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> if ... the entire universe is essentially one uniform mass, wouldn't that mass then collapse upon itself, in a kind of big bounce? de Sitter space is a vacuu

• adamisom 7y ago
  So the universe would be "de Sitting"
• cobbzilla 7y ago
  Is it not the case that energy has mass? Why wouldn’t this be equivalent to a uniform mass distribution?
  • klank 7y ago
    I think you might be getting them switched around. Mass is a form of energy. Energy is not a form of mass. Put another way, mass is always energy. Energy is not always mass.
    • cobbzilla 7y ago
      I remember some science show where they said a cup of hot water technically has slightly more mass than a cup of cold water, due to the mass contribution from the (higher/hotter) energy. Are you saying this is not the case? Would a near-absolute-zero cup of water (ice) have the same mass as a cup of boiling water?
      • raattgift 7y ago
        That's always a good question! Since we're talking about General Relativity (the top of the thread introduces the stress-energy tensor, and a particular solution to the Einstein Field Equations) let's talk in those terms. We can also talk in those terms because it is the more-fundamental theory from which Special Relativity (which gives the famous E=mc^2) can be derived. The "stress-energy tensor" determines the curvature of spacetime. It goes by other names: the matter tensor, the energy-momentum tensor, and so forth, largely depending on how one wants to interpret the tensor components in a particular problem. Three quick paragraphs with some reference to the mathematics: You can write the energy-momentum tensor's components as a 4x4 matrix where each cell represents a flux of momentum from one direction to another. Momentum which "hangs around" flowing only from the past to the future corresponds to the m in "E=mc^2" either very locally or more globally in a spacetime which has no gravitation whatsoever. Since a nonzero "m" represents a nonzero value of the energy-momentum tensor, and since the energy-momentum determines the curvature of spacetime, "E=mc^2" is really just an excellent approximation that is better when m is small. The fuller Special Relativity relation is E^2= (mc^2)^2 + (pc)^2 where we square to get rid of sign problems and where p represents linear momentum; if there is no momentum and the mass stays at the spatial origin (moving only in time) in our reference coordinates, then all but one component of the energy-momentum tensor is zero, and the one remaining one is totally determined by "m". If we (non-gravitationally) impart momentum onto our m then p becomes nonzero and so does at least one other component of the energy-momentum tensor. Summing up: mass sources energy-momentum, which determines curvature. Moving mass sources even more energy-momentum, and so greater curvature. In your question about the cup, let's apply a restriction: we use the same number of water molecules at all times. As we heat the ice or water, the motion of the water molecules relative to the centre of the coffee cup (or the overall centre-of-mass or the overall centre-of-momentum) increases. Increased motion means increased momentum. This in turn means a greater curvature is sourced by the heated molecules, or if you like, that there is a greater "active" gravitational interaction for the hot water than for the cold water or ice. "Active" in the sense of small things falling towards it. It virtually certianly also has an identically larger "passive" gravitational interaction, where "passive" is in the sense of falling towards an object that makes it seem like a very small thing. (We have excellent experimental evidence that passive and active interaction strengths are effectively identical, and the underlying theory demands it, ignoring some details about gravitational back-reaction which even experts won't want to think too hard about.) So if we use a very sensitive scale we can measure that the hot water with exactly N molecules of hot water weighs more than the exactly N molecules of cold water. Ignoring gravitation again, we see that E is greater in the hot water case, but because there is a larger average value for p (momentum) for each of the molecules. The "m" remains effectively the same in the hot and cold water. Rest mass (m_0) is properly determined by the count of water molecules when they are completely free of momentum (including internal momentum, right down to the momentum of photons, electrons, quarks and gluons), which gives you the answer to when E is lowest for the cup of exactly N molecules of water. (It's at absolute zero!) Back to gravitation: the minimum influence on the curvature of spacetime by the N molecules of water is when they are at absolute zero. The difference is very tiny at the sorts of temperatures you're asking about. However, if we could somehow confine the N molecules of water into a magical box of the volume of the coffee cup, and heat the water molecules to the point where the molecular bonds break, the atoms all ionize, the oxygen nuclei disintegrate into protons, the protons disintegrate into a quark-gluon plasma, and keep going through several orders of magnitude (wherein we may discover new subatomic physics!) then as we go the "active" and "passive" gravitational interactions of the confined matter will grow substantially. It will become very heavy (passive interaction: hard to hold above the lab's floor, assuming the ultra high energy stuff didn't vapourize everything around it) and noticeably start affecting the trajectories of ever larger things (dust, pencils, lab assistants...). In the extreme, we can add so much momentum to to the stress-energy-momentum tensor the magical container encloses that it collapses into a black hole. In essence this is what happens in the cores of neutron stars when they collapse into black holes: the internal pressure and temperature gets so high that an event horizon forms. The "container" is a few solar masses worth of incredibly dense "nuclear pasta" and other exotic stuff we don't know much about yet, and is more the size of the city the magical-container lab is in than the lab itself. One last thing: the total value of the stress-energy tensor at a given point is observer-independent -- an ultrarelativistic observer passing by the lab and a scientist standing still in the lab will both agree on the total stress-energy-momentum of each container of water. However, different observers may prefer to split the total tensor value into its sixteen components in different ways; indeed, any single observer is free to do so because the splitting is coordinate-dependent and one is free to select from an infinite set of systems of coordinates in describing spacetime or a small patch thereof. Because of this sort of coordinate- and observer- independence ("general covariance", technically) modern physics uses tensor fields as fundamental objects and either copes with the heavyweight mathematics that requires or reduces tensors by imposing special conditions on coordinates and on how coordinate-dependent vector-values are extracted from the tensors. Finally, the stress-energy-momentum tensor is contributed to by all the fields of matter, so this would (where not excluded for convenience) include the classical electromagnetic tensor field, or the fields of quantum electrodynamics, or the fields of the Standard Model of particle physics. The contribution to stress-energy for known matter is always non-negative. So when considering classical or quantum matter fields, where there are nonzero field-values there is nonzero stress-energy. In heating up the water molecules we are also creating [a] more photons and [b] higher-momentum photons. Photons carry nonzero momentum and so contribute to the "p" term in E^2=(mc^2)^2+(pc)^2 or more fundamentally they add to the total tensor-value of the energy-momentum tensor. I hope this is a helpful answer. PS: I should have said that absolute zero is probably not physically achievable in our universe, but to the extent things can get very close to absolute zero (attokelvins or colder) we can always describe a relatively-moving observer who will think the object is warmer than someone at rest with respect to it (and some ultrarelativistic observers might think it's rather hot, spraying out a thermal bath of kilokelvin photons!). Nobody knows what the quantum chromodynamics equivalent of absolute zero would be, so that's always happening, and the momenta of the quarks and gluons thus don't completely vanish.
        • Etheryte 7y ago
          As someone who isn't that familiar with the physics behind it, this comment was just as an intriguing read, if not even more than the original article. Thanks for taking the time to write an in-depth response.
• raattgift 7y ago
  > de Sitter space is a vacuum solution for Einstein's field equations. Yes, an exact solution. > What this means is that the stress-energy tensor is basically canceled out" everywhere by the uniform energy distribution No, it means that the stress-energy tensor is zero at every point in the whole spacetime. There is no matter to source any curvature; the curvature is specified by the theoretician writing down the vacuum solution. In Lorentzian 4-dimensional vacuum de Sitter spacetime ("dS" below) there is no useful concept of energy anywhere. The parameter \Lambda is incorporated directly into the Einstein curvature tensor and so is interpreted geometrically. One would have to perturb the dS vacuum in order to have a useful interpretation of \Lambda as a source of gravitational energy. Vacuum solutions can be useful in understanding physical systems in which gravitation is important. For example, one can add non-zero stress-energy by hand and use perturbation theory to study the consequences rigorously. Alternatively, one can "cut out" parts of a vacuum solution and work with the rest which closely resembles a physical system (such as deep space far from the matter in and around galaxies). As the expansion of the universe dilutes away the matter contributing to a non-zero stress-energy tensor, our universe will resemble the sparser parts of a de Sitter spacetime. However, from our perspective (i.e. with data in our sky) there are much better approximations at various scales than de Sitter spacetime perturbed by matter fields: there is a lot of obviously non-zero stress-energy under our feet, lighting up our sky day and night, and causing heat to flow from one place to another. > That means no gravity There can be extremely strong spacetime curvature in de Sitter space! In general we can time-orient dS: in one direction objects which only interact gravitationally and do not generate significant stress-energy will always converge, in the other they will always diverge, depending on the value of the \Lambda parameter, which is always positive for dS. If \Lambda is large, all such objects leave the others' causal cones much more quickly than they could, even in principle, in flat spacetime. If we perturb dS with a dense ball of nonrelativistic (i.e., slowly-moving) dust that feels only the gravitational interaction, the dust will tend to collapse if \Lambda is sufficiently small, and will locally approximate a Lemaître-Tolman-Bondi (LTB) collapsing dust metric, which in turn resembles a perturbation of vacuum Schwarzschild or Kerr (which are eternal black hole models; there is no matter to collapse). A scattering of such collapsing dust-balls with significant gaps between them has been studied by a number of relativists as a family of "swiss-cheese" cosmologies very similar to the one developed by Einstein & Strauss in 1945, who represented the balls of dust as already-collapsed vacuum Schwarzschild solutions, calling them "holes" in the "cheese" of the surrounding expanding spacetime. The Einstein-Strauss-de Sitter model approximated some features of our universe at intermediate ranges, but has since been superseded by the the Friedmann-Lemaître-Robertson-Walker (FLRW) model of the modern standard model of cosmology that among other things has representations of nonzero stress-energy that the vacuum Einstein & Strauss model did not, and which better matches observables in our sky than models that added more realistic matter fields to Einstein & Strauss vacuum. One feature of Einstein-Strauss-de Sitter style expanding swiss-cheese models is that a small "test" amount of sparse low-mass nonrelativistic dust scattered in the LTB "holes" will stay confined within the hole while the hole separates from everything else. However, an identical test dust scattered in the "cheese" part well outside of holes will never find its way into a hole. Ultimately the former test dust will converge while the latter will diverge. We can get highly similar results by perturbing the FLRW model. (The observable universe is well approximated by an almost FLRW model in which there are primordial density fluctuations: over-dense areas collapse gravitationally, comparably to the matter in the "holes" of a non-vacuum Einstein-Strauss-de Sitter universe.) Locally in all these cases, what we have are the tensor fields contributing to the Einstein Field Equations, and general covariance. We can talk about the values of these tensors at any given point (and its neighbourhood), but General Relativity does not lend itself to universal definitions of energy or even energy-density. We can reduce the tensors by using gauge fixing and other techniques, and "demote" the cosmological constant in FLRW or in dS into an energy. (That's really all Dark Energy is in the standard cosmology). In such a gauge we can talk about the work matter interactions (e.g. the electromagnetic interaction) does "against" dark energy. However, the real physics are in the tensors, and applying coordinates and selecting particular observers, while often able to make things much easier to calculate, can be highly misleading too. The idea that the cosmological constant is "cancelled" out by matter-matter gravitational interactions (and electromagnetic and nuclear interactions) is a statement that the geometrical nature of the cosmological constant is "demoted" into an extremely weak force field, and leads to all sorts of statements that work in that "demotion" context but not generally. Typical statements are that the solar system is expanding, or the galaxy or cluster is expanding, because of the tiny local force on matter by dark energy. It is better to retreat to "swiss-cheese" and say that the geometry of our solar system is not well-approximated by expanding metrics like dS or expanding Robertson-Walker, and that consequently it is a better fit to observations that in our "hole" (which the matter of our solar system, galaxy and cluster sources) the cosmological constant simply vanishes. Better still would be to advance to inhomogeneous metrics, but those are a subject of research beyond the scope of this comment. Finally, your last paragraph can also be read as a request to be directed to the concept of the Jeans Instability in the Expanding Universe. If in the very early universe there was truly uniform and dense distribution of stress-energy, the slightest perturbations would magnify, leading to structure formation. As I wrote above primordial fluctuations of an otherwise FLRW model approximates modern cosmological observations very well. (One could alternatively introduce dissipation, quantum uncertainties, and so forth, as a source of gravitational Jeans Instability causing the collapse or fragmentation of a uniform distribution of stress-energy).