University Physics IV · Special Relativity I · 2.5

The Boost in Standard Configuration & Matrix Form

Topic 2.4 argued a boost into existence; this one hands you the object itself. Written with ct beside x, the boost is a single symmetric 2 × 2 matrix of pure numbers, and the questions worth asking — how to undo it, what it leaves alone, when it applies at all — turn into questions about that matrix's entries.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Measure time in metres by carrying ct rather than t, and the transformation between two frames in relative motion along their shared x-axis collapses to two symmetric lines — ct′ = γ(ct − βx) and x′ = γ(x − βct), with y and z untouched — where β = v/c and γ = 1/√(1 − β²). Stack the two coordinates into a column and those lines are one matrix Λ(β), γ on the diagonal and −γβ off it, every entry a pure number. Three structural facts then follow by algebra instead of by argument.

Its determinant is γ²(1 − β²) = 1, so a patch of the (ct, x) plane keeps its area. Its inverse is Λ(−β), so reciprocity becomes an identity: only the sign of β changes, and a single γ serves both directions. And the invariance of (ct)² − x² is the one condition ΛᵀηΛ = η with η = diag(1, −1), which is what actually picks boosts out, since the Galilean shear has determinant 1 too.

The price of all that compactness is the fine print inside the phrase standard configuration: collinear relative motion, parallel axes, origins coincident at t = t′ = 0, and Einstein-synchronised clocks within each frame. Break any one of them and Λ(β) is not the transformation between your frames.

Simple definition: The Lorentz boost in standard configuration is the linear map that takes the coordinates (ct, x, y, z) of an event in one inertial frame to its coordinates in a second frame moving at constant speed v along the shared x-axis, leaving the transverse coordinates unchanged.
Example: At v = 0.6c, γ = 1.25: an event at ct = 400 m, x = 100 m in S is read in S′ as ct′ = 1.25(400 − 60) = 425 m and x′ = 1.25(100 − 240) = −175 m, with (ct′)² − x′² = (ct)² − x² = 1.5 × 10⁵ m².

Boost in standard configurationct′ = γ(ct − βx), x′ = γ(x − βct), y′ = y, z′ = z

Interchange the roles of ct and x and the first two lines turn into each other; the last two do nothing at all.

β = v/c and γ = 1/√(1 − β²) are pure numbers; ct and x are both lengths, in metres.

The same content as a matrix(ct′, x′)ᵀ = Λ(β)(ct, x)ᵀ, Λ(β) = [[γ, −γβ], [−γβ, γ]]

One object now carries the whole transformation, so a question about boosts becomes a question about a matrix.

Every entry is dimensionless, and Λ comes out symmetric only because the first slot holds ct and not t.

Unit determinantdet Λ = γ² − (γβ)² = γ²(1 − β²) = 1

A cheap arithmetic check on any Λ you have just written down, though not a proof that you have a boost.

At β = 0.6 this reads 1.25² − 0.75² = 1.5625 − 0.5625 = 1.

Inverse: reverse β, keep γΛ(β)⁻¹ = Λ(−β): ct = γ(ct′ + βx′), x = γ(x′ + βct′)

Reach for this line whenever the coordinates you are handed are the primed ones.

γ depends on β only through β², so reversing the relative velocity leaves it untouched.

The condition that defines a boostΛᵀηΛ = η, η = diag(1, −1) ⇔ (ct′)² − x′² = (ct)² − x²

Test a candidate matrix with the condition, and test a transformed pair of coordinates with the invariant itself.

η is the 1+1 Minkowski metric; for a symmetric [[a, −b], [−b, a]] the condition collapses to a² − b² = 1.

Low-speed limitγ = 1 + ½β² + … ⇒ x′ ≈ x − vt, t′ ≈ t − vx/c²

Galileo sits at β → 0, and the last term to disappear is the one carrying the simultaneity offset.

The series runs in β², so the first correction to γ is second order in v/c and dies quickly.

Carry ct, not t, and the matrix comes out clean

The transformation is usually first met as t′ = γ(t − vx/c²) and x′ = γ(x − vt). Written that way its coefficient matrix is [[γ, −γv/c²], [−γv, γ]]: the entries carry three different units, and the matrix is not symmetric. Replace the time coordinate by ct, measured in metres exactly like x, and the same content reads ct′ = γ(ct − βx) and x′ = γ(x − βct) with β = v/c. Every entry of Λ = [[γ, −γβ], [−γβ, γ]] is now a pure number, the matrix is symmetric, and interchanging ct with x interchanges the two equations. The conversion is easy to carry in your head: one metre of ct is the 3.3 ns light takes to cross it, so 1.00 μs is 300 m. Nothing physical has changed — this is bookkeeping — but it is why a boost ends up looking like a rotation rather than like an untidy pair of formulas.

The four assumptions inside standard configuration

Four conditions are packed into that phrase, and the derivation uses all four. S′ moves at constant velocity +v along the common x-axis; the two sets of axes are parallel; the origins coincide, so the event (0, 0, 0, 0) in S is the event (0, 0, 0, 0) in S′, which is what allows a linear map with no additive constant; and the clocks of each frame are Einstein-synchronised, because t is not a number until a synchronisation convention has fixed it. The transverse coordinates pass straight through, y′ = y and z′ = z, and that is a symmetry result rather than an omission: if a boost changed transverse lengths, two identical rods passing each other sideways would each have to be shorter than the other. Lose a condition and you lose the matrix. Origins that do not coincide add a translation, which is the Poincaré group and not a boost; a relative velocity that is not along a coordinate axis has to be handled by rotating into standard configuration, boosting, and rotating back.

One event, two readings: multiplying the column

Take β = 0.8, so γ = 1/√(1 − 0.64) = 1/0.6 = 5/3. An event is recorded in S at ct = 500 m, x = 400 m. Multiply: ct′ = (5/3)(500 − 0.8 × 400) = (5/3)(180) = 300 m, and x′ = (5/3)(400 − 0.8 × 500) = 0. The event sits at the spatial origin of S′, which is no accident — x = βct = 0.8 × 500 = 400 m is exactly the worldline that origin traces in S, and that worldline is the ct′ axis. Notice also that 300 m = 500 m ÷ γ. That is time dilation, but it showed up only because this event happens to satisfy x′ = 0; the matrix by itself never says t′ = t/γ. Check the arithmetic with the invariant: (ct)² − x² = 250 000 − 160 000 = 90 000 m², and (ct′)² − x′² = 90 000 − 0 = 90 000 m². Check every transformed pair this way — the interval is the cheapest error trap you have.

The inverse is the same matrix with β reversed

Solve the two lines for ct and x, or simply multiply: Λ(β)Λ(−β) has diagonal entries γ² − γ²β² = 1 and off-diagonal entries γ²β − γ²β = 0, so Λ(β)⁻¹ = Λ(−β), giving ct = γ(ct′ + βx′) and x = γ(x′ + βct′). Because γ is built from β², undoing a boost costs nothing but a pair of sign changes, and that is reciprocity turned into an identity: if S says S′ moves at +v, S′ says S moves at −v, and one γ covers both statements. Two warnings travel with it. Inverting is not transposing — Λ is already symmetric here, so Λᵀ = Λ, while Λ⁻¹ differs from Λ at every β other than zero. And the two signs have to flip together: pairing ct = γ(ct′ + βx′) with x = γ(x′ − βct′) mixes the two directions, and the resulting matrix has determinant γ²(1 + β²), which is 2.125 at β = 0.6. It is still an invertible linear map, but it is not a Lorentz transformation, because it does not leave (ct)² − x² alone.

Unit determinant is necessary, not sufficient

det Λ = γ² − (γβ)² = γ²(1 − β²) = 1 for every β < 1, which says a region of the (ct, x) plane keeps its area: the boost shears a square of events into a parallelogram of the same area. Useful, and cheap to check — but it does not identify a boost. Write the Galilean transformation in these same coordinates, ct′ = ct and x′ = x − βct, and its matrix [[1, 0], [−β, 1]] has determinant exactly 1 at any β whatsoever. The condition that does the identifying is ΛᵀηΛ = η with η = diag(1, −1), which is nothing but the demand that (ct′)² − x′² = (ct)² − x² for every event; a Euclidean rotation satisfies RᵀR = I instead, and that single minus sign inside η is the entire difference between turning a plane and boosting a frame. Apply the condition to the Galilean shear and you get [[1 − β², β], [β, −1]], which equals η only at β = 0. Apply it to Λ(β) and the γ² and γ²β² terms cancel exactly, returning η. So check the determinant to catch arithmetic slips, and check the interval to know you have a boost.

What the boost leaves alone, and where Galileo hides

Feed the matrix a light ray. The column (1, 1), meaning ct = x, comes back as γ(1 − β)(1, 1): the direction survives untouched and only the scale changes, by γ(1 − β) = √((1 − β)/(1 + β)), the longitudinal Doppler factor. The column (1, −1) is stretched by the reciprocal √((1 + β)/(1 − β)), and the two eigenvalues multiply to the determinant, 1; at β = 0.6 they are 0.5 and 2.0. That the two light-line directions are eigenvectors is the second postulate written in linear algebra. Now let β → 0. With γ = 1 + ½β² + …, a speed of 300 m s⁻¹ gives β = 1.0 × 10⁻⁶ and γ − 1 ≈ 5 × 10⁻¹³, so x′ = x − vt is right to roughly twelve figures. The piece that survives longest is the one Galileo never had, t′ = t − vx/c², which at that same speed across x = 1000 km amounts to 3.3 ns — small, but not small enough for a satellite navigation system to ignore.

Change one variable at a time

Make the relationship visible.

Interactive model

Boost speed β = v/c0.60 c

Event position x in S7.0 m

Event time ct in S3.0 m

Push β from 0.15 to 0.85 and watch both primed axes close on the 45° light line while the event dot never moves: at the default event ct′ falls through zero near β = 0.43 and turns negative, yet (ct′)² − x′² sits at −40 m² the whole way.

LORENTZ FACTOR γ1.250

ct′ IN S′-1.50 m

x′ IN S′6.50 m

(ct′)² − x′²-40.0 m²

Live interpretationLORENTZ FACTOR γ: 1.250. ct′ IN S′: −1.50 m. x′ IN S′: 6.50 m. (ct′)² − x′²: −40.0 m²

Catch the common trap

Explain before calculating.

Frames S and S′ are in standard configuration with β = 0.6, so γ = 1.25. An event is recorded in S′ at ct′ = 200 m, x′ = 0. What are its coordinates in S?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyFrames S and S′ are in standard configuration with v = 0.6c. A flashbulb fires at x = 600 m, t = 1.00 μs in S. Find the event's coordinates in S′, and check the answer with the invariant interval. Take c = 3.00 × 10⁸ m s⁻¹.

Convert the time to a length so both coordinates share a unit: ct = (3.00 × 10⁸ m s⁻¹)(1.00 × 10⁻⁶ s) = 300 m. With β = 0.6, γ = 1/√(1 − 0.36) = 1/0.8 = 1.25.
Time row: ct′ = γ(ct − βx) = 1.25(300 − 0.6 × 600) = 1.25(300 − 360) = −75 m, so t′ = −75/(3.00 × 10⁸) s = −0.25 μs.
Space row: x′ = γ(x − βct) = 1.25(600 − 0.6 × 300) = 1.25(600 − 180) = 525 m.
Check with the invariant: (ct)² − x² = 90 000 − 360 000 = −270 000 m², and (ct′)² − x′² = 5 625 − 275 625 = −270 000 m². They agree, so the arithmetic stands.
The negative time is not a slip. In S the flash is 1.00 μs after the origin event; in S′ it is 0.25 μs before it. Reordering is possible only because the interval is negative — the two events are spacelike separated, so no signal runs between them and no cause is reversed.

Answerct′ = −75 m, so t′ = −0.25 μs, and x′ = 525 m. Both frames return (ct)² − x² = −2.70 × 10⁵ m², and the reversal of time order is allowed because that interval is negative.

MediumFrames S and S′ are in standard configuration with β = 0.6. A detector reports an event in S′ at ct′ = 400 m, x′ = 300 m, y′ = 80 m, z′ = 0. Find the coordinates in S, say what happens to the transverse entries, and check the invariant in both the 1+1 and the full 1+3 form.

The primed coordinates are the ones given, so the inverse is what is needed. With γ = 1.25 and γβ = 0.75, Λ(−β) = [[1.25, 0.75], [0.75, 1.25]] acting on the column (ct′, x′).
Time row: ct = 1.25(400) + 0.75(300) = 500 + 225 = 725 m, which is the same as γ(ct′ + βx′) = 1.25(400 + 180) = 725 m.
Space row: x = 0.75(400) + 1.25(300) = 300 + 375 = 675 m, matching γ(x′ + βct′) = 1.25(300 + 240) = 675 m.
Transverse entries: y = y′ = 80 m and z = z′ = 0. No row of Λ touches them, which is why the boost needs only a 2 × 2 block even though an event has four coordinates.
Invariant, 1+1 part: (ct′)² − x′² = 160 000 − 90 000 = 70 000 m², and (ct)² − x² = 525 625 − 455 625 = 70 000 m². Full 1+3 form: subtract the same y² = 6 400 m² from each, giving 63 600 m² in both frames — the transverse term simply rides along, which is exactly what leaving y and z alone has to mean.
The interval is positive, so this event and the origin event are timelike separated and every inertial frame agrees on their order. Note also that Λ(−β) maps (0, 0) to (0, 0): that is the coincident-origins assumption showing itself, and without it the map would need an additive constant and would no longer be this matrix.

Answerct = 725 m, x = 675 m, y = 80 m, z = 0. Both frames return (ct)² − x² = 7.00 × 10⁴ m² and (ct)² − x² − y² − z² = 6.36 × 10⁴ m². The separation from the origin event is timelike, so no frame reverses the order of the two events.

HardA student hands you the matrix Λ = [[2.60, −2.40], [−2.40, 2.60]], acting on the column (ct, x), and claims it is a boost. Decide whether it is; if so find β; then apply it to the event ct = 0, x = 13.0 m and check the result by transforming back.

Necessary check first: det = (2.60)² − (2.40)² = 6.76 − 5.76 = 1.00. It passes, but that settles nothing on its own — the Galilean matrix [[1, 0], [−0.92, 1]] also has determinant 1.
The test that decides is ΛᵀηΛ = η with η = diag(1, −1). For a symmetric [[a, −b], [−b, a]] it reduces to a² − b² = 1, which is exactly what the determinant just gave, so for a matrix of this particular shape the two tests coincide and this one really is a boost.
Read the entries: γ = 2.60 and γβ = 2.40, so β = 2.40/2.60 = 12/13 = 0.923. Confirm: 1/√(1 − (12/13)²) = 1/√(25/169) = 13/5 = 2.60 ✓.
Apply it to the event: ct′ = 2.60(0) − 2.40(13.0) = −31.2 m, so t′ = −104 ns with c = 3.00 × 10⁸ m s⁻¹, and x′ = −2.40(0) + 2.60(13.0) = 33.8 m.
Interval check: (ct)² − x² = 0 − 169.0 = −169.0 m², and (ct′)² − x′² = 973.44 − 1142.44 = −169.0 m² ✓.
Transform back with Λ(−β) = [[2.60, 2.40], [2.40, 2.60]]: ct = 2.60(−31.2) + 2.40(33.8) = −81.12 + 81.12 = 0 ✓, and x = 2.40(−31.2) + 2.60(33.8) = −74.88 + 87.88 = 13.0 m ✓.

AnswerIt is a genuine boost, at β = 12/13 = 0.923. The event at ct = 0, x = 13.0 m in S is read in S′ as ct′ = −31.2 m (t′ = −104 ns) and x′ = 33.8 m, and Λ(−β) returns (0, 13.0 m) exactly.