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Q. 15.1

In the test of Hafele and Keating of flying atomic clocks around Earth (see Chapter 2), the gravitational redshift had to be considered. Calculate the effect and compare it with the special relativity time dilation effect. Assume the jet airplane travels 300 m/s and the circumference of Earth is about $4 \times 10^{7} m$.

Strategy The ratio $\Delta f / f$ will be equal to the time difference ratio $\Delta T / T$ measured by the clocks on Earth and on the jet airplanes. Because the airplane’s altitude is negligible in comparison to Earth’s radius, we use the simpler of the two equations for the gravitational redshift, Equation (15.3). We use the time dilation Equation (2.19) from Chapter 2 to determine the special relativity effect.

$T^{\prime}=\frac{T_{0}}{\sqrt{1-v^{2} / c^{2}}}$ (2.19)

$\frac{\Delta f}{f}=\frac{g H}{c^{2}}$ (15.3)

Verified Solution

Equation (15.3) gives us

$\frac{\Delta T}{T}=\frac{g H}{c^{2}}$

The value of H for the clock on the Earth’s surface is $r_{e}$, and the value for the flying airplane is $r_{e}+A$, where A is the altitude of the airplane, about 33,000 feet (10,000 m). The value of H is the difference in the height of the two clocks in Earth’s gravitational field g, and H = A. We neglect the change of g over this altitude, and $\Delta T / T$ becomes

$\frac{\Delta T}{T}=\frac{\left(9.8 m / s ^{2}\right)(10,000 m )}{\left(3 \times 10^{8} m / s \right)^{2}}=1.09 \times 10^{-12}$

The eastward and westward airplane trips took about T = 45 hours flying time. The difference in the two clocks due to the gravitational redshift is

\begin{aligned}\Delta T &=\left(1.09 \times 10^{-12}\right)(45 h )\left(\frac{3600 s }{1 h }\right) \\&=177 \times 10^{-9} s =177 ns\end{aligned}

A clock fixed on Earth will measure a flight time $T_{0}$ of

$T_{0}=\frac{4 \times 10^{7} m }{300 m / s }=1.33 \times 10^{5} s$

Because a clock in the airplane will run slowly, an observer on Earth will say the time measured on the airplane is $T=T_{0} \sqrt{1-\beta^{2}} \text { where } \beta=v / c=(300 m / s ) /\left(3 \times 10^{8} m / s \right)=10^{-6}$. The time difference is

$\Delta T=T_{0}-T=T_{0}\left(1-\sqrt{1-\beta^{2}}\right)$

Because $\beta$ is so small, we can use a power series expansion of the square root and ignore all terms smaller than $\beta^{2}$. The special relativity effect is

\begin{aligned}\Delta T &=T_{0}\left[1-\left(1-\beta^{2} / 2+\cdots\right)\right]=\frac{\beta^{2} T_{0}}{2} \\&=\frac{1}{2}\left(10^{-6}\right)^{2}\left(1.33 \times 10^{5} s \right)=6.65 \times 10^{-8} s =66.5 ns\end{aligned}

The gravitational time dilation effect of 177 ns is larger than the approximate 66.5-ns special relativity time dilation effect. There is also a third correction due to the rotation of Earth.