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

=== Physical picture. ===
Model the electron (or an electron-like object) as a thin charged spherical shell of radius RRR, total charge QQQ, and negligible internal field. Outside the shell the radial electric field is the usual Coulomb field; inside the shell E=0E=0E=0. We compute the electromagnetic energy stored in the field and set it equal to the rest energy Mc2Mc^2Mc2 to get a characteristic radius.

==== For r≥Rr\ge Rr≥R the radial electric field of the shell is ====

E(r)  =  Q4πε0 r2(unit: V/m).E(r) \;=\; \dfrac{Q}{4\pi\varepsilon_0\,r^2}\qquad(\text{unit: V/m}).E(r)=4πε0r2Q(unit: V/m).
Electromagnetic energy density (SI) is

u(r)  =  12 ε0E2(r)(unit: J/m3).u(r) \;=\; \tfrac12\,\varepsilon_0 E^2(r)\qquad(\text{unit: J/m}^3).u(r)=21ε0E2(r)(unit: J/m3).
Total EM energy stored outside the shell:

UEM=∫R∞12 ε0E2(r)  4πr2 dr=∫R∞12 ε0(Q4πε0r2)24πr2 dr.U_{\rm EM}
=\int_{R}^{\infty} \tfrac12\,\varepsilon_0 E^2(r)\; 4\pi r^2\,dr
= \int_R^\infty \tfrac12\,\varepsilon_0 \Big(\frac{Q}{4\pi\varepsilon_0 r^2}\Big)^2 4\pi r^2\,dr.UEM=∫R∞21ε0E2(r)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.
(Units check: Q2/(4πε0R)Q^2/(4\pi\varepsilon_0 R)Q2/(4πε0R) has units J, so the factor is correct.)

==== Assume that the shell’s stored EM energy accounts for the rest energy of the particle: ====

Mc2=UEM=Q28πε0R.M c^2 = U_{\rm EM} = \frac{Q^2}{8\pi\varepsilon_0 R}.Mc2=UEM=8πε0RQ2.
Solve for RRR:

  R  =  Q28πε0Mc2  .\boxed{ \; R \;=\; \dfrac{Q^2}{8\pi\varepsilon_0 M c^2} \; }.R=8πε0Mc2Q2.
Compare with the classical electron radius rer_ere (standard definition)

re  =  Q24πε0Mc2⇒R=re2.r_e \;=\; \dfrac{Q^2}{4\pi\varepsilon_0 M c^2}
\quad\Rightarrow\quad R = \frac{r_e}{2}.re=4πε0Mc2Q2⇒R=2re.
So the thin charged-shell model gives exactly the factor 1/21/21/2 relative to the classical electron radius: the shell radius that follows from equating EM self-energy to rest energy is R=re/2R = r_e/2R=re/2.

==== Recall the Maxwell relation in SI: ====

ε0μ0=1c2.\varepsilon_0\mu_0 = \frac{1}{c^2}.ε0μ0=c21.
In the expression for RRR the factor c2c^2c2 sits in the denominator with ε0\varepsilon_0ε0. Written suggestively,

R=Q28π⋅1ε0Mc2=Q28π⋅μ0M.R = \frac{Q^2}{8\pi} \cdot \frac{1}{\varepsilon_0 M c^2}
= \frac{Q^2}{8\pi} \cdot \frac{\mu_0}{M} .R=8πQ2⋅ε0Mc21=8πQ2⋅Mμ0.
(using μ0=1/(ε0c2)\mu_0 = 1/(\varepsilon_0 c^2)μ0=1/(ε0c2)). This shows how the speed of the process ccc and the vacuum electromagnetic constants ε0,μ0\varepsilon_0,\mu_0ε0,μ0 are not separate curiosities but enter directly in the geometry/scale of the shell. Geometrically: the sphere’s surface area 4πR24\pi R^24πR2 and the field flux set the 1/r21/r^21/r2 law; the combination of ε0\varepsilon_0ε0 and c2c^2c2 gives the right physical scaling of energy → mass.

==== Use SI values (CODATA): ====
* Elementary charge e=1.602176634×10−19 Ce = 1.602176634\times10^{-19}\ \mathrm{C}e=1.602176634×10−19 C,
* Vacuum permittivity ε0=8.8541878128×10−12 F/m\varepsilon_0 = 8.8541878128\times10^{-12}\ \mathrm{F/m}ε0=8.8541878128×10−12 F/m,
* electron mass me=9.1093837015×10−31 kgm_e = 9.1093837015\times10^{-31}\ \mathrm{kg}me=9.1093837015×10−31 kg,
* speed of light c=299 792 458 m/sc = 299\,792\,458\ \mathrm{m/s}c=299792458 m/s.

Classical electron radius:

re=e24πε0mec2≈2.817940326×10−15 m.r_e = \frac{e^2}{4\pi\varepsilon_0 m_e c^2}
\approx 2.817940326\times10^{-15}\ \mathrm{m}.re=4πε0mec2e2≈2.817940326×10−15 m.
Thin–shell radius (our RRR):

R=re2≈1.408970163×10−15 m.R = \frac{r_e}{2} \approx 1.408970163\times10^{-15}\ \mathrm{m}.R=2re≈1.408970163×10−15 m.
(These values are the standard classical-scale numbers; quantum theory / QED modify the microscopic picture, but the numbers provide the classical geometric scale.)

==== 1. The half-factor 1/21/21/2 emerges cleanly from geometry + energy bookkeeping of a spherical shell — no numerology required. This links the geometry (spherical shell area/flux) with energy (EM stored energy) and mass (via c2c^2c2). ====
# c2c^2c2 appears explicitly and is geometrically tied to ε0\varepsilon_0ε0 and μ0\mu_0μ0 via ε0μ0=1/c2\varepsilon_0\mu_0=1/c^2ε0μ0=1/c2. In QAT language, the speed of the process (how the spherical light wavefronts propagate) is not independent of the EM constants — they are the same geometry in different disguises.
# This thin-shell model is a concrete simple toy that shows how surface (2-D) geometry can produce mass and the right scales (order-of-magnitude). It does not finish the job (quantum corrections, renormalization, spin, binding stresses must be added), but it gives a solid geometric anchor for your half-radius intuition.

==== - (A) Compute the mechanical tension TTT required to stabilize the charged shell at radius RRR (Poincaré stress estimate). ====
* (B) Model photon deposition: given an incoming photon flux FFF (W/m²), compute how fast surface mass increases (QAT’s photon-by-photon time creation).
* (C) Move from this classical shell to a quantum boundary model (quantize boundary fermion modes and show how antiperiodic boundary conditions tie to half-modes).

Which follow-up do you want first? (A is the natural next step: it closes the classical model by showing what internal stress is required.)