I was wondering whether any provision is made in the NTP standard to synchronize time between machines travelling near the speed of light relative to each other. Does NTP even work in these conditions?

I suppose it might not be practical to want to do this, because you will be getting their time and not your own time, but suppose you wanted to know what their time is, very precisely - could you do it with NTP?

I'm not well-versed in either NTP or relativity, but it just seemed like an interesting thing to think about.

  • The caesium time standards around the world require relativistic correction to compensate the rotational speeds of the earth at different latitudes, but NTP standards do not need any such accuracy, because of the unpredictable transmission delays, which would be grossly exaggerated in a client travelling at near the speed of light.
    – AFH
    May 27 '15 at 10:08

According to Wikipedia,

NTP can usually maintain time to within tens of milliseconds over the public Internet, and can achieve better than one millisecond accuracy in local area networks under ideal conditions. Asymmetric routes and network congestion can cause errors of 100 ms or more.

The total time dilation effect experienced over a trip of distance D, at speed v, is

delta t ~ D/c v/c

This is the difference between the duration of a photon round trip between two observers, one fixed to the Earth and the other making the trip. Even for very large trips (say, an intercontinental flight of D= 10^4 km) on a fast jetliner (v = 1000 km/h), this only amounts to 10^(-7) seconds, which is way smaller than the accuracy aimed for by the NTP project.

Even for an interplanetary flight (D = 10^8 km, v = 10^4 km/h),the total error would be of order 3 msec.

Hence there is no need for any provision to be made within NTP for relativistic speeds. Such provisions are instead routinely made for interferometric observations as in the VLBI, or in accurate general relativistic experiments in space (see Gravity Probe B for an example) but this has nothing to do with what this site is all about.

For more info, I suggest you turn to https://physics.stackexchange.com/

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.