The Lorentz Transformation Equations

A muon is created high in the atmosphere and, $6.393\ \mathrm{\mu s}$ later, decays $1907\ \mathrm{m}$ lower down. Those are the laboratory's numbers. Feed them through the Lorentz transformation at $\beta = 0.995$ and something remarkable comes back: $t' = 0.6385\ \mathrm{\mu s}$ and $x' = 0$ — exactly zero. In its own frame the muon never moved at all. It was born, it waited $0.6385\ \mathrm{\mu s}$ by its own clock, and it died in the same place. Two frames, one event, two utterly different stories — and the transformation is the dictionary that makes them the same story.

Derivation of the Lorentz Transformation builds that dictionary from the postulates. This article takes it as given and turns it into a working toolkit. Written as a matrix acting on the four-vector $X = (ct, x)$ , the boost is $\Lambda(\beta) = \begin{pmatrix}\gamma & -\gamma\beta\\ -\gamma\beta & \gamma\end{pmatrix}$ , and almost everything you need follows from three facts about that matrix: its inverse is $\Lambda(-\beta)$ , its determinant is $1$ , and — the deep one — it satisfies $\Lambda^{\mathsf T}\eta\,\Lambda = \eta$ with $\eta = \mathrm{diag}(+1,-1)$ . That last identity is not a curiosity. It is the definition of a Lorentz transformation: the matrices that preserve the Minkowski metric. Interval invariance, $c^2t^2 - x^2 - y^2 - z^2$ , is the same statement written out in components, and the four-vector rule follows in a line. With the toolkit in hand the famous consequences drop out in two lines each. Set $\Delta x' = 0$ and you have time dilation; set $\Delta t = 0$ and you have length contraction; divide $\mathrm{d}x'$ by $\mathrm{d}t'$ and you have velocity composition, with $u = c \Rightarrow u' = c$ for every $v$ — the invariant speed is a fixed point, not merely a ceiling. Each of those results has an owner elsewhere in the track; here they are read off, not re-derived. And one small term deserves your attention before you meet the man whose name is on the equations: expand the boost to first order in $v$ and what survives is $t' \approx t - vx/c^2$ — precisely the "local time" that Hendrik Lorentz introduced in 1895 as a calculational fiction he did not believe in.

Key takeaways

The lorentz boost along $x$ is a matrix on $X = (ct,x)$ : $\Lambda(\beta) = \begin{pmatrix}\gamma & -\gamma\beta\\ -\gamma\beta & \gamma\end{pmatrix}$ , with $\beta = v/c$ and $\gamma = 1/\sqrt{1-\beta^2}$ . Transverse coordinates are untouched.
Three structural facts do all the work: $\Lambda(\beta)^{-1} = \Lambda(-\beta)$ , $\det\Lambda = \gamma^2(1-\beta^2) = 1$ , and $\Lambda^{\mathsf T}\eta\,\Lambda = \eta$ with $\eta = \mathrm{diag}(+1,-1)$ .
$\Lambda^{\mathsf T}\eta\Lambda = \eta$ IS the definition of a Lorentz transformation; interval invariance $c^2t'^2 - x'^2 - y'^2 - z'^2 = c^2t^2 - x^2 - y^2 - z^2$ is that identity written in components, and it holds identically in $v$ .
The consequences are read off in two lines each: $\Delta x' = 0 \Rightarrow \Delta t = \gamma\Delta\tau$ (time dilation); $\Delta t = 0 \Rightarrow L = L_0/\gamma$ (length contraction); $u' = (u-v)/(1-uv/c^2)$ , with $u = c \Rightarrow u' = c$ for every $v$ .
A boost slides events along the hyperbolae $c^2t^2 - x^2 = \text{const}$ : at $\beta = 0.6$ the event $(ct,x) = (5,3)$ goes to $(ct',x') = (4,0)$ , and $s^2 = 16$ on both sides.
Key idea: a Lorentz boost is a hyperbolic rotation of spacetime — the linear map that mixes $t$ and $x$ while leaving $c^2t^2 - x^2$ exactly where it was. Everything else in special relativity kinematics is a corollary of that one invariance.

The core insight

Stop thinking of the boost as a pair of equations to memorise and start thinking of it as a matrix with a symmetry. $\Lambda(\beta)$ preserves the Minkowski metric, $\Lambda^{\mathsf T}\eta\Lambda = \eta$ — and that single condition already implies $\det\Lambda = \pm1$ , that the inverse is $\eta\Lambda^{\mathsf T}\eta = \Lambda(-\beta)$ , and that the interval $c^2t^2 - x^2 - y^2 - z^2$ is the same number in every inertial frame. Geometrically it is a hyperbolic rotation: events slide along the hyperbolae of constant interval, never off them. Time dilation, length contraction and velocity composition are then not three separate discoveries but three ways of reading one matrix.

Where this lives today

Every high-energy physics experiment on Earth runs equation (1) millions of times a second. At the LHC, particles are produced in a collision and analysed in the laboratory frame, but the physics that governs the decay lives in the rest frame of the parent particle — so reconstruction code boosts four-momenta back and forth continuously, using exactly the matrix above. The invariant of equation (4), applied to the energy–momentum four-vector, is the invariant mass: $m^2c^4 = E^2 - p^2c^2$ . That single number, computed from the debris of a collision and independent of the frame in which it was measured, is how the Higgs boson was found — a bump at $125\ \mathrm{GeV}$ in a quantity that no boost can shift. The same arithmetic keeps satellites honest. A GPS satellite moves at about $3.874\ \mathrm{km/s}$ , giving $\gamma - 1 \approx 8.35\times10^{-11}$ and a special-relativistic clock loss of roughly $7.2\ \mathrm{\mu s}$ per day; the general-relativistic gain is larger and of the opposite sign, and the net correction is about $+38.6\ \mathrm{\mu s}$ per day. Left uncorrected, that drift would corrupt positions by several kilometres within a day. The clocks are pre-tuned before launch so that, once in orbit, they tick at the rate the ground expects — the Lorentz transformation, compiled into hardware.

The physical picture

Feed an event in, get an event out. That is all the transformation does — but the numbers it returns can be startling, and the muon is the cleanest demonstration in physics. In the laboratory the muon is a fast particle that travels almost two kilometres. Boost to its own frame and the journey vanishes: the muon sits still, and the mountain rushes past it. Same event, same physics, different bookkeeping.

Interactive Simulation

Lorentz Boost of an Event — drag β and watch a fixed event's coordinates transform between frames as the axes skew (the event stays put; only its description changes).

Figure

The boost as a machine. A real event goes in — the muon's decay, logged in the laboratory at ct = 1916.58 m and x = 1907 m — and the matrix Λ(0.995), with γ = 10.0125, returns ct′ = 191.42 m and x′ = 0.00 m. That x′ = 0 is exact, not rounded: in its own frame the muon dies precisely where it was born, after only 191.42/c = 0.6385 μs of its own time.

Now watch what a boost does to the whole spacetime diagram at once. It does not translate events and it does not rotate them in the Euclidean sense; it slides them along hyperbolae. Every event stays on the curve $c^2t^2 - x^2 = \text{constant}$ that it started on, and the value of that constant — the interval — is the one number both observers agree on. The axes scissor toward the $45^\circ$ light line, but the hyperbolae do not budge.

Figure

PUNCHLINE — the invariant hyperbola c²t² − x² = 16. The event E = (ct, x) = (5, 3) sits on it, since 25 − 9 = 16. Under a β = 0.6 boost it slides along the curve to the vertex V = (4, 0) — landing exactly on the t′ axis, so the primed observer says x′ = 0 and ct′ = 4. The interval is 16 on BOTH sides. A boost slides events along the hyperbolae; the interval is what never moves.

The core equations

Standard configuration: $S'$ moves at $+v$ along the $x$ -axis of $S$ , axes parallel, origins coinciding at $t = t' = 0$ ; $\beta \equiv v/c$ , $\gamma \equiv 1/\sqrt{1-\beta^2}$ . Scale time to a length, $x^0 = ct$ , so both entries of the column carry the same units, and the boost is a single matrix.

\begin{pmatrix} ct' \\ x' \end{pmatrix} = \Lambda(\beta) \begin{pmatrix} ct \\ x \end{pmatrix}, \qquad \Lambda(\beta) = \begin{pmatrix} \gamma & -\gamma\beta \\ -\gamma\beta & \gamma \end{pmatrix}, \qquad y' = y, \qquad z' = z \qquad (1)

where: $(ct,x)$ and $(ct',x')$ — one event, in $S$ and in $S'$ ; $\gamma\beta$ — the off-diagonal mixing coefficient. Physical reading: written out, the rows are $t' = \gamma(t - vx/c^2)$ and $x' = \gamma(x - vt)$ — the same equations Derivation of the Lorentz Transformation obtains from the postulates, now in a form you can multiply, invert and compose. The matrix is symmetric, its diagonal is $\gamma$ , and the whole of relativistic kinematics is hiding in the off-diagonal $-\gamma\beta$ . Two facts follow immediately, and both are used constantly.

\Lambda(\beta)^{-1} = \Lambda(-\beta) = \begin{pmatrix} \gamma & +\gamma\beta \\ +\gamma\beta & \gamma \end{pmatrix}, \qquad \det\Lambda = \gamma^{2}(1-\beta^{2}) = 1 \qquad (2)

where: $\Lambda(-\beta)$ — the boost of the same speed in the opposite direction. Physical reading: the inverse transformation is the forward one with $v \to -v$ — because $S$ moves at $-v$ as seen from $S'$ . Written out: $t = \gamma(t' + vx'/c^2)$ and $x = \gamma(x' + vt')$ . The unit determinant says a boost neither creates nor destroys spacetime volume; there is no hidden rescaling. Note the trap the warning block below dissects: $v \to -v$ is only half the rule — you must also swap which coordinates are primed. Now the identity that everything rests on. Introduce the Minkowski metric $\eta$ , and the boost's defining property is a single matrix equation.

\Lambda^{\mathsf T}\,\eta\,\Lambda = \eta, \qquad \eta = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \quad\text{(signature } +\,-\,-\,-\,) \qquad (3)

where: $\eta$ — the metric, $\mathrm{diag}(+1,-1,-1,-1)$ in full $3+1$ . Physical reading: this is the definition of a Lorentz transformation — the set of matrices preserving $\eta$ is the Lorentz group. Everything else is a corollary: taking determinants of (3) gives $(\det\Lambda)^2 = 1$ , and multiplying (3) by $\eta$ gives $\Lambda^{-1} = \eta\,\Lambda^{\mathsf T}\eta$ , which is $\Lambda(-\beta)$ . A boost is not "the equations that happen to work"; it is the linear map that leaves one quadratic form alone. Write (3) out in components and it becomes the statement students meet first.

c^{2}t'^{2} - x'^{2} - y'^{2} - z'^{2} \;=\; c^{2}t^{2} - x^{2} - y^{2} - z^{2} \;\equiv\; s^{2} \qquad (4)

where: $s^2$ — the spacetime interval, invariant under every boost and rotation. Physical reading: two observers disagree about the time between two events and about the distance between them, but they agree — exactly, to all orders in $v$ — about this combination. The same rule extends to any four numbers that transform with $\Lambda$ : they form a four-vector, and their square $\,(a^0)^2 - (a^1)^2 - (a^2)^2 - (a^3)^2$ is an invariant. What $s^2$ means geometrically belongs to The Invariant Spacetime Interval, and the four-vector formalism to Four-Vectors; what this article owns is the fact that the transformation preserves it. Finally, the read-off that catches the postulate coming back out.

u' = \frac{u - v}{1 - uv/c^{2}}, \qquad\text{so}\qquad u = c \;\Longrightarrow\; u' = \frac{c-v}{1-v/c} = c \quad\text{for every } v \qquad (5)

where: $u = \mathrm{d}x/\mathrm{d}t$ in $S$ ; $u' = \mathrm{d}x'/\mathrm{d}t'$ in $S'$ . Physical reading: divide the two rows of (1) in differential form and the $\gamma$ 's cancel. Setting $u = c$ returns $u' = c$ identically — the invariant speed is a fixed point of the composition law, not merely an upper bound. As a kinematic tool this formula is owned by the Relativistic Velocity Addition Formula and by Velocity Composition and the Speed-of-Light Invariant; here it is simply one more thing the matrix hands you.

Where the results come from

Derivation AThe matrix, its inverse, and the identity that defines a Lorentz transformation

The goal is to prove the three structural facts of equations (2) and (3) rather than assert them, and then to notice that the third one implies the other two. Most treatments present $\det\Lambda = 1$ and $\Lambda^{-1} = \Lambda(-\beta)$ as separate observations. They are not: both fall out of $\Lambda^{\mathsf T}\eta\Lambda = \eta$ , which is the real content — and which says, in one line, that a boost is whatever preserves the Minkowski metric.

Step 1 — the inverse. Multiply $\Lambda(-\beta)\Lambda(\beta)$ . The diagonal entries are $\gamma^2 - \gamma^2\beta^2 = \gamma^2(1-\beta^2)$ , and the off-diagonal entries are $\gamma^2\beta - \gamma^2\beta = 0$ . Since $\gamma^2(1-\beta^2) = 1$ by the definition of $\gamma$ , the product is $\mathbb{1}$ , so $\boxed{\Lambda(\beta)^{-1} = \Lambda(-\beta)}$ . Why it works: the cross terms cancel because the matrix is symmetric, and the diagonal is rescued by the one identity $\gamma^2(1-\beta^2) = 1$ — the same identity that fixed $\gamma$ in the first place.

Step 2 — the determinant. $\det\Lambda = \gamma\cdot\gamma - (-\gamma\beta)(-\gamma\beta) = \gamma^2(1-\beta^2) = \boxed{1}$ . Why it matters: a unit determinant means the boost preserves spacetime volume and contains no sneaky overall scale factor. Historically this is exactly the factor $l(v)$ that Lorentz left undetermined in 1904 and that Poincaré fixed at $1$ in 1905, using the group property.

Step 3 — the metric condition. Compute $\Lambda^{\mathsf T}\eta\Lambda$ with $\eta = \mathrm{diag}(+1,-1)$ . Because $\Lambda$ is symmetric, $\Lambda^{\mathsf T} = \Lambda$ , and the $(1,1)$ entry is $\gamma^2 - \gamma^2\beta^2 = 1$ ; the $(2,2)$ entry is $\gamma^2\beta^2 - \gamma^2 = -1$ ; the off-diagonal entries are $-\gamma^2\beta + \gamma^2\beta = 0$ . So $\Lambda^{\mathsf T}\eta\Lambda = \mathrm{diag}(1,-1) = \boxed{\eta}$ . Why this is the whole subject in one equation: the Lorentz group is by definition the set of $\Lambda$ satisfying this. A boost does not "cause" length contraction; it is a metric-preserving map, and contraction is one of the shadows it casts.

Step 4 — the corollaries, for free. Take the determinant of Step 3: $(\det\Lambda)^2\det\eta = \det\eta$ , hence $\det\Lambda = \pm1$ — and $+1$ for a boost continuously connected to the identity. Multiply Step 3 on the left by $\eta$ and use $\eta^2 = \mathbb{1}$ : $\eta\Lambda^{\mathsf T}\eta\Lambda = \mathbb{1}$ , so $\Lambda^{-1} = \eta\,\Lambda^{\mathsf T}\eta$ — which, worked out, is precisely $\Lambda(-\beta)$ . Why the logic is worth reordering: Steps 1 and 2 are not independent facts to memorise. They are consequences of Step 3, and Step 3 is the definition. *Verified in verify_aq-sr-lorentz-transformation-equations.py: sympy confirms $\Lambda$ reproduces $t' = \gamma(t-vx/c^2)$ and $x' = \gamma(x-vt)$ ; that $\Lambda(-\beta)\Lambda(\beta) = \mathbb{1}$ and $\Lambda^{-1} = \Lambda(-\beta)$ symbolically; that $\det\Lambda = \gamma^2(1-\beta^2) = 1$ ; that $\Lambda^{\mathsf T}\eta\Lambda = \eta$ entry by entry; and that the corollaries $\det\Lambda = \pm1$ and $\Lambda^{-1} = \eta\Lambda^{\mathsf T}\eta$ follow from the metric condition alone.*

Go deeper · Derivation B · 2 minInterval invariance for a general boost, and the four-vector rule

Derivation BInterval invariance for a general boost, and the four-vector rule

Equation (3) is a statement about a matrix; equation (4) is a statement about numbers. They are the same statement, and this card proves it by brute substitution — not to some order in $v/c$ , but identically in $v$ , for every event. That exactness is the point: the interval is not approximately conserved at low speed, it is exactly conserved at every speed, and that is what makes it the right thing to build a geometry out of.

Step 1 — substitute. Take $t' = \gamma(t - vx/c^2)$ , $x' = \gamma(x - vt)$ , $y' = y$ , $z' = z$ and form $c^2t'^2 - x'^2 - y'^2 - z'^2$ . The first two terms give $\gamma^2\big[(ct - \beta x)^2 - (x - \beta ct)^2\big]$ . Why this is the whole calculation: $y$ and $z$ ride through untouched, so the only work is in the $t$ – $x$ block.

Step 2 — expand, and watch the cross terms die. $(ct-\beta x)^2 - (x-\beta ct)^2 = c^2t^2 - 2\beta ctx + \beta^2x^2 - x^2 + 2\beta ctx - \beta^2c^2t^2$ . The $\pm 2\beta ctx$ terms cancel exactly, leaving $c^2t^2(1-\beta^2) - x^2(1-\beta^2)$ . Why the cancellation is not luck: it is the off-diagonal symmetry of $\Lambda$ — the same $-\gamma\beta$ appears in both rows, so the cross terms are equal and opposite.

Step 3 — collect. The bracket is $(1-\beta^2)(c^2t^2 - x^2)$ , and the prefactor is $\gamma^2 = 1/(1-\beta^2)$ . They cancel: $c^2t'^2 - x'^2 = \boxed{c^2t^2 - x^2}$ , and restoring $y' = y$ , $z' = z$ gives equation (4) in full. This holds for arbitrary $t, x, y, z$ and arbitrary $v$ — an identity, not an approximation. Why $\gamma^2(1-\beta^2) = 1$ keeps reappearing: it is the same identity that gave the inverse and the determinant. One relation, three theorems.

Step 4 — the four-vector rule. Nothing in Steps 1–3 used the fact that $(ct,x,y,z)$ were coordinates. Let $A = (a^0, a^1, a^2, a^3)$ be any four numbers that transform with the same $\Lambda$ . The identical algebra gives $(a^0)'^2 - (a^1)'^2 - (a^2)'^2 - (a^3)'^2 = (a^0)^2 - (a^1)^2 - (a^2)^2 - (a^3)^2$ . Such an object is a four-vector, and its square is a Lorentz invariant. Why this is the most productive sentence in relativity: it means energy–momentum, current density and the potentials all inherit an invariant for free, once you check they transform with $\Lambda$ . The formalism belongs to Four-Vectors; the licence to use it is proved here. *Verified in verify_aq-sr-lorentz-transformation-equations.py: sympy substitutes the boost into $c^2t^2-x^2-y^2-z^2$ and returns it unchanged, identically in $v$ ; repeats it for coordinate differences $(\Delta t, \Delta x, \Delta y, \Delta z)$ ; confirms the full $4\times4$ $\Lambda^{\mathsf T}\eta\Lambda = \eta$ with $\eta = \mathrm{diag}(+1,-1,-1,-1)$ and $\det\Lambda = 1$ ; and checks the four-vector rule for arbitrary symbolic components $(a^0,a^1,a^2,a^3)$ .*

Go deeper · Derivation C · 2 minThe consequences, read off in two lines each

Derivation CThe consequences, read off in two lines each

This card is deliberately short, because that is the claim: once you have the matrix, the celebrated effects of special relativity are not discoveries but substitutions. Each takes two lines. Each is owned in full by another article — this one shows only how quickly the toolkit yields them, and, crucially, exactly which condition you must impose to get each.

Step 1 — time dilation: set $\Delta x' = 0$ . Two ticks of one clock at rest in $S'$ happen at the same place in $S'$ . Use the inverse, $\Delta t = \gamma(\Delta t' + v\Delta x'/c^2)$ , and put $\Delta x' = 0$ : $\boxed{\Delta t = \gamma\,\Delta t' = \gamma\,\Delta\tau}$ . Why the condition matters: "one clock, one place" is what $\Delta x' = 0$ encodes. Impose it on the wrong equation and you get the wrong power of $\gamma$ . Owner: Time Dilation.

Step 2 — length contraction: set $\Delta t = 0$ . To measure a moving rod you must mark both ends at once in your frame. Use the forward equation $\Delta x' = \gamma(\Delta x - v\Delta t)$ with $\Delta t = 0$ : $\Delta x' = \gamma\Delta x$ . Here $\Delta x' = L_0$ is the proper length and $\Delta x = L$ the measured one, so $\boxed{L = L_0/\gamma}$ . Why the condition matters: simultaneity is the whole content of the effect. Mark the ends at different times and you can get any answer you like — including a longer rod. Owner: Length Contraction.

Step 3 — velocity composition: divide $\mathrm{d}x'$ by $\mathrm{d}t'$ . Differentiate (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 and divide top and bottom by $\mathrm{d}t$ : the $\gamma$ 's cancel and $\boxed{u' = \dfrac{u-v}{1 - uv/c^{2}}}$ . Why the $\gamma$ 's cancel: they are a common factor of both rows — velocity is a ratio, so any overall scale drops out. Owners: Relativistic Velocity Addition Formula and Velocity Composition and the Speed-of-Light Invariant.

Step 4 — the fixed point. Put $u = c$ in Step 3: $u' = \dfrac{c-v}{1-cv/c^2} = \dfrac{c-v}{1-v/c} = \dfrac{c(c-v)}{c-v} = \boxed{c}$ , for every $v$ with $|v| < c$ . Why this is more than a consistency check: it says $c$ is a FIXED POINT of the transformation, the one speed on which all observers agree. The postulate that was fed into the derivation comes back out of the finished machine — which is what it means for the theory to be self-consistent. *Verified in verify_aq-sr-lorentz-transformation-equations.py: sympy checks the inverse map really undoes the forward one; derives $\Delta t = \gamma\Delta\tau$ from $\Delta x' = 0$ ; derives $L = L_0/\gamma$ from $\Delta t = 0$ ; obtains $u' = (u-v)/(1-uv/c^2)$ by dividing $\mathrm{d}x'$ by $\mathrm{d}t'$ ; and confirms $u = c \Rightarrow u' = c$ symbolically and numerically at $\beta = 0, 0.6, 0.9, 0.995$ .*

Limiting cases

Case	Result	Meaning
$\beta = 0$	$\Lambda(0) = \mathbb{1}$	the frames coincide; the identity element of the group
$\Lambda(\beta)\Lambda(-\beta)$	$= \mathbb{1}$	the inverse boost is the reversed boost
$\det\Lambda$	$= \gamma^2(1-\beta^2) = 1$	no rescaling; spacetime volume is preserved
$\Lambda^{\mathsf T}\eta\Lambda$	$= \eta$	the definition of a Lorentz transformation
$u = c$	$u' = c$ for every $v$	the invariant speed is a fixed point, not just a ceiling
$\beta \ll 1$	$x' \approx x - vt$ , $t' \approx t - vx/c^2$	Galilean shear plus the surviving $-vx/c^2$ — Lorentz's "local time"

Common trap“The inverse is easy — just flip the sign of v anywhere you see it.”

Half true, and the half that is false will cost you marks. The rule is not " $v \to -v$ "; it is " $v \to -v$ and swap which coordinates are primed". The inverse of $x' = \gamma(x - vt)$ is $x = \gamma(x' + vt')$ — the primes have moved to the right-hand side. If you merely flip the sign and keep the unprimed coordinates on the right, writing $x'' = \gamma(x + vt)$ , you have not inverted anything: you have boosted to a different frame, one moving at $-v$ . Here it is concretely, with the numbers from worked example 2. The event $(ct,x) = (5,3)$ under $\Lambda(0.6)$ goes to $(4,0)$ . Apply $\Lambda(-0.6)$ to the primed event $(4,0)$ and you get back $(5,3)$ : correct, that is the inverse. But apply $\Lambda(-0.6)$ to the unprimed event $(5,3)$ and you get $(8.5, 7.5)$ — a perfectly legal boost (it still has $s^2 = 8.5^2 - 7.5^2 = 16$ , so it is still on the same hyperbola), but it is not the inverse of anything you did. Same matrix, different input, completely different meaning. The safe habit: never "flip a sign"; always write $X = \Lambda(-\beta)X'$ and check which column you are feeding it.

Worked examples

1 · Transforming a real event — the muon's own frame

A muon is created at $(t,x) = (0,0)$ and decays in the laboratory at $t = 6.393\ \mathrm{\mu s}$ , $x = 1907\ \mathrm{m}$ (the Frisch–Smith geometry: Mount Washington to Cambridge). It travels at $\beta = 0.995$ . Find the decay event's coordinates in the muon's own frame.

Step 1. Get $\gamma$ . $\gamma = 1/\sqrt{1 - 0.995^2} = 1/\sqrt{0.009975} = \boxed{10.0125}$ . Note the event lies on the muon's worldline, so $x = \beta c\,t$ exactly; that is what makes $t = 1907/(0.995c) = 6.393\ \mathrm{\mu s}$ .

Step 2. Transform the time. $t' = \gamma\!\left(t - \dfrac{vx}{c^2}\right) = \gamma\,t\,(1 - \beta^2) = \dfrac{t}{\gamma} = \dfrac{6.393}{10.0125} = \boxed{0.6385\ \mathrm{\mu s}}$ . The muon's own clock records only $0.6385\ \mathrm{\mu s}$ between birth and death — a little under a third of a mean lifetime ( $\tau_0 = 2.197\ \mathrm{\mu s}$ ), which is precisely why so many of these muons survive the trip.

Step 3. Transform the position. $x' = \gamma(x - vt) = \gamma\,(x - \beta c t) = \gamma\,(1907 - 1907) = \boxed{0}$ — exactly zero, not a small number. In its own frame the muon does not move; it is the mountain that comes to it.

Step 4. Check with the invariant. In the lab, $s^2 = (ct)^2 - x^2 = (1916.58)^2 - (1907)^2 = 3.664\times10^{4}\ \mathrm{m^2}$ . In the muon frame, $s^2 = (ct')^2 - 0 = (191.42)^2 = 3.664\times10^{4}\ \mathrm{m^2}$ . Identical, as equation (4) demands — and $\sqrt{s^2}/c = 0.6385\ \mathrm{\mu s}$ , which is the muon's own elapsed time. The interval, read on a timelike worldline, is the proper time.

2 · Sliding along the hyperbola, and inverting safely

At $\beta = 0.6$ , transform the event $(ct, x) = (5, 3)$ (units: metres, or any consistent unit). Show it lands on the $t'$ axis, verify the interval, and then use the inverse to get back.

Step 1. Build the matrix. $\gamma = 1/\sqrt{1-0.36} = 1/0.8 = 1.25$ and $\gamma\beta = 0.75$ , so $\Lambda(0.6) = \begin{pmatrix} 1.25 & -0.75 \\ -0.75 & 1.25\end{pmatrix}$ , with $\det = 1.25^2 - 0.75^2 = 1.5625 - 0.5625 = \boxed{1}$ .

Step 2. Apply it. $ct' = 1.25(5) - 0.75(3) = 6.25 - 2.25 = \boxed{4}$ and $x' = -0.75(5) + 1.25(3) = -3.75 + 3.75 = \boxed{0}$ . The event lands exactly on the $t'$ axis — which it had to, because $x = \beta\,ct$ here ( $3 = 0.6 \times 5$ ), and $x = \beta ct$ is the worldline of the $S'$ origin.

Step 3. Check the interval. Before: $s^2 = 5^2 - 3^2 = 25 - 9 = \boxed{16}$ . After: $s^2 = 4^2 - 0^2 = \boxed{16}$ . The event slid along the hyperbola $c^2t^2 - x^2 = 16$ from the point $(5,3)$ to that curve's vertex $(4,0)$ — the figure above, in arithmetic.

Step 4. Invert — carefully. $\Lambda(-0.6) = \begin{pmatrix} 1.25 & 0.75 \\ 0.75 & 1.25\end{pmatrix}$ . Feed it the primed event $(4,0)$ : $ct = 1.25(4) + 0.75(0) = 5$ and $x = 0.75(4) + 1.25(0) = 3$ . We are back at $\boxed{(5,3)}$ . Now the trap: feed the same matrix the unprimed event $(5,3)$ instead and you get $(1.25(5)+0.75(3),\ 0.75(5)+1.25(3)) = (8.5, 7.5)$ — still on the hyperbola ( $72.25 - 56.25 = 16$ ), still a legitimate boost, but a boost to a frame moving at $-0.6c$ , not the inverse of anything. The matrix is the same; only the column you feed it changed.

Physics of the portrait

The equations carry his name, and he never quite believed them. Lorentz spent his career defending an ether that his own mathematics was quietly dismantling, and the transformation that bears his name was, in his hands, a piece of scaffolding — a change of variables that made the equations of a moving system look like those of a stationary one. It fell to Einstein to say that the scaffolding was the building. Lorentz took that correction with a grace that made him, by common consent, the most beloved physicist of his generation.

Origins · 2 minHendrik Antoon Lorentz and how this was discovered

Hendrik Antoon Lorentz

1853–1928 · the man whose name is on the equations — and who did not believe what they meant

Born in Arnhem in 1853, Hendrik Antoon Lorentz was appointed to the newly created chair of theoretical physics at Leiden in 1878, aged twenty-four, and spent his life on the electron theory of matter. His 1895 Versuch</strong> introduced the quantity he called "local time", $t' = t - vx/c^2$ — and he introduced it explicitly as a calculational device, an auxiliary variable that made optical problems in moving bodies tractable. He did not regard it as the time that clocks actually read; true time belonged to the ether frame. By 1904 he had the full transformation, valid to all orders in $v/c$ , though he left an overall scale factor $l(v)$ undetermined; Poincaré fixed $l = 1$ in 1905, proved the transformations form a group, and gave them the name they still carry — the Lorentz transformation. His Nobel Prize was awarded in 1902, shared with Pieter Zeeman, for their work on the Zeeman effect — the splitting of spectral lines in a magnetic field. It was emphatically not for relativity, which in 1902 did not yet exist. He chaired the first Solvay Council in 1911 and most of those that followed, and from 1918 led the committee that computed how the Zuiderzee dam would alter the tides — a calculation whose predictions proved remarkably accurate. When he died in 1928 the Dutch telegraph and telephone services were suspended for three minutes; Einstein spoke at the graveside, calling him "the greatest and noblest man of our times".

History
1892
— Lorentz proposes the contraction of moving bodies and begins building his electron theory.
1895
— The Versuch introduces "local time" t′ = t − vx/c² as an auxiliary variable, together with the theorem of corresponding states to first order in v/c.
1899
— Lorentz extends the theorem of corresponding states and notices an effect amounting to time dilation.
1902
— Nobel Prize in Physics, shared with Pieter Zeeman, for the discovery and explanation of the Zeeman effect — not for relativity.
1904
— Lorentz publishes the full transformation, valid to all orders in v/c, leaving an overall factor l(v) undetermined.
1905
— Poincaré, in "Sur la dynamique de l'électron", names it the Lorentz transformation, proves the transformations form a group, and fixes l(v) = 1.
1905
— Einstein derives the same equations from two postulates, with no ether, and treats the primed time as the time that moving clocks read.
1908
— Minkowski recasts the transformation as the geometry of a four-dimensional spacetime with the invariant interval at its centre.
1911
— Lorentz chairs the first Solvay Council in Brussels.
1928
— Lorentz dies at Haarlem on 4 February; Einstein speaks at his graveside.

Going further

Texts. Rindler, Introduction to Special Relativity, 2nd ed., §§2–3 · Taylor and Wheeler, Spacetime Physics, 2nd ed., ch. 3 · Griffiths, Introduction to Electrodynamics, 4th ed., §12.1 · Hartle, Gravity, ch. 4 · H. A. Lorentz, "Electromagnetic phenomena in a system moving with any velocity smaller than that of light" (1904).

Provenance. The matrix form, the inverse, the determinant and the metric condition are standard (Rindler §§2–3; Griffiths §12.1; Hartle ch. 4). Every relation and every number is re-derived in `verify_aq-sr-lorentz-transformation-equations.py`, which confirms symbolically that $\Lambda(\beta)^{-1} = \Lambda(-\beta)$ , $\det\Lambda = 1$ , and $\Lambda^{\mathsf T}\eta\Lambda = \eta$ (in $2\times2$ and in full $4\times4$ ); that the interval is invariant identically in $v$ , for coordinates and for differences; the four-vector rule for arbitrary components; and the three read-offs ( $\Delta t = \gamma\Delta\tau$ , $L = L_0/\gamma$ , $u' = (u-v)/(1-uv/c^2)$ with $u = c \Rightarrow u' = c$ ). Numbers: $\gamma(0.995) = 10.0125$ , muon decay at $t = 6.393\ \mathrm{\mu s}$ , $x = 1907\ \mathrm{m}$ $\to$ $t' = 0.6385\ \mathrm{\mu s}$ , $x' = 0$ exactly, $ct = 1916.58\ \mathrm{m} \to ct' = 191.42\ \mathrm{m}$ ; $\gamma(0.6) = 1.25$ , $\gamma\beta = 0.75$ , $(5,3) \to (4,0)$ with $s^2 = 16$ both sides, inverse returns $(5,3)$ , and the $v \to -v$ trap gives $(8.5, 7.5)$ .

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

Why bother with the matrix at all? The two equations were fine.
Because the matrix is what lets you compose, invert and classify. Two boosts in a row are a matrix product, and the product is another boost — which is how you discover the group structure and the velocity-composition law without any new physics. The inverse is a matrix inverse, which removes the guesswork about signs and primes. And most importantly the matrix exposes $\Lambda^{\mathsf T}\eta\Lambda = \eta$ , which two loose equations hide completely; that identity is the doorway to four-vectors, to tensors, and to the whole apparatus of relativistic field theory.

If the interval is invariant, why do the coordinates change at all?
Because invariance of a combination is not invariance of its parts. Think of an ordinary rotation: it changes $x$ and $y$ separately while leaving $x^2 + y^2$ alone, and nobody finds that paradoxical. The Lorentz boost does the same thing with a minus sign: it changes $t$ and $x$ separately while leaving $c^2t^2 - x^2$ alone. Worked example 2 is exactly this — $(5,3)$ becomes $(4,0)$ , both coordinates change, and $25 - 9 = 16 - 0$ . The minus sign is what makes the invariant curves hyperbolae rather than circles, and it is the reason boosts can slide an event arbitrarily far along a curve without ever crossing the light cone.

In the muon example, is x′ = 0 exactly zero, or just very small?
Exactly zero, and the reason is structural rather than numerical. The decay event lies on the muon's own worldline, so its lab coordinates satisfy $x = \beta c\,t$ identically. Substituting into $x' = \gamma(x - vt) = \gamma(x - \beta c t)$ gives $\gamma(x - x) = 0$ — algebraically, with no rounding involved. That is simply the statement that in its rest frame the muon is at the origin at all times, which is what "rest frame" means. If you instead plug in the rounded $t = 6.393\ \mathrm{\mu s}$ , you will get a metre or two of residue; that residue is an artefact of the rounding, not physics.

Where has the relativity of simultaneity gone? I only see γ's.
It is sitting in the off-diagonal term, and it is easy to miss precisely because it looks so innocent. The time row of the matrix is $t' = \gamma(t - vx/c^2)$ , and that $-\gamma v x/c^2$ depends on position: two events with the same $t$ but different $x$ get different $t'$ . Concretely, two clocks a proper distance $L_0$ apart and synchronised in their own frame are found, from the ground, to be offset by $vL_0/c^2$ — for a $100\ \mathrm{m}$ carriage at $\beta = 0.6$ that is $200\ \mathrm{ns}$ , with the leading clock reading behind. Delete that one term and you are back to Galileo. The quantitative development belongs to Relativity of Simultaneity.

Does ΛᵀηΛ = η also allow rotations and reflections? Is a boost the only solution?
No, a boost is not the only solution — and that is a feature. The full set of matrices satisfying $\Lambda^{\mathsf T}\eta\Lambda = \eta$ is the Lorentz group, and it contains ordinary spatial rotations (which preserve $x^2+y^2+z^2$ and leave $t$ alone), boosts in any direction, and the discrete operations of parity ( $\mathbf{x} \to -\mathbf{x}$ ) and time reversal ( $t \to -t$ ). The boosts continuously connected to the identity, with $\det\Lambda = +1$ and the time direction preserved, form the proper orthochronous subgroup — the physically relevant one. So the metric condition is doing more work than just characterising boosts: it defines the entire symmetry group of spacetime.

The first-order term of the boost is t′ ≈ t − vx/c². Is that really Lorentz's "local time"?
Yes, and the coincidence is not one. Expand equation (1) to first order in $v$ : $\gamma = 1 + O(v^2)$ , so $x' \approx x - vt$ and $t' \approx t - vx/c^2$ . The spatial line is just the Galilean shear, and the time line carries the one genuinely new term — the same $t' = t - vx/c^2$ that Lorentz wrote down in the 1895 Versuch. The historical point is sharp: $\gamma$ is a second-order effect, so the first relativistic correction anyone was going to stumble upon was not time dilation at all, but the position-dependence of time. Lorentz found it, used it, and declined to believe it was time. Einstein's contribution was not a new equation on that line; it was taking the equation seriously.