<p>For simplicity, let's pick a standard star, of radius <span class="math-container">$R_\odot$</span> (you can easily generalize this for stars of any radius by picking <span class="math-container">$R_\odot$</span> to be the lower bound of star radius). </p>

<p>Let's split the universe into cubes. It would be much easier to use concentric shells, but later on it will be practical to have everything split into components of standard sizes. Each cube is of side <span class="math-container">$\ell$</span>, and volume <span class="math-container">$\ell^3$</span>. Within each cube, there is a probability distribution <span class="math-container">$P(N_\odot = n)$</span> to have <span class="math-container">$n$</span> stars within that cube. We're not requiring anything of that probability distribution, except</p>

<p><span class="math-container">\begin{equation}
E[N_\odot] = \rho_\odot \ell^3
\end{equation}</span></p>

<p>There is an average stellar density <span class="math-container">$\rho_\odot &gt; 0$</span> that is the same everywhere. </p>

<p>Now trace your line of sight from <span class="math-container">$0$</span> to infinity. This will go through an infinite number of such cubes. We can simplify the problem by having <span class="math-container">$\ell \approx R_\odot$</span>, so that we don't have to worry about whether the line passes through a star. The fact that this line will encounter no star is equivalent to the experiment of each of those cubes to have zero stars. As each such probability is independent, this is</p>

<p><span class="math-container">\begin{equation}
\prod_{n = 1}^\infty P(N_{n, \odot} = 0)
\end{equation}</span></p>

<p>We can set this as a Bernoulli trial. Zero stars is a failure with probability <span class="math-container">$p$</span>, more than zero star is a success with probability <span class="math-container">$q$</span>. The probability of zero star cannot be <span class="math-container">$1$</span>, since we assumed a non-zero density. Therefore, the probability becomes</p>

<p><span class="math-container">\begin{equation}
\lim_{n \to \infty} p^n = 0
\end{equation}</span></p>

<p>While it is not impossible, the probability of that line never hitting a star is of measure <span class="math-container">$0$</span>, and the probability of hitting one is of measure <span class="math-container">$1$</span>.</p>

CC BY-SA 4.0
2020-03-26T15:35:43.247Z
2020-03-26T15:35:43.247Z
Slereah
<p>From <a href="http://en.wikipedia.org/wiki/Olbers%27_paradox" rel="nofollow">Wikipedia</a>, here's Lord Kelvin's statement of Olbers' Paradox:</p>

<blockquote>
  <p>Were the succession of stars endless, then the background of the sky would present us a uniform luminosity, like that displayed by the Galaxy – since there could be absolutely no point, in all that background, at which would not exist a star.</p>
</blockquote>

<p>This is one form of Olbers' Paradox I've heard, which is that with infinitely many stars scattered in an infinite volume, every ray must eventually intersect a star. I call this the "strong form" because it seems to me stronger than the statement that infinite stars imply an infinite amount of light hitting every point, though not necessarily from every possible direction.</p>

<p>The strong form of the paradox doesn't hold if each star is a point mass and there are countably many stars, since the number of rays from any given point is uncountably infinite. Or imagine instead that stars are balls with small positive radius. If there is such a star at every lattice point in $R^3$ (an assumption that is consistent with my crude understanding of the cosmological principle), I can easily find a ray that intersects no star. I start from (0, 0, 0.5) and choose the ray in the direction of (1, 0, 0.5); the z coordinate will always be 0.5 and therefore my ray will never come close to a lattice point.</p>

<p>Is there another, palatable assumption we can add to make the strong form of Olbers' Paradox true, perhaps an argument that the distribution of stars must be more random than what I've described?</p>

CC BY-SA 3.0
2015-03-06T06:03:11.233Z
**I can easily find a ray that intersects no star. I start from (0, 0, 0.5) and choose the ray in the direction of (1, 0, 0.5); the z coordinate will always be 0.5 and therefore my ray will never come close to a lattice point.** This is not good enough, if we are going to resort to the type of reasoning you seem to want you need to show that there is a cone of finite cone angle of rays which never intersect a star. Or prove that the set of rays which do not intersect a star is not of measure zero.
Conrad Turner
CC BY-SA 3.0
2015-03-06T18:25:29.933Z
Gravitational lensing would make an infinite lattice appear uniformly infinitely bright. It is a strong statement to simply say: If there were an infinite number of stars, stationary with respect to us, and existing for a long enough time, then the sky would not be as dark as it is.
eshaya
<cosmology>
Assumptions necessary for the strong form of Olber's Paradox
kuzzooroo
# Assumptions necessary for the strong form of Olber's Paradox

(Asked by: username_0 on 2015-03-06)

From [Wikipedia](http://en.wikipedia.org/wiki/Olbers%27_paradox), here's Lord Kelvin's statement of Olbers' Paradox:

> Were the succession of stars endless, then the background of the sky would present us a uniform luminosity, like that displayed by the Galaxy – since there could be absolutely no point, in all that background, at which would not exist a star.

This is one form of Olbers' Paradox I've heard, which is that with infinitely many stars scattered in an infinite volume, every ray must eventually intersect a star. I call this the "strong form" because it seems to me stronger than the statement that infinite stars imply an infinite amount of light hitting every point, though not necessarily from every possible direction.

The strong form of the paradox doesn't hold if each star is a point mass and there are countably many stars, since the number of rays from any given point is uncountably infinite. Or imagine instead that stars are balls with small positive radius. If there is such a star at every lattice point in $R^3$ (an assumption that is consistent with my crude understanding of the cosmological principle), I can easily find a ray that intersects no star. I start from (0, 0, 0.5) and choose the ray in the direction of (1, 0, 0.5); the z coordinate will always be 0.5 and therefore my ray will never come close to a lattice point.

Is there another, palatable assumption we can add to make the strong form of Olbers' Paradox true, perhaps an argument that the distribution of stars must be more random than what I've described?

**Tags:** cosmology

> **Comments:**
> **username_1**: **I can easily find a ray that intersects no star. I start from (0, 0, 0.5) and choose the ray in the direction of (1, 0, 0.5); the z coordinate will always be 0.5 and therefore my ray will never come close to a lattice point.** This is not good enough, if we are going to resort to the type of reasoning you seem to want you need to show that there is a cone of finite cone angle of rays which never intersect a star. Or prove that the set of rays which do not intersect a star is not of measure zero.  
> **username_2**: Gravitational lensing would make an infinite lattice appear uniformly infinitely bright. It is a strong statement to simply say: If there were an infinite number of stars, stationary with respect to us, and existing for a long enough time, then the sky would not be as dark as it is.  

---

## 1 answers

### Answer by username_3 on 2020-03-26. Score: 7 | Accepted answer

For simplicity, let's pick a standard star, of radius $R\_\odot$ (you can easily generalize this for stars of any radius by picking $R\_\odot$ to be the lower bound of star radius).

Let's split the universe into cubes. It would be much easier to use concentric shells, but later on it will be practical to have everything split into components of standard sizes. Each cube is of side $\ell$, and volume $\ell^3$. Within each cube, there is a probability distribution $P(N\_\odot = n)$ to have $n$ stars within that cube. We're not requiring anything of that probability distribution, except

\begin{equation}
E[N\_\odot] = \rho\_\odot \ell^3
\end{equation}

There is an average stellar density $\rho\_\odot > 0$ that is the same everywhere.

Now trace your line of sight from $0$ to infinity. This will go through an infinite number of such cubes. We can simplify the problem by having $\ell \approx R\_\odot$, so that we don't have to worry about whether the line passes through a star. The fact that this line will encounter no star is equivalent to the experiment of each of those cubes to have zero stars. As each such probability is independent, this is

\begin{equation}
\prod\_{n = 1}^\infty P(N\_{n, \odot} = 0)
\end{equation}

We can set this as a Bernoulli trial. Zero stars is a failure with probability $p$, more than zero star is a success with probability $q$. The probability of zero star cannot be $1$, since we assumed a non-zero density. Therefore, the probability becomes

\begin{equation}
\lim\_{n \to \infty} p^n = 0
\end{equation}

While it is not impossible, the probability of that line never hitting a star is of measure $0$, and the probability of hitting one is of measure $1$.


---

