Question

An experimenter arranges to trigger two flashbulbs simultaneously, producing a big flash located at the origin of his reference frame and a small flash at $\mathbf{3}\mathbf{.}\mathbf{0}\mathit{k}\mathit{m}$. An observer moving at a speed of $\mathbf{0}\mathbf{.}\mathbf{250}\mathit{c}$in the positive direction of $\mathit{x}$also views the flashes. (a) What is the time interval between them according to her? (b) Which flash does she say occurs first?

Short answer

1. The time interval between the flashes is$2.58\times {10}^{-5}s$ .
2. The small flash occurs first in frame $S^{\text{'}}$.

Step-by-step solution

The time dilation equation

The time dilation equation states that $t^{\text{'}}=\frac{t-{\displaystyle \frac{\beta x}{c}}}{\sqrt{1-{\beta }^2}}$, where $t-\frac{\beta x}{c}$shows that time measured in the reference frame of inertial observer.

The time interval between the flashes 

Let$t_s$be the time coordinate and $x_s$be the position coordinate of small flash. So, the time in the frame$S^{\text{'}}$can be obtained using the formula ${t^{\text{'}}}_s=\frac{t_s-{\displaystyle \frac{\beta x_s}{c}}}{\sqrt{1-{\beta }^2}}$.

Let$t_b$be time coordinate of big flash and $x_b$be the position coordinate of the big flash. So, the time of the big flash in the frame$S^{\text{'}}$can be obtained using the formula .

${t^{\text{'}}}_b=\frac{t-{\displaystyle \frac{\beta x_b}{c}}}{\sqrt{1-{\beta }^2}}$

The time interval between the flashes can be obtained as follows:

$$\begin{gathered} ∆t={t^{\text{'}}}_b-{t^{\text{'}}}_s \\ =\frac{t_b-{\displaystyle \frac{\beta x_b}{c}}}{\sqrt{1-{\beta }^2}}-\frac{t_s-{\displaystyle \frac{\beta x_s}{c}}}{\sqrt{1-{\beta }^2}} \\ =\frac{1}{\sqrt{1-{\beta }^2}}\left[t_b-\frac{\beta x_b}{c}-t_s+\frac{\beta x_s}{c}\right] \end{gathered}$$

In frame , both the flashes occurs simultaneously. So, $t_b=t_s$. Then the above equation becomes:

$$\begin{gathered} ∆t^{\text{'}}=\frac{1}{\sqrt{1-{\beta }^2}}\left[-\frac{\beta x_b}{c}+\frac{\beta x_s}{c}\right] \\ =\frac{\beta }{c\sqrt{1-{\left(\beta \right)}^2}}\left[x_s-x_b\right] \\ =\frac{0.250}{3\times {10}^8\sqrt{1-{\left(0.250\right)}^2}}\left[30000\right] \\ =2.58\times {10}^{-5}s \end{gathered}$$

Thus, the time interval between both flashes is$2.58\times {10}^{-5}s$ .

Occurrence of flashes

Here, from the calculation is it is clear that $∆t^{\text{'}}$is positive, this means ${t^{\text{'}}}_b$is greater than ${t^{\text{'}}}_s$. This implies the small flash occurs first.

Thus, the small flash occurs first in the frame$S^{\text{'}}$ .