The conventionality thesis in physics concerns the *conventionality of simultaneity*, which states that the choice of a characteristic synchrony is a convention, not an observable. This arises because the speed of light in a vacuum can only be measured as a two-way speed, so the one-way speeds are either taken to be the same (the characteristic answer) or two speeds whose harmonic mean speed is the constant *c*. This is expressed as *c */ (1±*κ*), where *κ* is between 0 and 1 (see *here* and references), and the characteristic value for *κ* is 0.

As pointed out in *Lorentz generalized*, different observers (or travelers) may have different characteristic (modal) speeds for various reasons. If one accepts that the characteristic round-trip speed is a constant for all travelers (or observers), that restricts the characteristic one-way speeds but still allows different possibilities.

Let there be observer-travelers going in the same direction but in different vehicles (or trains, boats, etc.). Distinguish them by their frame of reference, unprimed or primed. Call their frames *S* and S’, their positions in space *r* and *r´*, in time *t* and *t´*, the actual speed of the second frame relative to the first* v,* and their reference travel speeds *a* and *b* respectively.

Consider only the path/trajectory followed, i.e., one dimension of space and time each. Then we have: *r = at* and *r´ = bt´* as time-space conversions for each frame. To proceed like the previous derivations of the Lorentz transform, let there be a factor *γ* for each equation:

*r´* = *γ *(*r – vt*) = *γt *(*a – v*) = *bt´*, and

*r* = *γ *(*r´ + vt´*) = *γt´ *(*b + v*) = *at*.

Multiply these together and divide out *tt´* to get:

*γ*² (*a – v*)(*b + v*) = * ab*, so that

*γ*² = *ab* / ((*a – v*)(*b + v*)).

Now let *a* = *c*/(1–*κ*) and *b* = *c*/(1+*κ*). Then

*γ*² = *c²* / ((*c + κv – v*)(*c + κv + v*)).

If *κ* = 0, then *a = b = c*, and there is only one reference speed for both traveler-observers, which is the requirement of the Lorentz transformation:

*r´ = γ* (*r – vt*) and *t´ = γ* (*t – rv/c²*) with *γ*² = 1 / (1 – v²/c²).

On the other hand, if *κ* → 1, then a → ∞ and b → c/2, so γ² = 1 / (1 + 2v/c) with

*r´ = γ* (*r – vt*) and *t´ = γ**t*, which is the Galilean transformation with a factor.

Thus two observer-travelers don’t have to agree on a characteristic one-way speed.