I've recently stumbled upon an alternative version of showing Olbers' Paradox analytically, namely the following problem from Arfken and Weber's Mathematical Methods for Physicists:

Assume a static universe in which the stars are uniformly distributed. Divide all space into shells of constant thickness; the stars in any one shell by themselves subtend a solid angle of $ω_0$. Allowing for the blocking out of distant stars by nearer stars, show that the total net solid angle subtended by all stars, shells extending to infinity, is exactly $4π$.

Now, I am familiar with the light intensity formulation of the paradox, but here I seem to have a bit of a problem working with solid angles. This is my attempt at the solution:

I assumed that every star on any spherical shell covers some constant fraction of the surface, call it $dS$. Then the solid angle that each star in some shell subtends is $d\omega_0(r)=\frac{dS}{r^2}$. Hence, for every spherical shell, the total sum of solid angles should be:

$$N_i \ d\omega_0=4\pi r_i^2dr_i \gamma \frac{dS}{r_i^2}=4\pi\gamma dr_i dS$$

where $N_i$ is the number of stars/shell, $4\pi r_i^2dr_i$ is the volume of the shell with thickness $dr_i$ and $\gamma$ is the number density of stars/volume of the shell. Now summing over all shells (and all stars, in essence), we have

$$\omega_{total}=Nd\omega_0=4\pi\gamma dS \int_0^\infty dr$$

where $N$ is the number of all stars and also noting $\gamma=\frac{N}{\iiint_V drdS}$, since the distribution is uniform. But this logic gets me nowhere and I think the reason is that I do not fully understand the question, namely what exactly is meant by

the stars in any one shell by themselves subtend a solid angle of $ω_0$. Allowing for the blocking out of distant stars by nearer stars

Also, this problem is under the chapter of convergence of series in the previously mentioned book, thus I assume these should pop up somewhere, but I can't see any apart from the sum over all shells.

Any help is much appreciated!

  • $\begingroup$ So it's pretty easy to show that each shell has a fraction $f$ shinning, and hence a fraction (of $4\pi$) $1-f$ blocking. Now add the shells up. There is no need to worry about individual stars. Also: try to pick symbols that don't clutter the problem--it just makes it harder to follow what you are doing. $\endgroup$
    – JEB
    Oct 1, 2018 at 17:33

1 Answer 1


Thanks to @JEB the answer is, of course, a lot simpler than I have set it out. Let me post it here.

Suppose each star covers some constant surface on each shell, call it $dS$. Then the solid angle subtended by that star is


The number of stars in each shell is related to:

$$N(r)=\gamma dV=\gamma4\pi r^2dr,$$

where, assuming equal distribution, the number density $\gamma$ is constant, and so is $dr$ by starting assumptions. Then the total solid angle subtended by all stars in any which shell is:

$$\omega=N(r)\omega_0(r)=4\pi\gamma drdS$$

Albeit $N$ and $\omega_0$ being functions of $r$, we now notice, that the sum of solid angles for all stars in any shell is a constant. Thus we define:

$$f:=\frac{\omega}{4\pi}=\gamma drdS$$

as being the constant fraction of $4\pi$ for stars in any shell. Hence, in each shell, $f$ contains stars, but assuming blocking out of distant stars by nearer stars, only $1-f$ of each shell is going to let through light from consecutive shells. Thus, summing these fractions over all shells:

$$f+f(1-f)+f(1-f)^2+f(1-f)^3+...=f\sum_{n=0}^\infty (1-f)^n$$

Since $\mid{1-f}\mid<1$, we are dealing with convergence of the geometric series and in the limit:

$$ f lim_{n\rightarrow\infty} \frac{1-(1-f)^n}{1-(1-f)}=f \frac{1}{f}=1 $$

Hence, $\omega=4\pi$, and the night sky should be bright with stars, given the assumptions of Olbers' paradox.

Again, thank you for pointing me in the right direction @JEB.


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