Manuscript X-E-ST@1 · X-E-ST.tex · lemma · 44614221

Lemma: Palatini first variation

E1: action, Palatini variation, and switched Noether identity › Exact geometric conventions · lines 518–558 · proof included or linked

X-E-ST@1#lem:palatini, X-E-ST@1#eq:palatini

For a compactly supported variation $δ q_{μν}$ ,

δ (J_{q} R) = - J_{q} G^{μν} δ q_{μν} + \partial_{ρ} (J_{q} W^{ρ}), W^{ρ} = q^{μν} δ Γ^{ρ}_{μν} - q^{ρ μ} δ Γ^{ν}_{ν μ} .

Consequently

\frac{δ S _{gr}}{δ q _{μν}} = \frac{J _{q}}{2 κ ^{2}} G^{μν} .

Proof. The difference of two connections is a tensor, so $δ Γ$ is tensorial. In normal coordinates, and hence tensorially,

δ R^{ρ}_{σ μν} = \nabla_{μ} δ Γ^{ρ}_{ν σ} - \nabla_{ν} δ Γ^{ρ}_{μ σ} .

Contraction and $\nabla q = 0$ give $q^{μν} δ R_{μν} = \nabla_{ρ} W^{ρ}$ . Furthermore,

δ J_{q} = \frac{1}{2} J_{q} q^{μν} δ q_{μν}, δ q^{μν} = - q^{μα} q^{ν β} δ q_{α β} .

Substitution in $δ (J_{q} q^{μν} R_{μν})$ gives (eq:palatini); compact variation removes the divergence from the Euler derivative.

Displayed formulas in this passage

X-E-ST@1#eq:palatini · line 521
$δ (J_{q} R) = - J_{q} G^{μν} δ q_{μν} + \partial_{ρ} (J_{q} W^{ρ}), W^{ρ} = q^{μν} δ Γ^{ρ}_{μν} - q^{ρ μ} δ Γ^{ν}_{ν μ} .$
unlabeled · line 531
$\frac{δ S _{gr}}{δ q _{μν}} = \frac{J _{q}}{2 κ ^{2}} G^{μν} .$
unlabeled · line 541
$δ R^{ρ}_{σ μν} = \nabla_{μ} δ Γ^{ρ}_{ν σ} - \nabla_{ν} δ Γ^{ρ}_{μ σ} .$
unlabeled · line 549
$δ J_{q} = \frac{1}{2} J_{q} q^{μν} δ q_{μν}, δ q^{μν} = - q^{μα} q^{ν β} δ q_{α β} .$