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So I definitely don’t think there is an ether for starters. I am reading electrodynamics by Griffith and he states that Einstein’s second postulate which says “The speed of light in vacuum is the same for all inertial observers, regardless of the motion of the source” disproves the ether. Griffiths then says the principle of relativity is not enough to disprove the ether.

My thought process is that the ether would imply an absolute reference frame which would violate the first postulate but Griffiths claims that is not true. I have been thinking about it a lot and I can’t seem to find a clear argument just from Einsteins two postulates which disproves the ether.

Can anyone give me a clear argument as to how this works? Also in this thought experiment I am not using the wave particle duality of light because then it’d be obvious.

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    $\begingroup$ You might find this helpful: physics.stackexchange.com/questions/613722/… $\endgroup$
    – Lambda
    Commented Apr 17, 2022 at 5:35
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    $\begingroup$ It follows from Einstein's theory of relativity that - from a third point of view - EM radiation is slower in the vicinity of a larger gravitational potential than in a lower-mass region. Ergo, the gravitational potential of the masses in space is a kind of ether. $\endgroup$
    – HolgerFiedler
    Commented Apr 22, 2022 at 14:37

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Nothing disproves the ether. Lorentz and Poincaré created a theory, equivalent to special relativity, that incorporated an unobservable ether. But Einstein's approach was simpler and clearer, especially with Minkowski's geometric interpretation, so that's what we use.

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    $\begingroup$ I see. So is it more that Einstein’s postulates eliminated the need for the ether so we just don’t use it? $\endgroup$
    – JosephSanders
    Commented Apr 16, 2022 at 23:13
  • $\begingroup$ yes............. $\endgroup$
    – Math Keeps Me Busy
    Commented Apr 17, 2022 at 1:35
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    $\begingroup$ Please see the famous Michelson–Morley experiment which is a strong evidence against the ether $\endgroup$
    – Jasoba
    Commented Apr 17, 2022 at 16:30
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    $\begingroup$ @JacobBach Ah, but it's not. The Lorentz-Poincaré theory of the ether accounts for the Michelson–Morley result just fine. $\endgroup$
    – John Doty
    Commented Apr 17, 2022 at 18:53

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