# Olbers' paradox in a closed space

I'm looking for a proper interpretation of the following, which is a variant of the Olbers' paradox.

Consider a spatially closed empty universe ($k = 1$) with a positive cosmological constant $\Lambda$ (closed deSitter universe). It is easy to solve the Friedmann-Lemaître-Robertson-Walker equations to find the scale factor : $$\tag{1} a(t) = \ell_{\Lambda} \cosh{(t/\ell_{\Lambda})},$$ where $\ell_{\Lambda} = \sqrt{\frac{3}{\Lambda}}$. The cosmological time doesn't have any boundary : $-\, \infty < t < \infty$ (no Big Bang, no Big Crunch). Yet, because of the expansion of space, there are two horizons, of the following proper distance ($t_0$ is the present time of the stationnary observer) : \begin{align} \mathcal{D}_{P}(t_0) &= a( t_0) \int_{-\, \infty}^{t_0} \frac{1}{a(t)} \; dt \\[12pt] &= \ell_{\Lambda} \arccos{\big(- \tanh{(t_0 / \ell_{\Lambda})} \big)} \cosh{(t_0 / \ell_{\Lambda})}, \tag{2} \\[12pt] \mathcal{D}_{E}(t_0) &= a( t_0) \int_{t_0}^{\infty} \frac{1}{a(t)} \; dt \\[12pt] &= \ell_{\Lambda} \Big( \pi - \arccos{\big(- \tanh{(t_0 / \ell_{\Lambda})} \big)} \Big) \cosh{(t_0 / \ell_{\Lambda})}. \tag{3} \end{align} The first distance represents the particle horizon, while the second distance represents the event horizon of that empty deSitter spacetime.

Now, it can be proved that the total apparent luminosity at time $t_0$ of all the "fake stars" uniformly distributed in any isotropic/homogeneous space is given by the following integral : $$\tag{4} \mathcal{I}(t_0) = \int I \, n \; d^3 x = L_{\text{e}} \, n_0 \int_{t_{\text{min}}}^{t_0} \frac{a(t)}{a(t_0)} \; dt,$$ where $L_{\text{e}}$ is the absolute luminosity (or power) of a single star, and $n_0$ is the current number density of stars at time $t_0$. In the case of the dust-like matter in an euclidian flat space ($k = 0$, $a(t) \propto t^{2/3}$ and $t_{\text{min}} = 0$), this integral converges and solves the Olbers paradox. But in the case of the empty deSitter space defined above, the scale factor (1) gives (because of $t_{\text{min}} = -\, \infty$) $$\tag{5} \mathcal{I} = \infty.$$ I'm puzzled by this result, since there is still a particle (or causality) horizon, which gives $\mathcal{D}_{P}(t_0) \rightarrow \ell_{\Lambda}$ when $t_0 \rightarrow -\, \infty$ (lower limit of expression (2)), and since the space section is closed (finite number of "fake stars") : $$\tag{6} N_{\text{tot}} = n(t) \, 2 \pi^2 a^3(t) = n_0 \, 2 \pi^2 a^3 (t_0) = cste.$$

So how should we interpret the result of an infinite total apparent luminosity of a finite number of stars in a closed universe, with a particle horizon ?

Note that the space is exponentially contracting during the interval $-\, \infty < t < 0$ (see the scale factor (1)), so I suspect this is a clue to the proper interpretation (the lower limit $t_{\text{min}} = -\, \infty$ is blowing up the integral (4)).

Also, note that the maximal instantaneous proper distance in the closed space is $\mathcal{D}_{\text{max}}(t) = \pi \, a(t)$.

I'm not sure anymore that I should use $t_{\text{min}} = -\, \infty$ in the integral (4) above. This may be a clue to the origin of my problem. Light may turn around several times in the closed universe before beeing detected by the observer, so this may produce the infinite luminosity if the stars are emitting light since infinity in the past. But should we take this into account ? Luminosity should be defined to collect all the light that is reaching a given point (the observer) at a given time $t_0$, so several turns around shouldn't be included, I believe. Each time the light is reaching the observer's location at some time, it counts for the luminosity at that time. See my comments below. What do you think ?

• You have finite number of stars, yes, but they emit light for infinite amount of time. – Kostya Oct 8 '16 at 18:42
• @Kostya, well, the thing that puzzles me the most is the causality horizon. It doesn't prevent the luminosity to blow up. I guess that the contraction of space in the past is concentrating the light rays in a way that makes the integral to diverge, is that the "proper interpretation" of this infinity ? – Cham Oct 8 '16 at 19:15
• @Kostya, your argument isn't valid : for a finite number of stars in a closed space, even if the stars are radiating from an infinite amount of time, the total luminosity at a given location and at a given time should be finite. The light emitted at different moments is simply reaching the observer at different moments. In the case of an eternal closed space (i.e. $a = \textit{cste}$), my calculation gives the following luminosity :$$\mathcal{I} = \pi \, L_\text{e} \, n_0 \, a \equiv \frac{N L_{\text{e}}}{2 \pi \, a^2}.$$ where $N$ is the total number of stars in that closed space. – Cham Oct 18 '16 at 16:21
• The light emitted at different moments is reaching the observer at different moments, correct. But at a given moment from observer's point of view, he receives the light emitted from many (infinite) moments in the past. He sees infinite past copies of the same star. – Kostya Oct 18 '16 at 19:56
• "Light may turn around several times in the closed universe before being detected by the observer" - this is true for Einstein static closed universe but not for de Sitter space. Light emitted a $t=-\infty$ would cross only half of the big circle by the time $t\to \infty$. – A.V.S. Feb 8 '18 at 4:58

I see two ways to better understand this divergence: (1) by considering spectral structure of observed radiation, (2) by considering finite number $N$ of "fake stars" distributed on a hypersphere rather than taking continuous limit.

First, let us look at the spectral content of observed radiation, while assuming continuous limit for the number of stars. This allows us to only concentrate on the 'when' the radiation was emitted ignoring the 'where' because of the homogeneity of space. For simplicity, let us assume that all the "fake stars" emit monochromatic (in their proper frame) radiation with frequency $\Omega$. If we have a static universe with $a(t)=\rm const$ with stars radiating for a finite time $t_\text{fin}$ with a total constant luminosity $W$, then the spectrum of radiation would be $f(\omega)=W t_\text{fin}\, \delta(\omega-\Omega)$. If we have evolving universe with some $a(t)$ that spectrum would be smeared, as now radiation emitted at different times would be blue or redshifted, according to the well known formula: $$\frac{\omega_\text{now}}{\omega_\text{then}}=\frac{a_\text{then}}{a_\text{now}}.$$

To further simplify situation, let us limit ourself with calculating spectrum only for the moment $t=0$ (the Big Bounce of this parametrization). Note, that although de Sitter space allows reparametrizations that could make any moment the Big Bounce, they do not leave the population of fake stars invariant. So, we have $a_\text{now}=a_0=\ell_\Lambda$, $\omega_\text{then}=\Omega$ and $a_\text{then}=a(t)=\ell_\Lambda \cosh(t/\ell_\Lambda)$. We also now have the frequency as observed at $t=0$ as a function of time when the radiation was emitted: $$\omega(t)=\Omega \cosh(t/\ell_\Lambda)\tag{(*)}$$ The energy emitted between moments $t$ and $t+dt$ gets blueshifted: $$dE_\text{now}=\frac{a(t)}{a_0} dE_\text{then}=\frac{a(t)}{a_0} W \,dt .$$ That energy could also be obtained from the spectrum of radiation $f(\omega)$: $$dE_\text{now}=f(\omega)d\omega = f(\omega(t))\frac{d\omega(t)}{dt} dt=f(\omega) \frac{\omega_0}{\ell_\Lambda} \sinh(t/\ell_\Lambda).$$

Equating these expressions we can obtain the desired spectrum of radiation: $$f(\omega)=\frac{\ell_\Lambda W}{\omega_0 \tanh(t/\ell_\Lambda)}=\frac{\omega\,\ell_\Lambda W}{\omega_0 \sqrt{\omega^2-\Omega^2}}.$$ We see that the spectrum diverges at $\omega=\Omega$, but this is expected: it corresponds to the universe contraction stopping at $t=0$, and total energy in the finite part of the spectrum remains finite.

However, we also see, that the spectrum $f(\omega)\to C >0$ when $\omega\to \infty$. This is our divergence: it means that the further away in time the radiation was emitted the more is gets blueshifted and so higher frequencies dominate in the diverging total energy. This is the de Sitter version of ultraviolet catastrophy.

Now secondly, we could consider that not only total luminosity at any given time but also the number of stars should be finite. If we assume, that there are $N$ point-like stars in the universe, spread over the hypersphere we could, in principle, calculate for any given moment and any given point when the radiation currently observed from a given star was emitted. And so the OP was mostly right in his comment:

for a finite number of stars in a closed space, even if the stars are radiating from an infinite amount of time, the total luminosity at a given location and at a given time should be finite.

The mostly part of this is the missing qualifier "upto a set of measure zero". Indeed for almost all points the times when stars emit radiation being observed presently would be finite and so blueshifting the spectra of each star could only produce finite total luminosity. However if we keep time constant (say $t=0$) and vary the point for such calculation are done then there would be points at which observed luminosity would be arbitrarily large, and there would be points where it would diverge. These are the points for which past lightcone contains one of the stars at infinity. And of course, the total amount of energy as integrated over the section $t=\rm const$ would be diverging.

For $t=0$ this zero measure set is quite simple to construct. Since light emitted at $t=-\infty$ would have travel exactly $\pi/2$ angle on the hypersphere by the moment $t=0$, then for any given star we would have a great circle equidistant from the star. And so this set is simply a union of $N$ such circles. At other times this set of infinite blueshift would be a set of circles (no longer great) with centers in the stars.

It would be interesting to illustrate this reasoning by combining first and second approaches and produce a simulated "night sky" displaying a sampling of stars with colors and brightness indicating spectrum shifts and luminosities.