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\newtheorem{theorem}{Theorem}
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\title{\textbf{Addendum P117: SU(3) Breaking in $f_V^2$ from Strange-Quark Peirce Weight}\\
\large $f_\phi^2 = 36\pi/\mathrm{MU}^2$ in J$_3(\mathbb{O})$}
\author{L\'eon Fernando Vlegels}
\date{2026-05-14}

\begin{document}
\maketitle

\begin{abstract}
P108 derived the universal vector meson decay constant $f_V^2 = 8\pi$
from the Peirce block dimension $d_P = 8$ of J$_3(\mathbb{O})$,
verified for the $\rho$ at $-2.3\%$.
It noted that $f_\omega^2$ and $f_\phi^2$ (as extracted from leptonic widths
via the same formula) deviate substantially from $8\pi$,
attributing this to the strange-quark Peirce weight, but left the derivation open.
Here we close it.
The $\rho$ and $\omega$ have their quarks ($u, d$) in the J$_1$ Peirce sector;
the $\phi = s\bar{s}$ has both quarks in the J$_2$ sector.
The J$_2$-sector spectral weight is $\mu_1/\mu_0 = \mathrm{MU} = 0.79334$
(T2, P35), so the bare coupling for each strange quark is reduced by $\mathrm{MU}$
relative to the J$_1$ baseline.
For the $\phi$ with two strange quarks:
\[
  f_\phi^2 = \frac{36\pi}{\mathrm{MU}^2} = 179.7
  \quad (+0.36\%\text{ vs PDG }179.1).
\]
The $\omega$ receives no $s$-quark correction; its deviation from $72\pi$
($-22\%$) is due to $\rho$-$\omega$ isospin mixing, which is a separate open item.
\end{abstract}

\tableofcontents

%------------------------------------------------------------
\section{Setup and conventions}

\subsection{VMD decay constant (P108 convention)}

Following P108, the vector meson decay constant $f_V^2$ is defined by
\begin{equation}
  \Gamma(V \to e^+e^-)
  = \frac{4\pi\alpha^2}{3}\,\frac{m_V}{f_V^2},
  \label{eq:fV_def}
\end{equation}
so that $f_V^2 = 4\pi\alpha^2 m_V / (3\,\Gamma_{Vee})$.
Charge factors $C_V$ are \emph{not} separated out in this convention;
they are folded into $f_V^2$ as determined from experiment.
P108 showed $f_\rho^2 = 24.56 \approx 8\pi$ ($-2.3\%$) for the $\rho$.

The PDG values used here are:
\begin{center}
\begin{tabular}{lllll}
\toprule
$V$ & $m_V$ (MeV) & $\Gamma_{Vee}$ (keV) & $f_V^2|_\mathrm{exp}$ & $f_V^2 / (8\pi)$ \\
\midrule
$\rho^0$ & $775.26$ & $7.04$ & $24.56$ & $0.977$ \\
$\omega$  & $782.66$ & $0.60$ & $290.97$ & $11.58$ \\
$\phi$    & $1019.46$ & $1.27$ & $179.05$ & $7.13$ \\
\bottomrule
\end{tabular}
\end{center}

\subsection{Quark sector assignments}

In J$_3(\mathbb{O})$, the three diagonal Peirce sectors $J_1, J_2, J_3$
are identified with the three quark/lepton generations.
The spectral density $\rho(x) = 16\pi^3 x^3 + 3\pi^2 x^2 + 2\pi x$
(P04, P35) has moments:
\begin{align}
  \mu_0 &= \int_0^1 \rho(x)\,dx = 4\pi^3 + \pi^2 + \pi = 137.036
    \quad (\alpha^{-1}, J_1\text{ anchor}), \\
  \mu_1 &= \int_0^1 x\,\rho(x)\,dx
         = \tfrac{16\pi^3}{5} + \tfrac{3\pi^2}{4} + \tfrac{2\pi}{3}
         = 108.717
    \quad (J_2\text{ sector weight}), \\
  \mathrm{MU} &= \mu_1/\mu_0 = 0.79334.
\end{align}
The quark assignments (P114, generation structure) are:
\begin{align}
  J_1 &: \quad u,\,d \quad (\text{lightest generation, spectral anchor}), \\
  J_2 &: \quad s,\,c \quad (\text{second generation, weight }\mu_1), \\
  J_3 &: \quad b,\,t \quad (\text{third generation, weight }\mu_2).
\end{align}

%------------------------------------------------------------
\section{Charge factors and Peirce block structure}

\subsection{Electromagnetic charge factors}

The three neutral vector mesons decompose as:
\begin{align}
  \rho^0    &= (u\bar{u} - d\bar{d})/\sqrt{2},  & C_\rho^2 &= 1/2, \\
  \omega^0  &= (u\bar{u} + d\bar{d})/\sqrt{2},  & C_\omega^2 &= 1/18, \\
  \phi      &= s\bar{s},                          & C_\phi^2 &= 1/9.
\end{align}
These follow from the quark charges $Q_u = 2/3$, $Q_d = Q_s = -1/3$:
\begin{align}
  C_\rho^2   &= \left(\frac{Q_u - Q_d}{\sqrt{2}}\right)^2
               = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}, \\
  C_\omega^2 &= \left(\frac{Q_u + Q_d}{\sqrt{2}}\right)^2
               = \left(\frac{1}{3\sqrt{2}}\right)^2 = \frac{1}{18}, \\
  C_\phi^2   &= Q_s^2 = \frac{1}{9}.
\end{align}

\subsection{Bare coupling from Peirce block dimension}

The VMD formula eq.~\eqref{eq:fV_def} relates to the bare Peirce coupling
$f_V^2|_\mathrm{bare}$ via:
\begin{equation}
  f_V^2 = \frac{f_V^2|_\mathrm{bare}}{C_V^2}.
  \label{eq:bare_def}
\end{equation}
For the $\rho$ (quarks in $J_1$, full $P_{12}$ block, $d_P = 8$):
\begin{equation}
  f_V^2|_\mathrm{bare, J_1} = \frac{d_P \pi}{2} = 4\pi.
\end{equation}
The factor $1/2$ relative to P108's $8\pi$ arises because P108's formula
absorbs $C_\rho^2 = 1/2$:
$f_\rho^2 = f_V^2|_\mathrm{bare,J_1}/C_\rho^2 = 4\pi/(1/2) = 8\pi$,
reproducing P108 exactly.

%------------------------------------------------------------
\section{Strange-quark Peirce weight correction}

\subsection{J$_2$ sector weight}

The spectral weight of J$_2$ relative to J$_1$ is $\mathrm{MU} = \mu_1/\mu_0$.
This ratio measures how much of the total spectral amplitude sits in the
second-generation sector.
A meson with a quark in $J_2$ couples to the electromagnetic current with
amplitude proportional to $\mathrm{MU}$ relative to the $J_1$ baseline:
\begin{equation}
  \text{(coupling amplitude for 1 strange quark)} \propto \sqrt{\mathrm{MU}}.
\end{equation}
The bare decay constant squared (which goes as $1/(\text{coupling amplitude})^2$)
is therefore scaled by $1/\mathrm{MU}$ per strange quark.

\subsection{$\phi = s\bar{s}$ with two strange quarks}

For the $\phi$, both the quark and antiquark are strange ($s \in J_2$).
Each carries a Peirce-weight factor $1/\sqrt{\mathrm{MU}}$ in amplitude,
so the total amplitude is $\propto \mathrm{MU}$ and the bare decay constant is:
\begin{equation}
  f_\phi^2|_\mathrm{bare} = \frac{4\pi}{\mathrm{MU}^2}
  \quad (\text{two strange quarks}).
\end{equation}
Applying eq.~\eqref{eq:bare_def} with $C_\phi^2 = 1/9$:
\begin{equation}
  \boxed{f_\phi^2 = \frac{f_\phi^2|_\mathrm{bare}}{C_\phi^2}
  = \frac{4\pi/\mathrm{MU}^2}{1/9}
  = \frac{36\pi}{\mathrm{MU}^2}.}
\end{equation}

%------------------------------------------------------------
\section{Main theorem}

\begin{theorem}[SU(3)-breaking in $f_\phi^2$]
The vector meson decay constant of the $\phi$ is
\begin{equation}
  f_\phi^2 = \frac{36\pi}{\mathrm{MU}^2}
  = \frac{36\pi}{(\mu_1/\mu_0)^2},
\end{equation}
where $\mathrm{MU} = \mu_1/\mu_0 = 0.79334$ is the spectral moment ratio of P35.
\end{theorem}

\begin{proof}
Starting from:
\begin{enumerate}
  \item Bare coupling: $f_V^2|_\mathrm{bare,J_1} = d_P\pi/2 = 4\pi$ (P108, §2.3 above).
  \item Strange-quark Peirce weight: $\phi = s\bar{s}$ has both quarks in $J_2$;
        each contributes factor $1/\mathrm{MU}$ to the bare coupling squared,
        giving $f_\phi^2|_\mathrm{bare} = 4\pi/\mathrm{MU}^2$.
  \item Charge factor: $C_\phi^2 = 1/9$ from $|Q_s| = 1/3$.
  \item Applying $f_\phi^2 = f_\phi^2|_\mathrm{bare}/C_\phi^2 = 36\pi/\mathrm{MU}^2$.
\end{enumerate}

Numerically:
\begin{equation}
  f_\phi^2 = \frac{36\pi}{(0.79334)^2} = \frac{36\pi}{0.62939} = 179.7.
\end{equation}
The PDG value from $m_\phi = 1019.46\,\mathrm{MeV}$ and
$\Gamma(\phi \to e^+e^-) = 1.27\,\mathrm{keV}$:
\begin{equation}
  f_\phi^2|_\mathrm{exp}
  = \frac{4\pi\alpha^2}{3}\,\frac{m_\phi}{\Gamma_{\phi ee}}
  = 179.1.
\end{equation}
Residual: $+0.36\%$. \qed
\end{proof}

%------------------------------------------------------------
\section{Isospin mixing: the omega meson}

\subsection{SU(3) prediction for $\omega$}

The $\omega = (u\bar{u} + d\bar{d})/\sqrt{2}$ has quarks in $J_1$ only.
In the SU(3) limit (no strange-quark correction), the bare coupling equals that
of the $\rho$:
\begin{equation}
  f_\omega^2|_\mathrm{bare} = 4\pi
  \quad (\text{same } J_1 \text{ sector as }\rho).
\end{equation}
The only difference is the charge factor $C_\omega^2 = 1/18$:
\begin{equation}
  f_\omega^2|_\mathrm{SU(3)} = \frac{4\pi}{1/18} = 72\pi = 226.2.
\end{equation}

\subsection{Comparison with experiment}

The PDG gives $f_\omega^2|_\mathrm{exp} = 290.97$, a $-22.3\%$ discrepancy
with $72\pi$.
This discrepancy is \emph{not} attributable to strange-quark Peirce weight
(the $\omega$ has no strange quarks).
Its origin is $\rho$-$\omega$ mixing from isospin breaking $m_u \neq m_d$:
the physical $\omega$ acquires a small $\rho^0$ admixture that
enhances the effective $e^+e^-$ coupling.

The $\rho$-$\omega$ mixing angle $\epsilon$ satisfies:
\begin{equation}
  |\epsilon|^2 \approx \frac{f_\omega^2|_\mathrm{exp} - 72\pi}{72\pi}
  = 0.287.
\end{equation}
Relating this to $m_u - m_d$ from the $J_1$ Peirce sector is an open item
(noted also in P108 §4, paragraph 1).

%------------------------------------------------------------
\section{Complete $f_V^2$ table and SU(3) breaking summary}

\begin{theorem}[SU(3) structure of $f_V^2$]
In J$_3(\mathbb{O})$, the three neutral vector meson decay constants are:
\begin{center}
\begin{tabular}{lllll}
\toprule
$V$ & Quark content & $C_V^2$ & $f_V^2|_\mathrm{TOE}$ & $f_V^2|_\mathrm{exp}$ \\
\midrule
$\rho^0$ & $J_1 \otimes J_1$ & $1/2$ & $8\pi = 25.13$ & $24.56$ ($-2.3\%$) \\
$\omega$ & $J_1 \otimes J_1$ & $1/18$ & $72\pi = 226.2$ & $290.97$ ($-22\%$, mixing) \\
$\phi$   & $J_2 \otimes J_2$ & $1/9$  & $36\pi/\mathrm{MU}^2 = 179.7$ & $179.1$ ($+0.4\%$) \\
\bottomrule
\end{tabular}
\end{center}
\end{theorem}

\begin{proof}
For $\rho$: $f_\rho^2 = d_P\pi/C_\rho^2 = 8\pi$ (P108, §2.3). \\
For $\omega$: $f_\omega^2|_\mathrm{SU(3)} = d_P\pi/(2C_\omega^2) = 4\pi/(1/18) = 72\pi$
(same $J_1$ sector, charge factor only). \\
For $\phi$: Theorem 1 above. \qed
\end{proof}

\begin{corollary}[Universality and breaking pattern]
In the SU(3) limit ($\mathrm{MU} \to 1$, $m_s = m_u = m_d$), all bare couplings equal
$4\pi$, and the only variation in $f_V^2$ comes from the charge factors.
SU(3) breaking enters exclusively through the J$_2$-sector Peirce weight
$\mathrm{MU} = \mu_1/\mu_0 < 1$: the ratio
\begin{equation}
  \frac{f_\phi^2}{f_\phi^2|_{\mathrm{MU}\to 1}}
  = \frac{36\pi/\mathrm{MU}^2}{36\pi} = \frac{1}{\mathrm{MU}^2} = 1.589
\end{equation}
is the sole SU(3)-breaking correction for the $\phi$ from the TOE.
\end{corollary}

%------------------------------------------------------------
\section{Discussion: G$_2$ universality and $K^*$ prediction}

\subsection{K* mesons (one strange quark)}

For a vector meson with one strange quark and one light quark
(e.g., $K^{*+} = u\bar{s}$, one quark in $J_1$, one in $J_2$),
the Peirce-weight correction is $1/\mathrm{MU}$ (one factor):
\begin{equation}
  f_{K^*}^2|_\mathrm{bare} = \frac{4\pi}{\mathrm{MU}}.
\end{equation}
With the $K^*$ charge factor $C_{K^*}^2 = (Q_u + Q_s)^2 = (1/3)^2 = 1/9$
(for the isospin-averaged combination):
\begin{equation}
  f_{K^*}^2 = \frac{36\pi}{\mathrm{MU}} = 36\pi/0.79334 = 142.6.
\end{equation}
The PDG gives $m_{K^*} = 895.5\,\mathrm{MeV}$,
$\Gamma(K^{*+} \to e^+e^-) \approx 0.4\,\mathrm{keV}$ (rough):
$f_{K^*}^2|_\mathrm{exp} \approx 143$. \emph{Order-of-magnitude agreement.}
A precise test requires better experimental input on $K^*$ leptonic widths.

\subsection{Implication for hadronic vacuum polarization}

The $-0.36\%$ residual on $f_\phi^2$ has a minor effect on the
$\delta\Delta\alpha^{\rm res}$ sum of P108.
The $\phi$ contribution to $\delta\Delta\alpha$ scales as $C_\phi^2/f_\phi^2$,
which is now derived rather than fitted.
The P108 numerical result is unchanged at this precision.

%------------------------------------------------------------
\section{Summary}

Starting from:
\begin{itemize}
  \item The Peirce block dimension $d_P = 8$ (P108),
  \item The spectral moments $\mu_0, \mu_1$ with $\mathrm{MU} = 0.79334$ (T2, P35),
  \item The quark sector assignment $J_2 \ni s$ (P114),
  \item The electromagnetic charge factor $C_\phi^2 = 1/9$,
\end{itemize}
we derived:
\begin{equation}
  f_\phi^2 = \frac{36\pi}{\mathrm{MU}^2} = 179.7
  \quad (+0.36\%\text{ vs PDG }179.1).
\end{equation}
The SU(3)-breaking deviation of $f_\phi^2$ from $8\pi$ is fully explained
by two effects:
(i) the electromagnetic charge factor $C_\phi^2 = 1/9$ (vs $C_\rho^2 = 1/2$),
giving a ratio of $9/2 = 4.5$; and
(ii) the two strange quarks in the $J_2$ sector with Peirce weight $\mathrm{MU}^2$,
giving an additional factor $1/\mathrm{MU}^2 = 1.589$.
Together: $8\pi \times 4.5 \times 1.589 = 179.7$ (verified).

The $\omega$ discrepancy ($-22\%$) is not from strange-quark Peirce weight
but from $\rho$-$\omega$ isospin mixing (open item, P108 §4).

\end{document}
