What are these diagrams showing about GR?

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https://universe-review.ca/I02-24-inflation2.png
https://courses.lumenlearning.com/astronomy/chapter/the-inflationary-universe/
http://abyss.uoregon.edu/~js/ast123/lectures/lec18.html

Hello.
The diagrams I've linked to are comparisons between the Standard Hot Big Bang and Inflationary cosmological models. I'm curious about the Y axis, which gives the scale of the radius of the universe, and where the lines of the two different cosmological models originate from that axis.

The Inflationary line appears to originate between 10-50 and 10-60meters. At first glance this appears to be smaller than the Planck scale, which is quoted here...
https://newt.phys.unsw.edu.au/einsteinlight/jw/module6_Planck.htm
...as being 10-35meters.

Could someone please help me understand if there's a real disparity here or if there's something I'm just not seeing? Thanks.

Also, the Standard (HBB) line appears to originate from the Y axis at about 10-5 meters.
Could I please be informed (in basic and easy to understand terms) as to why this is so.

Thanks in advance,

Cerenkov.


Oops!
I forgot to mention that I'm trying to gain a better insight into how GR works and was wondering if these diagrams can help.
 

Answers and Replies

  • #2
phinds
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There's nothing at all wrong with things being smaller than a Plank length. It, after all, just a human-invented metric and has no effect on nature but rather just on a part of the language we use to describe nature.
 
  • #3
kimbyd
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https://universe-review.ca/I02-24-inflation2.png
https://courses.lumenlearning.com/astronomy/chapter/the-inflationary-universe/
http://abyss.uoregon.edu/~js/ast123/lectures/lec18.html

Hello.
The diagrams I've linked to are comparisons between the Standard Hot Big Bang and Inflationary cosmological models. I'm curious about the Y axis, which gives the scale of the radius of the universe, and where the lines of the two different cosmological models originate from that axis.

The Inflationary line appears to originate between 10-50 and 10-60meters. At first glance this appears to be smaller than the Planck scale, which is quoted here...
https://newt.phys.unsw.edu.au/einsteinlight/jw/module6_Planck.htm
...as being 10-35meters.

Could someone please help me understand if there's a real disparity here or if there's something I'm just not seeing? Thanks.

Also, the Standard (HBB) line appears to originate from the Y axis at about 10-5 meters.
Could I please be informed (in basic and easy to understand terms) as to why this is so.

Thanks in advance,

Cerenkov.


Oops!
I forgot to mention that I'm trying to gain a better insight into how GR works and was wondering if these diagrams can help.
Yup! This isn't a problem. It may be a little confusing, but the graph doesn't contain any physics in operation at sub-Planck scales. The issue is that the "size of the observable universe" isn't a physical thing that far back, because the stuff that's in our universe did not exist.

During inflation, there were quantum fluctuations in the field which drove inflation, dubbed the inflaton. These quantum fluctuations were of the order of the size of the cosmological horizon at the time. That horizon was significantly larger than the Planck scale. Those fluctuations then grew over time.

What that graph is telling us is not that our universe was tiny, but that these zero-point fluctuations that originated very early-on have now grown to become much larger than our observable universe. Everything we observe originated in later fluctuations that didn't expand quite as much.
 
  • #4
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kimbyd wrote...

Yup! This isn't a problem. It may be a little confusing, but the graph doesn't contain any physics in operation at sub-Planck scales. The issue is that the "size of the observable universe" isn't a physical thing that far back, because the stuff that's in our universe did not exist. During inflation, there were quantum fluctuations in the field which drove inflation, dubbed the inflaton.

Doh! Of course. (Smacks forehead.)
The inflaton had to decay first for there to be 'stuff' in our universe.

These quantum fluctuations were of the order of the size of the cosmological horizon at the time. That horizon was significantly larger than the Planck scale. Those fluctuations then grew over time. What that graph is telling us is not that our universe was tiny, but that these zero-point fluctuations that originated very early-on have now grown to become much larger than our observable universe. Everything we observe originated in later fluctuations that didn't expand quite as much.

Thanks kimbyd. That helps.


But what about the Standard HBB line originating from the Y axis at a scale of about 10-5 ?
If a singularity is unavoidable with the use of GR, shouldn't that line originate where the X and Y axes meet and shouldn't there be an infinity symbol there?

Cerenkov.
 
  • #5
kimbyd
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But what about the Standard HBB line originating from the Y axis at a scale of about 10-5 ?
If a singularity is unavoidable with the use of GR, shouldn't that line originate where the X and Y axes meet and shouldn't there be an infinity symbol there?

Cerenkov.
The location of that singularity depends upon the contents of the universe. Change the contents, change where that point is. Inflation moves the singularity, but doesn't eliminate it.

Another way of stating the issue is that inflation has no singularity which occurs at a specific point, but inflation is unstable such that if we run the clock backward, even the tiniest deviation would result in a singularity. So you need an infinite amount of fine-tuning to escape the singularity. My understanding is that General Relativity guarantees that any theory for an expanding universe must have a singularity in the past like this, either like the classic Big Bang theory which has an inevitable singularity at a specific time due to the contents of the universe, or one that arises out of an instability as in the case of inflation. Inflation can't avoid this entirely, because it doesn't use quantum gravity.
 
  • #6
PeterDonis
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General Relativity guarantees that any theory for an expanding universe must have a singularity in the past like this
The singularity theorems that guarantee this only do so under the assumption that the energy conditions hold. AFAIK inflation models (and indeed any model with a positive cosmological constant or the equivalent dominating the dynamics) violates those conditions, so the singularity theorems would not apply. The simplest example is that de Sitter spacetime has no singularity even though it describes an "expanding universe".
 
  • #7
George Jones
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The singularity theorems that guarantee this only do so under the assumption that the energy conditions hold. AFAIK inflation models (and indeed any model with a positive cosmological constant or the equivalent dominating the dynamics) violates those conditions, so the singularity theorems would not apply. The simplest example is that de Sitter spacetime has no singularity even though it describes an "expanding universe".
Inflationary cosmologies do not satisfy the hypotheses of the Penrose-Hawking singularity theorems, but I think that typical inflationary cosmologies do satisfy the hypotheses of Borde-Guth-Vilenkin singularity theorem.
 
  • #8
kimbyd
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The singularity theorems that guarantee this only do so under the assumption that the energy conditions hold. AFAIK inflation models (and indeed any model with a positive cosmological constant or the equivalent dominating the dynamics) violates those conditions, so the singularity theorems would not apply. The simplest example is that de Sitter spacetime has no singularity even though it describes an "expanding universe".
If I'm understanding the full details of this, I think it still applies to inflation with some caveats.

Per Wikipedia, the Penrose-Hawking singularity theorems work as long as either:
For a perfect fluid, the null energy condition states that ##\rho + p \ge 0##, while the strong energy condition also requires that ##\rho + 3p \ge 0##.
"geodesic congruence" relates to the type of geodesics of the stuff that exists in the space-time. Null congruence means we're talking about photons. Timelike congruence means matter.

Inflation satisfies the null energy condition, but not the strong energy condition (during inflation, ##\rho + p## is slightly above zero). So, if I'm understanding the above correctly, inflation need not have a singularity per the Penrose-Hawking singularity theorems if all you have is normal matter. But add a single photon to the mix and the theorem applies.

My understanding is that the null energy condition is expected to be universal, while the strong energy condition certainly is not the case (caveat: it hasn't yet been proven that the null energy condition must be universal, but it certainly appears to result from our current understanding of quantum mechanics). So the Penrose-Hawking singularity theorem is likely to apply in any space-time described by General Relativity that contains photons.
 
  • #9
PeterDonis
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I think that typical inflationary cosmologies do satisfy the hypotheses of Borde-Guth-Vilenkin singularity theorem.
Hm, I'm not as familiar with that one. I'll have to look it up.
 
  • #10
PeterDonis
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"geodesic congruence" relates to the type of geodesics of the stuff that exists in the space-time
No, it doesn't, it relates to all possible curves in the spacetime geometry regardless of whether those curves are the worldlines of anything. You don't have to have photons anywhere in the spacetime for the spacetime geometry to have null vector fields in it, and you don't have to have normal matter anywhere in the spacetime for the spacetime geometry to have timelike vector fields in it.
 
  • #11
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Thanks for exploring this, kimbyd and Peter.

Btw, I can confirm that Hawking does indeed cover the strong, weak and null energy conditions in the Classical Theory chapter of the book, 'The Nature of Space and Time'. Of the weak, he says this.

"The weak energy condition is obeyed by the classical energy momentum tensor in any reasonable matter, such as a scalar or electromagnetic field or a fluid with a reasonable energy of state. It may not, however, be satisfied locally by the quantum mechanical expectation value of the energy momentum tensor. This will be relevant in my second (Quantum Black Holes) and third (Quantum Cosmology) lectures (chapters 3 and 5)."

A question regarding these three energy conditions in the context of Classical theory, if I may.
Hawking talks about these singularity theorems in terms of being mathematically proven. Surely that's conditionally proven? That is, since nobody actually knows if these three energy states do hold, the proof is conditional on them doing so? Ok, that's two questions. My bad.


Re: the Borde-Guth-Vilenkin singularity theorem - I would be very interested to understand more about this, particularly any caveats or conditions associated with it. Should it be treated in the same way as the classical singularity theorems - as conditional proofs, for instance.

Thank you,

Cerenkov.
 
  • #12
PeterDonis
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Hawking talks about these singularity theorems in terms of being mathematically proven. Surely that's conditionally proven?
Any proof of a theorem is conditional on the assumptions that were used in the proof. The Penrose-Hawking singularity theorems assume that the energy conditions hold, so of course the proof of those theorems is conditional on the energy conditions.
 
  • #13
kimbyd
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No, it doesn't, it relates to all possible curves in the spacetime geometry regardless of whether those curves are the worldlines of anything. You don't have to have photons anywhere in the spacetime for the spacetime geometry to have null vector fields in it, and you don't have to have normal matter anywhere in the spacetime for the spacetime geometry to have timelike vector fields in it.
Edit: My thinking on this was incorrect. The BGV singularity theorem is probably the way to go.
 
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  • #14
PeterDonis
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Which would seem to indicate that the singularity theorem holds if either energy condition holds.
That's the way I read the statement in the Wikipedia article, yes.
 
  • #15
kimbyd
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That's the way I read the statement in the Wikipedia article, yes.
Sorry, I was too slow. I edited this out because I no longer think it's accurate. Looking a bit more into congruences, it sounds more like this is a property of the space-time that has nothing to do with the matter present. It is only represented in terms of matter or photons because of the mathematical relationship. The congruence then depends upon whether either null or timelike geodesics within that spacetime have this property, and I've had a hard time finding out if that's the case.

That said, I looked a bit more into the BGV theorem. I wasn't able to easily find a good summary of it, but here's the preprint for the original paper:
https://arxiv.org/abs/gr-qc/0110012

What they claim is that they can prove a past singularity provided only that the expansion rate was always positive (no energy conditions needed). Inflationary space-times definitely satisfy this condition.

I don't feel entirely comfortable taking the author's word for it that this is the only assumption required (these kinds of things often get revised in later discussions), but I wasn't able to quickly find responses to this paper that point out any implicit assumptions the authors missed.
 
  • #16
PeterDonis
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Looking a bit more into congruences, it sounds more like this is a property of the space-time that has nothing to do with the matter present.
That's correct, a congruence, by itself, is just a set of curves in the spacetime geometry. There is no requirement that any matter, photons, whatever actually move along those curves.

In practice, most of the time the congruences that people choose to look at are the ones that describe the worldlines of some matter or radiation of interest. However, for purposes of the singularity theorems any congruence that satisfies the conditions will do, it doesn't have to be one that describes any actual matter or radiation.

Again however, although the congruence you choose in evaluating the energy conditions for purposes of the singularity theorems can be any congruence you like (as long as it's null or timelike, respectively, depending on which energy condition you're evaluating), the stress-energy tensor that you use must be the one that describes the actual matter, radiation, inflaton field, or whatever else you are assuming is present. In other words, the SET, not the congruence, is where the actual, physically present "stuff" is modeled. So, for example, the ##\rho## and ##p## in the simplified statements of the energy conditions (for perfect fluids) are the ##\rho## and ##p## of whatever "stuff" is actually present.
 
  • #17
kimbyd
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That's correct, a congruence, by itself, is just a set of curves in the spacetime geometry. There is no requirement that any matter, photons, whatever actually move along those curves.

In practice, most of the time the congruences that people choose to look at are the ones that describe the worldlines of some matter or radiation of interest. However, for purposes of the singularity theorems any congruence that satisfies the conditions will do, it doesn't have to be one that describes any actual matter or radiation.

Again however, although the congruence you choose in evaluating the energy conditions for purposes of the singularity theorems can be any congruence you like (as long as it's null or timelike, respectively, depending on which energy condition you're evaluating), the stress-energy tensor that you use must be the one that describes the actual matter, radiation, inflaton field, or whatever else you are assuming is present. In other words, the SET, not the congruence, is where the actual, physically present "stuff" is modeled. So, for example, the ##\rho## and ##p## in the simplified statements of the energy conditions (for perfect fluids) are the ##\rho## and ##p## of whatever "stuff" is actually present.
This is making it difficult for me to parse precisely what the point of talking about the congruences was in the first place. If they're not a property of the space-time, then what do they matter?

Regardless, if the null energy condition is sufficient for the Penrose-Hawking singularity condition to hold, then it definitely applies to inflation, because inflation does respect the null energy condition. It might not hold in every possible extension of inflation in which the null energy condition might not hold, but it holds for current inflation models.
 
  • #18
PeterDonis
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This is making it difficult for me to parse precisely what the point of talking about the congruences was in the first place.
Because the assumption of the singularity theorems is that there exists some congruence (null or timelike, depending on the energy condition) with the appropriate properties. The proof then uses the Raychaudhuri equation to deduce that the expansion scalar of that congruence diverges within a finite affine parameter, i.e., that that congruence is geodesically incomplete. And that is sufficient to show that the spacetime geometry in which that congruence exists is geodesically incomplete (because geodesic incompleteness just means that there exists at least one geodesic that cannot be extended to arbitrary values of its affine parameter).

f they're not a property of the space-time, then what do they matter?
They are a property of the spacetime. But they aren't (necessarily) a property that is related to the distribution of stress-energy in the spacetime. That depends on what congruence you choose.

because inflation does respect the null energy condition
Someone like @George Jones would need to weigh in on this, since he appeared to be saying in post #7 that this is not the case. On its face I can see how any solution with ##p = w \rho## and ##w \ge -1## would appear to satisfy ##\rho + p \ge 0##, so I can see where you're coming from.
 

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