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Chapter 9: Problem 1

Show that $$ z=A \mathrm{e}^{\mathrm{i}\left\\{\omega t-\left(k_{1} x+k_{2} y\right)\right\\}} $$ where \(k^{2}=\omega^{2} / c^{2}=k_{1}^{2}+k_{2}^{2}\) is a solution of the two- dimensional wave equation $$ \frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial y^{2}}=\frac{1}{c^{2}} \frac{\partial^{2} z}{\partial t^{2}} $$

Short Answer

Expert verified
The function satisfies the wave equation by matching both sides when derived and substituted.

Step by step solution

01

Identify the given function

We start with the function given in the problem: \( z = A \mathrm{e}^{\mathrm{i}\left(\omega t - (k_1 x + k_2 y)\right)} \). This is a complex exponential form of a wave, where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( k_1 \) and \( k_2 \) are the wave numbers in the \( x \) and \( y \) directions, respectively.
02

Compute the second partial derivative of z with respect to x

We compute \( \frac{\partial z}{\partial x} \) first: \( \frac{\partial z}{\partial x} = -\mathrm{i}k_1 A \mathrm{e}^{\mathrm{i}(\omega t - (k_1 x + k_2 y))} \).Then compute \( \frac{\partial^2 z}{\partial x^2} \): \( \frac{\partial^2 z}{\partial x^2} = -k_1^2 A \mathrm{e}^{\mathrm{i}(\omega t - (k_1 x + k_2 y))} \).
03

Compute the second partial derivative of z with respect to y

Similarly, compute \( \frac{\partial z}{\partial y} \): \( \frac{\partial z}{\partial y} = -\mathrm{i}k_2 A \mathrm{e}^{\mathrm{i}(\omega t - (k_1 x + k_2 y))} \).Then compute \( \frac{\partial^2 z}{\partial y^2} \): \( \frac{\partial^2 z}{\partial y^2} = -k_2^2 A \mathrm{e}^{\mathrm{i}(\omega t - (k_1 x + k_2 y))} \).
04

Compute the second partial derivative of z with respect to t

First compute \( \frac{\partial z}{\partial t} \): \( \frac{\partial z}{\partial t} = \mathrm{i}\omega A \mathrm{e}^{\mathrm{i}(\omega t - (k_1 x + k_2 y))} \).Then compute \( \frac{\partial^2 z}{\partial t^2} \): \( \frac{\partial^2 z}{\partial t^2} = -\omega^2 A \mathrm{e}^{\mathrm{i}(\omega t - (k_1 x + k_2 y))} \).
05

Substitute into the wave equation

Substitute the expressions into the left side of the wave equation: \( \frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} \) becomes \(-k_1^2 A \mathrm{e}^{\mathrm{i}(\omega t - (k_1 x + k_2 y))} - k_2^2 A \mathrm{e}^{\mathrm{i}(\omega t - (k_1 x + k_2 y))} = -(k_1^2 + k_2^2) A \mathrm{e}^{\mathrm{i}(\omega t - (k_1 x + k_2 y))} \).
06

Equate to the right side and verify

The right side \( \frac{1}{c^2} \frac{\partial^2 z}{\partial t^2} \) becomes \( \frac{1}{c^2} (-\omega^2 A \mathrm{e}^{\mathrm{i}(\omega t - (k_1 x + k_2 y))}) = -\frac{\omega^2}{c^2} A \mathrm{e}^{\mathrm{i}(\omega t - (k_1 x + k_2 y))} \).Since \( k^2 = \frac{\omega^2}{c^2} \) is equivalent to \( k_1^2 + k_2^2 = \frac{\omega^2}{c^2} \), the equation holds: \(-k_1^2 - k_2^2 = -\frac{\omega^2}{c^2}\). Therefore, the function is a solution.

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Key Concepts

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

two-dimensional waves
Two-dimensional waves are an extension of simple wave concepts into two spatial dimensions. Imagine waves traveling in a plane, like ripples spreading across a pond's surface. These waves can be described using mathematical equations that account for both the horizontal and vertical changes in the wave's structure. By understanding two-dimensional waves, we can model many real-world phenomena like sound and water waves.

In two-dimensional waves, the wave vector is represented by two components, typically denoted as \( k_1 \) and \( k_2 \). These represent the wave numbers in the \( x \) and \( y \) directions, respectively. The wave number corresponds to the number of wave cycles per unit distance.
  • Wave speed: The speed at which a wave propagates through a medium. It depends on the medium and the type of wave.
  • Wave equation: Describes how waves move in a given space. It includes parameters like frequency, amplitude, and direction.
  • A practical example: Sound waves in a room are a two-dimensional wave. They spread out in all available directions, generating complex patterns.
complex exponential wave solutions
Complex exponential wave solutions are a powerful tool in solving wave equations. They simplify the analysis of wave behavior over space and time. By using complex numbers, they encapsulate both the amplitude and phase information of a wave.

The given function \( z = A \mathrm{e}^{\mathrm{i}(\omega t - (k_1 x + k_2 y))} \) is an instance of this type of solution. Here:
  • \( A \) is the amplitude, representing the wave's strength.
  • \( \omega \) is the angular frequency, determining how many oscillations occur in a given time frame.
  • \( k_1 \) and \( k_2 \) are wave numbers along the \( x \) and \( y \) directions, showing how the wave varies in space.
The expression \( \mathrm{e}^{\mathrm{i}(\omega t - (k_1 x + k_2 y))} \) represents oscillations over time \( t \) and space \( x, y \). By using Euler's formula, these complex exponentials can be resolved into sine and cosine functions, simplifying calculations. These solutions make use of complex numbers for their elegant and efficient representation.

When working with complex exponential solutions, always remember:
  • They often simplify differential equations, making them easier to solve.
  • They incorporate both amplitude and phase, providing a full wave description.
  • They can easily be converted to real-valued solutions when needed.
partial differential equations
Partial differential equations (PDEs) are mathematical equations that involve functions of several variables and their partial derivatives. They're essential in describing various physical phenomena, especially when it comes to waves.

The two-dimensional wave equation presented here is a classic example of a PDE. It relates the changes in a wave function \( z(x, y, t) \) with respect to both space and time:\[\frac{\partial^{2} z}{\partial x^{2}} + \frac{\partial^{2} z}{\partial y^{2}} = \frac{1}{c^{2}} \frac{\partial^{2} z}{\partial t^{2}} \]
  • Left side: Involves spatial derivatives, capturing how the wave changes in direction \( x \) and \( y \).
  • Right side: Time derivative, accounting for the wave's change over time with factor \( 1/c^2 \).
To solve PDEs, we first compute each partial derivative needed, then substitute into the equation to verify a given solution, like the complex exponential form shown in the exercise.

The solution involves demonstrating equality between both sides of the PDE, achieved through careful calculation of derivatives and substituting constraints given by the wave's physical properties, such as \( k^2 = \frac{\omega^2}{c^2} \). In essence, solving a PDE involves:
  • Calculating partial derivatives to express changes.
  • Balancing both sides of the equation to confirm solutions.
  • Incorporating physical conditions or constraints specific to the problem.

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Most popular questions from this chapter

The tungsten filament of an electric light bulb has a temperature of \(\approx 2000 \mathrm{~K}\). Show that in this case \(\lambda_{m} \approx 14 \times 10^{-7} \mathrm{~m}\), well into the infrared. Such a lamp is therefore a good heat source but an inefficient light source.

The wave representation of a particle, e.g. an electron, in a rectangular potential well throughout which \(V=0\) is given by Schr枚dinger's time- independent equation $$ \frac{\partial^{2} \Psi}{\partial x^{2}}+\frac{\partial^{2} \Psi}{\partial y^{2}}+\frac{\partial^{2} \Psi}{\partial z^{2}}=-\frac{8 \pi^{2} m}{h^{2}} E \Psi $$ where \(E\) is the particle energy, \(m\) is the mass and \(h\) is Planck's constant. The boundary conditions to be satisfied are \(\psi=0\) at \(x=y=z=0\) and at \(x=L_{x}, y=L_{y}, z=L_{z}\), where \(L_{x}, L_{y}\) and \(L_{z}\) are the dimensions of the well. Show that $$ \Psi=A \sin \frac{l \pi x}{L_{x}} \sin \frac{r \pi y}{L_{y}} \sin \frac{n \pi z}{L_{z}} $$ is a solution of Schr枚dinger's equation, giving $$ E=\frac{h^{2}}{8 m}\left(\frac{l^{2}}{L_{x}^{2}}+\frac{r^{2}}{L_{y}^{2}}+\frac{n^{2}}{L_{z}^{2}}\right) $$ When the potential well is cubical of side \(L\), $$ E=\frac{h^{2}}{8 m L^{2}}\left(l^{2}+r^{2}+n^{2}\right) $$ and the lowest value of the quantized energy is given by $$ E=E_{0} \quad \text { for } \quad l=1, \quad r=n=0 $$ Show that the next energy levels are \(3 E_{0}, 6 E_{0}\) (three-fold degenerate), \(9 E_{0}\) (three-fold degenerate), \(11 E_{0}\) (three-fold degenerate), \(12 E_{0}\) and \(14 E_{0}\) (six-fold degenerate).

Planck's radiation law, expressed in terms of energy per unit range of wavelength instead of frequency, becomes $$ E_{\lambda}=\frac{8 \pi c h}{\lambda^{5}\left(e^{c h / \lambda k T}-1\right)} $$ Use the variable \(x=c h / \lambda k T\) to show that the total energy per unit volume at temperature \(T^{\circ}\) absolute is given by $$ \int_{0}^{\infty} E_{\lambda} \mathrm{d} \lambda=a T^{4} \mathrm{~J} \mathrm{~m}^{-3} $$ where $$ a=\frac{8 \pi^{5} k^{4}}{15 c^{3} h^{3}} $$ (The constant \(c a / 4=\sigma\), Stefan's Constant in the Stefan-Boltzmann Law.) Note that $$ \int_{0}^{\infty} \frac{x^{3} \mathrm{~d} x}{e^{x}-1}=\frac{\pi^{4}}{15} $$

An electromagnetic wave loses negligible energy when reflected from a highly conducting surface. With repeated reflections it may travel along a transmission line or wave guide consisting of two parallel, infinitely conducting planes (separation \(a\) ). If the wave is plane polarized, so that only \(E_{z}\) exists, then the propagating direction \(\mathbf{k}\) lies wholly in the \(x y\) plane. The boundary conditions require that the total tangential electric field \(E_{z}\) is zero at the conducting surfaces \(x=0\) and \(x=a .\) Show that the first boundary condition allows \(E_{z}\) to be written \(E_{z}=\) \(E_{0}\left(\mathrm{e}^{i k_{x} x}-\mathrm{e}^{-i k_{x} x}\right) \mathrm{e}^{\mathrm{i}\left(k_{x} y-u t\right)}\), where \(k_{x}=k \cos \theta\) and \(k_{y}=k \sin \theta\) and the second boundary condition requires \(k_{x}=n \pi / a\) If \(\lambda_{0}=2 \pi c / \omega, \lambda_{c}=2 \pi / k_{x}\) and \(\lambda_{g}=2 \pi / k_{y}\) are the wavelengths propagating in the \(x\) and \(y\) directions respectively show that $$ \frac{1}{\lambda_{c}^{2}}+\frac{1}{\lambda_{g}^{2}}=\frac{1}{\lambda_{0}^{2}} $$ We see that for \(n=1, k_{x}=\pi / a\) and \(\lambda_{c}=2 a\), and that as \(\omega\) decreases and \(\lambda_{0}\) increases, \(k_{\mathrm{y}}=k \sin \theta\) becomes imaginary and the wave is damped. Thus, \(n=2\left(k_{x}=2 \pi / a\right)\) gives \(\lambda_{c}=a\), the 'critical wavelength', i.e. the longest wavelength propagated by a waveguide of separation \(a\). Such cut-off wavelengths and frequencies are a feature of wave propagation in periodic structures, transmission lines and wave-guides.

An electromagnetic wave \((\mathbf{E}, \mathbf{H})\) propagates in the \(x\) -direction down a perfectly conducting hollow tube of arbitrary cross section. The tangential component of \(\mathbf{E}\) at the conducting walls must be zero at all times. Show that the solution \(\mathbf{E}=E(y, z) \mathbf{n} \cos \left(\omega t-k_{x} x\right)\) substituted in the wave equation yields $$ \frac{\partial^{2} E(y, z)}{\partial y^{2}}+\frac{\partial^{2} E(y, z)}{\partial z^{2}}=-k^{2} E(y, z) $$ where \(k^{2}=\omega^{2} / c^{2}-k_{x}^{2}\) and \(k_{x}\) is the wave number appropriate to the \(x\) -direction, \(\mathbf{n}\) is the unit vector in any direction in the \((y, z)\) plane.
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