Index Notation and the Metric Tensor Formalism

Take a perfectly ordinary four-vector, $A^{\mu} = (2,\ 1,\ 0,\ 0)$ , and boost it to $\beta = 0.6$ . Its components become $A'^{\mu} = (1.75,\ -0.25,\ 0,\ 0)$ . Now build two numbers out of it. The first is $\sum_{\mu}A^{\mu}A^{\mu} = 4 + 1 = \mathbf{5}$ — a sum of squares, the thing every instinct trained on Euclid reaches for. The second is $A^{\mu}A_{\mu} = 4 - 1 = \mathbf{3}$ — the same components, one index lowered. In the boosted frame the first is $1.75^2 + 0.25^2 = \mathbf{3.125}$ ; it has changed by $37\%$ . The second is $1.75^2 - 0.25^2 = \mathbf{3.0000}$ ; it has not changed at all. The first is not a scalar. It is a number that is computable in any frame and means something different in each — the physics equivalent of a typo that compiles. The second is a Lorentz invariant, and the only thing separating them is that one index was lowered before the sum. In the chapter's locked signature $(+,-,-,-)$ the whole machine is one array, $\eta_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)$ , and one rule: contract one index up with one index down, never two the same way. What the rule guarantees is not a convention but a theorem — the covariant index transforms with the inverse matrix, because the metric is preserved. Everything else, up to "any equation with matching free indices is automatically relativistic", is a corollary.

Key takeaways

Index notation labels four-vector components by $\mu = 0,1,2,3$ , with $x^{\mu} = (ct, x, y, z)$ , and the Einstein summation convention (1916) sums any index appearing once up and once down.
The metric tensor in this chapter's locked signature is $\eta_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)$ — timelike-positive. It is its own inverse ( $\eta^{\mu\alpha}\eta_{\alpha\nu} = \delta^{\mu}{}_{\nu}$ ), so lowering an index just flips the sign of the three spatial components: $A_{\mu} = (A^0, -\vec{A})$ .
Hence the only invariant two four-vectors admit: $A\cdot B = \eta_{\mu\nu}A^{\mu}B^{\nu} = A^{\mu}B_{\mu} = A^0B^0 - \vec{A}\cdot\vec{B}$ . In this signature $p^{\mu}p_{\mu} = +m^2c^2$ and $\Box = \partial^{\mu}\partial_{\mu} = \frac{1}{c^2}\partial_t^2 - \nabla^2$ — both flip in the mostly-plus convention, and mixing the two is the commonest source of an imaginary mass.
The owned theorem: the inverse matrix is not a new object — it is a lowered index. From $\Lambda^{\mathsf{T}}\eta\Lambda = \eta$ alone, defining $\Lambda_{\mu}{}^{\nu} \equiv \eta_{\mu\alpha}\Lambda^{\alpha}{}_{\beta}\eta^{\beta\nu}$ gives $\Lambda_{\mu}{}^{\nu} = (\Lambda^{-1})^{\nu}{}_{\mu}$ , hence the master identity $\Lambda^{\mu}{}_{\alpha}\,\Lambda_{\mu}{}^{\beta} = \delta_{\alpha}{}^{\beta}$ .
Corollary (manifest covariance). That identity makes every matched up-down contraction invariant, for tensors of any rank — so any equation whose free indices match on both sides is automatically Lorentz-covariant. You can check relativistic consistency by looking at the indices.
Key idea: index bookkeeping is a theorem, not a convention. Preserving the metric forces the lower index to transform by $\Lambda^{-1}$ , the two matrices then cancel inside every contraction, and Lorentz invariance becomes a syntax rule you can enforce with your eyes.

The core insight

A four-vector's components are frame-dependent; its contractions are not. The metric converts one into the other, and index position is the flag telling you which side of the machine you are on. Upper indices transform with $\Lambda$ , lower indices with $\Lambda^{-1}$ — not a definition someone chose, but what $\Lambda^{\mathsf{T}}\eta\Lambda = \eta$ forces. Once forced, the two matrices annihilate each other inside any up-down pair, and the whole content of "this equation is relativistically correct" collapses into "the free indices match".

Where this lives today

The formalism outgrew relativity within a decade. Its central promise — that an equation whose indices match is automatically valid in every coordinate system — is exactly what general relativity needs, and there $\eta_{\mu\nu}$ becomes a dynamical field $g_{\mu\nu}(x)$ while every rule in this article survives verbatim, with one addition: $\partial_{\mu}$ is replaced by the covariant derivative $\nabla_{\mu}$ , built from the connection Levi-Civita supplied in 1917. Einstein's field equations $G_{\mu\nu} = 8\pi G\,T_{\mu\nu}/c^4$ are a statement whose correctness of form you can verify with your eyes: two free lower indices on each side, none repeated illegally. In quantum field theory the discipline is load-bearing rather than decorative. Every Lagrangian is a Lorentz scalar by construction, and "by construction" means all indices contracted up-with-down: $\mathcal{L} = \tfrac12\partial^{\mu}\phi\,\partial_{\mu}\phi - \tfrac12 m^2\phi^2$ , or $-\tfrac14 F^{\mu\nu}F_{\mu\nu}$ . The Dirac equation $(i\gamma^{\mu}\partial_{\mu} - m)\psi = 0$ contracts a matrix-valued upper index against a derivative's lower one, and the signature sits in plain sight inside $\gamma^{\mu}\gamma^{\nu} + \gamma^{\nu}\gamma^{\mu} = 2\eta^{\mu\nu}$ . The same machinery reappears in machine learning as einsum, where the index string is the entire specification of the operation — the summation convention, rediscovered by people who had never heard of Ricci.

The physical picture

Euclidean geometry lets you be sloppy. With $\delta_{ij}$ as the metric, lowering an index changes nothing — $A^1 = A_1$ — so nobody bothers with the distinction and $\vec{A}\cdot\vec{B} = \sum_i A_iB_i$ works with every index in the same place. Minkowski geometry forbids this, and the reason is one array: $\eta_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)$ . Lowering an index with $\eta$ flips the sign of the three spatial components and leaves the time component alone: if $A^{\mu} = (A^0,\ \vec{A})$ then $A_{\mu} = (A^0,\ -\vec{A})$ . That one sign flip is the whole difference between a scalar and a number. Contract up with down and it produces the difference of squares that a boost preserves. Contract up with up and you get a sum of squares — a perfectly well-defined number in each frame, and a different number in each frame. The overview's case, unpacked: with $A^{\mu} = (2,1,0,0)$ , lowering gives $A_{\mu} = (2,-1,0,0)$ , so $A^{\mu}A_{\mu} = 4 - 1 = 3$ , and under a boost to $\beta = 0.6$ ( $\gamma = 1.25$ , $\gamma\beta = 0.75$ ) the components become $A'^{\mu} = (1.75, -0.25, 0, 0)$ with $A'^{\mu}A'_{\mu} = 1.75^2 - 0.25^2 = 3.0000$ — unmoved, while the illegal sum slides from $5$ to $3.125$ . The formalism is designed so that you cannot make this mistake without seeing it. In $A^{\mu}A_{\mu}$ the repeated index appears once up and once down — legal. In $\sum_{\mu}A^{\mu}A^{\mu}$ it appears twice up, and the summation convention pointedly refuses to apply: you had to write the $\sum$ by hand, which is the notation telling you that what you are computing is not a scalar. Index notation is a type system, and the up-down rule is its type check.

Figure

The index machine. LEFT: the metric η = diag(+1, −1, −1, −1) lowers an index by flipping the sign of the three SPATIAL components — Aᵘ = (A⁰, A⃗) becomes A_μ = (A⁰, −A⃗) — and it is its own inverse, so raising undoes it exactly: η^μα η_αν = δᵘ_ν. RIGHT: the only legal contraction pairs one UP index with one DOWN index. Up-with-down gives the difference of squares A⁰B⁰ − A⃗·B⃗, which every observer agrees on. Up-with-up gives a sum of squares, which nobody does

The core equations

The signature is the chapter's locked $(+,-,-,-)$ , timelike-positive, and every sign below flips in the mostly-plus convention. State your $\eta$ and stay in it.

x^{\mu} = (ct,\ x,\ y,\ z), \qquad \eta_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1), \qquad \mathrm{d}s^2 = \eta_{\mu\nu}\,\mathrm{d}x^{\mu}\mathrm{d}x^{\nu} = c^2\mathrm{d}t^2 - \mathrm{d}x^2 - \mathrm{d}y^2 - \mathrm{d}z^2 \qquad (1)

where: $\mu = 0,1,2,3$ ; the Einstein summation convention sums a repeated index appearing once up and once down. Physical reading: the chapter's convention, owned by The Minkowski Metric. Timelike $\mathrm{d}s^2 > 0$ , so $c\,\mathrm{d}\tau = +\sqrt{\mathrm{d}s^2}$ and proper time is real and positive. In $(-,+,+,+)$ this line reads $\mathrm{d}s^2 = -c^2\mathrm{d}t^2 + \ldots$ and every subsequent sign in this article flips with it.

A_{\mu} = \eta_{\mu\nu}A^{\nu} = (A^0,\ -\vec{A}), \qquad A^{\mu} = \eta^{\mu\nu}A_{\nu}, \qquad \eta^{\mu\alpha}\eta_{\alpha\nu} = \delta^{\mu}{}_{\nu} \qquad (2)

where: $\eta^{\mu\nu}$ — the inverse metric, numerically the same array; $\delta^{\mu}{}_{\nu}$ — the Kronecker delta. Physical reading: $\eta^2 = I$ , so raising undoes lowering exactly. Lowering flips only the spatial signs. This is a gift of flatness — in a curved spacetime the inverse metric is a genuinely different array — and it is what makes index gymnastics cheap.

\boxed{\;A\cdot B \;=\; \eta_{\mu\nu}A^{\mu}B^{\nu} \;=\; A^{\mu}B_{\mu} \;=\; A^0B^0 - \vec{A}\cdot\vec{B}\;} \qquad (3)

where: the only Lorentz scalar you can build from two four-vectors. Physical reading: a difference of squares, hence indefinite — $A\cdot A$ can be positive, negative or zero for nonzero $A$ . The zero-norm case is not a degeneracy but the photon: $k^{\mu} = (\omega/c,\ \vec{k})$ with $|\vec{k}| = \omega/c$ has $k\cdot k = 0$ while $k^{\mu} \neq 0$ . The four-vectors themselves are owned by Four-Vectors and Lorentz Invariance and Invariants; this article supplies the machinery.

\partial_{\mu} \equiv \frac{\partial}{\partial x^{\mu}} = \left(\frac{1}{c}\frac{\partial}{\partial t},\ \nabla\right), \qquad \partial^{\mu} = \left(\frac{1}{c}\frac{\partial}{\partial t},\ -\nabla\right), \qquad \Box \equiv \partial^{\mu}\partial_{\mu} = \frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \nabla^2 \qquad (4)

where: $\Box$ — the d'Alembertian. Physical reading: note the index position on $\partial_{\mu}$ . The gradient with respect to an upper-index coordinate is naturally a lower-index object — the chain rule, not a convention — and it catches everybody once. In $(-,+,+,+)$ the d'Alembertian is the exact negative of ours, which leaves $\Box\phi = 0$ unaffected and $\Box\phi = \rho$ ruined.

p^{\mu} = (E/c,\ \vec{p}), \qquad \boxed{\;p^{\mu}p_{\mu} = \frac{E^2}{c^2} - |\vec{p}|^2 = +\,m^2c^2\;} \qquad (5)

where: $m$ — the invariant mass. Physical reading: $+m^2c^2$ , not $-m^2c^2$ — the base article for this page prints the latter, because it was written in the other signature. Unpacked, this contraction is the energy–momentum relation of relativistic dynamics, and that unpacking belongs to Four-Momentum and Energy–Momentum Relation. Here it is an example of a contraction, and nothing more.

Deriving the structure

Derivation AThe metric machinery, rebuilt from scratch in the timelike-positive signature

The base article for this page is written entirely in $(-,+,+,+)$ — metric, scalar product, mass shell and d'Alembertian all carry the opposite sign to ours. Each is correct in its own house and wrong in ours, so nothing below is transcribed: the machinery is rebuilt from the interval up, and the four signs are the point of the exercise.

Step 1. Fix $\eta$ from the interval. The interval is $\mathrm{d}s^2 = c^2\mathrm{d}t^2 - \mathrm{d}x^2 - \mathrm{d}y^2 - \mathrm{d}z^2$ , and writing it as the quadratic form $\eta_{\mu\nu}\mathrm{d}x^{\mu}\mathrm{d}x^{\nu}$ with $x^0 = ct$ forces $\eta_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)$ . Why: $\eta$ 's entries are not a choice — they are read off an invariant that experiment fixed. The only freedom is an overall sign, and that is the convention.

Step 2. Invert it, and lower an index. $\eta$ is diagonal with entries $\pm1$ , so $\eta^2 = I$ : it is its own inverse, $\eta^{\mu\nu}$ has the same entries as $\eta_{\mu\nu}$ , and $\eta^{\mu\alpha}\eta_{\alpha\nu} = \delta^{\mu}{}_{\nu}$ . Lowering reads $A_{\mu} = \eta_{\mu\nu}A^{\nu} = (A^0, -A^1, -A^2, -A^3)$ . Why: three signs flip, one does not — get that backwards and you have silently changed convention. The round trip closes: raising again gives $\eta^{\mu\nu}A_{\nu} = (A^0, -(-A^1), \ldots) = A^{\mu}$ .

Step 3. Build the scalar product: $A\cdot B = \eta_{\mu\nu}A^{\mu}B^{\nu} = A^{\mu}B_{\mu} = A^0B^0 - \vec{A}\cdot\vec{B}$ — bilinear, symmetric, and indefinite. Why: indefiniteness is the physics. $A\cdot A = (A^0)^2 - |\vec{A}|^2$ is a difference of squares, so a nonzero four-vector can have zero norm — and those are exactly the photons, $k\cdot k = 0$ with $k^{\mu} \neq 0$ . No positive-definite metric admits any such vector.

Step 4. Do the two contractions the base gets wrong. The four-gradient $\partial_{\mu} = \partial/\partial x^{\mu} = (\tfrac1c\partial_t,\ \nabla)$ carries a lower index by the chain rule, so raising gives $\partial^{\mu} = (\tfrac1c\partial_t,\ -\nabla)$ and $\Box = \partial^{\mu}\partial_{\mu} = \frac{1}{c^2}\partial_t^2 - \nabla^2$ . The mass shell: $p^{\mu} = (E/c,\ \vec{p})$ gives $p^{\mu}p_{\mu} = E^2/c^2 - |\vec{p}|^2 = +m^2c^2$ . Why: the base prints both with the opposite sign, for the honest reason that it lives in the other convention. The wave operator's sign is invisible in $\Box\phi = 0$ and lethal in $\Box A^{\mu} = \mu_0 J^{\mu}$ ; the mass shell's sign is the difference between a real mass and an imaginary one. This is the mixing error the whole formalism exists to prevent. *Verified in verify_aq-sr-index-notation-metric-tensor.py: $\eta^2 = I$ and $\det\eta = -1$ ; the raise–lower round trip returns $A^{\mu}$ identically; $A\cdot B$ reduces to $A^0B^0 - \vec{A}\cdot\vec{B}$ ; the nonzero vector $(1,1,0,0)$ has zero norm; $\Box = \frac{1}{c^2}\partial_t^2 - \nabla^2$ and $p^{\mu}p_{\mu} = +m^2c^2$ (asserted not $-m^2c^2$ ); and $\eta^{\mu\nu}\eta_{\mu\nu} = 4$ , which is signature-INDEPENDENT and therefore useless as a convention check — a fact the script asserts explicitly.*

The owned result: index bookkeeping is a theorem, not a convention

Everyone is told that upper indices transform with $\Lambda$ and lower ones with $\Lambda^{-1}$ , and that the two therefore cancel. Almost nobody is shown why the lower one transforms with the inverse. It is not a definition. It is forced — by the one equation that defines a Lorentz transformation in the first place.

Go deeper · Derivation B · 2 minFrom the metric condition alone, the lowered index transforms by the inverse matrix

Derivation BFrom the metric condition alone, the lowered index transforms by the inverse matrix

Take $\Lambda^{\mathsf{T}}\eta\Lambda = \eta$ — the defining condition of a Lorentz transformation, owned by The Lorentz Transformation Equations and The Minkowski Metric — and add nothing. Define the lowered-index transformation matrix in the only way the formalism allows, by lowering and raising with $\eta$ . What comes out is the inverse matrix, exactly, and with it a single index identity from which every invariance statement in relativity follows at once, for tensors of any rank.

Step 1. Define, don't assume. A vector transforms as $A'^{\mu} = \Lambda^{\mu}{}_{\nu}A^{\nu}$ . Its lowered partner must transform as $A'_{\mu} = \eta_{\mu\alpha}A'^{\alpha} = \eta_{\mu\alpha}\Lambda^{\alpha}{}_{\beta}\eta^{\beta\nu}A_{\nu}$ , so the matrix acting on lower indices is, by construction and not by choice, \Lambda_{\mu}{}^{\nu} \;\equiv\; \eta_{\mu\alpha}\,\Lambda^{\alpha}{}_{\beta}\,\eta^{\beta\nu}. Why: there is only one way to build it — lower the row index, raise the column index — because $\eta$ is the only thing available. Whatever this object turns out to be, we were not free to pick it.

Step 2. Turn the metric condition into a matrix statement. From $\Lambda^{\mathsf{T}}\eta\Lambda = \eta$ , multiply on the left by $(\Lambda^{\mathsf{T}})^{-1}$ : $\eta\Lambda = (\Lambda^{\mathsf{T}})^{-1}\eta = (\Lambda^{-1})^{\mathsf{T}}\eta$ . Multiply on the right by $\eta$ and use $\eta^2 = I$ : \eta\,\Lambda\,\eta \;=\; (\Lambda^{-1})^{\mathsf{T}}. Why: three lines of matrix algebra, no physics added. And $\eta\Lambda\eta$ is precisely the array $\Lambda_{\mu}{}^{\nu}$ of Step 1 — lowering the first index and raising the second is left- and right-multiplication by $\eta$ .

Step 3. Read off the theorem. \boxed{\;\Lambda_{\mu}{}^{\nu} \;=\; \big[(\Lambda^{-1})^{\mathsf{T}}\big]_{\mu}{}^{\nu} \;=\; (\Lambda^{-1})^{\nu}{}_{\mu}\;} The lowered index transforms with the inverse matrix — and it does so because the metric is preserved, not because anyone decreed it. Why: if $\Lambda$ did not preserve $\eta$ , the lowered index would transform with some third, unrelated matrix, contractions would not cancel, and the whole apparatus would collapse. Index gymnastics works because the Lorentz group is defined by metric preservation. Nothing else would do.

Step 4. Extract the master identity. Contract $\Lambda^{\mu}{}_{\alpha}$ with $\Lambda_{\mu}{}^{\beta}$ over the up-down pair $\mu$ : \Lambda^{\mu}{}_{\alpha}\,\Lambda_{\mu}{}^{\beta} \;=\; \Lambda^{\mu}{}_{\alpha}\,(\Lambda^{-1})^{\beta}{}_{\mu} \;=\; (\Lambda^{-1}\Lambda)^{\beta}{}_{\alpha} \;=\; \delta_{\alpha}{}^{\beta}. Why: <strong>this</strong> is the engine. Whenever an upper and a lower index are summed, the two transformation matrices collapse into a Kronecker delta and vanish. It is a statement about the matrices themselves, so it does not care what they act on.

Step 5. Harvest manifest covariance. Take any tensor equation with matched free indices, say $T^{\mu\nu}{}_{\rho} = S^{\mu\nu}{}_{\rho}$ . Each index carries its own $\Lambda$ or $\Lambda^{-1}$ ; Step 4 says every contracted pair cancels, while every free index gets the same factor on both sides. So the equation holds in one frame iff it holds in all, and a fully contracted object like $A^{\mu}B_{\mu}$ or $F^{\mu\nu}F_{\mu\nu}$ is a Lorentz scalar at any rank: A'^{\mu}B'_{\mu} = \Lambda^{\mu}{}_{\alpha}A^{\alpha}\;\Lambda_{\mu}{}^{\beta}B_{\beta} = \delta_{\alpha}{}^{\beta}A^{\alpha}B_{\beta} = A^{\alpha}B_{\alpha}. Why: <strong>you can check relativistic consistency by looking at the indices.</strong> If the free indices match on both sides and every repeated index is one up and one down, the equation is Lorentz-covariant — no computation required. If they do not, it is not, and no algebra will save it. The syntax is the physics. *Verified in verify_aq-sr-index-notation-metric-tensor.py: $\eta\Lambda\eta = (\Lambda^{-1})^{\mathsf{T}}$ symbolically for a general boost (and numerically for a boost, a rotation and a boost–rotation product); $\Lambda^{\mu}{}_{\alpha}\Lambda_{\mu}{}^{\beta} = \delta_{\alpha}{}^{\beta}$ exactly; $A'^{\mu}B'_{\mu} = A^{\mu}B_{\mu}$ symbolically; and the rank-2 case $F'^{\mu\nu}F'_{\mu\nu} = F^{\mu\nu}F_{\mu\nu}$ checked on an explicit antisymmetric array, confirming the theorem holds beyond rank 1.*

Figure

PUNCHLINE — the price of one misplaced index, in numbers. Take Aᵘ = (2, 1, 0, 0) and Bᵘ = (3, 2, 0, 0) and boost to β = 0.6 (γ = 1.25): A′ᵘ = (1.75, −0.25, 0, 0), B′ᵘ = (2.25, 0.25, 0, 0). The LEGAL contraction Aᵘ B_μ = A⁰B⁰ − A⃗·B⃗ reads 6 − 2 = 4 before and 3.9375 + 0.0625 = 4.0000 after — invariant, exactly. The ILLEGAL sum Σ Aᵘ Bᵘ reads 6 + 2 = 8 before and 3.8750 after: it has lost more than half its value. Same components, same boost; one index down, and the number becomes physics

Limiting cases

Case	Result	Meaning
$\eta \to \delta_{ij}$ (Euclidean)	$V_i = V^i$	index position becomes invisible and $\sum_i A_iB_i$ is legal — why the habit is so hard to break
$\eta \to g_{\mu\nu}(x)$	general relativity	the constant array becomes a field; $g^{\mu\nu}$ is a genuinely different array and $\partial_\mu \to \nabla_\mu$
$(-,+,+,+)$	$A\cdot B \to -A\cdot B$	every scalar product, $\Box$ and $p^{\mu}p_{\mu}$ flips sign. No measurable number changes
$A\cdot A = 0$ , $A \neq 0$	null vector	impossible in Euclid; here it is the photon, $k^{\mu} = (\omega/c, \vec{k})$ with $\|\vec{k}\| = \omega/c$
$\eta^{\mu\nu}\eta_{\mu\nu}$	$= \delta^{\mu}{}_{\mu} = 4$	signature-independent — the same in both conventions, so it will never catch a signature error

Common trapThe base for this article is written in the OTHER signature — and here is the exact damage

The AllQFT source for Index Notation and the Metric Tensor is in $(-,+,+,+)$ throughout: its first equation card is literally $\eta_{\mu\nu} = \mathrm{diag}(-1,+1,+1,+1)$ , its interval is $\mathrm{d}s^2 = -c^2\mathrm{d}t^2 + \mathrm{d}x^2 + \ldots$ , its scalar product is $A_{\mu}B^{\mu} = -A^0B^0 + \vec{A}\cdot\vec{B}$ , its d'Alembertian is $\Box = -\frac{1}{c^2}\partial_t^2 + \nabla^2$ , and its mass shell is $p^{\mu}p_{\mu} = -m^2c^2$ . Every one of those is correct in its own convention and wrong in this chapter's. Nothing has been transcribed; the article above re-derives all of it in $(+,-,-,-)$ .

No numeric check catches a signature error, because each convention is internally consistent: $\eta^{\mu\nu}\eta_{\mu\nu} = 4$ in both, $\det\eta = -1$ in both, and every proper time and invariant mass comes out to the same number in both. What breaks is taking one line from one book and one line from another — then $c^2\mathrm{d}\tau^2 = -\mathrm{d}s^2$ meets $\mathrm{d}s^2 = +c^2\mathrm{d}t^2 - \ldots$ and your clock reads $4i$ seconds. The symptom of a mixed signature is an imaginary proper time or an imaginary mass. Declare your $\eta$ at the top of the page and never move. Scoping note: the base's third derivation card boxes $E^2 = (pc)^2 + (mc^2)^2$ . That belongs to relativistic dynamics (Four-Momentum, Energy–Momentum Relation), not to index notation. It appears here only as an instance of a contraction, $p^{\mu}p_{\mu} = +m^2c^2$ , and is not derived.

Worked examples

1 · The electron whose invariant mass survives the boost

An electron ( $mc^2 = 0.511\ \mathrm{MeV}$ ) moves at $\beta = 0.6$ . Write its four-momentum, contract it, and confirm the invariant.

Step 1 — identify. $\gamma = 1/\sqrt{1-0.36} = \mathbf{1.25}$ , so $E = \gamma mc^2 = 1.25 \times 0.511 = \mathbf{0.63875\ MeV}$ and $pc = \gamma\beta mc^2 = 0.75 \times 0.511 = \mathbf{0.38325\ MeV}$ ; hence $p^{\mu}c = (0.63875,\ 0.38325,\ 0,\ 0)\ \mathrm{MeV}$ .

Step 2 — lower the index. $p_{\mu}c = \eta_{\mu\nu}p^{\nu}c = (0.63875,\ -0.38325,\ 0,\ 0)\ \mathrm{MeV}$ — time untouched, spatial flipped.

Step 3 — contract. p^{\mu}p_{\mu}\,c^2 = (0.63875)^2 - (0.38325)^2 = 0.408002 - 0.146881 = \mathbf{0.261121\ MeV^2}, and $(mc^2)^2 = 0.511^2 = \mathbf{0.261121\ MeV^2}$ — identical to every digit carried. So $p^{\mu}p_{\mu} = +m^2c^2$ , positive in this signature, and the mass comes out real.

Step 4 — the boundary check. Do the illegal contraction instead. Adding the squares rather than subtracting them gives $\sum_{\mu}(p^{\mu}c)^2 = \mathbf{0.554882\ MeV^2}$ , square root $\mathbf{0.7449\ MeV}$ . That is not the electron's mass, nor anything's mass, and at $\beta = 0.9$ the same illegal quantity gives $\mathbf{1.5772\ MeV}$ — a "mass" that grows with the frame — while the legal contraction still returns $0.261121\ \mathrm{MeV}^2$ there, unmoved. One index, and a conserved quantity becomes a coordinate artefact. (In $(-,+,+,+)$ : $p^{\mu}p_{\mu} = -0.261121\ \mathrm{MeV}^2$ , $m = \sqrt{-p^{\mu}p_{\mu}}/c^2$ — same number, different square root. Mix them and you get $0.511i$ .)

2 · Check an equation for relativistic correctness without computing anything

Which of these can possibly be a Lorentz-covariant law? $\ \partial_{\mu}F^{\mu\nu} = \mu_0 J^{\nu}$ · $\ \partial_{\mu}F^{\mu\nu} = \mu_0 J_{\nu}$ · $\ \partial^{\mu}\partial_{\mu}A^{\nu} = \mu_0 J^{\nu}$ · $\ F^{\mu\nu}F^{\mu\nu} = \text{invariant}$ .

Step 1 — apply the syntax rule. By the theorem above: (i) every repeated index once up and once down, (ii) free indices matching on both sides in position as well as name.

Step 2 — grade them. $\partial_{\mu}F^{\mu\nu} = \mu_0J^{\nu}$ : repeated $\mu$ is down-and-up ✓, free $\nu$ is up on both sides ✓ — legal, and in fact one of Maxwell's equations. $\partial_{\mu}F^{\mu\nu} = \mu_0 J_{\nu}$ : free $\nu$ is up on the left and down on the right ✗ — illegal; the two sides transform with different matrices and no algebra will rescue it. $\partial^{\mu}\partial_{\mu}A^{\nu} = \mu_0J^{\nu}$ : repeated $\mu$ up-and-down ✓, free $\nu$ up on both sides ✓ — legal (the wave equation in Lorenz gauge). $F^{\mu\nu}F^{\mu\nu}$ : repeated indices both up, twice ✗ — not a scalar, despite looking like one; the invariant is $F^{\mu\nu}F_{\mu\nu}$ .

Step 3 — the boundary number. The last is the punchline figure's trap at rank $2$ , and the difference is not academic: $F^{\mu\nu}F_{\mu\nu} = 2(B^2 - E^2/c^2)$ is the same in every frame, while $\sum_{\mu\nu}(F^{\mu\nu})^2 = 2(B^2 + E^2/c^2)$ is a different number for every observer — which is why $E^2 - c^2B^2$ appears in every textbook and $E^2 + c^2B^2$ appears in none.

Step 4 — what the rule cannot do. It certifies the form, not the content: $\partial_{\mu}F^{\mu\nu} = 17\,\mu_0J^{\nu}$ is perfectly covariant and perfectly wrong. Index matching says an equation could be a law of physics. Experiment says whether it is.

Physics of the portrait

Origins · 2 minTullio Levi-Civita and how this was discovered

Tullio Levi-Civita

1873–1941 · Padua · Rome — the absolute differential calculus, and the mathematics Einstein could not have done without

Born in Padua on 29 March 1873, Levi-Civita studied there under Gregorio Ricci-Curbastro, and the two of them built the machinery this article is about. Ricci had been developing an "absolute differential calculus" — a way of writing geometric statements so that they hold in any coordinate system, with the indices doing the bookkeeping — and in 1900 the pair published the systematic account: Méthodes de calcul différentiel absolu et leurs applications, Mathematische Annalen 54, 125–201. It is the origin of tensor analysis as physicists use it, and of the up-down discipline in this article. It might have stayed a curiosity of Italian differential geometry. Instead, in 1912, a struggling Einstein turned to his friend Marcel Grossmann for the mathematics of a gravitational theory, and Grossmann pointed him at Ricci and Levi-Civita. Einstein learned the absolute differential calculus from their work, adopted it wholesale, and in the 1916 paper laying out general relativity introduced the labour-saving device that carries his name — the summation convention, which drops the $\sum$ on any repeated index. He and Levi-Civita corresponded intensively through 1915–1917. In 1917 Levi-Civita supplied the missing geometric idea, parallel transport (Nozione di parallelismo in una varietà qualunque, Rendiconti del Circolo Matematico di Palermo 42, 173), which turned the index calculus into genuine geometry and gave the connection that bears his name. Levi-Civita taught at Padua from 1898 and moved to Rome in 1918. In 1938 the Fascist racial laws stripped him, as a Jew, of his chair and of his membership of every Italian scientific academy. He died in Rome on 29 December 1941, largely isolated. The notation survived him completely: every equation on this page is written in it.

History

1900

— Ricci-Curbastro and Levi-Civita publish Méthodes de calcul différentiel absolu et leurs applications (Mathematische Annalen 54, 125–201): tensor analysis with up-down index bookkeeping.

1905

— Poincaré's Comptes Rendus note (140, 1504, 5 June) names the Lorentz transformation and proves the transformations form a group. It contains no four-dimensional formalism.

1906

— Poincaré's Palermo memoir (Rendiconti del Circolo Matematico di Palermo 21, 129–175; received 23 July 1905, published February 1906) introduces the four-dimensional treatment with x⁴ = ict, the invariant quadratic form, and the first four-vectors.

1908

— Minkowski's Raum und Zeit (Cologne, 21 September) makes four-dimensional geometry and its invariant quadratic form the language of relativity.

1912

— Marcel Grossmann points Einstein to the Ricci–Levi-Civita absolute differential calculus as the mathematics for a theory of gravitation.

1916

— Einstein's Die Grundlage der allgemeinen Relativitätstheorie (Annalen der Physik 49, 769) introduces the summation convention: a repeated index is summed and the sigma is dropped.

1917

— Levi-Civita introduces parallel transport (Rendiconti del Circolo Matematico di Palermo 42, 173), turning the index calculus into geometry.

Going further

Texts. Rindler, Introduction to Special Relativity, 2nd ed., §§7–8 · Griffiths, Introduction to Electrodynamics, 4th ed., ch. 12 · Schutz, A First Course in General Relativity, ch. 3 · Misner, Thorne & Wheeler, Gravitation, ch. 2–3 (in $(-,+,+,+)$ — read the warning card first). Primary: Ricci-Curbastro & Levi-Civita, Math. Ann. 54 (1900) 125; Einstein, Ann. Phys. 49 (1916) 769.

Provenance. The base article is written entirely in the mostly-plus signature $(-,+,+,+)$ — metric card, interval, scalar product, d'Alembertian and mass shell all carry the opposite sign to this chapter's locked convention. Nothing was transcribed; every equation was re-derived in $(+,-,-,-)$ , and the four sign changes are itemised in the warning card. Its third derivation card boxes $E^2 = (pc)^2 + (mc^2)^2$ — relativistic dynamics, out of band; named here as a forward pointer, not derived. What this article proves, and the base merely asserts, is that the covariant index transforms by $\Lambda^{-1}$ because the metric is preserved: from $\Lambda^{\mathsf{T}}\eta\Lambda = \eta$ alone, $\eta\Lambda\eta = (\Lambda^{-1})^{\mathsf{T}}$ , hence $\Lambda^{\mu}{}_{\alpha}\Lambda_{\mu}{}^{\beta} = \delta_{\alpha}{}^{\beta}$ , hence manifest covariance at any rank. Every printed number is recomputed in verify_aq-sr-index-notation-metric-tensor.py, which also asserts $p^{\mu}p_{\mu} = +m^2c^2$ , $\Box = \frac{1}{c^2}\partial_t^2 - \nabla^2$ , and that $\eta^{\mu\nu}\eta_{\mu\nu} = 4$ is signature-blind. Signature: $(+,-,-,-)$ , timelike-positive.

Builds on. Four-Vectors · The Minkowski Metric · The Spacetime Interval · Lorentz Invariance and Invariants · The Lorentz Transformation Equations. Leads to. Lorentz Invariance and Invariants · Causal Structure of Spacetime · Hyperbolic Geometry of Boosts · Four-Momentum.

Why can't I just put all the indices upstairs and insert minus signs by hand?

You can, and people did for about a decade, and it is why the early literature is so hard to read. The trouble is that the minus signs are not attached to components but to contractions — and once you have tensors of rank two or three, or derivatives, or products of several objects, there is no reliable way to remember where they go. The up-down convention outsources the bookkeeping to the notation: the metric has already put the sign in, so a legal contraction cannot be wrong. The habit worth breaking is writing $A^{\mu}A^{\mu}$ and thinking "sum over $\mu$ " — the summation convention deliberately does not fire on that, which is the notation refusing to compute a non-scalar for you.

Is η really a tensor if its components never change?

Yes — and the constancy is a special feature, not an exception. A $(0,2)$ tensor transforms as $\eta'_{\mu\nu} = \Lambda^{\alpha}{}_{\mu}\Lambda^{\beta}{}_{\nu}\eta_{\alpha\beta}$ , and that is literally the equation $\Lambda^{\mathsf{T}}\eta\Lambda = \eta$ . So $\eta$ is transforming like a tensor; it simply comes out numerically unchanged, exactly as $\delta_{ij}$ does under rotations. That numerical invariance is not a property of tensors in general — it is the defining property of the Lorentz group, turned around. In general relativity the allowed transformations are no longer Lorentz ones, the components do change, and $g_{\mu\nu}(x)$ becomes a dynamical field. The tensor was always a tensor; only the group shrank.

The four-gradient ∂_μ has a lower index. Why, when x^μ has an upper one?

The chain rule, not a convention. Under $x'^{\mu} = \Lambda^{\mu}{}_{\nu}x^{\nu}$ the derivative transforms as $\partial'_{\mu} = \frac{\partial x^{\nu}}{\partial x'^{\mu}}\partial_{\nu} = (\Lambda^{-1})^{\nu}{}_{\mu}\,\partial_{\nu}$ — with the inverse matrix, which by this article's theorem is exactly how a lower index transforms. So $\partial_{\mu}$ is naturally covariant and its index position is forced. The payoff: the four-divergence $\partial_{\mu}A^{\mu}$ is a scalar with no metric inserted by hand, and $\Box = \partial^{\mu}\partial_{\mu}$ needs exactly one raise — giving $\frac{1}{c^2}\partial_t^2 - \nabla^2$ here.

Does anything physical depend on the choice of signature?

Nothing whatever. Proper times, invariant masses, cross-sections, decay rates — every measurable number is identical in $(+,-,-,-)$ and $(-,+,+,+)$ . What differs is the sign of $\mathrm{d}s^2$ , of every scalar product, of $\Box$ and of $p^{\mu}p_{\mu}$ , and which square root extracts a physical quantity: here $c\Delta\tau = +\sqrt{s^2}$ and $m = \sqrt{p^{\mu}p_{\mu}}/c$ ; there $c\Delta\tau = \sqrt{-s^2}$ and $m = \sqrt{-p^{\mu}p_{\mu}}/c$ . The danger is never in using one convention — it is in silently using two, and the symptom is an imaginary mass or an imaginary proper time.

What does "manifestly covariant" buy me, if I could transform both sides by hand?

The ability to see correctness instead of computing it. Transforming by hand is feasible at rank 1 and hopeless beyond: a rank-3 tensor equation in four dimensions has $64$ components, each a sum of $64$ terms. The theorem above collapses all of it into a syntax rule, and once the rule holds the equation holds in every inertial frame with no computation at all. That is why theorists write Lagrangians rather than component equations: a Lorentz-invariant Lagrangian is one with no free indices, and you can check that by looking.

Why is η^μν η_μν = 4 a bad way to check my signature?

Because it comes out $4$ in both conventions, and that is no accident: $\eta^{\mu\nu}\eta_{\mu\nu} = \delta^{\mu}{}_{\mu} = 4$ is the trace of the identity in four dimensions, and it never touches the pattern of signs — only the fact that $\eta$ is its own inverse, true either way. It is a useful check that your raising and lowering are consistent; it is a useless check that you are in the right signature. For that you need a quantity whose sign the convention flips: $p^{\mu}p_{\mu} = +m^2c^2$ here and $-m^2c^2$ there; $U\cdot U = +c^2$ here and $-c^2$ there; a timelike $s^2 > 0$ here and $< 0$ there. The verify script asserts all three — because the one number everyone checks is the one number that cannot tell you.