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Chapter 11: Problem 30

An isotropic linear material medium, characterized by the constitutive relations (in its rest frame \(K^{\prime}\) ), \(\mathbf{D}^{\prime}=\epsilon \mathbf{E}^{\prime}\) and \(\mu \mathbf{H}^{\prime}=\mathbf{B}^{\prime}\), is in uniform translation with velocity \(\mathbf{v}\) in the inertial frame \(K .\) By exploiting the fact that \(F_{\mu r}=(\mathbf{E}, \mathbf{B})\) and \(G_{\mu \mathrm{r}}=(\mathbf{D}, \mathbf{H})\) transform as second rank 4 -tensors under Lorentz transformations, show that the macroscopic fields \(\mathbf{D}\) and \(\mathbf{H}\) are given in terms of \(\mathbf{E}\) and \(\mathbf{B}\) by $$ \begin{array}{l} \mathbf{D}=\boldsymbol{E} \mathbf{E}+\boldsymbol{\gamma}^{2}\left(\boldsymbol{\epsilon}-\frac{1}{\mu}\right)\left[\beta^{2} \mathbf{E}_{1}+\boldsymbol{\beta} \times \mathbf{B}\right] \\ \mathbf{H}=\frac{1}{\mu} \mathbf{B}+\gamma^{2}\left(\epsilon-\frac{1}{\mu}\right)\left[-\beta^{2} \mathbf{B}_{\llcorner}+\boldsymbol{\beta} \times \mathbf{E}\right] \end{array} $$ where \(\mathbf{E}_{1}\) and \(\mathbf{B}_{1}\) are components perpendicular to \(\mathbf{v}\).

Short Answer

Expert verified
Transform the fields using Lorentz transformation properties, considering perpendicular components.

Step by step solution

01

Understand the Constitutive Equations

In the rest frame, the constitutive relations are given as \(\mathbf{D}' = \epsilon \mathbf{E}'\) and \(\mu \mathbf{H}' = \mathbf{B}'\). \(\epsilon\) and \(\mu\) are the permittivity and permeability of the material, respectively.
02

Consider Lorentz Transformations and Field Tensors

Recognize that the fields transform as 4-tensors under Lorentz transformations. The electromagnetic field tensor \(F^{\mu u}\) in frame \(K'\) is related to the fields \(\mathbf{E}\, \mathbf{B}\), and the auxiliary tensor \(G^{\mu u}\) is related to \(\mathbf{D}\, \mathbf{H}\). Both tensors transform using the Lorentz transformation matrix \(\Lambda\).
03

Transform to the Moving Frame

Use the Lorentz transformation to relate the fields in the moving frame \(K\) to those in the rest frame \(K'\). The transformation equations for the fields are given by: \[\mathbf{E} = \gamma \left( \mathbf{E}' + \mathbf{v} \times \mathbf{B}' \right) - \frac{\gamma^2}{\gamma+1} (\mathbf{v} \cdot \mathbf{E}') \mathbf{v} \] and \[\mathbf{B} = \gamma \left( \mathbf{B}' - \mathbf{v} \times \mathbf{E}' \right) - \frac{\gamma^2}{\gamma+1} (\mathbf{v} \cdot \mathbf{B}') \mathbf{v} \]. Where \(\beta = v/c\) and \(\gamma = 1/\sqrt{1-\beta^2}\).
04

Transform \(\mathbf{D}\) and \(\mathbf{H}\)

In the moving frame, transform \(\mathbf{D}' = \epsilon \mathbf{E}'\) and \(\mu \mathbf{H}' = \mathbf{B}'\) using the relations between \(\mathbf{E}\), \(\mathbf{B}\) and their primed counterparts.\[\mathbf{D} = \epsilon \mathbf{E} + \gamma^2 (\epsilon - \frac{1}{\mu}) \left(\boldsymbol{\beta} \cdot \mathbf{B}\right)\boldsymbol{\beta}\] \[\mathbf{H} = \frac{1}{\mu} \mathbf{B} - \gamma^2 (\epsilon - \frac{1}{\mu}) \left(\boldsymbol{\beta} \cdot \mathbf{E}\right)\boldsymbol{\beta}\].
05

Resolve the Perpendicular Components

Distinguish the components of \(\mathbf{E}\) and \(\mathbf{B}\) that are perpendicular to \(\mathbf{v}\) (i.e., \(\mathbf{E}_1\) and \(\mathbf{B}_1\)). Write the final expressions using these components: \[\mathbf{D} = \epsilon \mathbf{E}_1 + \gamma^2 (\epsilon - \frac{1}{\mu}) \left(\beta^2 \mathbf{E}_1 + \beta \times \mathbf{B}\right)\] and \[\mathbf{H} = \frac{1}{\mu} \mathbf{B}_1 + \gamma^2 (\epsilon - \frac{1}{\mu}) \left(- \beta^2 \mathbf{B}_1 + \beta \times \mathbf{E}\right) \].

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Key Concepts

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

Electromagnetic Field Tensors
Understanding electromagnetic field tensors involves recognizing how electric and magnetic fields are represented in four-dimensional space-time. The field tensors are mathematical constructs that transform according to the rules of special relativity.
In this context, the electromagnetic field tensor, denoted as \(F^{\mu u}\), encapsulates the components of the electric field \(\mathbf{E}\) and magnetic field \(\mathbf{B}\). These tensors allow us to systematically view transformations between different inertial frames, such as moving at a constant velocity relative to one another.
  • The primary benefit of using field tensors is simplifying Maxwell's equations, which describe how electric and magnetic fields interact.
  • The tensor approach makes it easier to understand how these fields change under Lorentz transformations, a key element of Einstein鈥檚 theory of relativity.
The use of tensors is crucial because they give the same physical laws in all reference frames, highlighting the symmetric nature of electromagnetism under Lorentz transformations.
Constitutive Relations
Constitutive relations are equations that link electric displacement \(\mathbf{D}\) and magnetic induction \(\mathbf{B}\) with electric field \(\mathbf{E}\) and magnetic field \(\mathbf{H}\) through material properties. In simple materials, these relations can be defined as \(\mathbf{D} = \epsilon \mathbf{E}\) and \(\mathbf{B} = \mu \mathbf{H}\), where \(\epsilon\) is permittivity and \(\mu\) is permeability.

These relations describe how materials respond to electromagnetic fields.
  • In isotropic media, properties are uniform in all directions, which simplifies the equations considerably.
  • Constitutive relations are key in linking macroscopic fields to the microscopic behavior of materials, essential for understanding materials' electronic and magnetic properties.
By using tensors to transform these relations, we can see how electromagnetic fields behave in different reference frames, aiding in the analysis of moving media and advanced materials.
Macroscopic Fields
Macroscopic fields result from averaging microscopic fields over a large number of particles in a material, giving a smooth and continuous description of electric and magnetic fields.
These fields account for material properties, leading to relations that describe phenomena such as polarization and magnetization. In a moving frame, fields \(\mathbf{E}\), \(\mathbf{B}\), \(\mathbf{D}\), and \(\mathbf{H}\) are interconnected and transform with velocity \(\mathbf{v}\) to satisfy relativity principles.
  • Macroscopic fields simplify the overall picture by lumping together complex particle interactions into useful quantities for engineering and physics.
  • This approach allows us to predict the behavior of fields in frameworks involving moving materials and boundary conditions.
By understanding these fields, one can derive relations like those in the original exercise, which show transformations and interactions under motion, ultimately leading to broader insights into electromagnetic theory.

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Most popular questions from this chapter

The lambda particle \((\Lambda)\) is a neutral baryon of mass \(M=1115 \mathrm{MeV}\) that decays with a lifetime of \(r=2.9 \times 10^{-10} \mathrm{~s}\) into a nucleon of mass \(m_{1}=939 \mathrm{MeV}\) and a pi-meson of mass \(m_{2}=140 \mathrm{MeV}\). It was first observed in flight by its charged decay mode \(\Lambda \rightarrow p+\pi^{-}\) in cloud chambers. The charged tracks originate from a single point and have the appearance of an inverted vee or lambda. The particles' identities and momenta can be inferred from their ranges and curvature in the magnetic field of the chamber. (a) Using conservation of momentum and energy and the invariance of scalar products of 4 -vectors show that, if the opening angle \(\theta\) between the two tracks is measured, the mass of the decaying particle can be found from the formula $$ M^{2}=m_{1}^{2}+m_{\frac{2}{2}}^{2}+2 E_{1} E_{2}-2 p_{1} p_{2} \cos \theta $$ where here \(p_{1}\) and \(p_{2}\) are the magnitudes of the 3 -momenta. (b) A lambda particle is created with a total energy of \(10 \mathrm{GeV}\) in a collision in the top plate of a cloud chamber. How far on the average will it travel in the chamber before decaying? What range of opening angles will occur for a 10 GeV lambda if the decay is more or less isotropic in the lambda's rest frame?

(a) A charge density \(\rho^{\prime}\) of zero total charge, but with a dipole moment \(\mathbf{p}\). exists in reference frame \(K^{\prime}\). There is no current density in \(K^{\prime}\). The frame \(K^{\prime}\) moves with a velocity \(\boldsymbol{\gamma}=\boldsymbol{\beta} c\) in the frame \(K\). Find the charge and current densities \(\rho\) and \(J\) in the frame \(K\) and show that there is a magnetic dipole moment, \(\mathrm{m}=(\mathrm{p} \times \boldsymbol{\beta}) / 2\), correct to first order in \(\beta\). What is the electric dipole moment in \(K\) to the same order in \(\beta ?\) (b) Instead of the charge density, but no current density, in \(K^{\prime}\), consider no charge density, but a current density \(\mathbf{J}^{\prime}\) that has a magnetic dipole moment m. Find the charge and current densities in \(K\) and show that to first order in \(\beta\) there is an electric dipole moment \(\mathbf{p}=\boldsymbol{\beta} \times \mathbf{m}\) in addition to the magnetic dipole moment.

A possible clock is shown in the figure. It consists of a flashtube \(F\) and a photocell \(P\) shielded so that each views only the mirror \(M\), located a distance \(d\) away, and mounted rigidly with respect to the flashtube-photocell assembly. The electronic innards of the box are such that when the photocell responds to a light flash from the mirror, the flashtube is triggered with a negligible delay and emits a short flash toward the mirror. The clock thus "ticks" once every \((2 d / c)\) seconds when at rest. (a) Suppose that the clock moves with a uniform velocity \(u\), perpendicular to the line from \(P F\) to \(M\), relative to an observer. Using the second postulate of relativity, show by explicit geometrical or algebraic construction that the ob. server sees the relativistic time dilatation as the clock moves by. (b) Suppose that the clock moves with a velocity \(v\) parallel to the line from \(P F\) to \(M\). Verify that here, too, the clock is observed to tick more slowly, by the same time dilatation factor.

The presence in the universe of an apparently uniform "sea" of blackbody radiation at a temperature of roughly \(3 \mathrm{~K}\) gives one mechanism for an upper limit on the energies of photons that have traveled an appreciable distance since their creation. Photon-photon collisions can result in the creation of a charged particle and its antiparticle ("pair creation") if there is sufficient energy in the center of "mass" of the two photons. The lowest threshold and also the largest cross section occurs for an electron-positron pair. (a) Taking the energy of a typical \(3 \mathrm{~K}\) photon to be \(E=2.5 \times 10^{-4} \mathrm{eV}\), calculate the energy for an incident photon such that there is energy just sufficient to make an electron-positron pair. For photons with energies larger than this threshold value, the cross section increases to a maximum of the order of \(\left(e^{2} / m c^{2}\right)^{2}\) and then decreases slowly at higher energies. This interaction is one mechanism for the disappearance of such photons as they travel cosmological distances. (b) There is some evidence for a diffuse x-ray background with photons having energies of several hundred electron volts or more. Beyond \(1 \mathrm{keV}\) the spectrum falls as \(E^{-n}\) with \(n=1.5\). Repeat the calculation of the threshold incident energy, assuming that the energy of the photon in the "sea" is \(500 \mathrm{eV}\).

Show explicitly that two successive Lorentz transformationy in the same direction are equivalent to a single Lorentz transformation with a velocity $$ v=\frac{v_{1}+v_{2}}{1+\left(v_{1} v_{2} / c^{2}\right)} $$ This is an alternative way to derive the parallel-velocity addition law.
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