Relativistic Velocity Addition

Derivation

Relativistic Velocity Addition

D-095 Home PU-105 Threads light · symmetry Depends on Lorentz Transformation from the Two Postulates

Statement

Given a frame S′ moving with velocity v along the common x-axis of an inertial frame S, an object moving with velocity u′ along x as measured in S′ has velocity in S given by u = (u′ + v)⁄(1 + u′v⁄c²). This composition law is exact for all subluminal u′ and v; it returns c whenever either input equals c, and never produces a superluminal result from two subluminal inputs.

Why it matters

Galilean addition u = u′ + v is the most intuitive statement in kinematics and the first casualty of relativity. Replacing it with the correct law is what makes the invariance of the speed of light logically consistent rather than paradoxical: light emitted from a moving source still travels at c in every frame, and no chain of boosts of massive bodies can accelerate anything past c.

The law is also the kinematic backbone of the relativistic Doppler effect, aberration of starlight, the Fizeau drag experiment, and the interpretation of collider rapidities. It is one of the sharpest quantitative tests distinguishing Lorentz from Galilean physics.

Assumptions

The Lorentz transformation between S and S′ is already established.Without it we have no relation between the two frames' coordinates, and no composition law can be derived at all — only postulated.

S′ moves with constant velocity v along the shared x-axis; axes are parallel and clocks synchronised at the origin coincidence.A boost in a non-collinear direction couples the transverse components and the simple 1-D formula is replaced by the full velocity-transformation with a γ factor on the transverse parts.

The object's motion is along x so only one velocity component is nonzero.For general 3-D motion the transverse components transform differently (they pick up a 1⁄γ and share the same denominator), so the single-line formula applies only to the longitudinal component.

Velocity is defined as the ratio of coordinate differentials dx⁄dt in each frame, i.e. we compare a single worldline's tangent in two charts.If “velocity” instead meant a closing speed measured within one frame between two separate objects, the quantity is frame-internal and can legitimately approach 2c; conflating the two is the classic source of “faster than light” errors.

Derivation

x = γ(x′ + v t′), t = γ(t′ + v x′⁄c²), γ = 1⁄√(1 − v²⁄c²)

Write the inverse Lorentz transformation expressing S-coordinates in terms of S′-coordinates; this is the prior result we are permitted to assume. A

dx = γ(dx′ + v dt′), dt = γ(dt′ + v dx′⁄c²)

Take differentials. The transformation is linear with constant coefficients (γ and v are frame constants), so the differentials obey exactly the same relations as the coordinates. A

u = dx⁄dt = [γ(dx′ + v dt′)] ⁄ [γ(dt′ + v dx′⁄c²)]

Form the velocity in S as the ratio of the two differentials. This is the definition of instantaneous velocity along the worldline. A

u = (dx′ + v dt′) ⁄ (dt′ + v dx′⁄c²)

The common factor γ cancels identically between numerator and denominator — the composition law is independent of the boost's time-dilation factor. A

u = (dx′⁄dt′ + v) ⁄ (1 + (v⁄c²)·dx′⁄dt′)

Divide numerator and denominator by dt′ (nonzero for a timelike or any physical worldline). This exposes the S′-velocity dx′⁄dt′ as the only kinematic input. A

u = (u′ + v) ⁄ (1 + u′v⁄c²), u′ ≡ dx′⁄dt′

Identify u′ = dx′⁄dt′ as the object's velocity in S′. Because the differentials were exact and no expansion was made, this holds for all speeds, not just small ones. B

c − u = (c − u′)(c − v) · c ⁄ (c² + u′v)

Rearranging the result to isolate c − u shows the numerator is a product of two non-negative factors when u′, v ≤ c, so u ≤ c necessarily — the closure property is algebraic, not asymptotic. C

Result

u = (u′ + v) ⁄ (1 + u′v⁄c²)

Reading. To get the object's speed in the ground frame, add its speed in the moving frame to the frame's speed, then divide by a correction factor that grows as the two speeds approach c. The denominator is what keeps every composition of subluminal speeds subluminal and pins any speed equal to c at c.

Units check. Numerator u′ + v has units of m s⁻¹. In the denominator u′v⁄c² is (m s⁻¹)²⁄(m s⁻¹)² = dimensionless, so 1 + u′v⁄c² is dimensionless and u comes out in m s⁻¹. Consistent.

Limiting cases

Low speed (u′v « c²): denominator → 1, recovering Galileo's u ≈ u′ + v.
Light input (u′ = c): u = (c + v)⁄(1 + v⁄c) = c for any v — the invariance of c is built in.
Two half-light speeds (u′ = v = c⁄2): u = c⁄(1 + 1⁄4) = 0.8c, not c.
Equal and opposite (u′ = −v): u = 0, as required by symmetry.
Maximal case (u′ = v = c): u = 2c⁄2 = c; light plus light is still light.

Breaks when

Non-collinear boost or transverse motion. If u′ has components off the boost axis, the single-line formula fails; the transverse components transform as u_y = u′_y⁄[γ(1 + u′_xv⁄c²)] and must be handled separately.
Non-inertial frames. If S or S′ accelerates or rotates, there is no global Lorentz transformation, γ is not constant, and the derivation's differential step is invalid; only local (instantaneous-comoving-frame) application survives.
Phase or group velocities exceeding c. The law constrains signal/energy velocities of real worldlines; a phase velocity u′ > c can be fed in and yields nonsense as a “particle speed” because it is not the velocity of any matter.
Superluminal inputs generally. Feeding u′ > c can drive the denominator through zero (at u′v = −c²), producing infinite or negative u — a signal that tachyonic inputs violate the derivation's premises.

Failure modes

Adding directly: writing u = u′ + v and getting 1.6c for two 0.8c inputs — forgetting the denominator entirely.
Sign of v: using the direct transform coefficients (with −v) when the object is in S′, flipping the numerator sign. Always match the transform direction to which frame the velocity is measured in.
Squaring c wrong: writing u′v⁄c instead of u′v⁄c², which is not even dimensionless.
Closing-speed confusion: claiming two particles approaching at 0.9c each “violate relativity” at 1.8c. The 1.8c closing rate is measured in one frame between two objects and is perfectly legal; the formula answers a different question (one object's speed in the other's rest frame → 0.994c).
Applying to transverse components: using (u′_y + 0)⁄(1 + 0) and forgetting the 1⁄γ factor when u′_x ≠ 0.

Discussion

The composition law is not an ad hoc patch on Galileo; it is the statement that boosts along a line form a one-parameter group under the rapidity variable. Writing u = c tanh φ and v = c tanh ψ, the addition-of-tangents identity tanh(φ+ψ) = (tanhφ + tanhψ)⁄(1 + tanhφ tanhψ) reproduces the law exactly with φ + ψ as the composed rapidity. Rapidities simply add — boosts are hyperbolic rotations in the t-x plane, and the law is the “angle addition” for them.

This is why c is unreachable rather than merely large: u = c corresponds to φ → ∞, an infinite rapidity, so no finite sum of finite rapidities attains it. The speed of light plays the role of the boundary at infinity of velocity space, and the law's denominator is precisely the metric distortion that pushes that boundary out of reach.

The cancellation of γ in step 4 carries a physical message: velocity composition knows nothing about time dilation per se. The same γ appears in energy-momentum, but the velocity law is a pure ratio and is scale-free in γ. This is why the formula could be guessed from the invariance of c alone, before ever computing γ.

Geometrically, the set of subluminal velocities with this composition rule forms a model of hyperbolic space (the Beltrami–Klein model on the unit ball |u|⁄c < 1). The non-commutativity of non-collinear boosts — two perpendicular boosts compose to a boost plus a rotation, the Wigner rotation — is exactly the statement that this velocity space has negative curvature; the resulting Thomas precession is a direct kinematic consequence of the same composition law applied in two dimensions.

Common misconceptions. The law is not “you can never measure anything moving faster than c” — separation and closing speeds within one frame routinely exceed c. It is the sharper claim that the speed of one massive body in another massive body's rest frame stays below c. And light is invariant not because “you can't catch up to it” but because c is the algebraic fixed point of the composition map.

Worked examples

u = (u′ + v)⁄(1 + u′v⁄c²)

Rocket problem. A ship moves at v = 0.80c relative to Earth and fires a probe forward at u′ = 0.60c relative to the ship. Find the probe's speed in Earth frame. Set up symbols first. A

u = (0.60c + 0.80c)⁄(1 + (0.60c)(0.80c)⁄c²) = 1.40c⁄(1 + 0.48)

Insert numbers; the c² cancels the product of the two c-factors, leaving 0.48 in the denominator. A

u = 1.40c⁄1.48 = 0.9459c

Divide. Compare with the wrong Galilean answer 1.40c: the correction is substantial and keeps u < c. A

u ≈ 0.946c

Reading. Two large subluminal speeds compose to a speed close to but below c. Units check. All quantities in units of c; result dimensionless-in-c, i.e. m s⁻¹. In SI, u = 0.946 × 2.998×10⁸ = 2.84×10⁸ m s⁻¹.

u = (u′ + v)⁄(1 + u′v⁄c²), u′ = −c

Headlight problem. A car approaches at v = 30 m s⁻¹ and shines a beam toward you, so in the car's frame the light moves at u′ = −c (toward you). Confirm you measure c. A

u = (−c + v)⁄(1 + (−c)v⁄c²) = (v − c)⁄(1 − v⁄c)

Substitute u′ = −c symbolically before inserting v = 30 m s⁻¹. B

u = −c(c − v)⁄(c − v) = −c

Factor −c from numerator against (c − v)⁄c in the denominator; they cancel exactly, independent of v. B

u = −c (exactly, for any v)

Reading. You measure the beam at c regardless of the car's 30 m s⁻¹ — the everyday case where the correction is invisible yet mandatory. Units check. v and c both in m s⁻¹; the ratio (c−v)⁄(c−v) is dimensionless, leaving u in m s⁻¹.

Problems

Two particles head toward each other, each at 0.90c in the lab frame. What is the speed of one particle as measured in the rest frame of the other?
Solution
Work in the frame boosted to ride with one particle: v = 0.90c, the other approaches at u′ = 0.90c (same sign of relative approach). u = (0.90c + 0.90c)⁄(1 + 0.81) = 1.80c⁄1.81 = 0.9945c. The lab “closing speed” is 1.80c, legal because it is not any single object's speed; the physical relative speed is 0.994c < c.
A frame S′ moves at v = 0.60c. An object in S′ moves backward at u′ = −0.60c. Find u and interpret.
Solution
u = (−0.60c + 0.60c)⁄(1 + (−0.60)(0.60)) = 0⁄(1 − 0.36) = 0. The object is at rest in S: its backward motion in S′ exactly cancels the frame's forward motion, as symmetry demands.
Show algebraically that if |u′| < c and |v| < c then |u| < c.
Solution
Compute c − u = c − (u′+v)⁄(1+u′v⁄c²) = [c + u′v⁄c − u′ − v]⁄(1+u′v⁄c²). The numerator is (c − u′)(c − v)⁄c. Since c − u′ > 0 and c − v > 0 and the denominator 1 + u′v⁄c² > 0, we get c − u > 0. The mirror computation gives c + u > 0. Hence |u| < c. □
Fizeau drag: light travels through water (refractive index n = 1.33, so u′ = c⁄n) flowing at v = 5.0 m s⁻¹. To first order in v⁄c, find the lab speed of the light and identify the Fresnel drag coefficient.
Solution
u = (c⁄n + v)⁄(1 + v⁄(nc)). Expand: u ≈ (c⁄n + v)(1 − v⁄(nc)) ≈ c⁄n + v − v⁄n² = c⁄n + v(1 − 1⁄n²). The drag coefficient is 1 − 1⁄n² = 1 − 1⁄1.769 = 0.435. Numerically the added speed is 0.435 × 5.0 = 2.2 m s⁻¹ above c⁄n. This matched Fizeau's 1851 result and was a key pre-Einstein clue.
A ship moves at v = 0.50c. It launches a shuttle forward at 0.50c relative to itself; the shuttle then launches a pod forward at 0.50c relative to the shuttle. Find the pod's speed in the original frame, and check with rapidities.
Solution
First compose: u₁ = (0.50c + 0.50c)⁄(1 + 0.25) = 1.0c⁄1.25 = 0.80c. Compose again with 0.50c: u₂ = (0.50c + 0.80c)⁄(1 + 0.40) = 1.30c⁄1.40 = 0.9286c. Rapidity check: φ = arctanh(0.5) = 0.5493 each; total 3φ = 1.6479; tanh(1.6479) = 0.9286. Agreement confirms rapidities add. Note three 0.5c boosts give 0.93c, not 1.5c.