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Chapter 6: Problem 2

6.2 Consider electromagnetic radiation in a medium with dielectric constant \(\epsilon(\omega)\). Suppose the magnetio permeability \(\mu\) is 1. Then \(n(\omega)=[\epsilon(\omega)]^{1 / 2}\). According to the theory of relativity, no signal can propagate faster than \(c=3.0 \times 10^{10} \mathrm{~cm} / \mathrm{sec}\). What limitation does that put on the possible variation of \(\epsilon(\omega)\) with \(\omega ?\) [Assume \(\epsilon(\omega)\) is positive for all \(\omega .] \quad\)

Short Answer

Expert verified
\( \epsilon(\omega) \geq 1 \) for all \( \omega \).

Step by step solution

01

Understand the Problem Statement

We are given a medium with a dielectric constant \( \epsilon(\omega) \) and magnetic permeability \( \mu = 1 \). The refractive index, which determines the speed of light in the medium, is given by \( n(\omega) = [\epsilon(\omega)]^{1/2} \). Our goal is to determine how \( \epsilon(\omega) \) can vary with \( \omega \), ensuring that the speed of light in the medium is less than or equal to the speed of light in a vacuum, \( c = 3.0 \times 10^{10} \) cm/s, in order to obey the principle of relativity.
02

Define the Speed of Light in the Medium

In a medium, the speed of propagation of electromagnetic waves is given by \( v = \frac{c}{n(\omega)} \). Substituting \( n(\omega) = [\epsilon(\omega)]^{1/2} \), we get: \[ v = \frac{c}{[\epsilon(\omega)]^{1/2}} \]. Thus, the speed of light in the medium depends inversely on the square root of the dielectric constant.
03

Apply the Relativity Principle Constraint

According to the theory of relativity, no signal can propagate faster than \( c \). This means that \( v \leq c \). Replacing \( v \) with \( \frac{c}{[\epsilon(\omega)]^{1/2}} \), we get: \[ \frac{c}{[\epsilon(\omega)]^{1/2}} \leq c \]. Simplifying the inequality, we obtain \( [\epsilon(\omega)]^{1/2} \geq 1 \).
04

Solve the Inequality

To solve \( [\epsilon(\omega)]^{1/2} \geq 1 \), square both sides to remove the square root, leading to \( \epsilon(\omega) \geq 1 \). This implies that the dielectric constant \( \epsilon(\omega) \) must be greater than or equal to 1 for all \( \omega \) in order to comply with the relativity principle.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dielectric Constant
The dielectric constant, often represented as \( \epsilon(\omega) \), is a crucial property of materials when discussing electromagnetic waves in mediums. It measures how much a material can respond to electric fields, and thus affects how electromagnetic waves propagate through it. When light or any electromagnetic radiation enters a medium, the way it moves through the material depends greatly on \( \epsilon(\omega) \). A higher dielectric constant generally means that the material can store more electric energy, affecting the speed and wavelength of waves traveling through it.

In simple terms, \( \epsilon(\omega) \) indicates how easily a medium can be polarized by an external electric field. For our exercise, knowing that \( n(\omega) = [\epsilon(\omega)]^{1/2} \) allows us to link the dielectric constant to the refractive index, providing insights into how light speed is influenced in different materials.
Refractive Index
The refractive index, denoted as \( n(\omega) \), is a measure of how much the speed of light is reduced inside a material compared to a vacuum. It is fascinating because it ties directly to how much light bends or refracts when entering a different medium. For our exercise, we use the relation \( n(\omega) = [\epsilon(\omega)]^{1/2} \), indicating that the refractive index depends on the dielectric constant of the material at different frequencies or \( \omega \).

What this means is that by altering \( \epsilon(\omega) \), you effectively change how light travels through the medium. The refractive index tells us how much the light's speed decreases compared to its speed in a vacuum. Since \( n(\omega) \) is crucial for understanding optical properties, manipulating it through varying \( \epsilon(\omega) \) allows precise control over light behavior in advanced optics and communication technologies.
Speed of Light in Medium
The speed of light diminution within a medium is a fundamental physical concept. In our discussion, the speed of light in a medium \( v \) is described by the relationship \( v = \frac{c}{n(\omega)} \). Here, \( c \) is the speed of light in vacuum, approximately \( 3.0 \times 10^{10} \) cm/s. The introduction of a medium essentially slows down the light compared to its speed in a vacuum.

By substituting \( n(\omega) = [\epsilon(\omega)]^{1/2} \), we recognize that the speed \( v \) is inversely proportional to the square root of the dielectric constant. This means that as \( \epsilon(\omega) \) increases, the speed of light in the medium decreases. It鈥檚 a crucial relationship to understand because it influences how we design lenses, fiber optics, and various optical devices which depend on precise control of light propagation.
Relativity Principle Constraint
According to the theory of relativity, a signal or information cannot travel faster than the speed of light in a vacuum, \( c \). This fundamental principle constrains physical phenomena and must be considered when analyzing light in different mediums. In our scenario, this constraint implies \( v \leq c \), or equivalently, \( \frac{c}{[\epsilon(\omega)]^{1/2}} \leq c \). Simplifying this, we understand it as \( [\epsilon(\omega)]^{1/2} \geq 1 \), which when squared, implies \( \epsilon(\omega) \geq 1 \).

Essentially, the dielectric constant must remain at or above 1 for all frequencies \( \omega \). This ensures that light always travels at a speed equal to or lesser than \( c \) when inside that medium, averting any violations of the relativity principle. This constraint is pivotal not just for theoretical elegance but also for ensuring that the physical laws remain consistent when exploring new electromagnetic materials.

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Most popular questions from this chapter

Show that for light of index \(n(\lambda)\), $$ \frac{1}{v_{g}}=\frac{1}{v_{\varphi}}-\frac{1}{c} \lambda \frac{d n(\lambda)}{d \bar{\lambda}} $$ where \(\lambda\) is the vacuum wavelength of the light.

Show that the sum of two traveling harmonic waves \(A_{1} \cos \left(\omega t-k z+\varphi_{1}\right)\) and \(A_{2} \cos \left(\omega t-k z+\varphi_{2}\right)\) traveling in the \(+z\) direction and having the same frequency \(\omega\) is itself a harmonic traveling wave of the same kind. That is, the sum can be written in the form \(A \cos (\omega t-k z+\varphi)\). Find out how \(A\) and \(\varphi\) are related to \(A_{1}, A_{2}\), \(\varphi_{1}\), and \(\varphi_{2} .\) (Hint: The use of complex numbers or a rotating vector diagram helps immensely.)

(a) One way to produce an amplitude-modulated carrier wave is to pass a current \(I=I_{0} \cos \omega_{0} t\) oscillating at the carrier frequency \(\omega_{0}\) through a resistance \(R\) which is not constant but has a component that varies at the modulation frequency \(\omega_{\text {mod }}\), that is, \(R=R_{0}\left(1+a_{m} \cos \omega_{\text {mod }} t\right)\). (In a "carbon-granule" microphone the resistance is modulated by the motion of a diaphragm, which compresses the carbon granules that provide the resistance.) The voltage \(V=I R\) across the resistor is an amplitude-modulated carrier wave. Find the expression for \(V\) in terms of a stperposition of carrier (frequency \(\omega_{0}\) ), upper sideband (frequency \(\omega_{0}+\omega_{\text {mod }}\) ), and lower sideband (frequency \(\left.\omega_{0}-\omega_{\text {mod }}\right)\). (b) Alternatively, suppose we happen to start with two voltages, one oscillating at the carrier frequency, the other at the modulation frequency. The problem is this: How can you physically combine these two voltages, \(V_{0}=A_{0} \cos \omega_{0} t\), and \(V_{m}=A_{m} \cos \omega_{\text {mod }} t\) in such a way as to produce an amplitude-modulated carrier wave? First, suppose you merely superpose the two voltages, i.e., you put them both on the broadcasting antenna. Will this work? (c) Next, suppose that the voltages in part \((b)\), after being superposed, are then applied to the input of a voltage amplifier. (For example they may be applied between control grid and cathode of a radio tube.) Suppose that the amplifier is a linear amplifier, i.e., its output (for example the plate-to- cathode voltage of the tube) is proportional to the input. Will this work? (d) Finally, suppose that the amplifier output has both a linear and a quadratic component, as follows: $$ V_{\text {out }}=A_{1} V_{\text {in }}+A_{2}\left(V_{\text {in }}\right)^{2} $$ Let \(V_{\text {in }}=V_{0}+V_{m}\) as defined in part \((b)\). Show that because of the nonlinear quadratic term \(A_{2}\left(V_{i \mathrm{in}}\right)^{2}\) the amplifier output includes, among other things, an amplitudemodulated carrier wave, with modulation amplitude proportional to \(A_{m} .\) (e) The amplitude-modulated carrier wave in \((d)\) contributes Fourier components with frequencies \(\omega_{0}, \omega_{0}+\omega_{\mathrm{mad}}\), and \(\omega_{0}-\omega_{\text {mod }}\). What other frequency components are there in \(V_{\text {eut }}\) ? Make a diagram showing a complete frequency spectrum of the amplifier output. Describe how you could get rid of these other (undesired) components, using bandpass filters. Suppose that \(\omega_{\text {rued }}\) is small compared with \(\omega_{0}\). How selective do the filters have to be?

The best way to understand the difference between phase and group velocities is to make water wave packets. To make expanding circular wave packets having dominant wavelength 3 or \(4 \mathrm{~cm}\) or longer, throw a big rock in a pond or pool. To make straight waves (the two-dimensional analog of threedimensional plane waves) with wavelengths of several centimeters, float a stick across the end of a bathtub or a large pan of water. Give the stick about two swift vertical pushes with your hand. After some practice, you should see that for these packets the phase velocity is greater than the group velocity. (See Table \(6.1\), Sec. 6.2.) You will see little wavelets grow from zero at the rear end of the packet, travel through the packet, and disappear at the front. (It takes practice; the waves travel rather fast.) Another good method is to put a board at the end of a bathtub and tap the board. To make millimeter-wavelength waves (surface tension waves), use an eyedropper full of water. Squeeze out one drop and let it fall on your pan or tub of water. First let the drop fall from a height of only a few millimeters. This gives dominant wavelengths of only a few millimeters. To see that these waves really are due to surface tension, add some soap to the water and repeat the experiment. You should notice a decrease in the group velocity when you add the soap. (To see that the longer wavelength waves are not due to surface tension, you can repeat the experiment at long wavelengths.) To lengthen the dominant wavelength of the group, let the water drop fall from a greater height. Here is a way to see (without doing a difficult measurement) that millimeter waves have a faster group velocity than waves of a centimeter or so. Generate a packet that has both millimeter and centimeter waves by dropping a water drop from a height of a foot or 50 into a circular pan filled to the brim. (A coffee can works very well.) Drop the drop near the center of the circular pan. Notice that after reflection from the rim the group comes to a focus at a point that is conjugate to the point where the drop hit. (By two conjugate points we mean points located on a line through the center of the circle and lying at equal distances on either side of the center.) When the packet is passing through the conjugate focus, there is a transitory standing wave there (similar to the transitory standing wave you get when you shake a wave packet onto a slinky tied to a wall). This enables you to judge the average arrival time of the packet. Look to see if there is a difference in arrival times for short- wavelength contributions to the packet as compared with longer wavelength contributions. It is difficult to measure, but you can see the effect fairly easily. An experiment \(I\) have not yet tried is to find a smooth running stream with flow velocity roughly equal to the group velocity for reasonable wavelengths. One should be able to make wave packets that travel upstream at about the flow velocity, so that the packet remains nearly at rest in your reference frame (assuming you are wading, not floating with the current). Surely that would be a most pleasant way to study wave packets.

In Prob. \(2.31\) you derived the dispersion law for sawtooth shallow-water standing waves, obtaining the result \(v_{\mathrm{p}} \approx\) \(1.1 \sqrt{g h} .\) For sinusoidal shallow-water waves the result turns out to be \(v_{\varphi}=\sqrt{g h}\). Thus shallow-water waves are nondispersive. (The phase velocity does not depend on the wavelength.) Instead of standing waves we now consider shallow-water traveling wave packets. Since the waves are nondispersive, a single "solitary wave" or "tidal wave" will propagate without changing its shape (approximately). Such waves, called tsunami, can be excited by undersea earthquakes in the ocean. The average water depth in the deep ocean is about 5 kilometers: \(h=5 \times 10^{5} \mathrm{~cm}\). Tidal waves of horizontal length much longer than \(5 \mathrm{~km}\) are therefore "shallow- water" waves. Tsunami waves propagate in the deep ocean at a velocity $$ v=\sqrt{g h}=\sqrt{(980) 5 \times 10^{5}}=2.2 \times 10^{4}=220 \text { meter } / \text { see } $$ \(\simeq 495\) miles/hour, which is somewhat slower than a typical jet airplane. How long does it take such a tidal wave to propagate from Alaska to Hawaii? In 1883 the volcano Krakatoa blew up, creating the world's biggest explosion. (Krakatoa is located in Sunda Strait, between Sumatra and Java. An account of the explosion can be found in any encyclopedia.) Huge tidal waves and atmospheric waves were created. Recently it has been discovered that there are air traveling waves with velocity about \(220 \mathrm{~m} / \mathrm{sec}\). (Recall that ordinary sound velocity at \(0^{\circ} \mathrm{C}\) is \(332 \mathrm{~m} / \mathrm{sec}\). On the average the air is colder than that, so the velocity is less than that.) The existence of these air waves probably explains how the tidal water waves from Krakatoa appeared on the far sides of land masses that should have blocked the water waves. Apparently the tidal waves "jumped over" the land masses by coupling to the air waves having the same velocity (and same excitation time). See the article by F. Press and D. Harkrider, "Air-Sea Waves from the Explosion of Krakatoa" Science \(154,1325(\) Dec. 9,1966\()\). In the experiment make your own shallow-water tidal waves as follows: Take a square pan a foot or two long. Fill it with water to a depth of about \(\frac{1}{2}\) or \(1 \mathrm{~cm}\). Give the pan a quick nudge (or lift one end and drop it suddenly). You will create two traveling wave packets, one at the near end and one at the far end, traveling in opposite directions. Follow the bigger of the packets. Measure the velocity by timing the wave for as many pan lengths as you can (probably about four). A stopwatch helps. Alternatively, you can count out loud as the packet hits the walls, memorize the "musical tempo," and finally measure the tempo with an ordinary watch. How well do your results agree with \(v=\sqrt{g h}\). As the depth of the water increases, you will finally get to the point where the waves are not shallow-water waves. Then the dispersion relation gradually goes over to the deep-water gravitational-wave dispersion relation \(\omega^{2}=g k\), i.e.r $$ v_{w}=\lambda \nu=\sqrt{\frac{g \lambda}{2 \pi}} $$ (We shall derive this relation in Chap. \(7 .)\) Thus the wave packet will spread out and not maintain its shape. For sufficiently shallow water (less than \(1 \mathrm{~cm}\), roughly), the shape is maintained fairly well for several feet. Finally, make a traveling tidal wave in your bathtub by suddenly pushing the entire end of the tub water with a board. Measure the down and back time and thus measure the velocity. Is it \(\sqrt{g h}\). Notice the breakers!
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