Take a clock in space above the earth (assuming a Schwarzchild spacetime) that works by relaying a light signal a small distance radially; ticking each time the light signal returns. Compare this to the same clock at the same location and rotated such that the light signal goes side to side (along $\theta$ ). They produce different measurements. Why should one be preferred to the other? What answer is correct?
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One of the basic principles in relativity is that spacetime always looks locally flat. By this I mean that if you restrict your observations to a small region surrounding you the curvature becomes negligable. You can make the effects of curvature arbitrarily small by making the region you consider arbitrarily small.
The point of this is that for your clock to be reliable it must be small enough to fit within the region where spacetime is effectively flat. As long as the clock does fit inside this region the orientation of the clock maes no difference so the problem disappears.
If the clock is so big that the spacetime curvature is significant over its width then it is not a reliable clock. It is pointless to ask how the clock should be oriented because the clock shouldn't hav ebeen constructed in that way in the first place.
In practice this isn't an issue for GPS clocks. The difference in time between the Earth's surface and the GPS satellite is certainly measurable, but that's over a distance of about 20,000 km (i.e. the altitiude of the satellite). A satellite, and therefore the clock inside it, is of the order of one metre in size, or about one ten millionth of its altitude. Over such a small distance the time dilation is too small to be significant for GPS purposes, though it is actually measurable.
John Rennie
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Actually the clocks tick at the same rate. You probably forgot that distances do not stay the same when you rotate the clock.
(In non-Euclidean space turning a rod changes its length)
stuffu
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