91Ó°ÊÓ

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

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 ${t^{\text{'}}}_s=\frac{t_s-{\displaystyle \frac{\mathrm{β}{x}_s}{c}}}{\sqrt{1-{\mathrm{β}}^2}}$.

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 .

${t^{\text{'}}}_b=\frac{t-{\displaystyle \frac{\mathrm{β}{x}_b}{c}}}{\sqrt{1-{\mathrm{β}}^2}}$

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

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

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

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

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

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$S^{\text{'}}$ .