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research28 min studyLesson 861 in Physics

Decompose metric disturbances before solving their dynamics

Fix time, metric, Fourier, and transverse-mode conventions on a flat background.

Before you begin
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The question for today

What you’ll understand

  • Decompose spatial vectors and symmetric tensors at nonzero Fourier momentum.
  • Count scalar, transverse-vector, and transverse-traceless metric components.
  • Distinguish a kinematic decomposition from gauge fixing and physical degrees of freedom.

Cosmological perturbation theory compares an inhomogeneous spacetime with a homogeneous expanding background. Before applying field equations, classify disturbances by their transformation under spatial rotations: scalars, transverse vectors, and transverse-traceless tensors. These labels concern the background spatial geometry, not the number of indices on the original four-dimensional object. The classification makes linear equations manageable, but it does not itself remove coordinate freedom or impose Einstein constraints. A careful convention ledger prevents a decomposition formula from being mistaken for a statement about independently propagating particles.

Declare the background and the scalar metric

Use signature (−,+,+,+), natural units c=1, a dimensionless scale factor a, comoving spatial coordinates with length units, and conformal time η with dt=a dη. A prime means d/dη; H is the cosmic-time expansion rate and the conformal rate is aH. For the scalar sector write lapse perturbation A, shift potential B, spatial-curvature potential ψ, and spatial-shear potential E. The spatial metric contains one minus twice ψ, fixing a sign that later curvature variables must respect.

The unperturbed geometry is spatially flat. Fourier-expand a field with exp(+ik·x) and measure d³k/(2π)³, so a spatial derivative becomes +ik and the Laplacian becomes −k². Comoving k has inverse-length units, while its physical magnitude is k/a. The scale factor therefore converts physical and comoving wavelengths; a numerical Fourier wave number without this declaration cannot determine whether a mode is inside an expansion scale.

ds2=a2 ⁣[−(1+2A)dη2+2∂iB dη dxi+((1−2ψ)δij+2∂i∂jE)dxidxj],H=a′a=aHds^2=a^2\!\left[-(1+2A)d\eta^2+2\partial_iB\,d\eta\,dx^i+\big((1-2\psi)\delta_{ij}+2\partial_i\partial_jE\big)dx^idx^j\right],\qquad \mathcal H=\frac{a'}a=aH
This displayed metric retains scalar perturbations to first order only. A and ψ are dimensionless, B has length units, and E length-squared units. Vector and tensor terms are added separately.↔ Scroll wide equations horizontally.

Project parallel and perpendicular components

For any nonzero spatial wave vector, separate a vector into its projection along k and a transverse remainder. The longitudinal part is constructed from one scalar amplitude even though it carries a vector index. A symmetric spatial tensor similarly separates into scalar trace and longitudinal pieces, two transverse-vector amplitudes, and two transverse-traceless amplitudes. The decomposition uses inverse spatial derivatives, so boundary conditions and the homogeneous k=0 sector require separate treatment.

In the full perturbed metric there are four scalar functions, two divergence-free spatial vectors with two independent components each, and a transverse-traceless tensor with two components. These ten functions match a symmetric four-dimensional metric’s ten entries. They are not ten gravitational wave polarizations. Gauge transformations and constraint equations still remove or relate variables. At first order on an isotropic background, different rotation sectors evolve separately; products at higher order can source other sectors.

Vi∥=kikjk2Vj,Vi⊥=(δij−kikjk2)Vj,kihij=0,hii=0V_i^{\parallel}=\frac{k_i k_j}{k^2}V_j,\qquad V_i^{\perp}=\left(\delta_{ij}-\frac{k_i k_j}{k^2}\right)V_j,\qquad k_ih_{ij}=0,\quad h_{ii}=0
The projector assumes k≠0 and the Euclidean background spatial metric. Transverse-traceless h is a spatial metric perturbation; its dynamical interpretation also requires the gravitational equations.↔ Scroll wide equations horizontally.

A mode classification is not an evolution solver

For a wave vector along z, transversality forces the z row and column of a tensor mode to vanish. Tracelessness then leaves a plus-like diagonal amplitude and a cross-like off-diagonal amplitude in the xy plane. Rotating the transverse basis mixes these two descriptions without adding another polarization. Matter scalar perturbations can source scalar metric potentials, while anisotropic stress and vector momentum have their own sector-specific roles. Initial conditions should be assigned consistently with those sources and the constraints.

The vector lab isolates ordinary projections. Reset and set angle to 0° and magnitude to 2; predict components (2,0). Reset and set angle to 90° and magnitude to 2; predict (0,2). Components change while magnitude stays 2. This two-dimensional vector has no expanding metric, Fourier projector, gauge constraints, or gravitational polarization dynamics. It provides a bookkeeping analogy for resolving components, with time playback unrelated to cosmological evolution.

f(η,x)=∫d3k(2π)3fk(η)eik⋅x,∇2fk=−k2fkf(\eta,\mathbf x)=\int\frac{d^3k}{(2\pi)^3}f_{\mathbf k}(\eta)e^{i\mathbf k\cdot\mathbf x},\qquad \nabla^2f_{\mathbf k}=-k^2f_{\mathbf k}
For a real field, f at −k is the complex conjugate of f at k. The integral is an infinite-volume convention; a finite periodic box instead has discrete modes and volume factors.↔ Scroll wide equations horizontally.

Work it through

Project one mode and check a tensor

Take k=(0,0,2) in inverse-length units and a Fourier-vector amplitude V=(3,4,5). Find its longitudinal and transverse parts. Then check whether h=[[0.002,0.003,0],[0.003,−0.002,0],[0,0,0]] is transverse-traceless.

  1. The direction of k is z. Its longitudinal projector keeps only the z component, giving V∥=(0,0,5), while the transverse projector gives V⊥=(3,4,0). Their sum reconstructs V.

  2. The transverse magnitude is √(3²+4²)=5 and its dot product with k vanishes. The longitudinal magnitude is also 5; neither fact turns one sector into the other.

  3. The tensor has zero z row and column, so contraction with k vanishes, and its trace is 0.002−0.002=0. It has two independent transverse entries and is a valid small TT amplitude, without yet specifying its time evolution.

The result

V splits into (0,0,5) and (3,4,0); the supplied tensor satisfies both transverse and trace-free conditions.

The living notebook

Explore the model

Interactive lab

Interactive lab · vectors

Resolve a fixed magnitude

Reset and set angle=0°,magnitude=2; predict components (2,0). Reset and set angle=90°,magnitude=2; predict (0,2).

Resolve a fixed magnitudePlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. 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 forThis is one Euclidean 2D vector, not a metric decomposition, gauge fixing, or gravitational-wave calculation.

Pause & reason

Check your understanding

What does the scalar–vector–tensor decomposition alone accomplish?

Your turn

Try it yourself

At nonzero k along z, how many independent components remain in a symmetric transverse-traceless 3×3 tensor? Exhibit its general matrix form and explain why the homogeneous k=0 mode needs separate care.

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

Give me a hint

First eliminate the z row and column, then use the trace condition.

Compare with the solution

The general form is [[u,v,0],[v,−u,0],[0,0,0]], with two independent amplitudes u and v. At k=0 the longitudinal projector divides by zero and the mode has no preferred wave-vector direction; homogeneous background changes and boundary conventions must be handled separately rather than by substituting zero into the projector.

Keep the idea

Take it with you

  • State time, metric, and Fourier conventions before comparing perturbation formulas.
  • SVT classification precedes gauge fixing and physical mode counting.
  • Linear rotational decoupling does not extend automatically to nonlinear sources.

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