Manuscript X-A-conv@1 · X-A-conv.tex · section · c76fffa3

Scalar sign check

Independent action variation and the plateau Noether identity › Scalar sign check · lines 1823–1855

X-A-conv@1#eq:scalarNoethercheck

At first order,

ℓ_{sc, 1} = - \frac{1}{2} h^{μν} T_{μν} (η, ϕ) .

The off-shell identity

\partial_{μ} T^{μν} = (□_{η} ϕ) \partial^{ν} ϕ

gives

δ_{ξ}^{(0)} \int g ℓ_{sc, 1} = \int g (□_{η} ϕ) X_{ξ} ϕ + \int (\partial_{μ} g) T^{μν} ξ_{ν} .

Meanwhile the free scalar variation is

δ_{ξ}^{(1)} S_{sc, 0} = - \int (□_{η} ϕ) X_{ξ} ϕ .

Their sum is

\int (g - 1) (□_{η} ϕ) X_{ξ} ϕ + \int (\partial_{μ} g) T^{μν} ξ_{ν} .

Both terms vanish for $ξ \in g_{U}$ . This explicitly checks the scalar Euler and switch signs.

Displayed formulas in this passage

X-A-conv@1#eq:scalarNoethercheck · line 1848
$\int (g - 1) (□_{η} ϕ) X_{ξ} ϕ + \int (\partial_{μ} g) T^{μν} ξ_{ν} .$
unlabeled · line 1826
$ℓ_{sc, 1} = - \frac{1}{2} h^{μν} T_{μν} (η, ϕ) .$
unlabeled · line 1831
$\partial_{μ} T^{μν} = (□_{η} ϕ) \partial^{ν} ϕ$
unlabeled · line 1835
$δ_{ξ}^{(0)} \int g ℓ_{sc, 1} = \int g (□_{η} ϕ) X_{ξ} ϕ + \int (\partial_{μ} g) T^{μν} ξ_{ν} .$
unlabeled · line 1842
$δ_{ξ}^{(1)} S_{sc, 0} = - \int (□_{η} ϕ) X_{ξ} ϕ .$