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One of the ways to test time dilation effect are so – called Mössbauer rotor time dilation tests. In these tests experimenter places source in the center of the disk and absorber on the rim and vice versa. The disk rotates with very large angular velocity. Then experimenter measures received by absorber deviation of frequency (Transverse Doppler Shift) and measures amount of time dilation this way. It is well known, that measurement brings results in accordance with Lorentz transform for time dilation, i.e. rotating clock dilates.

Clock hypotesys claims, that clock dilates purely because of relative velocity and acceleration has no effect on that.

One interesting thing is that when source and absorber were placed on opposite sides of the disc (1963 Champeney and Moon time dilation test) there was no frequency shift. Well, if two clock dilate at the same magnitude they will not be able to measure dilation of each other.

In special relativity time dilation is reciprocal. Clock dilates not because of it‘s own motion through a medium, but because of relative motion to another one. It is always another clock dilates and SR does not admit that clock dilates because of actual motion.

One of problem is that rotating absorber detect not redshift, but blueshift of frequency. That means, that rotating clock actually dilates. If they place absorber in the centre and source on the rim, there is a redshift of frequency. So, it seems that rotating observer can measure only acceleration of clock in the centre.

Distance between clocks doesn‘t change. Do they move relatively to each other? Why clock at rest sees dilation of rotating clock? Why rotating clock sees acceleration of clock in the centre? Why clocks on opposite sides of the rim see neither dilation nor acceleration?

What makes rotating clock dilate?

How does all that fits into the Special Relativity?


marked as duplicate by John Rennie special-relativity Feb 23 '17 at 8:22

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  • $\begingroup$ I know very well, that rotating in ultracentifuge clock dilates. And clock in the center does not. I asked how that confirms Special Relativity, not General. I asked why clock dilates though distance is the same. Relative to WHAT it moves? There was no answer in that article. Moreower, answer there is simply wrong, because rotaing absorber and absorber at rest (in the center) measure blueshift and redshift respectivelly, i.e. there is no even smell of reciprocity of observations. $\endgroup$ – Albert Feb 23 '17 at 8:41
  • $\begingroup$ I think your answer there correctly states that we consider the task in the framework of the SR. A fortiori I don't understand why absorbers on opposite sides of the rim do not measure frequency shift (time dilation) and how a clock can dilate ACTUALLY but not relatively. Why tangential observer can measure dilation, but rotating only acceleration of clock in the center? I will be most grateful for your advice, if you have one. $\endgroup$ – Albert Feb 23 '17 at 9:05
  • $\begingroup$ Albert The absence of frequency shift for opposite placed source and receiver is a very important fact (which was unknown to me until now). Thanks for posting. But this phenomenon don't surprise me. See my answer to the question Is it accurate that light loses energy in the absence of gravity and gains energy in its presence $\endgroup$ – HolgerFiedler Feb 23 '17 at 9:15
  • $\begingroup$ I will look through, thank you @HolderFiedler! Yes, angles of emission and reception are the same. According to relativistic aberration formula received frequency will be the same as emitted. iopscience.iop.org/article/10.1088/0370-1328/85/3/317 ; iopscience.iop.org/article/10.1088/0370-1328/77/2/318/meta; nature.com/nature/journal/v202/n4934/abs/202787a0.html; nature.com/nature/journal/v199/n4895/abs/199739a0.html $\endgroup$ – Albert Feb 23 '17 at 9:24
  • $\begingroup$ @John Rennie I think it's not a duplicate of the link question. Furthermore it's a very interesting question. Unfortunately once a question is marked it is lost for attentions and even the censor usually doesn't come back to it. $\endgroup$ – HolgerFiedler Feb 23 '17 at 9:24