Two one-way characteristic speeds

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 , in time t and , 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:

= 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:

γ² (a – v)(b + v) = ab, so that

γ² = ab / ((a – v)(b + v)).

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

γ² = / ((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 γ² = 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´ = γt, which is the Galilean transformation with a factor.

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