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Four-Vectors: Objects That Transform Correctly

Meet the four-vector, the natural building block of relativity: a four-component object that transforms like the coordinates, carries an invariant `length`, and lets you write physics that looks the same in every frame.

What makes something a four-vector

The coordinates of an event, x^\mu=(ct,x,y,z) with \mu=0,1,2,3, transform under a boost by a definite matrix rule. A four-vector is any set of four numbers that transforms by that same rule. That is not a triviality: it is a strict membership test. If a quantity transforms like x^\mu, it is a four-vector and inherits all the geometry; if it does not, it is just four numbers travelling together and none of the theorems apply.

x'^{\mu} = \Lambda^{\mu}{}_{\nu}\, x^{\nu}

The transformation law, using the Einstein summation convention: a repeated index (here \nu) is summed over $0,1,2,3$. \Lambda^{\mu}{}_{\nu} is the Lorentz (boost/rotation) matrix.

This is a strict upgrade over Volume I's ordinary vectors. A 3-vector like velocity has three components and rotates under spatial rotations. A four-vector has four and transforms under the full Lorentz group — boosts included. The pattern to internalise: good relativistic physics is written as equations between four-vectors (and tensors), because such an equation, if true in one frame, is automatically true in all of them.

Indices up and down, and the invariant dot product

Four-vectors come with components carrying an upper index, A^\mu (called contravariant). The metric tensor \eta_{\mu\nu}=\mathrm{diag}(+1,-1,-1,-1) — the Minkowski metric — lets you lower an index to make the covariant components A_\mu. Lowering just flips the sign of the spatial parts.

A_{\mu} = \eta_{\mu\nu} A^{\nu} = (A^0,\,-A^1,\,-A^2,\,-A^3)

Lowering an index with the metric. Raising uses the inverse metric \eta^{\mu\nu}, which here is numerically the same.

The payoff is the invariant dot product. Contract an upper with a lower index and the result is a scalar — the same number in every frame:

A\cdot B \equiv \eta_{\mu\nu}A^{\mu}B^{\nu} = A^{\mu}B_{\mu} = A^0 B^0 - \vec A\cdot\vec B

The Minkowski inner product. Setting B=A gives the squared `length` A\cdot A=(A^0)^2-|\vec A|^2, a Lorentz invariant. The interval s^2 of Guide 1 is just x\cdot x.

A four-vector and its invariant magnitude. A boost tilts the vector's components — A^0 and A^1 each change — but the hyperbolic `length` A\cdot A is pinned to the same invariant hyperbola.

Four-velocity and proper time

To build a four-vector out of a particle's motion, we need to differentiate its position — but with respect to what? Coordinate time t is frame-dependent, so dx^\mu/dt is not a four-vector. The fix is to use the one clock everyone agrees on: the particle's own clock, ticking proper time \tau, defined by the interval along the worldline.

c^2\,d\tau^2 = ds^2 \;\Longrightarrow\; d\tau = \frac{dt}{\gamma} = dt\sqrt{1-v^2/c^2}

Proper time is the interval divided by c; it is the time elapsed on a clock carried along the worldline, and it is a Lorentz scalar.

Because \tau is a scalar and x^\mu is a four-vector, the derivative U^\mu=dx^\mu/d\tau is automatically a four-vector — the four-velocity. Working it out with d\tau=dt/\gamma:

U^{\mu} = \frac{dx^{\mu}}{d\tau} = \gamma\,(c,\ \vec v), \qquad U\cdot U = \gamma^2(c^2 - v^2) = c^2

The four-velocity. Its invariant magnitude is always c — every particle moves through spacetime at the speed of light, splitting that motion between time and space.