Four-Vectors and Invariants

Equivalent forms

A\cdot B = -A^0B^0 + \vec{A}\cdot\vec{B}

A\cdot A = -(A^0)^2 + |\vec{A}|^2 = \text{invariant}

Any quantity written as a four-vector automatically transforms correctly between frames, and its self-dot-product is a conserved invariant — the reason four-momentum bookkeeping is so powerful.

Symbol	Name	SI unit	Dimension	Range
$AB$	Invariant Productoutput Frame-independent Minkowski inner product	dimensionless	1	-100 – 100
$A0$	Time Component Timelike component A^0 of the four-vector	dimensionless	1	0 – 5
$As$	Spatial Magnitude Magnitude of the spatial part \|vec A\|	dimensionless	1	0 – 5

Dimensional analysis

A.B -> dimensionless (per component)

-A^0B^0 + vecA.vecB -> same dimension as the product of components

Consistent; the invariant carries whatever units the four-vector's components do.

\checkmark

Discovery

Hermann Minkowski / Arnold Sommerfeld · 1910

Minkowski introduced four-dimensional vectors; Sommerfeld's four-vector calculus turned them into the working tool physicists still use to make every relativistic formula manifestly invariant.

Research status: stable

Real-world applications

Invariant-mass reconstruction at the LHC
Manifestly covariant electromagnetism (four-potential, four-current)
Relativistic kinematics of scattering

Common misconceptions

A four-vector's 'length' is not the ordinary Euclidean length of its four numbers
The time component alone is not invariant — only the full inner product is
Spatial rotations are a special case of Lorentz transformations that leave A^0 fixed

Experimental verification

The four-momentum invariant p.

p = -(mc)^2

is used to reconstruct particle masses from decay products at every collider; the fact that the reconstructed mass is frame-independent is a daily confirmation.

Derivation

Under a Lorentz boost, A^0 and A^1 mix exactly like ct and x: A'^

0 = \gamma

(A^0 - beta A^1), A'^

1 = \gamma

(A^1 - beta A^0).

Substitute into -(A'^0)^2 + (A'^1)^2 and the

\gamma ^2(1-\beta ^2)=1

factor collapses everything back to -(A^0)^2 + (A^1)^2.

The inner product is therefore boost-invariant.

Limiting cases

A^

0 = |vec

A|⟶ A.

A = 0

A null four-vector — e.g. the four-momentum of a photon.

|vec A|

= 0

⟶ A.

A =

-(A^0)^2A purely timelike vector — e.g. four-velocity in the rest frame.

A^

0 = 0

⟶ A.

A = |vec

A|^2Purely spacelike.

What if…

What if A.

B = 0

for two timelike vectors?

They are 'orthogonal' in the Minkowski sense — e.g. a particle's four-velocity is always orthogonal to its four-acceleration.

What if I forget the minus sign?

You get a Euclidean dot product that changes between frames and every relativistic conservation law breaks.

1

Norm of a four-velocity

Given ·

A0:: gamma
As:: gamma*beta

Find · U.U

Steps

U. $U = -(\gamma c)^2 + (\gamma v)^2$
$= -\gamma ^2\,\mathrm{c}^2 (1 - \beta ^2)$
$\gamma ^2(1-\beta ^2) = 1$
U. $U = -c^2$

Result ·

U^0 = \gamma c

, |

vec U| = \gamma v

. U.

U = -(\gamma c)^2 + (\gamma v)^2 = -\gamma ^2\,\mathrm{c}^2 (1 - v^2/c^2) = -c^2

. Every four-velocity has the same invariant length, -c^2.

Energy–Momentum Four-Vector→Four-Velocity→Minkowski Metric→Invariant Mass (Energy-Momentum Relation)→