PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 26, Problem 96P

(a)

To determine

The time elapse between the emission of wavefronts by the source, in observer’s frame.

(a)

Expert Solution
Check Mark

Answer to Problem 96P

The time elapse between the emission of wavefronts by the source, in observer’s frame is Ts=γfs_.

Explanation of Solution

Write the expression for the proper time interval.

  Δt0=Ts                                                                                                              (I)

Here, Δt0 is the proper time interval, Ts is the period of wavefront emission.

Write the expression for the period of the wavefront emission.

  Ts=1fs                                                                                                               (II)

Here, fs is the frequency of emission.

Use equation (II) in (I).

  Δt0=1fs                                                                                                             (III)

Write the expression for the time elapse between the emission of wavefronts by the source, in observer’s frame.

  Ts=Δt                                                                                                              (IV)

Here, Ts is the time elapse between the emission of wavefronts by the source, in observer’s frame, and Δt is the time interval in observer’s frame.

Write the expression for the time interval corresponding to the proper time interval, in observer’s frame.

  Δt=γΔt0                                                                                                             (V)

Here, γ is the Lorentz factor.

Use equation (III) in (V).

  Δt=γfs                                                                                                                (VI)

Use equation (VI) in (IV).

  Ts=γfs                                                                                                               (VII)

Conclusion:

Therefore, the time elapse between the emission of wavefronts by the source, in observer’s frame is Ts=γfs_.

(b)

To determine

The time To between the arrival of the two wavefronts as measured by the observer.

(b)

Expert Solution
Check Mark

Answer to Problem 96P

The time To between the arrival of the two wavefronts as measured by the observer is To=γfs(1+vc)_.

Explanation of Solution

Write the expression for the time that the first wavefront described in the problem statement arrives at the observer.

  t1=dc                                                                                                                (VIII)

Here, t1 is the time that the first wavefront arrives at the observer, d is the distance between the source and the observer, and c is the speed of light.

Write the expression for the time of travel of the second wavefront as it arrives at the observer.

  t2=Ts+d+vTsc                                                                                                   (IX)

Here, t2 is the time of travel of the second wavefront as it arrives at the observer.

Write the expression for time To between the arrival of the two wavefronts as measured by the observer.

  To=t2t1                                                                                                         (X)

Use equation (VIII) and (IX) in (X).

  To=Ts+d+vTscdc                                                                                       (XI)

Use equation (VII) in (XI).

  To=γfs+dc+vγcfsdc=γfs(1+vc)                                                                                         (XII)

Conclusion:

Therefore, the time To between the arrival of the two wavefronts as measured by the observer is To=γfs(1+vc)_.

(c)

To determine

To show that the frequency detected by the observer is fo=fs1v/c1+v/c.

(c)

Expert Solution
Check Mark

Answer to Problem 96P

The frequency detected by the observer is fo=fs1v/c1+v/c.

Explanation of Solution

Write the expression for the frequency detected by the observer.

  fo=1To                                                                                                            (XIII)

Use equation (XII) in (XIII).

  fo=1γfs(1+vc)=fs1γ(1+v/c)                                                                                          (XIV)

Write the expression of Lorentz constant.

  γ=11v2/c2                                                                                                 (XV)

Use equation (XV) in (XIV) and simplify.

  fo=fs1v2/c2(1+v/c)=fs1v2/c2(1+v/c)2=fs(1+v/c)(1v/c)(1+v/c)2=fs1v/c1+v/c

Conclusion:

Therefore, the frequency detected by the observer is fo=fs1v/c1+v/c.

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Chapter 26 Solutions

PHYSICS

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