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Introduction to Electrodynamics

David J. Griffiths

Physics

4th edition Edition 路 613 pages

ISBN: 9780321856562

Chapter 12: Q21P (page 529)

The coordinates of event Aare $(x_{A}, 0, 0), t_{A,}$ and the coordinates of event B are $(x_{B}, 0, 0), t_{A}$ . Assuming the displacement between them is spacelike, find the velocity of the system in which they are simultaneous.

Short Answer

Expert verified

The velocity of the system is $v = (\frac{t_{B} - t_{A}}{x_{B} - x_{A}}) c^{2}$ .

Step by step solution

01

Expression for the Lorentz transformation equation:

Write the expression for the Lorentz transformation equation.

$t = Y (t - \frac{v}{c^{2}} x)$ 鈥︹€� (1)

Here, v is the velocity of the system, x is the displacement, and c is the speed of light.

02

Determine the velocity of the system:

Write equation (1) in terms of $鈭�$ .

$鈭� t = 纬 (鈭� t - \frac{v}{c^{2}} 鈭� x)$

Here, $鈭� t = 0$

Hence, the equation becomes,

$纬 (鈭� t - \frac{v}{c^{2}} 鈭� x) = 0 鈭� t - \frac{v}{c^{2}} 鈭� x = 0 鈭� t - \frac{v}{c^{2}} 鈭� x v = \frac{c^{2} 鈭� t}{鈭� x}$

As the coordinates of event A and B are given as $(x_{A}, 0, 0), t_{A}$ and $(x_{B}, 0, 0), t_{B}$ respectively, the velocity of the system will be,

$v = \frac{c^{2} (t_{B} - t_{A})}{(x_{B} - x_{A})} v = (\frac{t_{B} - t_{A}}{x_{B} - x_{A}}) c^{2}$

Therefore, the velocity of the system is $v = (\frac{t_{B} - t_{A}}{x_{B} - x_{A}}) c^{2}$ .

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Most popular questions from this chapter

(a) Charge $q_{A}$ is at rest at the origin in system $S$ ; charge $q_{B}$ flies at speed $v$ on a trajectory parallel to the $x$ axis, but at $y = d$ . What is the electromagnetic force on $q_{B}$ as it crosses the axis?

(b) Now study the same problem from system $\overset{鈫�}{S}$ , which moves to the right with speed . What is the force on when passes the axis? [Do it two ways: (i) by using your answer to (a) and transforming the force; (ii) by computing the fields in and using the Lorentz law.]

Show that the second equation in Eq. 12.127 can be expressed in terms of the field tensor $F^{渭谓}$ as follows:

localid="1654746948628" $\frac{鈭� F_{渭谓}}{鈭� x^{位}} + \frac{鈭� F_{谓位}}{鈭� x^{渭}} + \frac{鈭� F_{位渭}}{鈭� x^{谓}} = 0$

(a) What鈥檚 the percent error introduced when you use Galileo鈥檚 rule, instead of Einstein鈥檚, with $v_{AB} = 5 mi / h and v_{BC} = 60 mi / h$ and?

(b) Suppose you could run at half the speed of light down the corridor of a train going three-quarters the speed of light. What would your speed be relative to the ground?

(c) Prove, using Eq. 12.3, that if $v_{AB} < c and v_{BC} < c then v_{AC} < c$ Interpret this result.

A cop pulls you over and asks what speed you were going. 鈥淲ell, officer, I cannot tell a lie: the speedometer read $4 脳 10^{8} m / s$ .鈥� He gives you a ticket, because the speed limit on this highway is $2.5 脳 10^{8} m / s$ . In court, your lawyer (who, luckily, has studied physics) points out that a car鈥檚 speedometer measures proper velocity, whereas the speed limit is ordinary velocity. Guilty, or innocent?

An ideal magnetic dipole moment m is located at the origin of an inertial system $\overset{炉}{S}$ that moves with speed v in the x direction with respect to inertial system S. In $\overset{炉}{S}$ the vector potential is

$\overset{炉}{A} = \frac{渭_{0}}{4 蟺} \frac{\overset{炉}{m} 脳 \overset{炉}{\hat{r}}}{{\overset{炉}{r}}^{2}}$

(Eq. 5.85), and the scalar potential $\overset{炉}{V}$ is zero.

(a) Find the scalar potential V in S.

(b) In the nonrelativistic limit, show that the scalar potential in S is that of an ideal electric dipole of magnitude

$p = \frac{v 脳 m}{c^{2}}$

located at $\overset{炉}{O}$ .

See all solutions

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