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research28 min studyLesson 862 in Physics

Change the slicing without changing the invariant potentials

Track passive time shifts, spatial threading, and the conditions for equal potentials.

Before you begin
On this page

The question for today

What you’ll understand

  • Apply a declared first-order scalar coordinate transformation.
  • Verify Bardeen-potential invariance explicitly.
  • Distinguish longitudinal gauge, residual coordinate freedom, and anisotropic stress.

A perturbation compares quantities at events assigned the same background coordinates. Changing that assignment changes the perturbation even when the physical spacetime stays fixed. This gauge freedom is especially important for time slicing: a background density evolving in time produces an apparent density disturbance when neighboring clocks are relabeled. Gauge-invariant combinations cancel that coordinate dependence at the stated order. They do not remove real inhomogeneity, and their existence does not make every raw density contrast or metric coefficient directly observable.

Specify the direction of the coordinate shift

Keep the preceding scalar metric and use the passive coordinate transformation ηtilde=η+T and xitilde=xi+∂iL. Both generators are first-order small. Transforming the metric components gives a lapse shift involving both the conformal expansion and the time derivative of T, a curvature shift proportional to T, and threading changes in B and E. A homogeneous background scalar transforms differently from a spatial vector: its perturbation acquires minus its background time derivative times T.

The signs reverse if an author defines the gauge generator in the opposite direction. Comparing a formula for a time shift without comparing that definition can create a false disagreement. T has length units and L length-squared units here, so every transformation has the units of its target variable. These relations are first order: at second order, products and evaluation shifts enter, and applying only the linear rules would leave spurious residual terms.

A~=A−HT−T′,ψ~=ψ+HT,B~=B+T−L′,E~=E−L,δρ~=δρ−ρˉ′T\widetilde A=A-\mathcal HT-T',\qquad\widetilde\psi=\psi+\mathcal HT,\qquad\widetilde B=B+T-L',\qquad\widetilde E=E-L,\qquad\widetilde{\delta\rho}=\delta\rho-\bar\rho'T
The metric has scalar shift +2∂iB dηdxi. The generator convention is passive tilde-coordinate=old-coordinate+generator; sources using the opposite generator require a sign translation.↔ Scroll wide equations horizontally.

Cancel the generators algebraically

Define s=B−E′. Under the stated transformation, s gains T because the spatial threading contributions cancel. Combining s with the lapse and curvature perturbations cancels the remaining time-shift terms, giving the Bardeen potentials Φ and Ψ. Each combination is invariant under the declared first-order scalar coordinate changes. Their construction uses background expansion but does not require a matter equation of state or Einstein’s equations.

Choosing B=E=0 gives longitudinal, or conformal Newtonian, gauge for the nonzero modes where that choice is well defined. The two potentials then coincide with the corresponding metric coefficients. This is a convenient representation of invariant quantities, not an argument that the other gauges describe different physics. Synchronous gauge instead fixes lapse and shift conditions while permitting residual freedoms unless extra choices are supplied; numerical growing modes must be distinguished from those residual gauge modes.

s=B−E′,s~=s+T,Φ=A+Hs+s′,Ψ=ψ−Hss=B-E',\qquad\widetilde s=s+T,\qquad \Phi=A+\mathcal Hs+s',\qquad\Psi=\psi-\mathcal Hs
Φ is the invariant lapse potential and Ψ the invariant spatial-curvature potential in this notation. Some literature exchanges their names; the displayed metric and definitions identify them unambiguously.↔ Scroll wide equations horizontally.

Use dynamics only after constructing invariants

In general relativity the traceless spatial Einstein equation relates the difference between the two scalar potentials to anisotropic stress, with coefficients depending on the stress convention. A perfect fluid with negligible anisotropic stress permits Φ=Ψ in the scalar sector. Free-streaming radiation or other stress sources can invalidate that simplification. Gauge invariance alone does not force equality, and a difference between potential names in two papers is not evidence for modified gravity.

The spacetime lab offers a separate invariant comparison. Reset and set velocity β=0 and event time ct=1; predict transformed event (x′,ct′)=(0.6,1) and lab invariant I=(ct′)²−x′²=+0.64. Reset and set β=0.6 and ct=1; predict (0,0.8) and the same I. This timelike-positive lab invariant has the opposite sign to metric ds² in this module’s (−,+,+,+) signature. The finite Lorentz transformation illustrates component changes with an unchanged invariant; it does not calculate cosmological gauge shifts or Bardeen potentials.

dslong2=a2 ⁣[−(1+2Φ)dη2+(1−2Ψ)δijdxidxj],Φ=Ψ if scalar anisotropic stress vanishes in GRds^2_{\rm long}=a^2\!\left[-(1+2\Phi)d\eta^2+(1-2\Psi)\delta_{ij}dx^idx^j\right],\qquad \Phi=\Psi\ \text{if scalar anisotropic stress vanishes in GR}
The equality is a dynamical statement under the specified matter and gravity assumptions. The longitudinal metric shown omits vector and tensor perturbations.↔ Scroll wide equations horizontally.

Work it through

Verify two invariant numerical combinations

At one time choose conformal units with ℋ=2, s=0.003,s′=0.004,A=0.010,ψ=0.020. Apply T=0.001 and T′=0.002. Compute Φ and Ψ before and after the shift.

  1. Initially Φ=0.010+2×0.003+0.004=0.020 and Ψ=0.020−2×0.003=0.014. The metric coefficients alone do not equal these invariant values in a general gauge.

  2. The transformed values are Atilde=0.010−0.002−0.002=0.006, ψtilde=0.022, stilde=0.004, and stilde′=0.006. No background quantity changes at the order used in these combinations.

  3. Recombining gives Φtilde=0.006+2×0.004+0.006=0.020 and Ψtilde=0.022−2×0.004=0.014. The nonzero potential difference is unchanged and would require its own dynamical stress interpretation.

The result

The raw coefficients shift, while Φ=0.020 and Ψ=0.014 remain unchanged.

The living notebook

Explore the model

Interactive lab

Interactive lab · spacetime

Compare components and one invariant

Reset and set β=0,ct=1; predict (x′,ct′)=(0.6,1) and I=(ct′)²−x′²=+0.64. Reset and set β=0.6,ct=1; predict (0,0.8) and the same I=+0.64. This timelike-positive lab invariant has the opposite sign to metric ds² in the module’s (−,+,+,+) signature.

Compare components and one invariantA fixed x = 0.6 event is Lorentz transformed with c = 1. Playback highlights the relation between coordinate descriptions; it does not move the event. Current readings: Lorentz factor γ: 1.12; Event x′: 0.168; Event ct′: 0.817. All parameters are adjustable with the labeled controls below.−4.69−2.3502.354.69−2.2−1.101.12.2x or x′ · c = 1ct or ct′EE′ct² − x² = 0.64
Original eventTransformed eventLight cone
Lorentz factor γ
1.12
Event x′
0.168
Event ct′
0.817
t = 0 s

A fixed x = 0.6 event is Lorentz transformed with c = 1. Playback highlights the relation between coordinate descriptions; it does not move the event. 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 flat Lorentz transformation illustrates invariant bookkeeping only; it does not perform a cosmological gauge shift or calculate metric potentials.

Pause & reason

Check your understanding

Why may Φ and Ψ differ even though both are gauge invariant?

Your turn

Try it yourself

At one event let ℋ=3,ψ=0.020 and δρ/ρ′=0.004. Evaluate ζ=−ψ−ℋδρ/ρ′. Then use T=0.002 to find the shifted ψ and δρ/ρ′ and verify ζ again.

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 ψtilde=ψ+ℋT and (δρ/ρ′)tilde=δρ/ρ′−T.

Compare with the solution

Initially ζ=−0.020−3×0.004=−0.032. The shifted curvature coefficient is 0.026 and the density-to-background-derivative ratio is 0.002, giving ζtilde=−0.026−3×0.002=−0.032. This cancellation requires ρ′ nonzero; uniform-density slicing becomes ill defined when the chosen background density cannot label time.

Keep the idea

Take it with you

  • Declare whether a gauge generator maps old coordinates to new ones or conversely.
  • Bardeen combinations are invariant before imposing field equations.
  • Equal scalar potentials require a matter and gravity assumption beyond invariance.

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