I’ve discussed before how there may be two different one-way speeds of light since they are matters of convention. See here and here. This post is concerned with the transformation between two observers in such a case. I continue to work with the standard configuration, see here.
The standard transformation of reference frames begins with two frames in uniform relative motion along one axis (usually called x). Here we take the spatial axis to be the r-axis, which parallels the spatial axis of motion. Similarly, the temporal axis is taken to be the t-axis, which parallels the temporal axis of motion.
The notation here is indifferent as to the existence of other dimensions. If they exist, they are orthogonal to the direction of motion, whether spatial or temporal, and their corresponding values are the same for both frames. One can generalize the results here to other directions by rotation.
The two frames are differentiated by primed and unprimed letters. They coincide at time t = 0 and their relative speed is v. The one-way speed of light is c1 in one direction and c2 in the opposite direction such that the round-trip speed equals c, the standard speed of light in a vacuum. That is, the harmonic mean of c1 and c2 equals c, i.e., (1/c1 + 1/c2)/2 = 1/c.
The trajectory of a reference particle or wave that travels at the speed of light follows these equations in both frames:
r = c1t (or r/c1 = t) and r′ = c2t′ (or r′/c2 = t′).
Consider a point event such as a flash of light that is observed from each reference frame. How are its coordinates in each frame related?
The basic relations are: r′ = r − tv = r (1 – v/c1) and t′ = t (1 – v/c1). Next a factor, γ, is included in the transformation equations:
r′ = γ (r − vt) = γr (1 – v/c1), and
t′ = γt (1 – v/c1),
with equal values for the other corresponding primed and unprimed coordinates. The inverse transformations use the other speed of light:
r = γ (r′ + vt′) = γr′ (1 + v/c2), and
t = γt′ (1 + v/c2).
Multiply each corresponding pair together to get:
rr′ = γ²rr′ (1 – v/c1)(1 + v/c2).
Dividing out rr′ yields:
1 = γ2 (1 – v/c1)(1 + v/c2).
Solving for γ leads to:
γ = [(1 – v/c1)(1 + v/c2)]–1/2, which applies if |v| < |c1|.
This is the Lorentz transformation with two one-way speeds of light.
In the extreme case, the speed of light in one direction is infinite, which is the Galilean transformation (or its complement). In that case, the speed of light in the other direction would be c/2, which leads to
γ = (1 – 2v/c)–1/2.
r′ = γ (r − vt) = r (1 – 2v/c)1/2, and
t′ = γt (1 – 2v/c) = t (1 – 2v/c)1/2.