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Ten Dimensional NS NS Field Equations

Physical scope

Conditional source-free Euler–Lagrange equations of the adopted ten-dimensional tree-level two-derivative NS–NS action. No RR fields, localized or magnetic sources, higher-derivative corrections, string loops, or claim of empirical validity are included.

Statement

Let $U \subseteq R^{10}$ be an oriented chart with smooth string-frame Lorentzian metric $G_{M N}$ of signature $(- + \dots +)$ , smooth real dimensionless dilaton $Φ$ , and smooth antisymmetric potential components $B_{M N}$ , $M, N = 0, \dots, 9$ . Set $H_{M N P} = 3 \partial_{[M} B_{N P]}, f = e^{- 2 Φ}, X = G^{M N} \partial_{M} Φ \partial_{N} Φ, H^{2} = H_{M N P} H^{M N P},$ with indices raised by $G$ , Levi-Civita connection $\nabla$ , and $□ Φ = G^{M N} \nabla_{M} \nabla_{N} Φ$ . Define $D : = R + 4 □ Φ - 4 X - \frac{1}{12} H^{2} .$ Use the adopted local action

S = \frac{1}{2 κ_{0}^{2}} \int_{U} d^{10} x \sqrt{- G} e^{- 2 Φ} (R + 4 X - \frac{1}{12} H^{2})

and signed form norm $∣ F_{p} ∣_{G}^{2} = (1 / p!) F_{M_{1} \dots M_{p}} F^{M_{1} \dots M_{p}}$ . In length-valued coordinates, the adopted component normalization makes $B_{M N}$ dimensionless and $H_{M N P}$ of dimension length $^{- 1}$ . A configuration is stationary for this local variational model when the first variation of that action vanishes for every independent smooth compactly supported variation of $G^{M N}$ , $Φ$ , and $B_{M N}$ , through Lorentzian metrics. Then stationarity is equivalent to $R_{M N} + 2 \nabla_{M} \nabla_{N} Φ - \frac{1}{4} H_{M P Q} H_{N}^{P Q} = 0, \nabla_{M} ⁣ (e^{- 2 Φ} H^{M N P}) = 0, D = 0.$ The two-form equation is equivalently $d (e^{- 2 Φ} ⋆_{G} H_{3}) = 0$ , where the Hodge star uses the convention $α \land ⋆_{G} β = ⟨ α, β ⟩_{G} {v o l}_{G}$ fixed by the earlier supergravity conventions definition. Independently of stationarity, $d H_{3} = 0$ holds wherever the smooth potential $B_{2}$ is defined. These are conditional deductions in the adopted model.

Statement references

Ten Dimensional NS NS Supergravity Action

Supergravity Conventions And Normalizations

Facts & Assumptions

Given: The smooth fields on $U$ , the action postulate, and stationarity under all independent compactly supported variations stated above.

[F1]

The convention fixes the curvature, scalar wave operator, signed $p$ -form norm, $H_{3} = d B_{2}$ , and the oriented Lorentzian Hodge star by $α \land ⋆_{G} β = ⟨ α, β ⟩_{G} {v o l}_{G}$ (Supergravity Conventions And Normalizations).

[A1]

Adopt the local action $S = \frac{1}{2 κ_{0}^{2}} \int_{U} \sqrt{- G} e^{- 2 Φ} (R + 4 G^{M N} \partial_{M} Φ \partial_{N} Φ - \frac{1}{12} H_{M N P} H^{M N P}) d^{10} x$ with no source, RR, loop, or higher-derivative terms (Ten Dimensional NS NS Supergravity Action).

[F2]

For a compactly supported inverse-metric variation $q^{M N}$ , the first variation of $\int \sqrt{- G} f R d^{10} x$ , with fixed smooth $f$ , has coefficient $f (R_{M N} - \frac{1}{2} G_{M N} R) + G_{M N} □ f - \nabla_{M} \nabla_{N} f$ against $q^{M N}$ ; moreover, the chart integral of a compactly supported smooth coordinate divergence is zero (Compactly Supported Metric Variation of a Weighted Scalar-Curvature Action).

[F6]

The same inverse-metric variation changes the density by $δ \sqrt{- G} = - \frac{1}{2} \sqrt{- G} G_{M N} q^{M N}$ (Compactly Supported Metric Variation of a Weighted Scalar-Curvature Action).

[F3]

If continuous coefficient components pair to zero with every smooth compactly supported test vector under a positive continuous weight, then every component vanishes pointwise (Compactly Supported Tests Determine Continuous Local Coefficients).

[F4]

Every smooth compactly supported symmetric inverse-metric direction is realized by a smooth family of Lorentzian metrics for sufficiently small parameter (Compactly Supported Metric Directions Stay Lorentzian for Small Amplitude).

[F5]

The convention has $α^{'} > 0$ and $2 κ_{0}^{2} = (2 π)^{7} α^{' 4} > 0$ , so the common action coefficient is nonzero (Supergravity Conventions And Normalizations).

Proof

technique · direct; vary the inverse metric, dilaton, and two-form independently, integrate compactly supported derivatives by parts, and extract each continuous Euler coefficient with [F3]

1.1givenF1A1F6algebra

Put $f = e^{- 2 Φ}$ and $C_{0} = (2 κ_{0}^{2})^{- 1}$ . For an inverse-metric variation $q^{M N} = δ G^{M N}$ , [F6] gives $δ \sqrt{- G} = - \frac{1}{2} \sqrt{- G} G_{M N} q^{M N}$ . Also $δ X = q^{M N} \partial_{M} Φ \partial_{N} Φ$ and $δ H^{2} = 3 H_{M P Q} H_{N}^{P Q} q^{M N}$ , since each inverse metric contributes the same contraction by antisymmetry. The scalar-kinetic and three-form terms in the metric variation, divided by $C_{0} \sqrt{- G}$ , therefore have coefficient $f (4 \partial_{M} Φ \partial_{N} Φ - 2 G_{M N} X - \frac{1}{4} H_{M P Q} H_{N}^{P Q} + \frac{1}{24} G_{M N} H^{2})$ . The trace coefficient $1 / 24$ is $(- 1 / 12) (- 1 / 2)$ , and the tensor coefficient $- 1 / 4$ is $(- 1 / 12) 3$ .

1.2givenA1F1F2algebra

Hold $G, B$ fixed and vary $Φ$ by a compactly supported $δ Φ$ . Then $δ f = - 2 f δ Φ$ and $δ X = 2 \nabla^{M} Φ \nabla_{M} δ Φ$ , so, after factoring out $C_{0} \sqrt{- G}$ , the pre-integration-by-parts coefficient is $f (- 2 R - 8 X + \frac{1}{6} H^{2}) δ Φ + 8 f \nabla^{M} Φ \nabla_{M} δ Φ$ . The vector $f \nabla^{M} Φ δ Φ$ is compactly supported. The divergence integral identity in [F2] therefore gives $8 \int_{U} \sqrt{- G} f \nabla^{M} Φ \nabla_{M} δ Φ = - 8 \int_{U} \sqrt{- G} (□ Φ - 2 X) f δ Φ$ . Combining coefficients yields the dilaton variation $- 2 C_{0} \int_{U} \sqrt{- G} f D δ Φ d^{10} x$ .

1.3givenA1F1F2algebra

Hold $G, Φ$ fixed and vary the antisymmetric potential by compactly supported $δ B_{N P}$ . Smoothness and the coordinate definition of $d$ give $δ H_{M N P} = 3 \nabla_{[M} δ B_{N P]}$ . Antisymmetry makes the three terms equal in contraction, hence $δ S_{B} = - \frac{1}{2 κ_{0}^{2}} \int_{U} \sqrt{- G} \frac{f}{2} H^{M N P} \nabla_{M} δ B_{N P} d^{10} x = \frac{1}{2 κ_{0}^{2}} \int_{U} \sqrt{- G} \frac{1}{2} \nabla_{M} (f H^{M N P}) δ B_{N P} d^{10} x$ , by integration by parts for the compactly supported vector $f H^{M N P} δ B_{N P}$ and [F2].

2.1step 1.1F2algebra

By [F2], the weighted curvature term has metric coefficient $f (R_{M N} - \frac{1}{2} G_{M N} R) + G_{M N} □ f - \nabla_{M} \nabla_{N} f$ against $q^{M N}$ . Adding the matter coefficient from step 1.1 gives the full metric coefficient, after division by $C_{0} \sqrt{- G}$ .

2.2F1step 1.3algebra

Let $ε_{M_{1} \dots M_{10}}$ be the oriented metric volume tensor. The stated Hodge convention gives $(⋆_{G} H)_{M_{1} \dots M_{7}} = \frac{1}{3!} ε_{N_{1} N_{2} N_{3} M_{1} \dots M_{7}} H^{N_{1} N_{2} N_{3}}$ . Using $(d ω)_{A_{0} \dots A_{7}} = 8 \nabla_{[A_{0}} ω_{A_{1} \dots A_{7}]}$ , define $(δ K)_{N P} : = - \nabla^{M} K_{M N P}$ for a 3-form $K$ . Substitution of the epsilon expression and contraction in an oriented orthonormal frame, where raising the single time index changes the volume-tensor sign, gives $δ K = ⋆_{G} d ⋆_{G} K$ ; equivalently the Hodge-sign exponent $10 (3 + 1) + 1 + 1 = 42$ is even. Hence $[⋆_{G} d (f ⋆_{G} H)]_{N P} = - \nabla_{M} (f H^{M}_{N P})$ . The Hodge star is invertible, so the component equation is equivalent to $d (f ⋆_{G} H) = 0$ .

3.1step 2.1givenalgebra

The chain rule gives $f^{- 1} \nabla_{M} \nabla_{N} f = - 2 \nabla_{M} \nabla_{N} Φ + 4 \partial_{M} Φ \partial_{N} Φ$ and $f^{- 1} □ f = - 2 □ Φ + 4 X$ . Substitute these into step 2.1. The $\partial_{M} Φ \partial_{N} Φ$ terms cancel, and the metric-proportional terms combine to $- \frac{1}{2} f G_{M N} D$ . Thus the symmetric metric coefficient is $M_{M N} = f (R_{M N} + 2 \nabla_{M} \nabla_{N} Φ - \frac{1}{4} H_{M P Q} H_{N}^{P Q} - \frac{1}{2} G_{M N} D)$ .

4.1givenF3F4F5step 3.1step 1.2step 1.3

By [F4], every compactly supported symmetric inverse-metric test direction in the metric variation is realized by an admissible metric family. Stationarity therefore makes the first-variation pairings vanish for all such independent tests. Divide out the common nonzero factor $C_{0}$ by [F5]. For unordered index pairs $M \leq N$ , define test components $q^{M N}$ freely and coefficient components $A_{M N} : = M_{M M}$ if $M = N$ , and $A_{M N} : = 2 M_{M N}$ if $M < N$ ; then $M_{M N} q^{M N} = \sum_{M \leq N} A_{M N} q^{M N}$ . Thus [F3] applies and gives $M_{M N} = 0$ . The dilaton coefficient is $- 2 f D$ , tested as a one-component vector. For $N < P$ , antisymmetry gives $\frac{1}{2} T^{M N} δ B_{M N} = T^{N P} δ B_{N P}$ after summing the two ordered terms, where $T^{N P} : = \nabla_{M} (f H^{M N P})$ ; hence [F3] applied to these independent pair components gives $T^{N P} = 0$ . Since $f = e^{- 2 Φ} > 0$ , $f D = 0$ also gives $D = 0$ .

5.1step 3.1step 1.2step 1.3step 4.1algebra

With $D = 0$ , the metric equation $M_{M N} = 0$ from step 4.1 reduces, after division by $f > 0$ , to $R_{M N} + 2 \nabla_{M} \nabla_{N} Φ - \frac{1}{4} H_{M P Q} H_{N}^{P Q} = 0$ . Conversely, if this metric equation, $D = 0$ , and the two-form equation hold, the first-variation coefficients in steps 3.1, 1.2, and 1.3 all vanish, so every compactly supported first variation is zero.

6.1F1algebra∎

Finally $d H_{3} = d (d B_{2}) = 0$ : in components, antisymmetrizing the second partial derivatives of the smooth $B_{M N}$ gives zero because mixed partial derivatives commute. This is an identity of the local potential, independent of stationarity and not an additional Euler–Lagrange equation.

Remarks

The action is a source-free tree-level, leading two-derivative truncation. The equations are conditional variational consequences in that model; they are not empirical proofs and do not apply across localized or magnetic sources without additional current and boundary data.

Ten Dimensional NS NS Field Equations

Statement

Statement references

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources