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Too much density and the universe is closed, analogous to a sphere in four dimensions: you travel in a straight line and you end up where you started. Too little and you have a saddle: not sure about the destination if you travel in a straight line. Just the right amount and the topology is flat. The flat topology is infinite: you travel in a straight line forever.

If the topology is flat (and at this point all evidence indicates that it is to within 0.4%), then multiplying the critical density by an infinite amount of cubic meters gives you an infinite energy/stress.$$\rho_{CRIT}\space kg\space m^{-3}\times \infty\space m^3=\infty\space kg$$

Is that a reasonable interpretation?

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  • $\begingroup$ If space is flat, then according to FLRW not only the energy of the BB was infinite, but it was infinite in every arbitrary small volume of the initially infinite universe. This is a non-physical result indicating that FLRW fails. One problem with your question is that you are asking about BB, but your formula is not for BB. In the flat case the formula for BB would be $\infty\cdot\infty=\infty$. The second problem with the question is that BB did not "require" energy, instead BB produced (created) energy. Finally, flat space is not the same as a flat universe, because spacetime is not flat. $\endgroup$ – safesphere Jan 6 at 16:35
  • $\begingroup$ @safesphere - Thanks for the comment, I updated the question to make it more clear. When I said 'required', I meant, 'required' an infinite amount of energy to produce a flat topology. $\endgroup$ – Quarkly Jan 6 at 17:40
  • $\begingroup$ @DonaldAirey, what you calculate above is just the proper mass-energy of matter. It doesn't take into account the - negative - energy of the gravitational "field", which cannot be localised in General Relativity. The "total energy" of the universe could be 0, but there's no way we could give a physical sense to it, since the whole universe energy cannot be measured from "inside". I'll make this an answer. $\endgroup$ – Cham Jan 6 at 18:08
  • $\begingroup$ FWIW, a succinct way to state the difference between spaces of positive, zero, or negative (constant) curvature is via Playfair's axiom, a well-known alternate expression of Euclid's 5th axiom. In a plane of 0 curvature, given a line L & a point P not on L, there's 1 line through P that doesn't intersect L (thus the lines are parallel). In a plane with +ve curvature (i.e. a sphere), there are no parallels. In a plane of -ve curvature(a hyperbolic plane), an infinite number of lines pass through P that don't intersect L. $\endgroup$ – PM 2Ring Jan 7 at 14:03
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You have to be careful with the interpretation of energy when you don't know what sort of gravitational potential well might be involved. For example, you might find some amount $m c^2$ of rest-energy of matter when calculated in an inertial frame near but outside the horizon of a black hole, but in order to use this energy at some other location, you would first have to pull the matter up against gravity, expending energy $E$ in order to do so. After spending that $E$ you acquire just $m c^2$ at your location, so overall you have gained $(m c^2-E)$ and this will be small compared to $m c^2$ if the matter started out near a horizon. This is the sense in which gravitational binding energy is negative. When applied to the whole universe, this consideration makes a calculation of the type you are proposing questionable, because it is hard to say what physical meaning it has.

A better analogy is, perhaps, with the concept of escape velocity. A flat universe is one where the motion of matter everywhere is just enough to keep escaping from its own mutual gravity.

Finally, the topology of a mathematical space is not in one-to-one correspondence with the curvature, and in particular, if a space is flat it does not necessarily follow that it is infinite. There are a number of different topologies that are mathematically possible for a flat space, and some of them are bounded (i.e. not infinite). So this may apply to the physical universe too. We don't know.

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  • $\begingroup$ But if space if homogeneous and isotropic, then flatness still implies infinity, right? $\endgroup$ – pela Aug 8 at 8:47
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    $\begingroup$ @pela The topology introduces a distinction between global and local. A torus is locally isotropic but not globally isotropic. If one asserts both global and local isotropy and homogeneity then I agree: exact flatness implies infinite extent. But this situation is unstable in the sense that the tiniest departure from flatness can make the difference between infinite and finite. $\endgroup$ – Andrew Steane Aug 8 at 9:24
  • $\begingroup$ Thanks, then we agree :) $\endgroup$ – pela Aug 8 at 11:16
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What you calculated above ($\infty$, for a flat space universe) is just the proper mass-energy of matter. It doesn't take into account the - negative - energy of the gravitational "field" itself, which cannot be localised in General Relativity. The "total energy" of the universe could be 0, but there's no way we could give a physical sense to it, since the whole universe energy cannot be measured from "inside".

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  • $\begingroup$ There is no such a thing as negative energy. If it existed, it would be equivalent to negative mass. Energy is always positive. What you are referring to is the "negative potential energy" of the gravitational field. Because the gravitational potential itself is not observable, we can set it anywhere to any value for calculations. The most convenient way is to set it to zero at infinity. This makes the potential energy "negative" in formulas, but only as a result of the chosen convention. No energy is actually negative. This only means that you work against the field while moving farther away. $\endgroup$ – safesphere Jan 6 at 18:33
  • $\begingroup$ I don't understand how you can include potential energy in the total mass-energy of the universe. If you release energy by bring a mass from infinity to x, you must be storing an equal amount of energy from any objects you move that mass away from. Why isn't the net gravitation potential change for the universe zero in this scenario? $\endgroup$ – Quarkly Jan 7 at 14:04
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If it was completely flat, you wouldn't talk about volume. The universe isn't completely flat, it does have a volume (like a pancake, it's called 'flat', but it definitely has a volume). For a truly flat object, your formula would need a surface density and surface area, and you would get a finite answer.

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