Actual photon box trace and Planck energy/entropy flux

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.