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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?

mathmath12
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  • With clocks in relativity, there isn't a preferred one. – Kyle Kanos Jul 22 '15 at 20:21
  • I've heard scientists speak of "the" time above earth when talking about GPS etc. What time are they talking about? – mathmath12 Jul 22 '15 at 20:24
  • Most likely a relative time to earth-based time, cf this Physics.SE post. – Kyle Kanos Jul 22 '15 at 20:27
  • But would vary depending on the orientation of the clock in space? – mathmath12 Jul 22 '15 at 20:34
  • The relative time varies based on the distance from the massive body & the orbital speed. – Kyle Kanos Jul 22 '15 at 20:41
  • The wonderful thing about science is that observations are never wrong. Whatever you measure is the correct measurement. You can't say "no, that observation is wrong" because it was observed, it just is. This fact extends to measuring something from different perspectives. However you measure something, that's the correct measurement for that experiment. – Jim Jul 22 '15 at 20:51
  • It seems strange to me that you could design and clock or simply rotate a clock and get a different time measurement when everything i read about this implies that the result would hold for any clock at that location. – mathmath12 Jul 22 '15 at 21:02
  • "when everything i read about this implies that the result would hold for any clock at that location" Yes. Good. :) And now can you tell where you get the idea that results do not hold for all clocks? – stuffu Jul 22 '15 at 22:27

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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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  • THere is no rod turning in my example. The clock is stationary. Do you have a source for this? Or could you show me the math? This doesn't make sense to me. There is no dilation for one of the clocks. – mathmath12 Jul 22 '15 at 21:27
  • There sure is gravitational time dilation for a clock that is in a gravity field. This formula tells how much there is time dilation for any device that is a clock: https://en.wikipedia.org/wiki/Gravitational_time_dilation#Outside_a_non-rotating_sphere – stuffu Jul 22 '15 at 22:05