Is relativity of simultaneity just a convention? Lorentz transformations are well known to imply


*

*time dilation,

*length contraction, and

*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.
 A: 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.
A: 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.
