# An alternative version of Olbers' Paradox

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!

• 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. – JEB Oct 1 '18 at 17:33

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

$$\omega_0(r)=\frac{dS}{r^2}$$

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.