Constructed timelike smear and massless accelerated spectrum

Fix a compact set $K⋐I$ containing $\mathrm{supp}f$ and let $d={\min }_{\tau \in K}({\dot{\gamma }}^0-\mathrm{\mid }\dot{\mathit{\gamma }}\mathrm{\mid })>0$. For $\mathbf{k}=r\mathbf{n}$, $r>0$, $\mathrm{\mid }\mathbf{n}\mathrm{\mid }=1$, the phase is $r{\varphi }_{\mathbf{n}}(\tau )$ with ${\varphi }_{\mathbf{n}}^{\mathrm{\prime }}={\dot{\gamma }}^0-\mathbf{n}\cdot \dot{\mathit{\gamma }}\ge d$. All derivatives of $1\mathrm{/}{\varphi }_{\mathbf{n}}^{\mathrm{\prime }}$ are uniformly bounded on $K\times S^2$. Since $e^{ir\varphi }=(ir{\varphi }^{\mathrm{\prime }}{)}^{-1}{\mathrm{\partial }}_{\tau }{e}^{ir\varphi }$, repeated integration by parts gives $\mathrm{\mid }{K}_{\gamma }f(r\mathbf{n})\mathrm{\mid }\le C_N{r}^{-N}$ for every $N$ and $r\ge 1$, with $C_N$ bounded by finitely many test-function seminorms on $K$. Boundary terms vanish because $f$ is compactly supported. For $0<r\le 1$, $\mathrm{\mid }{K}_{\gamma }f(r\mathbf{n})\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{L^1(K)}$.

Put $\kappa =a\mathrm{/}c$ and $\mathrm{\ell }=c^2\mathrm{/}a$. Direct differentiation gives ${\dot{\gamma }}_a=(c\cosh (\kappa \tau ),c\sinh (\kappa \tau ),0,0)$, $\eta ({\dot{\gamma }}_a,{\dot{\gamma }}_a)=-c^2$, and ${\ddot{\gamma }}_a=(a\sinh (\kappa \tau ),a\cosh (\kappa \tau ),0,0)$, $\eta ({\ddot{\gamma }}_a,{\ddot{\gamma }}_a)=a^2$. Hence the curve is future unit timelike and the defined proper-acceleration magnitude satisfies ${\alpha }_a(\tau )=a$ for every $\tau$. With $s=\tau -{\tau }^{\mathrm{\prime }}$ and $v=(\tau +{\tau }^{\mathrm{\prime }})\mathrm{/}2$, the differences satisfy $X+Y=2\mathrm{\ell }{e}^{\kappa v}\sinh (\kappa s\mathrm{/}2),X-Y=2\mathrm{\ell }{e}^{-\kappa v}\sinh (\kappa s\mathrm{/}2),$ where $Y=\mathrm{\Delta }{x}^1$. Therefore $Y^2-X^2=-4{\mathrm{\ell }}^2{\sinh }^2(\kappa s\mathrm{/}2)$, and the regulated denominator factors exactly as $Y^2+(ϵ+iX{)}^2=(Y-X+iϵ)(Y+X-iϵ)\mathrm{.}$

In polar coordinates $d{\nu }_0=rdrd\mathrm{\Omega }\mathrm{/}(2(2\pi {)}^3)$. The low-frequency bound makes ${\int }_0^{1}r\mathrm{\mid }{K}_{\gamma }f{\mathrm{\mid }}^{2}dr$ finite; the estimate of 1.1 with, for example, $N=2$ makes ${\int }_1^{\mathrm{\infty }}r\mathrm{\mid }{K}_{\gamma }f{\mathrm{\mid }}^{2}dr$ finite, uniformly in $\mathbf{n}$. The same estimates bound the norm by finitely many seminorms of $f$ on $K$, so $K_{\gamma }{\mathrm{\mid }}_{{\mathcal{D}}_K(I)}$ is continuous for each compact $K⋐I$ and $K_{\gamma }f\in {\mathcal{H}}_0$. Since $K_{\gamma }$ is complex-linear, [F13] gives its continuity from the LF test-function topology on $\mathcal{D}(I)$ to the norm topology of ${\mathcal{H}}_0$. The Hilbert-space vector is therefore defined before it is used in a field expression.

The creation and annihilation formulas in [F2] define ${\mathrm{\Phi }}_{\gamma }(f)$ on the finite-particle domain because $K_{\gamma }f\in {\mathcal{H}}_0$. For real $f$, it equals $\sqrt{2}$ times the Segal field with vector $K_{\gamma }f$ on that domain. Separability [F6] meets the hypothesis of [F5], which gives its self-adjoint closure. Mutual adjointness [F10] makes the finite-domain field symmetric. The vacuum is annihilated by $a(K_{\gamma }f)$ by [F10]. Therefore ${\mathrm{\Phi }}_{\gamma }(g)\mathrm{\Omega }=a^{†}(K_{\gamma }g)\mathrm{\Omega }$ and symmetry of ${\mathrm{\Phi }}_{\gamma }(f)$ gives $⟨\mathrm{\Omega },{\mathrm{\Phi }}_{\gamma }(f){\mathrm{\Phi }}_{\gamma }(g)\mathrm{\Omega }{⟩}_2=⟨K_{\gamma }f,K_{\gamma }g{⟩}_2$. This defines only the tested field and makes no point-field assumption.

For $ϵ>0$ insert $e^{-ϵr}$ in the shell integral. With $X=\mathrm{\Delta }{x}^0$ and $R=\mathrm{\mid }\mathrm{\Delta }\mathbf{x}\mathrm{\mid }$, absolute convergence permits angular integration and Fubini [F9]: $W_{ϵ}(X,R)=\frac{1}{4{\pi }^{2}R}{\int }_0^{\mathrm{\infty }}{e}^{-(ϵ+iX)r}\sin (rR)dr=\frac{1}{4{\pi }^2(R^2+(ϵ+iX{)}^2)}\mathrm{.}$ The value at $R=0$ is the continuous limit. The integral identity follows by writing the sine as two exponentials and integrating each, since $\mathrm{Re}(ϵ+iX)>0$. For compact real tests $f,g$, the shell-paired expression converges to $⟨K_{\gamma }f,K_{\gamma }g{⟩}_2$: $e^{-ϵr\mathrm{/}2}{K}_{\gamma }f\to K_{\gamma }f$ in ${\mathcal{H}}_0$ by [F7], and Cauchy–Schwarz [F12] passes the inner products to the limit. Thus the damped kernel is an actual regulator for the constructed covariance, not a formal substitution of a worldline into a point field.

For compact real tests $f,g$, the linear change $v=(\tau +{\tau }^{\mathrm{\prime }})\mathrm{/}2$, $s=\tau -{\tau }^{\mathrm{\prime }}$ has Jacobian of absolute value one and takes $f(\tau )g({\tau }^{\mathrm{\prime }})$ to a compact smooth test $\phi (v,s)$. On the compact average-time projection of its support, write $A(v)=2\mathrm{\ell }{e}^{-\kappa v}>0$ and $B(v)=2\mathrm{\ell }{e}^{\kappa v}>0$. The factors $Y-X+iϵ$ and $Y+X-iϵ$ are respectively $-A(v)(x-iϵ\mathrm{/}A(v))$ and $B(v)(x-iϵ\mathrm{/}B(v))$, where $x=\sinh (\kappa s\mathrm{/}2)$. For positive $\alpha ,\beta$ the identity $\frac{1}{(x-i\alpha ϵ)(x-i\beta ϵ)}={\int }_0^1\frac{dt}{(x-i((1-t)\alpha +t\beta )ϵ{)}^2}$ follows by integrating the derivative of $[x-i((1-t)\alpha +t\beta )ϵ{]}^{-1}$ in $t$ and dividing by $i(\beta -\alpha )ϵ$ when $\alpha \mathrm{\ne }\beta$; for $\alpha =\beta$ both sides agree directly. It shows that both dampings have the same boundary. To justify that boundary, for a compact smooth test $\psi$ choose an even compact smooth $\chi$ equal to one near zero and write $\psi (x)=\psi (0)\chi (x)+x\rho (x)$, where $\rho$ is smooth and compactly supported. The constant term satisfies $\int \frac{\chi (x)}{x-i\lambda ϵ}dx=i\lambda ϵ\int \frac{\chi (x)}{x^2+(\lambda ϵ{)}^2}dx⟶i\pi$ for each $\lambda >0$, since its odd real part integrates to zero. The other term is $x\rho (x)\mathrm{/}(x-i\lambda ϵ)=\rho (x)+i\lambda ϵ\rho (x)\mathrm{/}(x-i\lambda ϵ)$; the last integral is $O(ϵ\mathrm{\mid }\log ϵ\mathrm{\mid })$ uniformly for $\lambda$ in a compact positive interval. Let ${\delta }_0$ denote evaluation at zero, $⟨{\delta }_0,\psi ⟩=\psi (0)$, and define the distributional derivative by $⟨T^{\mathrm{\prime }},\psi ⟩=-⟨T,{\psi }^{\mathrm{\prime }}⟩$ for compact smooth tests $\psi$, so $⟨{\delta }_0^{\mathrm{\prime }},\psi ⟩=-{\psi }^{\mathrm{\prime }}(0)$. Thus $(x-i0{)}^{-1}=\mathrm{PV}(1\mathrm{/}x)+i\pi {\delta }_0$, where $\mathrm{PV}(1\mathrm{/}x)$ acts by ${\lim }_{\delta \downarrow 0}{\int }_{\mathrm{\mid }x\mathrm{\mid }>\delta }\psi (x)\mathrm{/}xdx$. Differentiating this distributional identity gives $(x-i0{)}^{-2}=\mathrm{Pf}(1\mathrm{/}{x}^2)-i\pi {\delta }_0^{\mathrm{\prime }}$ and $⟨\mathrm{Pf}(1\mathrm{/}{x}^2),\psi ⟩={\lim }_{\delta \downarrow 0}({\int }_{\mathrm{\mid }x\mathrm{\mid }>\delta }\frac{\psi (x)}{x^2}dx-\frac{2\psi (0)}{\delta })\mathrm{.}$ Integration by parts on $\mathrm{\mid }x\mathrm{\mid }>\delta$ identifies this finite part with $-(\mathrm{PV}(1\mathrm{/}x){)}^{\mathrm{\prime }}$. The bounds are uniform when the positive damping coefficient ranges in a compact subinterval of $(0,\mathrm{\infty })$, as it does for $v$ in a compact set. To identify the pullback with the usual $s-i0$ boundary, note that $\sinh (bs-iϵb)=\sinh (bs)-iϵb\cosh (bs)+O({ϵ}^2)$. More generally, for a positive smooth regulator $\lambda (v,x)$, Taylor expand $\lambda ={\lambda }_0(v)+{\lambda }_1(v)x+x^{2}r(v,x)$. The linear part factors as $x-iϵ({\lambda }_0+{\lambda }_{1}x)=(1-iϵ{\lambda }_1)(x-\frac{iϵ{\lambda }_0}{1-iϵ{\lambda }_1})\mathrm{.}$ Its pole has positive imaginary part comparable to $ϵ$ and real displacement $O({ϵ}^2)$, so the preceding test-subtraction calculation gives the same boundary. The quadratic remainder changes the inverse square by at most $Cϵx^2\mathrm{/}(x^2+{ϵ}^2{)}^{3\mathrm{/}2}$ on a small interval; its integral is $O(ϵ\mathrm{\mid }\log ϵ\mathrm{\mid })$. Thus the smooth positive regulator, including the one induced by $\sinh (b(s-iϵ))$, has the same boundary as the constant positive regulator. The change $x=\sinh (\kappa s\mathrm{/}2)$ is orientation-preserving with positive derivative. Consequently the test-paired limit of step 3.2 is $C_a(s)=-\frac{1}{16{\pi }^2{\mathrm{\ell }}^2}{\mathrm{csch}}^2\text{\hspace{0.17em}⁣}\left(\frac{\kappa (s-i0)}{2}\right)=-\frac{a^2}{16{\pi }^2{c}^4}{\mathrm{csch}}^2\text{\hspace{0.17em}⁣}\left(\frac{a(s-i0)}{2c}\right)\mathrm{.}$ Thus $⟨K_{{\gamma }_a}f,K_{{\gamma }_a}g{⟩}_2$ equals $⟨C_a(s),\int f(t)g(t-s)dt⟩$. This proves stationarity and the stated covariance boundary from ordinary regulated shell pairings. To verify temperedness explicitly, set $q={\lim }_{ϵ\downarrow 0}\coth (\kappa (s-iϵ)\mathrm{/}2)$ in the distribution sense and $r(s)=\coth (\kappa s\mathrm{/}2)-2\mathrm{/}(\kappa s)$ for $s\mathrm{\ne }0$. The Taylor expansion of $\coth$ at zero extends $r$ smoothly there, and $r^{\mathrm{\prime }}(s)=-(\kappa \mathrm{/}2){\mathrm{csch}}^2(\kappa s\mathrm{/}2)+2\mathrm{/}(\kappa s^2)$ is bounded: its apparent singularity cancels at zero, and both terms are bounded at infinity. Since $1\mathrm{/}(s-i0)=\mathrm{PV}(1\mathrm{/}s)+i\pi {\delta }_0$, the decomposition is $q=(2\mathrm{/}\kappa )\mathrm{PV}(1\mathrm{/}s)+r+(2\pi i\mathrm{/}\kappa ){\delta }_0$. By the derivative definition and $q_{ϵ}\to q$ on compact smooth tests, $q_{ϵ}^{\mathrm{\prime }}\to q^{\mathrm{\prime }}$. Direct differentiation of $q_{ϵ}(s)=\coth (\kappa (s-iϵ)\mathrm{/}2)$ then gives $C_a=(a^2\mathrm{/}(8{\pi }^2{c}^4\kappa ))q^{\mathrm{\prime }}$. For $m,n\ge 0$ set $p_{m,n}(\phi )={\sup }_s(1+s^2{)}^{m\mathrm{/}2}\mathrm{\mid }{\phi }^{(n)}(s)\mathrm{\mid }$, a Schwartz seminorm. For $\phi \in \mathcal{S}(\mathbb{R})$, the symmetric principal-value formula gives $\mid ⟨\mathrm{PV}(1\mathrm{/}s),{\phi }^{\mathrm{\prime }}⟩\mid \le 2p_{0,2}(\phi )+C{p}_{2,1}(\phi ),$ because subtracting ${\phi }^{\mathrm{\prime }}(0)$ on $\mathrm{\mid }s\mathrm{\mid }<1$ bounds the local integral by $2\sup \mathrm{\mid }{\phi }^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\mid }$, while the $\mathrm{\mid }s\mathrm{\mid }\ge 1$ integral is bounded by $p_{2,1}{\int }_{\mathrm{\mid }s\mathrm{\mid }\ge 1}[\mathrm{\mid }s\mathrm{\mid }(1+s^2){]}^{-1}ds$. Also $\mathrm{\mid }\int r^{\mathrm{\prime }}(s)\phi (s)ds\mathrm{\mid }\le C{p}_{2,0}(\phi )$ and $\mathrm{\mid }⟨{\delta }_0^{\mathrm{\prime }},\phi ⟩\mathrm{\mid }\le p_{0,1}(\phi )$. Hence $\mathrm{\mid }⟨C_a,\phi ⟩\mathrm{\mid }\le C_{a,c}(p_{0,2}(\phi )+p_{2,1}(\phi )+p_{2,0}(\phi )+p_{0,1}(\phi )),$ a finite Schwartz-seminorm bound. By [F2], $C_a\in {\mathcal{S}}^{\mathrm{\prime }}(\mathbb{R})$, as required for its subsequent Fourier transform.

On the inertial curve, direct radial integration gives $W_{0,ϵ}(s)=\frac{1}{4{\pi }^2}{\int }_0^{\mathrm{\infty }}r{e}^{-(ϵ+ics)r}dr=\frac{1}{4{\pi }^2(ϵ+ics{)}^2}\mathrm{.}$ Its boundary is $C_0(s)=-1\mathrm{/}(4{\pi }^2{c}^2(s-i0{)}^2)$. Reparameterizing the massless shell by $\omega =cr$ gives $C_0(s)=(4{\pi }^2{c}^2{)}^{-1}{\int }_0^{\mathrm{\infty }}\omega e^{-i\omega s}d\omega$ as a tempered boundary distribution. Fourier transformation with $e^{-i\nu s}$ then yields $S_0(\nu )=\max (-\nu ,0)\mathrm{/}(2\pi c^2)$, including continuous value zero at $\nu =0$. Specifically, the inverse-Fourier identity from [F11], after $\nu =2\pi \xi$, gives ${\mathcal{F}}_{-}(e^{-i\omega s})(\nu )=2\pi {\delta }_0(\nu +\omega )$; integrating this identity against $\omega {\mathbf{1}}_{\omega >0}\mathrm{/}(4{\pi }^2{c}^2)$ yields the displayed regular density. Pairing first with a Schwartz test makes the $\omega$ integral absolutely convergent by rapid decay, so the interchange with the shell integral is justified.

Let $\alpha =a\mathrm{/}{c}^2>0$, which has units of inverse length, and use the supplier's length coordinate $u=cs$. The supplier boundary definition in [L1] and the covariance in step 4.1 give $C_a(s)={\pi }^{-1}{C}_{\alpha }(cs)$ as tempered distributions: their smooth kernels have the same hyperbolic argument and their coefficients differ by exactly $1\mathrm{/}\pi$; the scaling is positive, so it preserves the lower-boundary orientation. By the bilinear transposed Fourier convention [F8], for each $\phi \in \mathcal{S}(\mathbb{R})$ the dual pairing gives $⟨{\mathcal{F}}_{-,s}[C_{\alpha }(c\cdot )],\phi ⟩=⟨S_{\alpha }(\omega ),\phi (c\omega )⟩=⟨c^{-1}{S}_{\alpha }(\nu \mathrm{/}c),\phi ⟩$: indeed, $\widehat{\phi }(u\mathrm{/}c)=c\widehat{\phi (c\cdot )}(u)$ and the distribution pullback contributes $c^{-1}$ under $u=cs$. Including the covariance factor ${\pi }^{-1}$ therefore gives ${\mathcal{F}}_{-,s}[{\pi }^{-1}{C}_{\alpha }(c\cdot )](\nu )={\pi }^{-1}{c}^{-1}{S}_{\alpha }(\nu \mathrm{/}c)$. Now apply the exact conditional formula of draft supplier [L1]: $S_a(\nu )=\frac{1}{\pi c}\frac{\nu \mathrm{/}c}{2(e^{2\pi (\nu \mathrm{/}c)\mathrm{/}\alpha }-1)}=\frac{\nu }{2\pi c^2(e^{2\pi c\nu \mathrm{/}a}-1)}\mathrm{.}$ Its zero limit is $\alpha \mathrm{/}(4{\pi }^{2}c)=a\mathrm{/}(4{\pi }^2{c}^3)$, and its detailed-balance relation gives $S_a(-\nu )=e^{2\pi c\nu \mathrm{/}a}{S}_a(\nu )$ for $\nu >0$. The spectral conclusion remains conditional until the two batch-74 draft suppliers are accepted; the covariance construction and inertial transform above are local arguments.

Put $g_{\mathsf{T}}(t)=h(t\mathrm{/}\mathsf{T})$ and $A_{\mathsf{T}}(s)=\int g_{\mathsf{T}}(t)g_{\mathsf{T}}(t-s)dt$. Compact support [F6] and [F5] permit the change of variables and interchange of the two ordinary integrals. Setting $u=t$ and $v=t-s$, and using the reality of $h$, gives the negative-sign angular transform $\widehat{A_{\mathsf{T}}}(\lambda )={\int }_{{\mathbb{R}}^2}{g}_{\mathsf{T}}(u)g_{\mathsf{T}}(v)e^{-i\lambda (u-v)}dudv=\widehat{g_{\mathsf{T}}}(\lambda )\widehat{g_{\mathsf{T}}}(-\lambda )=\mathrm{\mid }\widehat{g_{\mathsf{T}}}(\lambda ){\mathrm{\mid }}^2={\mathsf{T}}^2\mathrm{\mid }\widehat{h}(\mathsf{T}\lambda ){\mathrm{\mid }}^2\mathrm{.}$ By [F2] and [F3], the inverse angular-frequency relation is $C(s)=(2\pi {)}^{-1}\int S(\omega )e^{i\omega s}d\omega$ in the distributional sense: it is the negative-sign $2\pi$ inverse after the change $\xi =\omega \mathrm{/}(2\pi )$. Pairing with $e^{-i\nu s}{A}_{\mathsf{T}}(s)$ gives the absolutely convergent formula $R_{\mathsf{T}}^{(C)}(\nu )=\frac{{\mathsf{T}}^2}{2\pi }{\int }_{\mathbb{R}}S(\omega )\mathrm{\mid }\widehat{h}(\mathsf{T}(\nu -\omega )){\mathrm{\mid }}^{2}d\omega \mathrm{.}$ Here $S$ has at most linear growth and $\widehat{h}$ is rapidly decreasing, so the integral is finite.

Since $h\in C_c^{\mathrm{\infty }}$ by [F6], each derivative of $\widehat{h}$ is obtained by transforming a compactly supported smooth function $(-iu{)}^{j}h(u)$; repeated integration by parts shows every such derivative decreases faster than any inverse power. Hence $\int \mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }\widehat{h}(z){\mathrm{\mid }}^{2}dz<\mathrm{\infty }$. Plancherel in [F4], followed by $z=2\pi \xi$, yields ${\int }_{\mathbb{R}}\mathrm{\mid }\widehat{h}(z){\mathrm{\mid }}^{2}dz=2\pi \mathrm{\parallel }h{\mathrm{\parallel }}_2^2\mathrm{.}$ Changing variables $z=\mathsf{T}(\nu -\omega )$ in 1.1 therefore gives $\frac{R_{\mathsf{T}}^{(C)}(\nu )}{\mathsf{T}\mathrm{\parallel }h{\mathrm{\parallel }}_2^2}=\frac{1}{2\pi \mathrm{\parallel }h{\mathrm{\parallel }}_2^2}{\int }_{\mathbb{R}}S(\nu -z\mathrm{/}\mathsf{T})\mathrm{\mid }\widehat{h}(z){\mathrm{\mid }}^{2}dz\mathrm{.}$

For $C=C_0$, [F1] gives $S_0(\nu )=\max (-\nu ,0)\mathrm{/}(2\pi c^2)$. On $\nu \le 0$ this is linear with slope $-1\mathrm{/}(2\pi c^2)$, on $\nu \ge 0$ it is zero, and for $x\le 0\le y$, $\mathrm{\mid }{S}_0(x)-S_0(y)\mathrm{\mid }=-x\mathrm{/}(2\pi c^2)\le (y-x)\mathrm{/}(2\pi c^2)$. Thus its global Lipschitz constant is $1\mathrm{/}(2\pi c^2)$. For $C=C_a$, the conditional formula in [F1] is smooth at zero by its Taylor expansion, has bounded derivative on every compact interval, has bounded derivative as $\nu \to +\mathrm{\infty }$ by exponential decay. Its derivative is $S_a^{\mathrm{\prime }}(\nu )=\frac{1}{2\pi c^2}\frac{e^{b\nu }-1-b\nu e^{b\nu }}{(e^{b\nu }-1{)}^2},b=2\pi c\mathrm{/}a\mathrm{.}$ The Taylor expansion at zero makes this derivative finite there; as $\nu \to +\mathrm{\infty }$ it decays exponentially times a polynomial, and as $\nu \to -\mathrm{\infty }$ it tends $-1\mathrm{/}(2\pi c^2)$. Thus it is bounded on both tails and on compact intervals, so both spectra are globally Lipschitz (the accelerated assertion is conditional on [L1]). Subtract $S(\nu )$ in step 2.1 and use $\mathrm{\mid }S(\nu -z\mathrm{/}\mathsf{T})-S(\nu )\mathrm{\mid }\le \mathrm{Lip}(S)\mathrm{\mid }z\mathrm{\mid }\mathrm{/}\mathsf{T}$. The definition of $K_h$ gives the stated error bound and its $1\mathrm{/}\mathsf{T}$ limit.

The accelerated spectrum is positive: for $\nu >0$ numerator and denominator in [F1] are positive, for $\nu <0$ they are both negative, and the zero limit is positive. Thus $r_{+},r_{-}>0$. By [A2], the second component of ${\pi }^{\mathrm{\prime }}=\pi Q$ gives $p^{\mathrm{\prime }}(t)=r_{+}(1-p(t))-r_{-}p(t)=r_{+}-(r_{+}+r_{-})p(t)$. Solving this scalar linear ODE gives $p(t)=p_{\ast }+(p(0)-p_{\ast })e^{-(r_{+}+r_{-})t}$ with $p_{\ast }=r_{+}\mathrm{/}(r_{+}+r_{-})$. For $t\ge 0$ seconds, this is a convex combination of $p(0)$ and $p_{\ast }\in (0,1)$, so it remains a probability. Apply [L1] at $\nu =E\mathrm{/}\mathrm{\hslash }>0$ to get $r_{-}\mathrm{/}{r}_{+}=S_a(-\nu )\mathrm{/}{S}_a(\nu )=e^{2\pi c\nu \mathrm{/}a}$, so $p_{\ast }=1\mathrm{/}(1+e^{2\pi cE\mathrm{/}(a\mathrm{\hslash })})$. The ratio $p_{\ast }\mathrm{/}(1-p_{\ast })$ is therefore $e^{-2\pi cE\mathrm{/}(a\mathrm{\hslash })}$. Comparing with its defining Boltzmann form gives $T_{\mathrm{e}\mathrm{f}\mathrm{f}}=\mathrm{\hslash }a\mathrm{/}(2\pi c{k}_B)$. This calculation concerns only the specified Markov model and its adopted rates.