I Calculating the probability that the Universe is finite

[PeterDonis]

Do I now have a correct interpretation of your view?

No.

I have already said, explicitly, more than once, that the calculation you did in the OP of this thread cannot answer the title question of this thread. The only question it can answer is what is the probability that the universe is finite under the assumption that the particular model in which the calculation was done is correct. But that is not what the title question of this thread is asking. The title question of this thread is asking about the probability that the universe is finite, period. And you cannot answer that question by any calculation that assumes one particular model, because we are not certain that that model is correct.

 

[Buzz Bloom]

Hi @PeterDonis:

you need to find a different word for #1, since "model" is used in standard scientific terminology to mean #2.

Thank you very much for the clarification. I am trying to find a phrase to clearly express the type (1) model. Here some ideas.

detailed model​

valued model​

numerical model​

numericized model​

quantitative model​

Do you have any suggestions or choices?

under the assumption that the particular model in which the calculation was done is correct

I am definitely OK with including an "assumption" such as the quote above. However, before I edit my summary or my Post #1 to to add an "assumption", I would like to better understand the meaning of "in which the calculation was done is correct". I interpret "model" in the quote to mean the #2 model, that is:

the Friedmann equation with its five variables on the RHS and no values specified for these five variables.​

I do not understand

"in which the calculation was done".​

I suggest that "with which ..." would be clearer.

As I understand the process, the calculation of values for the five variables is based on finding which combination of five specific values minimizes the sum of differences squared (with appropriate coefficients) between a database of values for other variables and calculated values for these other variables using the Friedman equation with the five specific values. Is this understanding correct?

Regards,
Buzz

 

[PeterDonis]

I am trying to find a phrase to clearly express the type (1) model.

How about "particular solution"? That would match common usage in science and math fairly well: you have a model that has some free parameters, and setting particular values for those free parameters gives you a particular solution of that model.

I am definitely OK with including an "assumption" such as the quote above.

You shouldn't be if your purpose is to answer the title question of this thread, because the title question of this thread is not asking for the probability that the universe is finite given that assumption.

I interpret "model" in the quote to mean the #2 model

Yes.

the calculation of values for the five variables is based on finding which combination of five specific values minimizes the sum of differences squared (with appropriate coefficients) between a database of values for other variables and calculated values for these other variables using the Friedman equation with the five specific values.

The values being compared with are not for "other variables". They are values for actual observed data. In other words, the differences squared whose sum is to be minimized are the differences between values for quantities that are directly observed (such as the redshifts of galaxies, their apparent luminosities, and their apparent angular sizes, the observed temperature of the CMB, etc.) and values for those same quantities that are calculated using the equations of the model. So, in the terminology I suggested above, we are finding the particular solution of the model that minimizes the sum of the squares of these differences.

 

[Buzz Bloom]

Hi @PeterDonis:

How about "particular solution"?

Thank you. I will use that instead of "model".

You shouldn't be if your purpose is to answer the title question of this thread, because the title question of this thread is not asking for the probability that the universe is finite given that assumption.

I accept that the title I used can be improved. To avoid starting this topic in a new thread, I suggest that I edit the summary to include an assumption. The assumption is that

Ω_k = 0.0007 +/- 0.0019 is a reasonably assumed Gaussian approximation of values for a mean and standard deviation of a probability distribution of the variable Ω_k.​

Is that OK?

The values being compared with are not for "other variables".

I guess I again used sloppy language. By "other variables" I meant the variables used in the database to be calculated using the Friedmann equation for each of many possible "particular solutions" used to calculate values for the database variables. Can you suggest an alternated phrase to use for "other variables"?

Regards,
Buzz

 

[PeterDonis]

I suggest that I edit the summary to include an assumption. The assumption is

...not what you wrote. The critical assumption, as I have said explicitly multiple times now, is that the "standard" model in which all five of the parameters you listed, including $\Omega_k$, are free parameters, to be estimated from the data, is correct. What you are calculating is the probability that the universe is finite given that assumption. But of course that avoids the actual hard question, which is, is that "standard" model the correct model? We don't know the answer to that, and we have no feasible way of even estimating a probability for it. That is what I have been saying all along.

Can you suggest an alternated phrase to use for "other variables"?

"Observed data" for the values to be compared with, and "predicted values for observed data" for the values to be calculated.

 

[Buzz Bloom]

But of course that avoids the actual hard question, which is, is that "standard" model the correct model?

Hi @PeterDonis:

Is it correct to describe "that standard model" as being the Friedmann equation which expresses the calculation of H(a) as a function of H₀ and the four Ωs? If so, what would be the criteria for determining whether the model is correct?

It seems to me that it is a similar criteria for the velocity V of a distant object equaling the product of H₀ and the distance D:

V = H₀ D .​

Isn't this equation without a specified value for H₀ a model? If so, what is the difference between the two models for the criteria of correctness based on the goodness of the fit to a database of observed data?

I get that there is an issue about whether or not the universe is flat, and whether it is possible to know if it is flat or is not flat. The "standard model" does not address that question. The probability of finiteness is based on the assumption that it is not flat. Is the failure to deal with the flatness question a failure of correctness?

By the way, physics requires that the values of Ω_r and Ω_m must be greater than zero, but Ω_k and Ω_Λ in principle could be negative, zero, or positive. The assumption that Ω_k is not zero is also similar to an assumption that Ω_Λ is not zero. The fit of the values of the four Ωs has a value for Ω_Λ much bigger than zero, so it not necessary to deal with the fact that it cannot be zero. Because Ω_k is close to zero, it prevents the sureness that it is not and cannot be zero.

I have made an edit to the summary. I hope you will find it to be acceptable. If not I would appreciate a suggetion from you regarding how to improve the summary.

Regards,
Buzz

 

[PeterDonis]

Is it correct to describe "that standard model" as being the Friedmann equation which expresses the calculation of H(a) as a function of H₀ and the four Ωs?

That plus the assumption that all five of those are free parameters, to be estimated from the data. By contrast, in the alternate model I proposed, $H$ and three of the $\Omega$s are free parameters, but $\Omega_k$ is fixed at $0$.

If so, what would be the criteria for determining whether the model is correct?

There aren't any criteria that would let you conclude with certainty that that model, or any model, is correct. Such things are always matters of subjective human judgment, which might end up being wrong.

It seems to me that it is a similar criteria for the velocity V of a distant object equaling the product of H₀ and the distance D:

V = H₀ D .​

No, because neither $V$ nor $D$ are free parameters in the model you are using. You don't estimate them from the data. You estimate $H_0$ from the data, and once you have decided on a particular value of $H_0$, the equation $V = H_0 D$ is true by definition for every comoving object, for appropriate definitions of $V$ and $D$ (neither of which are observables so there is no correspondence between this equation and any data).

I get that there is an issue about whether or not the universe is flat, and whether it is possible to know if it is flat or is not flat. The "standard model" does not address that question.

Yes, it only addresses the question of what the best-fit values of its free parameters are under the assumption that it is the correct model.

The probability of finiteness is based on the assumption that it is not flat.

No, it is based on the assumption that $\Omega_k$ is a free parameter, to be estimated from the data, so it is possible in principle for it to have any value consistent with the data.

Is the failure to deal with the flatness question a failure of correctness?

See my statement about criteria for correctness above.

By the way, physics requires that the values of Ω_r and Ω_m must be freater than zero, but Ω_k and Ω_Λ in principle could be negative zero or positive.

In principle, yes.

The assumption that Ω_k is not zero is also an assumption that Ω_Λ is not zero.

No, it isn't, at least not in principle. In principle it is possible for $\Omega_\Lambda$ to be zero if $\Omega_k$ is nonzero. The only constraint is that the sum of all the $\Omega$s must be $1$. That can be satisfied for either negative or positive $\Omega_k$ if $\Omega_\Lambda$ is zero. Indeed, until the discovery of the accelerating expansion of the universe in the 1990s, $\Omega_\Lambda$ was assumed to be zero by cosmologists, and that certainly did not stop them from considering the possibility of $\Omega_k$ not being zero. Of course, as you note, we now have a best-fit value of $\Omega_\Lambda$ that is positive with high confidence, so cosmologists now use that value routinely.

 

[PeterDonis]

I have made an edit to the summary.

What summary are you referring to?

 

[Buzz Bloom]

Hi @PeterDonis:

The summary that immediately follows the title of this thread.

Regards,
Buzz

 

[PeroK]

@Buzz Bloom I wanted to give you an example of why I believe what you're doing does not involve probabilities, as I would understand them. My example is this:

At one time there was an idea that perhaps a neutron is a proton plus an electron bound in some way. Let's say the estimated mass of the neutron is $939-940 MeV$; the proton $938-939MeV$; and the electron about $0.5MeV$. So, it's possible.

At this point, you ask what is the probability that the neutron is, indeed, a proton plus an electron? Your test is how likely it is that the mass of the neutron is precisely the sum of the masses of the proton and electron.

You do some calculations using normal distributions and come up with some number: $0.2$ or $0.5$ or whatever. And claim that is a probability. But, what you've calculated cannot be a probability. The approximate alignment of the masses gives a reason to investigate that as a potential theory, but it has nothing to do with well-defined probabilities.

All you have, at best, is a probability conditional on your theory being correct. And, in this case, if the theory is correct, then it's certain that the neutron is a proton plus electron, so again probabilities don't come into it.

One reason is that you haven't taken into account any other potential theory - and deciding probabilistically between theories is a different matter altogether. This is what the Bayesian analysis referenced early in this thread attempts to do.

 

[PeterDonis]

The summary that immediately follows the title of this thread.

Then no, your edit is not correct. I explained why in post #57.

 

[Buzz Bloom]

Then no, your edit is not correct. I explained why in post #57.

Hi @PeterDonis:

I read Post #57, and I thought I had captured the relevant assumptions in the three I put into the summary. Please tell me specifically which of the three I put into the summary are wrong, and which of your comments in Post # 57 are necessary for me to describe in the summary since they are the relevant assumptions to the meaningfulness of the probability calculation for finiteness.

Regards,
Buzz

 

[PeterDonis]

Please tell me specifically which of the three I put into the summary are wrong

I already told you in post #57. Specifically, the fifth quote from you in that post and my response to it. My response explains why assumption #3 in your revised summary is the wrong assumption.

 

[Buzz Bloom]

No, it is based on the assumption that Ωk is a free parameter, to be estimated from the data, so it is possible in principle for it to have any value consistent with the data.

The mathematical fact I was trying to address by assumption 3 is that any particular value has a zero probability. It is possible to calculate a non-zero probability for a range of values, but not for a single value. A flat universe requires that Ω_k is exactly zero. A finite hyper-spherical universe can have any value of Ω_k greater than zero. An infinite hyper-hyperbolic universe can have any value of Ω_k less than zero. This is why I read your comment #5 as not an assumption, but a conclusion that the probability of flatness is zero given the three assumptions I made.

If I need to phrase my assumption 3 better, please advise me.

Regards,
Buzz

 

[PeterDonis]

The mathematical fact I was trying to address by assumption 3 is that any particular value has a zero probability.

Yes, but that holds for every value in the distribution, not just $\Omega_k = 0$. So this "fact", if you take it literally, means that every possible value for $\Omega_k$ has zero probability, hence $\Omega_k$ cannot have any value at all. So this "fact" cannot possibly mean what you are asserting it to mean.

This is why I read your comment #5 as not an assumption

Then you read it incorrectly.

If I need to phrase my assumption 3 better, please advise me.

I already told you how to phrase it, in the specific part of post #57 that I already drew your attention to.

 

[PeterDonis]

This is why I read your comment #5 as not an assumption

Even though I explicitly used the words "based on the assumption that..." followed by the assumption. I don't see how I could possibly be any clearer that I was stating an assumption, and since that statement was immediately preceded by "No" in response to your statement of an assumption, I don't see how I could possibly be any clearer that I was correcting your statement of the assumption.

I think you are making this a lot harder than it needs to be. I think you need to stop trying to over-interpret what others are saying and pay more attention to the actual words being used.

 

[PeterDonis]

A flat universe requires that Ω_k is exactly zero. A finite hyper-spherical universe can have any value of Ω_k greater than zero. An infinite hyper-hyperbolic universe can have any value of Ω_k less than zero.

Even leaving aside the issue I raised in post #65, what you're describing here isn't an assumption, it's a logical argument. But what you wanted to put in the summary for this thread was assumptions, not logical arguments. The assumption that this logical argument is based on is still the one I stated in post #57, not the one you put in assumption #3 of the summary when you edited it.

 

[PeterDonis]

a conclusion that the probability of flatness is zero given the three assumptions I made

The third of your assumptions says that the probability of flatness is zero. So you can't conclude from it that the probability of flatness is zero. You already assumed it directly.

 

[Buzz Bloom]

Hi @PeterDonis:

I am quoting my assumption #3 together with your reading of it.

The third of your assumptions says that the probability of flatness is zero.

My #3
(3) that the universe is not and cannot be flat.

My assumption (3) needs to be rewritten because my intention is to make clear the assumption that the probability that the universe is flat is zero. My carelessness led me to what I wrote, and it does not say that, and it is in fact an error.

If I wrote the assumption (3) as: the probability that the universe is flat is zero, would that be OK?

Regards,
Buzz

 

[PeterDonis]

If I wrote the assumption (3) as: the probability that the universe is flat is zero, would that be OK?

No. The assumption is what I said in post #57 in response to this. I have already said so. More than once.

The statement in bold above is a conclusion, not an assumption--and, as I have already said in post #65, the logical argument you are using to get from the actual assumption being made (the one I stated in post #57 in response to the fifth quote from you in that post--I'm not going to quote it here since you have had more than enough time to go back and read what it actually says, which it appears you haven't done since you are unable to correctly state what it actually says) is questionable.

But regardless of the status of that logical argument, it is a logical argument, and the assumption you should be stating is the point from which that argument starts, not the point at which it ends. The end of a logical argument is not an assumption, it is a conclusion.

 

[PeterDonis]

And with that, this thread is closed. @Buzz Bloom you can PM me if you have further questions about what I have said. We have beat this topic more than enough for a public thread.