--- title: Four-Vectors and Index Notation module: Spacetime and the Lorentz Group moduleNumber: 2 lessonNumber: 2 order: 202 summary: > The index calculus that the rest of the course runs on. Contravariant and covariant components, the Minkowski metric as the machine that raises and lowers indices, and the Einstein summation convention are assembled into scalar products that are the same in every frame. The four-velocity and four-acceleration follow, together with the identity that the four-velocity has constant invariant length. topics: [Spacetime and the Lorentz Group] draft: false sources: - book: Schutz ref: "A First Course in General Relativity, Ch. 2 — Vector Analysis in Special Relativity" - book: Carroll ref: "Lecture Notes on General Relativity, §1 — Special Relativity and Flat Spacetime" --- The [interval](/relativity/spacetime-and-the-lorentz-group/minkowski-spacetime-and-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 $\d x^\mu$ do. Under a boost described by the matrix $\Lambda^\mu{}_\nu$, $$ V'^\mu = \Lambda^\mu{}_\nu\, V^\nu. $$ 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 $x^1$, $$ \Lambda^\mu{}_\nu = \begin{pmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}, \qquad \beta = \frac{v}{c},\quad \gamma = \frac{1}{\sqrt{1-\beta^2}}. $$ 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. > **Definition (Contravariant components).** The components $V^\mu$ with the > index upstairs are the contravariant components of a four-vector. They > transform with the boost matrix $\Lambda^\mu{}_\nu$ — the same law obeyed by > the displacement $\d x^\mu$. Displacement, velocity, momentum, and current are > naturally contravariant. ## The metric and covariant components Alongside the contravariant components sits a second, equivalent set. The **Minkowski metric** is the array $$ \eta_{\mu\nu} = \operatorname{diag}(-1, +1, +1, +1), $$ and it acts on a four-vector by summing away one index to produce the **covariant components**, $$ V_\mu = \eta_{\mu\nu}\, V^\nu. $$ 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, $\eta^{\mu\nu} = \operatorname{diag}(-1,+1,+1,+1)$ — are inverse operations, $$ V^\mu = \eta^{\mu\nu} V_\nu, \qquad \eta^{\mu\alpha}\eta_{\alpha\nu} = \delta^\mu{}_\nu, $$ 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. > **Definition (Covariant components).** The components $V_\mu = \eta_{\mu\nu}V^\nu$ > with the index downstairs are the covariant components. They transform with the > inverse-transpose boost matrix $\Lambda_\mu{}^\nu$, defined so that > $\Lambda_\mu{}^\alpha \Lambda^\nu{}_\alpha = \delta^\nu{}_\mu$. Gradients > $\partial_\mu = \partial/\partial x^\mu$ are naturally covariant. The gradient is the paradigm covariant object. From the chain rule $\partial/\partial x'^\mu = (\partial x^\nu/\partial x'^\mu)\,\partial/\partial x^\nu$, the derivative operator carries a lower index and transforms oppositely to the coordinates, matching what the downstairs position records. $$ % caption: The metric raises and lowers indices. Lowering with eta flips the sign % of the time component and leaves the space components alone; raising with the % inverse metric undoes it, returning the original contravariant four-vector. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} \node[draw=acc, fill=acc!10, minimum width=2.4cm, minimum height=0.9cm] (up) at (0,1.6) {up: V0, V1, V2, V3}; \node[draw=black, fill=black!5, minimum width=2.4cm, minimum height=0.9cm] (down) at (0,-1.6) {down: minusV0, V1, V2, V3}; \draw[->, black, thick] (1.35,1.3) to[bend left=32] node[right, black!70] {lower with eta} (1.35,-1.3); \draw[->, black, thick] (-1.35,-1.3) to[bend left=32] node[left, black!70] {raise with inverse eta} (-1.35,1.3); \end{tikzpicture} $$ ## 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 $$ V \cdot W = \eta_{\mu\nu} V^\mu W^\nu = V_\mu W^\mu = -V^0 W^0 + V^1 W^1 + V^2 W^2 + V^3 W^3. $$ Invariance is a one-line consequence of how the two index positions transform. Under a boost, $$ V'_\mu W'^\mu = (\Lambda_\mu{}^\alpha V_\alpha)(\Lambda^\mu{}_\beta W^\beta) = (\Lambda_\mu{}^\alpha \Lambda^\mu{}_\beta)\, V_\alpha W^\beta = \delta^\alpha{}_\beta\, V_\alpha W^\beta = V_\alpha W^\alpha, $$ 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 $V \cdot V > 0$. $$ % caption: The invariant dot product. Two frames read different components for % the four-vectors V and W, but the contraction with the metric returns one and % the same scalar in every frame. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} \node[draw=black, fill=black!5, minimum width=2.7cm, minimum height=1.0cm] (a) at (-2.6,1.4) {frame one: V, W}; \node[draw=black, fill=black!5, minimum width=2.7cm, minimum height=1.0cm] (b) at (2.6,1.4) {frame two: V, W}; \node[draw=acc, very thick, fill=acc!12, minimum width=3.2cm, minimum height=1.0cm] (c) at (0,-1.2) {same scalar VdotW}; \draw[->, black, thick] (a) -- (c); \draw[->, black, thick] (b) -- (c); \draw[<->, black, dashed] (a) -- node[above, black] {boost} (b); \end{tikzpicture} $$ ## 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, $$ \vec{e}_{(\mu)} \cdot \vec{e}_{(\nu)} = \eta_{\mu\nu}, $$ 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. $$ % caption: An orthonormal frame in Minkowski spacetime. The time basis vector has % squared length minus one and the space basis vectors have squared length plus % one; the dual basis is fixed by pairing to the Kronecker delta. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} \draw[black, dashed] (-2.4,-2.4) -- (2.4,2.4); \draw[black, dashed] (-2.4,2.4) -- (2.4,-2.4); \draw[->, black] (-2.6,0) -- (2.8,0) node[right] {x}; \draw[->, black] (0,-1.0) -- (0,3.0) node[above] {ct}; \fill[black] (0,0) circle (1.6pt); \draw[->, acc, very thick] (0,0) -- (0,1.6) node[left, midway] {e time}; \draw[->, black, very thick] (0,0) -- (1.6,0) node[below, midway] {e space}; \node[black, anchor=west] at (0.15,1.55) {squared length minus one}; \node[black, anchor=north] at (1.6,-0.15) {squared length plus one}; \end{tikzpicture} $$ ## 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 $\d x^\mu$ is a four-vector and $\d\tau$ is invariant. The **four-velocity** is $$ U^\mu = \frac{\d x^\mu}{\d\tau}. $$ Using $\d\tau = \d t/\gamma$, its components in terms of the ordinary three-velocity $\vec{v}$ are $$ U^\mu = \frac{\d x^\mu}{\d t}\frac{\d t}{\d\tau} = \gamma\,(c, \vec{v}) = (\gamma c, \gamma v_x, \gamma v_y, \gamma v_z). $$ The four-velocity is the unit tangent to the worldline, normalized in the Minkowski sense. Its invariant square is fixed once and for all: $$ U \cdot U = \eta_{\mu\nu}U^\mu U^\nu = \gamma^2(-c^2 + v^2) = -c^2 \gamma^2 \parens{1 - \frac{v^2}{c^2}} = -c^2. $$ > **Theorem (Normalization of the four-velocity).** Every massive particle's > four-velocity satisfies $U \cdot U = -c^2$, independent of speed or frame. The > four-velocity is a timelike unit vector (of "length" $c$) tangent to the > worldline. The normalization has an immediate corollary for acceleration. The **four-acceleration** is the proper-time derivative of the four-velocity, $$ A^\mu = \frac{\d U^\mu}{\d\tau}, $$ and differentiating the constant $U \cdot U = -c^2$ shows that four-velocity and four-acceleration are always Minkowski-orthogonal. > **Proof.** Differentiate $\eta_{\mu\nu}U^\mu U^\nu = -c^2$ with respect to > $\tau$. The metric components are constants, so > $$ > 0 = \frac{\d}{\d\tau}\parens{\eta_{\mu\nu}U^\mu U^\nu} > = 2\,\eta_{\mu\nu}\frac{\d U^\mu}{\d\tau}U^\nu = 2\, A \cdot U. > $$ > Hence $A \cdot U = 0$. Because $U$ is timelike, its orthogonal complement is > spacelike, so $A$ is a spacelike four-vector: $A \cdot A \ge 0$, with equality > only for unaccelerated motion. $$ % caption: The four-velocity is the unit tangent to the worldline and the % four-acceleration is Minkowski-orthogonal to it, pointing toward the concave % side of the curve. Their scalar product vanishes in every frame. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (-0.3,0) -- (4.4,0) node[right] {x}; \draw[->, black] (0,-0.3) -- (0,4.6) node[above] {ct}; % worldline \draw[black, very thick] (0.5,0.3) .. controls (1.0,1.8) and (2.0,2.4) .. (3.4,3.6); % point and tangent (four-velocity) + acceleration \fill[black] (1.85,2.32) circle (1.7pt); \draw[->, acc, very thick] (1.85,2.32) -- (2.75,3.05) node[right, midway, above] {U}; \draw[->, black, very thick] (1.85,2.32) -- (1.05,2.9) node[left] {A}; \node[black, anchor=west] at (2.6,1.7) {A orthogonal to U}; \end{tikzpicture} $$ The four-velocity, four-acceleration, and the four-momentum $p^\mu = mU^\mu$ that the [dynamics module](/relativity/relativistic-dynamics/four-momentum-force-and-accelerated-motion) 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 | Operation | Rule | Result | | --- | --- | --- | | Raise an index | $V^\mu = \eta^{\mu\nu}V_\nu$ | contravariant component | | Lower an index | $V_\mu = \eta_{\mu\nu}V^\nu$ | covariant component | | Contract a pair | $V_\mu W^\mu$ (one up, one down) | Lorentz scalar (invariant) | | Scalar square | $V\cdot V = \eta_{\mu\nu}V^\mu V^\nu$ | sign classifies the vector | | Four-velocity | $U^\mu = \d x^\mu/\d\tau = \gamma(c,\vec v)$ | timelike unit, $U\cdot U = -c^2$ | | Four-acceleration | $A^\mu = \d U^\mu/\d\tau$ | spacelike, $A\cdot U = 0$ | 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](/relativity/spacetime-and-the-lorentz-group/the-lorentz-group-and-rapidity) is the set of all matrices $\Lambda^\mu{}_\nu$ that preserve $\eta_{\mu\nu}$, and the next lesson works out its structure. [^schutz-vectors]: **Schutz**, _A First Course in General Relativity_, Ch. 2 — four-vectors and their components, the metric as the raising-and-lowering operator, the invariant scalar product, and the four-velocity with its normalization $U\cdot U = -c^2$ in the mostly-plus signature. [^carroll-index]: **Carroll**, _Lecture Notes on General Relativity_, §1 — the summation convention, contravariant and covariant indices, the distinction between vectors and dual vectors, and Lorentz-invariant contractions.