# Is relativity of simultaneity just a convention?

Lorentz transformations are well known to imply

1. time dilation,
2. length contraction, and
3. relativity of simultaneity.

This is prominently featured in any course on Special Relativity (SR), e.g. in Wikipedia article on SR. I perfectly understand how this follows, algebraically and geometrically, but I think relativity of simultaneity is very unlike the other two. I came to the understanding that time dilation and length contractions say something about the laws of Nature, as they can be tested: after all they explain the classic null results of Michelson-Morley, Kennedy-Thorndike, Møller rotor, etc. But that relativity of simultaneity only says something about clock synchronisation. I would like to know whether my understanding is correct.

To make it clearer what I am arguing, I would like to share an experiment with Galilean transformations, so as to get rid of time dilation and length contraction: I tried to tweak them to force relativity of simultaneity. My idea was to modify them so as to realise Einstein synchronisation in the moving frame. The results are the following transformations,

\begin{align} t' &= t - \frac{v}{c^2-v^2}x'\\ x' &= x - vt \end{align} \tag{T}

between a frame $F$ where light speed is the same in both direction and a frame $F'$, moving with respect to $F$ with a speed $v$, and where Einstein synchronisation is used (I give my demonstration below). These transformations exhibit relativity of simultaneity, since we can have $\Delta t=0$ and $\Delta t'\ne 0$ or vice-versa, but length and time are absolute as in normal Galilean relativity. The only difference with normal Galilean transforms, I argue, is the choice of clock synchronisation.

So am I correct that indeed relativity of simultaneity is just a product of clock synchronisation?

Proof that equations (T) implement Einstein synchronisation in the moving frame If a light signal is emitted from $x'=0$ toward position $x'_1$, reaching it at time $t'_1$ and then bouncing back toward $x'=0$, reached at time $t'_2$, then Einstein synchronisation posits that $t'_2=2t'_1$, i.e. in frame $F$, using transformations (T),

$$t_2 = 2\left(t_1-\frac{v}{c^2-v^2}(x_1-vt_1)\right).$$

which is indeed verified since light propagates at $c$ in $F$, and therefore

\begin{align} ct_1 &= x_1\\ c(t_2 - t_1) &= x_1 - vt_2 \end{align}

as the origin has advanced by $vt_2$ as the light signal comes back there.

• IMHO you had a good question, but ruined it by introducing unreal transformations "to make it clearer", but in reality making it a showstopper from the get-go. Following your argument is discouraging, because the results of unreal assumptions cannot be real in principle and cannot prove anything. My suggestion is to edit the question, drop everything from the words "To make it clearer" and present your case using the Lorentz transformations. Then people might find it more interesting and engaging. – safesphere Nov 13 '17 at 6:07
• A fair treatment would be like this: (1) relativity of time (2) relativity of distance (3) relativity of simultaneity. Time dilation and length contraction are not physical changes in the proper frame, but projections of the coordinates of the proper frame to the coordinates of the observer's frame. So the only thing they tell about the laws of nature is that the spacetime geometry is hyperbolic. Your difficulty may be in confusing "simultaneity" with "relativity of simultaneity". Relativity of simultaneity is a fact, but simultaneity "achieved" by the clock synchronization is its product. – safesphere Nov 13 '17 at 6:08
• @safesphere why objecting to tge hypothetical.example? If one wants to show that A does not necessaryly imply B, one can shoe a model where A holds and B not. Also strictly speaking Gallilei transfornation are also 'unreal' but does this mean one cannot use them for reasoning? – lalala Nov 13 '17 at 6:17
• @lalala You are welcome to indulge, but I just stopped reading when I saw the made up transformations. If there is a point to be made about the Lorentz transformations, why not make this point using the Lorentz transformations instead of making up some convoluted unrealistic example? The Galililean transformations are valid in the $c\rightarrow\infty$ limit just like the Lorentz transformations are valid in the flat spacetime limit. The transformations presented in the question are never valid. – safesphere Nov 13 '17 at 6:40
• @safesphere My point is that my transformations have the exact same physical content as Galilean transformations since they only differ by the choice of clock synchronisation. Besides in the limit $c\to+\infty$, we recover the usual Galilean transformations (look carefully at the transformations), as it should since then the infinite speed of light makes it the usual absolute synchronisation. – frapadingue Nov 13 '17 at 8:35

When you think of silly questions like these, the best thing to do in special relativity is think of the space/time analog, since space and time enjoy a certain symmetry.

The spatial analog of simultaneity ("happens at the same time") is "happens at the same place". For example, consider two observers at the same location as Earth -- but one stationary with respect to Earth and another moving (pretend that the Earth is an inertial reference frame). They both observe the extinction of the dinosaurs -- an event temporally separated from them.

The stationary/co-moving one measures this as having happened "at the same place" as himself, while the moving observer measures it as having happened somewhere far out in the distance.

Would you say this is just a convention? Depends on the convention in which you're defining "convention", I suppose, but if you're defining any co-ordinate measurements as just a convention, you could call everything except invariant quantities a convention.

• That "certain symmetry" between space and time you have in mind is the symmetry of Lorentz transformations, isn't it? But then I argue that those transformations rely on the hypothesis that Einstein synchronisation is in use. So that "certain symmetry" is actually the very same convention I was wondering about. – frapadingue Jun 8 '18 at 17:54
• Yes, showing that space and time are symmetric is equivalent to proving the relativity of simultaneity, but my answer doesn't provide the proof of relativity of simultaneity. From what I understood from your question, you weren't denying that simultaneity is relative, only that "simultaneity" was "just a convention". So regardless of whether you've proved yet that space and time are symmetric, you can still construct a spatial analog of simultaneity just to clear things up. – Abhimanyu Pallavi Sudhir Jun 9 '18 at 3:23

The validity of relativity of simultaneity is exactly the same as the validity of time dilation and length contraction. Consider how, in special relativity, time and space (or rather, each spatial dimension) are considered to be equivalent in the sense that they are just independent components of a vector. All special relativity does is specify that the length of this vector is to be invariant between inertial reference frames.

The Lorentz transformation can then help you to calculate what the individual values of these components may be depending on the frame you are in. In some frames, some of these components may be zero, including the one for time.

As Stephen Hawking wrote in "A Brief History of Time": "time is not absolute". That is to say, it doesn't mean much to claim that something happens at a given time. Meaning is only conveyed in describing the separation in time between two events. Special relativity generalizes this into the notion that the significant quantity between two events is the spacetime interval, and the Lorentz transformation can show that the time component of this interval can be zero, resulting in simultaneous events.