Physics theoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-11-05 (gpt-6.1-sol)

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Massless tadpole logarithm on a separating collar

Physical scope

Conditional infrared divergence from one nonzero massless separating-channel contraction; it neither computes a model's tadpole nor proves that a shifted stationary background exists.

Statement

Assume a separating closed collar has a zero-momentum massless scalar channel with $L_{0}^{m} = {\overset{ˉ}{L}}_{0}^{m} = a = \overset{ˉ}{a} = 1$ . Let $T_{ϕ}$ be the regulated integrated one-point coefficient on one component, and let $A_{o t h e r} (ϕ^{\lor})$ be the coefficient on the other component for the state dual to $ϕ$ . If their pairing contraction $C = T_{ϕ} A_{o t h e r} (ϕ^{\lor})$ is nonzero, the collar contribution contains $2 π C \int_{0}^{ϵ} d r / r$ , which diverges logarithmically as the cutoff at $r = 0$ is removed. A finite prescription must cancel/subtract this coefficient or change to a background in which the relevant one-point coefficient vanishes, if such a consistent stationary background exists. The theorem does not assert either remedy is available in a given model.

Facts & Assumptions

Given: The separating plumbing coordinate $q$ , the $d^{2} q = d x d y$ measure, the massless/intercept condition, a stated regulator for the component one-point coefficient, and the nonzero full contraction $C$ .

[F1]

Local string amplitude sewing and channel factorization: on the stated one-node local sewing input, the channel coefficient multiplies the collar factor; the massless level-matched state has $Δ = \overset{ˉ}{Δ} = 0$ .

[F2]

Plumbing family for punctured Riemann surfaces: the local collar is parameterized by $z w = q$ .

[A1]

Both component amplitudes and the nondegenerate pairing are defined with the same regulator and normalization; specifically $C \neq 0$ . A nonzero one-point coefficient alone would not imply this if the companion component coefficient vanished.

Proof

1.1givenF1F2A1algebra

For $q = r e^{i θ}$ , the massless state has no residual $q$ -power by [F1]. The collar density in [F1] is therefore $d^{2} q / ∣ q ∣^{2} = (d r / r) d θ$ . Integrating $θ$ from $0$ to $2 π$ gives $2 π d r / r$ , and the coefficient is exactly $C$ by [A1].

2.1F1A1step 1.1algebra∎

For a lower radial cutoff $δ > 0$ , the contribution is $2 π C \int_{δ}^{ϵ} d r / r = 2 π C (\log ϵ - \log δ)$ . As $δ ↓ 0$ , $\log δ \to - \infty$ ; since $C \neq 0$ , this has no finite limit. A finite prescription must remove the nonzero logarithmic coefficient, for example by a specified cancellation/subtraction or by a model-dependent background change that makes the relevant tadpole vanish. This conditional calculation does not prove the existence of such a background.

Remarks

This is an infrared collar statement under the supplied regulator. Witten's source notes also warn that other moduli-space integrals can be conditionally convergent; the local logarithm is not a claim about the full amplitude's convergence.

Depends on

Dependency tree · 2 levels

Sources

Edward Witten, Superstring Perturbation Theory via Super Riemann Surfaces: An Overview