Chapter 39, Problem 22P

(a)

To determine

Speed of the spaceship in frame $S$ as a function for boost 1 to boost 10.

Explanation of Solution

Given:

Spaceship that is at rest in frame $S$ is given speed increase boost 1 of $0.50c$ .

Spaceship is given further boost 2 of $0.50c$ after time $10\text{s}$ in new rest frame.

Process is continued indefinitely at $10\text{s}$ intervals.

Formula used:

Write the expression of equation for relativistic velocity addition

  $V_{j+1}=\frac{V_j+0.50c}{1+\frac{\left(0.50c\right)V_j}{c^2}}$

Here, $V_j$ is speed after $j\text{-th}$ boost and $V_{j+1}$ is speed after $\left(j+1\right)\text{-th}$ boost.

Simplify the above expression

  $V_{j+1}=\frac{\left(\frac{V_j}{c}\right)+0.50}{1+0.50\left(\frac{V_j}{c}\right)}$ …… (1)

Calculation:

Substitute 0 for $j$ and $0.000$ for $\left(\frac{V_0}{c}\right)$ in expression (1)

  $\begin{array}{c}{V}_{0+1}=\frac{\left(\frac{V_0}{c}\right)+0.50}{1+0.50\left(\frac{V_0}{c}\right)}\\ V_1=\frac{0.00+0.50}{1+0.50\left(0.00\right)}\\ V_1=0.50\end{array}$

For rest of the values of $j$ , speed is calculated in the table given below.

Number of boosts $j$	Speed after $j\text{-th}$ boost$\left(\frac{V_j}{c}\right)$
2	0.800
3	0.929
4	0.976
5	0.992
6	0.997
7	0.999
8	1.000
9	1.000
10	1.000

Draw a diagram to plot the speed in units of $c$ along $y$ axis and the number of boosts along $x$ axis

  

Conclusion:

Thus, the speed is calculated and represented graphically as a function of number of boosts.

(b)

To determine

Graphical representation of gamma factor.

Explanation of Solution

Given:

Spaceship that is at rest in frame $S$ is given speed increase boost 1 of $0.50c$ .

Spaceship is given further boost 2 of $0.50c$ after time $10\text{s}$ in new rest frame.

Process is continued indefinitely at $10\text{s}$ intervals.

Formula used:

Write the expression of gamma factor

  ${\gamma }_j=\frac{1}{{\left(1-{\left(\frac{V_j}{c}\right)}^2\right)}^{\frac{1}{2}}}$ …… (2)

Here, ${\gamma }_j$ is the gamma factor after $j\text{-th}$ boost

Calculation:

Substitute 0 for $j$ and $0.000$ for $\left(\frac{V_0}{c}\right)$ in expression (2)

  $\begin{array}{c}{\gamma }_0=\frac{1}{{\left(1-{\left(\frac{V_0}{c}\right)}^2\right)}^{\frac{1}{2}}}\\ =\frac{1}{{\left(1-{\left(0.000\right)}^2\right)}^{\frac{1}{2}}}\\ =1.00\end{array}$

For rest of the values of $j$ , gamma factor is calculated in the table given below.

Number of boosts $j$	Gamma factor after $j\text{-th}$boost ${\gamma }_j$
1	1.15
2	1.67
3	2.69
4	4.56
5	7.83
6	13.52
7	23.39
8	40.51
9	70.15
10	121.50

Draw a diagram to plot the gamma factor along $y$ axis and the number of boosts along $x$ axis

  

Conclusion:

Thus, the gamma factor is calculated and represented graphically as a function of number of boosts.

(c)

To determine

Number of required boosts until speed of ship in $S$ is greater than $0.999c$ .

Explanation of Solution

Introduction:

A spaceship that is at rest in frame $S$ is given speed increase boost 1 of $0.50c$ first. The spaceship is again given a boost 2 of $0.50c$ after time $10\text{s}$ in the new rest frame. The process is continued indefinitely at $10\text{s}$ intervals.

The graph of the speed after $j\text{-th}$ boost as a function of the number of boosts in Part (a) indicates that the speed of the spaceship is more than $0.999c$ after eight boosts as measured in the reference frame $S$ .

Conclusion:

Thus, the number of required boosts until speed of ship in $S$ is greater than $0.999c$ is 8.

(d)

To determine

Distance traveled by spaceship as well as average speed of spaceship between boost 1 and boost 6 in frame $S$ .

Explanation of Solution

Given:

Spaceship that is at rest in frame $S$ is given speed increase boost 1 of $0.50c$ .

Spaceship is given further boost 2 of $0.50c$ after time $10\text{s}$ in new rest frame.

Process is continued indefinitely at $10\text{s}$ intervals.

Formula used:

Write the expression of distance traversed by spaceship between boosts 1 and 6 in frame $S$

  $\Delta X=\left(10\text{s}\right)\left(V_1{\gamma }_1+V_2{\gamma }_2+V_3{\gamma }_3+V_4{\gamma }_4+V_5{\gamma }_5\right)$ …… (3)

Write the expression for elapsed time between boosts 1 and 6 in frame $S$

  $T=\left(10\text{s}\right)\left({\gamma }_1+{\gamma }_2+{\gamma }_3+{\gamma }_4+{\gamma }_5\right)$ …… (4)

Write the expression for average speed of spaceship in frame $S$ between boosts 1 and 6

  $U=\frac{\Delta X}{T}$ …… (5)

Calculation:

Substitute $0.5c$ for $V_1$ , $0.8c$ for $V_2$ , $0.929c$ for $V_3$ , $0.976c$ for $V_4$ , $0.992c$ for $V_5$ , 1.15 for ${\gamma }_1$ , 1.67 for ${\gamma }_2$ , 2.69 for ${\gamma }_3$ , 4.56 for ${\gamma }_4$ and 7.83 for ${\gamma }_5$ in expression (3)

  $\begin{array}{c}\Delta X=\left(10\text{s}\right)\left(\left(0.5c\right)\left(1.15\right)+\left(0.8c\right)\left(1.67\right)+\left(0.929c\right)\left(2.69\right)+\left(0.976c\right)\left(4.56\right)+\left(0.992\right)\left(7.83\right)\right)\\ =166c\cdot \text{s}\end{array}$ Substitute 1.15 for ${\gamma }_1$ , 1.67 for ${\gamma }_2$ , 2.69 for ${\gamma }_3$ , 4.56 for ${\gamma }_4$ and 7.83 for ${\gamma }_5$ in expression (4)

  $\begin{array}{c}T=\left(10\text{s}\right)\left(1.15+1.67+2.69+4.56+7.83\right)\\ =179\text{s}\end{array}$

Substitute $166c\cdot \text{s}$ for $\Delta X$ and $179\text{s}$ for $T$ in expression (5)

  $\begin{array}{c}U=\frac{\left(166c\cdot \text{s}\right)}{\left(179\text{s}\right)}\\ =0.927c\end{array}$

Conclusion:

Thus, the distance traveled by spaceship is $166c\cdot \text{s}$ and average speed of spaceship between boost 1 and boost 6 in frame $S$ is $0.927c$ .