{B}$

$$ {\hat{u}}_x\times \overrightarrow{B}=\int {\hat{u}}_x\times {\overrightarrow{B}}_0\kern0.2em (k)\kern0.22em {e}^{i\kern0.22em \left(k\kern0.1em x-\omega \kern0.22em t\right)}\;d\kern0.1em k=-\int \frac{\omega }{k\kern0.1em {c}^2}\;{\overrightarrow{E}}_0(k)\;{e}^{i\kern0.22em \left(k\kern0.1em x-\omega \kern0.22em t\right)}\;d\kern0.1em k=-\frac{1}{c}\;\overrightarrow{E} $$

• (e) We now consider the relations ${\hat{u}}_x\times \overrightarrow{E}=c\kern0.1em \overrightarrow{B},\kern0.48em {\hat{u}}_x\times \overrightarrow{B}=-\left(1/c\right)\;\overrightarrow{E}$ for real values of $\overrightarrow{E}$ and $\overrightarrow{B}$. By taking the results of part (c) into account, the first relation yields:

$$ \mid {\hat{u}}_x\times \overrightarrow{E}\mid =c\kern0.3em \mid \overrightarrow{B}\mid \kern0.48em \Rightarrow \kern0.36em \mid \overrightarrow{E}\mid =c\kern0.3em \mid \overrightarrow{B}\mid \kern0.48em \Rightarrow \kern0.48em E=c\kern0.1em B $$

We get the same result by using the second relation.

• 6. A lamp emits a huge number of monochromatic e/m waves of various frequencies ω and wave vectors $\overrightarrow{k}$. The e/m field in the surrounding empty space is the result of superposition of these waves and, for continuously varying ω and $\overrightarrow{k}$, it is written as

$$ {\displaystyle \begin{array}{l}\overrightarrow{E}\kern0.2em \left(\overrightarrow{r},t\right)=\iiint {\overrightarrow{E}}_0\left(\overrightarrow{k}\right)\;{e}^{i\kern0.2em \left(\overrightarrow{k}\cdot \overrightarrow{r}-\omega \kern0.22em t\right)}\;d\kern0.1em {k}_x\kern0.1em d\kern0.1em {k}_y\kern0.1em d\kern0.1em {k}_z\equiv \int {\overrightarrow{E}}_0\kern0.1em \left(\overrightarrow{k}\right)\kern0.22em {e}^{i\kern0.2em \left(\overrightarrow{k}\cdot \overrightarrow{r}-\omega \kern0.22em t\right)}\;d\kern0.1em \overrightarrow{k}\\ {}\overrightarrow{B}\kern0.2em \left(\overrightarrow{r},t\right)=\iiint {\overrightarrow{B}}_0\left(\overrightarrow{k}\right)\;{e}^{i\kern0.2em \left(\overrightarrow{k}\cdot \overrightarrow{r}-\omega \kern0.22em t\right)}\;d\kern0.1em {k}_x\kern0.1em d\kern0.1em {k}_y\kern0.1em d\kern0.1em {k}_z\equiv \int {\overrightarrow{B}}_0\kern0.1em \left(\overrightarrow{k}\right)\kern0.22em {e}^{i\kern0.2em \left(\overrightarrow{k}\cdot \overrightarrow{r}-\omega \kern0.22em t\right)}\;d\kern0.1em \overrightarrow{k}\end{array}} $$

where $\omega = ck\iff \omega /k=c\;\left(k=|\overrightarrow{k}|\right)$. As we know (Sect. 10.5), the monochromatic waves

$$ {\overrightarrow{E}}_0\left(\overrightarrow{k}\right)\;{e}^{i\kern0.2em \left(\overrightarrow{k}\cdot \overrightarrow{r}-\omega \kern0.22em t\right)},\kern0.72em {\overrightarrow{B}}_0\left(\overrightarrow{k}\right)\;{e}^{i\kern0.2em \left(\overrightarrow{k}\cdot \overrightarrow{r}-\omega \kern0.22em t\right)} $$

are solutions of the Maxwell equations in empty space if

$$ {\displaystyle \begin{array}{l}\;\overrightarrow{k}\cdot {\overrightarrow{E}}_0\left(\overrightarrow{k}\right)=0,\kern0.84em \overrightarrow{k}\cdot {\overrightarrow{B}}_0\left(\overrightarrow{k}\right)=0\\ {}\overrightarrow{k}\times {\overrightarrow{E}}_0\left(\overrightarrow{k}\right)=\omega\;{\overrightarrow{B}}_0\left(\overrightarrow{k}\right),\kern0.6em \overrightarrow{k}\times {\overrightarrow{B}}_0\left(\overrightarrow{k}\right)=-\frac{\omega }{c^2}\kern0.22em {\overrightarrow{E}}_0\left(\overrightarrow{k}\right)\end{array}} $$

• (a) Show that the $\overrightarrow{E}\kern0.2em \left(\overrightarrow{r},t\right)$ and $\overrightarrow{B}\kern0.2em \left(\overrightarrow{r},t\right)$ satisfy the wave equation in empty space.

• (b) Show that the $\overrightarrow{E}\kern0.2em \left(\overrightarrow{r},t\right)$ and $\overrightarrow{B}\kern0.2em \left(\overrightarrow{r},t\right)$ satisfy the Maxwell equations in empty space. (Remember that ε₀ μ₀ = 1/c² .)

• (c) Explain why the e/m radiation emitted by the lamp does not, in general, correspond to a plane e/m wave, thus does not exhibit the properties stated in Problem 5. What if in place of the ordinary lamp we had a laser-beam emitter?

• Solution:

• (a)
  $$ {\displaystyle \begin{array}{l}{\nabla}^2\overrightarrow{E}=\frac{\partial^2\overrightarrow{E}}{\partial \kern0.1em {x}^2}+\frac{\partial^2\overrightarrow{E}}{\partial \kern0.1em {y}^2}+\frac{\partial^2\overrightarrow{E}}{\partial \kern0.1em {z}^2}=-\int \left({k_x}^2+{k_y}^2+{k_z}^2\right)\;{\overrightarrow{E}}_0\left(\overrightarrow{k}\right)\;{e}^{i\kern0.2em \left(\overrightarrow{k}\cdot \overrightarrow{r}-\omega \kern0.22em t\right)}\;d\kern0.1em \overrightarrow{k}\\ {}\kern6.699996em =-\int {k}^2\kern0.1em {\overrightarrow{E}}_0\left(\overrightarrow{k}\right)\;{e}^{i\kern0.2em \left(\overrightarrow{k}\cdot \overrightarrow{r}-\omega \kern0.22em t\right)}\;d\kern0.1em \overrightarrow{k}\end{array}} $$

$$ \frac{\partial^2\overrightarrow{E}}{\partial \kern0.3em {t}^2}=-\int {\omega}^2\kern0.1em {\overrightarrow{E}}_0\left(\overrightarrow{k}\right)\;{e}^{i\kern0.2em \left(\overrightarrow{k}\cdot \overrightarrow{r}-\omega \kern0.22em t\right)}\;d\kern0.1em \overrightarrow{k} $$

(Note that the k² and ω² cannot be taken out of the corresponding integrals since they are not constant quantities!) Hence,

$$ {\nabla}^2\overrightarrow{E}-\frac{1}{c^2}\;\frac{\partial^2\overrightarrow{E}}{\partial \kern0.3em {t}^2}=\int \left(\frac{\omega^2}{c^2}-{k}^2\right)\;{\overrightarrow{E}}_0\left(\overrightarrow{k}\right)\;{e}^{i\kern0.2em \left(\overrightarrow{k}\cdot \overrightarrow{r}-\omega \kern0.22em t\right)}\;d\kern0.1em \overrightarrow{k}=0;\mathrm{similarly}\ \mathrm{for}\kern0.24em \overrightarrow{B} $$

• (b) The Maxwell equations in empty space are

$$ {\displaystyle \begin{array}{l}\overrightarrow{\nabla}\cdot \overrightarrow{E}=0\kern1.92em \overrightarrow{\nabla}\times \overrightarrow{E}=-\frac{\partial \kern0.1em \overrightarrow{B}}{\partial \kern0.2em t}\kern0.6em \\ {}\overrightarrow{\nabla}\cdot \overrightarrow{B}=0\kern1.9em \overrightarrow{\nabla}\times \overrightarrow{B}={\varepsilon}_0\kern0.1em {\mu}_0\kern0.24em \frac{\partial \kern0.1em \overrightarrow{E}}{\partial \kern0.2em t}=\frac{1}{c^2}\kern0.24em \frac{\partial \kern0.1em \overrightarrow{E}}{\partial \kern0.2em t}\end{array}} $$

By using Eqs. (10.41) and the relation

$$ \overrightarrow{\nabla}\;{e}^{i\kern0.2em \left(\overrightarrow{k}\cdot \overrightarrow{r}-\omega \kern0.22em t\right)}=i\;\overrightarrow{k}\;{e}^{i\kern0.2em \left(\overrightarrow{k}\cdot \overrightarrow{r}-\omega \kern0.22em t\right)} $$

we have:

$$ \overrightarrow{\nabla}\cdot \overrightarrow{E}=\int {\overrightarrow{E}}_0\kern0.2em \left(\overrightarrow{k}\right)\cdot \overrightarrow{\nabla}\;{e}^{i\kern0.2em \left(\overrightarrow{k}\cdot \overrightarrow{r}-\omega \kern0.22em t\right)}\kern0.22em d\kern0.1em \overrightarrow{k}=\int i\;\overrightarrow{k}\cdot {\overrightarrow{E}}_0\left(\overrightarrow{k}\right)\;{e}^{i\kern0.2em \left(\overrightarrow{k}\cdot \overrightarrow{r}-\omega \kern0.22em t\right)}\;d\kern0.1em \overrightarrow{k}\kern0.34em =0;\mathrm{similarly},\overrightarrow{\nabla}\cdot \overrightarrow{B}=0 $$

$$ {\displaystyle \begin{array}{l}\overrightarrow{\nabla}\times \overrightarrow{E}+\frac{\partial \kern0.1em \overrightarrow{B}}{\partial \kern0.2em t}=\int \left[\overrightarrow{\nabla}\kern0.1em {e}^{i\kern0.2em \left(\overrightarrow{k}\cdot \overrightarrow{r}-\omega \kern0.22em t\right)}\right]\times {\overrightarrow{E}}_0\kern0.2em \left(\overrightarrow{k}\right)\;d\kern0.1em \overrightarrow{k}\kern0.36em -\kern0.34em \int i\;\omega\;{\overrightarrow{B}}_0\left(\overrightarrow{k}\right)\;{e}^{i\kern0.2em \left(\overrightarrow{k}\cdot \overrightarrow{r}-\omega \kern0.22em t\right)}\;d\kern0.1em \overrightarrow{k}\\ {}\kern3.179999em =i\int d\kern0.1em \overrightarrow{k}\kern0.22em {e}^{i\kern0.2em \left(\overrightarrow{k}\cdot \overrightarrow{r}-\omega \kern0.22em t\right)}\kern0.3em \left[\overrightarrow{k}\times {\overrightarrow{E}}_0\kern0.2em \left(\overrightarrow{k}\right)-\omega\;{\overrightarrow{B}}_0\left(\overrightarrow{k}\right)\right]\kern0.36em =0\\ {}\overrightarrow{\nabla}\times \overrightarrow{B}-\frac{1}{c^2}\kern0.24em \frac{\partial \kern0.1em \overrightarrow{E}}{\partial \kern0.2em t}=i\int d\kern0.1em \overrightarrow{k}\kern0.22em {e}^{i\kern0.2em \left(\overrightarrow{k}\cdot \overrightarrow{r}-\omega \kern0.22em t\right)}\kern0.22em \left[\overrightarrow{k}\times {\overrightarrow{B}}_0\kern0.2em \left(\overrightarrow{k}\right)+\frac{\omega }{c^2}\;{\overrightarrow{E}}_0\left(\overrightarrow{k}\right)\right]\kern0.36em =0\end{array}} $$

• (c) The emitted radiation does not have the form of a plane e/m wave since the e/m field is not of the form $\overrightarrow{F}\left(\hat{\tau}\cdot \overrightarrow{r}-c\;t\right)$ with a fixed direction of $\hat{\tau}$ (this is in contrast to Prob. 5, where we had that $\hat{\tau}={\hat{u}}_x= const.$). A laser beam, on the other hand, is characterized by a high degree of directivity; therefore the corresponding e/m wave may be considered as an almost plane wave.

• 7. We will study the propagation of a plane e/m wave inside a conducting medium having constant values of ε, μ, σ.

  • (a) By using the Maxwell equations for this medium,

$$ \overrightarrow{\nabla}\cdot \overrightarrow{E}=0 $$

(1)

$$ \overrightarrow{\nabla}\cdot \overrightarrow{B}=0 $$

(2)

$$ \overrightarrow{\nabla}\times \overrightarrow{E}=-\frac{\partial \kern0.1em \overrightarrow{B}}{\partial \kern0.2em t} $$

(3)

$$ \overrightarrow{\nabla}\times \overrightarrow{B}=\mu \kern0.2em \sigma \kern0.1em \overrightarrow{E}+\varepsilon \kern0.1em \mu \kern0.24em \frac{\partial \kern0.1em \overrightarrow{E}}{\partial \kern0.2em t} $$

(4)

show that the fields $\overrightarrow{E}$ and $\overrightarrow{B}$ satisfy the modified wave equations:

$$ {\nabla}^2\overrightarrow{E}-\varepsilon \kern0.1em \mu \kern0.22em \frac{\partial^2\overrightarrow{E}}{\partial \kern0.2em {t}^2}-\mu\;\sigma \kern0.22em \frac{\partial \kern0.1em \overrightarrow{E}}{\partial \kern0.2em t}=0 $$

(5)

$$ {\nabla}^2\overrightarrow{B}-\varepsilon \kern0.1em \mu \kern0.22em \frac{\partial^2\overrightarrow{B}}{\partial \kern0.2em {t}^2}-\mu\;\sigma \kern0.22em \frac{\partial \kern0.1em \overrightarrow{B}}{\partial \kern0.2em t}=0 $$

(5′)

• (b) Show that the above wave equations admit solutions of the form

$$ \overrightarrow{E}\;\left(x,t\right)={\overrightarrow{E}}_0\;{e}^{-s\kern0.1em x}\kern0.1em {e}^{i\kern0.2em \left(k\kern0.1em x-\omega \kern0.22em t\right)},\kern0.84em \overrightarrow{B}\;\left(x,t\right)={\overrightarrow{B}}_0\;{e}^{-s\kern0.1em x}\kern0.1em {e}^{i\kern0.2em \left(k\kern0.1em x-\omega \kern0.22em t\right)} $$

(6)

(where ${\overrightarrow{E}}_0,{\overrightarrow{B}}_0$ are constant complex vectors) on the condition that

$$ {s}^2-{k}^2+\varepsilon \kern0.1em \mu \kern0.2em {\omega}^2=0\operatorname{}\mathrm{and}\operatorname{}\mu \kern0.2em \sigma \kern0.1em \omega -2\kern0.1em s\kern0.1em k=0 $$

(7)

By solving the system (7) for k and s, verify Eqs. (10.65). The solutions (6) describe a plane e/m wave traveling in the +x direction.

• (c) Show that the $\overrightarrow{E}$ and $\overrightarrow{B}$ are normal to the direction of propagation of the wave (transverse wave). [Hint: Substitute Eqs. (6) into (1) and (2).]

• (d) Show that the $\overrightarrow{E}$ and $\overrightarrow{B}$ are mutually perpendicular. [Hint: Write ${\overrightarrow{E}}_0=\tilde{E_0}\;{\hat{u}}_y$ where $\tilde{E_0}$ is a complex constant. By substituting the first of Eqs. (6) into (3) and by integrating for t, show that ${\overrightarrow{B}}_0=\frac{k+i\kern0.1em s}{\omega}\kern0.32em \tilde{E_0}\kern0.22em {\hat{u}}_z$.]

• (e) Show that the real fields are written:

$$ {\displaystyle \begin{array}{l}\overrightarrow{E}\;\left(x,t\right)={E}_0\;{e}^{-s\kern0.1em x}\;\cos\;\left(k\kern0.1em x-\omega \kern0.3em t+\alpha \right)\kern0.22em {\hat{u}}_y\\ {}\overrightarrow{B}\;\left(x,t\right)=\frac{\sqrt{k^2+{s}^2}}{\omega}\kern0.22em {E}_0\;{e}^{-s\kern0.1em x}\;\cos\;\left(k\kern0.1em x-\omega \kern0.3em t+\alpha +\varphi \right)\kern0.22em {\hat{u}}_z\end{array}} $$

(8)

where Ε₀ is real and where φ = arc tan (s/k). Check your result at the limit σ → 0 (non-conducting medium) where, at this limit, s → 0. [Hint: Put $\tilde{E_0}=\mid \tilde{E_0}\mid {e}^{i\kern0.1em \alpha }={E}_0\;{e}^{i\kern0.1em \alpha }$ and $k+i\;s=\mid k+i\;s\mid {e}^{i\kern0.1em \varphi }=\sqrt{k^2+{s}^2}\kern0.32em {e}^{i\kern0.1em \varphi }$, where tan φ = s/k .]

• Solution:

  • (a) We take the rot of (3) and (4) and we work as in Sect. 10.4.

  • (b) By substituting the first of Eqs. (6) into (5) and by noticing that ${\nabla}^2\overrightarrow{E}={\partial}^2\overrightarrow{E}/\partial \kern0.1em {x}^2$, we are led to a complex equation whose real and imaginary parts correspond to the first and the second of Eqs. (7), respectively. [Alternatively, we may substitute the second of Eqs. (6) into (5′).] Show that the solution of the system (7) for k and s yields Eqs. (10.65).

  • (c) We substitute Eqs. (6) into (1) and (2), noticing that

$$ \overrightarrow{\nabla}\cdot \overrightarrow{E}={\overrightarrow{E}}_0\cdot \overrightarrow{\nabla}\;{e}^{-s\kern0.1em x+i\kern0.2em \left(k\kern0.1em x-\omega \kern0.22em t\right)}=\left(-s+i\;k\right)\kern0.32em {\hat{u}}_x\cdot \overrightarrow{E},\kern0.84em \overrightarrow{\nabla}\cdot \overrightarrow{B}=\left(-s+i\;k\right)\kern0.32em {\hat{u}}_x\cdot \overrightarrow{B} $$

Hence, ${\hat{u}}_x\cdot \overrightarrow{E}=0,\kern0.6em {\hat{u}}_x\cdot \overrightarrow{B}=0$.

• (d) We set ${\overrightarrow{E}}_0=\tilde{E_0}\;{\hat{u}}_y$, so that the first of Eqs. (6) is written as

$$ \overrightarrow{E}=\tilde{E_0}\kern0.22em {e}^{-s\kern0.1em x}\kern0.1em {e}^{i\kern0.2em \left(k\kern0.1em x-\omega \kern0.22em t\right)}\kern0.22em {\hat{u}}_y $$

(9)

or, briefly, $\overrightarrow{E}=\tilde{E}\kern0.22em {\hat{u}}_y$. Substituting (9) into (3), we have:

$$ -\frac{\partial \kern0.1em \overrightarrow{B}}{\partial \kern0.2em t}=\overrightarrow{\nabla}\times \overrightarrow{E}=\left|\begin{array}{ccc}{\hat{u}}_x& {\hat{u}}_y& {\hat{u}}_z\\ {}\frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ {}0& \tilde{E}& 0\end{array}\right|=\kern0.34em \frac{\partial \tilde{E}}{\partial \kern0.1em x}\kern0.24em {\hat{u}}_z=\left(-s+i\;k\right)\;\tilde{E_0}\kern0.22em {e}^{-s\kern0.1em x}\kern0.1em {e}^{i\kern0.2em \left(k\kern0.1em x-\omega \kern0.22em t\right)}\kern0.22em {\hat{u}}_z $$

Integrating with respect to t ⇒

$$ \overrightarrow{B}=\frac{k+i\kern0.1em s}{\omega}\kern0.32em \tilde{E_0}\kern0.22em {e}^{-s\kern0.1em x}\kern0.1em {e}^{i\kern0.2em \left(k\kern0.1em x-\omega \kern0.22em t\right)}\kern0.22em {\hat{u}}_z $$

(10)

Then, by (6), ${\overrightarrow{B}}_0=\frac{k+i\kern0.1em s}{\omega}\kern0.32em \tilde{E_0}\kern0.22em {\hat{u}}_z.$

• (e) We set $\tilde{E_0}=\mid \tilde{E_0}\mid {e}^{i\kern0.1em \alpha }={E}_0\;{e}^{i\kern0.1em \alpha }$ and $k+i\;s=\mid k+i\;s\mid {e}^{i\kern0.1em \varphi }=\sqrt{k^2+{s}^2}\kern0.32em {e}^{i\kern0.1em \varphi }$, where tan φ = s/k ⇔ φ = arc tan (s/k). Substituting into (9) and (10), and then taking the real parts of $\overrightarrow{E}$ and $\overrightarrow{B}$, we get Eqs. (8).

In the limit σ → 0, the second of Eqs. (7) yields s → 0 while the first one reduces to ω/k = υ = (1/εμ)^1/2. Moreover, the phase difference tends to φ → 0 (that is, the fields $\overrightarrow{E}$ and $\overrightarrow{B}$ oscillate in phase) while the instantaneous values of the fields are related by Ε = υΒ. Therefore, in the limit of vanishing conductivity our results essentially reduce to those of Sect. 10.5 (this time for an arbitrary non-conducting medium instead of empty space).

• 8. We will study the polarization of a monochromatic plane e/m wave traveling in the +x direction. The electric field in this wave is written as

$$ \overrightarrow{E}\;\left(x,t\right)={\overrightarrow{E}}_0\kern0.2em {e}^{i\kern0.2em \left(k\kern0.1em x-\omega \kern0.22em t\right)} $$

(1)

where ${\overrightarrow{E}}_0$ is a constant complex vector. In the simplest case we set, as in Sect. 10.5,

$$ {\overrightarrow{E}}_0={\overrightarrow{E}}_{0,R}\;{e}^{i\kern0.1em \theta } $$

(2)

where ${\overrightarrow{E}}_{0,R}$ is a real vector, so that

$$ \overrightarrow{E}\;\left(x,t\right)={\overrightarrow{E}}_{0,R}\kern0.2em {e}^{i\kern0.2em \left(k\kern0.1em x-\omega \kern0.22em t+\theta \right)} $$

(3)

Taking the real part of $\overrightarrow{E}$, we find the real field

$$ \overrightarrow{E}\;\left(x,t\right)={\overrightarrow{E}}_{0,R}\;\cos\;\left(k\kern0.1em x\kern0.4em -\kern0.4em \omega \kern0.4em t+\theta \right) $$

(4)

We notice that, at a given position x = const., the tip of the vector $\overrightarrow{E}$ oscillates along the axis defined by the constant vector ${\overrightarrow{E}}_{0,R}$. We thus say that this wave is linearly polarized.⁵

The above situation, however, is a special one. To make the problem more general, let us return to the complex expression (1). As we know, the ${\overrightarrow{E}}_0$ and $\overrightarrow{E}$ are normal to the direction of propagation (i.e., normal to ${\hat{u}}_x$); these vectors, therefore, belong to the yz-plane. In place of (2) we now write:

$$ {\overrightarrow{E}}_0={E}_1\;{e}^{i\kern0.2em {\theta}_1}\;{\hat{u}}_y+{E}_2\;{e}^{i\kern0.2em {\theta}_2}\;{\hat{u}}_z $$

(5)

where the Ε₁ and Ε₂ are real scalar quantities.

• (a) Show that the real electric field is written:

$$ {\displaystyle \begin{array}{l}\kern5.039997em \overrightarrow{E}={E}_y\;{\hat{u}}_y+{E}_z\;{\hat{u}}_z\kern0.36em ;\\ {}{E}_y={E}_1\;\cos\;\left(k\kern0.1em x\kern0.4em -\kern0.4em \omega \kern0.4em t+{\theta}_1\right),\kern0.72em {E}_z={E}_2\;\cos\;\left(k\kern0.1em x\kern0.4em -\kern0.4em \omega \kern0.4em t+{\theta}_2\right)\end{array}} $$

(6)

That is, $\overrightarrow{E}$ is the superposition of two linearly polarized waves with mutually perpendicular polarizations and with a phase difference equal to (θ₁−θ₂) .

• (b) Show that, for θ₁ = θ₂ as well as for | θ₁–θ₂ | = π, the field $\overrightarrow{E}$ is linearly polarized.

• (c) Show that, for | θ₁–θ₂ | = π /2,

$$ {\left(\frac{E_y}{E_1}\right)}^2+\kern0.34em {\left(\frac{E_z}{E_2}\right)}^2=1 $$

(7)

We say that $\overrightarrow{E}$ is elliptically polarized , since the tip of the vector $\overrightarrow{E}$ describes an ellipse on the yz-plane. In particular, if Ε₁ = Ε₂ = Ε₀, relation (7) is written

$$ {E_y}^2+{E_z}^2={E_0}^2 $$

(8)

and the e/m wave is said to be circularly polarized.

• (d) For an arbitrary phase difference (θ₁–θ₂), show that

$$ {\left(\frac{E_y}{E_1}\right)}^2+\kern0.5em {\left(\frac{E_z}{E_2}\right)}^2-\kern0.4em 2\kern0.4em \left(\frac{E_y}{E_1}\right)\left(\frac{E_z}{E_2}\right)\kern0.22em \cos\;\left({\theta}_1-{\theta}_2\right)=\kern0.32em {\sin}^2\left({\theta}_1-{\theta}_2\right) $$

(9)

Demonstrate that (9) reduces to our previous special results by allowing the phase difference to take on the values θ₁–θ₂ = 0, ± π and ± π /2 .

• (e) Consider a linearly polarized wave in which the electric field is

$$ \overrightarrow{E}\;\left(x,t\right)=\left({E}_1\;{\hat{u}}_y+{E}_2\;{\hat{u}}_z\right)\kern0.22em \cos\;\left(k\kern0.1em x\kern0.4em -\kern0.4em \omega \kern0.4em t+\theta \right) $$

(10)

where Ε₁, Ε₂, θ are given real constants. Show that the corresponding magnetic field in the wave is

$$ \overrightarrow{B}\;\left(x,t\right)=\frac{1}{c}\kern0.24em \left({E}_1\;{\hat{u}}_z-{E}_2\;{\hat{u}}_y\right)\kern0.22em \cos\;\left(k\kern0.1em x\kern0.4em -\kern0.4em \omega \kern0.4em t+\theta \right) $$

(11)

where c = ω/k. [Hint: The $\overrightarrow{E}$ and $\overrightarrow{B}$ are normal to the direction of propagation +x, they oscillate in phase and they satisfy the relations (cf. Prob. 5)

$$ {\hat{u}}_x\times \overrightarrow{E}=c\kern0.1em \overrightarrow{B},\kern0.72em {\hat{u}}_x\times \overrightarrow{B}=-\left(1/c\right)\;\overrightarrow{E} $$

(12)

Assume that $\overrightarrow{B}=\left({B}_1\;{\hat{u}}_y+{B}_2\;{\hat{u}}_z\right)\kern0.22em \cos\;\left(k\kern0.1em x\kern0.4em -\kern0.4em \omega \kern0.4em t+\theta \right)$ and determine the B₁, B₂ .]

• Solution:

• (a) We substitute (5) into (1) and then take the real part of $\overrightarrow{E}$. We thus find (6).

• (b) For θ₁ = θ₂ = θ, Eq. (6) reduces to (4) with ${\overrightarrow{E}}_{0,R}={E}_1\;{\hat{u}}_y+{E}_2\;{\hat{u}}_z$; similarly for θ₁ = θ and θ₂ = θ + π, with ${\overrightarrow{E}}_{0,R}={E}_1\;{\hat{u}}_y-{E}_2\;{\hat{u}}_z$.

• (c) We set θ₁ = θ and θ₂ = θ + π/2. Then,

$$ {E}_y={E}_1\;\cos\;\left(k\kern0.1em x\kern0.4em -\kern0.4em \omega \kern0.4em t+\theta \right),\kern0.72em {E}_z=-{E}_2\;\sin\;\left(k\kern0.1em x\kern0.4em -\kern0.4em \omega \kern0.4em t+\theta \right) $$

By using the identity cos²A + sin²A = 1, we obtain (7).

• (d) We solve the system

$$ {E}_y={E}_1\;\cos\;\left(k\kern0.1em x\kern0.4em -\kern0.4em \omega \kern0.4em t+{\theta}_1\right),\kern0.72em {E}_z={E}_2\;\cos\;\left(k\kern0.1em x\kern0.4em -\kern0.4em \omega \kern0.4em t+{\theta}_2\right) $$

for cos(kx–ωt) and sin(kx–ωt) (with a little help from Trigonometry!) and we use the identity cos²A + sin²A = 1 . For θ₁–θ₂ = 0 or π, Eq. (9) yields:

$$ {\left(\frac{E_y}{E_1}\kern0.4em \mp \kern0.32em \frac{E_z}{E_2}\right)}^2=0\kern0.48em \Rightarrow \kern0.48em \frac{E_y}{E_z}=\pm \frac{E_1}{E_2} $$

This is precisely what we get from (6) when θ₁–θ₂ = 0 or π (linear polarization). For θ₁–θ₂ = π /2, Eq. (9) reduces to (7) (elliptical polarization).

• (e) In the first or the second of Eqs. (12), we substitute $\overrightarrow{E}$ from (10) and $\overrightarrow{B}$ from the expression given in the hint. By equating coefficients of similar unit vectors, we find that B₁ = − E₂ /c and B₂ = E₁ /c .

• 9. In Problem 8 we saw that the superposition of two linearly polarized e/m waves of equal amplitudes, the electric fields in which waves are mutually perpendicular and differ in phase by ± π /2, yields a circularly polarized wave. Show now that, conversely, by a suitable superposition of two circularly polarized waves we can construct a linearly polarized wave.

• Solution: We consider the general expression

$$ {\displaystyle \begin{array}{l}\kern5.039997em \overrightarrow{E}={E}_y\;{\hat{u}}_y+{E}_z\;{\hat{u}}_z;\\ {}{E}_y={E}_1\;\cos\;\left(k\kern0.1em x\kern0.4em -\kern0.4em \omega \kern0.4em t+{\theta}_1\right),\kern0.72em {E}_z={E}_2\;\cos\;\left(k\kern0.1em x\kern0.4em -\kern0.4em \omega \kern0.4em t+{\theta}_2\right)\end{array}} $$

We let Ε₁ = Ε₂ = Ε₀ and we call θ₁ = θ. For θ₂ = θ + π/2 we have the circularly polarized wave

$$ {\displaystyle \begin{array}{l}{\overrightarrow{E}}_L={E}_0\;\cos\;\left(k\kern0.1em x\kern0.4em -\kern0.4em \omega \kern0.4em t+\theta \right)\kern0.22em {\hat{u}}_y-{E}_0\;\sin\;\left(k\kern0.1em x\kern0.4em -\kern0.4em \omega \kern0.4em t+\theta \right)\kern0.22em {\hat{u}}_z\\ {}\kern0.84em ={E}_0\;\cos\;\left[\omega \kern0.4em t-\left(k\kern0.1em x\kern0.3em +\kern0.3em \theta \right)\right]\kern0.22em {\hat{u}}_y+{E}_0\;\sin\;\left[\omega \kern0.4em t-\left(k\kern0.1em x\kern0.3em +\kern0.3em \theta \right)\right]\kern0.22em {\hat{u}}_z\end{array}} $$

while for θ₂ = θ-π/2 we have the circularly polarized wave

$$ {\displaystyle \begin{array}{l}{\overrightarrow{E}}_R={E}_0\;\cos\;\left(k\kern0.1em x\kern0.4em -\kern0.4em \omega \kern0.4em t+\theta \right)\kern0.22em {\hat{u}}_y+{E}_0\;\sin\;\left(k\kern0.1em x\kern0.4em -\kern0.4em \omega \kern0.4em t+\theta \right)\kern0.22em {\hat{u}}_z\\ {}\kern0.94em ={E}_0\;\cos\;\left[\left(k\kern0.1em x+\theta \right)-\kern0.4em \omega \kern0.4em t\right]\kern0.22em {\hat{u}}_y+{E}_0\;\sin\;\left[\left(k\kern0.1em x+\theta \right)-\kern0.4em \omega \kern0.4em t\right]\kern0.22em {\hat{u}}_z\end{array}} $$

By adding the electric fields of these two waves, we find:

$$ \overrightarrow{E}={\overrightarrow{E}}_L+{\overrightarrow{E}}_R=2{E}_0\;\cos\;\left(k\kern0.1em x\kern0.4em -\kern0.4em \omega \kern0.4em t+\theta \right)\kern0.22em {\hat{u}}_y $$

which represents a linearly polarized wave. The wave ${\overrightarrow{E}}_L$ is called left-hand polarized since, at each point x, it can be pictured as a vector of magnitude Ε₀, rotating counterclockwise on the yz-plane (Fig. 10.14). Similarly, the wave ${\overrightarrow{E}}_R$ is right-hand polarized since it can be pictured as a vector rotating clockwise. [Notice that the angle ωt-(kx + θ) of the vector ${\overrightarrow{E}}_L$ with the y-axis increases with time, while the angle (kx + θ)–ωt of ${\overrightarrow{E}}_R$ with this axis decreases with time.]

Fig. 10.14

Left-handed and right-handed polarized e/m wave