The Tensor Transformation Law

A single arrow drawn on a curved surface does not care how you label the surface with numbers. Rotate your grid, stretch it, bend it into polar coordinates — the arrow still points where it always pointed. Yet the moment you write the arrow as a list of components, those numbers depend entirely on the grid you chose. The tensor transformation law is the precise rule that keeps the object fixed while its components rearrange: it is the grammar that lets physics be written once and read in every coordinate system at once. The setting comes straight from the prerequisite pieces of this chapter. Two overlapping charts, $x^{\mu}$ and $x'^{\mu}$ , cover the same patch of a smooth manifold and are stitched together by smooth, invertible transition functions (from aq-gr-atlases-transition-functions). Differentiating those functions produces the Jacobian matrix and its inverse. Tangent vectors, being directional derivatives (aq-gr-tangent-vectors-tangent-space), inherit the Jacobian; one-forms, being gradients dual to vectors (aq-gr-cotangent-space-one-forms), inherit the inverse Jacobian. Every richer object is a tensor product of these two, so its law is dictated slot by slot. The pattern is not bookkeeping. It is forced by a single demand: that a contraction — the pairing of an upper with a lower index — be the same number in every chart. That demand makes the two Jacobians mutual inverses, welds the contravariant/covariant distinction to the law itself, and is so decisive that most textbooks simply define a tensor as any array obeying it. Get this law and the whole apparatus of general relativity — metrics, curvature, the invariant volume element — becomes well posed.

Key takeaways

A tensor is a geometric object that exists independently of coordinates; its components are only its shadow in a chosen chart and must rearrange precisely when the chart changes.
Contravariant (upper) components transform with the Jacobian $\partial x'^{\mu}/\partial x^{\nu}$ ; covariant (lower) components transform with the inverse Jacobian $\partial x^{\nu}/\partial x'^{\mu}$ — the two matrices are inverses, which is exactly why contractions are invariant.
A general $(p,q)$ -tensor carries one Jacobian factor per upper index and one inverse-Jacobian factor per lower index, all evaluated at the same point.
The law is the operational definition of a tensor: an indexed array is a tensor if and only if it obeys this rule under every admissible coordinate change — the Christoffel symbols famously fail it.
Because the law is linear and homogeneous, a tensor that vanishes in one chart vanishes in all of them, which is why tensor equations encode coordinate-independent physics.
Multiplying by a power of the Jacobian determinant generalizes the rule to tensor densities, the ingredient behind $\sqrt{-g}\,\mathrm{d}^n x$ and integration on curved manifolds.
Key idea: attach one Jacobian factor per index — $\partial x'/\partial x$ for each upper index, $\partial x/\partial x'$ for each lower index, all at the same point, summed over repeats. An array is a tensor precisely when its components obey this rule in every chart.

The core insight

For each index attach one Jacobian factor — $\partial x'/\partial x$ for an upper (contravariant) index and $\partial x/\partial x'$ for a lower (covariant) index, all evaluated at the same point and summed over repeated indices; an object is a tensor precisely when its components obey this rule under every coordinate change.

Where this lives today

The transformation law is quietly load-bearing in modern gravitational-wave astronomy. When the LIGO and Virgo detectors recorded the first binary black-hole merger, GW150914, in 2015, the waveform templates they matched against came from numerical-relativity codes that integrate Einstein's equations on a computer. Those codes live on coordinates in different ways. The Einstein Toolkit evolves the field equations by finite differences on Cartesian adaptive-mesh grids in the "moving-puncture" gauge, whose coordinates let the singularities glide across the grid without crashing the run; its optional Llama infrastructure adds overlapping patches, and there a common choice is to keep tensor components in one global Cartesian basis and apply the Jacobians to the derivative operators instead. SpEC takes the other route: multidomain spectral grids — shells hugging each excised black hole, a different decomposition in the wave zone — with the system mapped between a comoving grid frame and an asymptotically inertial frame. Wherever such a scheme does re-express components in a new chart, they are carried by exactly the $(p,q)$ law of equation (5), with the Jacobians computed on the fly. A second use is diagnostic. The apparent blow-up of the Schwarzschild metric at $r = 2GM/c^2$ looks alarming until you transform to Eddington–Finkelstein or Kruskal coordinates and watch the components become finite, while curvature scalars — true invariants built by contraction — stay perfectly regular. Distinguishing a coordinate artifact from a physical singularity, the everyday business of any black-hole simulation, is nothing but the transformation law used as a truth-detector.

The physical picture

Draw one arrow at a point of a plane and lay two grids over it: an ordinary Cartesian mesh and a polar mesh of circles and rays. The arrow is untouched — but "how many steps east, how many north" is a different pair of numbers from "how far outward, how far around." The Jacobian is the dictionary that retranslates the first pair into the second so that both still describe the same arrow. Nothing physical happens in the translation; only the labels change.

Figure

One arrow, two grids: the same vector at the point x = 1, y = 1 reads as Cartesian components (Vˣ, Vʸ) = (1, 0) but polar components (Vʳ, Vᶿ) = (0.71, −0.50); the Jacobian ∂x′/∂x is the dictionary between them

Why do vectors and one-forms transform in opposite ways? Picture a vector as "how fast you move along a curve" and a one-form as "how densely the level surfaces of a function are stacked." Compress the coordinate spacing and a fixed physical displacement now spans more coordinate units — the vector's components grow. But that same compression means a fixed gradient crosses fewer level surfaces per coordinate unit — the one-form's components shrink. The two effects are exactly reciprocal, which is the whole reason their product survives the change of chart. The next figure makes that reciprocity a matter of arithmetic.

Figure

Rescaling the coordinate x′ = 2x doubles the vector component Vˣ from 1 to 2 but halves the one-form component ωₓ from 2 to 1, so their contraction Vˣωₓ = 2 is unchanged — the reciprocal Jacobian factors are precisely what make a scalar coordinate-independent

The mnemonic packs it: co goes below, contra goes above, and co transforms like the basis, contra like the coordinates. Basis vectors $\partial_{\mu}$ carry a lower index and follow the inverse Jacobian; the components $V^{\mu}$ carry an upper index and compensate with the Jacobian so that the assembled object $V = V^{\mu}\partial_{\mu}$ never moves. The payoff is enormous: a tensor equation true in one frame is true in all frames, because every term rearranges identically. That single fact is what lets Einstein's equations and Maxwell's equations be written in a form valid in every coordinate system simultaneously.

The core equations

Everything begins with the change of chart itself. Two coordinate systems cover the same patch, and each point carries two sets of labels related by a smooth, invertible map. This map is the raw material from which every Jacobian is built.

x'^{\mu} = x'^{\mu}(x^0, x^1, \dots, x^{n-1}) \qquad (1)

where: $x^{\nu}$ — old coordinates of a point; $x'^{\mu}$ — new coordinates of the same point; $n$ — dimension of the manifold (4 for spacetime). Physical reading: the map relabels points, it does not move them. Differentiating the transition map gives the Jacobian and its inverse. Because relabelling and un-relabelling must compose to the identity, the two matrices are forced to be inverses of one another — this is the hinge on which the entire law turns.

J^{\mu}{}_{\nu} \equiv \frac{\partial x'^{\mu}}{\partial x^{\nu}}, \qquad (J^{-1})^{\nu}{}_{\mu} \equiv \frac{\partial x^{\nu}}{\partial x'^{\mu}}, \qquad \frac{\partial x'^{\mu}}{\partial x^{\nu}}\,\frac{\partial x^{\nu}}{\partial x'^{\rho}} = \delta^{\mu}{}_{\rho} \qquad (2)

where: $J^{\mu}{}_{\nu}$ — Jacobian of the coordinate change; $(J^{-1})^{\nu}{}_{\mu}$ — inverse Jacobian, derivatives of old with respect to new; $\delta^{\mu}{}_{\rho}$ — Kronecker delta (1 if the indices agree, else 0). Physical reading: the inverse Jacobian undoes the Jacobian exactly, at every point. A contravariant vector is the prototype of an upper index. Its components pick up one Jacobian factor so that the geometric arrow they describe is unchanged.

V'^{\mu} = \frac{\partial x'^{\mu}}{\partial x^{\nu}}\, V^{\nu} \qquad (3)

where: $V^{\nu}$ — components in the old basis; $V'^{\mu}$ — components in the new basis. Physical reading: coordinate differentials $\mathrm{d}x^{\mu}$ transform this way, which is why displacements and velocities are contravariant. A covariant one-form is the prototype of a lower index. Its components pick up one inverse Jacobian factor — the opposite rule to the vector, and the source of every cancellation to come.

\omega'_{\mu} = \frac{\partial x^{\nu}}{\partial x'^{\mu}}\, \omega_{\nu} \qquad (4)

where: $\omega_{\nu}$ — components in the old dual basis $\mathrm{d}x^{\nu}$ ; $\omega'_{\mu}$ — components in the new dual basis $\mathrm{d}x'^{\mu}$ . Physical reading: gradients $\partial_{\mu}\phi$ transform this way, which is why gradients are covariant. A general $(p,q)$ -tensor is a tensor product of $p$ vectors and $q$ one-forms, so its law is simply one factor per slot: $p$ Jacobians upstairs, $q$ inverse Jacobians downstairs. Textbooks take this as the very definition of a tensor.

T'^{\mu_1\cdots\mu_p}{}_{\nu_1\cdots\nu_q} = \frac{\partial x'^{\mu_1}}{\partial x^{\alpha_1}}\cdots\frac{\partial x'^{\mu_p}}{\partial x^{\alpha_p}}\,\frac{\partial x^{\beta_1}}{\partial x'^{\nu_1}}\cdots\frac{\partial x^{\beta_q}}{\partial x'^{\nu_q}}\,T^{\alpha_1\cdots\alpha_p}{}_{\beta_1\cdots\beta_q} \qquad (5)

where: $p$ — number of upper indices; $q$ — number of lower indices; $T^{\alpha\cdots}{}_{\beta\cdots}$ — old components summed against all the Jacobian factors. Physical reading: each slot is independent; the rule for a big tensor is just the rule for a vector or one-form, repeated. The reason for the whole up/down machinery is that a contraction — an upper index summed against a lower one — is invariant. This is the statement that makes physical laws expressible in coordinate-free form.

S = V^{\mu}\,\omega_{\mu} = V'^{\mu}\,\omega'_{\mu} \qquad (6)

where: $S$ — the scalar produced by pairing a vector with a one-form; $V^{\mu}\omega_{\mu}$ — the sum over the repeated index. Physical reading: the Jacobian from $V$ and the inverse Jacobian from $\omega$ collapse to a Kronecker delta, leaving a number that no coordinate change can touch. Finally, the metric is a genuine $(0,2)$ -tensor: two lower indices, two inverse-Jacobian factors. This one equation is what lets you carry a line element between any two charts.

g'_{\mu\nu} = \frac{\partial x^{\alpha}}{\partial x'^{\mu}}\frac{\partial x^{\beta}}{\partial x'^{\nu}}\, g_{\alpha\beta} \qquad (7)

where: $g_{\alpha\beta}$ — metric components in old coordinates; $g'_{\mu\nu}$ — in new coordinates. Physical reading: proper distance $\mathrm{d}s^2 = g_{\mu\nu}\,\mathrm{d}x^{\mu}\mathrm{d}x^{\nu}$ comes out the same number in every chart, because the metric's two inverse Jacobians cancel the two Jacobians from the differentials.

Derivation AWhy the two Jacobians are inverse matrices

The entire transformation law leans on one fact: the Jacobian and the inverse Jacobian are genuine matrix inverses, so that a factor from an upper index and a factor from a lower index annihilate to a Kronecker delta. This is not assumed; it follows the instant you insist that relabelling a point and then un-relabelling it returns the point to itself. The goal here is to extract that identity from the chain rule alone, using nothing but the smoothness and invertibility of the transition functions the atlas already guarantees.

Step 1. Compose the two relabellings into the identity map on coordinates, $x^{\rho} = x^{\rho}\!\big(x'(x)\big)$ . Going to the new chart and back reproduces the original coordinate value exactly. Why: this is just the statement that the transition function is invertible, which the differentiable structure of the manifold provides.

Step 2. Differentiate both sides with respect to $x^{\nu}$ using the chain rule, giving $\dfrac{\partial x^{\rho}}{\partial x'^{\mu}}\dfrac{\partial x'^{\mu}}{\partial x^{\nu}} = \dfrac{\partial x^{\rho}}{\partial x^{\nu}}$ . The right-hand side is $\delta^{\rho}{}_{\nu}$ because distinct coordinates are functionally independent. Why: the coordinates $x^{\rho}$ are the independent variables, so one never depends on another — their mutual derivative is the identity matrix.

Step 3. Read off $\,(J^{-1})^{\rho}{}_{\mu}\,J^{\mu}{}_{\nu} = \delta^{\rho}{}_{\nu}$ , i.e. $J^{-1}J = \mathbb{1}$ . The same computation differentiating with respect to $x'$ gives $JJ^{-1} = \mathbb{1}$ . Why: an upper-index Jacobian and a lower-index inverse Jacobian, summed on a shared index, always collapse to $\delta$ — this single fact powers every cancellation in the article.

Go deeper · Derivation B · 1 minBuilding the general tensor law from vectors and one-forms

Derivation BBuilding the general tensor law from vectors and one-forms

A $(p,q)$ -tensor is not a fresh object with its own postulated rule; it is a multilinear machine assembled from $p$ vectors and $q$ one-forms glued by the tensor product. The goal here is to show that its transformation law is therefore forced, slot by slot, by the vector law (3) and the one-form law (4): write the tensor in a coordinate basis, transform each basis element by the rule it already obeys, and collect the Jacobians. What drops out is exactly equation (5), with no independent assumption — the general law contains nothing new.

Step 1. Expand the tensor in its coordinate basis, $T = T^{\alpha_1\cdots\alpha_p}{}_{\beta_1\cdots\beta_q}\,\partial_{\alpha_1}\otimes\cdots\otimes\partial_{\alpha_p}\otimes\mathrm{d}x^{\beta_1}\otimes\cdots\otimes\mathrm{d}x^{\beta_q}$ . The object $T$ itself is basis-independent; only its split into components and basis will change. Why: this is the geometric content — $T$ is fixed, and a change of chart merely redistributes it between components and basis elements.

Step 2. Transform each basis vector by inverting the chain-rule relation $\partial'_{\mu} = (\partial x^{\alpha}/\partial x'^{\mu})\,\partial_{\alpha}$ , which gives $\partial_{\alpha} = (\partial x'^{\mu}/\partial x^{\alpha})\,\partial'_{\mu}$ . Each upper slot therefore contributes a $\partial x'/\partial x$ . Why: basis vectors transform inversely to components, so pushing the transformation onto the basis reproduces the Jacobian that the component law (3) demands.

Step 3. Transform each basis one-form by differentiating the transition function, $\mathrm{d}x^{\beta} = (\partial x^{\beta}/\partial x'^{\nu})\,\mathrm{d}x'^{\nu}$ . Each lower slot contributes a $\partial x/\partial x'$ . Why: the dual basis follows the inverse-Jacobian rule (4), so every lower index inevitably carries an inverse Jacobian.

Step 4. Collect all the factors in front of the new basis $\partial'_{\mu_1}\otimes\cdots\otimes\mathrm{d}x'^{\nu_q}$ . Whatever multiplies that basis is, by definition, the new component array $T'^{\mu\cdots}{}_{\nu\cdots}$ — and it is precisely $\prod_i (\partial x'^{\mu_i}/\partial x^{\alpha_i}) \prod_j (\partial x^{\beta_j}/\partial x'^{\nu_j})\, T^{\alpha\cdots}{}_{\beta\cdots}$ . Why: setting $(p,q)=(1,0)$ recovers the vector law, $(0,1)$ the one-form law, and $(0,0)$ gives $\phi' = \phi$ — the general rule degrades correctly to every special case, matching MTW §3 and Carroll Ch. 1.

Go deeper · Derivation C · 1 minWhy the Christoffel symbols fail the tensor law

Derivation CWhy the Christoffel symbols fail the tensor law

Not every indexed array is a tensor, and the connection coefficients are the canonical counterexample: they are the objects one needs precisely because they are not tensorial. The goal here is to expose where the tensor law breaks. Demand that the covariant derivative of a vector be a genuine $(1,1)$ -tensor even though the plain partial derivative is not, transform the partial derivative honestly, and watch a second-derivative-of-the-coordinates term appear that no homogeneous tensor rule can produce. The connection must then carry an inhomogeneous piece to cancel that anomaly — proof that $\Gamma$ is not a tensor, yet a difference of two connections is.

Step 1. Posit the covariant derivative $\nabla_{\mu}V^{\lambda} = \partial_{\mu}V^{\lambda} + \Gamma^{\lambda}{}_{\mu\nu}V^{\nu}$ and require the left side to transform as a $(1,1)$ -tensor. Why: the partial derivative $\partial_{\mu}V^{\lambda}$ alone does not transform tensorially, so a correction term $\Gamma$ is introduced whose job is to restore tensor behaviour.

Step 2. Transform the partial derivative directly and find two terms, $\partial'_{\mu}V'^{\lambda} = \dfrac{\partial x^{\sigma}}{\partial x'^{\mu}}\dfrac{\partial x'^{\lambda}}{\partial x^{\rho}}\,\partial_{\sigma}V^{\rho} + \dfrac{\partial x^{\sigma}}{\partial x'^{\mu}}\dfrac{\partial^2 x'^{\lambda}}{\partial x^{\sigma}\partial x^{\rho}}\,V^{\rho}$ . The first term is the clean tensorial piece; the second, carrying a second derivative of the coordinates, is the anomaly. Why: the product rule hitting the Jacobian inside $V'^{\lambda} = (\partial x'^{\lambda}/\partial x^{\rho})V^{\rho}$ generates a derivative of the Jacobian — a term absent from any homogeneous $(1,1)$ law.

Step 3. For $\nabla'_{\mu}V'^{\lambda} = \dfrac{\partial x^{\sigma}}{\partial x'^{\mu}}\dfrac{\partial x'^{\lambda}}{\partial x^{\rho}}\,\nabla_{\sigma}V^{\rho}$ to hold cleanly, the inhomogeneous part of $\Gamma$ must exactly cancel the anomaly. Solving this requirement yields the connection's transformation law with its telltale extra term,

\Gamma'^{\lambda}{}_{\mu\nu} = \frac{\partial x'^{\lambda}}{\partial x^{\rho}}\frac{\partial x^{\sigma}}{\partial x'^{\mu}}\frac{\partial x^{\tau}}{\partial x'^{\nu}}\,\Gamma^{\rho}{}_{\sigma\tau} + \frac{\partial x'^{\lambda}}{\partial x^{\rho}}\frac{\partial^2 x^{\rho}}{\partial x'^{\mu}\partial x'^{\nu}}.

Why: the first group of factors is the honest tensor rule; the second, inhomogeneous group is what disqualifies $\Gamma$ as a tensor — and because it is independent of $\Gamma$ , it cancels in any difference $\Gamma_1 - \Gamma_2$ , which is therefore a genuine tensor. This matches Wald §3.1 and Carroll Ch. 3.

Limiting cases

The general $(p,q)$ law of equation (5) collapses to every familiar object by choosing $p$ and $q$ :

Case	Result	Meaning
Scalar $(0,0)$	$\phi'(x') = \phi(x)$	No Jacobian factors: a scalar field has the same value at the same physical point in every chart.
Vector $(1,0)$	$V'^{\mu} = \dfrac{\partial x'^{\mu}}{\partial x^{\nu}}\, V^{\nu}$	One Jacobian factor — the prototype contravariant rule, matching how $\mathrm{d}x^{\mu}$ transforms.
One-form $(0,1)$	$\omega'_{\mu} = \dfrac{\partial x^{\nu}}{\partial x'^{\mu}}\, \omega_{\nu}$	One inverse-Jacobian factor — the prototype covariant rule, matching how $\partial_{\mu}$ transforms.
Kronecker delta $(1,1)$	$\delta'^{\mu}{}_{\nu} = \dfrac{\partial x'^{\mu}}{\partial x^{\alpha}}\dfrac{\partial x^{\beta}}{\partial x'^{\nu}}\,\delta^{\alpha}{}_{\beta} = \delta^{\mu}{}_{\nu}$	Jacobian and inverse collapse: $\delta$ is the same numerical array in every chart — an invariant tensor.
Density, weight $w$	$\mathfrak{T}' = \left\vert \dfrac{\partial x}{\partial x'}\right\vert^{w}\dfrac{\partial x'^{\mu}}{\partial x^{\alpha}}\cdots\,\mathfrak{T}$	An extra power $w$ of the Jacobian determinant; $w=0$ recovers an ordinary tensor, $\sqrt{-g}$ has $w=1$ .

Common trapAny array that carries indices is a tensor

It is not. Indices are necessary but nowhere near sufficient. The Christoffel symbols $\Gamma^{\lambda}{}_{\mu\nu}$ , the plain partial derivative $\partial_{\mu}V^{\nu}$ of a vector field, and ordinary coordinate differences all wear indices yet violate the law — each picks up an inhomogeneous term built from second derivatives of the coordinates. The only test that matters is the transformation law itself. Two corollaries are worth memorizing. First, a genuine tensor that vanishes in one chart vanishes in every chart, because the law is linear and homogeneous in the components — this is what makes $T_{\mu\nu}=0$ a coordinate-independent statement. Second, the converse warning: an array that merely happens to be zero in one convenient frame tells you nothing in another unless you have already verified it is a tensor. Always check the law before you trust the indices.

Worked examples

1 · The flat metric in polar coordinates (confidence builder)

The Euclidean plane in Cartesian coordinates has metric $g_{\alpha\beta} = \delta_{\alpha\beta}$ , so $\mathrm{d}s^2 = \mathrm{d}x^2 + \mathrm{d}y^2$ . Use the $(0,2)$ law to find the metric in polar coordinates $x = r\cos\theta,\ y = r\sin\theta$ , and read off the value at $r = 2$ .

Step 1. Build the inverse Jacobian $\partial x^{\alpha}/\partial x'^{\mu}$ with $x'^{\mu} = (r,\theta)$ . The needed derivatives are $\partial x/\partial r = \cos\theta$ , $\partial x/\partial\theta = -r\sin\theta$ , $\partial y/\partial r = \sin\theta$ , $\partial y/\partial\theta = r\cos\theta$ . These are the four entries that carry Cartesian components into polar ones.

Step 2. Apply $g'_{\mu\nu} = \dfrac{\partial x^{\alpha}}{\partial x'^{\mu}}\dfrac{\partial x^{\beta}}{\partial x'^{\nu}}\,\delta_{\alpha\beta}$ , which since $g_{\alpha\beta}=\delta_{\alpha\beta}$ is just the sum of squared column entries. The $rr$ component is $\cos^2\theta + \sin^2\theta = \mathbf{1}$ ; the $\theta\theta$ component is $r^2\sin^2\theta + r^2\cos^2\theta = \mathbf{r^2}$ ; the cross term is $\cos\theta(-r\sin\theta) + \sin\theta(r\cos\theta) = \mathbf{0}$ .

Step 3. Assemble the result: $g'_{\mu\nu} = \mathrm{diag}(1,\,r^2)$ , i.e. $\mathrm{d}s^2 = \mathrm{d}r^2 + r^2\,\mathrm{d}\theta^2$ . At $r = 2$ the angular metric component is $g'_{\theta\theta} = \mathbf{4}$ , so a small angular step $\mathrm{d}\theta$ spans a physical length $2\,\mathrm{d}\theta$ — the familiar arc-length factor, produced automatically by the transformation law. The flat metric becomes diag(1, r²); no curvature was created, only a new chart.

2 · Transforming a vector and checking that its length is invariant

At the point $(x,y) = (1,1)$ take the unit east-pointing vector with Cartesian components $(V^x, V^y) = (1, 0)$ . Find its polar components and verify that its length is unchanged.

Step 1. Use the contravariant law $V'^{\mu} = (\partial x'^{\mu}/\partial x^{\nu})\,V^{\nu}$ with $x'^{\mu} = (r,\theta)$ . The relevant derivatives are $\partial r/\partial x = x/r = \cos\theta$ and $\partial\theta/\partial x = -y/r^2 = -\sin\theta/r$ . At $(1,1)$ we have $r = \sqrt{2}$ and $\theta = 45°$ .

Step 2. Contract with $(V^x, V^y) = (1,0)$ : $V^r = (x/r)\,V^x = 1/\sqrt{2} = \mathbf{0.707}$ and $V^{\theta} = (-y/r^2)\,V^x = -1/2 = \mathbf{-0.500}$ . Note that $V^{\theta}$ is a coordinate-basis component, not an arc-length component — negative because moving east at this point decreases the polar angle.

Step 3. Check the length with the polar metric from Example 1, $g'_{\mu\nu} = \mathrm{diag}(1, r^2)$ : $\lvert V\rvert^2 = g'_{rr}(V^r)^2 + g'_{\theta\theta}(V^{\theta})^2 = 1\cdot\tfrac12 + 2\cdot\tfrac14 = \tfrac12 + \tfrac12 = \mathbf{1}$ . The magnitude is exactly 1, matching the Cartesian value $(V^x)^2 + (V^y)^2 = 1$ . Components changed from (1, 0) to (0.707, −0.500), yet the invariant length stayed 1 — the metric's inverse Jacobians cancelled the vector's Jacobians.

3 · A quantity that is zero in one chart but not another: the Christoffel symbols

In the flat plane, Cartesian coordinates make every Christoffel symbol vanish, $\Gamma^{\rho}{}_{\sigma\tau} = 0$ . Compute $\Gamma^{r}{}_{\theta\theta}$ in polar coordinates and explain why this proves $\Gamma$ is not a tensor.

Step 1. With the Cartesian $\Gamma$ all zero, the connection law from the third derivation keeps only its inhomogeneous term: $\Gamma'^{\lambda}{}_{\mu\nu} = \dfrac{\partial x'^{\lambda}}{\partial x^{\rho}}\dfrac{\partial^2 x^{\rho}}{\partial x'^{\mu}\partial x'^{\nu}}$ , where $x'^{\mu} = (r,\theta)$ and $x^{\rho} = (x,y)$ . Only the second-derivative term can survive.

Step 2. For $\Gamma'^{r}{}_{\theta\theta}$ compute the second derivatives $\partial^2 x/\partial\theta^2 = -r\cos\theta$ and $\partial^2 y/\partial\theta^2 = -r\sin\theta$ , and the first derivatives $\partial r/\partial x = \cos\theta$ , $\partial r/\partial y = \sin\theta$ . Then $\Gamma'^{r}{}_{\theta\theta} = \cos\theta\,(-r\cos\theta) + \sin\theta\,(-r\sin\theta) = -r(\cos^2\theta + \sin^2\theta) = \mathbf{-r}$ .

Step 3. The result is decisive. If $\Gamma$ were a genuine tensor, its vanishing in the Cartesian chart would force it to vanish in every chart, including polar. Instead $\Gamma^{r}{}_{\theta\theta} = -r \neq 0$ . A tensor cannot be zero in one chart and nonzero in another; because $\Gamma$ manifestly is, it is not a tensor — the inhomogeneous term is exactly the culprit, and it is why curved-space calculus needs the covariant derivative.

Physics of the portrait

The transformation law is not the work of a single mind, but it has a clear architect: the man who built tensor calculus into a self-contained discipline and handed Einstein the exact tool he needed a generation later.

Origins · 1 minGregorio Ricci-Curbastro and how this was discovered

Gregorio Ricci-Curbastro

1853–1925 · founder of the absolute differential calculus — tensor analysis

Gregorio Ricci-Curbastro spent the 1880s and 1890s at the University of Padua turning scattered results of Riemann, Christoffel, and Lipschitz into a systematic calculus of quantities that transform correctly under any change of coordinates — what he called the absolute differential calculus, and what we now call tensor analysis. His central insight was that if you organize objects strictly by how their components respond to the Jacobian, you obtain a language in which equations automatically hold in every coordinate system, freeing physics from any privileged frame. In 1900, with his former student Tullio Levi-Civita, he published the definitive account, Méthodes de calcul différentiel absolu et leurs applications, in Mathematische Annalen — the paper that codified the covariant and contravariant transformation rules, the summation of Jacobian factors slot by slot, and the algebra of raising and lowering indices. For fifteen years the work was admired mostly by mathematicians. Then, between 1912 and 1915, Einstein and Marcel Grossmann realized that Ricci's calculus was precisely the machinery general relativity required. The Ricci tensor, the contraction of the curvature that sits on the left of Einstein's field equations, carries his name; Ricci-Curbastro authored the grammar in which the geometry of spacetime is still written today.

History

1854

— Riemann's habilitation lecture founds the intrinsic geometry of manifolds and the metric form

1869

— Christoffel studies the transformation of quadratic differential forms and defines his symbols

1900

— Ricci-Curbastro and Levi-Civita codify the absolute differential calculus of tensors

1913

— Einstein and Grossmann adopt tensor calculus for a generally covariant theory of gravity

1915

— Einstein casts general relativity in fully covariant tensor form

1917

— Levi-Civita gives the law geometric meaning through parallel transport

Going further

References. Misner, Thorne & Wheeler, Gravitation (1973), §3 · Wald, General Relativity (1984), Ch. 2 · Carroll, Spacetime and Geometry (2004), Ch. 1–2 · Schutz, A First Course in General Relativity (2009), Ch. 5 · Hartle, Gravity (2003), Ch. 20.

Provenance. Every transformation law and Jacobian identity here is standard and recomputable directly from the chain rule on transition functions; cross-check the general $(p,q)$ law against MTW §3, Wald Ch. 2, Carroll Ch. 1–2, and Schutz Ch. 5. The Christoffel inhomogeneous term and its derivation match Wald §3.1 and Carroll Ch. 3. The worked polar-coordinate results — $g'_{\mu\nu} = \mathrm{diag}(1, r^2)$ , the vector components $(V^r, V^{\theta}) = (0.707, -0.500)$ with invariant length 1, and $\Gamma^{r}{}_{\theta\theta} = -r$ — are elementary and were re-verified by direct differentiation. No empirical numbers enter; all relations are exact mathematical identities a reader can reproduce with a page of algebra.

Builds on: Smooth Manifolds and Coordinate Charts · Atlases, Transition Functions and Differentiable Structure · Tangent Vectors and the Tangent Space · Cotangent Space and One-Forms. Leads to: General Tensors and the Tensor Product · The Metric Tensor and the Line Element · Tensor Densities and Weights · The Invariant Volume Element.

Ask the tutor: Want to see the law in action — pick any coordinate change (polar, rotating, Schwarzschild-to-Kruskal) and I will transform a vector, the metric, or a Christoffel symbol with you, checking every Jacobian by hand.

Is a tensor really "the same in every coordinate system"?

The geometric object is the same; its components are not. A vector at a point is a single arrow in the tangent space, fixed once and for all. What changes when you switch charts is the list of numbers you use to name that arrow — and the transformation law is precisely the rule that changes the numbers so the arrow stays put. The everyday shorthand "tensors are coordinate-independent" means the object is invariant, and that any equation written purely between tensors holds in all charts at once. It never means the components are the same; in general they differ from chart to chart, as Example 2 shows the components going from (1, 0) to (0.707, −0.500) for one unchanged arrow.

Why do upper and lower indices transform with opposite matrices?

Because they must, if contractions are to be invariant. An upper index sits on a vector-like slot that follows the Jacobian $\partial x'/\partial x$ ; a lower index sits on a one-form-like slot that follows the inverse Jacobian $\partial x/\partial x'$ . When you contract one against the other, the two factors meet on a shared summed index and collapse to the Kronecker delta, leaving a coordinate-free number. If both indices transformed the same way, the factors would not cancel and the "scalar" would depend on your chart — a contradiction. So the opposite rules are not a convention but a requirement forced by demanding that pairing a vector with a one-form yield a genuine invariant.

How can I quickly tell whether an array is a tensor?

Apply the law and see if it holds with no leftover terms. Write the candidate in two charts and check that new components equal old components times the correct product of Jacobians and inverse Jacobians — one factor per index, nothing else. If an inhomogeneous piece appears — typically something with a second derivative of the coordinates, like $\partial^2 x/\partial x'\partial x'$ — the array is not a tensor. The Christoffel symbols and the plain partial derivative of a vector both fail exactly this way. A fast sanity check: a real tensor that is zero in one chart is zero in all charts, so if you can make an object vanish by a clever coordinate choice and it reappears elsewhere (as $\Gamma^{r}{}_{\theta\theta} = -r$ does), it cannot be a tensor.

If the Christoffel symbols are not tensors, why keep them?

Because their non-tensorial part is exactly what fixes the non-tensorial part of the partial derivative. The plain derivative $\partial_{\mu}V^{\nu}$ is not a tensor; it acquires an anomalous second-derivative term under a coordinate change. The connection $\Gamma$ acquires the same anomalous term with the opposite sign, so the combination $\nabla_{\mu}V^{\nu} = \partial_{\mu}V^{\nu} + \Gamma^{\nu}{}_{\mu\lambda}V^{\lambda}$ is a clean $(1,1)$ -tensor. In other words, $\Gamma$ is deliberately not a tensor: it is the correction that repairs differentiation on a curved manifold. And although $\Gamma$ itself is not tensorial, the difference of any two connections is, because the shared inhomogeneous term cancels.

Where does the Jacobian determinant enter, and what is a tensor density?

It enters when you integrate. Coordinate volume $\mathrm{d}^n x$ is not invariant — under a change of chart it picks up the Jacobian determinant, so a naive integral would depend on your coordinates. A tensor density of weight $w$ is an object whose law carries an extra factor $\lvert\partial x/\partial x'\rvert^{w}$ beyond the usual Jacobian factors. The star example is $\sqrt{-g}$ , a scalar density of weight 1: the metric determinant $g$ transforms with two powers of the Jacobian determinant, so its square root carries one. Multiplying $\sqrt{-g}$ by $\mathrm{d}^n x$ cancels the offending determinant and produces the invariant volume element $\sqrt{-g}\,\mathrm{d}^n x$ , which is what lets you integrate a scalar over curved spacetime unambiguously.

Do coordinate differentials dxᵘ and gradients ∂_μφ really transform oppositely?

Yes, and they are the cleanest prototypes of the two rules. A coordinate differential transforms as $\mathrm{d}x'^{\mu} = (\partial x'^{\mu}/\partial x^{\nu})\,\mathrm{d}x^{\nu}$ — one Jacobian factor, the contravariant law, which is why $\mathrm{d}x^{\mu}$ carries an upper index. A gradient transforms as $\partial'_{\mu}\phi = (\partial x^{\nu}/\partial x'^{\mu})\,\partial_{\nu}\phi$ — one inverse-Jacobian factor, the covariant law, which is why $\partial_{\mu}$ carries a lower index. Their contraction, the directional derivative $\mathrm{d}x^{\mu}\,\partial_{\mu}\phi = \mathrm{d}\phi$ , is a scalar, exactly as the invariance of contractions requires. These two everyday objects are the reason the whole up/down index notation was invented.