Four-Vectors and Index Notation

The interval singles out one combination of coordinate differences as frame-independent. To build physics from invariants rather than from frame-dependent components, every quantity is organized into objects that transform under a Lorentz boost the same way the coordinates do. Those objects are four-vectors, and manipulating them efficiently requires index notation: upstairs and downstairs indices, the metric that trades one for the other, and the summation convention that keeps the algebra compact.

Four-vectors and the summation convention

A four-vector is a set of four components $V^{\mu }=(V^0,V^1,V^2,V^3)$ that transform between inertial frames exactly as the coordinate differences $\mathbf{d}{x}^{\mu }$ do. Under a boost described by the matrix ${\Lambda }^{\mu }{}_{\nu }$,

The Einstein summation convention is in force: an index that appears once upstairs and once downstairs in a term is summed over its four values, and the summation sign is dropped. Here $\nu$ is summed, so the right side is ${\sum }_{\nu =0}^3{\Lambda }^{\mu }{}_{\nu }{V}^{\nu }$. A repeated index is a dummy index and can be renamed freely; an unrepeated one is a free index and must match on both sides of an equation. For the standard boost of speed $v$ along $x^1$,

The upper index of ${\Lambda }^{\mu }{}_{\nu }$ is the row and the lower index is the column; the staggered spacing keeps their order unambiguous, which matters once indices are raised and lowered.

The metric and covariant components

Alongside the contravariant components sits a second, equivalent set. The Minkowski metric is the array

and it acts on a four-vector by summing away one index to produce the covariant components,

Because $\eta$ is diagonal, this only flips the sign of the time component: $V_0=-V^0$ and $V_i=V^i$ for the spatial indices $i=1,2,3$. Lowering with ${\eta }_{\mu \nu }$ and raising with its matrix inverse ${\eta }^{\mu \nu }$ — which in these coordinates has the same entries, ${\eta }^{\mu \nu }=\mathrm{diag}(-1,+1,+1,+1)$ — are inverse operations,

with ${\delta }^{\mu }{}_{\nu }$ the Kronecker delta, equal to $1$ when $\mu =\nu$ and $0$ otherwise. The metric is the single object that converts between the two index positions; raising an already-upstairs index or lowering an already-downstairs one is undefined.

The gradient is the paradigm covariant object. From the chain rule $\partial /\partial x^{\prime \mu }=(\partial x^{\nu }/\partial x^{\prime \mu })\partial /\partial x^{\nu }$, the derivative operator carries a lower index and transforms oppositely to the coordinates, matching what the downstairs position records.

Scalar products and invariants

Pairing an upstairs index with a downstairs one and summing yields a Lorentz scalar, a single number that is the same in every inertial frame. The scalar product of two four-vectors is

Invariance is a one-line consequence of how the two index positions transform. Under a boost,

using the defining property ${\Lambda }_{\mu }{}^{\alpha }{\Lambda }^{\mu }{}_{\beta }={\delta }^{\alpha }{}_{\beta }$. Every contraction of matched upper and lower indices is invariant for the same reason. The squared magnitude of a four-vector is the special case $V\cdot V$, and its sign classifies the vector exactly as the interval classified event separations:

• Timelike if $V\cdot V<0$,
• Null if $V\cdot V=0$,
• Spacelike if $V\cdot V>0$.

Basis vectors and the dual basis

The component notation rests on a choice of basis. Write ${\vec{e}}_{(\mu )}$ for the four basis vectors of an inertial frame, one pointing along each coordinate axis, so that a four-vector is $V=V^{\mu }{\vec{e}}_{(\mu )}$. The metric records the scalar products of the basis vectors,

which states that the time basis vector has ${\vec{e}}_{(0)}\cdot {\vec{e}}_{(0)}=-1$ and the three space basis vectors are unit and mutually orthogonal. A basis with these products is orthonormal in the Minkowski sense.

The covariant components belong to a dual basis ${\vec{e}}^{(\mu )}$ defined by ${\vec{e}}^{(\mu )}\cdot {\vec{e}}_{(\nu )}={\delta }^{\mu }{}_{\nu }$, so that $V_{\mu }$ are the components of the same vector on the dual basis. The two bases coincide up to the sign flips carried by $\eta$; the distinction becomes substantive in curved spacetime, where the coordinate basis is not orthonormal and the metric entries are functions of position.

Four-velocity and four-acceleration

Differentiating a particle's worldline $x^{\mu }(\tau )$ with respect to its own proper time — a Lorentz scalar — gives a four-vector, because $\mathbf{d}{x}^{\mu }$ is a four-vector and $\mathbf{d}\tau$ is invariant. The four-velocity is

Using $\mathbf{d}\tau =\mathbf{d}t/\gamma$, its components in terms of the ordinary three-velocity $\vec{v}$ are

The four-velocity is the unit tangent to the worldline, normalized in the Minkowski sense. Its invariant square is fixed once and for all:

The normalization has an immediate corollary for acceleration. The four-acceleration is the proper-time derivative of the four-velocity,

and differentiating the constant $U\cdot U=-c^2$ shows that four-velocity and four-acceleration are always Minkowski-orthogonal.

The four-velocity, four-acceleration, and the four-momentum $p^{\mu }=m{U}^{\mu }$ that the dynamics module builds on them are all four-vectors, so their scalar products are invariants that can be evaluated in whichever frame is simplest and then trusted in all others. This is the practical payoff of the index calculus: reduce a frame-dependent question to an invariant, compute the invariant once, and read the answer in any frame.

Summary of the index rules

A well-formed relativistic equation has the same free indices, in the same positions, on both sides; every summed index appears once up and once down. Checking this index balance catches most algebra errors before any numbers are substituted. The Lorentz group is the set of all matrices ${\Lambda }^{\mu }{}_{\nu }$ that preserve ${\eta }_{\mu \nu }$, and the next lesson works out its structure.