How do you figure? Newtonian gravity assumes space is flat, you can't get a curved universe out of it. And we haven't directly measured the matter/energy density of the universe, since most of it is dark matter and energy: we derive these quantities from measurements of the CMB which assume a flat universe, which we know for other reasons, and from the motion of high redshift objects. If it is so easy then show me.
IMHO the best way to approach this issue is to consider the gravitational field of a planet, along with a void at the centre of the planet. The planet is surrounded by what we call space-time curvature. As a result your motion is affected, and you fall down. However at the centre of the void we have a small region where space-time is "flat". There is no gμv gradient here, and no energy-density gradient. There is no discernible gravity, you "float" instead of falling anywhere. Now imagine that this void is the whole universe. At a very large scale once you've discarded a few inconvenient things like galaxies, you have a region of uniform energy-density. Hence there is no overall gravitational field or curvature.
^an interesting approach yet I think you have one misconception. you're talking about the curvature caused by placing matter on the spacetime fabric, we're talking about the curvature of the geometric object on which the spacetime fabric is stretched and embeded on.
Regions of uniform energy-density still create curvature, in fact the standard cosmological models all assume a uniform energy density distribution across the whole universe. There isn't really a difference. Although perhaps you are arguing that the cosmological constant doesn't really count as matter/energy and instead is an intrinsic modification of spacetime geometry? I suppose you could argue that, but the cosmological constant can also be (and is more often) interpreted as a contribution to the energy-momentum tensor, and counted as an intrinsic energy associated with a volume of space, or else some constant energy-density scalar field permeating all of space.
With respect camilus, once we set galaxies and all other matter aside, the fabric (for want of a better word) is three-dimensional space, and it isn't actually stretched over something. It has its instrinsic stress-energy, and stress is akin to pressure. So IMHO space itself is the geometric object, with an innate "pressure" synonymous with energy density. If it's uniform light will travel in straight lines, hence there would be no spacetime curvature and thus a flat universe. Can you give me a reference for that please? No problem, I'm aware of the Lambda CDM model.
Can you quantify this or are you just making up a narrative for otherwise unsupported opinion? Quantify things is important because there's many different notions of curvature. Intrinsic, extrinsic, Gaussian, principle, Riemannian, Ricci (tensor and scalar). The Schwarzchild black hole metric is 'flat' in the sense of R=0. Likewise you can do cosmological analysis using R=0, as that doesn't imply \(R_{ab}\) or \(R^{a}_{bcd}\) are zero as they are stronger conditions (since \(R_{ab} = R^{c}_{acb}\) zero Riemannian curvature tensor is stronger than zero Ricci tensor). A torus is flat , as can be seen by its definition in terms of a quotient of a Euclidean plane, but its embedding in \(\mathbb{R}^{3}\) has non-zero \(R^{a}_{bcd}\). All non-trivial isotropic homogeneous solutions to the field equations are counter examples to that claim.
http://en.wikipedia.org/wiki/Closed_universe#FLRW_model_of_the_universe Not that you can believe everything on wikipedia, but there. And an extract: "The homogeneous and isotropic universe allows for a spatial geometry with a constant curvature. One aspect of local geometry to emerge from General Relativity and the FLRW model is that the density parameter, Omega (Ω), is related to the curvature of space. Omega is the average density of the universe divided by the critical energy density, i.e. that required for the universe to be flat (zero curvature). The curvature of space is a mathematical description of whether or not the Pythagorean theorem is valid for spatial coordinates. In the latter case, it provides an alternative formula for expressing local relationships between distances: * If the curvature is zero, then Ω = 1, and the Pythagorean theorem is correct. * If Ω > 1, there is positive curvature, and * if Ω < 1 there is negative curvature;"
Thanks kurros, I have seen that, but I don't know where it comes from. I take my cue for this from Einstein's Leyden address where he described a gravitational field in terms of inhomogeneous space: "This space-time variability of the reciprocal relations of the standards of space and time, or, perhaps, the recognition of the fact that “empty space” in its physical relation is neither homogeneous nor isotropic, compelling us to describe its state by ten functions (the gravitation potentials gμν)..." There also this interesting paper: http://arxiv.org/abs/physics/0204044. It isn't peer-reviewed, but the quotes look robust. Imagine yourself as a light beam moving through a universe where the energy density is uniform. There is no inhomogeneity in the metric to give you any form of curvilinear motion or geodesic. There is no Riemann tensor, and spacetime is flat. Curved space is of course something different to curved spacetime, but the light beam simply travels straight, it doesn't bend around in a great circle or do anything exotic. So I'm left saying: if space is uniform, where's the curvature? Alphanumeric: The Schwarzschild metric is in no way 'flat'. See http://en.wikipedia.org/wiki/Schwar...c.29_formulations_of_the_Schwarzschild_metric. If it was, light would go past that black hole as straight as a die. Changing the coordinate system doesn't transform this away.
This is precisely why I commented on the need to be precise about that curvature means. The way to derive the Schwarzchild metric via Birkhoff's theorem is to solve the equations \(R_{ab} = 0\). As a result the metric is flat in the Ricci tensor and Ricci scalar (since \(R = g^{ab}R_{ab}\)) sense by construction. I did make it clear I was referring to the Ricci scalar R by making the indices explicit on the Ricci and Riemman tensors, as well as explaining it. Clearly your grasp of the specifics is not as good as you might want to portray it. I've said it many times to you before, without a quantitative grasp of the specifics its easy to be led astray by faulty 'intuition'. A torus is Euclidean flat, as seen by its quotient definition but when embedded in higher dimensions it will have have non-zero \(R^{a}_{bcd}\). A Calabi-Yau manifold is Ricci flat yet they are extremely curled up within string theory space-time models. What does that have to do with what I said? By definition a tensor expression will only be zero if a given (valid) coordinate system if its zero in all (valid) coordinate systems. Or were you just trying to appear to be familiar with the subject, hoping that if you quote the basic defining property of tensors that you might seem to understand them? That was the tactic Reiku/NeoNo.1/Green Destiny used, as definitions are easy to find and spout. The fact you didn't understand my comment about different curvatures demonstrated the point I was making, its important to be precise with definitions and statements. In addition you not replying to my request you quantify your statements further supports the point I was making, your lack of ability to do quantified work is a serious hindrance. But then I suppose that's what you get when you learn about curvature and tensors from reading Wikipedia and not by actually doing anything with them. :facepalm:
A corollary to what AN is saying is that, when the space is Ricci flat like the Schwarzschild metric, then the Einstein field equations imply \(T_{\mu \nu} = 0\) because \(R_{\mu \nu} = R = 0\). Hence, no pressure, stress or energy density.
I disregarded your comment because it isn't relevant. If on a large scale we have uniform space in the universe, then spacetime is flat, and light travels in straight lines. There is no curvature, and that's the end of story. Please try to focus on this simple point rather than digressing into the Schwarzschild metric, higher dimensions, Calabi-Yau manifolds, and string theory space-time models. Prometheus: the average energy-density of the universe isn't zero. There is pressure. Hence the universe expands. But there is no sense in which it is curved, and we have no evidence that it is infinite.
Did I say any of what you assumed I did? (Clue: No). What I said was that a manifold that is Ricci flat is a solution to the EFE's in vacuum - No energy, matter or stress. That's not to say the FRW manifold is Ricci flat (it's not - the FRW models of cosmology assume that you have a non zero energy density and pressure, but no shear stress. In other words, the energy momentum tensor is diagonal). There are three different types of universe that are of the FRW type, and none of them have zero curvature, as I presume you can show since you seem to be so keen pronounce on cosmology. This is really the simplest type of cosmology there is that is truly scientific. Modern measurements of the CMB show that the average density of the universe is below the critical value so the universe is open. That means the universe will go on expanding for ever, so there certainly is evidence the universe is infinite.
So that's another irrelevant comment that distracts from the very simple point. Good. Don't mistake mathematical models for reality. There's only one type of universe. The universe. No, you show me the curvature in homogeneous space where light travels in straight lines. You can't, because there isn't any. That isn't scientific at all. Thirteen point seven billion years ago the average energy density of the universe was enormous. But it didn't collapse back in on itself. Because in a universe where space is homogeneous, there is no curvature, no hypersphere, and no gravity either. Why are you having so much difficulty with this? It is just so crushingly simple. A gravitational field is present where space is inhomogeneous. Einstein told us that. So a gravitational field is a stress-energy density or pressure gradient. And as a result light moves in curved lines to give us what we call curved spacetime. No pressure gradient, no curved spacetime. Hence a homogeneous universe is of necessity flat. Not so. We have evidence that the universe began expanding 13.7 billion years ago. But we have absolutely no evidence that the universe is infinite.
Wow, you really have trouble with the incredibly complicated skill of reading. You're not honestly going to tell me that you neither know nor can look up what a corollary is? This type of argument is so stupid I don't know where to begin with it. Whether you like it or not, the universe as we observe it evolves according to the mathematical rules of some theories that we are working on understanding. Since you're having trouble with this extraordinarily simple point, I will write it again, including all the pauses that I left out last time where I hoped you'd be able you wipe your own dribble. The FRW metric is the maths that describes the early universe's evolution. When you, y'know, actually try to extract some predictions out of this because, y'know, that's what science does, you find there are three possible outcomes - a universe that is either open, closed or flat. Now before you jump on this with the moronic attitude that flat means there is no curvature at all, don't because flat here means the spatial sections are flat. The universe is one of these, and which one depends on the energy momentum tensor of the universe. Current experiments tell us we are in an open universe. Hell, I can set up coordinates in flat space where light doesn't move in a straight line. That aside, spaces that are conformally flat but have a non zero curvature like the Poincare patch of AdS will allow light to travel in straight lines (in appropriate coordinates). Checking to see if light moves in a straight line is not a good test of curvature. What's not scientific? Making observations, doing experiments and developing theories that fit the observations, then extracting predictions from the theories and testing them in more experiments and observations, then improving the theories and repeating the whole thing over and over? Seriously man, look up the scientific method. It's really not much more than a less sarcastic version of the above. In the FRW models the universe is assumed to be homogeneous (and isotropic) and gravity still does stuff, the universe still has curvature in any of the three solutions. Just because you say if doesn't that doesn't mean it doesn't. Unfortunately for all of us, you can just make stuff up and have it be true. If that was the case then my publication record would be incredible, but sadly we have the scientific method to hold back our careers. Damn. Well, there has been a lot of coverage of the accelerating expansion of the universe. If the universe's expansion is accelerating then the universe will continue to expand for ever, making the universe infinite.
Ok that paper is relatively long and I don't have time to read it all, if there is something in particular in there you are referring to a page number(s) would be nice. From my very quick skim it seemed to be more concerned with interpretations of gravity and questions like whether a homogeneous gravitational field counted as a gravitational field since it doesn't induce tidal forces, etc, not our questions regarding the large scale shape of the universe. Now, back to the spacetime filled homogeneously with stuff. You say "Imagine yourself as a light beam moving through a universe where the energy density is uniform. There is no inhomogeneity in the metric to give you any form of curvilinear motion or geodesic. There is no Riemann tensor, and spacetime is flat." I feel certain the FLRW spacetime has a non-vanishing Riemann tensor. In fact it must not vanish, because in fact light beams don't remain parallel, not in general. If \(\Omega=1\) then yes light beams move in straight lines, but if \(\Omega>1\) or \(\Omega<1\) then light beams which are initially parallel will either diverge or converge, although I can't remember which \(\Omega\) results in what. In fact this has been used to verify that the universe is indeed very close to flat... oh wait, that was what the OP was referring to, full circle indeed. I can't find the original paper though, damn. Anyway the point was that this isn't some "imaginary" curvature, it produces perfectly real physical effects, and yet the energy density of the spacetime is uniform.
It is, as demonstrated by the fact you were talking about something different because you were unaware of multiple terminologies. Was an explicit example of your misunderstanding too subtle? In other words you don't want to talk about the details because it exposes you don't understand the actual physics details. Instead you want to stick to the vague wordy discussions, allowing you to just Wiki search and pass off parroting words you don't understand as 'knowledge'. If you plan to break into the mainstream you shouldn't be shying away from the details, you'll be required to stand up to scrutiny if your work is to be taken seriously by mainstream researchers. But then that's the reason it's not, you have no details to defend.
Page 20 is probably the best place to focus on. It is important, and it is related to the discussion. People talk about a gravitational field in terms of spacetime curvature, forgetting that Einstein said that a gravitational field is inhomogeneous space. So if space is homogeneous, the spacetime curvature has gone. You can't have a homogeneous gravitational field because it's a contradiction in terms, it's homogeneous inhomogeneous space. This is the whole point. If space is homogeneous, spacetime is flat. Light beams don't converge, and they don't diverge. There's no evidence of anything other than a flat universe. Note that it's the energy density of space rather than spacetime, and that curved spacetime is the result of a gradient in this energy density. Also note that curved spacetime is different to curved space.
Can you provide a source for that quote of Einstein please. False. The most commonly used space-time metric in cosmology is the FRW metric. It's derived from the assumptions of isotropy and homogeneity in space-time yet it can have non-zero curvature (in both the Ricci and Riemannian senses). Isotropy means that it doesn't matter which way you look from the position the observation is unchanged. Homogeneity means that if you change your location then you end up with much the same. And since you're so fond of saying that if someone disagrees with you they should explain themselves if you disagree with me then please demonstrate that the FRW metric's derivation is wrong, ie state precisely the step(s) in the derivation which are false and demonstrate it. False, as just mentioned. Yes, constant zero curvature everywhere is an example of a space-time with the properties you just stated but homogeneity doesn't preclude the possibility of constant non-zero curvature in the Ricci scalar sense and since the Ricci scalar is zero if the Riemann curvature tensor is zero then the space has non-zero curvature in the Riemannian sense too. Both AdS and dS space-times are examples. Previously you said "There is no curvature, and that's the end of story. " and I corrected you with an example where that wasn't the case. Now you've had another one provided to you. Anyone whose done any cosmology or GR will have come across these examples. Since you have avoided doing anything quantitative you've been just guessing as to the details. Doing this for your own 'work' is one thing but to state things about well known mainstream things is just daft. How many times are you and I going to do this little dance of you stating 'facts' about things you don't understand and then having explicit counter examples put in front of you, which you then ignore? If you'd spent less time reading pop science books and just making up waffle about Relativity+ and a bit more time reading the details of the things you talk about (and sometimes profess expert knowledge in) you'd not be making so many mistakes. Come on Farsight, you know you haven't a grasp of this stuff in anything more than a superficial level and you know you'll get called on things which you just make up in regards to relativity or quantum mechanics, so why do it? It baffles me that people like yourself or Reiku come to physics forums you know to be frequented by people who didn't sleep through physics class, proceed to just make things up and then get all defensive when someone points out you're incorrect. /edit I've just noticed Kurros mentioned the FRW metric too and its even in your quote of him. This demonstrates you aren't familiar with it and you didn't look it up when it was mentioned. Seriously, at least have the good sense to check out the facts when someone mentions something you don't understand.
Can you please provide a link that explains these statements. According to everything I know about Cosmology the shape of the universe has been observed to be as close to flat as we can currently measure. I'm not disagreeing with your statement out of hand but it does fly in the face of what I know about current observations. http://map.gsfc.nasa.gov/universe/uni_shape.html This margin of error leaves the door open for your assessment. I just want to see where you got your information from. A truly flat universe rests on a razors edge and any variation one way or the other will determine the ultimate fate of the universe. . P.S. I'm sure you already know this but both a flat universe and an open universe will expand forever. In othet words, flat does not mean static thanks to dark energy. The only differnce between the two is the rate in which the expansion of the universe accelerates as observed over vast distances (a.k.a., time).
That's back to front. The universe evolves the way that it does because space and energy are what they are. We attempt to express it mathematically. Wrong. Current experiments tell us we are in a flat universe. See WMAP Your coordinates aren't real. That straightline motion of light is. I don't know how else I can say this: get real. I adhere to it. You're living in cloud cuckoo land. How many more times do I have to say it? If space is homogeneous and isotropic, there is no gravity. Einstein said it, not me. He said a gravitational field was inhomogeneous space. It isn't me making it up here. We have no evidence to suggest it's infinite now, and if it's finite now it will always be finite.
