Finite density at the BB preinflation epoch implies a finite universe?

[King Solomon]

TL;DR Summary

The density of the universe was 10^93 kg/cm^3 at 10^-43 (secs) after the Big Bang...doesn't that imply the universe is finite, since 10^-43 occurs before the inflationary epoch?

http://astronomy.swin.edu.au/cosmos/B/Big+Bang

D (density) = m/V

At t = 10^-47, D = 10^93 kg/cm ^3, r = 10^-57 meters, V = 4pi(r^3)/3 which is about (4/3)pi(10^-171)(meters^3)

t = 10^-47 precedes the inflationary epoch at t = 10^-35, this is important since this implies all matter at the big bang remains casually connected at this time.

D = 10^93 kg/cm^3 = 10^99 kg/m^3 = 10^102 g/m^3

DV = (10^102 g/m^3) * (4pi/3)(10^-171 m^3)

m = DV

m = (4/3pi) * 10^-69 g... yes, that was the mass of the universe according to this paper...less than a proton.
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Anyway...let's assume that the density was indeed 10^(102) gm^3 and that due to parameters of which I am not aware, the mass of the universe at time = 10^-47 was m = 10^58grams (made up number for this example), and let's call this mass X.

The mass of the observable universe today is roughly 10^56 grams, let's call this mass Y.

This would imply (in this fictional example) that Y/X is 1/100, or that our observable universe contains 1% of the mass entire universe, and under the assumption of homogeneity, that our observable universe is about 1/100th the size of the entire universe...and also... that there exists an exterior shape to our universe that compromises of all matter (some massive concave polyhedron) in the event that we live in an open or flat universe.

The moment we declare that the mass of the entire universe is finite, and that we do not live in a closed universe...then we have also declared that there exists an exterior to the entirety of the universe (beyond our observable part).

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If I am wrong, then may I ask...how could the universe have a finite density in a very small volume if the mass of the universe was not finite?

If the mass of the universe was finite at the Big Bang, then it must have remain finite, correct?

If the the mass of the entire universe (beyond the observable universe) is finite, and we live an an open/flat universe, then there is strict limit to the number of atoms/energy/particles/etc under the assumption of e = mc^2, correct?

If this strict limit exists, then if all mass were converted to the maximum number of allowable protons (mass of proton = 1.67 *10^-27kg), then we we would have a finite number of protons, and therefore there would some polyhedron (3D shape) that utilized every proton at one of its vertices, and therefore the universe has an exterior shape, correct (assuming the entirety of the universe is open/flat)?

"Grünbaum, B., Hamiltonian polygons and polyhedra, Geombinatorics, 3 (1994) 83-89."
Given a set S of n ≥ 3 points in a space (not all in line in or in the same plane) it is well known that it is always possible to polyhedronize S, i.e., construct a simple polyhedron, P such that the vertices of P are precisely the given points in S, this was proven in 1994 by Gr¨unbaum.

 

[PAllen]

Finite density says nothing at all about size or mass. There is one diagram in your link which misleadingly implies finite volume. Instead, this is causally connected radius, not total radius. None of the discussion in the link talks about finite volume. With this correction, all of your reasoning and calculations are irrelevant.

 

[Orodruin]

In addition, the paragraph quoting $10^{-47}$ seconds (actually, t^-47 seconds, but with reasonable interpretations) is just saying that you cannot reasonably extrapolate the Big Bang back to t = 0.

 

[PeterDonis]

t = 10^-47 precedes the inflationary epoch at t = 10^-35, this is important since this implies all matter at the big bang remains casually connected at this time.

This is not correct. Matter at sufficiently large spatial separations is not causally connected. This is true at all times in any FRW spacetime model. All inflation does is to make "sufficiently large spatial separations" larger than the size of our observable universe, so all the matter we can see now was causally connected in the far past.

 

[Orodruin]

Matter at sufficiently large spatial separations is not causally connected. This is true at all times in any FRW spacetime model.

In the Milne model, where $a(t) \propto t$, this is not true as $\int dt/t$ diverges as the lower boundary tends to zero, meaning that any set of comoving coordinates is going to be within the past light-cone of any event for all $t < t_p$, where $t_p \propto e^{-\Delta x}$. This is not surprising as the Milne model is essentially coordinates in the interior of the future light cone for an event E in Minkowski space and this means that E is going to be in the interior of the past lightcone of all events in the region described by the FRW coordinates of the Milne model.

 

[PeterDonis]

In the Milne model, where $a(t) \propto t$, this is not true

Yes, you're right, the Milne model is an edge case here.

 

[Orodruin]

Yes, you're right, the Milne model is an edge case here.

The Milne universe was just an example that came to mind. You can also have a flat universe containing some strange ideal fluid with $-1 < w \leq -1/3$. Such a universe would have a scale factor growing as $a \propto t^\gamma$, where $\gamma \geq 1$. The same argument holds there, but for $\int dt/t^\gamma$.

 

[PeterDonis]

You can also have a flat universe containing some strange ideal fluid with $-1 < w \leq -1/3$. Such a universe would have a scale factor growing as $a \propto t^\gamma$, where $\gamma \geq 1$. The same argument holds there, but for $dt/t^\gamma$.

Ok, but this is also an edge case, at least in the sense that it doesn't describe anything like our actual universe, correct?

 

[Orodruin]

Ok, but this is also an edge case, at least in the sense that it doesn't describe anything like our actual universe, correct?

This depends on what actually happens before reheating. I am not sure whether or not there are inflationary models dominated by an equation of state parameter different from -1 (this would inflate the universe but not remove the singularity).

 

[PeterDonis]

This depends on what actually happens before reheating. I am not sure whether or not there are inflationary models dominated by an equation of state parameter different from -1

Hm, ok.

 

[Orodruin]

Hm, ok.

To put it differently, in order to have inflation, you need a phase with w < -1/3. Whatever is driving inflation must eventually decay through reheating (or we would never enter radiation domination), eg, at the end of slow roll of the inflaton field.

In the case of -1/3 > w > -1, you do not remove the initial singularity, but you do manage to causally connect the entire universe. For w = -1 you actually remove the past singularity.

 

[King Solomon]

Finite density says nothing at all about size or mass. There is one diagram in your link which misleadingly implies finite volume. Instead, this is causally connected radius, not total radius. None of the discussion in the link talks about finite volume. With this correction, all of your reasoning and calculations are irrelevant.

Yet the total mass of the universe in this casual radius amounts to less than a proton. What am I missing? According to those graphs any observer in our observable universe should live in a nearly empty universe.

 

[Orodruin]

Yet the total mass of the universe in this casual radius amounts to less than a proton. What am I missing? According to those graphs any observer in our observable universe should live in a nearly empty universe.

It does not. You have used the early volume with inflation (which is teeny tiny) with the density without inflation.

 

[King Solomon]

It does not. You have used the early volume with inflation (which is teeny tiny) with the density without inflation.

I do have an unrelated question to the OP, but I'd rather not make thread on it.

Why aren't complex numbers used to explain the curvature/warp/contraction of space in general and special relativity?

Wouldn't it be simpler to use the imaginary (lateral) components of each of three spatial dimensions? Most commonly I see 3D + T (three spatial dimensions + time) and 4D (four spatial dimensions), but I've never seen an article or publication that invokes our three spatial dimensions alongside their natural lateral components (the imaginary/lateral components of a dimension).

In fact, I've only ever witnessed complex numbers used (in physics) to describe alternating currents when I young teenager, and I remained surprised to this day that I never saw them used since (in physics), aside from the description of "tachyons."

 

[PeterDonis]

I've only ever witnessed complex numbers used (in physics) to describe alternating currents when I young teenager, and I remained surprised to this day that I never saw them used since (in physics), aside from the description of "tachyons."

I assume that by "physics" you mean "classical physics", since complex numbers are used extensively in quantum physics.

 

[PeterDonis]

the imaginary (lateral) components of each of three spatial dimensions?

What does this mean?

 

[King Solomon]

What does this mean?

What does this mean?

s² = x² + y² + z² - c²t²

Let's simplify to:

s² = w² - c²t²

We're looking at the spacetime interval, which forms a continuous family of concentric hyperbolas in the direction time.

When I see these things, I wonder why complex numbers were not used instead.

The complex numbers must be an inherit part of our universe. The real numbers cannot exist in isolation. For any real (x,y,z) there must be an (ix, iy, iz) component), even if that component is zero.

It seems to me that the "warping of space' due to gravity is setting the imaginary (lateral) component to a non-zero value.

 

[PeterDonis]

We're looking at the spacetime interval, which forms a continuous family of concentric parabolas in the direction time.

Hyperbolas, not parabolas.

The complex numbers must be an inherit part of our universe. The real numbers cannot exist in isolation. For any real (x,y,z) there must be an (ix, iy, iz) component), even if that component is zero.

It seems to me that the "warping of space' due to gravity is setting the imaginary (lateral) component to a non-zero value.

This is personal speculation, which is off limits for discussion here at PF. Further discussion of it will cause this thread to be closed and you to receive a warning.

 

[King Solomon]

I assume that by "physics" you mean "classical physics", since complex numbers are used extensively in quantum physics.

And yet I've never seen them used to describe general relativity, where the "warping of space" would seem to be the most natural opportunity to invoke them.

I just want to know why they are not used. That's all.

 

[PeterDonis]

the "warping of space" would seem to be the most natural opportunity to invoke them.

It might seem that way to you, but evidently it doesn't seem that way to, well, all the physicists who have ever worked on GR.

I just want to know why they are used.

You mean why they aren't used? Because, as I said just now, no physicist who has worked on GR has seen any reason to do so.

If you are looking for an explicit argument from a relativity physicist explaining why they see no need to use complex numbers, I doubt you'll find one. I have never seen one. I suspect it doesn't even occur to physicists that this is a question they need to answer. GR works just fine with real numbers and there is no obvious physical phenomenon like quantum interference that points to a need for complex numbers.

 

[PeterDonis]

the "warping of space" would seem to be the most natural opportunity to invoke them.

You might want to consider that, since nobody else appears to see it that way, perhaps your own way of looking at it is mistaken in some fashion. For example, your talk of an "imaginary lateral component" looks like nonsense to me. Either that or you are making the (fairly common) mistake of trying to visualize spacetime as embedded in a higher dimensional manifold, and the spacetime curvature described by GR as extrinsic curvature due to that embedding. There are particular cases where that viewpoint can be somewhat helpful, but it is not how GR standardly views spacetime curvature. Spacetime curvature is intrinsic curvature of the manifold, and can be described and modeled and used for calculations without ever having to postulate any embedding in a higher dimensional manifold. Therefore, Occam's razor says there is no such higher dimensional manifold.

 

[King Solomon]

You might want to consider that, since nobody else appears to see it that way, perhaps your own way of looking at it is mistaken in some fashion. For example, your talk of an "imaginary lateral component" looks like nonsense to me. Either that or you are making the (fairly common) mistake of trying to visualize spacetime as embedded in a higher dimensional manifold, and the spacetime curvature described by GR as extrinsic curvature due to that embedding. There are particular cases where that viewpoint can be somewhat helpful, but it is not how GR standardly views spacetime curvature. Spacetime curvature is intrinsic curvature of the manifold, and can be described and modeled and used for calculations without ever having to postulate any embedding in a higher dimensional manifold. Therefore, Occam's razor says there is no such higher dimensional manifold.

I do remember reading an article that claimed and (also showed) how differential geometry was used to remove complex numbers from special relativity.

However on the topic of general relativity, I have never seen them used (and/or removed by simplification) at all at any point in history.

That being said, you and I have a philosophical difference concerning complex numbers. I do not see "three real dimensions" and three extrinsic "imaginary dimensions." I see only three dimensions, which are only "perpendicular" at rest in proper time. The "imaginary" or "lateral" component represents the skews the orthogonality of the dimensions.

http://bandtechnology.com/PolySigned/NonOrthogonal/index.html

I just personally believe that would be a simpler way to expression "warped" space.

 

[PeterDonis]

I do remember reading an article that claimed and (also showed) how differential geometry was used to remove complex numbers from special relativity.

Can you give a reference? This sounds extremely dubious to me.

I do not see "three real dimensions" and three extrinsic "imaginary dimensions." I see only three dimensions, which are only "perpendicular" at rest in proper time. The "imaginary" or "lateral" component represents the skews the orthogonality of the dimensions.

Then I am back to "nonsense". Non-orthogonality is not a property of "dimensions". It's a property of vectors. And non-orthogonality of vectors does not mean there is an "imaginary" or "lateral" component to them. It just means they're not orthogonal. That's all there is to it.

Also, there are four (real) dimensions in spacetime, not three.

 

[King Solomon]

Can you give a reference? This sounds extremely dubious to me.

Neither of these are the articles I read, but they discuss the same topic:


https://en.wikipedia.org/wiki/Split-complex_number#History

I cannot recall the title/url/author of the article I read nearly five years ago.

 

[PeterDonis]

they discuss the same topic

These aren't complex numbers, they're split-complex numbers. Yes, they were discovered in the 1800s, but no one that I'm aware of thought of trying to use them in special relativity prior to the time the Wikipedia article refers to, the late 20th century.

 

[PeterDonis]

they discuss the same topic

Btw, the answer to the question that heads the reddit article is that "describing the Minkowski norm" is only one of many things we want the numbers we use in relativity to do. Focusing on that one thing, being able to reproduce the minus sign in $x^2 - t^2$, to the exclusion of all others, is not a good way to make progress.

 

[PeterDonis]

The OP question has been answered, and a thread tangent has turned into personal speculation. Thread closed.