Relativistic Velocity Addition

You are on a train gliding at 100\ \text{km/h} and you walk forward down the aisle at 5\ \text{km/h}. How fast are you moving over the ground? 105\ \text{km/h}, of course — you just add the speeds. This simple addition is one of the most trusted rules in all of physics, and for trains and walkers it is right to spectacular precision. But it is not exactly true, and near the speed of light it fails spectacularly. If it were exact, you could fire a fast probe from a fast rocket, add the speeds, and sail straight past c — but nothing ever does.

The Lorentz transformation tells us precisely how velocities really combine, and the answer has a built-in guardian: no matter how you stack speeds, the result never reaches c unless something was already moving at c — in which case it stays at c in every frame. This page derives the rule, shows the guardrail in action, and watches ordinary addition emerge as the slow-speed approximation it always was.

The velocity-addition formula

Suppose an object moves at velocity u along the x-axis as measured in frame S. Frame S' moves at velocity v along the same axis. What velocity u' does S' measure? Divide the Lorentz transformation of \mathrm{d}x' by that of \mathrm{d}t' and the \gamma's cancel, leaving:

Transforming a velocity between frames. u' = \dfrac{u - v}{1 - \dfrac{uv}{c^2}}. The Galilean answer u' = u - v sits on top; the whole novelty is the denominator.
Combining two velocities (the "addition" form). If a body moves at u' in S', its velocity back in S is u = \dfrac{u' + v}{1 + \dfrac{u'v}{c^2}} — flip the sign of v, exactly as for the coordinate transform.
The speed limit is automatic. Feed in any u \le c and v \le c and the result is always \le c, with equality only if one input already equals c.

That denominator, 1 - uv/c^2, is the entire story. When u and v are small compared with c, the product uv/c^2 is a vanishingly tiny number, the denominator is essentially 1, and we are back to plain subtraction (or addition). But as the speeds climb toward c, that denominator swells and reins the answer back in, keeping it forever short of the speed of light.

The guardrail, drawn

Fix a boost speed v with the slider, then read across: the horizontal axis is the object's speed u (as a fraction of c) and the vertical axis is the combined speed. The straight Galilean line u + v sails right past 1 (the speed of light) as if there were no limit at all. The curved relativistic sum starts along the same line at low speed but bends over and presses against the ceiling u = c without ever breaking through. Crank v up: the Galilean line rockets away, but the true curve stays trapped below the light line.

Worked examples

Example 1 — two fast ships. A mother ship flies past Earth at v = 0.5c and fires a probe forward at u' = 0.8c relative to the ship. Naively the probe should do 1.3c in Earth's frame. Relativity says

u = \frac{u' + v}{1 + u'v/c^2} = \frac{0.8c + 0.5c}{1 + (0.8)(0.5)} = \frac{1.3c}{1.4} \approx 0.929c.

Under the cosmic limit, not over it. The naive 1.3c was a mirage of ordinary addition.

Example 2 — light stays at light speed. Now the ship (still at v = 0.5c) switches on a headlamp, so u' = c. What does Earth measure?

u = \frac{c + 0.5c}{1 + (1)(0.5)} = \frac{1.5c}{1.5} = c.

Exactly c — the boost cancels perfectly. This is the second postulate popping straight out of the algebra: light travels at c in every frame, no faster and no slower.

Example 3 — a head-on approach. Two protons rush toward each other, each at 0.9c in the lab. How fast does one approach the other, in the rest frame of the second proton? Take u = 0.9c (one proton) and transform to the frame moving at v = -0.9c (the other):

u' = \frac{u - v}{1 - uv/c^2} = \frac{0.9c - (-0.9c)}{1 - (0.9)(-0.9)} = \frac{1.8c}{1.81} \approx 0.994c.

Not 1.8c but a whisker under c. Even two near-light beams closing head-on approach each other slower than light.

Watch out — this is the tempting escape hatch, and it is bolted shut. Surely, you think, if I add 0.9c and 0.9c I get close to c, so a few more boosts must push me over? Try it: combining 0.9c with 0.9c gives 1.8/1.81 \approx 0.994c; add another 0.9c and you reach 0.9998c; and so on. Each boost pushes you a fraction of the remaining gap to c, so you approach it like a runner forever halving the distance to a wall. You get arbitrarily close and never arrive.

Mathematically, the combination law is built so that speeds never quite add — they add a quantity called rapidity that does add plainly, but which maps onto a speed through a tanh function that saturates at c. So you can pour in infinite rapidity and only ever asymptote to the light barrier. The lesson to carry: "just add the speeds" is a low-speed convenience, not a law, and it has no power to break the cosmic speed limit.

A subtle but important "no." Suppose in your frame two ships fly apart, one at 0.8c to the left and one at 0.8c to the right. The gap between them, as measured by you, really does grow at 1.6c — and that is allowed, because no single object or signal is moving at 1.6c; it is just two things receding in your one frame (a "closing speed", not a physical velocity). But ask how fast the right ship sees the left ship approaching — that is a physical relative velocity in a single frame, and the addition formula gives (0.8 + 0.8)/(1 + 0.64) = 1.6/1.64 \approx 0.976c, safely under c. The distinction — closing speed in one frame versus relative velocity measured in one object's own frame — is where most "I broke light speed!" arguments quietly go wrong.