# 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!

1. Cool post. So the special relativity effect here is only about 15%, so this is not a good example for time dilation in special relativity. Another common example that is frequently cited is the twin paradox, which is also a general relativity problem. Do you know of any special relativity problem involving time dilation which can be used as an example?

Comment by flamendialis — June 1, 2008 @ 11:06 am

2. Hi flamendialis,

First off, thanks for the reply.

There is a very interesting problem in special relativity which has a real world application. Muons, as you may know, are high speed (relativistic) particles that are similar to electrons (they’re both types of leptons, actually). They are also constantly bombarding Earth.

The interesting thing is that they only have a lifetime of about 2.2 microseconds before they decay into lighter particles. So even moving at the speed of light, on average they will travel less than a kilometer, much less than the height of Earth’s atmosphere. Of course, this is only the mean lifetime, so some will still hit the Earth, and this number can be calculated. But measurements show that this number is much smaller than the actual number of muons hitting Earth.

What’s happening is that, since the muons are moving so close to c, their “internal” clocks are moving slower than one on Earth. So 2.2 microseconds on Earth is actually a much shorter time to the muon, and 2.2 microseconds to the muon is much longer on Earth.

Another way to look at the problem is that the size of the atmosphere is contracted due to its high speed; you’ll get exactly the same answer. This shows that time dilation and length contraction are actually manifestations of the same thing in special relativity.

If you’re interested in a more quantitative description of the muon problem, you can visit here.

Comment by Kevin — June 1, 2008 @ 3:27 pm

3. Hi Kevin,
Thanks for the reply. I had forgotten about the time dilation for the muons, but the length contraction aspect for the atmosphere is something that I hadn’t thought of. By the way, I had been maintaining that there was no practical application for length contraction as we cannot accelerate a particle which is physically large to a significant fraction of the speed of light.

Also, I had been wary of looking at what a contracted object _looks_ like, as light is fundamental to measurement. But according to Terrell, a sphere looks like a sphere, and a cube is rotated! It is very tempting to draw a scale of length L and then one of length L/\gamma to show what the stationary observer sees after discussing length contraction. I see that most text books do not have this caveat, causing potential embarrassment.

The general analysis of 3-d bodies under Lorentz transformations looks like a job for a computer program. Do you know of any good ones? A quick web search did not turn up much.

Comment by flamendialis — June 1, 2008 @ 9:25 pm

4. Hi again,

Relativity does very strange things. It had brought to my attention that spheres remain spheres under the transformation!

In response to your question, this may be something like what you’re looking for. It’s free, there are lots of objects to be imported, and you can make your own shapes from external programs.

Comment by Kevin — June 2, 2008 @ 7:40 pm

5. Found this post from Dr. Pion’s site. As a physics student and an occasional geocacher, this is quite a cool thing to read about!

Comment by Matt — June 6, 2008 @ 7:15 pm

6. A hat tip to Chad at Uncertain Principles for linking to this article. I don’t get over here very often. Very nice article, but let’s separate theory and experiment.

The size of the effect is irrelevant to whether this is a “good” example of the effect of both SR time dilation and the GR correction term, both separately and together. What makes it “good” is that both the space and ground based clocks are exceptionally accurate and stable so the signal (38 micro s) is tested to very high precision. An error of that scale corresponds to some huge position error on earth, so it had better be right.

As I understand it, the first GPS satellite could be switched between the desired ground frequency and the frequency required for it to work if relativity was correct. (Apparently, as the story goes, there were some unconvinced folks in the Air Force.) It did, and this makes it an actual twin experiment that, because the speed is so much higher than in the cases studied using aircraft, does show the effect for those “twin” clocks very clearly.

Of course, Dr. Pion knows all about the importance of time dilation when trying to transport a pion from its production area, through a target, and on through a magnetic spectrometer. Experiments with pions and muons were the first tests of the time predictions of SR.

Comment by CCPhysicist — June 7, 2008 @ 3:10 pm

7. Oh, yes, the Relativity FAQ includes a category on experimental tests that has a section for “twin” experiments:

Comment by CCPhysicist — June 7, 2008 @ 3:14 pm

8. Hi Matt and CCPhysicist,

Thanks for posting. Indeed, the resolution power of the GPS satellites depends on the precision of clocks. It is my understanding that, roughly speaking, precision to 1 nanosecond translates to about a foot of precision (the speed of light is very close to one foot per nanosecond). So to be off 38 microseconds per day amounts to the satellites being off by about 38,000 feet per day, or just over 7 miles! The more and more accurate we make our clocks, the more and more accurate GPS systems can become.

Comment by Kevin — June 8, 2008 @ 11:38 am

9. flamendialis, there is a neat computer program which does this. It’s called lightspeed, and you can find it here:
http://lightspeed.sourceforge.net/

Comment by Porges — June 9, 2008 @ 8:56 am

10. Neat post.

I’m confused about the special relativity part. Wouldn’t the *orbiting* satellite be in a constantly accelerating reference frame, and not in constant linear motion away from / towards an observer?

Is it possible for the special relativistic effects to even accumulate; wouldn’t it “un-do” the effects on the second half of an orbit?

Comment by Andy — June 9, 2008 @ 10:52 pm

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12. @Andy: The relevant value is the speed — the direction doesn’t come into play.

Comment by Tom — June 16, 2008 @ 11:00 am

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