Actual photon box trace and Planck energy/entropy flux

The number of integer vectors with $\mathrm{\mid }n{\mathrm{\mid }}_{\mathrm{\infty }}=j$ is $(2j+1{)}^3-(2j-1{)}^3=24j^2+2$. On this shell $j\le \mathrm{\mid }n\mathrm{\mid }\le \sqrt{3}j$, so $q_n\le e^{-\beta aj}$ and ${ϵ}_n^p{q}_n\le (a\sqrt{3}{)}^p{j}^p{e}^{-\beta aj}$. The shell contributions are bounded by $C_p{j}^{p+2}{e}^{-\beta aj}$ for a finite constant $C_p$; the ratio of consecutive terms tends to $e^{-\beta a}<1$, so their series converges.

Since $\mathrm{\mid }n\mathrm{\mid }\ge 1$ on the nonzero lattice and equality occurs, $\sup q_n=e^{-\beta a}<1$; the $p=0$ case of step 1.1 gives $\sum q_n<\mathrm{\infty }$, and each $p\ge 0$ gives the asserted energy-weighted sum.

For every $n\mathrm{\ne }0$, ${ϵ}_n\mathrm{/}(e^{\beta {ϵ}_n}-1)={ϵ}_n{q}_n\mathrm{/}(1-q_n)\le (1-e^{-\beta a}{)}^{-1}{ϵ}_n{q}_n$. Summing and applying step 1.1 with $p=1$ proves finiteness; a second polarization multiplies each nonnegative sum by two.

On $[-\frac{1}{2},\frac{1}{2}]$ define the continuous periodic representative $F(y)=(2\pi y{)}^2$. Its normalized Fourier coefficients are $c_n=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }{x}^2{e}^{-inx}dx$ after the change $x=2\pi y$. Direct integration gives $c_0={\pi }^2\mathrm{/}3$; for $n\mathrm{\ne }0$, two integrations by parts give $c_n=2(-1{)}^n\mathrm{/}{n}^2$ because the first boundary term cancels at $x=\pm \pi$ and ${\int }_{-\pi }^{\pi }{e}^{-inx}dx=0$.

The normalized squared norm is $(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }{x}^{4}dx={\pi }^4\mathrm{/}5$. Under the stated Countable Choice assumption [F1], Parseval [F2] therefore gives ${\pi }^4\mathrm{/}5=\mathrm{\mid }{c}_0{\mathrm{\mid }}^2+{\sum }_{n\mathrm{\ne }0}\mathrm{\mid }{c}_n{\mathrm{\mid }}^2={\pi }^4\mathrm{/}9+8{\sum }_{n=1}^{\mathrm{\infty }}{n}^{-4}$, whence ${\sum }_{n=1}^{\mathrm{\infty }}{n}^{-4}={\pi }^4\mathrm{/}90$.

For $m\ge 1$ and $p=0,1,2,3$, put $J_p(m)={\int }_0^{\mathrm{\infty }}{x}^p{e}^{-mx}dx$. Direct integration gives $J_0(m)=1\mathrm{/}m$; integration by parts on $[0,R]$ and then $R\to \mathrm{\infty }$ gives $J_p(m)=p{J}_{p-1}(m)\mathrm{/}m$, since $R^p{e}^{-mR}\to 0$ (for example, $e^{mR}\ge (mR{)}^{p+1}\mathrm{/}(p+1)!$). Hence $J_3(m)=6\mathrm{/}{m}^4$ and $J_2(m)=2\mathrm{/}{m}^3$. The finite geometric identity and its limit give $(e^x-1{)}^{-1}={\sum }_{m\ge 1}{e}^{-mx}$ for $x>0$. The partial sums $x^3{\sum }_{m\le N}{e}^{-mx}$ are nonnegative measurable and increase pointwise; the monotone-convergence theorem [F3] for the integral defined in [F4] gives the first integral as ${\sum }_{m\ge 1}{J}_3(m)=6{\sum }_{m\ge 1}{m}^{-4}={\pi }^4\mathrm{/}15$.

For $0<z<1$, the geometric series is uniformly convergent on $[0,z]$, so integrating its partial sums gives $-\log (1-z)={\sum }_{m\ge 1}{z}^m\mathrm{/}m$. At $z=e^{-x}$ the nonnegative partial sums $x^2{\sum }_{m\le N}{e}^{-mx}\mathrm{/}m$ increase pointwise to the second integrand. Applying [F3] to the integral defined in [F4] and using step 3.1 gives its integral as ${\sum }_{m\ge 1}{J}_2(m)\mathrm{/}m=2{\sum }_{m\ge 1}{m}^{-4}={\pi }^4\mathrm{/}45$. The integral ratio is therefore $1\mathrm{/}3$.

For a continuous compactly supported $h:{\mathbb{R}}^3\to \mathbb{R}$, the cubes $Q_{n,d}=dn+[-d\mathrm{/}2,d\mathrm{/}2{]}^3$ tile ${\mathbb{R}}^3$ up to boundaries of volume zero. Uniform continuity of $h$ bounds $\mathrm{\mid }{d}^3{\sum }_{n}h(dn)-\int h\mathrm{\mid }$ by the volume of one fixed enlarged support box times the modulus of continuity at $\sqrt{3}d\mathrm{/}2$, which tends to zero. Taking $h(x)=\psi (\mathrm{\mid }x\mathrm{\mid })$ proves the compactly supported radial lattice limit. To rewrite its integral when $\psi$ is signed, apply [F2] separately to the nonnegative radial functions $h_{+}(x)=\max (h(x),0)$ and $h_{-}(x)=\max (-h(x),0)$; both are continuous and compactly supported, so their integrals are finite. Subtracting the resulting identities and using $\sigma (S^2)=4\pi$ from [F3] gives ${\int }_{{\mathbb{R}}^3}h(x)d^{3}x=4\pi {\int }_0^{\mathrm{\infty }}{r}^2\psi (r)dr$. Omitting $n=0$ changes the sum by $d^3\psi (0)\to 0$.

For $0<r\le 1$, $f(r)\le 1$ and $1-e^{-r}\ge r\mathrm{/}2$, so $0\le g(r)\le C(1+\mathrm{\mid }\log r\mathrm{\mid })$. Formulae [F2] and [F3] give the small-ball integral bound because, for $0<\delta \le 1$, ${\int }_0^{\delta }{r}^2(1+\mathrm{\mid }\log r\mathrm{\mid })dr={\delta }^3(4\mathrm{/}9+\mathrm{\mid }\log \delta \mathrm{\mid }\mathrm{/}3)$. For the discrete sum with $0<d\mathrm{\mid }n\mathrm{\mid }\le \delta$, put $j=\mathrm{\mid }n{\mathrm{\mid }}_{\mathrm{\infty }}$; then $dj\le \delta$ and $d\mathrm{\mid }n\mathrm{\mid }\ge dj$. Group the shells by $dj\in (\delta 2^{-\mathrm{\ell }-1},\delta 2^{-\mathrm{\ell }}]$. If such a group is nonempty, then $d\le \delta 2^{-\mathrm{\ell }}$ and its shell indices obey $j\le J_{\mathrm{\ell }}:=\lfloor \delta 2^{-\mathrm{\ell }}\mathrm{/}d\rfloor$. By [F1], the number of lattice points in shells up to $J_{\mathrm{\ell }}$ is ${\sum }_{j\le J_{\mathrm{\ell }}}(24j^2+2)\le C{J}_{\mathrm{\ell }}^3\le C(\delta 2^{-\mathrm{\ell }}\mathrm{/}d{)}^3$. Every point in the selected small-ball subset of this group has $d\mathrm{\mid }n\mathrm{\mid }\ge dj>\delta 2^{-\mathrm{\ell }-1}$ and $d\mathrm{\mid }n\mathrm{\mid }\le \delta \le 1$, so its entropy value is at most $C(1+\mathrm{\mid }\log \delta \mathrm{\mid }+\mathrm{\ell })$. Multiplication by the cell volume $d^3$ bounds this group's contribution by $C{\delta }^3{2}^{-3\mathrm{\ell }}(1+\mathrm{\mid }\log \delta \mathrm{\mid }+\mathrm{\ell })$. Summing the convergent geometric series over $\mathrm{\ell }\ge 0$ gives $C^{\mathrm{\prime }}{\delta }^3(1+\mathrm{\mid }\log \delta \mathrm{\mid })$ uniformly in $d$.

The function $f$ is continuous at zero since ${\lim }_{r\downarrow 0}r\mathrm{/}(e^r-1)=1$, and for $r\ge 1$ it is bounded by $C(1+r)e^{-r}$. For $0<d\le 1$ and $x\in Q_{n,d}$, $\mathrm{\mid }\mathrm{\mid }dn\mathrm{\mid }-\mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }\le \sqrt{3}\mathrm{/}2$. Thus, for $R\ge 2$, each cube with $\mathrm{\mid }dn\mathrm{\mid }>R$ lies in $\mathrm{\mid }x\mathrm{\mid }>R-\sqrt{3}\mathrm{/}2>1$, and the exponential bound yields ${\sum }_{\mathrm{\mid }dn\mathrm{\mid }>R}{d}^{3}f(d\mathrm{\mid }n\mathrm{\mid })\le C^{\mathrm{\prime }}{\int }_{\mathrm{\mid }x\mathrm{\mid }>R-\sqrt{3}\mathrm{/}2}(1+\mathrm{\mid }x\mathrm{\mid })e^{-\mathrm{\mid }x\mathrm{\mid }}{d}^{3}x$. The integral tail tends to zero as $R\to \mathrm{\infty }$, uniformly in $0<d\le 1$. Apply step 1.1 on compact cutoffs and then this uniform tail bound to obtain the asserted $f$ limit and integrability.

For each fixed $d>0$, only finitely many lattice points have $d\mathrm{\mid }n\mathrm{\mid }<1$, and there are only finitely many shells with $dj<1$. On every remaining shell, $dj\ge 1$, the Euclidean norm satisfies $j\le \mathrm{\mid }n\mathrm{\mid }\le \sqrt{3}j$, and the exponential bound gives $f(d\mathrm{\mid }n\mathrm{\mid }),g(d\mathrm{\mid }n\mathrm{\mid })\le C(1+dj)e^{-dj}$. The shell count [F1] therefore bounds the shell contribution by $C{j}^2(1+dj)e^{-dj}\le C{j}^3(1+dj)e^{-dj}$. The comparison series converges by the ratio test, since the ratio of consecutive terms tends to $e^{-d}<1$. For $r\ge 1$, $-\log (1-e^{-r})\le e^{-r}\mathrm{/}(1-e^{-r})$, so $g(r)\le C(1+r)e^{-r}$; the cube estimate of step 2.1 then gives uniformly vanishing large-radius tails for the integral and the lattice sum. On each annulus $\delta \le \mathrm{\mid }x\mathrm{\mid }\le R$, $g(\mathrm{\mid }x\mathrm{\mid })$ is continuous and the compact Riemann-sum argument of step 1.1 applies with a continuous radial cutoff equal to one on that annulus and zero near the origin and outside a larger ball. Letting $\delta \downarrow 0$ and $R\to \mathrm{\infty }$ using the small-ball bound of step 1.2 and these tail bounds proves the $g$ limit and its finiteness.

For $i=(n,s)$ with $n\mathrm{\ne }0$, $\mathrm{\mid }n\mathrm{\mid }\ge 1$, so ${ϵ}_i\ge 2\pi \mathrm{\hslash }c\mathrm{/}L>0$. Thus $q_i\le e^{-2\pi \beta \mathrm{\hslash }c\mathrm{/}L}<1$, and [F3] gives ${\sum }_{i\in I}{q}_i<\mathrm{\infty }$ with the required strict supremum bound.

Put $d=2\pi \beta \mathrm{\hslash }c\mathrm{/}L$ and $f(r)=r\mathrm{/}(e^r-1)$ with its continuous extension $f(0)=1$. Then ${\omega }_n=d\mathrm{\mid }n\mathrm{\mid }\mathrm{/}(\beta \mathrm{\hslash })$ and ${ϵ}_n=k_{B}Td\mathrm{\mid }n\mathrm{\mid }$. The function $r\mapsto f(r)\psi (r\mathrm{/}(\beta \mathrm{\hslash }))$ is continuous and compactly supported, so the signed radial-limit clause of [F4] gives its lattice sum limit as $4\pi {\int }_0^{\mathrm{\infty }}{r}^{2}f(r)\psi (r\mathrm{/}(\beta \mathrm{\hslash }))dr$. Converting $L^{-3}=(d\mathrm{/}(2\pi \beta \mathrm{\hslash }c){)}^3$ gives the radial integral with coefficient $k_{B}T\mathrm{/}({\pi }^2(\beta \mathrm{\hslash }c{)}^3)$. The substitution $r=\beta \mathrm{\hslash }\omega$ multiplies the integrand by $(\beta \mathrm{\hslash }{)}^4$ and this coefficient becomes $\mathrm{\hslash }\mathrm{/}({\pi }^2{c}^3)$ because $k_{B}T\beta =1$. This is exactly the stated spectral energy measure.

The countable mode set has the specified enumeration, [F1] supplies its bosonic occupation basis and countability, and [F6] supplies exactly ${\mathrm{A}\mathrm{C}}_{\omega }$. Applying the actual batch-45 draft interface [F2] with $\mu =0$ identifies its self-adjoint diagonal multiplier and domain with $H$ and $D(H)$ here. Its trace conclusion gives $Z_L={\prod }_{n\mathrm{\ne }0}(1-q_n{)}^{-2}<\mathrm{\infty }$ and the positive trace-one state with diagonal probabilities $p_m$. The same hypotheses hold at every inverse temperature $\gamma >0$, in particular at $\gamma =\beta \mathrm{/}2$, by [F3].

Using $f$ from step 1.2, $u_L=2k_{B}T[(2\pi {)}^3(\beta \mathrm{\hslash }c{)}^3{]}^{-1}{d}^3{\sum }_{n\mathrm{\ne }0}f(d\mathrm{\mid }n\mathrm{\mid })$. By [F4] the limit is $2k_{B}T[(2\pi {)}^3(\beta \mathrm{\hslash }c{)}^3{]}^{-1}{\int }_{{\mathbb{R}}^3}f(\mathrm{\mid }x\mathrm{\mid })d^{3}x$. Formula [F7] makes this $k_{B}T[{\pi }^2(\beta \mathrm{\hslash }c{)}^3{]}^{-1}{\int }_0^{\mathrm{\infty }}{r}^3\mathrm{/}(e^r-1)dr$. Applying [F5] and $\beta =(k_{B}T{)}^{-1}$ gives $u={\pi }^2{k}_B^4{T}^4\mathrm{/}(15{\mathrm{\hslash }}^3{c}^3)$.

Let $J_N$ exhaust the enumerated modes by finite initial subsets and set $E_N(m)={\sum }_{i\in J_N}{ϵ}_i{m}_i$. By [F2], finite-sum linearity and its one-mode means give ${\sum }_m{p}_m{E}_N(m)={\sum }_{i\in J_N}{ϵ}_i{q}_i\mathrm{/}(1-q_i)$. The nonnegative configuration sum is the supremum over finite sets of configurations; every finite set of finite-support configurations is supported in some $J_N$, so taking the supremum over $N$ yields $U_L={\sum }_m{p}_{m}E(m)=2{\sum }_{n\mathrm{\ne }0}{ϵ}_{(n,1)}{q}_n\mathrm{/}(1-q_n)$. Since $q_i\le q_{\ast }<1$, [F3] bounds this by $(1-q_{\ast }{)}^{-1}{\sum }_i{ϵ}_i{q}_i<\mathrm{\infty }$. For every integer $r\ge 1$, $y^r{e}^{-\beta y}\le C_{r,\beta }{e}^{-\beta y\mathrm{/}2}$ for $y\ge 0$, where $C_{r,\beta }={\sup }_{y\ge 0}{y}^r{e}^{-\beta y\mathrm{/}2}<\mathrm{\infty }$. Summing and using $Z_L(\beta \mathrm{/}2)<\mathrm{\infty }$ from step 2.1 gives ${\sum }_m{p}_{m}E(m{)}^r<\mathrm{\infty }$; the zeroth moment equals one. Put $b_m=p_{m}E(m)\ge 0$. Their finite sum $U_L$ bounds ${\sup }_m{b}_m$, so $B_L$ is bounded on all ${\mathrm{\ell }}^2$. Its finite diagonal truncations have finite rank, and their operator-norm errors are bounded by the summable tails ${\sum }_{m\text{ omitted}}{b}_m\to 0$. Thus [F9] makes $B_L$ compact, and its summable nonnegative diagonal eigenvalues make it positive trace class with trace ${\sum }_m{b}_m=U_L$. On $D(H)$ its coordinate action is ${\rho }_{L}H$; the finite-support vectors are dense, so the bounded extension is unique. Moreover ${\sup }_{m}E(m)p_m<\mathrm{\infty }$ implies ${\sum }_{m}E(m{)}^2{p}_m^2\mathrm{\mid }{a}_m{\mathrm{\mid }}^2<\mathrm{\infty }$ for every $a\in {\mathrm{\ell }}^2$. Hence ${\rho }_{L}a\in D(H)$ and $H{\rho }_L=B_L$ everywhere. This justifies exactly the trace notation in the Statement.

The vacuum contributes one to $Z_L$, so $Z_L\ge 1$, and $-\log p_m=\log Z_L+\beta E(m)\ge 0$. Also $-\log (1-q_i)\le q_i\mathrm{/}(1-q_{\ast })$, so [F3] makes the logarithmic mode sum finite and the logarithm of the convergent product equals $2{\sum }_{n\mathrm{\ne }0}-\log (1-q_n)$. Summing $-p_m\log p_m$ using ${\sum }_m{p}_m=1$ and the finite mean from step 3.1 gives $S_L=k_B(\log Z_L+\beta U_L)$. Substituting the product logarithm and the mode-mean sum gives the displayed finite positive series.

Set $g(r)=f(r)-\log (1-e^{-r})$. The exact entropy series in step 4.1 gives $s_L=2k_B[(2\pi {)}^3(\beta \mathrm{\hslash }c{)}^3{]}^{-1}{d}^3{\sum }_{n\mathrm{\ne }0}g(d\mathrm{\mid }n\mathrm{\mid })$. The singular-sum result [F4] proves convergence despite the logarithmic singularity at zero. Formula [F7] turns the limiting integral into $4\pi$ times the sum of the two integrals in [F5]; their ratio is $1\mathrm{/}3$. Thus $s=(4\mathrm{/}3)k_B^4{T}^3{\pi }^2\mathrm{/}(15{\mathrm{\hslash }}^3{c}^3)=4u\mathrm{/}(3T)$.

By the angular surface element in [F8], the outward-hemisphere integral of the normal velocity factor is ${\int }_0^{2\pi }{\int }_0^{\pi \mathrm{/}2}\cos \theta \sin \theta d\theta d\phi =\pi$; the full sphere has measure $4\pi$, so the isotropic average of the positive normal component is $1\mathrm{/}4$. Under the explicitly defined angular densities in [A2], the one-sided fluxes are $F=cu\mathrm{/}4$ and $F_S=cs\mathrm{/}4$. Substituting the limits from steps 2.2 and 5.1 gives the stated $\sigma$, $F=\sigma T^4$, and $F_S=(4\mathrm{/}3)F\mathrm{/}T$. These are deductions in the defined model only.

Substitution of $r_H=2GM\mathrm{/}{c}^2$ gives $A_H=16\pi G^2{M}^2\mathrm{/}{c}^4$, $\kappa =c^2\mathrm{/}(4GM)$, $T_{\mathrm{\infty }}=\mathrm{\hslash }{c}^3\mathrm{/}(8\pi GM{k}_B)$, and $S_B(M)=4\pi k_{B}G{M}^2\mathrm{/}(\mathrm{\hslash }c)+S_0$. Differentiating the last expression yields $T_{\mathrm{\infty }}d{S}_B\mathrm{/}dM=c^2$.

By [F1] the effective-area power is $P=\eta A_H\sigma T_{\mathrm{\infty }}^4=\eta \mathrm{\hslash }{c}^6\mathrm{/}(15360\pi G^2{M}^2)$. The adopted balance equation [A1] is therefore $M^{\mathrm{\prime }}=-\alpha M^{-2}$ with $\alpha =\eta \mathrm{\hslash }{c}^4\mathrm{/}(15360\pi G^2)>0$.

Along any positive solution, $(M^3{)}^{\mathrm{\prime }}=3M^2{M}^{\mathrm{\prime }}=-3\alpha$, hence the initial condition forces $M(t{)}^3=M_0^3-3\alpha t$. The positive cube root is differentiable and satisfies the ODE until it reaches the positive cutoff, giving the stated solution and time $\tau =(M_0^3-M_{\min }^{})\mathrm{/}(3\alpha )$. This also proves uniqueness among positive solutions because every such solution has the same cube.

Integrating $P=-c^2{M}^{\mathrm{\prime }}$ from $0$ to $\tau$ gives ${\int }_0^{\tau }Pdt=c^2(M_0-M_{\min })$. Differentiating $S_B(M(t))$ and using step 1.1 and $M^{\mathrm{\prime }}=-P\mathrm{/}{c}^2$ yields $S_B^{\mathrm{\prime }}=-P\mathrm{/}{T}_{\mathrm{\infty }}$.

The entropy flux in [F1], multiplied by the stated area $A_{\mathrm{e}\mathrm{f}\mathrm{f}}$, is $S_{\mathrm{r}\mathrm{a}\mathrm{d}}^{\mathrm{\prime }}=(4\mathrm{/}3)P\mathrm{/}{T}_{\mathrm{\infty }}$. Thus $(S_B+S_{\mathrm{r}\mathrm{a}\mathrm{d}}{)}^{\mathrm{\prime }}=P\mathrm{/}(3T_{\mathrm{\infty }})\ge 0$ and integration gives the two endpoint entropy formulas. Since $S_B(M)=4\pi k_{B}G{M}^2\mathrm{/}(\mathrm{\hslash }c)+S_0$ has positive quadratic coefficient and $M_0>M_{\min }>0$, its difference is positive and the offset cancels. No physical generalized-second-law conclusion follows beyond this stipulated model.