100 Years of Relativity Space-Time Structure: Einstein and Beyond ...

Relativity in the Global Positioning System 263The time transformation t = t ′ in Eqs. (3) is deceivingly simple. Itmeans that in the rotating frame the time variable t ′ is really determinedin the underlying inertial frame. It is an example of coordinate time. Asimilar concept is used in the GPS.Now consider a process in which observers in the rotating frame attemptto use Einstein synchronization (that is, the principle of the constancy ofthe speed of light) to establish a network of synchronized clocks. Lighttravels along a null world line so we may set ds 2 = 0 in Eq. (4). Also, it issufficient for this discussion to keep only terms of first order in the smallparameter ω E r ′ /c. Then solving for (cdt ′ ),cdt ′ = dσ ′ + ω Er ′2 dφ ′. (5)cThe quantity r ′2 dφ ′ /2 is just the infinitesimal area dA ′ z in the rotatingcoordinate system swept out by a vector from the rotation axis to the lightpulse, and projected onto a plane parallel to the equatorial plane. Thus,the total time required for light to traverse some path is∫path∫dt ′ =pathdσ ′c+ 2ω Ec∫path2 dA ′ z . [ light ] (6)Observers fixed on the earth, who were unaware of earth rotation, woulduse just ∫ dσ ′ /c to synchronize their clock network. Observers at rest inthe underlying inertial frame would say that this leads to significant pathdependentinconsistencies, which are proportional to the projected areaencompassed by the path. Consider, for example, a synchronization processthat follows earth’s equator once around eastwards. For earth, 2ω E /c 2 =1.6227 × 10 −21 s/m 2 and the equatorial radius is a 1 = 6, 378, 137 m, sothe area is πa 2 1 = 1.27802 × 1014 m 2 . Then, the last term in Eq. (6) is ofmagnitude 2πω E a 2 1 /c2 = 207.4 ns.From the underlying inertial frame, this can be regarded as the additionaltravel time required by light to catch up to the moving referencepoint. Simple-minded use of Einstein synchronization in the rotating framegives only ∫ dσ ′ /c, and thus leads to a significant error. Traversing theequator once eastward, the last clock in the synchronization path would lagthe first clock by 207.4 ns. Traversing the equator once westward, the lastclock in the synchronization path would lead the first clock by 207.4 ns.In an inertial frame a portable clock can be used to disseminate time.The clock must be moved so slowly that changes in the moving clock’s ratedue to time dilation, relative to a reference clock at rest on earth’s surface,are extremely small. On the other hand, observers in a rotating frame who