The Universe has boundary?

Well anything smaller than a photon is going to be hard for us to track. So let's start with a photon and cut it in half, then cut one of those halves in half, and so on forever. The half keeps getting smaller. It's impossible not to have something left when you never take more than half of it away at any time is there? That doesn't sound like a limit to me. The only limit is our ability to see the smallness getting smaller.

 

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How would a flat space be a sphere? A sphere has constant positive curvature. In fact the FRW metric allows for a spherical universe, it is different from the flat case. Furthermore the sphere would be a hyper sphere since a sphere, in common parlance, pertains to the \(S^{2}\) case, which is 2 dimensional and we're in a universe with (at least) 3 spatial directions.

Your description makes the assumption that we see the photon because it has some size, like its a sphere of small radius. Quantum field theory regards photons are points, you can't cut them in half, so the notion of cutting particles into smaller bits and then reaching some size you can't then detect is stepping outside the current realms of physics. The photon's interactions with electrons (in a detector like our eyes or machine) is not a function of its 'size'.

 

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I think that if the cosmological constant, sometimes called the vacuum energy density of the universe, is 0, that means that the shape of the universe in terms of GR is flat as opposed to open or closed.

If that was the case, then if the universe is finite, flat, and expanding, it could be spherical.

However I think that the latest measurements show a positive cosmological constant so the expansion is accelerating and the universe will expand forever according to GR.

In that case it could be finite, open, expanding, and it could be spherical.

 

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Not possible, since spheres are not flat, as I just said. There are 3 archetypical cases, the sphere (positive constant curvature), a flat plane (zero curvature) and a hyperbolic space (negative constant curvature).

Furthermore you need to specify whether you talk about the universe's space-time shape or the universe's spatial shape. The latter requires you consider a particular instant in time, which also is coordinate dependent. The former takes into account the universe's structure throughout its existence.

The metric in question is the FRW metric, \(-ds^{2} = -dt^{2} + a(t)^{2}d\Sigma^{2}\). This is obviously a coordinate dependent expression and the distinction between what is space and what is time is coordinate dependent (something Farsight has a problem understanding) but in those coordinates the expansion of space is handled by \(a(t)\), which affects the spatial region, whose non-dilated line element is \(d\Sigma^{2} = \frac{dr^{2}{1-kr^{2}} + r^{2} d\Omega_{2}\) as a whole.

The second term, \(r^{2}d\Omega_{2}\) is the metric for a 2 dimensional sphere of radius r. This doesn't mean the shape of the universe is a sphere, only that you can define a sphere within it easily. This is a point Magneto failed to grasp recently. The shape \(d\Sigma^{2}\) describes depends on the \(dr^{2}\) coefficient, which itself is determined by \(kr^{2}\), which is effectively the radius of curvature of the system. k is the normalised curvature and thus, as Wiki says, k=-1 is a hyperbolic space, k=0 is flat and k=+1 is spherical. If observations are telling us the universe is flat then k=0 and you have a Euclidean metric for the undilated part of the 3 dimensional submanifold.

The universe being of finite age and once being very small and hot doesn't mean it has spherical structure.

 

A spherical universe can be flat in terms of the cosmological constant. Flat is not flat as on a plane. Flat is flat as when the CC is zero.


wiki: http://en.wikipedia.org/wiki/Shape_of_the_Universe

 

The CC being zero everywhere doesn't give you necessarily a flat universe. The Friendman Equations follow from the Einstein field equations and the FRW metric. As commented on that page the Ricci scalar curvature is \(R = 6\left( \frac{\ddot{a}}{a} + \left(\frac{\dot{a}}{a}\right)^{2} + \frac{kc^{2}}{a^{2}} \right) = 6 \left( \dot{H}+2H^{2} + \frac{kc^{2}}{a^{2}} \right)\). If you put in the expressions for H and its derivatives you don't get something purely \(O(\Lambda)\). In fact you can absorb \(\Lambda\) into pressure and density to remove it from explicitly entering into the equations.

'Flat' is a measure of curvature, not of energy density. This is why there's talk about how the curvature due to the gravity of the universe can be cancelled out by the curvature due to \(\Lambda\). Two balanced non-zero effects give a zero result.

 

It is true that a zero CC might not be perfectly flat due to fluctuations in vacuum density. But a zero is almost perfectly flat. It is flat if you say all space is at a zero CC though there can be localities in space where there is slight curvature even if the whole is a zero CC.

 

If the ejectae from the BB/inflation follows patterns I've seen when exploding a spherical explosive, it certainly may not be a spherical result. Differences in explosive density and detonation pattern have shown to me that some sectors got more material flung one way than another, not in an exact sphere. Clumping and voids occured, and the position of the detonator and the det itself seemed to make a difference. If the Universe had similar inconsistencies as my experiments, a non~spherical result is possible perhaps even probable.
Unfortunately the police tagged me before I could finish my research. I was using Benzly peroxide as it detonates easily. It's a bit poisionous too if anyone was thinking of trying it.

 

Wow. Well if we are going to predict the physical shape in three spatial dimensions, and considering energy density fluctuations during inflation and expansion, then it would be very unlikely that it would be physically spherical if it was finite. If it is infinite it is boundless.

 

LOL....

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I have to try.

 

And of course, your explosions were carried out in a perfect vacuum in a gravity free enviroment? Uhhh... you mean the weren't? You mean the have no physical realtionship to the conditions of the BB?

Oh, well, never mind....

 

You are so dishonest! You are the one that has failed to grasp the basics of GR, and you were embarassed in front of your friends!

You try and sound smart, this is a Con!

If a sphere is very, very large then any observer on the surface will experience his world as flat. This is similar to the Surface of the Earth. The earth appears flat to us local humans, however the earth is completely an oblate spheroid. This is why Euclidean and Non-Euclidean explanations are sometimes confusing. If the sphere is small enough the sum of the angles for a triangle on the surface will be greater than 180 degrees and this is Non-Euclidean. If the sphere is very large then, Flat and Euclidean could be considered synonymous, and why a 2-Sphere could also be described by a 3-Sphere.

Euclidean Description - See Generalizations

 

That's got nothing to do with it, I wasn't considering quantum fluctuations. 'Flat' in regards to the FRW metric refers to \(R = g^{ab}R^{c}_{acb} = 0\). It is possible for the CC to be non-zero but R to be zero. The curvature due to the gravity of all the stuff in the universe makes R want to be positive. The curvature due to dark energy makes R want to be negative and they appear to be cancelling out.

Do you live in a little world of your own? The last thing you tried to get me for was a typo while I demonstrated you can't do coordiante transformations. You've stopped replying to that thread. Everything I corrected you on you've failed to retort and I'll provide references or more specific details if requested by you or anyone else.

Wow, what kind of insulated world of your own do you live in!?

No, I don't have to try, it comes naturally because I paid attention in school.

The definition of a manifold is that it locally looks Euclidean, hence why you can cover a manifold in an atlas of charts of \(\mathbb{R}^{N}\).

While it is possible to choose a small section of a sphere and blow it up into an approximately Euclidean region (which is what a map of a city does, even though its located on a sphere) it is not possible to do the whole of the sphere with a Euclidean chart. Nor can you cover an entire flat plane with a spherical chart. The FRW metric isn't about our little corner of the universe (ie the visible universe), it is about the entire universe and thus if it is correct its spherical form means that the universe isn't flat.

Firstly I've already corrected you before about how 'flat' and 'Euclidean' are not synonymous. Minkowski space-time is not Euclidean but it is flat. Flat means \(R^{a}_{bcd} = 0\). Euclidean means \(ds^{2} = \delta_{ab}dx^{a}dx^{b}\). The former implies the latter but not vice versa.

Secondly a 2-sphere is 2 dimensional. A 3-sphere is 3 dimensional. They are not the same because they are different dimensionality. The clue is in the names. Unless you think 2=3, which wouldn't surprise me in the slightest given your completely detached view of how your defence of your 'work' went.

Feel free to go back to that thread and continue it.

 

It is the fact that some call the Minkowski Space-time pseudo-Euclidean. Which I consider Euclidean for simple GR; however it can also be considered Non-Euclidean; the difference has to do with sign convention whether (dt) or (-dt).

Euclidean Description - See Generalizations

This is the dishonesty that I am speaking of; you know I am not saying that 2=3.

Why would I want to discuss physics with someone that is not interesed in intellectual exchange; but is more looking to "prove" that you can fool people about your abilities. I am not interested in this.

 

You can consider it chocolate cheesecake and called it Fred but that doesn't make it so.

It's not a convention, the convention is in the overall sign of the metric. There must be a sign difference between the temperal coefficient and the spatial one. This is invariant under coordinate transformations (Sylwester's law of inertia). And the sign you refer to is \(+dt^{2}\) vs \(-dt^{2}\).

Come on now, this is like the hyper sphere thing. Do you think I'm unaware of generalisations of geometry? My thesis was on generalised geometry. Rather than offering sound bites of other people perhaps you could engage in discussion yourself?

You might not have said that explicitly but you mixed spaces of different dimensionality.

I'm not the one avoiding peer review and who made claims about his work and who charges people to read his work. As for intellectual exchange, you avoided answering questions, couldn't provide references, couldn't provide derivations and couldn't acknowledge when I categorically proved you wrong. You couldn't even do a Cartesian to spherical coordinates transformation!

You're making excuses for your failure to justify anything you claimed and to avoid facing up to your mistakes. You skipped over the bits of my reply where I pointed out you couldn't do transformations and how I understood the definition of 'manifold'. You clearly know you're avoiding particular parts of my posts, to cherry pick what you think you can give a response to. You did it all the time in the discussion thread. This illustrates you know you're being dishonest. I replied to all the points you raised and I offered to expand on any of my corrections and explanations. That offer still stands. But I suppose its a wiser decision on your part not to continue with that thread, you just dig yourself in deeper.

 

Uh, what is R again?

Are you saying that the CC and vacuum energy density are not the same thing? Are you saying we can or we can't refer to the whole universe in terms of a particular value for the CC? It sounds like the CC and R have to be considered from what you are saying; is that true?

 

since the theory of big bang, that means the univervce started to expand, and still is
however, there may be other universes, that also have been created after the bigbang, like a large web of univerces, and each univerce, is a large web of galaxies and etc... and all are connected somehow, however, what is after the univerce or after the univerces? nothing non? like, the nothing that don't exist, how can the univerce expand, if there are like boundaries, of, the non exist, i mean, since the univerce, is just the univerce, and nothing more after it, how can it expand on a space, that don't even exist, so if after a univerce, there's another univerce, like a hall web, would that web be infinit?

infinit, is kinda a measure created by us, to say the, too imaginary numbers and spaces, that we can't know it's size, or even get close of, numbers that we can't even imagibe, non our computers or super computers
the univerce have a limit or not, i'm not sure, i vote for that it have a limit, and after that limit, idk, what if the univerces web, are a part of a bigger univerce, somekind of univerce, like if univerces are atoms in that bigger univerce, or, just, a web and a part of that mega univerce, like in our univerce, galaxies and stars and black holes, and etc...

forms like a web in the space, that are all held by gravity forces in a very precise cycle
if the gravity in our univerce, hold everything together and orgenized, it must have a cycle, that make it all connected, so it can hold it together, like, 5 balls attached with each other with a line, that is circular, a closed circle somehow, if the univerce is infinite, means, there is infnit of galaxies and etc... (i don't mean high imaginary number, i mean, infinite, by, doesnt have a number at all, the number of galaxies increases in an imaginary number all the time, what makes like, unlimited number, but, even in that case, it is a number, and it is a closed cycle, i mean, everything is attached, since it reproduces, means, since always new galaxies are created, and created, means, created from what is existing, and get created and still attached to what exists, what makes it also a cycle)

everything must be attached to each other to hold together, and don't make a self destruction to the univerce (and we can't know if our univerce itself is distructing itself, who knows, why not)

so, infinite? what is infinite? (if not an imaginary numbers and measure we can never know, or, a number that is growing in an imaginary speed all the time, what makes it infinite, because it don't stop growing)


there would be three scenarios for the univerce with space expantion
either to get all ripped and destroyed and the gravity force get defeated by the accelerating speed of expantion
or the univerce, reach a limit of expantion, and start to shrink back again, to the first position, like if there was a force that attach it back when the speed of exoantion get weaker than the force of attaching
or, just continue expanding, and, gravity still attach the galaxies together, and each galaxy get, lost somewhere, and the galaxies start to get firther and firther from each other, when it dies and turn into a black hole, it deseppears, or, another galaxy get borned in it's place, and it continues, but, however, do you think it can just, continue like that, for ever? and nothing happends? (maybe finally everything is destroyed, all the mater get back to the form of atoms flying around here and their, far from each other by the extreme mega expansion of the univerce, and the story ends their

 

You must have a short memory, and and must have forgotten that I did the Cartesian to spherical coordinates transformation!

Or are you being totally "Dishonest" again trying to con people into believing that you are smart!

 

http://www.sciforums.com/showpost.php?p=2735460&postcount=52

When it comes to something which requires understanding and not a copy and paste you fell flat on your face and didn't acknowledge your mistake.

 

My problem with your mathematical statement was that it did not appear that you were describing the correct frame of reference.


\(ds^{2} = dx_s^{2} + dy_s^{2} + dz_s^{2}\)

The above does not equal:

\(ds^{2} \Not= dx^{2} + dy^{2} + dz^{2}\)

but is actually equal to

\(ds^{2} = dr^{2} + r^{2}(d\theta^{2} + \sin^{2}\theta \, d\phi^{2})\)

\(ds^{2} = dx^{2} + dy^{2} + dz^{2} + r^{2}(d\theta^{2} + \sin^{2}\theta \, d\phi^{2})\)

\(dx_s^{2} + dy_s^{2} + dz_s^{2} = dx^{2} + dy^{2} + dz^{2} + (x^{2} + y^{2} + z^{2})(d\theta^{2} + \sin^{2}\theta \, d\phi^{2})\)


I will give you this one as being correct, only because you use a lot of generalizations and shortcuts. So I do understand how you can make the math statement that you made in the post; I am a little more exacting, because I want a result that is closed solution that I can calculate and get a result; this type of result is not that important to you. And this is common with math people, not so much with physics people.