Given a mass (energy) density of $\rho$, a ball of radius $R$ has a mass of $M(R)=4\pi/3\cdot R^3\cdot\rho$. The Schwarzschild Radius of this mass in dependence of $R$ is $$ r_s(R) = \frac{2GM(R)}{c^2} = \frac{2G\cdot 4\pi\cdot R^3\cdot\rho}{c^2\cdot 3} = \frac{8\pi G R^3\rho}{3 c^2} $$

Now we can ask, given a fixed $\rho$, when does the Schwarzschild radius get larger than $R$, i.e. when do we have $r_s(R)\ge R$? That would be if \begin{align} \frac{8\pi G R^3\rho}{3 c^2} &\ge R\\ R^2 &\ge \frac{3c^2}{8\pi G\rho} \end{align} For the average density of the universe Wikipedia lists $\rho=9.9\times 10^{-27} \text{kg}/\text{m}^3$. We also have $G=6.7\times 10^{-11} \text{m}^2/(\text{s}^2\text{kg})$ and $c=299792458\, \text{m}/\text{s}$. Inserting the values we get that a volume with radius $R$ and the mass density of the universe passes its Schwarzschild radius if

\begin{align} R^2 &\ge \frac{3\cdot 299792458^2}{8\pi \cdot 6.7\times 10^{-11} \cdot 9.9\times 10^{-27}} \text{m}^2\\ R^2 &\ge 1.62\times 10^{52} \text{m}^2 \\ R &\ge 1.27 \times 10^{26} \text{m}\\ \text{or in light years}\\ R &\ge 13.44\times 10^9 \text{Ly} \end{align}

Question 1: Where did I screw up above? This seems to contradict this answer whereby the Schwarzschild Radius of the whole mass of the universe is around $500\times 10^9 \text{Ly}$? (I did this in three different ways and always got the avobe result. :-/)

Edit Question 2: So far nobody pointed out formal errors in the above, so I dare to ask: how do we explain away that any sufficiently large sub-ball of the universe has a radius smaller than its Schwarzschild Radius?

  • $\begingroup$ I dont see anything wrong particularly...there is only 1 order of difference which seems small to me. Most likely he just rounded some numbers. $\endgroup$
    – seVenVo1d
    Oct 5, 2023 at 17:39
  • $\begingroup$ From the linked answer: "Firstly we should note that the universe as a whole is not described by the Schwarzschild metric, so the Schwarzschild radius of the universe is a meaningless concept." Doesn't that answer your question? $\endgroup$ Oct 6, 2023 at 9:47
  • $\begingroup$ Only the first one in that my computation above may be similarly valid as the one in the cited answer. Which leaves Question 2. 14 billon light years is only a small part of the universe. And a ball of this radius, watched from the outside, seems to contain enough mass to be smaller than its Schwarzschild Radius. What is it that potentially makes the Schwarzschild metric completely inapplicable? Sure, the mass and energy in this ball may rotate and move and whatnot, but does this mean it is not even usable as an order of a handful of magnitudes estimate? $\endgroup$
    – Harald
    Oct 6, 2023 at 9:53
  • $\begingroup$ For that you don't need Schwarzschild but Schwarzschild de Sitler $\endgroup$
    – Yukterez
    Oct 6, 2023 at 10:44

1 Answer 1


Schwartzschild solution applies ONLY in a stationary spherically symmetric vacuum. The universe is not stationary and not a vacuum. So what did you get wrong? Actually everything, the complete idea. You can't apply the Schwartzschild solution to the Einstein equations for the universe because the universe doesn't satisfy the same conditions


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