0
$\begingroup$

If one can estimate the mass of our universe, might one also calculate its ultimate size, (after it fully expands), based on the Schwarzschild formula for the radius of a black hole? Please note that this simple formula $$R=\frac{2GM}{c^2}$$ informs us that larger black holes are less dense. So, if it turns out that our universe is in fact a gigantic black hole, then one would expect it to be extremely low-density!

$\endgroup$

closed as unclear what you're asking by called2voyage Jul 15 '16 at 13:16

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

1
$\begingroup$

If I understand the question the answer is, no, you can't estimate the final size of the Universe by calculating the schwarzschild radius of the universe.

The mass of the Universe is not known, and may be infinite.

It is not know if the observable universe will ever stop expanding, however the evidence suggests that it won't, and so won't have an "ultimate" size.

The mass of the observable universe is of the order $10^{53}kg$, mostly comprised of dark energy and dark matter.

A black hole with that mass would have a radius of $2MG/c^2 = 10^{26}m$, which as has been observed is on the same order of magnitude as the (comoving) radius of the observable universe. Whether this has any cosmological implications is not clear. It doesn't imply that the Universe is a black hole. It doesn't say anything about the Universe at all, as this calculation was for the observable universe. It doesn't say anything about the "ultimate size" of the universe or the observable universe (which is expected to continue expanding without limit)

$\endgroup$
  • $\begingroup$ The "mass of the observable universe" is much greater than $10^{53}$ kg - you have not included dark matter/energy. $\endgroup$ – Rob Jeffries Mar 25 '18 at 3:45

Not the answer you're looking for? Browse other questions tagged or ask your own question.