Editing Openai/6897769e-4ee4-800f-aba5-69cca34f701c (section)

=== Consider a thin charged spherical shell of radius RRR and total charge QQQ. The electric field outside the shell is E(r)=Q4πε0r2E(r)=\dfrac{Q}{4\pi\varepsilon_0 r^2}E(r)=4πε0r2Q for r≥Rr\ge Rr≥R; inside E=0E=0E=0. The electromagnetic energy stored in the field is ===

UEM  =  ∫R∞12ε0E2⋅4πr2 dr  =  ∫R∞12ε0(Q4πε0r2)24πr2 dr.U_{\rm EM} \;=\; \int_{R}^{\infty}\tfrac{1}{2}\varepsilon_0 E^2 \cdot 4\pi r^2\,dr
\;=\; \int_{R}^{\infty}\tfrac{1}{2}\varepsilon_0 \Big(\frac{Q}{4\pi\varepsilon_0 r^2}\Big)^2 4\pi r^2\,dr.UEM=∫R∞21ε0E2⋅4πr2dr=∫R∞21ε0(4πε0r2Q)24πr2dr.
Evaluate the integral:

UEM  =  Q28πε0  ∫R∞1r2 dr  =  Q28πε0R.U_{\rm EM} \;=\; \frac{Q^2}{8\pi\varepsilon_0}\;\int_R^\infty\frac{1}{r^2}\,dr
\;=\; \frac{Q^2}{8\pi\varepsilon_0 R}.UEM=8πε0Q2∫R∞r21dr=8πε0RQ2.
Set this electromagnetic energy equal to the rest energy mc2m c^2mc2. For an electron, Q=eQ=eQ=e and m=mem=m_em=me:

mec2  =  e28πε0R⇒R  =  e28πε0mec2.m_e c^2 \;=\; \frac{e^2}{8\pi\varepsilon_0 R}
\quad\Rightarrow\quad 
R \;=\; \frac{e^2}{8\pi\varepsilon_0 m_e c^2}.mec2=8πε0Re2⇒R=8πε0mec2e2.
The classical electron radius is usually defined as

re  =  e24πε0mec2.r_e \;=\; \frac{e^2}{4\pi\varepsilon_0 m_e c^2}.re=4πε0mec2e2.
So the thin shell model gives

R=re2.\boxed{R = \frac{r_e}{2}.}R=2re.
So: a purely classical thin charged shell whose EM self-energy accounts for the entire rest mass would have radius re/2r_e/2re/2. That exactly produces the factor 1/2 you’ve been thinking about. (Historically classical EM mass models like Lorentz–Abraham ran into problems, but the numerical half-factor is real for this simple thin-shell model.)

Why this matters for QAT: your geometric half-radius idea matches a classical electromagnetic self-energy calculation. That is a strong hint that boundary geometry (a shell) naturally produces a factor ½ in energetics/scale.