I wonder about something. How big would the event horizon of the universe be when all matter was infinitely packed together?
What you are asking about depends on your estimate of the total mass of the universe. Nobody has a scientific estimate of the total mass of the universe. So just pick one, making sure that it is bigger than the estimated mass of the observable piece of it that we know about. The actual mass could be hundreds of times bigger, or infinitely bigger, than the mass of the observable piece of it. At least we know for a fact that it's bigger, but we don't know how much. what you are asking about is big crunch---or quantum gravity bounce---conditions there is some (quantum cosmology) theory that has been developed to deal with extreme condtions like that and it still needs to be tested. Here is a sample http://www.slac.stanford.edu/spires...+date+>+2005&FORMAT=WWW&SEQUENCE=citecount(d) these are the most highly cited papers published since 2005 in quantum cosmology Narrowing it down to even more recent, here are the most highly cited quantum cosmology papers published since 2006 http://www.slac.stanford.edu/spires...+date+>+2006&FORMAT=WWW&SEQUENCE=citecount(d) =================== To respond to your question in terms of these papers, they deal with both cases where the universe has infinite matter/spatial extent and where it has finite matter and finite spatial extent In no case do you get a Schwarzschild black hole. Space tends to collapse along with matter when you are talking about the universe as a whole, so it tends to be coextensive with it. If you have finite matter and you pack it down, then space gets very small too. The idea of a Schwarzschild black hole radius in this case is meaningless. If the universe has collapsed there is no space for the radius to stick out into! So you need to reformulate your question to be about something that makes sense. One way is to ask what the MAXIMUM DENSITY is that can be achieved in a big bounce type collapse. In the topcited quantum cosmology papers they give results on this. From running computer models in many different cases they consistently get a bounce when the density reaches about 80 percent of Planck density. If I remember right that is around 10^{93} times the density of water. So if you want to get an idea of how big the universe is at the moment of a big bounce, all you need to do is chose an estimate for the total mass of the universe. Then dividing by that density will tell you the volume. (at the moment of bounce). Just for fun, let's find out what Planck density is.
you can look up the formula for Planck density in wikipedia it is [tex]\frac{c^5}{\hbar G^2}[/tex] the fun thing is to be able to calculate this yourself, instead of having to look it up in a handbook so we are talking about how to evaluate c^5/(hbar*G^2) one way to do that is just to TYPE IT INTO GOOGLE because google has a calculator that knows the value of c, the speed of light, and G, newton constant, and hbar, Planck constant so the google calculator will do it for you if you type in "c^5/hbar/G^2" and press search, what you get is 5 x 10^{96} kilogram/cubic meter and that is 5 x 10^{93} metric tonne/cubic meter which is 5 x 10^{93} times the density of water ============== what they have been finding repeatedly (with both analytical models and numerical models in the computer) is that the bounce happens at about 80 percent of that density. that is when quantum effects overwhelm classical gravity and make gravity repell instead of attract----so it can't collapse any further. remember these models are just being developed and need to be tested---nor is it clear how this will be done---but you are asking about extreme condition so this is the best response I know to give. ============== the google calculator understands c^5/hbar/G^2 as an instruction meaning take the 5-th power of the speed of light, and divide it by hbar, and then divide by the square of the Newton G constant. you can get it to calculte all sorts of other stuff. it knows the mass of the electron and the mass of the earth etc etc.
I think the argument in 1000 pop cosmology/physics books is the following. If you see e.g. this 'latest' supernova event or anything now it is because the light has arrived now. The event is in the past. Last year you didn't see it though it had already happened long ago, becausethe light had not arrived. The farther back in time you go the less you can see. So in the very early Universe very little of what we can see now could have been seen, i.e. the bulk of the now visible universe was not in any kind of physical communication or contact. So it could not e.g. even up its temperature between different parts the way matter in contact does. Yet the cosmic background indicates near-uniformity of temperature of an early universe. So that was reckoned to be a difficulty, called the 'horizon problem', in cosmology fitting the facts. The idea of the 'inflationary universe', not initially invented for explaining that did seem to explain it nicely along with other facts, which was one reason for its rapid acceptance.
I may have misunderstood the question. I guessed that you were talking about a black hole event horizon. That is, the Schwarzschild radius, that would correspond to a certain mount of mass (the total mass of the universe). That is a very large distance that we don't have any way to get a handle on. The Schw. radius is proportional to the mass, and we only have a lower bound on the mass---it has to be at least as big as the observable universe mass. Is this what you were asking about Khursed? If not, you could explain the question a bit more. The black hole "event horizon" does not have to do with the horizons involved in the "horizon problem" that motivates inflation scenarios. Those are different horizons. I'm not sure the "horizon problem" comes up in bounce cosmologies because they provide more time for different corners of the universe to communicate and get in causal contact. But that and the inflation business is really a separate issue from the original question, I think.
Ok, very good info guys, thank you. I was just wondering, what would be the size of the event horizon for a black hole the mass of the universe? I know many of the laws of physics break down when we talk about those conditions, I'm also aware its impossible to even predict most of the effects and conditions related to the moment 0 of the big bang. I was just curious, to know for example, how big the event horizon of a black hole the mass of the universe would be. I got the idea when I wondered exactly how big was the universe after the "inflation" moment. I've read that the universe was X big, X being equal to whatever the current best guess of the inflation was. Also, I have often read how sophisticated the universe was to have expended at just the right speed so as to not recolapse, and so as to not be flung so fast out that gravity couldn't start forming today's stars, galaxy etc... using this <sorry can't post url yet> Submitting a 3E52kg mass for the universe according only to the mass of the visible stars. It gives me a Schwarzschild radius of 4.7 billion light years for a universe mass black hole. Obviously, it seems my calculation are wrong. I have a hard time imagining such a big number for the Schwarzschild radius of a black hole the mass of the universe being so big. I was trying to figure out what was the absolute minimum size of the universe likely to be after inflation to avoid collapsing into a singularity. I'm not sure how to plug the dark matter into that calculation however. So I'll leave it at that for now.
The mass of the universe is unknown, but we know about its density. We can therefore reformulate you question as "what would be the size of the event horizon for a black hole the mass of a sphere of radius [tex]r[/tex] with density equal to the density of the universe?" This is easy to calculate. Assume a spatially flat universe with density equal to the critical density: [tex]\rho = \frac{3 H^2}{8 \pi G}[/tex] With this density a sphere of radius [tex]r[/tex] contains a mass ([itex]M = 4/3 \, \pi r^3 \rho[/itex]): [tex]M = \frac{r^3 H^2}{2G}[/tex] Insert this into the formula for the Schwarzschild radius [itex]r_s = 2GM / c^2[/itex] [tex]r_s = \frac{r^3 H^2}{c^2}[/tex] The Schwarzschild radius will depend on the size of the sphere you are considering of course. The nice results is however that there will be one single radius of the sphere for which the Schwarzschild radius will be equal to the size of the sphere you are considering: [tex]r_s = \frac{r^3 H^2}{c^2}[/tex] [tex]r = \frac{c}{H}[/tex] This is the Hubble radius with current value 13.7 Gly. I.e. the Schwarzschild radius of the Hubble sphere is equal to its own radius. This is a consequence from the fact that for a flat universe the escape velocity from a sphere of radius [tex]r[/tex] is equal to the recession velocity.