Undergrad Lambda CDM Model and Spatial Flatness

[timmdeeg]

[Moderator's note: Spun off from previous thread due to topic/level change.]

Since, according to our best current model, the universe is and always has been spatially infinite, there were plenty of such places at the early times you mention.

Which current model? Doesn't "spatially infinite" imply Euclidean flatness which as far as I'm aware of requires $\Omega = 1$? The theories of inflation don't predict that. They predict a value for $\Omega$ which is very close to $1$ but not a value which equals $1$, because during inflation the growth of $a$ is finite.
Isn't is more likely from this point of view that the universe isn't spatially infinite?
Shouldn't we say at least we don't know?

 

[PeterDonis]

Which current model?

The Lambda CDM model.

Doesn't "spatially infinite" imply Euclidean flatness

For spatial slices, yes.

which as far as I'm aware of requires $\Omega = 1$?

It requires $\Omega = 1$ to be within the current error bars of the data, yes. Right now that is the case. There are some cosmologists who believe it will cease to be the case as our data becomes more precise and the error bars narrow. But others don't think so. If that ever does happen, obviously our best current model would need to be re-evaluated.

The theories of inflation don't predict that. They predict a value for $\Omega$ which is very close to $1$ but not a value which equals $1$, because during inflation the growth of $a$ is finite.

I don't know that this is true for all inflation models, though it is true for some of them, yes.

Isn't is more likely from this point of view that the universe isn't spatially infinite?

For purposes of this thread, it doesn't really matter, because in inflationary models where $\Omega$ is not precisely equal to $1$, the spatial extent of the universe is still so much larger than the extent of our observable universe that it is effectively infinite when you're thinking about things like whether there are events at early times, like the emission of the CMBR or light emitted by early galaxies, from which light signals would not yet have reached us.

Of course you could turn this around and argue that the fact that we are still seeing light signals from such early events--that not all such light signals have passed us by now--does not establish that the universe must be spatially infinite, only that its spatial extent is much, much larger than the spatial extent of our observable universe.

 

[timmdeeg]

I don't know that this is true for all inflation models, though it is true for some of them, yes.

I should have shown more precisely to what I take reference.

$(\Omega^{-1}-1)\rho{a}^2=-\frac{3c^2}{8\pi{G}}k$

It doesn't really make a difference which inflation model we are talking about. During inflation the expansion, the growth of $a$ is large but finite and hence $(\Omega^{-1}-1)$ can't be zero as it should in order to yield $\Omega = 1$. So this equation seems to predict that the universe is spatially finite. Please clarify where I'm misled.

However if correct, don't we have the good old flatness problem again then?(*) Regarding the above euclidean flatness seems to require $\Omega = 1$ before inflation started. I haven't seen this reasoning somewhere else which makes me suspicious. If true the inflation still explains a lot about the CMB.

EDIT (*) No, the flatness problem had to do with the observation that the universe is almost flat.

 

[PeterDonis]

During inflation the expansion, the growth of $a$ is large but finite

Yes.

and hence $(\Omega^{-1}-1)$ can't be zero

No. If $\Omega = 1$ exactly, then $k = 0$ for any value of $a$. That's what the equation says. So this equation by itself cannot tell you whether the universe is spatially finite or spatially infinite.

Also, you seem to be mistakenly assuming that if the universe is spatially infinite, then the growth of $a$ must be infinite. That's not correct. A spatially infinite universe still has a finite scale factor and scale factor growth.

 

[timmdeeg]

No. If $\Omega = 1$ exactly, then $k = 0$ for any value of $a$. That's what the equation says. So this equation by itself cannot tell you whether the universe is spatially finite or spatially infinite.

Agreed. But it tells me that the universe can't be infinite due to inflation thereby neglecting the case $k = 0$. So if you say "to our best current model, the universe is and always has been spatially infinite" you seem to mean the case $k = 0$. My point is that I don't see that the Lambda CDM model predicts this case.

 

[PeterDonis]

it tells me that the universe can't be infinite due to inflation thereby neglecting the case $k = 0$.

Inflation is perfectly compatible with $k = 0$. Maybe you need to read this again:

you seem to be mistakenly assuming that if the universe is spatially infinite, then the growth of $a$ must be infinite. That's not correct. A spatially infinite universe still has a finite scale factor and scale factor growth.

 

[timmdeeg]

Inflation is perfectly compatible with $k = 0$.

Yes, but I didn't say otherwise. I only said that if we have the case that $k$ isn't $0$ then the inflation can't produce a spatially infinite universe.

Could you elaborate how the Lambda CDM model predicts a spatially infinite universe?

 

[PeterDonis]

I didn't say otherwise

It sure seemed like you were saying that $k = 0$ is not possible in any inflation model, which is saying otherwise.

I only said that if we have the case that $k$ isn't $0$ then the inflation can't produce a spatially infinite universe.

Well, of course not, since $k > 0$ by definition means a spatially finite universe. (I assume you actually meant $k > 0$, not $k \neq 0$, since a $k < 0$ universe, like $k = 0$, is spatially infinite.) But that's not at all the same as saying that inflation requires $k > 0$. It doesn't.

Could you elaborate how the Lambda CDM model predicts a spatially infinite universe?

Um, by having $k = 0$? (More precisely, by having $\Omega = 1$, which implies $k = 0$.)

Look at the list of "fixed parameters" on this page:

https://en.wikipedia.org/wiki/Lambda-CDM_model

 

[timmdeeg]

It sure seemed like you were saying that $k = 0$ is not possible in any inflation model, which is saying otherwise.

I said "neglecting the case $k = 0$", "not possible" wasn't meant. Perhaps I've a problem with correct wording. Of course inflation doesn't exclude a certain sign of $k$.

I assume you actually meant $k > 0$

Yes, thanks for correcting.

Look at the list of "fixed parameters" on this page:
https://en.wikipedia.org/wiki/Lambda-CDM_model

The author discusses the measurement

1.0023

and mentions "flat or almost flat". I still can't see that the model predicts that our universe is spatially infinite. In my opinion we will never be able to establish $\Omega = 1$ experimentally.

 

[PeterDonis]

I still can't see that the model predicts that our universe is spatially infinite.

The model explicitly says that $\Omega = 1$ exactly in the model. It's a fixed parameter. That's what I pointed you at in the Wikipedia article. So the model by definition predicts $k = 0$ and a spatially infinite universe.

In my opinion we will never be able to establish $\Omega = 1$ experimentally.

Of course not, since we can never make measurements to infinite accuracy. But that doesn't mean a model can't adopt $\Omega = 1$ as a fixed parameter. That's what the Lambda CDM model does.

If we ever establish that $\Omega$ cannot equal $1$ exactly (because the error bars from observation exclude that value), then our current Lambda CDM model will be falsified and we will need to re-evaluate what our best current model of the universe is. But that hasn't happened, so the Lambda CDM model is our best current model.

 

[PeterDonis]

The author discusses the measurement

The measurement is not the model. The model is a mathematical model that makes predictions. The measurement data is what you compare the predictions with.

 

[timmdeeg]

The model explicitly says that $\Omega = 1$ exactly in the model. It's a fixed parameter. That's what I pointed you at in the Wikipedia article. So the model by definition predicts $k = 0$ and a spatially infinite universe.

Yes the model predicts that. But didn't you say that our our universe is spatially infinite in #2? Doesn't this claim require the knowledge that regarding our universe $\Omega = 1$ ?

 

[PeterDonis]

didn't you say that our our universe is spatially infinite in #2?

No. Go back and read the post again, carefully.

 

[timmdeeg]

No. Go back and read the post again, carefully.

I understood "Since, according to our best current model, the universe is and always has been spatially infinite, there were plenty of such places at the early times you mention." such that you mean our universe.

 

[PeterDonis]

I understood "Since, according to our best current model, the universe is and always has been spatially infinite, there were plenty of such places at the early times you mention." such that you mean our universe.

See the bolded text above.

 

[timmdeeg]

See the bolded text above.

My reasoning is that the model can't predict anything unless we have a definitive value of $k$.

A definite value of $k$ can only be fixed empirically though and only in principle, e.g. if the size of the universe isn't too large.

To my knowledge $k = 1$ can't be excluded, so we have to leave the question open as to whether or not the universe is spatially infinite.

Kindly clarify where I am wrong.

 

[PeterDonis]

My reasoning is that the model can't predict anything unless we have a definitive value of $k$.

The model does have a definite value of $k$, $k = 0$.

A definite value of $k$ can only be fixed empirically though

Wrong. A model can assume any definite value of $k$ that it wants. You are confusing experimental data with model assumptions and parameters. They're not the same. If we could never make models without having empirical measurements to fix all the parameters, we could not do science.

Also, you have already said (and I agreed) that we can't ever fix an exact value of any quantity empirically, because our measurements always have some finite error. But that would mean we could never fix empirically a definite value of $k$, or any parameter, which, by your reasoning, would mean we could never make models at all.

Kindly clarify where I am wrong.

You seem to have a flawed understanding of how scientific models work. See above.

 

[PeterDonis]

To my knowledge $k = 1$ can't be excluded, so we have to leave the question open as to whether or not the universe is spatially infinite.

You are still failing to address what I actually said. What I actually said (in the post that originally spawned this discussion) is that according to our best current model, the universe is spatially infinite. That is a true statement. It is still true even when we recognize that our best current model, like all scientific models, is not necessarily final and might have to be changed in the future when we have more data and more accurate data then we have now.

You seem to be confusing two different statements: the statement that our best current model, which says $k = 0$, might have to be changed in the future, and the statement that our best current model doesn't actually say $k = 0$. The first statement is true. But all of your posts are trying to assert the second statement, which is false.

 

[timmdeeg]

The model does have a definite value of $k$, $k = 0$.

The model does have also a definitive value of $k$, $k = 1$.
Why do you have a preference for $k = 0$?

 

[PAllen]

The model does have also a definitive value of $k$, $k = 1$.
Why do you have a preference for $k = 0$?

No, the model doesn't directly specify a value for k, because k=0 exactly follows from the fixed parameters (i.e. those assumed, not empirically determined). The validity of the assumptions is then tested by ongoing observations.

 

[PeterDonis]

The model does have also a definitive value of $k$, $k = 1$.

No, $k = 0$, $\Omega = 1$.

Why do you have a preference for $k = 0$?

I have said nothing at all about any preference of mine. I have merely said, repeatedly, what the Lambda CDM model says.

 

[timmdeeg]

No, the model doesn't directly specify a value for k, because k=0 exactly follows from the fixed parameters (i.e. those assumed, not empirically determined). The validity of the assumptions is then tested by ongoing observations.

Do you mean the assumption $\Omega = 1$?

If yes, we then assume euclidean flatness. But we don't know for sure, so we can't claim it. It's just an assumption, how can we ever be sure? It is also possible that we live in a large sphere. This was my main point.

Another question, doesn't $\Omega = 1$ require $k = 0$ at the big bang (before inflation started)?

 

[timmdeeg]

I have said nothing at all about any preference of mine. I have merely said, repeatedly, what the Lambda CDM model says.

I understand, thank you for your patience.

 

[PeterDonis]

If yes, we then assume euclidean flatness. But we don't know for sure, so we can't claim it.

Which is why I said, according to our best current model, the universe is spatially infinite. That was the whole point of my putting in that qualifier.

doesn't $\Omega = 1$ require $k = 0$ at the big bang (before inflation started)?

Yes.

 

[timmdeeg]

Thanks and sorry for some misunderstandings, my fault.

 

[timmdeeg]

doesn't $\Omega = 1$ require $k = 0$ at the big bang (before inflation started)?

Yes.

Hm, after rethinking, $k = 0$, seems to require extreme fine tuning which isn't too much liked by physicists. I wonder if from this point of view assuming euclidean geometry of the universe is not really realistic.

 

[PeterDonis]

after rethinking, $k = 0$, seems to require extreme fine tuning which isn't too much liked by physicists. I wonder if from this point of view assuming euclidean geometry of the universe is not really realistic.

That is a completely separate question from the topic of this thread, which is what the Lambda CDM model says. If you want to discuss whether the Lambda CDM model, which says $k = 0$, involves fine-tuning, and you can find references where that topic is discussed, you are welcome to start a new thread on it based on those references.

This thread is closed.