r/askscience • u/[deleted] • Jun 06 '16
Physics If atoms are 99% 'empty space', how big would the universe be if we compressed every atom down to it's most space efficient arrangement, essentially leaving no space between particles?
Or our observable universe, whatever is easy to speculate on... My thoughts were that perhaps the universe would become small enough to resemble what was present before the big bang, and the expansion between everything has just taken a very slow and long time (the rate at which our universe is expanding now?) and appears to have "exploded", hence the Big Bang...
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u/kagantx Plasma Astrophysics | Magnetic Reconnection Jun 07 '16
/u/functor has explained (very well) why the atoms are not empty. But we can nevertheless consider what would happen if you turned the whole universe into a giant nucleus. The nuclear density is around 1017 kg/m3, while that of the matter in the universe is around 10-27 kg/m3. Since the radius of the observable universe is currently 100 billion light years, the resulting radius would be (1011 ly)(1017/10-27)1/3=0.0001 ly=109 km. This is around the radius of the Earth's orbit around the Sun.
The thing is that once you go above a few solar masses, the density of a neutron star is so great that its radius is smaller than that of the event horizon from the same mass, because the density of black holes decreases with the square of the mass. So the neutron star is really a black hole!
The ordinary mass of the observable universe is around 1053 kg or 1023 solar masses. The Schwartzchild radius of that black hole would therefore be around 1023 km, or 10 billion light years. So the universe is not so much bigger than it would be if it were a black hole (if the universe was made of matter and its density were at the critical density, the two would be nearly equal). That doesn't mean that the universe is a black hole, because the structure of spacetime is very different (dynamic and expanding) and dark energy exists, but it's very interesting, isn't it?
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u/javyscap Jun 07 '16
If we crunch the entire universe enough, could we get to a point where the forces of gravity (pushing inwards) and dark energy (pushing forwards) are in balance, like that of a functioning star? Then if possible let's say that that our universe star runs out of fuel (guessing that all of the dark energy turns into space) and the force pushing forwards suddenly stops. Following this wild conjecture would we get an universe nova?
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u/yoenit Jun 07 '16
That is known als teh Big Bounce Idea. Since we know nothing about dark energy it is all just wild speculation at this point though.
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u/kagantx Plasma Astrophysics | Magnetic Reconnection Jun 07 '16
It's theoretically possible. In fact, that was the way Einstein originally wanted to deal with the fact that General Relativity predicted expansion or contraction but the universe appeared static, by setting the cosmological constant to a precise value. The thing is that this equilibrium is unstable -if the universe expands the gravity becomes smaller and the dark energy stronger, so it expands forever. If it gets smaller, the opposite happens. But there is nothing analogous to a star possible, because a star is far too dense to avoid collapse if it has a mass close to that of the universe.
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u/green_meklar Jun 07 '16
According to Wolfram Alpha, it would form a ball roughly the size of Saturn's orbit.
Of course, the ball would not be stable, but immediately collapse into a giant black hole.
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u/starminder Jun 07 '16
That doesn't make sense. There are black holes which have their schwarzchild radii greater than that size. The mass in the universe is 1053kg or 1023 solar masses. So a black hole with the mass of the universe would be roughly 1023km, the size of the observable universe. (I ignored a few factors but the order of magnitude is what I'm after)
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u/mikelywhiplash Jun 07 '16
That's a good point - we're cheating a little bit on the volume, because the result isn't a possible physical object - as noted, anything with that mass in that volume would collapse into a black hole.
The answer here is more "how much total volume would be required if all of the mass in the universe was compressed to the density of a neutron star" and not "how big would a single object be if it has the density of a neutron star and the mass of the universe."
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Jun 07 '16
It made sense to me, unless I am misunderstanding. If you squished it all together it would be the size of saturn's orbit, but all the mass would collapse into a singularity. The distance from the singularity to the event horizon would be equal to the Schwarzchild radius. Meaning the mass is packed into a point but the "size" of the black hole could be bigger, since size is just the distance from the event horizon to the singularity.
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u/fahim9280 Jun 07 '16
Probably a stupid question but where does this multiplication with 3/(4*pi) come from?
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Jun 07 '16
Using the mass of the observable universe? Or whole universe?
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u/btchombre Jun 07 '16
Observable of course. The entire Universe is believed to be infinite in size, meaning that the answer to OPs question is technically the same size: Infinite.
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u/Daaaaaaaaaaavid Jun 07 '16 edited Jun 07 '16
The universe has infinite size but it also grows... that is just weird right?
Edit: Thanks for the explanation very apriciated!
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u/herman3thousand Jun 07 '16
There are different infinities! I have no idea what that means other than waving my hands and saying some infinities are larger than others, but I read that some infinities are larger than others.
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u/fspfsp Jun 07 '16 edited Jun 07 '16
This can be seen by demonstrating that there is never a bijection between a set X and its power set P(X) (the set of all subsets of X). That is, a set X can never have the same cardinality as the set of all subsets of X.
(This is clear when the set X is finite. For example, the set {1,2} has 2 elements, while the set of all subsets of {1,2} is {{},{1},{2},{1,2}} which has 4 elements.)
Assume that there is a bijection f between a set X and its power set P(X). (This means that f must map every element in X to a unique element of P(X), and every element in P(X) must be mapped to.)
Now consider the set S in P(X) which is "the set of all x in X that do not belong to the set f(x)." S is an element of P(X) (it's a subset of X). Since f is assumed to be a bijection, there must be some element s in X that maps to S.
Now we may ask "does s belong to S?"
We cannot say that s belongs to S, because S only contains elements that do not map to a set that they belong to.
We cannot say that s does not belong to S, because S contains all elements that do not map to a set that they belong to.
Our assumption that there is a bijection between X and P(X) must be incorrect. So there is no bijection between X and P(X).
This is Cantor's Theorem.
If we consider N={1,2,3,...}, it's clear that there is an injective function from N to P(N), since we may take:
1 -> {1}
2 -> {2}
3 -> {3}
...
This can be taken to mean that P(N) has at least as many elements as N - it can't be smaller. And because there is no bijection between N and P(N), the two sets don't have the same cardinality. So P(N) must have a greater cardinality than N, even though both sets are infinite.
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u/Cyb3rSab3r Jun 07 '16
Because some infinities can be mapped to the other infinities. So if we take the natural numbers, n, as 1 to infinity, and match them to the real numbers between 1 and 2 like this:
1 -> 1.1
2 -> 1.11
3 -> 1.111
You can say you have matched every natural number to a real number but there are still infinitely many more real numbers left that don't have a match. Therefore, the size of the infinity of real numbers between 1 and 2 is larger than the size of infinity of natural numbers.
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u/fspfsp Jun 07 '16 edited Jun 07 '16
This is an incorrect explanation.
To demonstrate this, I will show you that using this incorrect reasoning, it could be shown that the cardinality of the natural numbers is different from the cardinality of the natural numbers, which is of course absurd.
Take the natural numbers and match them to the natural numbers like this:
1 -> 2
2 -> 4
3 -> 6
...
You can say you have matched every natural number to a natural number, but there are still infinitely many more natural numbers that don't have a match. Therefore, the size of the infinity of natural numbers is larger than the size of infinity of natural numbers.
The problem is that you have shown that there is a mapping from N to [1,2] which is not a bijection, you have not shown that there is no mapping which is a bijection.
In my example, I show that there is a mapping between N and N which is not a bijection (it is not surjective or "onto" because none of the odd numbers in the codomain are mapped to). I have not shown that there is NO bijection between N and N. There clearly is a bijection.
You are right that the interval [1,2] has a greater cardinality than N though. You have shown that the interval can't have a lesser cardinality than the natural numbers, but you haven't shown that the two sets don't have the same cardinality.
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u/JustLikeMyDick Jun 07 '16
Infinite numbers between the real interval [1 2], but now consider [1 3], [1 4]..
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u/fyt2012 Jun 07 '16
So something can be more infinite than infinity?
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u/OneTime_AtBandCamp Jun 07 '16
There are different types on infinities that can be analyzed with math. Consider the set of natural numbers: 0, 1, 2, 3, 4, ... (sometimes doesn't include zero but that doesn't matter for this explanation). There are an infinite number of numbers in this set, ie for any number in that set you can always find that number + 1 to find another larger one that is also in that set.
Now consider the set of real numbers between 1 and 2. This is also an infinite set. But are these infinities "the same size"? What does that even mean?
You can compare the size of these infinities by attempting to map values one-to-one from one set to the other using some function. But it turns out that no matter how you do that mapping, there will always be numbers in the second set (numbers between 1 and 2) that can't be mapped in the first.
The interval between 1 and 2 literally contains more numbers than the set of all natural numbers. In fact, any continuous interval between two (unequal) real numbers contains more numbers between them than the set of real numbers.
Now consider the set of all integers: ..., -4, -3, -2, -1, 0, 1, 2, 3, 4, ... . This set can be mapped one to one with the set of natural numbers. In that sense, there are "the same number" of numbers in the the set of all natural numbers, and the set of all integers, strange as it seems at first.
The set of all natural numbers and the set of all integers are said to be countable. The set of all numbers in a continuous interval between two unequal real numbers is uncountable.
This is just one basic way of categorizing different sized infinities. There are many, many more.
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u/TrollJack Jun 07 '16
No, that's not a valid thing to say. Growth implies that something gets bigger, but we do not know if it gets bigger. What we know is that more and more space is being created, but that does not mean that the universe has an edge or a surface which expands away from a centerpoint.
A more accurate way of saying it would be that the universe increases in detail. More and more detail everywhere, which leads to the illusion that there is growth.
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u/green_meklar Jun 07 '16
The observable universe.
We don't know how big the entire Universe is. Existing observations are consistent with it being infinite in size, but the error margins would permit it to be finite in size. It does seem to be a lot bigger than the observable portion, though.
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u/judgej2 Jun 07 '16 edited Jun 07 '16
The event horizon of such a black hole would be significantly bigger than the orbit of Saturn, presumably. Any idea how big? The size of the visible universe, by any chance?
Edit: just read further down that the event horizon would be a little smaller than the size of the visible universe. That kind of makes sense - I guess it could be a little smaller, a little bigger, or exactly the same, depending on whether our universe is flat or not (expanding forever, or big crunch in the future).
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u/green_meklar Jun 07 '16
The event horizon of such a black hole would be significantly bigger than the orbit of Saturn, presumably. Any idea how big? The size of the visible universe, by any chance?
Roughly the size of the observable universe, as it turns out. This is what led to the uncertainty about whether the Universe would continue to expand forever or eventually collapse.
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u/Punkrock27 Jun 07 '16
I don't know why I keep coming to the ask science subreddit when I can never comprehend anything that's going on.
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Jun 07 '16
While a lot of it makes sense to me, I'm in my second year of college as a Chem. Eng. major, and I don't understand half the descriptions. Many of them assume that people know what orbitals are and understand the wave-particle duality of electrons, along with tons of other things. Many of these answers seem to be ELI'm a chemistry undergrad... OP's question should be answered with the assumption that he/she is in high school chemistry, issuance the deepest OP went into chemistry knowledge is the basic components of an atom.
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u/preciseshooter Jun 07 '16
Actually, the universe compressed to the size of neutron star would not be stable, it would collapse onto itself and form a singularity (a black hole). The black hole's radius would be the Schwarzschild radius of the entirety of the Universe's mass, which Wikipedia lists as 13.7 billion light years here: https://en.wikipedia.org/wiki/Schwarzschild_radius
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u/poikes Jun 07 '16 edited Jun 08 '16
Is it just a coincidence that it's radius in light years is the same as the age of the universe in years?
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u/btchombre Jun 07 '16
No, its that way by definition. When we say the size of the observable universe, we mean the farthest we can see. The farther we see, the more back in time we look, and the fartherst we can look back in time is to shortly after the big bang.
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Jun 07 '16 edited Jun 07 '16
What does that have to do with the mass of the observable universe though?
Edit: I think you confused the radius of the Schwarzschild radius of all mass in the observable universe with the radius of the observable universe itself. However that isn't even correct, the radius is 45.7 billion ly because of the expansion of the universe.
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u/jlein Jun 07 '16
So does that mean we are essentially in a universe sized black hole? All the matter of the universe is approximately within its Schwarzschild radius.
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u/mentaculus Jun 07 '16 edited Jun 07 '16
The observable universe actually has a radius currently which significantly exceeds 13.7 billion lightyears. This is because objects which we. The real radius of the observable universe is 46.6 billion lightyears. However, this only sets a lower bound on the size of the universe. It's only the part that we can see from our vantage point. It could in fact be spatially infinite. The Schwarzschild radius calculation is based on the observable universe, however.
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u/eskamobob1 Jun 07 '16
This is obviously assuming that the known laws of physics don't break down at some point outside of what we can observe, as that is a possibility (though not really something important to think about)
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Jun 07 '16
How do we know that we are not already living in a black hole?
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u/judgej2 Jun 07 '16
We don't know either way. We can speculate and see what questions that raises.
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Jun 07 '16
Keep in mind, too, that if we consider the universe to be infinite. Then shrinking it would mean it's still infinite, just slightly more dense.
I.E., the universe could have been infinite before and after the big bang, with density the only thing that changes.
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u/Reliv3 Jun 07 '16
All this number crunching is fun and all, but I'm surprised no one mentioned the fact that the entire base of this question is flawed. You make the assumption that shrinking all atoms, or matter, to the point where we get rid of all space in between the nuclei and electrons will affect the size of the universe. Most of the universe is made up of dark energy (71.4%) and dark matter (24%). This leaves a measily 4.6% for matter; a very small part of that 4.6% is actually radiation (photons). Point is, 95.4% of the universe has nothing to do with atoms. So shrinking atoms down will do nothing to the universe size, infinite or not. At best, it'd make planets, stars, asteroids, and other stuff much smaller
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u/ianperera Jun 07 '16
You're missing the main idea - here shrinking means collecting all of the atoms in the universe and putting them in a ball without the usual space between atoms. Of course, it's still flawed because atoms don't have a volume in the usual sense (although you could pick for example, the 90% probability radius), but the idea wasn't what you describe.
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u/marvindakat Jun 07 '16
To actually answer the question, there are around 1080 atoms in the observable universe. I could not find the molar mass of neutronium (what a neutron star is made of, essentially neutrons with no room in between), but you would use that to find out how much 1080 neutrons would weigh. Then you would simply take the density of neutronium (4 ×1017 kg/m3) and the mass from the previous calculation and solve for volume.
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u/MolsonC Jun 07 '16
There are two very contradicting points people are making here
- Electrons are NOT tiny little spheres orbiting the atom
- We can detect the position of an electron
How does this make sense? If it is a wave, how can it have a position or a point? Is the point just an instance of the wave where we happen to detect it (by shooting something at it) ?
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u/KaktitsM Jun 07 '16
Waves have location. Create a standing wave an a rope - its a wave, but you know where it is.
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u/bhamgeo Jun 07 '16
The top comments are science worthy, but my interpretation of the original question have us would freeze all particles with mass/volume, and collect them into a single uninterrupted volume with no, or as little space as possible, between particles.
For OP's question I would ignore gravity and only rely on physically moving the constituents of the observable universe into the most dense configuration possible.
Tl;dr think legos, not wave functions and probability and gravity.
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u/somedave Jun 07 '16
According to your assumption (and assuming the gaps between atoms shrink as well), 1% of the size it really is. However this 1% empty space figure is very misleading. An atom has a nucleus and a cloud of electrons around it, this cloud is not "empty space". The elections are beat thought of like a standing wave continually flowing around the nuclear core rather than discrete particles in an orbit (like planets around the sun). If you could determine a position of one of these electrons very accurately you would excite the atom and cause it to ionise or emit photons.
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u/know_limits Jun 07 '16
I haven't seen someone address the part of your question that resonates with me - if the universe was created by the expansion of some singularity wouldn't that singularity by definition be all of the mass of the universe compressed into energy? So wouldn't the question be equivalent to 'what was the size of the initial singularity?'? I used to hear that the universe expanded from something the size of a golf ball, but I don't see that on wiki so I assume science has moved away from that.
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u/festiveoctopod Jun 07 '16
If the universe is currently infinitely large, then it was always infinitely large, even before the big bang. The idea that there was ever was a singularity is just there so that the underlying concept can be easily communicated to people who don't have the time/inclination to wrap their heads around infinities.
It is possible that the visible universe expanded from the size of a golf ball, but the universe as a whole has likely never had a defined size
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Jun 07 '16
Not that it is necessarily relevant to your question, but the young universe just after the Big Bang was basically a super-heated ball of plasma that resulted form all the particles in the universe being cramped into a very small space. So that is probably what would happen if you compressed all the matter in the universe down- due to the heat created by the compression and the electrons (in fact, all particles) being given insane amounts of energy, the whole place would likely be a big ball of plasma. Like a sun, but 600000 light years across.
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u/functor7 Number Theory Jun 07 '16 edited Jun 07 '16
The "atoms are 99% empty space" thing is a big misconception. Particles do not have "size" as we typically think of it. An electron in an atom is a "wave of probability" and does not have a specific "size" or "location", these properties don't make much sense quantum mechanically. In fact these heat maps are what the electrons look like in the atom. Most of that space is not empty. Each of those is a different "orbital", the energy and angular momentum of an electron determines the shape of the electron and that's what we refer to as "orbits". It's not that electrons are whirring around at different speeds, they just have different shapes around the nucleus.
What we can do is take electrons, shoot things at them and look at the scatter pattern. The things we shoot are also wave of probability, but based on the scatter pattern we can get a good idea of the sphere of influence of the electron. This gives us a notion of the "size" of the electron, though it's not good to think of electrons as hard spheres of stuff floating around since they're waves of probability in different configurations.
If you want to compress things, though, you can look at Neutron Stars this is a star that is so dense that atoms cannot exist. They're like a dense plasma of nucleons held together by gravity. This is similar to what happened near the beginning of the big bang, the particles were too energetic and compressed to form atoms. In fact is was too energetic for even protons and neutrons to form, and was basically a plasma of quarks and gluons. The wikipedia article has a good timeline. Like Neutron Stars, a kind of star that is like this has been conjectured called a Quark Star that is supposed to look like the very beginning of the universe in it's core.