University Physics IV · Special Relativity I · 2.9
Relativistic Velocity Transformation
One division of the boost yields three things: the rule itself, a ceiling no chain of boosts can climb, and a hard line between what a particle does and what a coordinate difference does. The lesson derives all three, then holds them against Fizeau's flowing water and a fast pion's gamma rays.
Build the model
Connect the measurement to the mechanism.
A velocity is a ratio of two coordinate differences, and a boost transforms both of them, so the velocity law is not an extra postulate — it is the Lorentz transformation differentiated. Take two neighbouring events on one particle's worldline, write dx′ = γ(dx − v dt) and dt′ = γ(dt − v dx/c²), divide, and γ cancels: u′ₓ = (uₓ − v)/(1 − uₓ v/c²). Transverse components keep their numerators, since dy′ = dy, but inherit the same transformed dt and so carry an extra 1/γ.
Everything the rule is famous for lives in that denominator. Put uₓ = c and numerator and denominator share the factor (c − v), leaving u′ₓ = c for every boost: the second postulate reappearing as algebra rather than as an assumption. Feed it two sub-luminal speeds and the output stays sub-luminal, guaranteed by the identity 1 − u′²/c² = (1 − v²/c²)(1 − u²/c²)/(1 − uₓ v/c²)², whose right-hand side cannot change sign.
What it costs is the habit of adding. Velocities are no longer summable: only rapidities add, and only for collinear boosts. It also costs a distinction nobody needed before — the law transforms one object's velocity between two frames, and says nothing about the rate at which a third frame watches the gap between two objects close, which is a coordinate bookkeeping rate, free to exceed c, and routinely does.
- Simple definition
- The relativistic velocity transformation converts a velocity measured in one inertial frame into the velocity a boosted frame measures, by transforming the displacement and the time separately and then dividing, rather than by subtracting the frame speed.
- Example
- A ship recedes from Earth at 0.60c and fires a probe forward at 0.50c relative to itself; Earth measures (0.50 + 0.60)/(1 + 0.30) = 0.846c, not 1.10c.
Turns a lab-frame velocity into the boosted frame's, and its denominator is what keeps every result short of c.
uₓ measured in S; S′ moves at v along +x; both in m s⁻¹ or as multiples of c
Sideways velocity changes although sideways length does not: the ruler is untouched, the clock is not.
γ = 1/√(1 − v²/c²) is built from the frame speed v, never from the particle's u
The form for a probe launched from a moving ship, or for light inside flowing water.
the same map with v → −v, since S moves at −v as seen from S′
Speeds do not add but rapidities do, so no finite chain of boosts can ever reach β = 1.
β = u/c and the rapidity φ are both dimensionless; φ is unbounded, tanh φ is not
The right side never changes sign, so sub-luminal maps to sub-luminal, and u = c maps exactly to u′ = c.
u and u′ are full speeds in any direction; the denominator is squared, so positive
The first is a rate a third frame assigns to a gap; the second is what A's own instruments read for B.
uA and uB both measured in the same frame S, motion collinear; only uᵣₑₗ is a body's speed
Differentiate the boost; do not invent a rule
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.
The denominator is the simultaneity offset, cashed in
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.
Transverse velocity changes though transverse length does not
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′ = uy/[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.
c is a fixed point, and everything under it stays under
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.
One object's velocity, or the gap between two
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.
What the rule has already been measured against
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.
Change one variable at a time
Make the relationship visible.
Push the particle slider to uₓ = ±1 and then sweep the boost: the marker will not leave the corner, because light is a fixed point of the map. Then set uₓ = 0.95c and see how far the curve has bent away from the diagonal — that gap is the whole departure from Galileo.
u′ₓ IN S′-0.143 c
GALILEAN uₓ − v-0.10 c
DENOMINATOR0.700
TRANSVERSE u′y/uy1.143
Live interpretationu′ₓ IN S′: −0.143 c. GALILEAN uₓ − v: −0.10 c. DENOMINATOR: 0.700. TRANSVERSE u′y/uy: 1.143
Catch the common trap
Explain before calculating.
Two probes leave a station on opposite headings, each at 0.75c as measured by the station. What speed does probe A's onboard radar measure for probe B?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA ship travels away from Earth at 0.60c along the +x axis and launches a probe straight ahead at 0.50c relative to the ship. What speed does Earth measure for the probe, and how far is that from the classical answer?
- Name the frames. S is Earth; S′ is the ship, moving at v = +0.60c along +x. The probe's velocity is given in S′ as u′ₓ = +0.50c, so the inverse form is the one to use.
- uₓ = (u′ₓ + v)/(1 + u′ₓ v/c²) = (0.50c + 0.60c)/(1 + (0.50)(0.60)) = 1.10c/1.30.
- uₓ = 0.846c. In SI units, 0.84615 × 2.998 × 10⁸ = 2.54 × 10⁸ m s⁻¹.
- Compare: the classical sum is 1.10c, so the whole correction sits in the denominator 1.30. The relativistic result is 77% of the classical one — at these speeds the correction is not a refinement, it is the answer.
Answeruₓ = 1.10c/1.30 = 0.846c ≈ 2.54 × 10⁸ m s⁻¹. Nothing crosses c, and no term was capped by hand; the denominator did it.
MediumIn the lab frame S a particle has velocity components uₓ = 0.40c and uy = 0.60c. A detector frame S′ moves at v = 0.80c along +x. Find both velocity components in S′, the particle's speed there and its direction, then verify the speed using the ceiling identity.
- Compute the shared denominator once: 1 − uₓ v/c² = 1 − (0.40)(0.80) = 0.68. It appears in both components.
- Longitudinal: u′ₓ = (0.40c − 0.80c)/0.68 = −0.40c/0.68 = −0.588c. The particle now moves backwards in S′, as expected since the frame outruns it along x.
- Transverse: γ = 1/√(1 − 0.64) = 1/0.60 = 1.667, so u′y = 0.60c/(1.667 × 0.68) = 0.60c/1.133 = 0.529c.
- Speed: |u′| = c√(0.588² + 0.529²) = c√0.6263 = 0.791c, up from |u| = c√(0.16 + 0.36) = 0.721c in the lab.
- Direction: the vector is (−0.588, +0.529), so θ′ = 180° − arctan(0.529/0.588) = 180° − 42.0° = 138.0°, against 56.3° in the lab.
- Check with the identity: (1 − 0.64)(1 − 0.52)/0.68² = (0.36)(0.48)/0.4624 = 0.3737, which is 1 − u′²/c², giving u′²/c² = 0.6263 and u′ = 0.791c as found.
Answeru′ = (−0.588c, +0.529c), speed 0.791c at 138.0° from the +x axis. Both components and the direction changed, and the speed rose from 0.721c — while staying under c.
HardLight travels at c/n through still water with n = 1.333. The water is pumped along the beam at V = 7.00 m s⁻¹. Find the light's speed in the laboratory, identify the drag coefficient, and estimate the fringe shift a Fizeau interferometer would show with 3.00 m of water in each arm at λ = 546 nm.
- In the water's rest frame S′ the light has u′ = c/n = 2.99792458 × 10⁸/1.333 = 2.2490057 × 10⁸ m s⁻¹, and S′ moves at v = +7.00 m s⁻¹ in the lab, so use the inverse form.
- u = (c/n + V)/(1 + V/(nc)). Here V/(nc) = 7.00/(1.333 × 2.998 × 10⁸) = 1.752 × 10⁻⁸, so the denominator departs from 1 only in the eighth decimal place.
- Expand to first order in V: u ≈ c/n + V(1 − 1/n²). With n² = 1.7769 and 1/n² = 0.5628, the drag coefficient is 1 − 1/n² = 0.4372.
- So the water drags the light forward by 0.4372 × 7.00 = 3.06 m s⁻¹, not the full 7.00 and not zero. The exact quotient gives u = 224 900 571.7 m s⁻¹ against c/n = 224 900 568.6 m s⁻¹ — the same 3.06 m s⁻¹.
- Send a second beam against the flow. With ℓ = 3.00 m of water per arm, Δt = ℓ/(c/n − 3.06) − ℓ/(c/n + 3.06) ≈ 2ℓ(3.06)/(c/n)² = 18.36/(5.058 × 10¹⁶) = 3.63 × 10⁻¹⁶ s.
- As an optical path that is cΔt = 109 nm, or 109/546 = 0.20 of a fringe; reversing the pump swings the pattern the other way, so the displacement actually watched for is 0.40 fringe — the few-tenths-of-a-fringe scale Fizeau was working at in 1851.
Answeru = c/n + V(1 − 1/n²) = 224 900 571.7 m s⁻¹, a drag of 3.06 m s⁻¹ from water moving at 7.00 m s⁻¹. The Fresnel coefficient 0.437 is just this transformation to first order, and it shows up as a 0.40-fringe reversal shift.