advanced28 min studyLesson 802 in Physics

A tensor stays fixed while its component arrays change

Transform tangent and cotangent components and verify invariant contractions under a shear.

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What you’ll understand

  • Derive vector and covector component transformations from dual bases.
  • Transform a metric with two inverse Jacobians.
  • Check invariant contractions without confusing coordinate lengths with components.

Tensor notation is useful because it distinguishes a geometric object from the array used to represent it. When coordinates change, basis vectors change and the components compensate. Covectors compensate differently from vectors, so copying the same matrix onto every index produces wrong answers. We will derive the two transformation laws and check a shear example in which all the component lists change while a contraction and a vector norm remain invariant.

Tangent vectors and covectors use dual bases

At a point, a tangent vector can be viewed as a directional derivative. The coordinate basis is ∂μ, and its dual one-form basis is dxμ. The chain rule changes the derivative basis with an inverse Jacobian and the differential basis with the forward Jacobian. Requiring the full vector and one-form to remain the same then gives forward-Jacobian vector components and inverse-Jacobian covector components. These transformations concern components at the same event, not transporting a vector from one event to another.

A scalar differential df is a useful covector example. Its components are partial derivatives of f and act on a vector to return a directional derivative. Under a smooth coordinate transformation, the contraction is unchanged because the Jacobians cancel. A mixed tensor combines one factor for each index. Index placement therefore encodes which basis compensation is required. Contraction pairs one upper and one lower index; summing two upper component arrays as though both were Cartesian does not define an invariant without an additional metric.

Vμ′=∂xμ′∂xνVν,ωμ′=∂xν∂xμ′ων,ωμ′Vμ′=ωμVμV^{\mu\prime}=\frac{\partial x^{\mu\prime}}{\partial x^\nu}V^\nu,\qquad \omega_{\mu\prime}=\frac{\partial x^\nu}{\partial x^{\mu\prime}}\omega_\nu,\qquad \omega_{\mu\prime}V^{\mu\prime}=\omega_\mu V^\mu
The Jacobians are evaluated on the chart overlap at one event. Units follow the chosen coordinate units and the vector parameter; for dimensionless f and length coordinates, df components have units m⁻¹. The contraction is a scalar with its own physical units.↔ Scroll wide equations horizontally.

The metric is a tensor rather than a permanent identity matrix

The metric is a bilinear map on tangent vectors, so its two lower indices each acquire an inverse Jacobian. A coordinate shear of an ordinary Euclidean plane creates off-diagonal metric components even though the plane remains flat. The squared norm is computed with this transformed metric, not by summing the new coordinate components squared. Raising an index also requires the inverse metric. In Lorentzian spacetime the same bookkeeping applies, with the chosen signature supplying the time-direction sign.

The metric can change its component appearance drastically under a coordinate change, but its nondegenerate signature is preserved by an invertible transformation. Determinants transform with the square of the inverse Jacobian; a coordinate volume element transforms with the opposite Jacobian factor. These facts later combine into an invariant volume density. A tensor equation valid in one chart remains valid in any overlapping valid chart, whereas a statement that some connection coefficients vanish need not survive: connections have an additional inhomogeneous transformation term.

gμ′ν′=∂xρ∂xμ′∂xσ∂xν′gρσ,gμ′ν′Vμ′Vν′=gρσVρVσg_{\mu\prime\nu\prime}=\frac{\partial x^\rho}{\partial x^{\mu\prime}}\frac{\partial x^\sigma}{\partial x^{\nu\prime}}g_{\rho\sigma},\qquad g_{\mu\prime\nu\prime}V^{\mu\prime}V^{\nu\prime}=g_{\rho\sigma}V^\rho V^\sigma
For length coordinates the metric components can be dimensionless; angular coordinates instead give some metric components length factors. The transformation law does not assume a Euclidean signature, although the worked shear uses a positive-metric spatial plane.↔ Scroll wide equations horizontally.

Coordinate invariance is an algebraic consistency check

It is helpful to distinguish a passive coordinate change from an active diffeomorphism moving fields or points. Both can be written using related Jacobians, but their interpretation and evaluation points must be tracked. For our component examples the event and geometric tensor are held fixed while its coordinates change. A failed contraction check usually indicates a wrong inverse, a transposition error, or components evaluated at different points. It is not evidence that the scalar itself depends on the observer’s chosen coordinate notation.

Use the matrix lab to inspect the forward shear J=[[1,b],[0,1]]. Reset and set both controls to a=1,b=0, predicting determinant one. Reset and set a=1,b=1, again predicting determinant one but unequal singular values about 1.618 and 0.618. Current geometry is identity at Reset and interpolates under Play or Step; readouts always describe the selected final matrix. The lab transforms Euclidean geometry, not tensor indices automatically. The reciprocal covector and metric transformations must be calculated manually, and area preservation alone does not imply preservation of every ordinary component norm.

Transform a vector and its dual pairing

On a spatial plane with length coordinates, set x′=x+y and y′=y. A tangent whose parameter has length units has components V=(2,1), and a covector has ω=(3,4) m⁻¹. Find the new components and check ω(V).

  1. The forward Jacobian is J=[[1,1],[0,1]], so V′=JV=(3,1). The vector remains the same tangent despite the changed array.

  2. The covector column transforms with J⁻ᵀ=[[1,0],[−1,1]], giving ω′=(3,1) m⁻¹. Using J on both arrays would be incorrect.

  3. The old contraction is 3×2+4×1=10 m⁻¹. The new one is 3×3+1×1=10 m⁻¹, confirming the inverse compensation.

The result

The new arrays are V′=(3,1) and ω′=(3,1) m⁻¹, while the scalar pairing remains 10 m⁻¹.

Explore the model

Interactive lab

Interactive lab · matrix

Inspect a coordinate shear and its unequal stretches

Reset and set BOTH controls to a=1,b=0, predicting det=1. Reset and set a=1,b=1, predicting det=1 and singular values 1.618,0.618. Geometry starts at identity and interpolates under Play/Step; readouts refer to final J. Covectors and metrics need manual reciprocal transformations.

Inspect a coordinate shear and its unequal stretchesThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA shear preserves area but changes component norms unless the metric is transformed with the coordinates.

Check your understanding

How does a covector column transform under a forward vector-component Jacobian J?

Try it yourself

For the same shear, the original Euclidean metric is I. Find the primed metric and the norm squared of V′=(1,1). Recover the corresponding original vector as a check.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use g′=J⁻ᵀJ⁻¹ and include both off-diagonal terms in the quadratic form.

Compare with the solution

J⁻¹=[[1,−1],[0,1]], giving g′=[[1,−1],[−1,2]]. The norm squared is 1−2+2=1. The original vector is J⁻¹(1,1)=(0,1), whose Euclidean norm squared is also one. Summing the primed components squared would give two and would ignore the transformed metric.

Take it with you

  • Vector and covector components compensate their bases with reciprocal Jacobians.
  • Contractions remain invariant when every index is transformed correctly.
  • A nonorthogonal coordinate change also changes the metric component matrix.

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