University Physics IV · Special Relativity I · 2.4

Deriving the boost from symmetry

Special relativity is usually sold as a consequence of light. It is not. Homogeneity, isotropy, and the demand that boosts form a group deliver the whole transformation with one constant left blank — and light's only job is to fill it in.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Ask what a coordinate change between inertial frames is allowed to look like before asking anything about light. Homogeneity says the map cannot know where or when it is applied, so its coefficients are constants and it is linear; tracking the moving frame's origin ties two of them together and leaves x′ = a(v)(x − vt). Isotropy plus the relativity principle say that inverting the map must be the same as reversing v, which forces the time row to carry the same factor — call it γ(v) — and the determinant to be exactly 1. One free function survives, and demanding that two boosts compose into a third — group closure — collapses it to a single universal constant K with units s² m⁻²: t′ = γ(t − Kvx), γ = (1 − Kv²)^(−1/2), velocities combining as (v₁ + v₂)/(1 + Kv₁v₂).

Symmetry has written everything except one number. Three worlds remain. K < 0 makes boosts behave like rotations, so composing them can reverse the order of cause and effect, and causality throws it out. K = 0 is Galileo, exactly.

K > 0 supplies one invariant speed V = K^(−1/2) that no boost can reach, and the second postulate is nothing grander than the claim that light travels at it. The cost is the assumptions — flat spacetime, genuinely inertial frames, no gravity — and the fact that no amount of symmetry can hand you K. Only a measurement can.

Simple definition: The Lorentz boost is the unique linear map between inertial frames allowed by homogeneity, isotropy and group closure, fixed up to one universal constant K whose value only experiment supplies.
Example: With K = 1.111 × 10⁻¹⁷ s² m⁻² and v = 1.80 × 10⁸ m s⁻¹, Kv² = 0.360 and γ = (1 − 0.360)^(−1/2) = 1.250; set K = 0 instead and the same v gives γ = 1 exactly — Galileo, from the same formula.

Linear ansatz, origin trackedx′ = a(v)(x − vt), t′ = b(v)t + d(v)x

Homogeneity kills every x- and t-dependence, leaving three functions of v alone.

x, x′ in m; t, t′ in s; a, b dimensionless; d in s m⁻¹

Reciprocity closes the matrixa(v) = b(v) ≡ γ(v), det = 1, d = −γKv

Isotropy makes γ even and forces the determinant to 1, so one unknown survives.

K ≡ (1 − γ⁻²)/v², units s² m⁻², still allowed to depend on v

The one-parameter familyx′ = γ(x − vt), t′ = γ(t − Kvx), γ = (1 − Kv²)^(−1/2)

Every kinematics symmetry permits, indexed by one number and nothing else.

K now a constant in s² m⁻²; needs Kv² < 1 on the K > 0 branch

Closure and velocity additionv₃ = (v₁ + v₂)/(1 + Kv₁v₂), γ₃ = γ₁γ₂(1 + Kv₁v₂)

Demanding boost ∘ boost = boost is exactly what makes K independent of v.

speeds in m s⁻¹; the product Kv₁v₂ is dimensionless

The invariant speedV = K^(−1/2), (V − v)/(1 − KVv) = V

One speed is a fixed point of the addition law, so no boost can reach it.

K > 0 branch only; K = 1/c² gives V = 2.998 × 10⁸ m s⁻¹

The Galilean limitK → 0: γ → 1, t′ = t, γ − 1 ≈ ½Kv²

Galileo is not wrong, it is K = 0 — which is why it went unchallenged.

at v = 3.0 × 10⁴ m s⁻¹, Kv² = 1.0 × 10⁻⁸ so γ − 1 = 5 × 10⁻⁹

Homogeneity makes the map linear

A change of inertial coordinates must not care where or when it is applied — that is what homogeneity of space and time means. If the coefficients of the map depended on x or t, two identical rods would transform differently at different places and the frame would have a preferred location. So the coefficients are constants and the map is affine; letting the origins coincide at t = t′ = 0 removes the constant terms and leaves it linear. The same conclusion arrives from dynamics: a free particle draws a straight line in (x, t) in every inertial frame, and a map defined on all of spacetime that carries every straight line to a straight line is affine — the fractional-linear maps that also preserve lines blow up on some hyperplane, which homogeneity forbids. Now track the origin of the moving frame. The event x′ = 0 must happen at x = vt, which forces the first row to read x′ = a(v)(x − vt). The second row is still wide open, t′ = b(v)t + d(v)x, and nothing about light has been used.

Isotropy forces reciprocity, and reciprocity forces det = 1

Reversing the roles of the two frames must undo the boost: the inverse of a boost by v is a boost by −v. Reflect space, x → −x, and v → −v with it, so a and b are even functions of v while d, which multiplies a length inside a time, is odd. Now invert the matrix explicitly. With Δ = det = a(b + vd), the inverse gives x = (b x′ + a v t′)/Δ and t = (a t′ − d x′)/Δ, while the −v boost reads x = a(x′ + vt′) and t = bt′ − dx′. Match coefficient by coefficient. The t′ term in x demands av = av/Δ, so Δ = 1 outright; the x′ term then demands a = b ≡ γ(v). The unit determinant is forced, not assumed, and no sign choice was available. Then γ(γ + vd) = 1 fixes d = (1 − γ²)/(γv), which is tidier written as d = −γKv with K ≡ (1 − γ⁻²)/v². Exactly one unknown function of v is left.

Closure collapses the family to one constant K

Write the boost as a matrix acting on the column (t, x): M(v) = γ[[1, −Kv], [−v, 1]]. Let K still depend on v, and compose two boosts, v₁ then v₂. The product has diagonal entries 1 + K₂v₁v₂ and 1 + K₁v₁v₂. But a boost matrix has equal diagonal entries, so if the product is to be a boost at all — group closure, without which the relativity principle is empty — then K₁ = K₂ for every pair of speeds. K is one universal constant with units s² m⁻². The off-diagonal entries then hand over v₃ = (v₁ + v₂)/(1 + Kv₁v₂) and γ₃ = γ₁γ₂(1 + Kv₁v₂). Feed the now-constant K back into K = (1 − γ⁻²)/v² and solve: γ = (1 − Kv²)^(−1/2). The entire transformation is written, and exactly one number in it is unknown. A note on names: some accounts call the invariant speed itself K, so that Galileo is K → ∞; here K is the coefficient multiplying vx in the time row, the invariant speed is K^(−1/2), and Galileo is K → 0.

Three branches, and causality removes one

Three worlds are left, one per sign of K. K = 0 gives γ = 1 identically, t′ = t and x′ = x − vt: Galileo, exactly, with velocities simply adding. K < 0 is the Euclidean branch; write K = −1/κ², so γ = (1 + v²/κ²)^(−1/2) is less than 1 and moving clocks run fast. Set v = κ tan θ and boosts compose by adding θ, so two boosts of κ each reach θ = 90° and an infinite velocity, and larger angles flip the sign. Worse, for a signal with dx = u dt and u > 0, the transformed interval dt′ = γ dt(1 + uv/κ²) turns negative for any v more negative than −κ²/u; since v is unbounded on this branch such a frame always exists, so every cause-and-effect pair can be reversed. That is the exclusion — causality, not experiment. K > 0 survives, carrying exactly one speed V = K^(−1/2) that every frame agrees on.

The second postulate is one measurement

On the surviving branch symmetry has produced an invariant speed but not its value. Any experiment returning a finite invariant speed sets K. The second postulate is the historically convenient one: light travels at the invariant speed, so V = c = 2.998 × 10⁸ m s⁻¹ and K = 1/c² = 1.113 × 10⁻¹⁷ s² m⁻². Note carefully what this does and does not claim. It does not make relativity a theory about light. If the photon carried a small mass, light would travel a shade below the invariant speed, c would still be the invariant speed, and every formula in this unit would stand unchanged — only the road in would need repaving. Modern Ives–Stilwell tests on stored ion beams bound any deviation from the derived γ at a few parts in 10⁸, and what they measure is K, through the rate of a moving clock, not the speed of light.

What the derivation assumes, and what it cannot give

The argument is cheap but not free. It assumes flat spacetime and globally defined inertial coordinates, so gravity — which makes homogeneity fail from place to place — sits outside it, and that is where general relativity begins. It assumes isotropy, which is what allowed γ to be even and what forces the transverse coordinates to go untouched, y′ = y and z′ = z. And it delivers one boost direction: the full symmetry needs rotations added to make the Lorentz group and translations added to make the ten-parameter Poincaré group. What it cannot do is name K, choose a branch without a causality axiom, or say anything whatever about dynamics. Momentum, mass–energy and the transformation of fields are separate arguments built on top of this kinematics, not consequences of it.

Change one variable at a time

Make the relationship visible.

Interactive model

Constant k = Kc²1.00

Speed β = v/c0.60

Slide k to 0: the curve flattens onto the horizontal line and every clock keeps Galilean time. Push k to −1 and 1/γ rises above 1 — moving clocks run fast — and with β at 0.95 the composed speed β ⊕ β reads 19.5c. That is the branch causality throws out.

1/γ = √(1 − Kv²)0.800

γ1.250

Kv²0.360

β ⊕ β0.882 c

Live interpretation1/γ = √(1 − Kv²): 0.800. γ: 1.250. Kv²: 0.360. β ⊕ β: 0.882 c

Catch the common trap

Explain before calculating.

Misconception check

The usual telling makes invariant c the input: assume light's speed is the same in all frames, grind the algebra, collect the boost. Students then conclude that relativity is a theory about light, and that a photon with a tiny mass would bring the structure down. It would not. Symmetry alone delivers x′ = γ(x − vt) and t′ = γ(t − Kvx) with one blank constant; the light postulate fills in K = 1/c² and does nothing else. A massive photon would only change which particle happens to move at V — the invariant speed, and every formula built on it, would stay exactly where symmetry put them. The related slip is treating γ(v) = γ(−v) as obvious. It is not — it is the isotropy of space, an assumption you are entitled to test, and it is the step that forces the determinant to 1.

Symmetry alone — homogeneity, isotropy, and closure of boosts under composition — delivers x′ = γ(x − vt) and t′ = γ(t − Kvx) with γ = (1 − Kv²)^(−1/2). What is the status of the constant K at that point in the argument?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyTake the derived family x′ = γ(x − vt), t′ = γ(t − Kvx) with γ = (1 − Kv²)^(−1/2). Using K = 1/c² with c = 3.00 × 10⁸ m s⁻¹, evaluate Kv² and γ for v = 1.80 × 10⁸ m s⁻¹, confirm that the determinant is 1, then repeat for a satellite at v = 7.50 × 10³ m s⁻¹.

K = 1/c² = 1/(3.00 × 10⁸ m s⁻¹)² = 1.111 × 10⁻¹⁷ s² m⁻². The one blank the symmetry argument left is now a number.
Kv² = 1.111 × 10⁻¹⁷ × (1.80 × 10⁸)² = 1.111 × 10⁻¹⁷ × 3.24 × 10¹⁶ = 0.360. It is dimensionless, as it must be to sit inside 1 − Kv².
γ = (1 − 0.360)^(−1/2) = (0.640)^(−1/2) = 1/0.800 = 1.250.
Determinant: det = γ²(1 − Kv²) = 1.250² × 0.640 = 1.5625 × 0.640 = 1.000, exactly as reciprocity demanded — no extra input was needed.
Satellite: Kv² = 1.111 × 10⁻¹⁷ × (7.50 × 10³)² = 1.111 × 10⁻¹⁷ × 5.625 × 10⁷ = 6.25 × 10⁻¹⁰, so γ − 1 ≈ ½Kv² = 3.1 × 10⁻¹⁰.

AnswerKv² = 0.360 and γ = 1.250 at 1.80 × 10⁸ m s⁻¹, with det = 1.000. At 7.50 × 10³ m s⁻¹, Kv² = 6.25 × 10⁻¹⁰ and γ − 1 = 3.1 × 10⁻¹⁰: Galileo is the K → 0 corner of the same family.

MediumCompose two collinear boosts of v₁ = 0.500c and v₂ = 0.800c using v₃ = (v₁ + v₂)/(1 + Kv₁v₂). Find v₃ and γ₃ on the Lorentz branch K = 1/c², check them against the closure prediction γ₃ = γ₁γ₂(1 + Kv₁v₂), then repeat the composition on the Euclidean branch K = −1/c² and say what goes wrong.

Lorentz branch: Kv₁v₂ = (0.500)(0.800) = 0.400, so v₃ = (0.500 + 0.800)c/1.400 = 1.300c/1.400 = 0.9286c. Two sub-c boosts have not reached c.
γ₁ = (1 − 0.250)^(−1/2) = 1.1547 and γ₂ = (1 − 0.640)^(−1/2) = 1.6667.
γ₃ directly: β₃ = 13/14 = 0.92857, so β₃² = 0.86224 and γ₃ = (0.13776)^(−1/2) = 2.694. The closure prediction γ₁γ₂(1 + Kv₁v₂) = 1.1547 × 1.6667 × 1.400 = 2.694 agrees, so the product really is a boost.
Euclidean branch K = −1/c²: now Kv₁v₂ = −0.400, so v₃ = 1.300c/0.600 = 2.167c. Nothing caps the result — and v₁ = v₂ = c would give 2c/0, two finite boosts composing to an infinite one.
On the Lorentz branch c is a fixed point of the map: (c − v)/(1 − v/c) = c for every v, so composing sub-luminal boosts approaches c and never crosses it. The Euclidean branch has no fixed point at all, which is the pathology causality objects to.

Answerv₃ = 0.9286c with γ₃ = 2.694 on the K = 1/c² branch, matching γ₁γ₂(1 + Kv₁v₂) = 2.694. The K = −1/c² branch returns 2.167c and has no invariant speed.

HardA precision experiment compares two inertial frames in relative motion at v = 2.000 × 10⁷ m s⁻¹ and finds the boost's time-row coefficient to be γ = 1.002230. Using only the symmetry result γ = (1 − Kv²)^(−1/2), extract K and name its branch, convert it to an invariant speed, then state the precision that would be needed to see the same effect on an airliner at 250 m s⁻¹.

Invert the symmetry result rather than assuming a value: γ⁻² = 1 − Kv², so K = (1 − γ⁻²)/v². No light has entered the argument at any point.
γ² = 1.002230² = 1.0044650, so γ⁻² = 0.9955549 and 1 − γ⁻² = 0.0044451.
v² = (2.000 × 10⁷)² = 4.000 × 10¹⁴ m² s⁻², so K = 0.0044451 ÷ 4.000 × 10¹⁴ = 1.1113 × 10⁻¹⁷ s² m⁻². K > 0: the Lorentz branch, neither Galileo nor Euclidean.
Invariant speed V = K^(−1/2) = (1.1113 × 10⁻¹⁷ s² m⁻²)^(−1/2) = 3.000 × 10⁸ m s⁻¹. The experiment has found c without ever looking at light.
Airliner: the same K at v = 250 m s⁻¹ gives Kv² = 1.1113 × 10⁻¹⁷ × 6.25 × 10⁴ = 6.95 × 10⁻¹³, so γ − 1 ≈ ½Kv² = 3.5 × 10⁻¹³. Separating this world from the K = 0 world at 250 m s⁻¹ needs γ to roughly thirteen decimal places.

AnswerK = 1.111 × 10⁻¹⁷ s² m⁻², positive, so the Lorentz branch; V = K^(−1/2) = 3.000 × 10⁸ m s⁻¹. At 250 m s⁻¹ the same K gives γ − 1 = 3.5 × 10⁻¹³, so telling it from Galileo needs 13-figure timing.