Unused Cycles

May 30, 2008

Physics of GPS relativistic time delay

So I realized today I’ve been posting quite a bit about GNU/Linux and mathematics, but I haven’t really done much with physics. So here’s my first physics post!

This problem is actually one assigned in the undergraduate general relativity course I took in the spring 2008. It’s from James B. Hartle’s book Gravity: An Introduction to Einstein’s General Relativity, chapter 6, problem 9.

A GPS satellite emits signals at a constant rate as measured by an onboard clock. Calculate the fractional difference in the rate at which these are received by an identical clock on the surface of Earth. Take both the effects of special relativity and gravitation into account to leading order in 1/c^2 . For simplicity, assume the satellite is in a circular equatorial orbit, the ground-based clock is on the equator, and that the angle between the propagation of the signal and the velocity of the satellite is 90° in the instantaneous rest frame of the receiver.

The problem is very simplified as to make the calculations doable at the undergraduate level. Thus it is using the simplified Geometric Newtonian gravity, that is, the line element given by

\displaystyle ds^2=-\left(1+\frac{2\Phi}{c^2}\right)(c dt)^2+\left(1-\frac{2\Phi}{c^2}\right)(dx^2+dy^2+dz^2).

One could, of course, use Schwarzchild coordinates (and this is, in effect, what I did for the general relativistic part). The solution is broken into three parts: orbital information, the special relativistic effects, and the general relativistic effects.

Orbital Information

The key here (that’s not given in the problem) is that the time t that it takes satellites to orbit Earth is 12 hours. Recalling that the speed of an orbit in Newtonian gravity is v_{\text{orbit}}=\sqrt{\frac{GM}{R}}, where R is the radius of the orbit and M is the mass of the object being orbited, we get that

\begin{array}{rcl}\displaystyle \frac{2\pi R}{t}&\displaystyle=&\displaystyle \sqrt{\frac{GM}{R}}\vspace{0.3 cm}\\\displaystyle R&\displaystyle=&\displaystyle \sqrt[3]{\frac{GMt^2}{4\pi^2}}\vspace{0.3 cm}\end{array}

Plugging in the appropriate values gives R=26,605 km and v_\text{orbit}=3871.0 m/s.

Special Relativistic Effects

Let A denote the ground observer and B denote the satellite. Then we can use time dilation to get that

\begin{array}{rcl}\displaystyle\frac{\Delta \tau_{B,S}}{\Delta\tau_{A,S}}&\displaystyle=&\displaystyle\frac{1}{\gamma}\vspace{0.3 cm}\\&\displaystyle=&\displaystyle \sqrt{1-v_\text{orbit}^2/c^2}\vspace{0.3 cm}\\&\displaystyle=&\displaystyle 0.999999999917.\end{array}

General Relativistic Effects

For general relativistic effects, note that the frequencies are related by

\displaystyle \frac{\Delta\tau_{B,G}}{\Delta\tau_{A,G}}=\frac{\omega_A}{\omega_B}=\sqrt{\frac{g_{00}\left[(\vec{x}(B)\right]}{g_{00}\left[(\vec{x}(A)\right]}}=\sqrt{\frac{1+\frac{GM}{Rc^2}}{1+\frac{GM}{R_\oplus c^2}}}.

The derivation of this formula is beyond the scope of chapter 6 and uses Killing Vectors and photon geodesics introduced in a later chapter. However, an approximation to this result is given in equation 6.12 of Hartle. Plugging in the appropriate results gives the ratio to be 1+5.2873 x 10-10, very nearly 1.

Putting things together, the whole shift is

\begin{array}{rcl}\displaystyle\frac{\omega_A}{\omega_B}=\frac{\Delta\tau_B}{\Delta\tau_A}&\displaystyle=&\displaystyle(0.999999999917)(1+5.2873\times 10^{-10})\vspace{0.3 cm}\\\displaystyle&=&\displaystyle 1+4.4573\times 10^{-10}.\end{array}

As a check, suppose that a day passes on Earth (in other words, \Delta\tau_A = 24 hours = 86,400 s). Then for the satellite, (86,400 s)(4.4573 x 10-10) = 38.511 μs more have passed every day. According to Wikipedia, this number is 38 μs.

Also according to Wikipedia, the desired frequency on Earth is \omega_A=10.23 MHz. This leaves \omega_B to be 0.0045598 Hz less (compare this to Wikipedia’s claim of 0.0045700 Hz less).

The problem of course is not as simple as it was made to be. Difficulties arise when one takes into account that Earth is rotating (so the metric is more complicated), that the observer is not necessarily orbiting in the same plane as the satellite, and the orbit is not perfectly circular. However, these corrections are minute and don’t affect the problem very much.

Hope this was an interesting read!

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