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Chapter 4: Problem 9

In ordinary three-space, Ohm's law is \(\mathbf{j}=\sigma \mathbf{E}\) where \(\mathbf{j}\) is the current, \(\mathbf{E}\) the electric field, and \(\sigma\) the conductivity. Assuming \(\sigma\) is a scalar, write a four-tensor form of Ohm's law using the four-current \(j_{\mu}\), the Maxwell field tensor \(F_{\mu v}\) and the four-velocity of the conducting element \(U_{\mu}\). Remember, a tensor equation that reduces to the correct expression in any frame (e.g., the rest frame of the conducting element) is correct in all frames.

Short Answer

Expert verified
The four-tensor form of Ohm's law is \( j_{\mu} = \sigma F_{\mu\nu} U^{\nu} \).

Step by step solution

01

Identify Given Quantities

We start by identifying the given quantities in the problem. Ohm's law in three-space is given as \( \mathbf{j} = \sigma \mathbf{E} \), where \( \mathbf{j} \) is the current, \( \mathbf{E} \) is the electric field, and \( \sigma \) is the conductivity. We are tasked with finding the four-tensor form of Ohm's law.
02

Define Four-Current

The four-current is defined as \( j_{\mu} = (c \rho, \mathbf{j}) \), where \( \rho \) is the charge density and \( c \) is the speed of light. This four-vector represents the conserved current in spacetime, generalizing three-current and charge density into a single entity.
03

Introduce Four-Velocity

The four-velocity \( U_{\mu} \) is defined as \( U_{\mu} = (\gamma c, \gamma \mathbf{v}) \), where \( \gamma \) is the Lorentz factor given by \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \). In the rest frame of the conducting element, \( U_{\mu} = (c, 0) \).
04

Define Maxwell Field Tensor

The Maxwell field tensor is defined as \( F_{\muu} \), combining the electric field \( \mathbf{E} \) and magnetic field \( \mathbf{B} \) into a single antisymmetric tensor. In three-space, the electric field components are expressed using the field tensor as \( E_i = F_{0i} \).
05

Derive Tensor Form of Ohm's Law

To express Ohm's law in tensor form, use the relation \( j_{\mu} = \sigma F_{\muu} U^{u} \). This equation captures the same physical relationship as the original Ohm's law in three-space but extended to four-dimensional spacetime. It ensures the relation holds in all inertial frames.
06

Confirm Frame Consistency

Check if the derived tensor equation reduces to the classical Ohm's law in the rest frame. With \( U^u \) having only the temporal component \( U^0 = c \), the equation becomes \( j_{i} = \sigma F_{0i} = \sigma E_{i} \), confirming consistency with \( \mathbf{j} = \sigma \mathbf{E} \).

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Key Concepts

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

Ohm's law
Ohm's law is a fundamental principle in electrical engineering and physics. It relates the current flowing through a conductor to the electric field and the material's conductivity. In three-dimensional space, Ohm's law is expressed as \( \mathbf{j} = \sigma \mathbf{E} \):
  • \(\mathbf{j}\) is the current density (amount of charge flowing per unit area per unit time).
  • \(\mathbf{E}\) is the electric field (force per unit charge).
  • \(\sigma\) is the conductivity of the material, measuring how well it conducts electricity.
This law indicates that the current density is proportional to the electric field, with the constant of proportionality being the conductivity. In essence, it tells us how easily electrical current can flow through a material when an electric field is applied.
Maxwell field tensor
The Maxwell field tensor is an essential concept that merges electric and magnetic fields into one entity. It is highly beneficial in formulating the laws of electromagnetism in relativistic contexts.
  • The tensor is denoted by \( F_{\mu v} \), a 4x4 antisymmetric matrix.
  • It includes components of the electric field \( \mathbf{E} \) and the magnetic field \( \mathbf{B} \).
The field tensor simplifies equations by treating electric and magnetic fields equally in spacetime. For example, the electric field components correspond to \( F_{0i} \) elements, while magnetic field components are linked to spatial combinations like \( F_{ij} \). This unification allows for a more straightforward and universal application of electromagnetic theory across different frames of reference.
Four-current
The concept of four-current unifies charge density and current density into a single four-vector. This is a critical step when moving from three-dimensional to four-dimensional descriptions in physics:
  • Four-current is represented as \( j_{\mu} = (c \rho, \mathbf{j}) \).
  • \( \rho \) is the charge density, and \( c \) is the speed of light.
  • In the spatial components, \( \mathbf{j} \) is the regular current density.
It encapsulates the idea that current and charge density are not separate in spacetime but part of a collective entity, making it easier to apply laws like conservation of charge universally. This seamless integration facilitates expressing laws like Ohm's law in a four-dimensional framework, showing consistency across all frames.
Four-velocity
Four-velocity is a relativistic concept used to describe the state of motion of an object in spacetime:
  • Defined as \( U_{\mu} = (\gamma c, \gamma \mathbf{v}) \), where \( \gamma \) is the Lorentz factor.
  • \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \) adjusts for relativistic speeds, ensuring consistency with the theory of relativity.
In the rest frame of an object, its four-velocity simplifies to \( U_{\mu} = (c, 0) \). This term becomes crucial when describing the motion of charge carriers in materials, allowing Ohm's law to be expressed consistently across all frames of reference. Four-velocity thus seamlessly integrates velocity into the framework of spacetime, enhancing the universality of physical laws.

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Most popular questions from this chapter

Show that \(U_{\mathrm{em}}^{2}-c^{-2} \mathbf{S}^{2}\) is a Lorentz scalar, where \(U_{\mathrm{em}}\) is the free-space electromagnetic energy density and \(\mathbf{S}\) is the Poynting vector.

A rocket starts out from earth with a constant acceleration of \(1 \mathrm{~g}\) in its own frame. After 10 years of its own (proper) time it reverses the acceleration, and in 10 more years it is again at rest with respect to the earth. After a brief time for exploring, the spacemen retrace their journey back to earth, completing the entire trip in 40 years of their own time. a. Let \(t\) be earth time and \(x\) be the position of the rocket as measured from earth. Let \(\tau\) be the proper time of the rocket and let \(\beta=\) \(c^{-1} d x / d t\). Show that the equation of motion of the rocket during the first phase of positive acceleration is $$ \gamma^{3} \frac{d^{2} x}{d t^{2}}=g $$ b. Integrate this equation to show that $$ \beta=\frac{g t / c}{\sqrt{(g t / c)^{2}+1}} . $$ c. Integrating again, show that $$ x=\frac{c^{2}}{g}\left[\sqrt{(g t / c)^{2}+1}-1\right] . $$ d. Show that the proper time is related to earth time by $$ \frac{g t}{c}=\sinh \left(\frac{g \tau}{c}\right) $$ so that $$ x=\frac{c^{2}}{g}\left[\cosh \left(\frac{g \tau}{c}\right)-1\right] . $$ e. How far away do the spacemen get? f. How long does their journey last from the point of view of an earth observer? Will friends be there to greet them when they return? Hint: In answering parts (e) and (f) you need only the results for the first positive phase of acceleration plus simple arguments concerning the other phases. g. Answer parts (e) and (f) if the spacemen can tolerate an acceleration of \(2 g\) rather than \(1 g\).

A particle (rest mass \(m\) ) initially at rest absorbs a photon of energy \(h \nu\) and converts this energy into increased internal energy (say, heat). The particle has increased its rest mass to \(m^{\prime}\) and moves with some velocity \(v^{\prime}\). a. Setting up the conservation of energy and momentum, show that $$ \frac{m}{m^{\prime}}=\left(1+\frac{2 h v}{m c^{2}}\right)^{-1 / 2} $$ b. By considering the appropriate Lorentz transformations, show that if the particle had been moving initially and absorbed a photon of energy \(h v\), this same equation for the ratio of the initial and final rest masses holds with \(\nu^{\prime}\) replacing \(\nu\), where \(\nu^{\prime}\) is given by the Doppler formula.

In astrophysics it is frequently argued that a source of radiation which undergoes a fluctuation of duration \(\Delta t\) must have a physical diameter of order \(D \leqslant c \Delta t\). This argument is based on the fact that even if all portions of the source undergo a disturbance at the same instant and for an infinitesimal period of time, the resulting signal at the observer will be smeared out over a time interval \(\Delta t_{\min } \sim D / c\) because of the finite light travel time across the source. Suppose, however, that the source is an optically thick spherical shell of radius \(R(t)\) that is expanding with relativistic velocity \(\beta \sim 1, \gamma \gg 1\) and energized by a stationary point at its center. By consideration of relativistic beaming effects show that if the observer sees a fluctuation from the shell of duration \(\Delta t\) at time \(t\), the source may actually be of radius $$ R<2 \gamma^{2} c \Delta t, $$ rather than the much smaller limit given by the nonrelativistic considerations. In the rest frame of the shell surface, each surface element may be treated as an isotropic emitter. This latter argument has been used to show that the active regions in quasars may be much larger than \(c \Delta t \sim 1\) light month across, and thus avoids much energy being crammed into so small a volume.

Consider the stress-energy tensor for an electromagnetic field $$ T^{\mu \nu} \equiv \frac{1}{4 \pi}\left(F^{\mu \alpha} F^{\nu}{ }_{\alpha}-\frac{1}{4} \eta^{\mu \nu} F^{\alpha \beta} F_{\alpha \beta}\right) $$ where \(F^{\alpha \beta}\) and \(\eta^{\mu}\) are the electromagnetic field tensor and Minkowski metric, respectively. a. Show that \(T^{\mu \nu}\) is traceless: \(T^{\mu}{ }_{\mu}=0\). b. Show that in free space \(T^{\mu \nu}\) is divergenceless: \(T^{\mu \nu},{ }_{\nu}=0\).
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