Why Planck's Radiation Law Cannot be Derived from an Exponential Redshift

An exponential redshift has been associated with the emergence of a temperature in the presence of a horizon [cf. T. Pandmanabhan, "Thermodynamical Aspects of Gravity: New Insights" arXiv:0911.5004] More specifically, it is the power spectrum whose amplitude is given by the Fourier transform of the exponential redshift when the temperature is given by the so-called Unruh temperature for which the acceleration of an observer is proportional to the absolute temperature. Suppose a light signal is sent out in time $t_1$ and received back in time $t_2$. If the signal is reflected back at $t_r$, the hyperbolic velocity for the out-bound journey is \[(1)\qquad\qquad \bar{v}=c\ln\frac{t_r}{t_1},\] while it will have the same velocity for the in-bound journey, \[(2)\qquad\qquad\bar{v}=c\ln\frac{t_2}{t_r}.\] Taken together, (1) and (2) imply that the reflection time will be the geometric average of the two times, \[(3)\qquad\qquad t_r=\sqrt{t_1t_2}.\] Now, according to Einstein, reflection will occur at the arithmetic mean time \[(4)\qquad\qquad t_E=\frac{1}{2}\left(t_1+t_2\right),\] when the light signal is at \[(5)\qquad\qquad r_E=\frac{1}{2}c\left(t_2-t_1\right).\] The ratio of (5) to (4) defines the Euclidean velocity $v=r_E/t_E$. Then rearranging the ratio of (4) to (5) gives \[(6)\qquad\qquad t_2=\left(\frac{1+v/c}{1-v/c}\right)t_1,\] which upon introducing the geometric mean (3) gives \[(7)\qquad\qquad t_2=\sqrt{\left(\frac{1+v/c}{1-v/c}\right)}t_r,\] so that the hyperbolic velocity (2) will be given by the logarithm of the longitudinal Doppler shift, \[(8)\qquad\qquad \bar{v}=c\ln\frac{t_2}{t_r}=c\ln\sqrt{\left(\frac{1+v/c}{1-v/c}\right)}.\] Consider now a train of photons, the first of which passes the observer $r$ at $t_r$ and observer at $2$ at $t_2$. Successive photons will pass these observers at $t_r+\lambda_r/c$ and $t_2+\lambda_2/c$, respectively, where the $\lambda_i$ are their wavelengths. Their ratio must also give the same hyperbolic velocity (2), and this implies that \[\frac{t_2}{t_1}=\frac{t_2+\lambda_2/c}{t_r+\lambda_r/c},\] or equivalently, \[(9)\qquad\qquad \frac{\lambda_2}{\lambda_r}=\frac{t_2}{t_r}=e^{\bar{v}/c}.\] In terms of their frequencies, we get the exponential redshift, \[(10)\qquad\qquad \nu_2=\nu_r\cdot e^{-\bar{v}/c}.\] We now want to consider the Fourier transform of the red-shift, but not in the form (10). Rather, we introduce time according to \[(11)\qquad\qquad \frac{\Omega t}{h}=\frac{\bar{v}}{c},\] where $\Omega$ has units of energy, and $h$ is Planck's constant. So we are led to consider the Fourier transform of the "double exponential" distribution \[(12)\qquad\qquad f(\omega)=\int_{-\infty}^{\infty}\exp\left(-e^{-\Omega t}\right)e^{i\omega t}dt.\] Introducing the change of variable, $z=\exp(-\Omega t)$, and "analytically continuing to imaginary values of $z$", give \[(13)\qquad\qquad f(\omega)=\frac{1}{\Omega}\int_0^{\infty}z^{-i\omega/\Omega-1}e^{-z}dz=\frac{1}{\Omega}\Gamma(-i\omega/\Omega),\] where $\Gamma$ is the Gamma function of an imaginary argument. If we consider the power spectrum of (13), $|f(\omega)|^2$, and use \[(14)\qquad\qquad\Gamma(i\omega/\Omega)\cdot\Gamma(-i\omega/\Omega)=\frac{\pi\Omega}{\omega}\frac{2}{e^{\pi\omega/\Omega}-e^{-\pi\omega/\Omega}},\] to evaluate it, we come out with \[(15)\qquad\qquad \omega e^{-\pi\omega/\Omega}|f(\omega)|^2=\frac{2\pi}{\Omega}\frac{1}{e^{2\pi\omega/\Omega}-1}.\] Expression (15) will look like Planck's radiation law if we define the absolute temperature as \[\qquad \qquad k T=\frac{\Omega}{2\pi},\] where $k$ is Boltzmann's constant. In view of (11) this means the temperature is proportional to an acceleration, \[(16)\qquad\qquad kT=\frac{\bar{v}}{t}\cdot\frac{\hbar}{c}=\frac{a\hbar}{c}=\beta^{-1},\] where $\hbar$ is Planck's constant divided by $2\pi$, and we define a (hyperbolic?) acceleration $a=\bar{v}/t$. Thus, we can write (14) in the form \[(17)\qquad\qquad \omega e^{-\beta\omega/2}|f(\omega)|^2=\frac{\beta}{e^{\beta\omega}-1}.\] The conclusion now follows: "waves propagating from a region near the horizon will undergo exponential redshift. An observer detecting this exponentially redshifted radiation at late times ($t\rightarrow\infty$), originating from a region close to [the horizon] will attribute to this radiation a Planckian power spectrum given by Eq. [17]. This result lies at the foundation of associating a temperature with a horizon." The $\Omega$ stands either for the observer's acceleration or the "surface gravity of the horizon." There is much to criticize here. First, and foremost, red-shifts arising from the Doppler effect occur at constant velocity. Acceleration plays no role in redshifts, and, therefore, a temperature cannot be introduced which is proportional to the acceleration of an observer. We would like to believe that a body has a temperature independent of the acceleration of an observer who is holding a thermometer. Apart from this, the big question remains: Why on earth (or elsewhere) should a redshift induce a blackbody radiation spectrum? The formation of blackbody radiation requires a cavity that is kept at constant temperature and a piece of charcoal which will absorb and emit radiation at all frequencies. The property that the Gamma function that was used in the derivation, (14), is symmetric in exponentially increasing and decreasing functions of $\beta\omega$. What has made it unsymmetric? [In fact, in Padmanabhan's article, Eq. (23), there should be a term $e^{\beta\omega/2}$ in the numerator of the expression for the power spectrum, coming from the last term in (22), which will not cancel the factor $e^{-\beta\omega/2}$ in the denominator that is necessary to give it its Planckian form.] Why is there a frequency on the left-hand side and an inverse temperature on the right-hand side? As such it will not obey Wien's displacement law that temperature should always appear in the ratio $\omega/T$. Morever, (17), will not give the equipartition law in the limit as $T\rightarrow\infty$.