Exact ideal emitter mass ODE and entropy accounting

Assume the exact Planck energy and coarse entropy flux formulas of Actual photon box trace and Planck energy/entropy flux. Fix model constants $G,c,\mathrm{\hslash },k_B>0$ in SI units, with units respectively ${\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, $\mathrm{m}{\mathrm{s}}^{-1}$, $\mathrm{J}\mathrm{s}$, and $\mathrm{J}{\mathrm{K}}^{-1}$. Fix a normalization of the asymptotic time and energy reference, a dimensionless fraction $0<\eta \le 1$, an initial mass $M_0>0$ in kilograms, a cutoff $0<M_{\min }<M_0$, and an entropy offset $S_0\in \mathbb{R}$ in joules per kelvin. Let $M(t)>0$ be a $C^1$ mass function on an interval containing $0$ and its first cutoff time, with $M(0)=M_0$. For each $M>0$ define the model quantities $r_H=\frac{2GM}{c^2},A_H=4\pi r_H^2,\kappa =\frac{c^2}{4GM},T_{\mathrm{\infty }}=\frac{\mathrm{\hslash }c\kappa }{2\pi k_B},S_B(M)=\frac{k_B{c}^3{A}_H}{4G\mathrm{\hslash }}+S_0\mathrm{.}$ Here $r_H$ is in metres, $A_H$ in square metres, $\kappa$ is the fixed-reference inverse-length parameter in inverse metres, $T_{\mathrm{\infty }}$ is in kelvins, and $S_0$ is a fixed entropy offset in joules per kelvin. Adopt an effective far-field emitter with area $A_{\mathrm{e}\mathrm{f}\mathrm{f}}=\eta A_H$ and power $P=A_{\mathrm{e}\mathrm{f}\mathrm{f}}\sigma T_{\mathrm{\infty }}^4$ in watts, where $\sigma$ is the coefficient in Actual photon box trace and Planck energy/entropy flux. Adopt the mass-balance equation $c^2{M}^{\mathrm{\prime }}(t)=-P(t)$, with $t$ in seconds measured by the fixed asymptotic time parameter; terminate the model at the first time $M(t)=M_{\min }$. Then the exact solution until that cutoff is $M(t)=(M_0^3-3\alpha t{)}^{1\mathrm{/}3},\alpha =\frac{\eta \mathrm{\hslash }{c}^4}{15360\pi G^2},\tau =\frac{M_0^3-M_{\min }^{}}{3\alpha }\mathrm{.}$ The parameter $\alpha$ has units $\mathrm{k}{\mathrm{g}}^3{\mathrm{s}}^{-1}$. The model exports energy $c^2(M_0-M_{\min })$ in joules through cutoff. Define its coarse exported entropy by $S_{\mathrm{r}\mathrm{a}\mathrm{d}}(t)={\int }_0^t(4\mathrm{/}3)P(t^{\mathrm{\prime }})\mathrm{/}{T}_{\mathrm{\infty }}(t^{\mathrm{\prime }})d{t}^{\mathrm{\prime }}$, in joules per kelvin. Then $S_B(M_0)-S_B(M_{\min })>0.$ $S_{\mathrm{r}\mathrm{a}\mathrm{d}}(\tau )=\frac{4}{3}[S_B(M_0)-S_B(M_{\min })]\mathrm{.}$ $\mathrm{\Delta }(S_B+S_{\mathrm{r}\mathrm{a}\mathrm{d}})=\frac{1}{3}[S_B(M_0)-S_B(M_{\min })]\ge 0,$ where $\mathrm{\Delta }(S_B+S_{\mathrm{r}\mathrm{a}\mathrm{d}})=S_B(M_{\min })+S_{\mathrm{r}\mathrm{a}\mathrm{d}}(\tau )-S_B(M_0)$ and $S_{\mathrm{r}\mathrm{a}\mathrm{d}}(0)=0$. This is an exact conditional calculation in the fixed-reference effective-area model. It is not a theorem about a material patch of a null horizon, an exact time-dependent Einstein solution, quantum backreaction, reference-independent Stefan emission, or a universal generalized second law.