Manuscript XI-A@1 · XI-A.tex · theorem · b8211ecc

Theorem: Metric Euler coefficients

Metric Euler derivative through coupling order two › Exact metric derivative and its expansion · lines 668–709 · proof included or linked

XI-A@1#thm:euler, XI-A@1#eq:eexact, XI-A@1#eq:eexp, XI-A@1#eq:egf

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With $e^{μν} = δ S_{inv} / δ γ_{μν}$ , the exact formal derivative and its order-two expansion are

e^{μν} = \frac{2}{κ} - q G^{μν} (q) - \frac{κ}{2} g - q T_{q}^{μν}, = 2 H_{1}^{μν} + κ (2 H_{2}^{μν} - \frac{1}{2} g T_{0}^{μν}) + κ^{2} (2 H_{3}^{μν} - \frac{1}{2} g J_{1}^{μν}) + O (κ^{3}) .

The gauge-fixing contribution is separately

e_{gf}^{μν} = - 2 \partial^{(μ} D^{ν)} + η^{μν} \partial_{ρ} D^{ρ}, \partial_{μ} e_{gf}^{μν} = - □ D^{ν} .

Thus $δ S_{tot} / δ γ = e + e_{gf}$ . There are no $\partial g$ terms in (eq:eexact) or (eq:eexp).

Proof. Combine the Einstein variation in Proposition prop:FP, Lemma lem:vertex, and $δ q = κ δ γ$ . The flat scalar density is independent of $γ$ . Equations (eq:Hdef) and (eq:J1) give the expansion. Varying $D_{ν} D^{ν}$ and integrating by parts gives (eq:egf); taking its divergence gives the last identity. No integration by parts in a metric variation of the scalar density is needed because that density has no metric derivatives.

Displayed formulas in this passage

XI-A@1#eq:eexact · line 673
$e^{μν} = \frac{2}{κ} - q G^{μν} (q) - \frac{κ}{2} g - q T_{q}^{μν}, = 2 H_{1}^{μν} + κ (2 H_{2}^{μν} - \frac{1}{2} g T_{0}^{μν}) + κ^{2} (2 H_{3}^{μν} - \frac{1}{2} g J_{1}^{μν}) + O (κ^{3}) .$
XI-A@1#eq:eexp · line 673
$e^{μν} = \frac{2}{κ} - q G^{μν} (q) - \frac{κ}{2} g - q T_{q}^{μν}, = 2 H_{1}^{μν} + κ (2 H_{2}^{μν} - \frac{1}{2} g T_{0}^{μν}) + κ^{2} (2 H_{3}^{μν} - \frac{1}{2} g J_{1}^{μν}) + O (κ^{3}) .$
XI-A@1#eq:egf · line 687
$e_{gf}^{μν} = - 2 \partial^{(μ} D^{ν)} + η^{μν} \partial_{ρ} D^{ρ}, \partial_{μ} e_{gf}^{μν} = - □ D^{ν} .$

Theorem: Metric Euler coefficients

Displayed formulas in this passage

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