Chapter 26, Problem 15P

(a)

To determine

The length measured in the rest frame of the particle, if the length of the field between goal lines in the earth frame is $91.5\text{m}$.

Answer to Problem 15P

The length measured in the rest frame of the particle is $79\text{m}$ .

Explanation of Solution

Since the length of the field between the goal lines in the rest frame of earth is $91.5\text{m}$, proper length is $91.5\text{m}$.

Write the expression for the length of field between the goal lines as measured in the rest frame of particle.

  $L=\frac{L_0}{\gamma }$                                                                                                                      (I)

Here, $L_0$ is the proper length, $L$ is the length of field between the goal lines as measured in the rest frame of particle and $\gamma$ is the Lorentz factor.

Write the expression for the Lorentz factor.

  $\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Here, $v$ is the speed of cosmic ray particle and $c$ is the speed of light in vacuum.

Substitute $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ for $\gamma$ in equation (I) to get $L$.

  $L=L_0\sqrt{1-\frac{v^2}{c^2}}$                                                                                                         (II)

Conclusion:

Substitute $91.5\text{m}$ for $L_0$ and $0.50c$ for $v$ in equation (II) to get $L$.

  $\begin{array}{c}L=\left(91.5\text{m}\right)\sqrt{1-\frac{{\left(0.50c\right)}^2}{c^2}}\\ =91.5\text{m}\left(0.86\right)\\ =79.24\text{m}\\ \approx 79\text{m}\end{array}$

Therefore, the length measured in the rest frame of the particle is $79\text{m}$ .

(b)

To determine

The time the particle takes to go from one goal line to the other according to Earth observers.

Answer to Problem 15P

The time the particle takes to go from one goal line to the other according to Earth observers is $610\text{ns}$ .

Explanation of Solution

The speed of the particle is constant with respect to earth. Thus, the time can be found using the equation $\Delta x=v\Delta t$, where $\Delta x$ is the distance between the goal lines in frame of earth and $\Delta t$ is the time taken by the cosmic ray to move from one goal line to another.

Write the expression for the time taken by the particle to move from one goal line to another relative to earth observer.

  $\Delta t=\frac{\Delta x}{v}$                                                                                                                 (III)

Substitute $91.5\text{m}$ for $\Delta x$ and $0.50c$ for $v$ in equation (III) to get $\Delta t$.

  $\Delta t=\frac{91.5\text{m}}{0.50c}$

Conclusion:

Substitute $3.00\times {10}^8\text{m}/\text{s}$ for $c$ in above equation to get $\Delta t$.

  $\begin{array}{c}\Delta t=\frac{91.5\text{m}}{0.50\left(3.00\times {10}^8\text{m}/\text{s}\right)}\\ =6.1\times {10}^{-7}\text{s}\times \frac{{10}^9\text{ns}}{1\text{s}}\\ =610\text{ns}\end{array}$

Therefore, the time the particle takes to go from one goal line to the other with respect to Earth observers is $610\text{ns}$ .

(c)

To determine

The time the particle takes to go from one goal line to the other in the rest frame of the particle.

Answer to Problem 15P

The time the particle takes to go from one goal line to the other in the rest frame of the particle is $530\text{ns}$ .

Explanation of Solution

In rest frame of the particle, the distance travelled is $79\text{m}$ and speed is $0.50c$.

Write the expression for the time required to travel between goal lines in the rest frame of the particle.

  $\Delta t=\frac{\Delta x}{v}$                                                                                                                  (IV)

Here, $\Delta x$ is the distance between the goal lines in rest frame the particle and $\Delta t$ is the time taken by the cosmic ray to move from one goal line to another in rest frame of particle.

Conclusion:

Substitute $79\text{m}$ for $\Delta x$ and $0.50c$ for $v$ in equation (IV) to get $\Delta t$.

  $\Delta t=\frac{79\text{m}}{0.50c}$

Substitute $3.00\times {10}^8\text{m}/\text{s}$ for $c$ in above equation to get $\Delta t$.

  $\begin{array}{c}\Delta t=\frac{79\text{m}}{0.50\left(3.00\times {10}^8\text{m}/\text{s}\right)}\\ =5.3\times {10}^{-7}\text{s}\times \frac{{10}^9\text{ns}}{1\text{s}}\\ =530\text{ns}\end{array}$

Therefore, the time the particle takes to go from one goal line to the other with respect to rest frame of the particle is $530\text{ns}$ .