Derivation of the Lorentz Transformation

Take two boosts of $0.6c$ , one after the other. Common sense adds them and reports $1.2c$ . The right answer is $0.8824c$ — and not because something resists at high speed, or because a force runs out. The reason is arithmetic. The transformations relating inertial frames form a group, and multiplying two of them gives $\beta_3 = (\beta_1+\beta_2)/(1+\beta_1\beta_2)$ , which cannot escape $(-1,1)$ however often it is applied. Four such boosts give $0.9922c$ ; ten give $0.999998c$ . Never $c$ . The speed limit is not a fact about engines. It is a fact about a group. This article derives the Lorentz transformation three times. The classical route takes five steps, each justified by a named principle: homogeneity forces the map to be linear; the motion of the $S'$ origin fixes $x' = \gamma(v)(x-vt)$ ; relativity plus isotropy force reciprocity and make $\gamma$ even; the light postulate pins $\gamma = 1/\sqrt{1-\beta^2}$ . The second route composes two boosts and reads off the group law. The third answers the sharpest objection a student can raise — you assumed $c$ was invariant, so of course you got an invariant $c$ . In 1910 Vladimir Ignatowski showed that relativity, homogeneity, isotropy and group closure alone, with no mention of light, already force $\Gamma = 1/\sqrt{1-v^2/K^2}$ , with $K$ a universal constant of the dimensions of a speed — and closure is what makes $K$ the same for every frame. $K = \infty$ gives Galileo; $K$ finite gives Lorentz. That is all the second postulate really claims: that $K$ is finite. Experiment says $K = c$ . The sibling article The Lorentz Transformation Equations puts the finished map to work; this one builds it.

Key takeaways

The Lorentz transformation is fixed almost entirely by symmetry: linearity from homogeneity, $x' = \gamma(v)(x-vt)$ from the motion of the $S'$ origin, and reciprocity ( $\gamma$ even, inverse at $-v$ ) from the principle of relativity plus isotropy.
The light postulate does exactly one job: the rays $x = \pm ct$ give $A = D$ and $B = -vD/c^2$ ; reciprocity then forces $D^2(1-\beta^2) = 1$ , i.e. $\gamma = 1/\sqrt{1-\beta^2}$ . The result is $t' = \gamma(t-vx/c^2)$ , $x' = \gamma(x-vt)$ , $y' = y$ , $z' = z$ — the Galilean map as $c \to \infty$ .
Boosts form a one-parameter group, $\beta_3 = (\beta_1+\beta_2)/(1+\beta_1\beta_2)$ . Since $1-\beta_3 = (1-\beta_1)(1-\beta_2)/(1+\beta_1\beta_2) > 0$ , no stack of boosts reaches $c$ ; in the rapidity $\theta = \operatorname{artanh}\beta$ the law is literally additive.
Ignatowski (1910): drop the light postulate and closure still delivers $\Gamma = 1/\sqrt{1-v^2/K^2}$ with a universal invariant speed $K$ .
Key idea: the transformation is not a concession extracted from light — it is what the principle of relativity looks like once a universal invariant speed is allowed to be finite. Closure forces such a speed to exist and to be frame-independent; only experiment says its value is $c$ . The second postulate is the empirical claim that $K < \infty$ .

The core insight

Symmetry does almost all the work: homogeneity makes the map linear, the motion of the $S'$ origin makes it $x' = \gamma(v)(x-vt)$ , and relativity plus isotropy make $\gamma$ even and the inverse the same map at $-v$ . One function is then undetermined, and one physical input fixes it. Supply the light postulate and you get $\gamma = 1/\sqrt{1-\beta^2}$ ; supply nothing but the demand that two boosts compose into a third, and you still get $\Gamma = 1/\sqrt{1-v^2/K^2}$ with a universal $K$ . Whether $K$ is infinite (Galileo) or finite (Lorentz) is the one thing symmetry cannot decide — and the one thing the second postulate is for.

Where this lives today

The group law is the arithmetic of every accelerator on Earth. At the LHC a $7\ \mathrm{TeV}$ proton has $\gamma = E/mc^2 = 7000/0.938272 = 7460.5$ , so $1-\beta = 8.98\times10^{-9}$ — about $2.7\ \mathrm{m/s}$ short of light. Two such protons meet head-on, and while naive addition reports $1.99999998c$ , the group law returns a relative speed still below $c$ . Rapidity, by contrast, simply adds along the beam, so a rapidity difference is boost-invariant — which is why every LHC detector paper quotes rapidity rather than velocity. The second home is deeper: the classification of particles. Once boosts, rotations and spacetime translations are known to form a group — the Poincaré group — one can ask what its irreducible unitary representations are — and Wigner's answer, in 1939, is that they are labelled by exactly two numbers, mass and spin. The reason an electron has a mass and a spin, and no other label of that kind, is that those are the labels the Poincaré group's representations come with — without the translations there would be no mass label at all.

The physical picture

An event is a point in spacetime; two inertial observers label it with different numbers, and the transformation is the dictionary between the labellings. The question is how much of that dictionary can be written down before any physics is put in — and the answer is: nearly all of it. Homogeneity says no place or moment is special, so the coefficients cannot depend on where you are: the map is linear. Isotropy says no direction is special, so a $180^\circ$ turn, flipping both $x$ and $v$ , must map the law onto itself. Relativity says neither observer is special, so the inverse dictionary is the forward one with $v \to -v$ . Three symmetries, and the map is down to a single unknown function.

Figure

The logical scaffold. Each symmetry kills a degree of freedom, and the equation at each node is what survives. The light postulate enters at ONE place only — node 4 — and does one job: it fixes the value of the invariant speed. The right-hand rail is the Ignatowski route, which skips that node entirely and still lands on the same shape, with an undetermined universal constant K in place of c.

Now compose two dictionaries. Boost from the platform to a train at $\beta_1$ , then from the train to a trolley at $\beta_2$ ; the net map must itself be a legal boost, because platform and trolley are both inertial frames. That single demand — closure — is stronger than it looks: the composite velocity is $\beta_3 = (\beta_1+\beta_2)/(1+\beta_1\beta_2)$ , and $1-\beta_3$ factorises into $(1-\beta_1)(1-\beta_2)/(1+\beta_1\beta_2)$ , a product of positive numbers. It can shrink toward zero. It can never reach it. The light barrier is a theorem about multiplication.

Figure

PUNCHLINE — doubling a speed. The Galilean rule (grey) doubles it honestly: 0.6 → 1.2, straight through the light barrier. The group law (blue) returns 2β/(1+β²) and SATURATES: 0.6 ⊕ 0.6 = 0.8824, and even 0.99 ⊕ 0.99 = 0.999949. The curve approaches c and never touches it — closure of the group is what makes c unreachable, and no finite stack of boosts ever gets there.

The core equations

Standard configuration: $S'$ moves at $+v$ along the $x$ -axis of $S$ , axes parallel, origins coinciding at $t = t' = 0$ ; $\beta \equiv v/c$ . Nothing below is assumed — all of it is derived next. Homogeneity makes the map linear, $t' = At + Bx$ and $x' = Ct + Dx$ ; the $S'$ origin (the locus $x' = 0$ , which traces $x = vt$ ) kills $C$ ; the light postulate and reciprocity finish the job.

t' = \gamma\!\left(t - \frac{vx}{c^{2}}\right), \qquad x' = \gamma\,(x - vt), \qquad y' = y, \qquad z' = z, \qquad \gamma = \frac{1}{\sqrt{1-\beta^{2}}} \qquad (1)

where: $(t,x)$ and $(t',x')$ — one event, labelled in $S$ and in $S'$ ; $c = 299\,792\,458\ \mathrm{m/s}$ (exact); $\gamma \ge 1$ , with equality only at $v = 0$ . Physical reading: against Galileo's $t' = t$ , the term $-\gamma vx/c^2$ is position-dependent, so clocks synchronised in $S$ are not synchronised in $S'$ — simultaneity is forced by the same algebra that produces $\gamma$ , not bolted on afterwards. Let $c \to \infty$ and both novelties vanish at once. Write (1) as a matrix on $X = (ct,x)$ and compose two of them: the product is a third boost, whose velocity is not the sum.

\Lambda(\beta_2)\,\Lambda(\beta_1) = \Lambda(\beta_3), \qquad \beta_3 = \frac{\beta_1+\beta_2}{1+\beta_1\beta_2}, \qquad \theta_3 = \theta_1 + \theta_2 \ \ \text{with}\ \ \theta \equiv \operatorname{artanh}\beta \qquad (2)

where: $\Lambda(\beta)$ — the boost matrix; $\theta$ — the rapidity. Physical reading: the boosts are closed, with identity $\Lambda(0)$ and inverse $\Lambda(-\beta)$ — a one-parameter group. Velocities do not add, so velocity is the wrong label for it; rapidities do, and $\beta = \tanh\theta$ can never leave $(-1,1)$ . The hyperbolic geometry belongs to Rapidity and the Hyperbolic Form of Boosts, and $\beta_3$ as a kinematic tool to the Relativistic Velocity Addition Formula and Velocity Composition and the Speed-of-Light Invariant. Finally, delete the light postulate entirely and demand only relativity, homogeneity, isotropy and closure.

t' = \Gamma\!\left(t - \frac{vx}{K^{2}}\right), \qquad x' = \Gamma\,(x - vt), \qquad \Gamma = \frac{1}{\sqrt{1 - v^{2}/K^{2}}} \qquad (3)

where: $K$ — a universal constant of the dimensions of a speed, the same for every frame. Physical reading: this is (1) with $K$ where $c$ stood, obtained without ever uttering the word "light". Symmetry gives the existence and frame-independence of $K$ , never its value: $K = \infty$ collapses (3) to the Galilean map, $K$ finite makes it Lorentzian. Experiment says $K = c$ — and that, nothing more, is the content of the second postulate.

Where the results come from

Derivation AThe five steps, from the postulates

The goal is to derive equation (1) rather than assert it, with every step tied to a named physical principle and none to algebraic convenience. Watch which input kills which freedom: symmetry alone kills three of the four unknown constants, and the famous light postulate is responsible for exactly one. That accounting is what makes Derivation C possible.

Step 1 — homogeneity gives LINEARITY. No point of spacetime is privileged, so the coefficients cannot depend on position or time: $t' = At + Bx$ , $x' = Ct + Dx$ with $A,B,C,D$ constant, and coincident origins remove any constant term. Why this is physics, not convenience: coordinate-dependent coefficients bend straight worldlines, so a free particle would coast in one frame and accelerate in the other.

Step 2 — the motion of the $S'$ origin fixes $C$ . The set $x' = 0$ traces $x = vt$ in $S$ , so $0 = Ct + Dvt$ for all $t$ , giving $C = -vD$ and $x' = D(x-vt)$ . Name the survivor $\gamma(v) \equiv D$ . Why it is nearly empty: Galileo agrees too — it says only that $S'$ moves at $v$ .

Step 3 — relativity and isotropy give RECIPROCITY. Relativity makes the inverse map the forward map with $v \to -v$ . Isotropy adds that no direction is preferred: reflecting $x \to -x$ also sends $v \to -v$ , and the Step-2 law must map onto itself — which holds if and only if $\gamma(-v) = \gamma(v)$ . So $\gamma$ is even. Why it matters: evenness permits the sign choice in Step 4, and says neither observer is "really" moving.

Step 4 — the light postulate fixes $A$ and $B$ ; reciprocity then fixes $D$ . A flash at the common origin goes as $x = ct$ in $S$ and must go as $x' = ct'$ in $S'$ , giving $D(c-v) = c(A+Bc)$ ; the backward ray $x = -ct$ gives $D(c+v) = c(A-Bc)$ . Add: $A = D$ . Subtract: $B = -vD/c^2$ . Writing the map as $M(v) = D\begin{pmatrix} 1 & -v/c^2 \\ -v & 1\end{pmatrix}$ , its inverse is $M(-v)$ with the same $D$ (Step 3), and $M(-v)M(v) = D^2(1-v^2/c^2)\,\mathbb{1}$ , so $M(-v)M(v) = \mathbb{1}$ demands $D^2(1-\beta^2) = 1$ . The positive root, selected by $\gamma(0) = 1$ , is $\boxed{\gamma = 1/\sqrt{1-\beta^{2}}}$ . Why two rays: one ray is one equation for two unknowns — it takes both to separate $A$ from $B$ .

Step 5 — the transverse coordinates are untouched. Write $y' = \kappa(v)y$ ; relativity gives $y = \kappa(-v)y'$ , so $\kappa(v)\kappa(-v) = 1$ , and isotropy makes $\kappa$ even, so $\kappa^2 = 1$ ; continuity with $\kappa(0) = 1$ selects $\kappa = +1$ . Hence $y' = y$ , $z' = z$ . Why this cannot be waved through: if transverse lengths changed, two frames would disagree about whether a moving rod fits through a ring of equal rest diameter — and whether it scratches is absolute. *A symbolic check confirms the constant Jacobian, recovers $C = -vD$ , forces $D$ even under the parity flip, solves the two light-ray equations for $A = D$ and $B = -vD/c^2$ , obtains $\gamma = 1/\sqrt{1-\beta^2}$ from $M(-v)M(v) = \mathbb{1}$ , confirms $\kappa^2 = 1$ , and verifies $c^2t'^2-x'^2 = c^2t^2-x^2$ identically in $v$ .*

Go deeper · Derivation B · 2 minThe group law — closure and the speed limit

Derivation BThe group law — closure and the speed limit

Two boosts in succession must compose into a single legal boost, because the first and last frames are perfectly good inertial frames. That demand — closure — is the only input to this card, and it delivers the composition law and a two-line proof that $c$ is unreachable using no energy, mass or force. Its second job — discarding the light postulate altogether — is Derivation C.

Step 1 — split off the square root. On $X = (ct,x)$ , equation (1) reads $\Lambda(\beta) = \gamma\begin{pmatrix}1 & -\beta\\ -\beta & 1\end{pmatrix}$ . Treat the rational core and the scalar $\gamma$ separately, so the composition is exact rational arithmetic. Why bother: the square root is the only delicate step; isolating it turns "it works out" into a proof.

Step 2 — multiply the cores. $\begin{pmatrix}1 & -\beta_2\\ -\beta_2 & 1\end{pmatrix}\begin{pmatrix}1 & -\beta_1\\ -\beta_1 & 1\end{pmatrix} = (1+\beta_1\beta_2)\begin{pmatrix}1 & -\beta_3\\ -\beta_3 & 1\end{pmatrix}$ , with $\beta_3 = \dfrac{\beta_1+\beta_2}{1+\beta_1\beta_2}$ . The prefactor is indeed $\gamma_3$ : squaring turns that claim into the exact rational identity $\dfrac{(1+\beta_1\beta_2)^2}{(1-\beta_1^2)(1-\beta_2^2)} = \dfrac{1}{1-\beta_3^2}$ , with both sides positive for $|\beta_1|,|\beta_2| < 1$ . So $\Lambda(\beta_2)\Lambda(\beta_1) = \Lambda(\beta_3)$ exactly, and with identity $\Lambda(0)$ and inverse $\Lambda(-\beta)$ the boosts form a one-parameter abelian group. Why this is most of the theorem: the composite has the boost SHAPE — equal diagonal, symmetric off-diagonal — before any square root is touched.

Step 3 — the speed limit, in one line. $1-\beta_3 = \dfrac{(1-\beta_1)(1-\beta_2)}{1+\beta_1\beta_2}$ and $1+\beta_3 = \dfrac{(1+\beta_1)(1+\beta_2)}{1+\beta_1\beta_2}$ . If $|\beta_1|,|\beta_2| < 1$ , every factor on every right-hand side is strictly positive — so $-1 < \beta_3 < 1$ . Always. Why this is the deepest sentence here: nothing dynamical was used, yet two sub-luminal boosts compose to a sub-luminal boost as pure ARITHMETIC — and $\beta_1 = 1$ gives $\beta_3 = 1$ for every $\beta_2$ , so light is not merely unreachable but unmovable.

Step 4 — rapidity makes the law additive. Put $\beta = \tanh\theta$ ; then $\tanh(\theta_1+\theta_2) = \dfrac{\tanh\theta_1+\tanh\theta_2}{1+\tanh\theta_1\tanh\theta_2}$ is the composition law, so $\theta_3 = \theta_1+\theta_2$ : rapidities add, and $\beta = \tanh\theta \to 1$ as $\theta \to \infty$ without ever arriving. Why this is the right variable: it turns the group into the additive group of the real line. *A symbolic check multiplies the cores and returns $(1+\beta_1\beta_2)$ times the core at $\beta_3$ , proves $\gamma_3^2 = [\gamma_1\gamma_2(1+\beta_1\beta_2)]^2$ , confirms $\Lambda(\beta_2)\Lambda(\beta_1) = \Lambda(\beta_3)$ on a $6\times5$ grid to $10^{-12}$ with $|\beta_3| < 1$ throughout, verifies identity/inverse/commutativity, factorises $1 \mp \beta_3$ , checks $\beta_1 = 1 \Rightarrow \beta_3 = 1$ , and confirms rapidity additivity.*

Go deeper · Derivation C · 2 minThe transformation with NO light postulate

Derivation CThe transformation with NO light postulate

Closure's second job is bigger than its first. Keep relativity, homogeneity and isotropy, keep the demand that two boosts compose into a boost — and delete the light postulate entirely: the transformation is still forced into Lorentz shape, with a universal invariant constant $K$ whose value is the only thing left for experiment. That is Ignatowski's theorem of 1910.

Step 1 — now throw the light postulate away. Keep linearity, $x' = \Gamma(v)(x-vt)$ and the evenness of $\Gamma$ , but leave the time line wide open: $t' = \mathcal{D}(v)\,t + \mathcal{C}(v)\,x$ . Impose reciprocity, $M(-v)M(v) = \mathbb{1}$ , on $M(v) = \begin{pmatrix}\mathcal{D} & \mathcal{C}\\ -\Gamma v & \Gamma\end{pmatrix}$ . The $(2,1)$ entry of the product is $\Gamma v(\mathcal{D}-\Gamma)$ , which vanishes for $v \neq 0$ only if $\mathcal{D} = \Gamma$ ; the $(2,2)$ entry then forces $\mathcal{C} = (1-\Gamma^2)/(\Gamma v)$ , non-zero unless $\Gamma = 1$ . Why this is already remarkable: with no word about light, reciprocity shows $t'$ MUST depend on $x$ unless $\Gamma \equiv 1$ — absolute time is refuted by any $\Gamma \neq 1$ , not by light.

Step 2 — name the constant. Define $K$ by writing $\mathcal{C} \equiv -\Gamma v/K^2$ : a definition, not an assumption, giving the forced coefficient a name with the dimensions of a speed. Rearranging $K^2 = \Gamma^2v^2/(\Gamma^2-1)$ gives $\Gamma^2(K^2-v^2) = K^2$ , i.e. $\Gamma = 1/\sqrt{1-v^2/K^2}$ — equation (3). Why nothing is smuggled in: at this stage $K$ could still differ from frame to frame; it is merely a relabelling of $\mathcal{C}(v)$ .

Step 3 — THE THEOREM: closure forces $K$ to be universal. Give frame 1 the constant $K_1$ and frame 2 the constant $K_2$ , and compose. The product's diagonal entries come out as $1 + v_1v_2/K_2^2$ and $1 + v_1v_2/K_1^2$ . But a boost of the same form has equal diagonal entries — that is what the shape means. So the composite is a legal boost if and only if $K_2 = K_1$ : $K$ is one and the same number for the whole universe. Why this is the punchline: a universal invariant speed is not an extra postulate — it is a THEOREM, proved by the demand that two boosts make a boost.

Step 4 — read off the two possible worlds. With $K$ universal, $v_3 = (v_1+v_2)/(1+v_1v_2/K^2)$ and $|v_3| < K$ always, by the Step-3 factorisation of Derivation B. If $K = \infty$ : $\Gamma \to 1$ , $t' \to t$ , $x' \to x-vt$ — the Galilean transformation, absolute time, no ceiling. If $K$ is finite: the Lorentz transformation, with $K$ an unreachable invariant speed. Symmetry cannot choose; measurement can, and it says $K$ is finite and light travels at it, so $K = c$ . Why this reframes the second postulate: not that "light is special", but that $K < \infty$ — one empirical number, which light realises because the photon is massless. *A symbolic check extracts $\mathcal{D} = \Gamma$ and $\mathcal{C} = (1-\Gamma^2)/(\Gamma v)$ from $M(-v)M(v) = \mathbb{1}$ , rearranges to $\Gamma^2(K^2-v^2) = K^2$ , and shows the composite's diagonals agree only for $K_2 = K_1$ — the universality theorem; it confirms $|v_3| < K$ for $K = 1, 3, c$ and recovers the Galilean map as $K \to \infty$ .*

Limiting cases

Case	Result	Meaning
$c \to \infty$ (equivalently $K \to \infty$ )	$\gamma \to 1$ , $t' \to t$ , $x' \to x-vt$	the Galilean transformation — the map this one replaces
$v = 0$	$\Lambda(0) = \mathbb{1}$	the identity element of the group
a light ray, $u = c$	$u' = c$ for every $v$	$c$ is a fixed point of the composition law, not merely a bound
any $\|\beta_1\|,\|\beta_2\| < 1$	$\|\beta_3\| < 1$ , since $1-\beta_3 = \dfrac{(1-\beta_1)(1-\beta_2)}{1+\beta_1\beta_2} > 0$	the speed limit is group-theoretic, not dynamical
$N$ boosts of $0.6c$	$\beta_N = \dfrac{4^N-1}{4^N+1} < 1$ for every finite $N$	$0.600,\ 0.8824,\ 0.9692,\ 0.9922,\ \dots \to 1$ , never $= 1$

Common trap“The derivation is circular — you assumed light travels at c in both frames, so of course you got a transformation in which light travels at c in both frames.”

This is the best objection in the subject, and it is half right: the light postulate is an assumption, it does appear in Step 4, and the transformation does come out with $c$ invariant. What is wrong is the inference that the argument is therefore empty. Count what each input bought — homogeneity bought linearity, the $S'$ -origin condition bought $x' = \gamma(x-vt)$ , and relativity plus isotropy bought reciprocity and an even $\gamma$ . That is three of the four unknowns, none from light; the postulate bought only the value of the last. Derivation C shows you can decline to buy even that: relativity, homogeneity, isotropy and closure by themselves force $\Gamma = 1/\sqrt{1-v^2/K^2}$ with a universal $K$ — no light, no Maxwell. The transformation was there all along, waiting for one number. The content of the second postulate is not "light is special" but " $K$ is finite".

Worked examples

1 · Stacking boosts: how close to c can you get?

A rocket leaves Earth at $0.6c$ . It launches a second rocket at $0.6c$ relative to itself, which launches a third at $0.6c$ relative to the second, and so on. What is the $N$ -th rocket's speed relative to Earth?

Step 1. The group law (2) gives $\beta_2 = \dfrac{0.6+0.6}{1+0.36} = \dfrac{1.2}{1.36} = \dfrac{15}{17} = \boxed{0.8824}$ — not the $1.2c$ that adding would give, and crucially below $c$ .

Step 2. Find the pattern with rapidity. Rapidities add, so $N$ boosts of $0.6c$ carry rapidity $N\theta$ with $\theta = \operatorname{artanh}(0.6) = \tfrac12\ln 4 = \boxed{\ln 2}$ , exactly. Hence $\beta_N = \tanh(N\ln 2)$ , and in exponentials $\beta_N = \dfrac{2^N-2^{-N}}{2^N+2^{-N}} = \boxed{\dfrac{4^N-1}{4^N+1}}$ .

Step 3. Evaluate. $N=1$ : $3/5 = 0.600000$ . $N=2$ : $15/17 = 0.882353$ — Step 1 confirmed. $N=3$ : $63/65 = 0.969231$ . $N=4$ : $255/257 = \boxed{0.992218}$ . $N=10$ : $0.999998$ . Every one is strictly below 1 for every finite $N$ , because numerator and denominator differ by exactly $2$ . The shortfall $1-\beta_N = 2/(4^N+1)$ shrinks fourfold per rocket and never reaches zero.

Step 4. Cross-check at the top end. With $\beta_1 = \beta_2 = 0.99$ : $\dfrac{1.98}{1.9801} = \boxed{0.999949}$ . By rapidity, $\operatorname{artanh}(0.99) = 2.6467$ , doubled is $\ln 199 = \boxed{5.2933}$ , and $\tanh(5.2933) = 0.999949$ . (Note $2 \times 2.6467 = 5.2934$ is not the way to get this: doubling a rounded rapidity and rounding again spoils the last place.)

2 · The transformation hands back the postulate it was built from

Send a photon along $+x$ at $u = c$ in $S$ . Compute its speed $u'$ in $S'$ from the derived transformation, for arbitrary $v$ and then for $v = 0.9c$ .

Step 1. Differentiate. From (1), $\mathrm{d}x' = \gamma(\mathrm{d}x - v\,\mathrm{d}t)$ and $\mathrm{d}t' = \gamma(\mathrm{d}t - v\,\mathrm{d}x/c^2)$ . Take the ratio, divide top and bottom by $\mathrm{d}t$ , write $u = \mathrm{d}x/\mathrm{d}t$ : the factors of $\gamma$ cancel identically, leaving $u' = \dfrac{u-v}{1-uv/c^{2}}$ .

Step 2. Put in $u = c$ : $u' = \dfrac{c-v}{1-v/c} = \dfrac{c(c-v)}{c-v} = \boxed{c}$ — for every $v$ with $|v| < c$ . The $(c-v)$ upstairs cancels the $(c-v)$ hiding downstairs; nothing was assumed about $v$ .

Step 3. Check at $v = 0.9c$ . Naively the photon should be nearly overtaken and crawl away at $0.1c$ . Instead $u' = \dfrac{c-0.9c}{1-0.9} = \dfrac{0.1c}{0.1} = \boxed{c = 2.998\times10^{8}\ \mathrm{m/s}}$ : the numerator drops to a tenth, and so does the denominator.

Step 4. The invariant speed is not merely permitted by the transformation; it is a fixed point of the composition law, since (2) gives $\beta_3 = 1$ whenever $\beta_1 = 1$ , for any $\beta_2$ . That is far stronger than "nothing exceeds $c$ ", and it is why the derivation is self-consistent rather than circular. In Derivation C the logic runs the other way: closure forces the fixed point to exist, and only its value has to be looked up.

Physics of the portrait

The portrait is not a person but an equation, and deliberately so, because no one person owns it. The transformation was written down, in whole or in part, by at least five people over eighteen years, most of whom thought it meant something quite different from what it turned out to mean. Einstein's contribution was not the algebra — that was already on the page — but the belief that the time coordinate it produced was time.

Origins · 2 minLorentz transformation and how this was discovered

Lorentz transformation

1887–1910 · Voigt · Lorentz · Larmor · Poincaré · Einstein · Ignatowski

The equation arrived before its meaning did. Woldemar Voigt (1887), in a paper on the Doppler principle, applied to the wave equation a transformation equivalent to a Lorentz boost combined with an overall rescaling; Lorentz acknowledged the priority in 1909. Hendrik Antoon Lorentz built it up in stages — the contraction in 1892, the "local time" $t' = t - vx/c^2$ of the 1895 Versuch (an auxiliary variable for optical problems, explicitly not real time), and the full transformation by 1904. Joseph Larmor reached it in 1897 — the first to see that it entails a time dilation — and presented it more completely in Aether and Matter (1900). Henri Poincaré, in Sur la dynamique de l'électron (1905), gave it the name that stuck and proved the transformations form a group, fixing the undetermined scale factor $l(v)$ at $1$ . Albert Einstein (1905) derived the same equations from the relativity principle and the constancy of $c$ — no ether, no dynamics — and took the primed time seriously as the time read by moving clocks. Finally Vladimir Ignatowski (1910) closed the loop: the group structure alone forces the form, leaving only the invariant constant's value to experiment. The algebra was never the hard part. Believing it was.

History

1887

— Voigt, in "Über das Doppler'sche Princip", applies to the wave equation a transformation equivalent to a Lorentz boost up to an overall scale factor; Lorentz acknowledged the priority in 1909.

1889

— FitzGerald proposes the length contraction to explain the Michelson–Morley null result.

1892

— Lorentz, independently, proposes the same contraction.

1895

— Lorentz's Versuch introduces "local time" t′ = t − vx/c², an auxiliary variable he does not regard as real time.

1897

— Larmor obtains the transformation to second order in v/c and is the first to see that it implies a time dilation.

1900

— Larmor's Aether and Matter presents the transformation more completely.

1904

— Lorentz publishes the full transformation, valid to all orders.

1905

— Poincaré, in "Sur la dynamique de l'électron", names it the Lorentz transformation, proves the transformations form a group, and fixes the factor l(v) = 1.

1905

— Einstein derives it from the relativity principle and the constancy of light, with no ether and no dynamical hypothesis about matter.

1908

— Minkowski recasts it as a rotation-like symmetry of four-dimensional spacetime.

1910

— Ignatowski derives the transformation from relativity, homogeneity, isotropy and group closure, WITHOUT the light postulate; the invariant speed emerges as an undetermined universal constant.

1910

— Varićak develops the hyperbolic (rapidity) form of the boosts.

Going further

Texts. Rindler, Introduction to Special Relativity, 2nd ed., §2 · Taylor and Wheeler, Spacetime Physics, 2nd ed., ch. 3 · Griffiths, Introduction to Electrodynamics, 4th ed., §12.1 · W. v. Ignatowsky, Phys. Zeitschr. 11 (1910) 972 · J.-M. Lévy-Leblond, Am. J. Phys. 44 (1976) 271.

Note on a rounding point: the doubled rapidity is sometimes quoted as $5.2934$ — twice the rounded value $2.6467$ — but the exact value is $\ln 199 = 5.2933$ , which is what is printed above.

Builds on. The Two Postulates of Special Relativity · The Lorentz Factor Gamma · Einstein Clock Synchronization · Galilean Transformations. Leads to. The Lorentz Transformation Equations · The Invariant Spacetime Interval · Relativity of Simultaneity · Time Dilation · Length Contraction · Rapidity and the Hyperbolic Form of Boosts.

If three of the four constants come from symmetry, why is the light postulate treated as the revolutionary idea?

Because it is the one that is false in Newtonian physics: the Galilean transformation satisfies homogeneity, isotropy and relativity perfectly well. The postulate merely chooses between the two worlds those symmetries leave open, $K = \infty$ or $K$ finite — formally a small choice, physically everything, since $K = \infty$ says there is a universal "now" and a finite $K$ says there is not.

Where exactly does relativity of simultaneity enter the derivation? I never saw it assumed.

It was never assumed — it is output. Step 4's light rays forced $B = -vD/c^2$ , and $B$ is the coefficient of $x$ in $t' = At + Bx$ ; a non-zero $B$ makes the primed time depend on the unprimed position, so two events at equal $t$ but different $x$ give $\Delta t' = -\gamma v\Delta x/c^2 \neq 0$ . The postulate-free route sharpens it: reciprocity alone forces $\mathcal{C} = (1-\Gamma^2)/(\Gamma v)$ , non-zero unless $\Gamma \equiv 1$ . The consequences belong to Relativity of Simultaneity.

Why must the transformation be linear? Couldn't a curved map also send inertial frames to inertial frames?

No. A free particle traces a straight worldline in any inertial frame, and a map carrying every straight line to a straight line is a collineation — over the reals the only continuous ones are affine, and coincident origins remove the shift. Linearity is therefore a theorem, not a convenience. Homogeneity gives it independently: no event is privileged, so the coefficients cannot depend on where you evaluate them.

What actually goes wrong if I try K² < 0 in the Ignatowski derivation?

You get a consistent group — the Euclidean rotations — but a world with no causal structure. The law becomes $v_3 = (v_1+v_2)/(1-v_1v_2/k^2)$ , whose denominator can vanish, so two finite boosts compose to an infinite one; and boosts turn periodic, so every pair of events can have its time-order reversed. That destroys causality, excluding the case on physical, not mathematical, grounds. So closure forces a universal $K^2$ , with $K^2 > 0$ (Lorentz), $K^2 = \infty$ (Galileo) and $K^2 < 0$ (Euclidean, acausal) the three possibilities.

Did Ignatowski really do all this in 1910, or is this a modern reconstruction?

Both. His 1910 paper genuinely was the first attempt to get the transformation from the relativity principle by group theory alone, and he did reach it with an undetermined invariant speed. But his identification of that speed with $c$ was not clean — he appealed to the contraction of a moving electrostatic field, importing electrodynamics through the back door, and later commentators (Torretti most sharply) found the supplementary hypotheses inconsistent. Rigorous versions came later, notably Lévy-Leblond (1976). The group argument gives the existence and frame-independence of $K$ , nothing more; its value is experimental.

The group law says velocities do not add. So what does add?

Rapidity, $\theta = \operatorname{artanh}\beta$ , exactly: $\theta_3 = \theta_1+\theta_2$ , with no correction at any order. It is the group's natural coordinate, and it makes the speed limit trivial, since $\beta = \tanh\theta$ lies in $(-1,1)$ for every real $\theta$ . Because $\theta$ is unbounded you can always add more rapidity even as $\beta$ crawls toward $1$ : each $0.6c$ boost adds exactly $\ln 2$ , forever, while $\beta$ climbs $0.600 \to 0.882 \to 0.969 \to 0.992$ .