Question:

Physicists conduct an experiment to measure time dilation of an atomic clock in a flying airplane. If the velocity of the airplane is 960 km/h, find the time dilated of the atomic clock from the laboratory flame. (You do not need to consider the effect from the general theory of relativity.)

Answer:

We consider one direction of motion, say, \(x\) direction. The Lorentz transformations for the displacement and time are

\[  x’ = \frac{-vt+x}{\sqrt{1-\beta^2}}  \]

\[  t’ = \frac{t-(v/c^2)x}{\sqrt{1-\beta^2}}  \]

where \(\beta=v^2/c^2\). Suppose that the clock is placed at the origin of moving flame, \(S’\). Then, one measures the time in laboratory flame, \(S\). The origin in \(S’\) indicates that \(x’=0\). Thus, we have

\[  0 = \frac{-vt+x}{\sqrt{1-\beta^2}} \quad \rightarrow \quad x = vt  \]

Plug it in the transformation for time.

\[  t’ = \frac{t-(v/c^2)(vt)}{\sqrt{1-\beta^2}}=\frac{t(1-\beta^2)}{\sqrt{1-\beta^2}}=\sqrt{1-\beta^2}t  \]

Since \(v\ll c \rightarrow \beta \ll 1\), we can use the approximated expression with a Taylor expansion;

\[  t’ = \left(1-\frac{1}{2}\beta^2\right)t  \]

Now, convert the velocity of the airplane into m/s. Note that 1 km = 1000 m and 1 hour = 3600 s.

\[  960 \ \mathrm{km/h} = 960 \times 1000 \div 3600 = 266.7 \ \mathrm{m/s}  \]

Therefore, we have the time dilation compared with the lab frame.

\[  t’= \left\{1-\frac{1}{2}\left(\frac{266.7}{3.00\times 10^8}\right)^2\right\}t=(1-3.95\times 10^{-13})t  \]

This means: While one second elapses in lab frame, the clock in the airplane only elapses \(1-3.95\times 10^{-13}\) seconds.