Global positioning systems compare signals from multiple satellites in order to determine the position of an object. For the computation to be accurate, one must be able to compare the rate at which signals are emitted with the rate at which they are received. Equivalently, the clocks on the satellites must be synchronized with clocks on the ground. This synchronization turns out to require both special and general relativity!

We present here an order-of-magnitude computation to illustrate the need for relativistic corrections. Let $dt$ denote an infinitesimal time interval on the Earth's surface, and let $d\tau$ denote an infinitesimal time interval on the satellite. These two quantities will differ due to the time dilation between moving reference frames (special relativity), as well as due to the difference in the gravitational field at different altitudes (general relativity).

We first consider time dilation. The speed of a typical satellite is given by \begin{equation} \tanh\beta = \frac{v}{c} \approx 1.3\times10^{-5} \end{equation} Time dilation leads to an Earth observer thinking that the satellite clock runs slow by a factor of $\cosh\beta$, that is, \begin{equation} d\tau = \frac{1}{\cosh\beta} \>dt \end{equation} But for such small speeds we have \begin{equation} \frac{1}{\cosh\beta} = \sqrt{1-\tanh^2\beta} \approx 1-\frac12\tanh^2\beta \approx 1-0.8\times10^{-10} \end{equation} In other words, special relativity leads to a time dilation effect of roughly one part in $10^{10}$.

Earth's gravitational field can be modeled by the Schwarzschild metric; $dt$ now refers to “far-away” time, which is a good approximation to time on the ground. Ignoring the satellite's motion, the relevant part of the line element is just \begin{equation} d\tau \approx \sqrt{1-\frac{2m}{r}}\,dt \end{equation} Estimating Earth's mass as $4 \hbox{mm}$, its radius as $6000 \hbox{km}$, and the radius of the satellite orbit as $26,000 \hbox{km}$, we obtain \begin{equation} \sqrt{1-\frac{2m}{r}} \approx 1-\frac{m}{r} \approx 1-1.7\times10^{-10} \end{equation} which is larger than the special relativistic effect, but of the same order of magnitude.

Why do errors on the order of one part in $10^{10}$ matter? Because they accumulate. There are roughly $10^5$ seconds in a day, so the daily error is roughly $10^4$ nanoseconds. Although a few nanoseconds might not seem like much, light travels on the order of $1 \hbox{km}$ in that amount of time. If relativistic effects, both special and general, were not taken into account, the resulting error in location would increase by roughly $1 \hbox{km}$ per day, which is definitely enough to matter.

The approximations used here are rather rough, not only in terms of the accuracy of the data used, but also in the assumptions made. There is a difference between clocks at infinity and clocks on the Earth's surface, the Earth's rotation must be taken into account, there are Doppler effects, and so forth. Nonetheless, the principle is sound: general relativity matters to global positioning systems.