This is an original research work of author, published here; 22 November 2011, 8:26 am.
The image above shows earth at an altitude of 20200 + 9.08 km. The red line I drew is a straight line of 696 km, 3-D straight line — perhaps non-Euclidean, near the CERN – Gran Sasso Baseline, parallel to the latter, which is not straight and about 733 km long.
— Isn’t that a reason to worry?
This altitude is where the GPS satellites are “falling towards earth”.
Why should different baseline ends not introduce any error, to synchronization?
Also if we imagine 13.7 km/s at this altitude, the satellite is almost not moving. It takes 2 minutes for a deviation of 1 degree angle on earth surface — or anywhere, which is “49000 time of flights” of a neutrino bunch from CERN to Gran-Sasso.
That is, 3.54375 E-7 rad / time-of-flight.
The earth spin is 3.000823 E-2 rad / time-of-flight — earth spin being 7.292 E-5 / s.
I have shown that static earth effect is less than 1 pico-sec — special-relativistic-effect is small and negative, compared to general-relativistic-effect.
Spinning earth effect being 1% order of static earth, and earth spinning speed being E-5 orders higher, than satellite angular speed, again the “1% general relativistic effect of 1 pico-sec” will preside over the special relativistic effect, of the satellite.
So, the non-inertial effect of earth, will be ~1% of static earth, which is 1 picosecond. In other words, the total relativistic effect, of the GPS satellite, considering even earth-spin will be, within, about, 1-pico-sec.
We need not worry about this.
Here is a diagram, that will make it more concise;
I suspect it still has some calculation mistakes, although that does not change the idea, I will fix this later, lets the check next diagram, with correct calculation, error in Earth-spin and satellite spin, tof — time of flight, but satellite spin is already present, in static earth calculation, that is, the 8 meters / tof, has already produced an effect, less than 1 pico-second, which was static earth gravitational potential, on frequency emitted by satellite.
At any point B, there are two considerations.
i. Static earth weak-field gives 1 ps and gives opposing effect, 0 < effect < 1 pico-sec.
ii. Spinning earth (lets say any de-Sitter rotation) 1% of static earth from principle of equivalence.
hence spinning earth gives 0.01 ps and remaining (from earth spin and satellite angular rate, latter is of earth spin, latitude effect is null, since angular speed is free from altitude) that is: 0.5 km/s vs 0<effect<0.01 ps, at any point B.
A completely new thinking: on earth any point can be considered to suffer from general and special relativistic effects and I have shown for GPS satellites, from their weak-field approximation, the static-earth for any point, has the highest effect, about 1 pico-secs, the special relativity effect which comes from beta = E-5 order therefore, negative and less than the static effect. Thus total effect here is ~1 pico-sec.
Diagram: The corrected earth and satellite spin, at any point on surface, the equatorial projections and trajectory around equator. This diagram is slated to be improved upon for better clarity.
The calculations shown above, in the diagram;
The corrected earth and satellite spin at any point on surface, equatorial projections and trajectory around equator. The distance 733 kms is not equatorial distance. The 60 cms earth spin correction that OPERA does is because of this reason, there is angles whose cosine or sines are less than 1 for 733 kms distance.
IF we can determine the unknown longitude x, we can adjust the additional distance, we have, in our calculation and subtract it from 733 kms or whatever we have used all along, consistently. Then we will see a more correct result ,whether OPERA is doing anything wrong or not. Hint: this triangle is a right angle triangle on the surface of earth, by construction.
— What a cool web-site I found, in this respect, you can plug in the latitude, longitude, and you can get x and distances, check here.
Note: coordinates in lat. and long. shown above, are from google earth, for locations, which are giving distance, within few km, so I need to plug in actual-quoted lat. and long. from OPERA paper, later.
A completely new thinking:
On earth, any point can be considered to suffer from general and special relativistic effects and I have shown for GPS satellites, from their weak-field approximation, the static earth for any point has the highest effect about 1 pico-secs, the special relativity effect, which comes from beta = E-5 order therefore, negative and less than the static effect.
Thus total effect here is ~1 pico-sec.
Then comes earth spinning, which from principle of equivalence, has a 1% acceleration effect, hence equivalent static earth effect, is, 0.01 pico-sec. The special relativity effect here, will again be less, than this and negative, to the general relativity effect, because at any point on earth, the beta of satellite motion, is, E-6 order to beta of earth-spin.
The altitude of satellite does not matter, as long as the signal from it, reaches earth continuously, which is of nano-second range, at-least, at any point, time of flight being 2.43 mili-sec. Note that the nano-sec is not playing any special role here, just that, if we start with such precision and not do anything silly we would not blow it up.
— The flyby satellites are hyperbolic and their effects vanish, at very large distances, from perigee, what remains is just static and spinning earth effects, at earth surface where radars are placed.
Suggestions for OPERA;
One thing OPERA might have to do is; divide their 3 years data, in a “Julian time arrow”, and place the actual time-coordinate, during which neutrinos were released, from CERN or registered at Gran-Sasso, wrt all the satellite position-coordinate, that were synchronizing, the procedure on earth. This n-tuple, if available, for the entire data, will be helpful in seeing, if there is any deviation, from time-synchronization, as expected, by a nano-second consideration, that they claim. I assume such a n-tuple, would be available, in the total data they have.