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=== Assistant: Wonderful — let’s do the classic pillbox calculation step-by-step, but with the QAT story in the foreground so the phys… === Wonderful — let’s do the classic pillbox calculation step-by-step, but with the QAT story in the foreground so the physics stays intuitive. I’ll keep the math direct and the geometry explicit (use h/2πh/2\pih/2π where relevant). At the end I’ll summarize how this ties to Faraday and the Third Law in QAT language. ==== • A single spherical electron shell sits at radius r=r0r=r_0r=r0. ==== • Bulk EM potential Aμ(x)A_\mu(x)Aμ(x) with field-strength Fμν=∂μAν−∂νAμF_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\muFμν=∂μAν−∂νAμ. Bulk Maxwell action is unchanged. • Boundary spinor Ψ(t,Ω)\Psi(t,\Omega)Ψ(t,Ω) lives on the shell; Noether U(1) gives a surface current jsurfα(t,Ω)j^\alpha_{\rm surf}(t,\Omega)jsurfα(t,Ω) (indices α\alphaα run over shell coordinates (t,θ,φ)(t,\theta,\varphi)(t,θ,φ)). That surface current is what appears in the JμAμJ_\mu A^\muJμAμ coupling: Sint=∫Sr02d3σ Aα jsurfα.S_{\rm int}=\int_{S^2_{r_0}} d^3\sigma\;A_\alpha\,j^\alpha_{\rm surf}.Sint=∫Sr02d3σAαjsurfα. • In 3+1 bulk coordinates the surface source appears as a delta layer: Jμ(x) = Jbulkμ(x)+δ(r−r0) jsurfμ(t,Ω).J^\mu(x) \;=\; J^\mu_{\rm bulk}(x) + \delta(r-r_0)\,j^\mu_{\rm surf}(t,\Omega).Jμ(x)=Jbulkμ(x)+δ(r−r0)jsurfμ(t,Ω). This is the source we plug into Maxwell’s equations. Maxwell (differential form, SI): ∇⋅E=ρε0,∇×B=μ0J+μ0ε0∂tE\nabla\cdot\mathbf{E}=\frac{\rho}{\varepsilon_0},\qquad \nabla\times\mathbf{B}=\mu_0\mathbf{J}+\mu_0\varepsilon_0\partial_t\mathbf{E}∇⋅E=ε0ρ,∇×B=μ0J+μ0ε0∂tE and the homogeneous pair ∇⋅B=0,∇×E=−∂tB.\nabla\cdot\mathbf{B}=0,\qquad \nabla\times\mathbf{E}=-\partial_t\mathbf{B}.∇⋅B=0,∇×E=−∂tB. ==== Take a tiny cylindrical pillbox that straddles the spherical shell at a chosen point: ==== * Cylinder axis aligned with the local normal n^\hat nn^ (radial direction). * Top face just outside the shell (out''), bottom face just inside (in''). * Let face area be AAA and radial thickness h→0h\to0h→0. The curved side area is proportional to hhh and will vanish as h→0h\to0h→0. We integrate Maxwell’s equations over this pillbox and then take the limit h→0h\to0h→0. ==== Integrate ∇⋅E=ρ/ε0\nabla\cdot\mathbf{E}=\rho/\varepsilon_0∇⋅E=ρ/ε0 over the pillbox volume VVV: ==== ∫V(∇⋅E) dV=1ε0∫Vρ dV.\int_V (\nabla\cdot\mathbf{E})\,dV = \frac{1}{\varepsilon_0}\int_V \rho\,dV.∫V(∇⋅E)dV=ε01∫VρdV. Left side → flux through faces (curved side vanishes as h→0h\to0h→0): Eout ⋅n^ A−Ein ⋅n^ A=1ε0∫Vρ dV.\mathbf{E}_{\rm out}\!\cdot \hat n\,A - \mathbf{E}_{\rm in}\!\cdot \hat n\,A = \frac{1}{\varepsilon_0}\int_V \rho\,dV.Eout⋅n^A−Ein⋅n^A=ε01∫VρdV. The volume integral of ρ\rhoρ picks up the delta layer: ∫Vρ dV=∫V[ρbulk+δ(r−r0)σ(Ω,t)]dV⟶σ(Ω,t) A(assuming no bulk charge inside pillbox).\int_V \rho\,dV = \int_V \big[\rho_{\rm bulk} + \delta(r-r_0)\sigma(\Omega,t)\big] dV \longrightarrow \sigma(\Omega,t)\,A \quad\text{(assuming no bulk charge inside pillbox).}∫VρdV=∫V[ρbulk+δ(r−r0)σ(Ω,t)]dV⟶σ(Ω,t)A(assuming no bulk charge inside pillbox). Therefore, dividing by AAA: n^ ⋅(Eout−Ein) = σ(θ,φ,t)ε0 (Gauss jump)\boxed{\;\hat n\!\cdot(\mathbf{E}_{\rm out}-\mathbf{E}_{\rm in}) \;=\; \frac{\sigma(\theta,\varphi,t)}{\varepsilon_0}\; } \tag{Gauss jump}n^⋅(Eout−Ein)=ε0σ(θ,φ,t)(Gauss jump) This is the exact relation: change in normal EEE equals surface charge density (ledger entry: surface charge). QAT interpretation: σ\sigmaσ is the surface-charge density produced by absorbed photon energy changing the boundary occupation. It is precisely the quantity recorded in the surface ledger. ==== Integrate ∇×B=μ0J+μ0ε0∂tE\nabla\times\mathbf{B}=\mu_0\mathbf{J}+\mu_0\varepsilon_0\partial_t\mathbf{E}∇×B=μ0J+μ0ε0∂tE over the pillbox volume and apply Stokes’ theorem to the tiny loop that runs around the pillbox edge. More simply, take the tangential components and integrate across the small height: ==== In the limit h→0h\to0h→0 the dominant contribution is from the tangential integral across the top and bottom faces. One obtains n^×(Bout−Bin)=μ0K∥(θ,φ,t)+μ0ε0n^×(Eout−Ein)⋅(order h).\hat n\times(\mathbf{B}_{\rm out}-\mathbf{B}_{\rm in}) = \mu_0 \mathbf{K}_\parallel(\theta,\varphi,t) + \mu_0\varepsilon_0 \hat n\times \big(\mathbf{E}_{\rm out}-\mathbf{E}_{\rm in}\big) \cdot(\text{order }h).n^×(Bout−Bin)=μ0K∥(θ,φ,t)+μ0ε0n^×(Eout−Ein)⋅(order h). The displacement term is order hhh and drops out as h→0h\to0h→0. Hence, n^×(Bout−Bin) = μ0 j∥(θ,φ,t) (Ampeˋre jump)\boxed{\;\hat n\times(\mathbf{B}_{\rm out}-\mathbf{B}_{\rm in}) \;=\; \mu_0\,\mathbf{j}_\parallel(\theta,\varphi,t)\; } \tag{Ampère jump}n^×(Bout−Bin)=μ0j∥(θ,φ,t)(Ampeˋre jump) where j∥\mathbf{j}_\parallelj∥ is the tangential surface current (the spatial part of jsurfμj^\mu_{\rm surf}jsurfμ). QAT interpretation: j∥\mathbf{j}_\parallelj∥ is produced by charge motion along the spherical boundary after absorption (decoherence and redistribution); it is the agent that radiates (emission) and rearranges local field lines. ==== The 4D conservation law ∂μJμ=0\partial_\mu J^\mu=0∂μJμ=0 with Jμ=Jbulkμ+δ(r−r0)jsurfμJ^\mu = J^\mu_{\rm bulk} + \delta(r-r_0) j^\mu_{\rm surf}Jμ=Jbulkμ+δ(r−r0)jsurfμ implies, after integrating across the thin radial interval, the surface continuity equation: ==== ∂tσ(θ,φ,t)+∇∥ ⋅j∥(θ,φ,t) = − n^ ⋅Jbulk∣r=r0 (Surface continuity)\boxed{\; \partial_t \sigma(\theta,\varphi,t) + \nabla_{\parallel}\!\cdot \mathbf{j}_\parallel(\theta,\varphi,t) \;=\; -\,\hat n\!\cdot \mathbf{J}_{\rm bulk}\big|_{r=r_0}\; } \tag{Surface continuity}∂tσ(θ,φ,t)+∇∥⋅j∥(θ,φ,t)=−n^⋅Jbulkr=r0(Surface continuity) Right-hand side is the normal bulk current flux into the shell (positive for absorption). This is the accounting equation: change of surface charge = net tangential outflow + normal inflow. QAT read: an arriving photon contributes a normal flux term n^⋅Jbulk\hat n\cdot\mathbf{J}_{\rm bulk}n^⋅Jbulk which increases σ\sigmaσ (absorption); later, surface currents j∥\mathbf{j}_\parallelj∥ can redistribute or radiate energy away (emission). ==== Start from Poynting theorem in differential form: ==== ∂tuem+∇⋅S=−E⋅Jbulk.\partial_t u_{\rm em} + \nabla\cdot\mathbf{S} = -\mathbf{E}\cdot\mathbf{J}_{\rm bulk}.∂tuem+∇⋅S=−E⋅Jbulk. Integrate over the pillbox and take h→0h\to0h→0. The integral of ∇⋅S\nabla\cdot\mathbf{S}∇⋅S becomes the net flux through the top and bottom faces: n^⋅Sout−n^⋅Sin=−∫VE⋅J dV−∫V∂tuem dV.\hat n\cdot\mathbf{S}_{\rm out} - \hat n\cdot\mathbf{S}_{\rm in} = -\int_V \mathbf{E}\cdot\mathbf{J}\,dV - \int_V \partial_t u_{\rm em}\,dV.n^⋅Sout−n^⋅Sin=−∫VE⋅JdV−∫V∂tuemdV. In the thin-shell limit, define usurfu_{\rm surf}usurf as the surface energy density stored in the shell degrees of freedom. Then the balance reduces to the surface energy equation (we wrote before): ∂tusurf=−n^⋅S∣r=r0−E∥⋅j∥+Qint (Surface energy)\boxed{\;\partial_t u_{\rm surf} = -\hat n\cdot\mathbf{S}\big|_{r=r_0} - \mathbf{E}_\parallel\cdot\mathbf{j}_\parallel + \mathcal{Q}_{\rm int}\;} \tag{Surface energy}∂tusurf=−n^⋅Sr=r0−E∥⋅j∥+Qint(Surface energy) This is the ledger: Poynting flux into the shell plus work by tangential fields changes the shell’s internal energy (absorption); emission is the reverse. ==== Imagine an incoming photon wavefront strikes the shell locally: ==== # Poynting flux n^⋅S>0\hat n\cdot\mathbf{S}>0n^⋅S>0 into the pillbox increases usurfu_{\rm surf}usurf. # The continuity equation records a positive normal flux −n^⋅Jbulk∣r0-\hat n\cdot\mathbf{J}_{\rm bulk}\big|_{r_0}−n^⋅Jbulkr0 which increases σ\sigmaσ. # The Gauss jump (3) updates the normal EEE accordingly; the Ampère jump (4) relates any induced surface current to a change in tangential BBB. # Later, the shell may re-emit: surface currents arrange so that outgoing Poynting flux is negative (energy leaves), and σ\sigmaσ falls consistent with the continuity equation. Every step follows the pillbox equations — Maxwell is never altered, only the localized source terms (the ledger entries) change. ==== • Faraday: flux changes (local ∂tB\partial_t\mathbf{B}∂tB produced by surface current changes) induce tangential electric fields around loops on the sphere (Faraday’s law). The pillbox tells you exactly how a tiny patch’s surface current and charge change the field-lines passing through that patch — the precise, quantitative realization of Faraday’s lines of force. ==== • Third Law (QAT read): if microscopic transitions freeze (no normal flux, n^⋅Jbulk→0\hat n\cdot\mathbf{J}_{\rm bulk}\to0n^⋅Jbulk→0, and j∥→0\mathbf{j}_\parallel\to0j∥→0), then ∂tσ=0\partial_t\sigma=0∂tσ=0 and ∂tusurf=0\partial_t u_{\rm surf}=0∂tusurf=0. The pillbox equations then reduce to static jump conditions and no Poynting exchange — the dynamical ledger stops being written. That’s the “locking” picture. ==== Pillbox jump/ledger equations (exact): ==== n^ ⋅(Eout−Ein)=σε0 ,n^×(Bout−Bin)=μ0j∥ \boxed{\;\hat n\!\cdot(\mathbf{E}_{\rm out}-\mathbf{E}_{\rm in})=\frac{\sigma}{\varepsilon_0}\;,\qquad \hat n\times(\mathbf{B}_{\rm out}-\mathbf{B}_{\rm in})=\mu_0\mathbf{j}_\parallel\;}n^⋅(Eout−Ein)=ε0σ,n^×(Bout−Bin)=μ0j∥ ∂tσ+∇∥ ⋅j∥=− n^ ⋅Jbulk∣r=r0 \boxed{\;\partial_t\sigma + \nabla_{\parallel}\!\cdot\mathbf{j}_\parallel = -\,\hat n\!\cdot\mathbf{J}_{\rm bulk}\big|_{r=r_0}\;}∂tσ+∇∥⋅j∥=−n^⋅Jbulkr=r0 ∂tusurf=−n^ ⋅S∣r=r0−E∥⋅j∥+Qint \boxed{\;\partial_t u_{\rm surf} = -\hat n\!\cdot\mathbf{S}\big|_{r=r_0} - \mathbf{E}_\parallel\cdot\mathbf{j}_\parallel + \mathcal{Q}_{\rm int}\;}∂tusurf=−n^⋅Sr=r0−E∥⋅j∥+Qint Each equation follows from integrating Maxwell (and Poynting) over the little pillbox that straddles the shell; the delta-supported surface current jsurfμj^\mu_{\rm surf}jsurfμ comes from the JμAμJ_\mu A^\muJμAμ boundary coupling (Noether current of the boundary action). Maxwell’s differential equations keep their form — QAT’s content sits inside the surface source terms and the dynamical rules that update σ\sigmaσ, j∥\mathbf{j}_\parallelj∥, and usurfu_{\rm surf}usurf. If you’d like next I can: • write this derivation as a LaTeX snippet you can paste into your notes or video description, or • make a simple labeled diagram (pillbox crossing a spherical shell with arrows showing n^⋅S\hat n\cdot\mathbf{S}n^⋅S, j∥\mathbf{j}_\parallelj∥, normal flux) suitable for a slide. Which do you prefer?
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