Chapter 25, Problem 57QAP

To determine

(a)

Travel time of each photon

Answer to Problem 57QAP

Travel time of each photon is ${10}^{-7}\sec$.

Explanation of Solution

Given:

Length of the ship, $60\text{m}$

Speed of the spacecraft, $v=0.6c$

Formula used:

The relation between time, distance and speed is given as

  $v=\frac{d}{t}$

Calculation:

Since the Gaar is standing in the middle of the ship, therefore distance cover by each photon is $30\text{m}$. We know that photon moves with the speed of light, therefore time required by each photon to strike is calculated as

  $\begin{array}{c}t=\frac{d}{v}\\ =\frac{30}{3\times {10}^8}\\ =1\times {10}^{-7}\sec \end{array}$

Conclusion:

Travel time of each photon is ${10}^{-7}\sec$.

To determine

(b)

According to you which photon will register first

Answer to Problem 57QAP

Backward photon will register first strike.

Explanation of Solution

Given:

Length of the ship,

  $60\text{m}$

Speed of the spacecraft, $v=0.6c$

Calculation:

Since we are in stationary reference frame, therefore the photon that is fired in the backward direction will have the relative velocity of $\sqrt{1-\frac{v^2}{c^2}}$ while the forward electron will have the relative velocity of $0.4c$, therefore in our reference backward electron will strike first.

Conclusion:

Backward photon will register first strike.

To determine

(c)

Difference in the striking times of the photons

Answer to Problem 57QAP

Difference in the striking times of the photons is

  $1.5\times {10}^{-7}\sec$.

Explanation of Solution

Given:

Length of the ship, $60\text{m}$

Speed of the spacecraft, $v=0.6c$

Formula used:

The time coordinate is calculated as

  $t=\frac{1}{\sqrt{1-\left(\frac{v^2}{c^2}\right)}}\left(t\text{'}-\frac{vx\text{'}}{c^2}\right)$

Here,

  $v$is the velocity of the rocket

  $c$is the velocity of light

Calculation:

Let the backward photon will strike first in the person reference frame. Now in the pilot reference plane the left bulb turns on at $x\text{'}=0andt\text{'}=0$

The time coordinate is calculated as

  $t=\frac{1}{\sqrt{1-\left(\frac{v^2}{c^2}\right)}}\left(t\text{'}-\frac{vx\text{'}}{c^2}\right)$

Plugging the values in the above equation

  $\begin{array}{c}t=\frac{1}{\sqrt{1-\left(\frac{0.60c^2}{c^2}\right)}}\left(0-\frac{0.60c\times 0}{c^2}\right)\\ =0\end{array}$

The time coordinate for the forward photon is

  $t=\frac{1}{\sqrt{1-\left(\frac{v^2}{c^2}\right)}}\left(t\text{'}-\frac{vx\text{'}}{c^2}\right)$

Plugging the values in the above equation

  $\begin{array}{c}t=\frac{1}{\sqrt{1-\left(\frac{0.60c^2}{c^2}\right)}}\left(0-\frac{0.60c\times 60}{c^2}\right)\\ =\frac{1}{\sqrt{1-\left(\frac{{0.60}^2}{1}\right)}}\left(0-\frac{0.60\times 60}{3\times {10}^8}\right)\\ =-1.5\times {10}^{-7}\sec \end{array}$

Conclusion:

Difference in the striking times of the photons is

  $1.5\times {10}^{-7}\sec$.