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Wrapped Nil D4 DBI Reduction And Large Field Potential

Physical scope

The leading two-derivative geometric DBI contribution for one Abelian probe D4 wrapped on the stated unit u2 circle, with constant dilaton, zero pulled-back B2 and F_A, fixed moduli, homogeneous u1(t), negligible backreaction, and u1 well above the asymptotic crossover. It is not the full stabilized potential or an inflationary observation.

Statement

Under the geometric DBI assumptions in [A1], integrating over the unit $u_{2}$ -circle and expanding only for $ϵ = α^{'} L_{u_{1}}^{2} {\dot{u}}_{1}^{2} ≪ 1$ gives $S_{D B I} = \int d^{4} x \sqrt{- g_{4, s}} (\frac{1}{2} K (u_{1}) {\dot{u}}_{1}^{2} - V_{D B I} (u_{1}) + O (ϵ^{2} V_{D B I})),$ $V_{D B I} = \frac{L_{u} \sqrt{β}}{(2 π)^{4} g_{s t r} α^{' 2}} {(1 + \frac{M^{2} L_{x}^{2} u_{1}^{2}}{β L_{u}^{2}})}^{1 / 2}, K = \frac{L_{u}^{3}}{(2 π)^{4} g_{s t r} α^{'} \sqrt{β}} {(1 + \frac{M^{2} L_{x}^{2} u_{1}^{2}}{β L_{u}^{2}})}^{1 / 2} .$ The canonical coordinate is fixed by $ϕ (0) = 0$ and $ϕ^{'} (u) = \sqrt{K (u)}$ . For $u ≫ u_{c r i t} : = \sqrt{β} L_{u} / (M L_{x})$ , $ϕ (u) \sim \frac{M^{1 / 2} L_{u} L_{x}^{1 / 2}}{6 π^{2} \sqrt{g_{s t r} α^{'}} β^{1 / 2}} u^{3 / 2}, V_{D B I} (u) \sim \frac{M L_{x}}{(2 π)^{4} g_{s t r} α^{' 2}} u,$ so $V_{D B I} (ϕ) \sim μ_{10 / 3} ϕ^{2 / 3}$ , with $μ_{10 / 3} = {(\frac{3}{2})}^{2 / 3} (2 π)^{- 8 / 3} M^{2 / 3} β^{1 / 3} \frac{L_{x} / L}{α^{' 5 / 3} g_{s t r}^{2 / 3}}, L_{u}^{2} L_{x} = L^{3} .$ In these units $u_{1}, u_{c r i t}, L_{u}, L_{x}, L, β, M,$ and $g_{s t r}$ are dimensionless, $[α^{'}] = L^{2}$ , and $ϵ$ is dimensionless; consequently $[V_{D B I}] = L^{- 4}$ , $[K] = L^{- 2}$ , $[ϕ] = L^{- 1}$ , and $[μ_{10 / 3}] = L^{- 10 / 3}$ . These are the geometric DBI terms only; no claim is made about omitted Chern–Simons, stabilization, loop, higher-curvature, or backreaction contributions.

Statement References

Silverstein and Westphal, §2.3

Facts & Assumptions

Given: A single Abelian D4 probe with $W_{5} = X_{4} \times S^{1}$ , $X_{4} = I \times Σ_{3}$ , smooth homogeneous $u_{1} (t)$ , fixed background moduli, and the full ten-dimensional product metric in the cited input items. Write $g_{s t r} = e^{Φ_{\infty}}$ for the constant dimensionless string coupling, and take $d s_{4, s}^{2} = - d t^{2} + h_{i j} (t, x) d x^{i} d x^{j}$ with smooth positive-definite $h$ .

[F1]

Type II Dp Brane DBI WZ Action: the leading DBI density is $- μ_{4} e^{- Φ} \sqrt{- \det (ι^{*} G^{(s)} + F)}$ , with $μ_{4} = (2 π)^{- 4} α^{' - 5 / 2}$ , and the worldvolume combination is $F = 2 π α^{'} F_{A} - ι^{*} B_{2}$ . This proof uses only the DBI clause and the explicitly assumed zero- $F$ configuration.

[F2]

Silverstein Westphal Nil Background Data For Wrapped D4 Reduction: $L_{u_{1}} = L_{u} / \sqrt{β}$ , $L_{u_{2}} = L_{u} \sqrt{β}$ , $g_{s t r} = e^{Φ_{\infty}}$ , and $L_{u}^{2} L_{x} = L^{3}$ .

[F3]

Wrapped D4 Embedding In Nil Coordinates: the wrapped coordinate has period one and the worldvolume is $X_{4} \times S_{u_{2}}^{1}$ .

[F4]

Nil Wrapped D4 Pullback Metric And Determinant: for smooth positive-definite $h_{i j}$ in $d s_{4, s}^{2} = - d t^{2} + h_{i j} d x^{i} d x^{j}$ , $\sqrt{- \det (ι^{*} G^{(s)})} = \sqrt{- g_{4, s}} \sqrt{α^{'} (L_{u_{2}}^{2} + M^{2} L_{x}^{2} u_{1}^{2})} \sqrt{1 - α^{'} L_{u_{1}}^{2} {\dot{u}}_{1}^{2}}$ .

[A1]

Work at constant dilaton $Φ = Φ_{\infty}$ with $ι^{*} B_{2} = 0$ and $F_{A} = 0$ , in the two-derivative small-velocity regime $α^{'} L_{u_{1}}^{2} {\dot{u}}_{1}^{2} ≪ 1$ . Use $u_{1} ≫ u_{c r i t}$ only for the displayed large- $u_{1}$ asymptotic, and omit the separate Chern–Simons and stabilization-sector terms.

Proof

1.1givenF1F3F4A1algebra

Set $F = 0$ by [A1] and [F1]. Insert [F4] into the DBI action, use $e^{- Φ} = g_{s t r}^{- 1}$ by [A1], and integrate $u_{2} \in [0, 1]$ , whose coordinate length is one by [F3]. The exact homogeneous density is $- \sqrt{- g_{4, s}} ((2 π)^{4} g_{s t r} α^{' 2})^{- 1} \sqrt{L_{u_{2}}^{2} + M^{2} L_{x}^{2} u_{1}^{2}} \sqrt{1 - α^{'} L_{u_{1}}^{2} {\dot{u}}_{1}^{2}}$ .

2.1givenF2step 1.1algebra

By [F2], $L_{u_{2}}^{2} = β L_{u}^{2}$ and $L_{u_{1}}^{2} = L_{u}^{2} / β$ . Substituting these into step 1.1 gives $- \sqrt{- g_{4, s}} ((2 π)^{4} g_{s t r} α^{' 2})^{- 1} \sqrt{β L_{u}^{2} + M^{2} L_{x}^{2} u_{1}^{2}} \sqrt{1 - α^{'} L_{u}^{2} {\dot{u}}_{1}^{2} / β}$ .

3.1givenA1F1F2step 2.1algebra

For $x = α^{'} L_{u}^{2} {\dot{u}}_{1}^{2} / β$ with $0 \leq x ≪ 1$ , Taylor's formula gives $\sqrt{1 - x} = 1 - x / 2 + O (x^{2})$ . Separating constant and quadratic terms in step 2.1 gives the stated $V_{D B I} (u_{1})$ , $K (u_{1})$ , and order- $ϵ^{2} V_{D B I}$ remainder. Positivity of the moduli and $g_{s t r}$ makes $K > 0$ . Since the background parameters are dimensionless and $[α^{'}] = L^{2}$ by [F2], these formulas give $[V_{D B I}] = L^{- 4}$ , $[K] = L^{- 2}$ , and dimensionless $ϵ$ .

4.1givenF2step 3.1algebra

Define $ϕ (u) = \int_{0}^{u} \sqrt{K (s)} d s$ . Since $K (s) > 0$ , this is strictly increasing on $u \geq 0$ , and the chain rule yields ${\dot{ϕ}}^{2} = K (u) {\dot{u}}^{2}$ . Taking the square root of the displayed $K$ gives $ϕ^{'} (u) = L_{u}^{3 / 2} β^{- 1 / 4} ((2 π)^{2} \sqrt{g_{s t r} α^{'}})^{- 1} (1 + M^{2} L_{x}^{2} u^{2} / (β L_{u}^{2}))^{1 / 4}$ , the exact canonical normalization in quadrature form. Since $u$ is dimensionless and $[K] = L^{- 2}$ by step 3.1, $[ϕ] = L^{- 1}$ .

5.1givenstep 3.1step 4.1algebra

For $u ≫ u_{c r i t} = \sqrt{β} L_{u} / (M L_{x})$ , the parenthesis in step 4.1 is asymptotic to $M^{2} L_{x}^{2} u^{2} / (β L_{u}^{2})$ . Thus $ϕ^{'} (u) \sim M^{1 / 2} L_{u} L_{x}^{1 / 2} u^{1 / 2} / ((2 π)^{2} \sqrt{g_{s t r} α^{'}} β^{1 / 2})$ , whose integral gives the displayed $u^{3 / 2}$ coefficient. The potential in step 3.1 satisfies $V_{D B I} \sim M L_{x} u / ((2 π)^{4} g_{s t r} α^{' 2})$ .

6.1givenA1F2step 5.1algebra∎

Set $C = M^{1 / 2} L_{u} L_{x}^{1 / 2} / (6 π^{2} \sqrt{g_{s t r} α^{'}} β^{1 / 2})$ , so step 5.1 gives $u_{1} \sim (ϕ / C)^{2 / 3}$ . Substitution in its potential yields coefficient $(M L_{x} / ((2 π)^{4} g_{s t r} α^{' 2})) C^{- 2 / 3} = (3 / 2)^{2 / 3} (2 π)^{- 8 / 3} M^{2 / 3} β^{1 / 3} (L_{x} / L_{u})^{2 / 3} / (g_{s t r}^{2 / 3} α^{' 5 / 3})$ . By [F2], $L_{u}^{2} L_{x} = L^{3}$ implies $(L_{x} / L_{u})^{2 / 3} = L_{x} / L$ , giving the stated $μ_{10 / 3}$ ; its units are $L^{- 10 / 3}$ , so $μ_{10 / 3} ϕ^{2 / 3}$ has the potential units $L^{- 4}$ . This remains only the geometric DBI term under [A1].

Wrapped Nil D4 DBI Reduction And Large Field Potential

Statement

Statement References

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources