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IIA D0 and First Kaluza-Klein Momentum Mass Match

Physical scope

Conditional equality of the leading single-D0 rest energy and the first Kaluza-Klein momentum magnitude on a flat asymptotic IIA/M-theory circle in hbar=c=1 units; assumes the stated D0/KK sector identification and says nothing about completeness or exact duality.

Statement

Assume ℏ=c=1, positive string length ℓs=α′, positive dimensionless coupling gs, and a flat asymptotic circle of radius R11=gsℓs with angle θ∼θ+2π. Put y=R11θ and define the nth Kaluza-Klein mode by its single-valued Fourier wave einθ with momentum operator p11=−i∂y. For a free massless eleven-dimensional mode, adopt the flat-space mass shell E2=∣p∣2+p112, where p∈R9 is the momentum in the noncompact spatial directions. Define its ten-dimensional rest mass to be the nonnegative rest energy mKK,n:=E∣p=0≥0, so the mass shell gives mKK,n=∣p11(n)∣. Adopt the leading Abelian D-brane tension formula Tp=μp/gs with μp=((2π)pα′(p+1)/2)−1, and identify the single unit D0 sector with the first Kaluza-Klein mode. Then

T0=mKK,1=∣p11(1)∣=1gsℓs=1R11.

This conditional mass match proves neither a complete Kaluza–Klein tower, full M/IIA equivalence, nor a nonperturbative M-theory construction.

Facts & Assumptions

Given: The positive circle radius, positive string length and coupling, periodic angle, leading D-brane normalization, free massless eleven-dimensional mass shell, and adopted unit D0/first-KK sector identification in the statement.

[F1]

The adopted M/IIA circle dictionary states R11=gsℓs with gs,ℓs>0 (M Theory Type IIA Circle Limit Dictionary).

[F2]

The leading Abelian Dp-brane action uses α′>0, μp=((2π)pα′(p+1)/2)−1 and Tp=μp/gs (Type II Dp Brane DBI WZ Action).

[A1]

The asymptotic eleven-dimensional mode is free and massless, so its flat-space energy satisfies E2=∣p∣2+p112; in the ten-dimensional rest frame p=0, its rest mass is ∣p11∣. The single D0/first-mode sector identification is adopted, not derived.

Proof

technique · direct arithmetic
1.1givenA1F1algebra

Periodicity makes einθ single-valued exactly for n∈Z. With y=R11θ, applying p11=−i∂y gives p11einθ=(n/R11)einθ. Thus the unit mode has ∣p11(1)∣=1/R11=1/(gsℓs) by [F1], with inverse-length units.

1.2givenF2algebra

Set p=0 in [F2]. Then (2π)0=1 and α′(0+1)/2=α′=ℓs, so μ0=1/ℓs and T0=μ0/gs=1/(gsℓs).

2.1givenA1step 1.1algebra

In the ten-dimensional rest frame p=0, [A1] reduces the eleven-dimensional mass shell to E2=p112. The positive rest energy is therefore mKK,1=∣p11(1)∣=1/R11 by step 1.1.

3.1givenA1step 2.1step 1.2algebra∎

Steps 2.1 and 1.2 give T0=mKK,1=∣p11(1)∣=1/(gsℓs)=1/R11. This equality is conditional on the sector identification in [A1]; it neither proves that identification nor any tower-completeness or duality claim.

Depends on

Dependency tree · 2 levels

Sources