Physics theoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-11-04 (gpt-6.1-sol)

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Planck coefficient from unwarped and warped products

Physical scope

Coefficient of the two-derivative Einstein–Hilbert term for fixed product, fixed internal-only warped, or block-diagonal variable-volume product metrics. Flux/source actions, higher-curvature terms, loop corrections, and dynamical dilaton terms are excluded. A constant dilaton appears only in the stated ten-dimensional frame conversion.

Statement

In natural units $ℏ = c = 1$ , let $d, n \geq 1$ , $D = d + n$ , take $G_{D}$ to be a declared $D$ -dimensional Einstein-frame metric, and let $κ_{D} > 0$ be the coupling in that frame's Einstein–Hilbert action, with $[κ_{D}^{2}] = (l e n g t h)^{D - 2}$ . Assume that action is $S_{D} = \frac{1}{2 κ_{D}^{2}} \int_{M_{d} \times Y_{n}} d^{d + n} X \sqrt{- G} R [G],$ and take either a fixed product metric $G_{D} = g_{d} \oplus h_{Y}$ , where $M_{d}$ is smooth Lorentzian and $Y_{n}$ is smooth, compact, Riemannian, and boundaryless, or the following internal-only warped metric with smooth $A, B : Y_{n} \to R$ : $d s_{D}^{2} = e^{2 A (y)} g_{d} + e^{2 B (y)} {\hat{g}}_{Y} .$ For the fixed product, ${Vol}_{E} (Y) : = \int_{Y} d^{n} y \sqrt{\det h_{Y}}$ and the coefficient of the $d$ -dimensional Einstein–Hilbert term is $\frac{1}{2 κ_{d}^{2}} = \frac{{Vol}_{E} (Y)}{2 κ_{D}^{2}}, κ_{d}^{- 2} = {Vol}_{E} (Y) κ_{D}^{- 2} .$ For $d > 2$ , define the reduced Planck scale by $M_{d, P l}^{d - 2} : = κ_{d}^{- 2}$ , taking the positive root when $κ_{d}^{- 2} > 0$ . In $d = 4$ this gives $M_{4, P l}^{2} = κ_{4}^{- 2} = (8 π G_{N})^{- 1}$ under the convention $κ_{4}^{2} = 8 π G_{N}$ , and hence $M_{4, P l}^{2} = {Vol}_{E} (Y) / κ_{D}^{2}$ . The internal-curvature integral contributes a separate term without external curvature.

For the warped metric, the coefficient of the external Einstein–Hilbert term is $\frac{1}{2 κ_{D}^{2}} \int_{Y} d^{n} y \sqrt{\det {\hat{g}}_{Y}} e^{(d - 2) A + n B} .$

For a block-diagonal variable-volume product metric $G_{D} (x, y) = g_{J} (x) \oplus γ_{Y} (x, y)$ , let $g_{J}$ be a smooth Lorentzian metric and let $γ_{Y} (x, \cdot)$ be a smooth family of positive-definite metrics on the compact boundaryless $Y_{n}$ , with no mixed components. Define $V_{J} (x) : = \int_{Y} d^{n} y \sqrt{\det γ_{Y} (x, y)} .$ The coefficient of $R [g_{J}]$ is $V_{J} (x) / (2 κ_{D}^{2})$ in the Jordan metric. If $d > 2$ , $V_{J}$ is positive and smooth, and $V_{*} > 0$ is a fixed reference volume, the Weyl rescaling $g_{E} = (V_{J} / V_{*})^{2 / (d - 2)} g_{J}$ makes the coefficient constant, $V_{*} / (2 κ_{D}^{2})$ , and also produces modulus-derivative terms. All metrics and couplings in each formula are in the same declared frame.

Statement references

Kaluza–Klein product and warped reduction data

Supergravity Conventions And Normalizations

Facts & Assumptions

Given: A smooth compact boundaryless $Y_{n}$ , the stated Einstein–Hilbert action and product metric, or the displayed warped metric; $V_{*} > 0$ is a fixed reference volume.

[F1]

Kaluza–Klein product and warped reduction data: the fixed product, internal-only warped, and block-diagonal variable-volume ansätze are defined, including $V_{J} (x)$ and the Weyl rescaling for $d > 2$ .

[F2]

Supergravity Conventions And Normalizations: in ten dimensions $G^{(E)} = e^{- Φ / 2} G^{(s)}$ and, when the constant asymptotic dilaton is absorbed, $κ_{10, p h y s}^{2} = e^{2 Φ_{\infty}} κ_{0}^{2}$ with $2 κ_{0}^{2} = (2 π)^{7} α^{' 4}$ . The coupling convention is not changed during a frame conversion.

[A1]

Each coefficient calculation uses the displayed Einstein–Hilbert action and metrics in one common frame; other action terms are excluded. External boundary terms vanish or are discarded under the declared boundary conditions.

[A2]

When $V_{J}$ varies, it is positive and smooth, $d > 2$ , and the external Weyl factor is smooth; its boundary term is discarded under the stated boundary conditions.

[A3]

For the constant-dilaton frame conversion in step 3.1, set $D = 10$ , $d = 4$ , $n = 6$ , $Φ = Φ_{0} = Φ_{\infty}$ constant, and use $κ_{10, p h y s}^{2} = e^{2 Φ_{0}} κ_{0}^{2}$ . This conversion does not include derivatives or dynamics of the dilaton.

Proof

1.1givenF1A1algebra

For $G_{D} = g_{d} \oplus h_{Y}$ , block diagonality gives $\det G_{D} = (\det g_{d}) (\det h_{Y})$ and hence $\sqrt{- \det G_{D}} = \sqrt{- \det g_{d}} \sqrt{\det h_{Y}}$ . The mixed Christoffel symbols vanish because the two blocks depend on separate coordinates; the remaining Christoffel symbols are those of $g_{d}$ and $h_{Y}$ . Contracting their block-diagonal curvature gives $R [G_{D}] = R [g_{d}] + R [h_{Y}]$ .

1.2givenF1A1algebra

For the internal-only warped metric, block determinants give $\sqrt{- \det G_{D}} = e^{d A + n B} \sqrt{- \det g_{d}} \sqrt{\det {\hat{g}}_{Y}}$ . The external-curvature part of $R [G_{D}]$ is $e^{- 2 A} R [g_{d}]$ ; warp-derivative terms carry no factor $R [g_{d}]$ . Multiplication therefore leaves $e^{(d - 2) A + n B}$ , proving the warped coefficient.

1.3givenF1A1A2algebra

For the block-diagonal variable-volume metric $G_{D} (x, y) = g_{J} (x) \oplus γ_{Y} (x, y)$ , its determinant gives $\sqrt{- \det G_{D}} = \sqrt{- \det g_{J}} \sqrt{\det γ_{Y}}$ . In the scalar curvature, the term containing the external scalar curvature is $R [g_{J}]$ ; terms involving $x$ -derivatives of $γ_{Y}$ contain no $R [g_{J}]$ . Integrating the coefficient over $Y$ therefore gives $V_{J} (x) / (2 κ_{D}^{2})$ .

2.1givenA2step 1.3algebra

Put $Ω = (V_{J} / V_{*})^{1 / (d - 2)}$ and $g_{E} = Ω^{2} g_{J}$ . The curvature transformation has $R [g_{J}] = Ω^{2} (R [g_{E}] + terms with derivatives of Ω)$ and $\sqrt{- \det g_{J}} = Ω^{- d} \sqrt{- \det g_{E}}$ . Thus the coefficient of $R [g_{E}]$ is $V_{J} Ω^{2 - d} = V_{*}$ ; derivatives of $Ω$ remain as scalar kinetic terms after the boundary term is treated.

2.2givenA1step 1.1algebra

Substituting these identities into $S_{D}$ and integrating over $Y$ yields $S_{D} = \frac{{Vol}_{E} (Y)}{2 κ_{D}^{2}} \int_{M_{d}} \sqrt{- g_{d}} R [g_{d}] + \frac{1}{2 κ_{D}^{2}} (\int_{Y} \sqrt{h_{Y}} R [h_{Y}]) \int_{M_{d}} \sqrt{- g_{d}} .$ The first term is the stated external Einstein coefficient; the second contains no external curvature and is a separate potential/cosmological contribution. For $d = 4$ , matching $1 / (2 κ_{4}^{2})$ to $M_{4, P l}^{2} / 2$ gives the Planck formula.

3.1givenF2A3algebra∎

Under [A3], put $V_{E} : = {Vol}_{E} (Y)$ and let $V_{s}$ be the volume in the string metric. The constant rescaling gives $V_{s} = e^{3 Φ_{0} / 2} V_{E}$ and $\sqrt{- \det g_{4, s}} R [g_{4, s}] = e^{Φ_{0} / 2} \sqrt{- \det g_{4, E}} R [g_{4, E}]$ . Consequently, for the fixed coupling $κ_{10}$ in the adopted Einstein action, its external curvature density obeys $\frac{V_{E}}{2 κ_{10}^{2}} \sqrt{- \det g_{4, E}} R [g_{4, E}] = \frac{V_{s} e^{- 2 Φ_{0}}}{2 κ_{10}^{2}} \sqrt{- \det g_{4, s}} R [g_{4, s}]$ . In particular, if that action uses $κ_{10} = κ_{0}$ , then $V_{s} e^{- 2 Φ_{0}} / (2 κ_{0}^{2}) = V_{s} / (2 κ_{10, p h y s}^{2})$ is the string-frame coefficient. Converting its external density back gives the Einstein-frame coefficient $V_{s} e^{Φ_{0} / 2} / (2 κ_{10, p h y s}^{2}) = V_{E} / (2 κ_{0}^{2})$ . If the adopted Einstein action instead uses $κ_{10} = κ_{10, p h y s}$ , its string-frame coefficient is $V_{s} e^{- 2 Φ_{0}} / (2 κ_{10, p h y s}^{2})$ ; the coupling is held fixed in each conversion. A varying dilaton adds terms and is excluded.

Planck coefficient from unwarped and warped products

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources