Relativistic velocity transformation

Velocity is dx/dt, and the boost has already told you what happens to dx and to dt. Take two neighbouring events on one particle's worldline. The boost gives dx′ = γ(dx − v dt), dy′ = dy, dz′ = dz and dt′ = γ(dt − v dx/c²). Form the ratio dx′/dt′: the γ appears once above and once below and cancels, leaving (dx − v dt)/(dt − v dx/c²). Divide numerator and denominator by dt, which is legitimate because both events lie on a timelike worldline, and every dx/dt becomes uₓ, giving u′ₓ = (uₓ − v)/(1 − uₓ v/c²). No new physics entered; the whole law is the boost applied twice and one division. Check the limit before trusting it. When |uₓ v| ≪ c² the denominator is 1 to first order and u′ₓ → uₓ − v, which is Galileo. For two cars at 30 m s⁻¹ each, the correction term uₓ v/c² is 1.0 × 10⁻¹⁴ — which is why the classical rule survived two centuries of laboratory work unchallenged.

Look at where the denominator comes from: dt′ = γ(dt − v dx/c²). That second term is the relativity of simultaneity — clocks in S′ spread along x disagree with S about what "now" means, and the disagreement is proportional to dx. A particle sitting still in S has dx = 0, contributes no offset, and transforms exactly as Galileo says: u′ₓ = −v, with no correction at any boost speed whatever. Only a particle that moves in x accumulates the offset, and the faster it moves the larger it grows. Take v = 0.80c and a particle at uₓ = 0.50c. The denominator is 1 − 0.40 = 0.60, the numerator is 0.50c − 0.80c = −0.30c, and u′ₓ = −0.50c where Galileo predicted −0.30c, so his answer falls 40% short of the right one and the whole shortfall is produced by the clocks. Note also that for |uₓ| ≤ c and |v| < c the product uₓ v/c² can never reach 1, so the denominator stays positive and the map never blows up.

dy′ = dy, so nothing at all happens to a transverse displacement — but the time it is divided by has changed, and that is enough. u′_y = dy/dt′ = u_y/[γ(1 − uₓ v/c²)], and the γ here is built from the frame speed v. Using the particle's own γ is the commonest slip in this calculation. Work one case. In S a particle moves purely sideways at u = (0, 0.80c); boost to S′ at v = 0.60c, so γ = 1.25 and the denominator 1 − uₓ v/c² is exactly 1 because uₓ = 0. Then u′ₓ = −0.60c and u′_y = 0.80c/1.25 = 0.64c. The speed is √(0.36 + 0.4096) = 0.877c, larger than the 0.80c it had in S, and the direction has swung from straight up the y-axis to 133.2° from the +x axis. Both magnitude and direction transform; the only thing preserved is the ceiling.

Set uₓ = c and the rule returns (c − v)/(1 − v/c) = c(c − v)/(c − v) = c, for every v short of c. Light is a fixed point of the map, which is exactly the second postulate — except that here it is a consequence of the boost rather than an input to this step. For motion in any direction the guarantee is the identity 1 − u′²/c² = (1 − v²/c²)(1 − u²/c²)/(1 − uₓ v/c²)². Both bracketed factors on the right are positive for sub-luminal u and v, and the squared denominator is positive, so 1 − u′²/c² cannot go negative: u < c in one frame means u′ < c in all of them. Rapidity says the same thing in one line. With β = tanh φ, collinear boosts add rapidities, and tanh is asymptotic to 1 however large its argument. Two boosts of 0.9c compose to 0.99448c, and ten successive boosts of 0.5c — each performed by the crew the previous one left behind — compose to (3¹⁰ − 1)/(3¹⁰ + 1) = 59048/59050 = 0.999966c, not 5c.

The rule has two inputs and they are not interchangeable: v is the velocity of one frame relative to another, and uₓ is a particle's velocity measured in the first of those frames. It is not a machine for combining two velocities measured in the same frame. When a station sees probes leaving at 0.9c in opposite directions, the station's own account is that the distance between them grows at 1.8c. That number is correct and forbids nothing: d(Δx)/dt belongs to a pair of worldlines, and nothing — no object, no signal, no energy — travels at it. The relativistic question is a different one: what does probe A measure? Change frames, and the answer is (0.9 + 0.9)/(1 + 0.81) = 0.99448c. The same distinction covers a laser on Earth sweeping its spot across the Moon: at r = 3.84 × 10⁸ m and ω = 1.0 rad s⁻¹ the spot moves at 3.8 × 10⁸ m s⁻¹, about 1.28c, because a different photon arrives at each place and nothing is carried along the surface.

Fizeau, 1851. Light travels at c/n in still water; run the water at V along the beam and the inverse form gives (c/n + V)/(1 + V/(nc)), which to first order is c/n + V(1 − 1/n²). That bracket is a drag coefficient, 0.437 for water at n = 1.333, so the medium carries the light forward by well under half its own speed — neither the full V a rigidly dragged medium would give nor the zero a stationary one would. Fizeau measured that fraction by interferometry half a century before the transformation that explains it, and Fresnel had already fitted the same coefficient to an aether model that did not survive; the third worked example below puts numbers on it. The other end of the range is the fixed point itself. At CERN in 1964 a group led by Alväger let neutral pions travelling at β = 0.99975 decay into gamma rays and timed those photons over a flight path: they arrived at 2.9977 × 10⁸ m s⁻¹, within about a part in 10⁴ of c, and not at the 6.0 × 10⁸ m s⁻¹ that any theory adding the source's speed to c would demand. Slow flow and a near-light source therefore test the same denominator at its two extremes.