The Olber's paradox

During the 19th century, there were many issues relating to cosmology. The standard model of cosmology of this era was roughly defined by an infinite and eternal, roughly homogeneous universe.

The Olber's paradox

The Olber's paradox is rather simple to understand. If there are an infinite number of stars uniformly distributed, and the universe is infinitely old, statistically speaking, wherever you look, there should be a star. On the other hand, the night sky is mostly black.

There's a variety of methods we can use to show this in a more rigorous manner.

The simplest we can consider is the homogeneous case. For every point in space, that point radiates some constant luminosity. If we assume furthermore that this is static, the light of every star will reach any point in the universe, with intensity

\begin{equation} L_r = \frac{L}{4\pi r^2} \end{equation}

At any point, we therefore have that the perceived luminosity is

\begin{eqnarray} L_{\mathrm{tot}} &=& \int_{S} \int_{0}^{\infty} L_r r^2 \sin \theta dr d\theta d\varphi\\ &=& L \int_{0}^{\infty} dr \end{eqnarray}

The luminosity is obviously divergent. Trying to salvage this by considering stars of varying luminosity doesn't help here. If we switch$L$ for $L(r, \theta, \varphi)$, with a lower bound $m$, corresponding to the lowest luminosity of a star, we have

\begin{eqnarray} L_{\mathrm{tot}} &=& \int_{S} \int_{0}^{\infty} \frac{L(r, \theta, \varphi)}{4\pi r^2} r^2 \sin \theta dr d\theta d\varphi\\ &=& \frac{1}{4\pi} \int_{S} \int_{0}^{\infty} L(r, \theta, \varphi) \sin \theta dr d\theta d\varphi\\ &\geq& \frac{1}{4\pi} \lim_{R \to \infty} m \frac{4}{3} \pi R^3 \end{eqnarray}

As long as the lower bound of luminosity is superior to zero, it will always diverge.

A slightly more realistic model is to consider the stars to be on a grid. If every star is on a lattice of length $\ell$, at position $(i, j, k)$, with the observer at $(0,0,0)$, then the total luminosity is

\begin{eqnarray} L_{\mathrm{tot}} &=& \sum_{i,j,k \in \mathbb{Z}^3 \setminus \{ 0 \}} \frac{L}{4\pi \ell^2 (i^2 + j^2 + k^2)}\\ &=& \frac{L}{4\pi \ell^2} \sum_{i,j,k \in \mathbb{Z}^3 \setminus \{ 0 \}} \frac{1}{(i^2 + j^2 + k^2)} \end{eqnarray}

For constant $j, k \neq (0,0)$, this sum is

\begin{eqnarray} \sum_{i \in \mathbb{Z}} \frac{1}{(i^2 + j^2 + k^2)} &=& \frac{\pi \coth(\pi \sqrt{j^2 + k^2})}{\sqrt{j^2 + k^2}}\\ &\geq& \frac{\pi}{\sqrt{j^2 + k^2}} \end{eqnarray}

Let's now consider the sum over $j$ such that $k \neq 0$.

\begin{eqnarray} \sum_{j \in \mathbb{Z}} \sum_{i \in \mathbb{Z}} \frac{1}{(i^2 + j^2 + k^2)} &=& \sum_{j \in \mathbb{Z}}\frac{\pi \coth(\pi \sqrt{j^2 + k^2})}{\sqrt{j^2 + k^2}}\\ &\geq& \pi (\frac{1}{k} + 2 \lim_{n \to \infty} \sum_{j = 1}^n \frac{1}{\sqrt{j^2 + k^2}})\\ &\geq& \pi (\frac{1}{k} + 2 \lim_{n \to \infty} \sum_{j = 1}^n \frac{1}{j \sqrt{1 + \frac{k}{j}^2}})\\ &\geq& \pi (\frac{1}{k} + 2 \lim_{n \to \infty} \sum_{j = 1}^n \frac{1}{j \sqrt{1 + k^2}})\\ &\geq& \frac{\pi}{k} (1 + 2 \lim_{n \to \infty} \sum_{j = 1}^n \frac{1}{j})\\ \end{eqnarray}

As the sum of $j^{-1}$ is divergent, and we haven't even finished summing over all of our stars, the luminosity is therefore divergent. There again, we can attempt to have varying luminosities, but the sum will always be superior to the sum using the lower bound of the luminosity and, as long as it is superior to zero, it will also diverge.

Statistical models

Now let's consider the case of a universe with a random (but uniform) distribution of stars. For any volume of space, we have some probability $P(N_{\odot} = n)$ to have $n$ stars in that volume, such that the average number in a volume is constant. In other words,

\begin{eqnarray} E[N_\odot] = \rho_\odot V \end{eqnarray}

The expectation value is then some stellar density $\rho_\odot$. Let's split our universe into cubes, such that the observer is in the center of the cube $(0,0,0)$ at $[-\ell/2, \ell/2]^3$. Our probability in one of those cubes then becomes

\begin{eqnarray} E[N_\odot] = \rho_\odot \ell^3 = \bar{N}_\odot \end{eqnarray}

Within a cube $(i,j,k)$, the distance to our observer is

\begin{equation} r \in [\frac{\ell}{\sqrt{2}} \sqrt{i^2 + j^2 + k^2}, \frac{\ell}{\sqrt{2}} \sqrt{(|i| + 1)^2 + (|j| + 1)^2 + (|k| + 1)^2}] \end{equation}

Any star in the cube $(i,j,k)$ will therefore have a minimal and maximal luminosity of

\begin{eqnarray} L_{\min(i,j,k)} &=& \frac{L}{4\pi \frac{\ell}{\sqrt{2}} r_{\max}} \\ L_{\max(i,j,k)} &=& \frac{L}{4\pi \frac{\ell}{\sqrt{2}} r_\min} \end{eqnarray}

If we have $n$ stars in that cube, the total luminosity will then be

\begin{equation} L_{n(i,j,k)} \in [n L_{\min(i,j,k)}, n L_{\max(i,j,k)}] \end{equation}

Now let's consider a shell of such cubes. We will pick the boundary of every cube in $[-d, d]^3$, which is simply every cube with coordinates of the form $(\pm d, j, k)$, $(i, \pm d, k)$, $(i, j, \pm d)$, for $i,j,k \in [-n, n]$. That boundary has exactly

\begin{equation} (2d)^3 - (2(d-1))^3 = 3d^2 - 3d - 1 \end{equation}

cubes. The radius within this shell is within the interval

\begin{equation} r \in [\frac{\ell}{\sqrt{2}} d, \sqrt{3}\frac{\ell}{\sqrt{2}} (d+1)] \end{equation}

and similarly, the luminosity of a single star is within

\begin{eqnarray} L_{d\min} &=& \frac{L}{4\pi \frac{\ell}{\sqrt{2}} r^2_{d\max}} \\ L_{d\max} &=& \frac{L}{4\pi \frac{\ell}{\sqrt{2}} r^2_{d\min}} \end{eqnarray}

What we need to compute is the probability of the total luminosity to be bounded. This is, for any $K$,

\begin{eqnarray} P(L_{\mathrm{tot}} < K) &=& P(\sum_{d = 1}^\infty L_{d} < K) \\ &=& P(\sum_{d = 1}^\infty \sum_{i = d \wedge j = d \wedge k = d} L_{(i,j,k)} < K) \end{eqnarray}

First we're going to have to look into the probabilities of sums of random variables. For the sum $Z = \sum_i X_i$, we are looking for every values of $X_i$ such that $Z = K$. For a finite number $n$ of $X$, this is the hypersurface on $[0,K]^n$ defined by that equation. The cumulative probability distribution to $K$ will be the area under that hypersurface. We do not need to worry about the actual integral on that domain, since it will obviously be inferior to the integral performed on the cube $[0, K]^n$ itself. In other words,

\begin{eqnarray} P(Z \leq K) &\leq& \int_{[0,k]^n} f_{\times_i X_i} (x) dx \end{eqnarray}

If they are independent, we have

\begin{eqnarray} P(Z \leq K) &\leq& \int_{[0,k]^n} \prod_{i = 1}^n (f(x_i) dx_i)\\ &\leq& \prod_{i = 1}^n \int_{[0,k]} (f(x_i) dx_i)\\ &\leq& \prod_{i = 1}^n P(X_i \leq K) \end{eqnarray}

As long as this integral is inferior to $1$ for a cofinite number, this will be zero. In our case, this will be

\begin{eqnarray} P(L_{\mathrm{tot}} < K) &\leq& \prod_{d = 1}^\infty P( L_{d} < K) \end{eqnarray}

Now, if we assume that each cube in the shell has $N_{a}$ stars, $a \leq 3d^2 - 3d - 1$,

\begin{eqnarray} P(L_{\mathrm{tot}} < K) &\leq& \prod_{d = 1}^\infty P( \sum_a^{3d^2 - 3d - 1} L_{a} < K)\\ \end{eqnarray}

For every cube $a$, we have a random variable of the number of stars, $N_a$. The total luminosity is then simply $N_a L_a$.

\begin{eqnarray} P(L_{\mathrm{tot}} < K) &\leq& \prod_{d = 1}^\infty P( \sum_a^{3d^2 - 3d - 1} N_a L_{a} < K)\\ \end{eqnarray}

By picking the closest possible distance in our shell, we can remove $L_a$ from the sum, and simply get

\begin{eqnarray} P(L_{\mathrm{tot}} < K) &\leq& \prod_{d = 1}^\infty P(L_{d\max} \sum_a^{3d^2 - 3d - 1} N_a < K)\\ &\leq& \prod_{d = 1}^\infty P(\frac{2L}{4\pi \ell^2 d^2} \sum_a^{3d^2 - 3d - 1} N_a < K)\\ \end{eqnarray}

By once again picking a larger function, we're going to recast everything inside to a more pleasant form

\begin{eqnarray} P(L_{\mathrm{tot}} < K) &\leq& \prod_{d = 1}^\infty P(L_{d\max} \sum_a^{3d^2 - 3d - 1} N_a < K)\\ &\leq& \prod_{d = 1}^\infty P(\frac{1}{3 d^2} \sum_a^{3d^2} N_a < \frac{\ell^2 K}{L})\\ \end{eqnarray}

We can recognize here something close to the law of large numbers. Its exact value isn't too important, but we wish to make sure that it will always be $< 1$, so that the infinite product will itself converge to zero. It would in fact converge to $1$ if the average value of $N$ was $0$, but as it is not, it will more likely converge to zero, via the $0-1$ theorem. As the mean isn't zero,

\begin{eqnarray} P(\frac{1}{3 d^2} \sum_a^{3d^2} N_a < \frac{\ell^2 K}{L}) &=& \end{eqnarray} \begin{eqnarray} P(L_{\mathrm{tot}} < K) &\leq& \prod_{d = 1}^\infty P((3d^2 - 3d - 1) (\sum_a n_a) L_{d\max} < K)\\ \end{eqnarray}

To simplify things a bit, we have that

\begin{eqnarray} (3d^2 - 3d - 1) L_{d\max} &=& (3d^2 - 3d - 1) \frac{L}{6 \pi \ell^2 (d+1)^2} \end{eqnarray}

We can pick a function that is always superior to preserve our inequality. Such a function is

\begin{eqnarray} \frac{L}{\ell^2} \end{eqnarray}

We then have the much simpler inequality

\begin{eqnarray} P(L_{\mathrm{tot}} < K) &\leq& \prod_{d = 1}^\infty P((\sum_a n_a) \frac{L}{\ell^2}< K)\\ \end{eqnarray}

Now using the law of large numbers, we have that, for $\varepsilon > 0$,

\begin{eqnarray} \lim_{d \to \infty} P(|\frac{1}{d}(\sum_a^d n_a) - \bar{n} | > \varepsilon) &=& 0\\ &=& \lim_{d \to \infty} P(|(\sum_a^d n_a) - d\bar{n} | > d\varepsilon) \\ &\leq& \end{eqnarray}

Reusing the formula for sums of independent random variables,

\begin{eqnarray} P(L_{\mathrm{tot}} < K) &\leq& \prod_{d = 1}^\infty \prod_{a = 0}^{3d^2 - 3d - 1} P(n_a \frac{L}{\ell^2}< K)\\ \end{eqnarray}

This is roughly our cumulative distribution. We can use Markov's inequality here, which can be shown by considering

\begin{eqnarray} E[n_a] &=& \sum_{n < K} n p(n) + \sum_{n \geq K} n p(n) \\ &\geq& \sum_{n < K} n p(n)\\ &\geq& \end{eqnarray}

And therefore the probability of having a bounded total luminosity is zero, as long as neither $N_{\odot}$ or $L$ are zero.

Newtonian cosmology

Our big problem regarding Olber's paradox is the distribution of our stars. Why are we assuming that the distribution of mass in the universe is roughly uniform? This is related to the stability of cosmology in Newtonian mechanics.

First, let's consider a case that would be much easier to work out. A universe where, outside of a ball of radius $R$, the mass distribution is zero.


Before general relativity came along, quite a wide variety

Last updated : 2020-03-27 02:34:24
Tags : physics , classical-mechanics , cosmology , thermodynamics