Chapter 39, Problem 20P

(a)

To determine

Verification of the statement that the time separation $\left(t_2{}^{\prime }-t_1{}^{\prime }\right)$ is $\gamma \left(T-vD/c^2\right)$ in frame $S^{\prime }$ .

Explanation of Solution

Introduction:

Events 1 and 2 are separated by a distance $D$ of $\left(x_2-x_1\right)$ and a time $T$ of $\left(t_2-t_1\right)$ in reference frame $S$ . Inverse Lorentz transformation equation is employed to determine the desired time interval.

Write the expression for the inverse time transformation for event 2

  $t_2{}^{\prime }=\gamma \left(t_2-\frac{v{x}_2}{c^2}\right)$

Here, $t_2$ is time coordinate and $x_2$ is space coordinate in frame $S$ , $t_2{}^{\prime }$ is time coordinate in frame $S^{\prime }$ , $\gamma$ is the gamma factor and $c$ is speed of light in vacuum.

Write the expression for the inverse time transformation for event 1

  $t_1{}^{\prime }=\gamma \left(t_1-\frac{v{x}_1}{c^2}\right)$

Here, $t_1$ is time coordinate and $x_1$ is space coordinate in frame $S$ , $t_1{}^{\prime }$ is time coordinate in frame $S^{\prime }$ .

Write the expression for the time interval between the two events in frame $S^{\prime }$

  $\Delta t^{\prime }=t_2{}^{\prime }-t_1{}^{\prime }$

Substitute $\gamma \left(t_2-\frac{v{x}_2}{c^2}\right)$ for $t_2{}^{\prime }$ and $\gamma \left(t_1-\frac{v{x}_1}{c^2}\right)$ for $t_1{}^{\prime }$ in above expression

  $\begin{array}{c}\Delta t^{\prime }=\gamma \left(t_2-\frac{v{x}_2}{c^2}\right)-\gamma \left(t_1-\frac{v{x}_1}{c^2}\right)\\ =\gamma \left(\left(t_2-t_1\right)-\frac{v}{c^2}\left(x_2-x_1\right)\right)\end{array}$

Substitute $D$ for $\left(x_2-x_1\right)$ and $T$ for $\left(t_2-t_1\right)$ in above expression

  $\Delta t^{\prime }=\gamma \left(T-\frac{vD}{c^2}\right)$   ...... (1)

Conclusion:

Thus, it is verified that the time separation $\left(t_2{}^{\prime }-t_1{}^{\prime }\right)$ is $\gamma \left(T-vD/c^2\right)$ in frame $S^{\prime }$ .

(b)

To determine

Verification of the statement that the events can be simultaneous in frame $S^{\prime }$ only if $D$ is greater than $cT$ .

Explanation of Solution

Introduction:

Events 1 and 2 are separated by a distance $D$ of $\left(x_2-x_1\right)$ and a time $T$ of $\left(t_2-t_1\right)$ in reference frame $S$ .Inverse Lorentz transformation equation is employed to determine the desired relation.

Since the events occur at the same time in frame $S^{\prime }$ therefore

  $\Delta t^{\prime }=0$

Substitute $\gamma \left(T-\frac{vD}{c^2}\right)$ for $\Delta t^{\prime }$ in above expression

  $\gamma \left(T-\frac{vD}{c^2}\right)=0$

Simplify the above expression for $D$

  $D=\frac{cT}{\left(\frac{v}{c}\right)}$

Since $v<c$ therefore $D>cT$ from the above expression.

Conclusion:

Thus, it is verified thatthe events can be simultaneous in frame $S^{\prime }$ only if $D$ is greater than $cT$ .

(c)

To determine

Verification of the statement that $t_2{}^{\prime }$ is greater than $t_1{}^{\prime }$ in all reference frames if $D$ is less than $cT$ .

Explanation of Solution

Introduction:

Events 1 and 2 are separated by a distance $D$ of $\left(x_2-x_1\right)$ and a time $T$ of $\left(t_2-t_1\right)$ in reference frame $S$ . One event causes another event.

Since $D>cT$ and $v<c$ therefore from expression (1)

  $\Delta t^{\prime }>0$

Substitute $\left(t_2{}^{\prime }-t_1{}^{\prime }\right)$ for $\Delta t^{\prime }$ in above expression

  $\left(t_2{}^{\prime }-t_1{}^{\prime }\right)>0$

Rearrange the above expression

  $t_2{}^{\prime }>t_1{}^{\prime }$

Conclusion:

Thus, it is verified that $t_2{}^{\prime }$ is greater than $t_1{}^{\prime }$ in all reference frames if $D$ is less than $cT$ .

(d)

To determine

Verification of the statement that the effect precedes the cause in a reference frame moving with speed $v$ less than $c$ .

Explanation of Solution

Introduction:

Events 1 and 2 are separated by a distance $D$ of $\left(x_2-x_1\right)$ and a time $T$ of $\left(t_2-t_1\right)$ in reference frame $S$ . A signal is sent with speed $c^{\prime }$ greater than that of light.

Write the expression of space interval $D$ in this case

  $D=c^{\prime }T$

Substitute $c^{\prime }T$ for $D$ in expression (1)

  $\Delta t^{\prime }=\gamma \left(T-\frac{v\left(c^{\prime }T\right)}{c^2}\right)$

Rearrange the above equation

  $\Delta t^{\prime }=\gamma T\left(1-\left(\frac{v}{c}\right)\left(\frac{c^{\prime }}{c}\right)\right)$

Since $c^{\prime }>c$ therefore $\Delta t^{\prime }$ can be negative from the above expression which implies that the effect can precede the cause in this case.

Conclusion:

Thus, it is verified that the effect precedes the cause in a reference frame moving with speed $v$ less than $c$ .