Chapter 1, Problem 23P

An observer in frame S sees lightning simultaneously strike two points 100 m apart. The first strike occurs at x₁ = y₁ = z₁ = t₁ = 0 and the second at x₂ = 100 m, y₂ = z₂ = t₂ = 0. (a) What are the coordinates of these two events in a frame S′ moving in the standard configuration at 0.70c relative to S? (b) How far apart are the events in S′? (c) Are the events simultaneous in S′? If not, what is the difference in time between the events, and which event occurs first?

To determine

To coordinates of the two events in a frame $S^{\prime }$.

Answer to Problem 23P

The coordinate is $S^{\prime }$ frame is

$\begin{array}{l}{x^{\prime }}_1=\gamma \left(x_1-v{t}_1\right)=0\\ {y^{\prime }}_1=y_1=0\\ {z^{\prime }}_1=z_1=0\\ {t^{\prime }}_1=\gamma \left[t_1-\left(\frac{v}{c^2}\right)x_1\right]=0\end{array}$

And

$\begin{array}{l}{x^{\prime }}_2=140\text{m}\\ {y^{\prime }}_2=0\\ {z^{\prime }}_2{}_1=0\\ {t^{\prime }}_1=-0.33\mu \text{s}\end{array}$

Explanation of Solution

In S frame, Event 1 coordinates  are $x_1=y_1=z_1=t_1=0$ and Event 2 coordinates are $x_2=100\text{mm,}{y}_2=z_2=t_2=0$

To obtain these coordinate in $S^{\prime }$ frame moving at 0.70c with respect to S, use Lorentz transformation.

In $S^{\prime }$ frame Event 1 coordinates are

$\begin{array}{l}{x^{\prime }}_1=\gamma \left(x_1-v{t}_1\right)=0\\ {y^{\prime }}_1=y_1=0\\ {z^{\prime }}_1=z_1=0\\ {t^{\prime }}_1=\gamma \left[t_1-\left(\frac{v}{c^2}\right)x_1\right]=0\end{array}$

Here, $\gamma$ is the Lorentz factor and v is the speed of $S^{\prime }$ frame.

Lorentz factor,

$\gamma ={\left[1-\frac{v^2}{c^2}\right]}^{-1/2}$

Event 2 coordinates are

$\begin{array}{c}{x^{\prime }}_2=\gamma \left(x_2-v{t}_2\right)\\ ={\left[1-\frac{v^2}{c^2}\right]}^{-1/2}\left(x_2-v{t}_2\right)\end{array}$

${y^{\prime }}_2={z^{\prime }}_2=0$

$\begin{array}{c}{t^{\prime }}_2=\gamma \left[t_2-\left(\frac{v}{c^2}\right)x_2\right]\\ ={\left[1-\frac{v^2}{c^2}\right]}^{-1/2}\left[t_2-\left(\frac{v}{c^2}\right)x_2\right]\end{array}$

Conclusion:

Substitute 0.70c for v and 100 m for $x_2$ in ${x^{\prime }}_2$ and simplify.

$\begin{array}{c}{x^{\prime }}_2=\gamma \left(x_2-v{t}_2\right)\\ ={\left[1-\frac{v^2}{c^2}\right]}^{-1/2}\left(x_2-v{t}_2\right)\\ {x^{\prime }}_2={\left[1-\frac{{\left(0.70c\right)}^2}{c^2}\right]}^{-1/2}\left[100\text{m}-\left(0.70c\right)\left(0\right)\right]\\ =140\text{m}\end{array}$

Substitute 0.70c for v and 100 m for $x_2$ in ${t^{\prime }}_2$ and simplify.

$\begin{array}{c}{t^{\prime }}_2={\left[1-\frac{{\left(0.70c\right)}^2}{c^2}\right]}^{-1/2}\left[0-\left(\frac{0.70c}{c^2}\right)\left(100\text{m}\right)\right]\\ =\frac{\left(1.4\right)\left(-0.70\right)\left(100\text{ m}\right)}{3.00\times {10}^8\text{m}/\text{s}}\\ =-0.33\mu \text{s}\end{array}$

Therefore, the coordinate is $S^{\prime }$ frame is

$\begin{array}{l}{x^{\prime }}_1=\gamma \left(x_1-v{t}_1\right)=0\\ {y^{\prime }}_1=y_1=0\\ {z^{\prime }}_1=z_1=0\\ {t^{\prime }}_1=\gamma \left[t_1-\left(\frac{v}{c^2}\right)x_1\right]=0\end{array}$

And

$\begin{array}{l}{x^{\prime }}_2=140\text{m}\\ {y^{\prime }}_2=0\\ {z^{\prime }}_2{}_1=0\\ {t^{\prime }}_1=-0.33\mu \text{s}\end{array}$

To determine

The distance apart the events in $S^{\prime }$

Answer to Problem 23P

The distance apart the events in is 140 m.

Explanation of Solution

The expression for distance of separation is

$\Delta x^{\prime }={x^{\prime }}_2-{x^{\prime }}_1$

Here, $\Delta x^{\prime }$ is the distance of separation.

Substitute 0 for ${x^{\prime }}_1$ and 140 m for ${x^{\prime }}_2$.

$\begin{array}{c}\Delta x^{\prime }=140\text{m}-0\text{m}\\ =140\text{m}\end{array}$

Conclusion:

Therefore, the distance apart the events in $S^{\prime }$ is 140 m.

To determine

Whether the events simultaneous in $S^{\prime }$. If not the difference in time between the events and which occurs first.

Answer to Problem 23P

The events are not simultaneous in $S^{\prime }$ , the difference in time between the event is $0.33\mu \text{s}$ and Negative sign indicates event 2 occurred earlier than event 1.

Explanation of Solution

The expression for difference in time is

$\Delta t^{\prime }=\Delta {t^{\prime }}_2-\Delta {t^{\prime }}_1$

Here, $\Delta t^{\prime }$ is the difference in time

Conclusion:

Substitute 0 for $\Delta {t^{\prime }}_1$ and $-0.33\mu \text{s}$ for $\Delta {t^{\prime }}_2$.

$\begin{array}{c}\Delta t^{\prime }=-0.33\mu \text{s}-0\\ =-0.33\mu \text{s}\end{array}$

Negative sign indicates event 2 occurred earlier than event 1.

Therefore, the events are not simultaneous in $S^{\prime }$ , the difference in time between the event is $0.33\mu \text{s}$ and Negative sign indicates event 2 occurred earlier than event 1.