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Texts on relativity, either special or general, talk conspicuously about rates of clocks but I quite don't understand this: does relativity theory assign some absolute meaning to the rate of a clock, or is it that - explicitly or implicitly - we are always comparing the rates of two clocks? As an example, this quote from Einstein's popular exposition Relativity: The Special and General Theory:

"in every gravitational field, a clock will go more quickly or less quickly, according to the position in which the clock is situated (at rest)"

"Thus the clock goes more slowly if set up in the neighbourhood of ponderable masses." (A. Einstein, The Foundation of the General Theory of Relativity, §22)

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marked as duplicate by Qmechanic Aug 28 '17 at 20:05

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    $\begingroup$ Not clear. What kind of "absolute meaning" are you suggesting? Are you asking if time flows at a constant rate in the universe? $\endgroup$ – sammy gerbil Aug 28 '17 at 19:50
  • $\begingroup$ @sammy gerbil I was not suggesting, I was just asking, because I wanted to be sure, although it seems to me almost obvious that there is no such absolute meaning and that we can only talk of the rate of a clock by comparison to another clock. $\endgroup$ – Fernando Passos Aug 29 '17 at 20:08
  • $\begingroup$ Einstein's words "more or less quickly" suggest that a comparison is being made, that the passage of time is relative. It is still not clear what you mean by "absolute meaning to the rate of a clock." You need to provide more explanation, or a better quote. Please look at some of the questions on the right or quoted in the headline to check if your question has already been asked and to refine what you mean. If you want your question to be re-opened, you will need to edit it to provide more explanation and address the charge that it has already been asked. $\endgroup$ – sammy gerbil Aug 29 '17 at 22:28
  • $\begingroup$ The rate of a clock is the number of times it ticks per second. Obviously this notion is frame-dependent. $\endgroup$ – WillO Aug 30 '17 at 0:46
  • $\begingroup$ @WillO Your definition is not frame-dependent - it is circular. For you will need to define "second" somehow, and the only way is to define it as a certain number of ticks of the clock. $\endgroup$ – Fernando Passos Aug 31 '17 at 12:54