Reference (http://download.library1.org/get/1547064EDA9124039512188DCF7D80E9/Martin%20Gardner-Relativity%20Simply%20Explained-Dover%20Publications%20%281997%29.pdf) Page number 27 and onwards

I recently learned that Ether wind theory and Special theory of Relativity predict same experimental outcomes and that there was no way to prove or disprove the existence of ether. And that Einstein gave a theory where ether was not necessary but still retained all the experimental outcomes.

I still can't get around this as in STR if I and you are in relative motion, I'll shrink for you and you'll shrink for me, but in Ether theory, if I am at rest w.r.t ether and you are moving, then you will shrink for me, but I will expand for you. So one of us will know for sure that which theory is correct. If this is true then how can Ether theory be identical with SRT, if this is false, how so?

Note: When I say Ether theory, I am considering Ether theory with Lorentz-FitzGerald space-time contraction, which is caused by "Ether wind"

  • $\begingroup$ Thanks for responding, I tried no question tries to address this particular concept. $\endgroup$ – VARUN.N RAO Jun 16 '18 at 17:37
  • $\begingroup$ My doubt is Ether wind theory and STR are said to be identical when it comes to predicting outcomes in an experiment, I just tried disprove it by giving a counter example $\endgroup$ – VARUN.N RAO Jun 16 '18 at 17:38
  • $\begingroup$ The counter example being length contraction is mutual in STR unlike in Ethere Wind Theory, where one sees contraction and other sees expansion. $\endgroup$ – VARUN.N RAO Jun 16 '18 at 17:39
  • $\begingroup$ I agree with Countto10 - it would be helpful if you identified the source of the statements in your 1st paragraph, if possible with a link to an accessible text. Also that you should look at Related questions to see if they solve your problem. $\endgroup$ – sammy gerbil Jun 16 '18 at 17:40
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    $\begingroup$ Possible duplicate of How did we disprove aether wind? or About the Ether Theory acceptance $\endgroup$ – sammy gerbil Jun 16 '18 at 17:45

In the Lorentz Ether theory the same thing takes place, we see the other guy as being shorter and slower than you are, if all observers synchronize clocks in their reference frames by Einstein signalling method, a.k.a Einstein synchronization.


That means, that moving observer is actually shorter and slower, than one that is at rest in the Ether. However, if moving observer synchronizes clocks in his frame by Einstein technique, he would measure, that clock a rest is ticking $\gamma$ times slower and measuring rod at rest is $\gamma$ times shorter than his own.

Please read the following papers:

“Full simulation of SR by classical mechanics”. All effects and paradoxes of the SR are simulated in aquatic medium (universal frame), including time dilation, Lorentz transformation, relativistic velocity addition, twin paradox, mathematical formalism e.t.c. https://arxiv.org/abs/1201.1828

Chapter 3.5.5 - A Comparison between Lorentz's Ether Theory and Special Relativity in the Light of the Experiments of Trouton and Noble https://sites.google.com/a/umn.edu/micheljanssen/home/papers

That makes these theories empirically equivalent, as the article explains: https://en.wikipedia.org/wiki/Lorentz_ether_theory

However, measuring technique by means of Einstein – synchronized clocks is not the only way to measure time dilation. Actual experiments have been conducted, and in these experiments researchers measure time dilation by means of the Transverse Doppler Effect.


Imagine that two observers rotate around a source a radiation in a circumference of very large diameter, let’s say many light years in diameter, so they are almost inertial. If they send a beam of light through the center from one point of the circumference to the opposite one, they would measure no time dilation, i.e. any Doppler shift.


Neither rotating ones, nor purely inertial ones that momentarily coincide with them would see any Doppler shift, since their clocks dilate at the same magnitude.


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  • $\begingroup$ Excellent, thank you so much, this was the through answer I was expecting $\endgroup$ – VARUN.N RAO Jun 16 '18 at 17:56

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