Relativistic velocity transformation

Take two events a moment apart on one particle's worldline: in S they are separated by dt, dx and dy. The boost of the previous topic gives dx′ = γ(dx − v dt), dy′ = dy and dt′ = γ(dt − v dx/c²). A velocity component is a ratio of two of these, so form dx′/dt′. The γ appears above and below and cancels; divide top and bottom by dt, which is legitimate because dt > 0 along any particle's worldline, and each dx/dt becomes uₓ: u′ₓ = (uₓ − v)/(1 − uₓ v/c²). Check the pieces. uₓ v/c² is (m s⁻¹)(m s⁻¹)/(m s⁻¹)², dimensionless, as anything subtracted from 1 must be. Let c → ∞ and the denominator is 1, returning Galileo's uₓ − v. For a bullet at 900 m s⁻¹ fired from a jet at 250 m s⁻¹ the correction term is 250 × 900/c² = 2.5 × 10⁻¹², beyond any speed measurement made at those speeds. The relativistic law is the whole law; the classical one is its c → ∞ shadow.

Read dt′ = γ(dt − v dx/c²) again. The second term is the relativity of simultaneity: two events separated by dx in S are assigned an extra time offset v dx/c² by S′, before the dilation by γ. A particle at rest in S has dx = 0, gets no offset, and transforms exactly as Galileo says, u′ₓ = −v, whatever the boost. A particle that moves in x accumulates the offset, and the faster it moves the further the denominator departs from 1. Take v = 0.60c and a particle at uₓ = 0.90c. Galileo says 0.30c. The denominator is 1 − 0.54 = 0.46, so u′ₓ = 0.30c/0.46 = 0.652c, more than twice the classical figure: the clocks of S′ tick out the particle's progress on a schedule S does not share. The denominator cannot vanish for a real particle, because |uₓ| ≤ c and |v| ⟨c give |uₓ v/c²| < 1, so the law never divides by zero and the bracket never changes sign.

dy′ = dy: a transverse displacement is untouched by the boost. But the time it is divided by is dt′, not dt, so u′_y = dy/dt′ = u_y/[γ(1 − uₓ v/c²)]. Two features matter. The γ is the frame's, built from v, never from the particle's own speed; using the particle's γ is the commonest error in this calculation. And the bracket is the same one as in the longitudinal component, so compute it once and use it for every component. Try u = (0.50c, 0.50c) and v = 0.50c: the bracket is 1 − 0.25 = 0.75, γ = 1/√0.75 = 1.155, and u′ₓ = 0 while u′_y = 0.50c/(1.155 × 0.75) = 0.577c. In S′ the particle moves purely sideways at 0.577c, slower than its 0.707c in S, so the law can lower a speed as readily as raise one. Divide the components and the direction transforms too: tan θ′ = u_y/[γ(uₓ − v)], which for u = c becomes the aberration of light, the next topic's business.

Put uₓ = c: (c − v)/(1 − v/c) = c(c − v)/(c − v) = c for every |v| < c. The speed of light is a fixed point of the map, the second postulate returned as a line of algebra. For motion in any direction, square and add the components and the identity 1 − u′²/c² = (1 − u²/c²)(1 − v²/c²)/(1 − uₓ v/c²)² follows. Every factor on the right is positive for u < c and v < c, so the left side is positive: the sign of 1 − u²/c² is invariant, and no boost lifts a particle over c. Chained boosts obey the same rule, most cleanly through the ratio K = (1 + β)/(1 − β), which multiplies under collinear composition. Two stages each boosting by 0.80c have K = 9 apiece, so K₁₂ = 81 and β = 80/82 = 0.976; three stages give 728/730 = 0.9973; ten give β = (9¹⁰ − 1)/(9¹⁰ + 1) = 0.99999999943, closer to c than any timing could resolve, yet not c. K grows without bound; β does not.

Two particles leave a source in opposite directions, each at 0.80c in the lab. The lab writes x_A = 0.80ct and x_B = −0.80ct, so the separation grows at d(x_A − x_B)/dt = 1.60c. This is correct and violates nothing: it is the rate at which a coordinate difference changes, and no object, signal or energy travels at it. Feeding 0.80c and 0.80c straight into the law would be a category error, because the law wants one frame velocity and one particle velocity measured in that frame. The relativistic question is what A measures for B. A's rest frame moves at v = +0.80c relative to the lab, and B has uₓ = −0.80c there, so u′ₓ = (−0.80 − 0.80)c/(1 + 0.64) = −1.60c/1.64 = −0.976c, the same 80/82 the composition ratio produced, because the two questions are the same boost. Closing rates up to 2c, a laser spot swept across the Moon, the crossing point of two blades: all are coordinate rates that carry nothing from place to place.

Most errors here are labelling, not algebra, so fix a procedure. Name S and S′, and let v be the velocity of S′ measured in S, sign included. Decide which frame the given velocity belongs to: a velocity in S goes into the direct form with (uₓ − v); a velocity given in S′ goes into the inverse form with (u′ₓ + v), the same law with v → −v. Compute the bracket 1 − uₓ v/c² once and use it for every component, and compute γ from v alone. For example, a frame moving at v = −0.40c, backwards along x, that watches a particle with uₓ = +0.20c has a bracket of 1 + 0.08 = 1.08 and u′ₓ = (0.20 + 0.40)c/1.08 = 0.556c. Then test: is the result below c, does it shrink to Galileo when both speeds shrink, does uₓ = v return u′ₓ = 0 as a comoving particle must, and does uₓ = c return c? A sign slipped into either bracket fails at least one of these four.