An engineer in a train moving toward the station with a velocity \(v=0.60 c\) lights a signal flare as he reaches a marker \(1.0 \mathrm{km}\) from the station (according to a scale laid out on the ground). By how much time, on the stationmaster's clock, does the arrival of the optical signal precede the arrival of the train?

To calculate the time taken for the train to cover the 1.0 km distance, we will use the formula: time = distance/velocity. Since the train is moving at a significant fraction of the speed of light, we should take into account relativistic effects and use the Lorentz factor to calculate the time dilation. The Lorentz factor is given by: $$\gamma = \frac{1}{\sqrt{1 - (\frac{v}{c})^2}}$$ where \(v\) is the train's velocity and \(c\) is the speed of light. Now let's calculate the Lorentz factor and time dilation. $$\gamma = \frac{1}{\sqrt{1 - (\frac{0.60c}{c})^2}} = \frac{1}{\sqrt{1 - 0.36}} = \frac{1}{\sqrt{0.64}} = \frac{1}{0.8} = 1.25$$ Now that we have the Lorentz factor, we find the time taken by the train to reach the station in the stationmaster's frame of reference using time dilation: $$t_\text{train} = \gamma t_0 = 1.25\frac{d}{v} = 1.25\frac{1.0\,\text{km}}{0.60c} = 1.25\frac{10^3\,\text{m}}{0.60(3\times10^8\,\text{m/s})} \approx 6.94\times10^{-6}\,\text{s}$$

Since the optical signal is light, it travels at the speed of light (c). To calculate the time it takes for the signal to cover the 1.0 km distance, we will use the same formula as before (time = distance/velocity). $$t_\text{signal} = \frac{d}{c} = \frac{1.0\,\text{km}}{c} = \frac{10^3\,\text{m}}{3\times10^8\,\text{m/s}} \approx 3.33\times10^{-6}\,\text{s}$$

The time difference between the arrival of the optical signal and the train is the difference between the time taken by the train and the time taken by the optical signal: $$\Delta t = t_\text{train} - t_\text{signal} = 6.94\times10^{-6}\,\text{s} - 3.33\times10^{-6}\,\text{s} \approx 3.61\times10^{-6}\,\text{s}$$ So, the optical signal arrives at the station \(3.61\times10^{-6}\) seconds before the train according to the stationmaster's clock.