Submitted:
01 April 2026
Posted:
02 April 2026
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Abstract
This paper critiques the established loss of simultaneity in special relativity which comes from Minkowski diagrams. Einstein's original thought experiment, with a train (observer M’), an embankment (observer M) and simultaneous lightnings, will become our test. For our purpose, lightnings will become photons. By applying the two postulates of special relativity (speed of light and principle of relativity), the paper shows that simultaneity should be observed by both observers. This would imply a superposition of some kind, as the photons meet simultaneously M and M’ while they are not at the same position. By using Lorentz invariance (therefore pure calculation), the conclusion of simultaneity for both observers will be confirmed. The superposition could be a quantum superposition. The conclusion is that Minkowski diagrams with their oblique coordinates are probably correct but, lacking the idea of superposition, fail to fully describe special relativity.
Keywords:
simultaneity
; spacetime
; Lorentz transformation
; Minkowski diagrams
; special relativity
; quantum superposition
1. Introduction
This paper questions the well-established loss of simultaneity for a moving reference with special relativity. Papers presenting the return of simultaneity are difficult to find. In addition, it would be difficult to publish such papers, as few journals would accept such a claim. John Field’s (2009) paper presents an excellent review.
This work starts with the study of the history of special relativity. We will question Einstein’s chosen example for the relativity of simultaneity. Then, we will show a difference between Minkowski spacetime diagram (1909) and the two postulates at the origin of special relativity. That difference will be confirmed by some calculations involving Lorentz invariance and Lorentz transformation (1904).
There seem to be many Minkowski spacetimes (M is for Minkowski in kappa-M, M-M, half M, M-Bouligand, Brunn-M, quantum M, φ-shell M, etc.); that is the first sign that Minkowski is not satisfactory. Questioning Minkowski is not new; for example, the geometric structure of Minkowski spacetime is discussed in (Zilber,2025). Lorentz invariance violation is a large subject as it also involves, for example, the breaking of CPT symmetry (an important part of quantum mechanics). Shantanu Desai has written a review relevant for our case (2023), and if Giulia Gubitosi is correct, no violation has been observed (2022). But if the idea of violation exists, it means Lorentz transformation is not perfect either (or still misunderstood?). Minkowski’s work will be shown as correct but incomplete, and there will be no need to question Lorentz invariance/transformation in this work.
If papers on invariance violation and Minkowski spacetime are numerous, there does not appear to be one reaching similar conclusions to this paper; the closest is probably (Ganesan,2024) but that is about quantum theories.
2. The History of the Loss of Simultaneity with Special Relativity
In Section 9 of Relativity, Einstein (1916) uses the example of two simultaneous bolts of lightning at equal distance from the observer M on the embankment as on Figure 1. The bolts strike when the observer M’ is at the same place as observer M. After the strikes, observer M receives the two signals simultaneously, while observer M’ has moved with the train since the strikes, and s/he receives one signal and then the other. The conclusion would be the loss of simultaneity for observer M’.
Is Einstein’s scenario a logical way to observe simultaneity? Let’s replace light by the sound of the strikes. The biggest advantage of sound waves is that, because the speed of sound is small compared to the speed of light, special relativity can be ignored. With sound (and no wind), common sense tells us that it wouldn’t matter if you are immobile or not. As long as you are in the middle when sound waves arrive, you would observe simultaneity. So with Einstein's scenario, M’ is not in the middle and receives the sound wave from B before the sound wave from A; as with light. So the non-simultaneity observation for M’ is expected; but why is this example used to describe as a loss of simultaneity? The only possible reason is that simultaneity was expected with special relativity as we will see now.
With the principle of relativity, no inertial frame of reference should be preferred; M is immobile but M’ can consider itself as immobile therefore should observe simultaneity. Minkowski will destroy this argument as we will see later on.
Einstein told us that the speed of light is constant for M and M’. Using M coordinates, AM’=BM’ at the moment of the strikes, M knows that the speed of light is relative to him but also relative to M’. Therefore for M’, the speed of photon A equals the speed of photon B; because AM’=BM’, M’ must observe simultaneity even from the viewpoint of M. This is the key argument, and it will be confirmed in the Appendix A. This argument has been overlooked by Minkowski despite the fact that, with Minkowski diagram, AM’=BM’ with M and M’ coordinates (see Figure 2).
To show the apparent loss of simultaneity, a Minkowski diagram is needed as in Figure 2.
What Minkowski has done is to represent what M is observing; M thinks that M’ receives the wave from B before the wave from A. That is correct with sound, and indeed he could have drawn the same diagram for sound; Minkowski diagrams may not be specific to special relativity. Minkowski’s reasoning is as follows: A and B for M are on the x axis; they are not on the x’ axis so they are not simultaneous for M’. With this reasoning there is no preferred reference because M’ immobile can be drawn in a similar way; but simultaneity for M’ will involve points A’ and B’ which are different to points A and B. For Minkowski, M and M’ don’t share the same spacetime therefore the difference between A and A’, and B and B’. The speed of light is the same for M and M’ so apparently Minkowski is also in agreement with the universality of the speed of light. Minkowski used Lorentz transformation and deduced the oblique coordinates (the t’1A and t’1B) as can be seen in the Appendix A. Minkowski is indeed very convincing.
But, with the speed postulate alone, photons A and B should meet at the same time M and M’ even if M’ is no longer in the middle. In 1908 when Minkowski presented his diagrams, photons A and B at two places at the same time was simply impossible. No coherent quantum theory with quantum superposition was written before 1925 (Bohr,1928). Photons are particles so they can be in a state of superposition, and in the Appendix A, we are going further than Minkowski by showing that photons A and B are at the same time at M and M’; simultaneity for M’ is back.
The appendix of this paper is essential. It is based on Lorentz transformation, so it is arithmetic which has been made as easy as possible with all the necessary comments to help you along the way; so please take time to read it. Hopefully you will agree that simultaneity is back. With Minkowski, simultaneity for M’ requires photons A’ and B’ which do not have the same spacetime story as A and B. But A’ and B’ different to A and B is a mathematical trick. With Einstein’s scenario, there is no A’ and B’; as AM’ = BM’ when lightnings strike, that is enough to show that M’ should observe simultaneity as confirmed in the Appendix A.
The result of the appendix is presented in Figure 3. The oblique coordinates of Minkowski have disappeared on the figure, but the appendix shows that the angle of x’ axis could be correct. The problem is that there would be two additional scales on the same x’ axis: one for photon A and a different one for photon B. Both dashed lines joining M’ represent the speed of light because of those two different scales. Also, according to the Appendix, photons A and B reach M and M’ at the same time, as described with the horizontal dotted lines.
3. Discussion
Please do check all this as I am not impartial: to explain the EPR paradox (Danis,2025), I needed simultaneity. While writing (Danis,2025), for me, Minkowski was simply a visualisation of the two original postulates. When I finally understood Minkowski, I realised the contradiction with Einstein’s postulates, hence this paper.
Minkowski is not wrong, but there is something missing, which exists in Lorentz transformation.
Time dilation is still a problem in my mind. In the Appendix A, the photons meet after 2s for M. M needs to observe time dilation on the clock of M’ in Figure 3 But the proper time of M’ is also 2s when photons meet. M’ observes time dilation on the clock of M. This doesn’t make sense. How can both time dilations be represented in one diagram? What is the time dilation observed by M on the clock of M’? The t’1A and t’1B are part of the solution as in the Appendix A; but t’1A and t’1B are time intervals, not time dilation. But I have the feeling that M observes two different time dilations for the clock of M’. If that is correct what is time dilation? I had presented an identical problem with a photon clock in (Danis,2024).
Similar to t’1A and t’1B which are important to reach simultaneity, there must be x’1A and x’1B which would shift A and B along the x axis to give back the 45° angle for light and reinstate Minkowski diagrams. Variables x’1A and x’1B will probably be involved but it is not so simple and that wasn’t my priority for this paper.
It seems the origin of quantum superposition could be found in Lorentz transformation; and it would be a spatial and temporal superposition as suggested in (Danis,2025). My feeling is that time dilation is the reason for quantum superposition; but the logic is not apparent. Is quantum superposition real? Or is it a trick due to time dilation? And by trick, I mean superposition is real to us because we have only one clock.
I don’t like the term fundamental theory because what is fundamental today may not be so tomorrow; but for me, Lorentz transformation is today a fundamental theory of physics. It is a 1D theory that can be applied to 3D, it should be at the foundation of quantum mechanics; the paper (Danis,2024) is far from perfect, but arguments for the 1D idea and the history of Lorentz transformation are presented.
Simultaneity is back because M and M’ share the same point when bolts strike. Because of that same point, it seems there is superposition. Is Lorentz to be applied only to quantum mechanics?
Please do come back to me, and leave a comment. Is there a mistake and where? Is it an important mistake? Do I jump to conclusions too fast and if so, where? Is there any observation that contradicts this work? Is it possible that I am on to something important? Whilst I don’t ask you to validate my work, it would be helpful to know that you haven’t found any mistake. I would greatly appreciate discussion, and not only with professionals.
4. Conclusion
If this paper is correct, simultaneity is back, Minkowski is incomplete, special relativity could explain quantum mechanics.
Acknowledgments
Thanks to Andreas for clarifying that A and A’ are not identical with Minkowski, a small remark with such an important impact.
Appendix A
Spacetime intervals
We try to check that time AM’ and BM’ are equal. We will use relative units (already used by many physicists): speed is 1 instead of c and distance is 1 instead of c time 1s. We will have equations where seemingly time=distance; c is the factor to transform time into a distance. M’ is moving at 0.99c (therefore 0.99) towards B (faster than on Figure 2&3). Variables x, y, z and t are for M; x’, y’, z’ and t’ are for M’. The question is when (and where) A and B meet M’ seen from M. Spacetime interval squared with relative units (therefore the usual c2 in front of t2-t1 has disappeared) is:
ds2 = (x2-x1)2 + (y2-y1)2 + (z2-z1)2 – (t2-t1)2
When the events are dictated by the speed of light, ds2 and ds’2 should be zero by definition. It is used to check that Lorentz transformation is correct.
A.1. Spacetime Interval for AM’
M is at x=0 (the origin), point A is at -2c (therefore x1A = -2), B is at +2. I prefer +2 and -2 because 1/1=1 which could become confusing.
Starting event (event 1): x1A = -2 y1 = 0 z1 = 0 and t1A = 0 (photon leaves A)
Final event (event 2): x2A = ? y2 = 0 z2 = 0 and t2A = ? (photon reaches M’)
We don’t know where M’ is when the photon reaches her/him. The relation between x2A and t2A is obviously the speed of light: t2A = 2+ x2A in relative coordinates. 2+ x2A is the distance covered by photon A to reach M’.
ds2 = (x2A + 2)2 + 0 + 0 – (2+ x2A)2 = 0
The same two events for M’ give:
x’1A = ? y’1 = 0 z’1 = 0 and t’1A = ? (photon leaves A)
x’2A = ? y’2 = 0 z’2 = 0 and t’2A = ? (photon reaches M’)
Lorentz transformation gives:
t’1A = γ (t1A - (v.x1A)) = γ (2v)
t’2A = γ (t2A - (v.x2A)) = γ ( 2+(1-v) x2A)
x’1A = γ (x1A - v.t1A) = γ (-2 - 0) = -2 γ
x’2A = γ (x2A - v.t2A) = γ ((1-v)x2A – 2v )
ds’2 = ( γ ((1-v)x2A – 2v+2 ))2 + 0 + 0 – (γ ( 2+(1-v) x2A -2v))2 = 0
The value of x’1A was expected (that is the special relativity length contraction), please note that the value of t’1A is not zero despite that fact that t1A = 0. Check Figure 2 and you will see the first clue for Minkowski to transform M’ coordinates into oblique coordinates.
A.2. Spacetime Interval for BM’
We keep the same origin: M is at zero.
Starting event: x1B = 2 y1 = 0 z1 = 0 and t1B = 0 (photon leaves B)
Final event: x2B = ? y2 = 0 z2 = 0 and t2B = ? (photon reaches M’)
As previously, x2B and t2B are linked by the speed of light: t2B = 2 -x2B
ds2 = (x2B -2)2 + 0 + 0 – (2-x2B)2 = 0
The same two events for M’ give:
x’1B = ? y’1 = 0 z’1 = 0 and t’1B = ? (photon leaves B)
x’2B = ? y’2 = 0 z’2 = 0 and t’2B = ? (photon reaches M’)
Lorentz transformation gives:
t’1B = γ (t1B - (v.x1B)) = γ (-2v)
t’2B = γ (t2B - (v.x2B)) = γ ((2-x2B) - (v.x2B)) = γ (2-(1+v)x2B )
x’1B = γ (x1B - v.t1B) = γ (2 - 0) = 2 γ
x’2B = γ (x2B - v.t2B) = γ ( x2B - 2v + v.x2B ) = γ (-2v + (1+v)x2B )
ds’2 = (γ (-2v+(1+v)x2B -2))2 + 0 + 0 – (γ (2-(1+v)x2B +2v)2 = 0.
Again, note that the value of t’1B is not zero despite that fact that t1B = 0. Check Figure 2 and you will see the second clue for Minkowski to transform M’ coordinates into oblique coordinates.
Minkowski used also x’1A and x’1B; with t’1A and t’1B he had enough information to create his diagrams and his maths. For him the only x2B = 0 and t2B = 2. So there are no x’2A, x’2B, t’2A and t’2B.
A.3. Joining the Dots i.e., Linking A.1 and A.2.
This part is where we will go further than Minkowski, because we believe that M’ reaches also the conclusion of simultaneity. That is why, in the part above, variables t’2A, t’2B, x’2A and x’2B are unknown (where photons meet M’). Minkowski would have never tried to determine those variables. Today, they are deemed possible only because superposition seems to exist and because we believe in the two postulates.
The distance x2A is from M to M’ in M reference, by adding the distance from point A to M we obtain the distance travelled by photon A. Back to the distance x2A which is from M to M’ in M reference, the distance 2-x2B has been travelled by photon B from point B to M’ in M reference. The distance from M to B is 2; therefore
x2A + (2-x2B) = 2 that is x2A = x2B
It means M’ is at x2A = x2B in M reference when the photons meet. We must find a value with those equations:
x2B = x2A.
t’1A = γ (2v)
t’2A = γ (2+(1-v) x2A)
t’1B = γ (-2v)
t’2B = γ (2-(1+v)x2B )
x’2A = γ ((1-v)x2A – 2v )
x’2B = γ (-2v + (1+v)x2B )
If I assume there is simultaneity (therefore superposition, the step Minkowski couldn’t do), (t’2A - t’1A)= (t’2B – t’1B), and if x2B is replaced by x2A, we can find one value of x2A.
(t’2A - t’1A) = γ (2+(1-v) x2A - 2v) = γ (2-(1+v)x2A +2v) = (t’2B – t’1B)
x2A = 2v = 1.98
The work so far means that, for M’, the distance travelled by photon A equals the distance travelled by photon B. For M; photon A has travelled (2+1.98) and photon B has travelled 0.02 but in M’ reference those distances are equal. Lorentz transformation is extremely clever in the sense that the deformation of spacetime seems to be different in front or behind M’. Similarly, in a photon clock, the deformation of spacetime seems to be different depending on the direction of the photon relative to the motion of the clock.
This result tells us that the place where photon A, photon B and observer M’ meet is 1.98 away from M in M reference. How long does it take for M’ to travel 1.98c? 2s in M reference. Thus, from M reference, we can deduce that M’ is at the right place and observes simultaneity. M’ can consider her/himself immobile, 2s is what is needed for the photons to reach M’. We assumed that (t’2A - t’1A)= (t’2B – t’1B) because of the speed of light relative to M’, I believe this result justifies the assumption but a rigorous demonstration will be required. Not only do M and M’ observe simultaneity but they observe it at the same time.
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Figure 1.
Train and embankment as used by Einstein.
Figure 2.
Minkowski spacetime diagram. note: t’1A is the time interval between the t’=0 of M’ (x’ axis) and the strike A, t’1B is between strike B and t’=0; those time intervals are used in the Appendix A.
Figure 2.
Minkowski spacetime diagram. note: t’1A is the time interval between the t’=0 of M’ (x’ axis) and the strike A, t’1B is between strike B and t’=0; those time intervals are used in the Appendix A.
Figure 3.
Lorentz spacetime diagram for the two scenarios. Note. The arrow with M’ is the world line in M reference, not necessarily the t’ axis. All dashed lines represent the trajectory of a photon. A change of scale is necessary to account for the different angles.
Figure 3.
Lorentz spacetime diagram for the two scenarios. Note. The arrow with M’ is the world line in M reference, not necessarily the t’ axis. All dashed lines represent the trajectory of a photon. A change of scale is necessary to account for the different angles.
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