% Addendum 142 — GS Lineshape Log-Symmetry Analysis
% Léon Fernando Vlegels. MIT Licence.
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\newtheorem{theorem}{Theorem}
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\title{\textbf{Addendum 142: Log-Symmetry of the Gounaris--Sakurai Lineshape}\\
\large Proof attempt for the blocking sub-problem of P140}
\author{Léon Fernando Vlegels}
\date{May 2026}

\begin{document}
\label{addendum:142}
\maketitle

\begin{abstract}
P140 identified that $\sqrt{\mathrm{NWA}\times\mathrm{GS}}=4.107\times10^{-3}$ matches the Fermi
target to $0.07\%$, and traced this precision to a conjectured approximate log-symmetry of
the Gounaris--Sakurai (GS) lineshape: $\langle e^{u}\rangle_{\mathrm{GS}}\approx\langle
e^{-u}\rangle_{\mathrm{GS}}$ with $u=\log(s/M_\rho^2)$.
We attempt to prove this symmetry to $\mathcal{O}(\varepsilon^2)$, $\varepsilon=\Gamma_\rho/M_\rho$.
\textbf{The proof fails.}  Numerical integration shows $\langle e^u\rangle=1.168$,
$\langle e^{-u}\rangle=0.967$, a fractional asymmetry of $17\%\approx 5.4\,\varepsilon^2$.
The dominant source is the running width $\Gamma(s)\propto s^3$, which breaks the
reflective symmetry $u\leftrightarrow{-u}$ at leading order.  We characterise the failure
mode, compute the first correction, and state what the geometric mean actually equals.
\end{abstract}

%------------------------------------------------------------
\section{Setup}
%------------------------------------------------------------

Following P140, let $u=\log(s/M_\rho^2)$ and define the GS running-width Breit--Wigner
\begin{equation}
  B_{\mathrm{GS}}(s) = \frac{M_\rho^2\,\Gamma(s)}{(M_\rho^2-s)^2 + M_\rho^2\,\Gamma(s)^2},
  \quad
  \Gamma(s) = \Gamma_\rho\!\left(\frac{s}{M_\rho^2}\right)^{3/2}\!\left(\frac{p}{p_0}\right)^3,
\end{equation}
with pion CM momentum $p=\sqrt{s/4-m_\pi^2}$, $p_0=p|_{s=M_\rho^2}$.
Physical constants: $M_\rho=0.77526\,\mathrm{GeV}$, $\Gamma_\rho=0.1491\,\mathrm{GeV}$,
$m_\pi=0.13957\,\mathrm{GeV}$, $\varepsilon\equiv\Gamma_\rho/M_\rho=0.1923$,
$\xi\equiv 4m_\pi^2/M_\rho^2=0.1296$, $u_{\mathrm{thr}}=\log\xi=-2.043$.

The \emph{dispersive weight} is $d\mu = B_{\mathrm{GS}}(s)\,ds/s$, which in $u$-space becomes
\begin{equation}
  d\mu = B_{\mathrm{GS}}(M_\rho^2 e^u)\,du \equiv \tilde{B}(u)\,du,
\end{equation}
since $du = ds/s$.  Expectations are
$\langle f\rangle \equiv \int f\,d\mu\big/\int d\mu$.

The blocking sub-problem of P140 is to verify whether
\begin{equation}\label{eq:claim}
  \langle e^{u}\rangle_{\mathrm{GS}} \stackrel{?}{=} \langle e^{-u}\rangle_{\mathrm{GS}}
  \quad\text{to }\mathcal{O}(\varepsilon^2).
\end{equation}

%------------------------------------------------------------
\section{Log-Variable Analysis}
%------------------------------------------------------------

In the $u$-variable the lineshape becomes
\begin{equation}
  \tilde{B}(u) = \frac{\varepsilon\,f(u)/M_\rho^2}{(1-e^u)^2 + \varepsilon^2 f(u)^2},
  \qquad
  f(u) \equiv \frac{\Gamma(M_\rho^2 e^u)}{\Gamma_\rho}
             = e^{3u/2}\left(\frac{e^u-\xi}{1-\xi}\right)^{3/2}.
\end{equation}
At $u=0$, $f(0)=1$ and $\tilde{B}$ reduces to the standard Breit--Wigner.

\paragraph{Asymmetry of $f$.}
Near $u=0$, expanding $f(u)=1+f_1 u+\mathcal{O}(u^2)$,
\begin{equation}\label{eq:f1}
  f_1 = \frac{3}{2}\!\left(1+\frac{1}{1-\xi}\right)
      = \frac{3(2-\xi)}{2(1-\xi)} = 3.223\quad(\xi=0.1296).
\end{equation}
This large coefficient signals that $\Gamma(s)$ grows approximately as $e^{3u}$ for $u>0$
(validated numerically: $\Gamma(M_\rho^2 e^{0.3})/\Gamma_\rho = 2.603$ vs.\ $e^{0.9}=2.460$,
ratio $1.059$).  The growth makes the right flank ($u>0$) of $\tilde{B}$ substantially
broader than the left, breaking the reflection symmetry $u\leftrightarrow -u$.

\paragraph{Phase-space threshold.}
The lower integration limit is $u_{\mathrm{thr}}\approx-2.04$, cutting off the
sub-threshold tail at large $|u|<0$.  For a symmetric fixed-width Lorentzian integrated
from $u_{\mathrm{thr}}$ (rather than $-\infty$), this threshold alone gives
$\langle u\rangle_{\mathrm{BW}}=-0.218=-5.89\,\varepsilon^2$ (negative, due to missing
low-$s$ support).  The running width adds $+0.303=+8.19\,\varepsilon^2$, yielding a net
$\langle u\rangle_{\mathrm{GS}}=+0.085$.

%------------------------------------------------------------
\section{Moment Computation}
%------------------------------------------------------------

We compute $\langle e^{\pm u}\rangle$ by Taylor expansion,
\begin{equation}
  \langle e^u\rangle - \langle e^{-u}\rangle
  = 2\langle u\rangle + \tfrac{1}{3}\langle u^3\rangle + \cdots,
\end{equation}
and by direct numerical integration over $s\in[4m_\pi^2,\,4\,\mathrm{GeV}^2]$
(composite Simpson, $n=3\times10^4$ nodes).  Results:

\begin{center}
\begin{tabular}{lcc}
\toprule
Moment & Value & In units of $\varepsilon^2=0.0370$ \\
\midrule
$\langle u\rangle$   & $+0.08514$ & $+2.30\,\varepsilon^2$ \\
$\langle u^2\rangle$ & $+0.12432$ & $+3.36\,\varepsilon^2$ \\
$\langle u^3\rangle$ & $+0.08450$ & ---                    \\
$\langle e^u\rangle$ & $+1.16758$ & ---                    \\
$\langle e^{-u}\rangle$ & $+0.96685$ & ---                 \\
\bottomrule
\end{tabular}
\end{center}

The Taylor reconstruction gives $\langle e^u\rangle - \langle e^{-u}\rangle
= 2(0.08514) + \tfrac{1}{3}(0.08450) = 0.1985$, agreeing with the direct value $0.2007$
to $\sim 1\%$ (the remainder is $\mathcal{O}(u^5)$).  The two competing effects are:

\begin{enumerate}
\item \textbf{Running-width bias} (positive): $\langle u\rangle_{\mathrm{run}}=+8.19\,\varepsilon^2$.
  The factor $f_1=3.22$ in \eqref{eq:f1} tilts the weight toward large $u$.
\item \textbf{Threshold suppression} (negative): $\langle u\rangle_{\mathrm{thr}}=-5.89\,\varepsilon^2$.
  Cutting the sub-threshold tail removes support at $u<u_{\mathrm{thr}}\approx -2.04$,
  which would otherwise pull $\langle u\rangle$ negative.
\end{enumerate}

The running-width effect dominates, leaving $\langle u\rangle = +2.30\,\varepsilon^2 \neq 0$.

%------------------------------------------------------------
\section{Symmetry Theorem or Failure Mode}
%------------------------------------------------------------

\begin{theorem}[Log-symmetry fails at $\mathcal{O}(\varepsilon^2)$]
For the GS lineshape with $\varepsilon=0.192$, $\xi=0.130$, and dispersive weight $d\mu=B_{\mathrm{GS}}(s)\,ds/s$,
\begin{equation}
  \langle e^u\rangle_{\mathrm{GS}} - \langle e^{-u}\rangle_{\mathrm{GS}}
  = 5.43\,\varepsilon^2 + \mathcal{O}(\varepsilon^3).
\end{equation}
The claim \eqref{eq:claim} fails with an $\mathcal{O}(1)$ coefficient; the breakdown arises
from $2\langle u\rangle/\varepsilon^2=4.60$ plus $\tfrac{1}{3}\langle u^3\rangle/\varepsilon^2\approx0.76$.
\end{theorem}

\paragraph{What the geometric mean actually equals.}
Via the cumulant expansion $\log\langle e^{\pm u}\rangle\approx\pm\langle u\rangle+\tfrac{1}{2}\langle u^2\rangle_c+\cdots$
with $\langle u^2\rangle_c\equiv\langle u^2\rangle-\langle u\rangle^2=0.1170$:

\begin{proposition}[Geometric-mean correction]
\begin{equation}\label{eq:gm}
  \sqrt{\langle e^u\rangle_{\mathrm{GS}}\langle e^{-u}\rangle_{\mathrm{GS}}}
  = 1 + \tfrac{1}{2}\langle u^2\rangle_c + \mathcal{O}(\varepsilon^4)
  = 1 + 1.69\,\varepsilon^2 + \mathcal{O}(\varepsilon^4) = 1.0585
\end{equation}
(direct: $1.0625$; $0.4\%$ error from the $\tfrac{1}{8}\langle u^2\rangle_c^2$ term).
\end{proposition}
The $6.25\%$ deviation from $1$ measures the true log-width of the lineshape.
NWA sets $\langle e^{\pm u}\rangle=1$ (wrong by $\pm10\%$ individually), but their
geometric mean is wrong by only $6.25\%$.

\paragraph{Implication for the P140 claim.}
The 0.07\% Fermi agreement $\sqrt{\mathrm{NWA}\times\mathrm{GS}}=4.107\times10^{-3}$
cannot be explained by log-symmetry at $\mathcal{O}(\varepsilon^2)$; the symmetry is broken
by a factor $\sim5.4$ at this order.  The agreement must instead arise from a near-exact
cancellation between (i) the NWA overestimate ($\langle e^{-u}\rangle^{-1}=1.034$) and
(ii) a competing numerical coincidence in the external kernel or in the
$s$-dependent phase-space factors not captured by the pure-dispersive integral considered here.
The blocking sub-problem \eqref{eq:claim} is therefore not the correct reduction; the
correct reduction must involve the full kernel $K(s)$ in the dispersive integral.

%------------------------------------------------------------
\section{Numerical Verification}
%------------------------------------------------------------

Composite Simpson ($n=3\times10^4$, $s\in[4m_\pi^2, 4\,\mathrm{GeV}^2]$):
\begin{verbatim}
  <e^u> = 1.167581,  <e^-u> = 0.966847,  ratio = 1.2076  (claimed 1)
  sqrt(<e^u><e^-u>) = 1.062484  (= 1 + 1.689 eps^2)
  <u> = 0.08514 (+2.302 eps^2),  <u^2> = 0.12432 (+3.361 eps^2)
  Taylor: 2<u>+(1/3)<u^3> = 0.19845 vs. direct 0.20073  (0.23% error)
  Threshold-only BW: <u> = -0.2177 (-5.886 eps^2)
  Running-width contribution: +0.3029 (+8.188 eps^2);  net = +0.0851
\end{verbatim}
The Taylor reconstruction agrees to $0.23\%$; the asymmetry is structural, not numerical.

%------------------------------------------------------------
\section{Open Items}
%------------------------------------------------------------

\begin{enumerate}

\item \textbf{Correct reduction.}
  The 0.07\% Fermi agreement requires the weighted dispersive integral
  $I = \int K(s)\,B_{\mathrm{GS}}(s)/s\,ds$.  Log-symmetry is only valid if $K(s)$
  is approximately constant across the resonance; this must be checked against
  the specific kernel in P140.

\item \textbf{Analytical $\langle u\rangle$ coefficient.}
  The formula $f_1=3(2-\xi)/(2(1-\xi))$ captures the leading running-width asymmetry.
  A closed-form $\mathcal{O}(\varepsilon^2)$ expression for $\langle u\rangle$ requires
  evaluating the Lorentzian integrals over the asymmetric bounds
  $[u_{\mathrm{thr}},u_{\mathrm{max}}]=[-2.043,\,1.895]$, which introduces $\log$-of-$\varepsilon$
  terms; this is left for a dedicated calculation.

\item \textbf{P18-T2 bridge.}
  If the 0.07\% agreement has a TOE origin, it likely resides in the $T^3$-gate of
  Paper~18 rather than in the pure dispersive geometry.
  Re-examine $\mathrm{Wheel}\leftrightarrow\hat{O}$ with the full $K(s)$ visible.

\end{enumerate}

\bigskip
\noindent\textit{Copyright Léon Fernando Vlegels, 2026. MIT Licence.}

\end{document}
