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Chapter 3: Problem 30

Transformations of \(\lambda\) and \(I\), Consider a composite line charge consisting of several kinds of carriers, each with its own velocity. For each kind, labeled by \(k\), the linear density of charge measured in frame \(F\) is \(\lambda_{k}\), and the velocity is \(\beta_{k} c\) parallel to the line. The contribution of these carriers to the current in \(F\) is then \(I_{k}=\lambda_{k} \beta_{k} c\). How much do these \(k\)-type carriers contribute to the charge and current in a frame \(F^{\prime}\) that is moving parallel to the line at velocity \(-\beta c\) with respect to \(F ?\) By following the steps we took in the transformations in Fig. \(5.22\), you should be able to show that $$ \lambda_{k}^{\prime}=\gamma\left(\lambda_{k}+\frac{\beta I_{k}}{c}\right), \quad I_{k}^{\prime}=\gamma\left(I_{k}+\beta c \lambda_{k}\right) $$ If each component of the linear charge density and current transforms in this way, then so must the total \(\lambda\) and \(I\) : $$ \lambda^{\prime}=\gamma\left(\lambda+\frac{\beta I}{c}\right), \quad I^{\prime}=\gamma(I+\beta c \lambda) $$ You have now derived the Lorentz transformation to a parallelmoving frame for any line charge and current, whatever its composition,

Short Answer

Expert verified
The Lorentz transformation shows that the charge density and current density transform as \[ \lambda_{k}^{\prime}=\gamma\left(\lambda_{k}+\frac{\beta I_{k}}{c}\right) \] and \[ I_{k}^{\prime}=\gamma\left(I_{k}+\beta c \lambda_{k}\right) \] respectively for each type of carrier in the moving frame \(F'\). The total charge and current undergo a similar transformation to give \[ \lambda^{\prime}=\gamma\left(\lambda+\frac{\beta I}{c}\right) \] and \[ I^{\prime}=\gamma(I+\beta c \lambda) \].

Step by step solution

01

Understanding the Contributions of Various Carriers

We have several kinds of carriers each represented by \(k\), each possessing different velocities. In the frame \(F\), the linear density of the charge for \(k\) type carriers is \(\lambda_{k}\), and the velocity is \(\beta_{k}c\). Their contribution to the current in \(F\) is represented by \(I_{k}=\lambda_{k}\beta_{k}c\). The original question asks us how these \(k\)-type carriers would contribute in a different frame \(F'\), which moves parallel to the line with velocity \(-\beta c\) as compared to \(F\).
02

Construct the Lorentz Transformations

Applying the Lorentz transformation, we get the contributions to charge and current in frame \(F'\) as: \[ \lambda_{k}^{\prime}=\gamma\left(\lambda_{k}+\frac{\beta I_{k}}{c}\right) \] and \[ I_{k}^{\prime}=\gamma\left(I_{k}+\beta c \lambda_{k}\right) \]. Here, \( \lambda_{k}^{\prime} \) and \( I_{k}^{\prime} \) represent the linear charge density and current density in frame \(F'\) respectively.
03

Find Total Transformation

The total \(\lambda\) and \(I\) (being a sum of individual \(\lambda_{k}\) and \(I_{k}\)) in frame \(F'\) also transform in a similar way, giving us \[ \lambda^{\prime}=\gamma\left(\lambda+\frac{\beta I}{c}\right) \], and \[ I^{\prime}=\gamma(I+\beta c \lambda) \]. This constitutes the Lorentz transformation for any line charge and current.

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Key Concepts

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

Relativity
Relativity is a fundamental concept in physics that explains how the laws of physics apply across different reference frames, particularly when objects are moving at high speeds relative to each other.
For example, light travels at the same speed for all observers, regardless of their motion, according to Einstein's theory of special relativity.
In our context, relativity helps us understand how electric charges and currents behave when observed from different moving frames. This involves transformations that keep the physics consistent across these frames.
Line Charge
A line charge refers to a distribution of electric charge along a line, which can be an actual wire or an imaginary line in space.
This linear distribution is characterized by a linear charge density, denoted as \(\lambda\).
The linear charge density is the amount of charge per unit length and is crucial when calculating electric fields and potential created by line charges. In our scenario, each type of charge carrier has its specific linear charge density, influencing how the system behaves.
Current Density
Current density is a measure of electric current per unit area, but in the case of line charges, it's related to the linear flow of charge along the line.
It may depend on both the linear charge density and the velocity of charge carriers.
For example, in frame \(F\), the current density contribution from the kth type carrier is given by \(I_k = \lambda_k \beta_k c\). This gives an idea of how much charge is flowing past a point per unit time.
Linear Density
Linear density in this context is the linear charge density, which denotes how charge is distributed along a length.
It鈥檚 represented by \( \lambda \), and it's crucial for calculating related quantities like the electric field or the current in the system.
In transformations between frames, it's essential to adjust the linear density according to how the observers are moving relative to each other.
Reference Frames
Reference frames are perspectives from which measurements are made, even when in motion.
In physics, especially relativity, different observers may be in different reference frames moving relative to each other.
For instance, in our problem, frame \(F'\) moves with a velocity \(-\beta c\)\ relative to frame \(F\). It involves applying the Lorentz transformation to understand how quantities like charge and current are perceived in different frames.

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Most popular questions from this chapter

A capacitor consists of two parallel rectangular plates with a vertical separation of \(2 \mathrm{~cm}\). The east-west dimension of the plates is \(20 \mathrm{~cm}\), the north-south dimension is \(10 \mathrm{~cm}\). The capacitor has been charged by connecting it temporarily to a battery of \(300 \mathrm{~V}\). What is the electric field strength between the plates? How many excess electrons are on the negative plate? Now give the following quantities as they would be measured in a frame of reference that is moving eastward, relative to the laboratory in which the plates are at rest, with speed \(0.6 c:\) the three dimensions of the capacitor; the number of excess electrons on the negative plate; the electric field strength between the plates. Answer the same questions for a frame of reference that is moving upward with speed \(0.6 c\).

Capacitance coefficients for shells : ? A capacitor consists of two concentric spherical shells. Label the inner shell, of radius \(b\), as conductor 1 ; and label the outer shell, of radius \(a\), as conductor \(2 .\) For this two-conductor system, find \(C_{11}\), \(C_{22}\), and \(C_{12}\)

Electron beam A beam of \(9.5\) megaelectron-volt (MeV) electrons \((\gamma \approx 20)\), amounting as current to \(0.05 \mu \mathrm{A}\), is traveling through vacuum. The transverse dimensions of the beam are less than \(1 \mathrm{~mm}\), and there are no positive charges in or near it. (a) In the lab frame, what is the average distance between an, electron and the next one ahead of it, measured parallel to the beam? What approximately is the average electric field strength 1 cm away from the beam? (b) Answer the same questions for the electron rest frame.

Attenuator chain Some important kinds of networks are infinite in extent. Figure \(4.49\) shows a chain of series and parallel resistors stretching off endlessly to the right. The line at the bottom is the resistanceless return wire for all of them. This is sometimes called an attenuator chain, or a ladder network. The problem is to find the "input resistance," that is, the equivalent resistance between terminals \(A\) and \(B\). Our interest in this problem mainly concerns the method of solution, which takes an odd twist and which can be used in other places in physics where we have an iteration of identical devices (even an infinite chain of lenses, in optics). The point is that the input resistance (which we do not yet know - call it \(R\) ) will not be changed by adding a new set of resistors to the front end of the chain to make it one unit longer. But now, adding this section, we see that this new input resistance is just \(R_{1}\) in series with the parallel combination of \(R_{2}\) and \(R\) Use this strategy to determine \(R\). Show that, if voltage \(V_{0}\) is applied at the input to such a chain, the voltage at successive nodes decreases in a geometric series. What should the ratio of the resistors be so that the ladder is an attenuator that halves the voltage at every step? Obviously a truly infinite ladder would not be practical. Can you suggest a way to terminate it after a few sections without introducing any error in its attenuation?

Moving perpendicular to a wire At the end of Section \(5.9\) we discussed the case where a charge \(q\) moves perpendicular to a wire. Figures \(5.25\) and \(5.26\) show qualtatively why there is a nonzero force on the charge, pointing in the positive \(x\) direction. Carry out the calculation to show that the force at a distance \(\ell\) from the wire equals \(q v l / 2 \pi \epsilon_{0} \ell c^{2}\). That is, use Eq. (5.15) to calculate the force on the charge in its own frame, and then divide by \(\gamma\) to transform back to the lab frame. Notes: You can use the fact that in the charge \(q\) 's frame, the speed of the electrons in the \(x\) direction is \(v_{0} / \gamma\) (this comes from the transverse-velocity-addition formula). Remember that the \(\beta\) in Eq. \((5.15)\) is the velocity of the electrons in the charge's frame, and this velocity has two components. Be careful with the transverse distance involved. There are many things to keep track of in this problem, but the integration itself is easy if you use a computer (or Appendix K).
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