Moving emitters and absorbers of photons
Abstract
Emission
for equal directions u´ = c results in u = (c + v) / (1 + (c v)/c²) = [(c + v) / (c + v)] ∙ c = c
and for opposite directions u´ = -c (with v ≠ c) u = (-c + v) / (1 + (-c v)/c²) = [(-c + v) / (c - v)] ∙ c = -c .
If the emitter moves in the positive direction and emits a photon in the coordinate origin a five-dimensional Minkowski-diagram [1] looks like figure 1.

With u´= -c the photon moves opposite to the emitter, i.e. on the angle bisector above the negative x´-axis. This angle bisector has moved in a plane orthogonal to the framework S being rotated by the angle φ . As this photon in framework S´ moves at the speed -c on the appropriate angle bisector which is part of this orthogonal plane, it´s orthogonal projection to S is the angle bisector there too (red, left side). That means, that for the observer too the velocity of the photon always is -c .
Absorption
If a body absorbs Photons their velocity in it´s rest system always is the light speed, no matter if the emitter moves towards the absorber or opposite, i.e. if the velocities “add” or “subtract”. If the directions of emitter and photon are equal the speed v of the emitter be positive, if they are opposite negative, see fig. 2.
Fig. 2: Absorber A, Photon P and moving emitter E in the rest system of the absorber
From (1) follows for u´ = c
u = (c + v) / (1 + (c v/c²)) = (c + v) / ((c + v) / c) = c .
The result is independent of v, so of it´s sign too:
If v < 0 (-c < v < 0) the emitter moves to the left. Therefore it´s worldline in a Minkowski-diagram runs in the 2nd quadrant, above the negative part of the x-axis [1]. As it is the t´-axis as well (x´= const. = 0) and as the x´-axis because of the units s and Ls always is mirror-symmetrical to the angle bisector t = x (t´ = const. = 0) the x´-axis runs in the 4th quadrant, see fig. 3.
Fig. 3: An emitter moves away with v < 0 to the left (blue worldline) from the resting absorber and sends a photon to the right to the absorber.
tan (β/2) = tan 45° ∙ cos ψ ,
cos ψ = tan (β/2) .
According to fig. 3 applies β/2 = 45° - |α| , but because here always α < 0 , applies β/2 = 45° + α ,
and so cos ψ = tan (45° + α) . (2)
With tan α = v/c follows cos ψ = tan (π/4 + arctan (v/c))
or ψ = arccos [ tan (π/4 + arctan (v/c)) ] . (3)
cos ψ = (1 + tan α) / (1 – tan α)
cos ψ = (1 + v/c) / (1 – v/c)
or ψ = arccos [ (1 + v/c) / (1 – v/c) ] (4)
(with v < 0) .
arctan (u/c) = -arctan (v/c)
and from that ψ(u) = φ(v) .
Oppositely equal velocities have equal rotation angles.
Summary
The paradoxical phenomena that the velocity of light always is the same although the emitter or the absorber move – and that even at opposite directions – become descriptively understandable within a five-dimensional spacetime. This is a further (see [2]) indication of the existence of a fifth dimension.
References
[1] Metzler Physik, J.B. Metzlersche Verlagsbuchhandlung, Stuttgart, 1988, S. 348