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Zero boundary metric value does not kill EH derivative flux

Example

Work in boundaryless ambient M=R4 and integrate the local divergence over the half-space subdomain H={x3≥0}. Give M the flat metric η=diag⁡(−1,1,1,1), coordinates (x0,x1,x2,x3), and outward conormal −dx3 at its boundary. All coordinates have length units and choose a length ℓ>0. Take dimensionless cutoffs χ∈Cc∞(R3) and ρ∈Cc∞(R) with χ≥0, χ=1 near 0∈R3, and ρ=1 near 0∈R. Define

f(x):=x3ℓχ(x0,x1,x2)ρ(x3),hab:=fηab.

Then hab=0 on ∂H, but its normal derivative is nonzero there. For the Einstein–Hilbert summand with coefficient c3/(16πGN), the local variation formula in Compact Einstein–Hilbert variation and SI coupling has a nonzero boundary flux for this variation. The diagram records only the chosen half-space and normal; the proof is the component calculation below.

x3=0+@3nout=¡@3H:x3¸0hj@H=0;@3hj@H6=0

Facts & Assumptions

Given: The half-space chart and flat metric above, compact cutoffs χ,ρ, and the metric variation gs=η+sh for sufficiently small s.

[F1]

The first variation of the Einstein–Hilbert density is δ(wR)=wGabqab+∂a(wVa), where Va=gbdδΓadb−gabδΓddb and δΓabc=12gad(∇bhcd+∇chbd−∇dhbc). (Compact Einstein–Hilbert variation and SI coupling)

[F2]

Curvature and the Einstein tensor are defined from the metric-selected connection and its curvature; for the constant metric η, its Christoffel symbols and curvature vanish. (The Riemann curvature operator and the derived Ricci, scalar and Einstein tensors)

[F3]

Nonnegative smooth cutoffs with compact support and a plateau equal to one near a point exist in Euclidean coordinate balls. (A smooth bump between concentric Euclidean balls)

[A1]

The cutoffs are smooth, compactly supported, nonnegative where stated, and equal to 1 on neighborhoods of the origin and x3=0 respectively. The first variation is interpreted as a compactly supported action difference on the half-space.

[A2]

The geometric supplier chain carries the ACω convention stated in The Axiom of Countable Choice (ACω); the local cutoff calculation makes no further choice.

Verification

1.1A1F3algebra

By [F3], choose nonnegative compactly supported dimensionless χ equal to one near the origin and dimensionless ρ equal to one near zero. The factor x3/ℓ gives hab∣x3=0=0. At the boundary, ∂3f=χ/ℓ and ∂αf=0 for α=0,1,2; in particular, at the origin ∂3f=1/ℓ. Since f is bounded and nonzero, gs=(1+sf)η has the same Lorentzian signature whenever ∣s∣<1/(2∥f∥∞). Thus the metric value is fixed at the boundary while its normal derivative is not.

2.1A2F1step 1.1algebra

At the flat background, substituting hab=fηab into [F1] gives δΓabc=12(δca∂bf+δba∂cf−ηbc∂af). Contracting the two terms in Va gives ηbdδΓadb=−∂af and ηabδΓddb=2∂af, hence Va=−3∂af. On x3=0, V3=−3χ/ℓ.

3.1A1F1F2step 2.1algebra

Since Gab=0, the linearized Einstein–Hilbert density is just ∂aVa by [F1]. The tangential integrals vanish because χ is compactly supported, and the upper normal endpoint contributes zero because ρ is compactly supported. The fundamental theorem of calculus on x3∈[0,∞) therefore gives, after multiplying by the action coefficient, the boundary contribution −c316πGN∫x3=0V3 dx0dx1dx2=3c316πGNℓ∫R3χ dx0dx1dx2>0. This is the outward flux for conormal −dx3.

4.1step 1.1step 3.1algebra∎

The nonzero first-variation boundary term in step 3.1 occurs despite hab=0 on ∂H. Therefore fixing only boundary metric values does not justify discarding the Einstein–Hilbert derivative flux.

Remarks

Depends on

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