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Introduction to Electrodynamics
Found in: Page 508
Introduction to Electrodynamics

Introduction to Electrodynamics

Book edition 4th edition
Author(s) David J. Griffiths
Pages 613 pages
ISBN 9780321856562

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Short Answer

(a) What’s the percent error introduced when you use Galileo’s rule, instead of Einstein’s, withvAB=5 mi/h and vBC=60 mi/hand?

(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 ifvAB<c and vBC <c then vAC<cInterpret this result.

(a) The percent error is 6.7×1014

(b) The speed of a person relative to the ground is 10c11.

(c) The result is obtained as vAC <Cand interpreted as if the person’s relative speed to the train and the relative speed of a train with respect to the ground is less than c, the person’s speed relative to the ground will also be less than c. So, it is impossible to travel with the speed of light or greater than it.

See the step by step solution

Step by Step Solution

Step 1: Expression for Galileo’s and Einstein's addition rule: 

Write the expression for Einstein’s addition rule.

vAC=vAB+vBC1+(vABvBCC2) …… (1)

Write the expression for Galileo’s addition rule.

vAC=vAB+vBC …… (2)

Step 2: Determine the percent error:+-

(a)

Write the expression for the velocity of A with respect to C (according to Einstein).

vE=vAB+vBC1+(vABvBCC2)

Substitute the value of equation (2) in the above expression.

vE=vG1+(vABvBCC2)vE=vG1+vABvBCC2-1

Use binomial expansion and neglect the higher-order terms.

vE=vG1-(vABvBCC2)

Subtract the above equation from.vG

vG-vE=vG-vG1-vABvBCC2vG-vE=vGvABvBCC2vG-vEvG=vABvBCC2

For the percent error, the above equation becomes,

vG-vEvG=vABvBCC2×100

Substitute all the values of vABand vBCin the above expression.

vG-vEvG=5 mi/h60 mi/h3×108 m/s×2.237 mi/h1 m/s2×100vG-vEvG=3006.7×108 m/s2×100vG-vEvG=6.7×10-16×100vG-vEvG=6.7×10-14

Therefore, the percent error is6.7×10-14.

Step 3: Determine the speed of a person relative to the ground:

(b)

Based on the given problem, the value ofvABand will be,

and .

vAB=c2and vBC=3c4.

Substitute the value ofvABandvBC in equation (1) to calculate the velocity of a person relative to the ground.

vACc2+3c41+c23c4c2vAC=1.25c1+38vAC=10c11

Therefore, the speed of a person relative to the ground is.

Step 4: Show thatVAC <c and interpret the obtained result:

(c)

Let’s assume,

β=vACcβ1=vABcβ2=vBCc

Substitute the value ofvABandvBCin equation (1).

βc=cβ1+cβ21+cβ1cβ2c2 β=β1+β21+β1β2

Squaring both sides,

β2=β1+β221+β1β22β2=β12+β22+2β1β21+β12+β22+2β1β2β2=1-1-β121-β221+β1β22β2=1-

From the above obtained value,β2should be less than 1. Hence,

β<1vACC<1 vAC<1

The above result implies that if the person’s relative speed to the train and the relative speed of a train with respect to the ground is less than c, the person’s speed relative to the ground will also be less than c. So, it is impossible to travel with the speed of light or greater than it.

Therefore, the result is obtained as vAC<Cand interpreted as if the person’s relative speed to the train and the relative speed of a train with respect to the ground is less than c, the person’s speed relative to the ground will also be less than c. So, it is impossible to travel with the speed of light or greater than it.

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