1
$\begingroup$

Will it suck the entire universe in? What will the black hole look like to us, assuming we do not immediately get sucked in?

$\endgroup$
  • $\begingroup$ A requirement of a black hole is that the escape velocity at it's event horizon is c. Dark energy might make that impossible. The vast distance and stretching of space and red-shifting of distant objects might make universe sized black holes impossible. (just throwing that out there). $\endgroup$ – userLTK Mar 24 '18 at 7:15
  • $\begingroup$ It wouldn't be better to consider how the formula for the Schwartzschikd radius looks like? As it is your question seem to suggest implications such as "can be our universe the interior of a black hole" and the like. And even conceptual and or semantic trouble as noted by a user. $\endgroup$ – Alchimista Mar 26 '18 at 9:56
  • $\begingroup$ It wouldn't be better to consider how the formula for the Schwartzschikd radius looks like? As it is your question seem to suggest implications such as "can be our universe the interior of a black hole" and the like. And even conceptual and or semantic trouble as noted by a user. $\endgroup$ – Alchimista Mar 26 '18 at 9:56
1
$\begingroup$

According to measurements of the cosmic microwave background the universe is geometrically flat - which means that the mass/energy density of the universe is close to the "critical value" of $\sim 10^{-26}$ kg/m$^{3}$.

The radius of the observable universe is 46.6 billion light years, so the mass/energy contained within it is equivalent to $3.6\times 10^{54}$ kg.

The Schwarzschild radius of a black hole is $2GM/c^2$. If the mass/energy of the universe is spherically symmetric then its Schwarzschild radius is 560 billion light years and thus larger than the observable universe.

Note though that the Schwarzschild solution in General Relativity is static. The universe is definitely not static.

$\endgroup$
  • $\begingroup$ Does that mean the Schwarzschild solution in GR is merely a close approximation, given that the accelerating expansion of space due to dark energy applies not just to the universe as a whole but also to smaller regions of space where gravity is the dominant factor? $\endgroup$ – Chappo Mar 25 '18 at 0:38
  • $\begingroup$ @chappo (a) Yes - the Schwarzschild metric can only be an approximation for real astrophysical black holes - since they have to form. (b) Your point about expansion is incorrect when applied to gravitationally bound structures - they do not experience expansion. $\endgroup$ – Rob Jeffries Mar 25 '18 at 3:48
  • $\begingroup$ I respectfully suggest on point (b) that the error is yours. Gravitationally bound structures do "experience" expansion: they're not somehow exempt from the FLRW metric. It would be more accurate to say that expansion is negligible at all but the very largest scales. For example, at the scale of Earth's orbit, the rate of expansion is a mere 1.2 mm per hour. Nonetheless, negligible is not the same as non-existent. $\endgroup$ – Chappo Mar 25 '18 at 11:38
  • $\begingroup$ @chappo No, it is much more complicated than that (and even if it were not, then 1 km/century, would imply $\dot{a}/a \simeq 10^{-8}$/century, that is completely ruled out by current estimates of $\sim 10^{-14}$/century.) arxiv.org/pdf/1108.0246.pdf . You could also usefully look at this physics.stackexchange.com/questions/70047/… $\endgroup$ – Rob Jeffries Mar 25 '18 at 12:57
  • $\begingroup$ Wouldn't your argument as treating the universe we are in as the interior of a black hole? Wouldn't this a self conflicting situation? Or in other words, is the interior of a black hole supposed to have a flat space too? If not than we can take any mass as big as what we think and calculate the S radius. Perhaps is the question that is highly pictorial. It should be "what is the formula for Schwarzschild radius?". $\endgroup$ – Alchimista Mar 26 '18 at 9:52
0
$\begingroup$

Before I answer this, it's important to correct a few assumptions:

(1) we can sit outside a universe-as-black-hole. This is impossible: since the universe includes everything that exists, then by definition we must be within it, so we can't look at it from the "outside".

(2) a black hole "sucks" matter into it. It doesn't, any more than a large star "sucks" matter. If the Sun somehow collapsed and became a black hole (it can't, this is just a thought experiment), all the planets would continue to orbit pretty much as usual, since the Sun's mass wouldn't have changed.

Now, to your core question:

If a black hole has a mass of a universe what will be the volume of it?

The Schwarzschild radius is the radius defining the event horizon of a Schwarzschild black hole. If we take the mass of the observable universe as roughly $10^{53}$ kg, then using the formula $$R=\frac{2GM}{c^2}$$ the Schwarzschild radius of this mass is 15.7 billion light years [NB: by comparison, the comoving distance to the edge of the observable universe is about 46.6 billion light years]. The volume is then easily calculated as 1.6 x $10^{31}$ cubic light years or roughly $10^{79}$ $m^3$.

For comparison, this is less than 4% of the volume of the observable universe.


EDIT:

Wikipedia's "quick facts" on the observable universe give the mass as $10^{53}$ kg, but the body of the article contains the following qualification:

The mass of the observable Universe is often quoted as $10^{50}$ tonnes or $10^{53}$ kg. In this context, mass refers to ordinary matter and includes the interstellar medium (ISM) and the intergalactic medium (IGM). However, it excludes dark matter and dark energy. This quoted value for the mass of ordinary matter in the Universe can be estimated based on critical density. The calculations are for the observable universe only as the volume of the whole is unknown and may be infinite.

My calculations are based on the mass of ordinary matter in the observable universe, representing 4.9% of the total "mass/energy" derived from the observed critical density and volume. Rob's answer includes dark matter (26.8% of total mass/energy) and dark energy (68.3% of total mass/energy). Both answers are thought experiments, since it's not possible to have a black hole with the mass of the universe within our universe.

In a comment on the main question, userTLK makes an additional valid point that "the escape velocity at [a black hole's] event horizon is c. Dark energy might make that impossible. The vast distance and stretching of space and red-shifting of distant objects might make universe sized black holes impossible."

$\endgroup$
  • $\begingroup$ You haven't answered the core of the question, which is to justify the figure of $10^{53}$ kg. I think it may be considerably in error. $\endgroup$ – Rob Jeffries Mar 24 '18 at 8:58
  • $\begingroup$ I took the mass of observable Universe from Wikipedia, and a similar mass and radius were given in an answer in PhysicsSE. But Rob, if you’re aware of a more reputable source please let me know (and the mass quoted). $\endgroup$ – Chappo Mar 24 '18 at 10:48
  • $\begingroup$ You need to read your sources more carefully. Wikipedia does not say that $10^{53}$ kg is the "mass of the observable universe". $\endgroup$ – Rob Jeffries Mar 24 '18 at 10:54
  • $\begingroup$ Ah yes, I see that it gives 10^53 kg in several places as “mass of the observable universe” but then qualifies this (later) as the mass of “ordinary matter”. Thanks for the correction. I’ll amend my answer accordingly. $\endgroup$ – Chappo Mar 24 '18 at 13:04
  • $\begingroup$ @RobJeffries: I also relied on James K's answer here, which says "The mass of the observable universe is of the order 10^53kg, 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". This error must have flown under your radar? $\endgroup$ – Chappo Mar 25 '18 at 0:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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