Manuscript XI-A-ST@1 · XI-A-ST.tex · section · c3f0929d

The quadratic normalization

The signed action and its Euler derivatives › The quadratic normalization · lines 586–691

XI-A-ST@1#rest:quadratic, XI-A-ST@1#eq:FP, XI-A-ST@1#eq:Glin, XI-A-ST@1#eq:egf, XI-A-ST@1#eq:quadraticoperator, XI-A-ST@1#rest:signs, XI-A-ST@1#eq:TT, XI-A-ST@1#eq:Newtonmetric, XI-A-ST@1#eq:Newton

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The bulk quadratic coefficient of (eq:EH), modulo a total divergence, is

\begin{split}
 S_{\mathrm{FP}}[\gamma]=\frac12\int_{M_T}\Big[
 &\partial_\lambda\gamma_{\mu\nu}
                       \partial^\lambda\gamma^{\mu\nu}
 -2(\partial_\mu\gamma^{\mu\nu})
                       (\partial^\rho\gamma_{\rho\nu})\\
 &+2(\partial_\mu\gamma^{\mu\nu})\partial_\nu\tau
 -\partial_\lambda\tau\partial^\lambda\tau
 \Big]\mathrm d^4x.
 \label{eq:FP}
\end{split}
Its Euler derivative is \(2G_{\mathrm{lin}}^{\mu\nu}(\gamma)\), where
\begin{equation}
 G_{\mathrm{lin}}^{\mu\nu}(\gamma)
 =
 -\frac12\Box\bar\gamma^{\mu\nu}
 +\partial^{(\mu}D^{\nu)}
 -\frac12\eta^{\mu\nu}\partial_\rho D^\rho.
 \label{eq:Glin}

Parentheses include the factor $1/2$ .

The separate gauge-fixing derivative is

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

Consequently the quadratic gauge-fixed operator is

2 G_{lin}^{μν} (γ) + e_{gf}^{μν} (γ) = - □ γ ˉ^{μν} .

Proof. These are XI-A :FP, :Glin, and :egf CheungXIA. The relevant action variation is

δ S_{EH} = \frac{2}{κ ^{2}} \int ρ_{q} G^{μν} δ q_{μν} d^{4} x .

Since $δ q = κ δ γ$ , its leading Euler coefficient is $2 G_{lin}$ . Compact variation of (eq:gf) gives (eq:egf); addition gives (eq:quadraticoperator). Lemma lem:transfer transports these variational pairings without changing the action sign.

In particular, this handoff is $- □ P_{tr}$ , not X-E's differently normalized metric operator. A shared de Donder name does not identify them.

two retained sign diagnostics

For

γ_{0 μ} = 0, \partial_{i} γ_{ij} = 0, δ^{ij} γ_{ij} = 0,

the Fierz–Pauli density is

L_{FP}^{TT} = \frac{1}{2} γ ˙_{ij} γ ˙_{ij} - \frac{1}{2} \partial_{k} γ_{ij} \partial_{k} γ_{ij} .

For the auxiliary static unswitched comparison with $T^{00} = ρ \geq 0$ , $T^{0 i} = T^{ij} = 0$ , where $ρ$ is smooth, compactly supported, and nonzero, put $H = q - η$ . The linear equation has the comparison

H_{00} = 2 Φ_{N}, H_{ij} = 2 Φ_{N} δ_{ij}, H_{0 i} = 0,

with

Δ Φ_{N} = \frac{κ ^{2}}{8} ρ, Φ_{N} (x) = - \frac{κ ^{2}}{32 π} \int_{R^{3}} \frac{ρ ( y )}{∣ x - y ∣} d^{3} y < 0.

Thus $κ^{2} = 32 π G_{N}$ has the attractive sign.

Displayed formulas in this passage

XI-A-ST@1#eq:FP · line 592

\begin{split}
 S_{\mathrm{FP}}[\gamma]=\frac12\int_{M_T}\Big[
 &\partial_\lambda\gamma_{\mu\nu}
                       \partial^\lambda\gamma^{\mu\nu}
 -2(\partial_\mu\gamma^{\mu\nu})
                       (\partial^\rho\gamma_{\rho\nu})\\
 &+2(\partial_\mu\gamma^{\mu\nu})\partial_\nu\tau
 -\partial_\lambda\tau\partial^\lambda\tau
 \Big]\mathrm d^4x.
 \label{eq:FP}
\end{split}
Its Euler derivative is \(2G_{\mathrm{lin}}^{\mu\nu}(\gamma)\), where
\begin{equation}
 G_{\mathrm{lin}}^{\mu\nu}(\gamma)
 =
 -\frac12\Box\bar\gamma^{\mu\nu}
 +\partial^{(\mu}D^{\nu)}
 -\frac12\eta^{\mu\nu}\partial_\rho D^\rho.
 \label{eq:Glin}

XI-A-ST@1#eq:Glin · line 592

\begin{split}
 S_{\mathrm{FP}}[\gamma]=\frac12\int_{M_T}\Big[
 &\partial_\lambda\gamma_{\mu\nu}
                       \partial^\lambda\gamma^{\mu\nu}
 -2(\partial_\mu\gamma^{\mu\nu})
                       (\partial^\rho\gamma_{\rho\nu})\\
 &+2(\partial_\mu\gamma^{\mu\nu})\partial_\nu\tau
 -\partial_\lambda\tau\partial^\lambda\tau
 \Big]\mathrm d^4x.
 \label{eq:FP}
\end{split}
Its Euler derivative is \(2G_{\mathrm{lin}}^{\mu\nu}(\gamma)\), where
\begin{equation}
 G_{\mathrm{lin}}^{\mu\nu}(\gamma)
 =
 -\frac12\Box\bar\gamma^{\mu\nu}
 +\partial^{(\mu}D^{\nu)}
 -\frac12\eta^{\mu\nu}\partial_\rho D^\rho.
 \label{eq:Glin}

XI-A-ST@1#eq:egf · line 616
$e_{gf}^{μν} = - 2 \partial^{(μ} D^{ν)} + η^{μν} \partial_{ρ} D^{ρ}, \partial_{μ} e_{gf}^{μν} = - □ D^{ν} .$
XI-A-ST@1#eq:quadraticoperator · line 625
$2 G_{lin}^{μν} (γ) + e_{gf}^{μν} (γ) = - □ γ ˉ^{μν} .$
XI-A-ST@1#eq:TT · line 665
$L_{FP}^{TT} = \frac{1}{2} γ ˙_{ij} γ ˙_{ij} - \frac{1}{2} \partial_{k} γ_{ij} \partial_{k} γ_{ij} .$
XI-A-ST@1#eq:Newtonmetric · line 676
$H_{00} = 2 Φ_{N}, H_{ij} = 2 Φ_{N} δ_{ij}, H_{0 i} = 0,$
XI-A-ST@1#eq:Newton · line 682
$Δ Φ_{N} = \frac{κ ^{2}}{8} ρ, Φ_{N} (x) = - \frac{κ ^{2}}{32 π} \int_{R^{3}} \frac{ρ ( y )}{∣ x - y ∣} d^{3} y < 0.$
unlabeled · line 636
$δ S_{EH} = \frac{2}{κ ^{2}} \int ρ_{q} G^{μν} δ q_{μν} d^{4} x .$
unlabeled · line 659
$γ_{0 μ} = 0, \partial_{i} γ_{ij} = 0, δ^{ij} γ_{ij} = 0,$

The quadratic normalization

Displayed formulas in this passage

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