Collapse ray tracing and greybody separation

Fix an upper endpoint $U_H\in \mathbb{R}$ and a scale $A>0$, both measured in seconds, and let $k>0$ be measured in seconds${}^{-1}$. Let $u\in \mathbb{R}$ be the outgoing coordinate, measured in seconds, and define $U(u)=U_H-A{e}^{-ku}\mathrm{.}$ Then $U$ is a smooth increasing diffeomorphism from $\mathbb{R}$ onto $(-\mathrm{\infty },U_H)$, with derivative $U^{\mathrm{\prime }}(u)=Ak{e}^{-ku}>0$; $U_H$ is only the upper-endpoint label in this channel. Let $\chi >0$ be a fixed conversion factor with units of length per second, and put $\mathrm{\ell }(u)=\chi U(u)$ and ${\mathrm{\ell }}_H=\chi U_H$. For $g\in C_c^{\mathrm{\infty }}(\mathbb{R};\mathbb{C})$, let $u_{\mathrm{\ell }}$ be the smooth inverse of $\mathrm{\ell }$ and define $f_g(x)=g(u_{\mathrm{\ell }}(x))$ for $x<{\mathrm{\ell }}_H$, and $f_g(x)=0$ for $x\ge {\mathrm{\ell }}_H$. The support of $g$ is compact, so $f_g\in C_c^{\mathrm{\infty }}(\mathbb{R};\mathbb{C})$. Define the channel current $j(g)=J(f_g)$ using the Fock-vacuum chiral current $J$, and set $W(g,h)=⟨\mathrm{\Omega },J(\overline{f_g})J(f_h)\mathrm{\Omega }{⟩}_2\mathrm{.}$ The associated two-point distribution is the diffeomorphism pullback of the supplied length-coordinate kernel $C_{\mathrm{\ell }}(X-Y)=-\frac{1}{4\pi }(X-Y-i0{)}^{-2}$; in seconds coordinates its kernel is $K(u,v)=U^{\mathrm{\prime }}(u)U^{\mathrm{\prime }}(v)[-\frac{1}{4\pi }(U(u)-U(v)-i0{)}^{-2}]\mathrm{.}$ Here $i0$ is the lower boundary obtained by testing the regularizations $-\frac{1}{4\pi }(s-iϵ{)}^{-2}$ as $ϵ\downarrow 0$, and the pullback is defined by the smooth diffeomorphism, not by pointwise evaluation on the diagonal. The tested form is of positive type: $W(g,g)\ge 0$ for every $g$.

For real-valued tests $g_1,\dots ,g_n$, write $M_n(g_1,\dots ,g_n)=⟨\mathrm{\Omega },J(f_{g_1})\cdots J(f_{g_n})\mathrm{\Omega }{⟩}_2\mathrm{.}$ This Fock-vacuum field is Gaussian in the precise sense that $M_n=0$ for odd $n$, while for even $n$ it is the sum over pair partitions $\pi$ of $\{1,\dots ,n\}$ of ${\prod }_{(i,j)\in \pi ,i<j}{M}_2(g_i,g_j)$, with each contraction ordered as displayed. This is an exact one-dimensional test-coordinate channel; it does not assert existence of four-dimensional collapse scattering.

Positive chiral current and exponential test-coordinate model

The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)

Schwartz/distribution spaces and Hilbert Fock conventions

Bosonic and fermionic Fock operators with exact domains

A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions

In the channel defined above, let $s=u-v$ and let $\omega$ be angular frequency in seconds${}^{-1}$. Its two-point distribution in centered time is $C_k(s)=-\frac{k^2}{16\pi }{\mathrm{csch}}^2(\frac{k}{2}(s-i0))\mathrm{.}$ Its Fourier spectrum, using ${\widehat{C}}_k(\omega )={\int }_{\mathbb{R}}{C}_k(s)e^{-i\omega s}ds$ in the distributional sense, is $S_k(\omega )=\frac{\omega }{2(e^{2\pi \omega \mathrm{/}k}-1)}(\omega \mathrm{\ne }0),S_k(0)=\frac{k}{4\pi }\mathrm{.}$ Thus the precise detailed-balance relation at strip width $\theta =2\pi \mathrm{/}k$ seconds is $S_k(-\omega )=e^{\theta \omega }{S}_k(\omega )$ for every $\omega >0$. No measured temperature is assigned.

Separately, for each fixed $\omega >0$, give $\mathbb{C}$ the second-variable-linear Hilbert inner product $⟨z,w{⟩}_{\mathbb{C}}=\bar{z}w$, so $1\in \mathbb{C}$ is a unit vector, and put $\mathcal{F}={\mathcal{F}}_s(\mathbb{C})$. Write $\mathrm{\mid }m⟩$ for the normalized vector in its $m$-particle sector, with $\mathrm{\mid }0⟩=\mathrm{\Omega }$; the annihilation map satisfies $a_{\mathbb{C}}(1)\mathrm{\mid }m⟩=\sqrt{m}\mathrm{\mid }m-1⟩$ on finite-particle vectors. Its maximal closed domain is $D(a_{\mathbb{C}}(1))=\{{\sum }_{m\ge 0}{\psi }_m\mathrm{\mid }m⟩:{\sum }_{m\ge 0}m\mathrm{\mid }{\psi }_m{\mathrm{\mid }}^2<\mathrm{\infty }\}\mathrm{.}$ Let $\mathrm{\Gamma }:(0,\mathrm{\infty })\to [0,1]$ be any pointwise transmissivity function; no measurability is needed because every assertion below is mode-by-mode. Every square root below denotes the unique nonnegative real square root, which exists for each nonnegative argument in the usual Dedekind-complete ordered field $\mathbb{R}$. Let $\mathcal{H}=\mathcal{F}\widehat{\otimes }\mathcal{F}$ with product basis $\mathrm{\mid }m,n⟩=\mathrm{\mid }m⟩\otimes \mathrm{\mid }n⟩$. The tensor-factor annihilation operators $a=a_{\mathbb{C}}(1)\otimes I$ and $e=I\otimes a_{\mathbb{C}}(1)$ act on basis vectors by $a\mathrm{\mid }m,n⟩=\sqrt{m}\mathrm{\mid }m-1,n⟩$ and $e\mathrm{\mid }m,n⟩=\sqrt{n}\mathrm{\mid }m,n-1⟩$, with a zero result when the corresponding index is zero, and have domains $D(a)=\{{\sum }_{m,n\ge 0}{c}_{mn}\mathrm{\mid }m,n⟩:{\sum }_{m,n\ge 0}m\mathrm{\mid }{c}_{mn}{\mathrm{\mid }}^2<\mathrm{\infty }\},D(e)=\{{\sum }_{m,n\ge 0}{c}_{mn}\mathrm{\mid }m,n⟩:{\sum }_{m,n\ge 0}n\mathrm{\mid }{c}_{mn}{\mathrm{\mid }}^2<\mathrm{\infty }\}\mathrm{.}$ On their common domain $D:=D(a)\cap D(e)$ define, for each $\omega$, the two output operators by the following linear transformation, with $D(b_{\omega })=D(d_{\omega })=D$: $b_{\omega }=\sqrt{\mathrm{\Gamma }(\omega )}a+\sqrt{1-\mathrm{\Gamma }(\omega )}e\mathrm{.}$ The complementary output is $d_{\omega }=-\sqrt{1-\mathrm{\Gamma }(\omega )}a+\sqrt{\mathrm{\Gamma }(\omega )}e$, also defined on $D$. Let ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}=\mathrm{span}\{\mathrm{\mid }m,n⟩:m,n\in {\mathbb{N}}_0\}$ be the algebraic finite-particle subspace. It is contained in $D$ and invariant under $a,e$ and their creation operators; on it define $b_{\omega }^{†}=\sqrt{\mathrm{\Gamma }(\omega )}{a}^{†}+\sqrt{1-\mathrm{\Gamma }(\omega )}{e}^{†}$ and $d_{\omega }^{†}=-\sqrt{1-\mathrm{\Gamma }(\omega )}{a}^{†}+\sqrt{\mathrm{\Gamma }(\omega )}{e}^{†}$ as the corresponding creation combinations. Let ${\psi }_{\omega }\in D(a_{\mathbb{C}}(1))$ be a normalized signal vector and put the environment in its unit vacuum vector ${\mathrm{\Omega }}_e=\mathrm{\mid }0⟩$. Then ${\mathrm{\Psi }}_{\omega }:={\psi }_{\omega }\otimes {\mathrm{\Omega }}_e$ lies in $D$: if ${\psi }_{\omega }={\sum }_m{\psi }_m\mathrm{\mid }m⟩$, the $a$-domain sum is ${\sum }_{m}m\mathrm{\mid }{\psi }_m{\mathrm{\mid }}^2<\mathrm{\infty }$ and the $e$-domain sum is zero. Put $n_{\omega }=\mathrm{\parallel }{a}_{\mathbb{C}}(1){\psi }_{\omega }{\mathrm{\parallel }}^2<\mathrm{\infty }$ and define the output occupation by the quadratic form ${\mathfrak{n}}_{\mathrm{o}\mathrm{u}\mathrm{t}}(\omega ):=\mathrm{\parallel }{b}_{\omega }{\mathrm{\Psi }}_{\omega }{\mathrm{\parallel }}^2$. Then ${\mathfrak{n}}_{\mathrm{o}\mathrm{u}\mathrm{t}}(\omega )=\mathrm{\Gamma }(\omega )n_{\omega }$. For the Planck occupation function $n_{\theta }(\omega )=(\exp (\theta \omega )-1{)}^{-1}$ with width $\theta >0$ measured in seconds, one has $n_{\theta }(\omega )>0$ for every $\omega >0$. Consequently, a filter preserving this same width at every positive frequency must satisfy $\mathrm{\Gamma }(\omega )n_{\theta }(\omega )=n_{\theta }(\omega )$ and hence $\mathrm{\Gamma }(\omega )=1$ at every $\omega$. Frequency-dependent filtering can nevertheless preserve Planck form at another width: ${\mathrm{\Gamma }}_{\ast }(\omega )=(\exp (\theta \omega )+1{)}^{-1}$ lies in $(0,1)$ and varies with $\omega$, while ${\mathrm{\Gamma }}_{\ast }(\omega )n_{\theta }(\omega )=n_{2\theta }(\omega )$. This is an algebraic channel statement; it does not identify $\mathrm{\Gamma }$ with a four-dimensional black-hole scattering coefficient.

Specified chiral collapse channel

Positive chiral current and exponential test-coordinate model

Single-pole thermal contour and exact detailed balance

Nonnegative real square roots exist uniquely

The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)

Schwartz/distribution spaces and Hilbert Fock conventions

Bosonic and fermionic Fock operators with exact domains

The real exponential function and the number $e$ by a power series

The exponential addition formula $\exp (x+y)=\exp (x)\exp (y)$

The exponential function is smooth and $(\exp {)}^{\mathrm{\prime }}=\exp$

The exponential is positive and satisfies $\exp (-x)=1\mathrm{/}\exp (x)$

Mean value theorem on a closed real interval

The diagram represents the two input annihilation modes and the real orthogonal transformation to the output modes. Its matrix entries and the vacuum-environment condition are fixed by the Statement; the occupation and commutator claims are proved algebraically below.