8 Neil Ashby
The time transformation 𝑡 = 𝑡
′
in Eqs. (3) is deceivingly simple. It means that in the rotating
frame the time variable 𝑡
′
is really determined in the underlying inertial frame. It is an example
of coordinate time. A similar concept is used in the GPS.
Now consider a process in which observers in the rotating frame attempt to use Einstein syn-
chronization (that is, the principle of the constancy of the speed of light) to establish a network of
synchronized clocks. Light travels along a null worldline, so we may set 𝑑𝑠
2
= 0 in Eq. (4). Also,
it is sufficient for this discussion to keep only terms of first order in the small parameter 𝜔
E
𝑟
′
/𝑐.
Then
(𝑐𝑑𝑡
′
)
2
−
2𝜔
E
𝑟
′2
𝑑𝜑
′
(𝑐𝑑𝑡
′
)
𝑐
− (𝑑𝜎
′
)
2
= 0, (5)
and solving for (𝑐𝑑𝑡
′
) yields
𝑐𝑑𝑡
′
= 𝑑𝜎
′
+
𝜔
E
𝑟
′2
𝑑𝜑
′
𝑐
. (6)
The quantity 𝑟
′2
𝑑𝜑
′
/2 is just the infinitesimal area 𝑑𝐴
′
𝑧
in the rotating coordinate system swept
out by a vector from the rotation axis to the light pulse, and projected onto a plane parallel to the
equatorial plane. Thus, the total time required for light to traverse some path is
path
𝑑𝑡
′
=
path
𝑑𝜎
′
𝑐
+
2𝜔
E
𝑐
2
path
𝑑𝐴
′
𝑧
. [light] (7)
Observers fixed on the earth, who were unaware of earth rotation, would use just
𝑑𝜎
′
/𝑐 for
synchronizing their clock network. Observers at rest in the underlying inertial frame would say that
this leads to significant path-dependent inconsistencies, which are proportional to the projected
area encompassed by the path. Consider, for example, a synchronization process that follows
earth’s equator in the eastwards direction. For earth, 2𝜔
E
/𝑐
2
= 1.6227 × 10
−21
s m
−2
and the
equatorial radius is 𝑎
1
= 6,378,137 m, so the area is 𝜋𝑎
2
1
= 1.27802 ×10
14
m
2
. Thus, the last term
in Eq. (7) is
2𝜔
E
𝑐
2
path
𝑑𝐴
′
𝑧
= 207.4 ns. (8)
From the underlying inertial frame, this can be regarded as the additional travel time required by
light to catch up to the moving reference point. Simple-minded use of Einstein synchronization in
the rotating frame gives only
𝑑𝜎
′
/𝑐, and thus leads to a significant error. Traversing the equator
once eastward, the last clock in the synchronization path would lag the first clock by 207.4 ns.
Traversing the equator once westward, the last clock 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 rate due 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 attempt this, find that the proper time elapsed on the portable clock is affected by earth’s
rotation rate. Factoring Eq. (4), the proper time increment 𝑑𝜏 on the moving clock is given by
(𝑑𝜏)
2
= (𝑑𝑠/𝑐)
2
= 𝑑𝑡
′2
1 −
𝜔
E
𝑟
′
𝑐
2
−
2𝜔
E
𝑟
′2
𝑑𝜑
′
𝑐
2
𝑑𝑡
′
−
𝑑𝜎
′
𝑐𝑑𝑡
′
2
. (9)
For a slowly moving clock, (𝑑𝜎
′
/𝑐𝑑𝑡
′
)
2
≪ 1, so the last term in brackets in Eq. (9) can be neglected.
Also, keeping only first order terms in the small quantity 𝜔
E
𝑟
′
/𝑐 yields
𝑑𝜏 = 𝑑𝑡
′
−
𝜔
E
𝑟
′2
𝑑𝜑
′
𝑐
2
(10)
Living Reviews in Relativity
http://www.livingreviews.org/lrr-2003-1
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