Suppose you are driving your car on a business trip and are traveling at 30 m/s. Your boss, who is waiting at your destination, expects the trip to take 5.0 h. When you arrive late, your excuse is that your car clock registered the passage of 5.0 h but that you were driving fast and so your clock

Question

Suppose you are driving your car on a business trip and are traveling at 30 m/s. Your boss, who is waiting at your destination, expects the trip to take 5.0 h. When you arrive late, your excuse is that your car clock registered the passage of 5.0 h but that you were driving fast and so your clock ran more slowly than your boss’s clock. If your car clock actually did indicate a 5.0-h trip, how much time passed on your boss’s clock, which was at rest on the Earth?

Accepted Answer

We begin by calculating γ from Equation 39.8:

[latex] \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} [/latex] (39.8).

[latex] \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\frac{1}{\sqrt{1-\frac{\left(3 imes 10^{1} m / s ight)^{2}}{\left(3 imes 10^{8} m / s ight)^{2}}}} [/latex].

[latex] =\frac{1}{\sqrt{1-10^{-14}}} [/latex].

If you try to determine this value on your calculator, you will probably obtain γ= 1. However, if we perform a binomial expansion, we can more precisely determine the value as

[latex] \gamma=\left(1-10^{-14} ight)^{-1 / 2} \approx 1+\frac{1}{2}\left(10^{-14} ight)=1+5.0 imes 10^{-15} [/latex].

This result indicates that at typical automobile speeds, γ is not much different from 1.
Applying Equation 39.7, we find Δt, the time interval measured by your boss, to be

[latex] \Delta t=\frac{\Delta t_{p}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\gamma \Delta t_{p} [/latex] (39.7).

[latex] \Delta t=\gamma \Delta t_{p}=\left(1+5.0 imes 10^{-15} ight)(5.0 h ) [/latex].

[latex] =5.0 h +2.5 imes 10^{-14} h =5.0 h +0.09 ns [/latex].

Your boss’s clock would be only 0.09 ns ahead of your car clock. You might want to think of another excuse!

Question 39.2: Suppose you are driving your car on a business trip and are ...

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