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I there any problem with assuming that Lorentz transformations hold only One-way from a preferred frame, let’s say stationary with respect to CMB. Still speed of light is constant and laws of physics are the same in every inertial frame, only the clocks in CMB are fastest comparing to the clocks in other frames. Is there an experiment that shows this cannot be the case or is there a theoretical contradiction?

Moji Ghadimi
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    https://en.wikipedia.org/wiki/Lorentz_ether_theory – John Doty Oct 21 '22 at 01:20
  • If $E$ is your preferred frame, and if $\phi,\psi$ are transformations that take $E$ to the frames $F$ and $G$, then $\phi\circ\psi^{-1}$ is a transformation that takes $G$ to $F$. – WillO Oct 21 '22 at 01:36
  • Is there even such a thing as a CMB frame? The CMB is a snapshot of matter at roughly T_cosmological = 300,000 years. Why would all of that matter have been "at rest" against each other with exception of the expansion? And what about dark matter? Do we have any evidence that the dark matter frame, if there is such a thing, isn't moving relative to the matter frame? – FlatterMann Oct 21 '22 at 02:32
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    As a complement to @FlatterMann's comment, why would you expect any frame centered on any event that is part of the CMB to extend to a coordinate system that is defined here and now,? Coordinate systems are local and the CMB is very far away. – WillO Oct 21 '22 at 04:19
  • The wiki page above says: "Lorentz covariance doesn't provide any experimentally verifiable distinctions between LET and SR". They mainly focus on the speed of the light, but it can also be tested with having a very fast rocket (relative to earth that we can assume is close to the preferred frame) and another that has the same speed at the start and then accelerates down to be stationary with respect to earth and accelerates up again to join the other rocket. – Moji Ghadimi Oct 21 '22 at 23:42
  • If SR is correct the accelerating rocket’s clock will be behind the other one but if the LET is correct, it will be the other way around. Until that experiment is done, I like to stick with LET and do not use a hand wavy solution for twin paradox. – Moji Ghadimi Oct 21 '22 at 23:42
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    @MojiGhadimi that is wrong. LET predicts the same as SR in that case too. You should work out the math. You don’t have your LET correct – Dale Oct 22 '22 at 12:03
  • @FlatterMann The CMB frame is well-defined. Please see https://physics.stackexchange.com/q/25928/123208 – PM 2Ring Oct 24 '22 at 03:03
  • @PM2Ring I am aware what the surface of last scattering is. What I doubt is that all the matter in that region was expanding isotropically. The residuals are obviously smaller than the motion that causes the dipole, but that doesn't mean that they don't exist. Neither is it certain that the CMB has to be the same frame as the dark matter frame or the neutrino background frame. All of that rests, at least right now, on the assumption of homogeneity and isotropy. – FlatterMann Oct 24 '22 at 03:13
  • @FlatterMann Ok, those objections are valid. It may be very difficult for us to ever directly measure the cosmic background neutrino flux, let alone determine its momentum. We may have better luck with the dark matter, via its effects on the CMB. OTOH, as I said in https://physics.stackexchange.com/a/411082/123208 the CMB does look fairly uniform, once we account for our proper motion, although I must admit that a bit of circular reasoning is involved in determining all the components of that motion. ;) – PM 2Ring Oct 24 '22 at 03:27
  • @MojiGhadimi you made a mistake in your algebra. Your equation 3) is incorrect. If you solve for and in equation 1) then you get equation 2), not equation 3) – Dale Oct 24 '22 at 03:29
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    Thanks @Dale I remove the edit and accept your answer. – Moji Ghadimi Oct 24 '22 at 03:32

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I there any problem with assuming that Lorentz transformations hold only One-way from a preferred frame

Yes, there is a rather large problem: the math contradicts that assumption.

From the math of the Lorentz transform it is just a little algebra to get the inverse transform. The inverse transform is also a Lorentz transform. So the Lorentz transform unambiguously holds both ways.

Dale
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  • When Veratsium made a video on this, and launched a zillion one-way speed of light questions on Internet forums, I looked into it: there is a modification (a generalization, really) of the Lorentz transform that makes observations indistinguishable from an isotropic $c$ with nominal LT's. – JEB Oct 21 '22 at 04:19
  • You can argue that the mathematical inverse should hold, only if the situation is symmetric that is assuming relativity and absenvce of preferred frame. If it is asymmetric, it is perfectly fine to use it one way. Now this means that if we naively apply Lorentz transformation between two non-preferred frames b and c, we will find discrepancies with experiments and from there we can infer what the preferred frame should be but all the experiments and physical laws will be the same in any single frame. – Moji Ghadimi Oct 21 '22 at 21:16
  • In a hypothetical universe that clocks slow down by Lorentz factor if we move with respect to a preferred frame, one-way Lorentz time transformation is the only way to go. Even though the other frames cannot find out about this internally but if they compare with others they can find out. Also, if the other frames want to use their own clocks but report the same magnitude of speed with respect to the preferred frame they also need to assume length contraction. – Moji Ghadimi Oct 21 '22 at 21:43
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    @MojiGhadimi no, that is absolutely not correct. The inverse transform is simply a matter of math, no physical assumptions are involved. You calculate the inverse transform mathematically and if it is the same then it is symmetric. If is is not the same then it is not symmetric. Your objection does not hold. However, it is possible that the postulate of relativity does not hold in which case the laws of physics will not be the same, but the inverse of the Lorentz transform is a Lorentz transform as a mathematical fact. – Dale Oct 21 '22 at 21:56
  • I am saying is exactly the same thing. If postulate of relativity is not correct, we must not require reverse Lorentz transformation to hold or even to hold between two non-preferred frames. But we can still have a situation that speed of light is the same in every frame but time and length transformation only hold one-way and we still have not done any experiment to distinguish between the two. (see the rocket experiment under the question.) – Moji Ghadimi Oct 22 '22 at 01:37
  • We are not saying the same thing. There is the issue of symmetry which is a pure mathematical fact and whether or not the transformation is inertial which is a matter of experiment. If you find that it “holds” which often means that it is inertial, then the reverse transform necessarily “holds” because the symmetry is a mathematical fact. You appear to misunderstand LET both here and in your other comments – Dale Oct 22 '22 at 12:50
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When you include the CMB, you're dealing with general relativity and something like the FLRW metric. It's not special relativity, in which there is simply no way, ever, to define a preferred rest frame.

Many have looked at the CMB and claimed it defines a preferred rest frame, and it does. At a point. The problem is, that rest frame depends on position.

For instance, we can define our CMB relative rest frame here on Earth, but over at Andromeda (0.9 mega parsecs away), "their" rest frame is moving at 66 km/s relative to us.

So: no universal restframe.

JEB
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  • It is a fair point that considering the inflation, it is not clear what frame is “preferred” but what if the only real effect is that clocks slow down if we are moving with respect to the local CMB stationary frame. Then a “moving” frame has to assume length contraction too to report same magnitude of relative speed with respect to the “preferred” frame and if they do that, they also report speed c for a photon that has speed c in the preferred frame (even on-way speed). But time-dilatation is not symmetric. – Moji Ghadimi Oct 21 '22 at 23:54
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As Dale has said, the Lorentz transforms are essentially two-way. I wanted to make a further point, which is that you must misunderstand Special Relativity if you believe that clocks in one inertial frame run faster than clocks in another.

EDIT- to address the points you have made in comments, you must misunderstand LET too, since you have stated in your question that clocks 'are fastest' in the preferred frame, and in your comment on JEB's question you say that 'clocks slow down' if they are moving relative to the preferred frame, and in your comments on Dale's answer you mention clocks slowing down if they move relative to the preferred frame. Your ideas are fundamentally flawed. In SR and in LET all good clocks tick off time at a rate of one second per second. What causes the time dilation effect is not that clocks are impaired in some way by their motion so that they under-report the passage of time- what causes it is that time in one frame is systematically out of synch with time in another. That is true both in SR and in LET. The lack of synchronisation is inherently two way.

So let us suppose there was an absolute preferred frame, and Alice is sitting in it at rest, and the LET applies. Now assume Alice is passed in turn by Bob and Carol who are coasting through the preferred frame at some speed. If Alice compares her watch firstly with the time on Bob's clock and secondly with the time on Carol's, she will see that the time difference between the two encounters is longer, according to the clocks of Bob and Carol, than it is according to her own watch. IE, her watch is time dilated and appears to be running slow, even though she is sitting at rest in the preferred frame.

Marco Ocram
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  • I don't believe in that, but I also do not like the argument for the twin paradox that we stick to the frame that is inertial. Using asymmetry in the accelerating part to justify a discrepancy created by the inertial part of the travel seems logically off to me. – Moji Ghadimi Oct 21 '22 at 21:17
  • Imagine two fast jets flying together around the earth. They synchronise their clocks and then one of them lands and flies again to join the other one when it is back. I think it is more like that if they compare their clocks still the flying “more initial jet” records less passing of time. Now I like to figure out if there is any consistent mathematical description for this situation if this hypothetical result happens. Also see my comments for Dale. – Moji Ghadimi Oct 21 '22 at 21:18
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    You misunderstand LET too. See my edited answer. – Marco Ocram Oct 22 '22 at 06:15