Lorentz Transformation of a Four-Vector

Source

Lecture: 13 — Four-velocity Vector
Simulation: Simulation 1 — Lorentz transformations on vectors (mode: guided-inquiry)
Concept: A Lorentz boost changes a four-vector's components between inertial frames but preserves its Minkowski magnitude — the Minkowski-geometry analogue of a rotation preserving Euclidean length.
Serves: Define four-vectors and their transformation properties; relate four-velocity to ordinary velocity; describe how four-velocity transforms under Lorentz transformations.
Inquiry: Under a boost, what changes and what stays the same? Does invariance hold for timelike, spacelike and null vectors?
Misconception: That because the components change under a boost, the vector's "size" must change too — probed structurally by the prediction.

Pedagogy principles applied

ID	Principle	How it shows up here
A1	One concept	Every element serves "a boost changes components but preserves U·U".
A2	Layers earn place	Four-velocity, worldline and matrix layers are OFF by default.
A3/A4	Formal off by default	The Lorentz boost matrix lives in the ⚛︎ The Physics sheet, closed on load.
B3	Define every symbol	v/c (boost velocity), γ≡1/√(1−(v/c)²), γ_u≡1/√(1−u²), U⁰ time, U¹ space — defined in the readouts and in ⚛︎ The Physics.
C1	Numbers referenced	U⁰,U¹,v/c,γ,U·U,U′ are each read/compared by a prompt; speed selector hidden.
C2	Frame-tagged numbers	Every readout carries "frame S" or "frame S′".
D1	Diagram primary	The Minkowski canvas fills the hero; readouts are compact overlays.
D3	Vector identifiable	Bold blue arrow with arrowhead and adjacent label U.
D5	No clipping	Tip clamped to [−4,4]; labels bounds-checked against canvas edges.
D6	Incremental layers	Each toggle adds one layer; prior layers de-emphasise (lighten).
D7	No gating	All sliders, Play, toggles and the prediction are live from load.
D8	Labels not prose	Canvas carries t, x, U, E₁, E₂; sentences live in side panels.
E1	Honest animation	Play moves the particle along its worldline at constant velocity; v/c never auto-sweeps.
F2	Conceptual prediction	Predict asks qualitatively whether the magnitude changes; numeric check is separate.
G2	Invariant labelled	Magnitude readout marked "Lorentz invariant" with identical values in both frames.
G5	Live numbers in text	Prompts show the current v/c and U through live spans.
G7	Displacement operational	Worldline layer marks E₁,E₂ and names Δx = E₂ − E₁.
J1	One notation	η = diag(−1,+1) throughout; contravariant superscript indices.
J3	Correct model	Oblique (parallel-to-S′-axis) projections; u′ = U′¹/U′⁰ = (u−v/c)/(1−u·v/c).

Teach me

Speed

Guided Inquiry

1 · A four-vector and two frames

Let’s get our bearings. The diagram is the t–x plane with c = 1, and the dashed 45° lines are the paths of light. The blue arrow is a four-vector U, with components U⁰ (time) and U¹ (space) read off frame S’s axes. The orange axes belong to a second frame S′ moving at v/c along x; at v/c = 0 they lie on S’s. A boost along x leaves y and z alone, so the whole transformation lives in these two components.

2 · What a boost moves

Let’s start with the simplest question. The two frames coincide at v/c = 0; the boost is currently v/c = 0.00. Raise it and keep your eye on the blue arrow U — what do you think it does?

3 · One number that survives

Now for the big one. A rotation changes a vector’s components but leaves x² + y² fixed. At v/c = 0.60 both components of U read differently in S′. So which of these do you think is the same number in both frames?

Take a look: the top-right strip now reports U′⁰, U′¹ and U′·U′ — the same vector read on the orange axes.

4 · A vector outside the light lines

Let’s push this idea further. U⁰ = 1 and U¹ = 3 are set for you, so the tip sits outside the 45° lines and U·U = +8.00. Sweep v/c across its whole range and watch U′⁰ and U′¹. What do you think — which holds?

5 · Reading a velocity in the new frame

One last calculation. U is now constrained to a four-velocity, U = γ_u(1, u), with u = 0.50 and the boost at v/c = −0.50. The ordinary velocity in S′ is u′ = U′¹/U′⁰. Make a prediction — what is it?

6 · Components, and the thing they describe

Let’s put names to what you’ve found. The boost is a single linear map, U′ = Λ(v/c) U, and its inverse is Λ(−v/c) — the same matrix with the sign of v flipped, which is the formal statement that neither frame is preferred.

Components U⁰, U¹ are frame-dependent bookkeeping; U·U = −(U⁰)² + (U¹)² is the physics. Try it for yourself: the Minkowski magnitude layer draws the curve of every vector sharing this value, and ⚛︎ The Physics carries the matrix.

Listen

Controls

▾

Four-vector U (frame S)

U⁰ [time component] 2.00

U¹ [space component] 1.00

Ordinary velocity u (four-velocity mode)

u [ordinary velocity, |u| < c] 0.50

Boost v/c

v/c [boost velocity] 0.60 · γ=1.25

Key idea

▾

Components U⁰, U¹ are frame-dependent. The Minkowski magnitude U·U = −(U⁰)² + (U¹)² is frame-independent — the invariant.

Current: U = (2.00, 1.00), U·U = -3.00 (timelike)

Check your calculation

▾

Given U = (2.00, 1.00) and v/c = 0.60, predict U′ in S′, then Check.

U′⁰

U′¹

Lorentz boost matrix (S → S′)

v/c (boost velocity), γ ≡ 1/√(1−(v/c)²).

Component transformation

U⁰ = time component, U¹ = space component (contravariant, superscript indices).

Minkowski metric & invariant magnitude

η = diag(−1, +1). U·U is the same number in every frame — a Lorentz invariant.

Four-velocity & velocity addition

U·U = −1. Ordinary velocity recovered as u′ = U′¹/U′⁰.

S′ axes on the diagram

t′-axis slope c/v, x′-axis slope v/c — both tilt symmetrically toward the 45° light cone. Calibration ticks placed where each axis crosses −t² + x² = ±n².

The Physics

Lorentz transformation of a four-vector · what this sim shows, the physics behind it, and every control

At a glance

This sim draws one four-vector U on a spacetime diagram and reads its two components against two sets of axes, those of frame S and those of a frame S′ moving past it. It shows what a Lorentz boost does to U, and what it leaves alone.

What you'll learn

What a boost changes when you move from frame S to frame S′, and what it does not.
How to compute the components U′⁰ and U′¹ from U⁰, U¹ and v/c.
How the quantity U·U behaves under a boost, and what its sign tells you.
How the four-velocity of a moving object is built from its ordinary velocity u.

One arrow U and two sets of axes: S (grey) and S′ (orange). Dashed: the light lines.

What's on screen

Blue arrow U: the four-vector, from the origin to the point (x, t) = (U¹, U⁰). Drag its tip to change it.
Grey axes: frame S.
Orange axes: frame S′, with dashed orange lines showing how U reads on them.
Light lines: x = ±t, at 45°.
Two strips on the canvas: U in S (top left) and U′ in S′ (top right), each with its own U·U line.

See itBoost to v/c = 0.60, with S′ axes on. Watch for: the orange axes tilting away from the grey ones, and the two strips giving different components.

The controls

Press ⌖ on a row to see where that control sits in the sim.

Control	What it changes	Range
U⁰	The time component of U in frame S.	−4 to 4 opens at 2
U¹	The space component of U in frame S.	−4 to 4 opens at 1
v/c	The boost: the velocity of S′ relative to S, as a fraction of c, along x. The readout also shows γ.	−0.95 to 0.95 opens at 0
S′ axes (layer)	Shows or hides the orange S′ axes, the dashed projections, and the S′ strip.	on / off opens on
Minkowski magnitude (layer)	Draws the violet hyperbola through the tip of U: every vector on it has the same U·U as U.	on / off opens off
Constrain to four-velocity (layer)	Replaces the U⁰ and U¹ sliders by one slider u, with \|u\| < c. U becomes γ_u(1, u), and the green curve U·U = −1 appears. Dragging the arrow then changes u only.	u: −0.95 to 0.95 opens at 0.50, layer off
Worldline context (layer)	Marks two events, E₁ at the origin and E₂ at the tip of U, and labels the separation between them.	on / off opens off
Frame strips	The components of U in S, and of U′ in S′, each with its own U·U line.	on canvas
Key idea	The current U, its U·U, and whether U is timelike, spacelike or null.	panel
Check your calculation	Type your own U′⁰ and U′¹ for the current U and v/c. The sim marks them right or wrong and shows the arithmetic.	panel opens collapsed
▶ Play / ⏸ Pause	Runs a small yellow marker along U, from the origin to its tip. It is drawn only for a timelike U with U⁰ > 0.	top bar
Speed	How fast the marker moves.	0.25× to 4×
↻ Reset	Restarts the mode you are in from its beginning.	top bar

Fixed: units with c = 1, so a speed v/c is a pure number and time and space are measured in the same unit. The boost is always along x. The sim's U·U uses the signature (−, +): U·U = −(U⁰)² + (U¹)².

A boost changes the axes, not the vector

U is a fixed arrow in spacetime. Frame S′ is a second set of axes, and the boost only decides how those axes are drawn on S's diagram.

The S′ time axis is the line x = (v/c) t, and the S′ space axis is the line t = (v/c) x. Both lean toward the same light line, by the same angle.
The light lines stay at 45° for every v/c. All observers agree on them.
The sign of v/c sets which light line the axes lean toward. The inverse boost is the boost with −v/c.

See itBoost to v/c = 0.60 with U fixed. Watch for: the blue arrow staying put while the orange axes lean.

See itSweep v/c from −0.60 to +0.60. Watch for: both orange axes closing on a light line from opposite sides, and swinging across when v/c changes sign.

The new components mix the old ones

U′⁰ and U′¹ are the components of the same arrow read along the S′ axes. They come from one linear map, with γ = 1/√(1 − (v/c)²):

No boost changes one component alone: each new component uses both old ones.
γ ≥ 1, but the bracket can shrink or change sign. For U = (2, 1) and v/c = 0.60, γ = 1.25 and U′ = (1.75, −0.25).

U⁰ U¹ u v/c

These sliders move the sim too. u drives U only while Constrain to four-velocity is on.

To practise, work out U′ by hand and enter it in the sim's Check your calculation panel.

See itReplay your current U and v/c, with the prediction first. Watch for: the U′ strip matching the numbers on the card.

The number that survives the boost

Both components change, and so does the ordinary sum of squares: for U = (2, 1) at v/c = 0.60 it is 5.00 in S and 3.13 in S′. The combination with a minus sign does not change.

U = (2.00, 1.00) at v/c = 0.60. The same arrow, read on two sets of axes.

Why it works. Expand −(U′⁰)² + (U′¹)²: the cross terms cancel, and what is left is γ²(1 − (v/c)²) times U·U, with γ²(1 − (v/c)²) = 1.
U·U can be negative. It is not the square of a length, because the time part carries the minus sign.

See itBoost U = (2, 1) to v/c = 0.30, 0.90, then 0.60, with the violet hyperbola on. Watch for: U′ changing at every step while U·U and U′·U′ stay equal.

Timelike, spacelike and null stay that way in every frame

The value of U·U sorts vectors into three kinds, and a boost never moves a vector from one kind to another.

U·U < 0, timelike: U points inside the light lines. It can be the four-velocity of an object.
U·U = 0, null: U lies on a light line. U = (a, a) becomes (γ(1 − v/c) a, γ(1 − v/c) a), still on the light line.
U·U > 0, spacelike: U points outside the light lines. For U = (1, 3), U·U = +8 in every frame, and |U′¹| stays larger than |U′⁰|.

A spacelike U cannot be a four-velocity, since U¹/U⁰ would exceed 1 in size. Its components still change: for U = (1, 3), U′⁰ is positive up to v/c = 1/3, then negative.

See itBoost U = (1, 3) to v/c = 0.20, 0.50, then 0.95. Watch for: U′⁰ changing sign while U′·U′ stays at +8.

Four-velocity is the timelike vector with U·U = −1

For an object moving at ordinary velocity u in S, the four-velocity is U = dx/dτ with τ the object's proper time. In units with c = 1:

U·U = −1 for every u. The speed shows up in the direction of U, never in its magnitude.
The green curve is every four-velocity: γ_u = 1, 1.15 and 2.29 for u = 0, 0.50 and 0.90.
The boost acts on U like on any four-vector, which gives the new velocity u′ = U′¹/U′⁰ = (u − v/c)/(1 − u v/c). For u = 0.50 and v/c = −0.50, u′ = 0.80, not 1.00.

See itConstrain to four-velocity and sweep u = 0, 0.50, 0.90. Watch for: U⁰ and U¹ growing while U·U stays at −1.00.

See itSet u = 0.50 and v/c = −0.50. Watch for: the S′ strip giving U′¹ / U′⁰ = 0.80.

Where it matters

Mass. The four-momentum is P = mU, so P·P = −m² (c = 1) in every frame. Energy and momentum depend on the frame; this combination does not.
Cause and effect. For two events separated by U, the time order is the same in every frame if U is timelike or null. If U is spacelike it is not: for U = (1, 3), the order reverses for v/c > 1/3.
Adding velocities. A boost mixes U⁰ and U¹, so velocities combine by u′ = (u − v/c)/(1 − u v/c), and speeds below c never combine to reach c.

Go deeper: a null vector and the Doppler factor

Set U⁰ = U¹ = 2 and v/c = 0.60 in the sim. The result is U′ = (1, 1), and the strip reads U′·U′ = 0.
In general U′ = γ(1 − v/c)(a, a) = √((1 − v/c)/(1 + v/c)) (a, a). At v/c = 0.60 the factor is 0.5.
For light moving along x, the same factor rescales its frequency between the frames: the relativistic Doppler effect.

Summary

In one line. A boost changes the axes you read a four-vector against, so its components change, while U·U = −(U⁰)² + (U¹)² stays the same.

Key points

U is one arrow. Frame S′ is a second set of axes; the light lines never move.
U′⁰ = γ(U⁰ − (v/c)U¹) and U′¹ = γ(U¹ − (v/c)U⁰): each component uses both old ones.
U·U is the same in every frame, and so is its sign: timelike, null or spacelike.
A four-velocity is γ_u(1, u), with U·U = −1.

Common mix-ups

"After the boost the vector is different, because the numbers are different." The arrow is the same. Only its components on the new axes differ.
"U·U is a length, so it can't be negative." With signature (−, +), timelike vectors have U·U < 0. The sum of squares is what changes.
"Velocities just add and subtract." For u = 0.50 and v/c = −0.50, u′ = 0.80, not 1.00.

Things to try on your own

Set U⁰ = U¹ = 2, sweep v/c, and watch U′ stay on the light line while its size changes.
Work out U′ by hand for any U and v/c, then enter it in Check your calculation.
Set U = (1, 3), turn on Minkowski magnitude, and sweep v/c from 0 to 0.95. Note where U′⁰ crosses zero.