I Calculating the probability that the Universe is finite

[PeterDonis]

the calculation of a Friedmann model

Is not the relevant calculation, because what you mean by "a Friedmann model" here is one particular Friedmann model with particular unknown values for all the parameters, and you're trying to estimate what the parameters are from the data. In other words, you're not comparing models, you're fixing the model and estimating its parameters, and you're assuming that all of the parameters you are estimating are free, to be estimated from the data.

But if I propose a different model, which is still described by a Friedmann equation but in which there is a constraint that forces one particular parameter, $\Omega_k$, to have the exact value zero always and everywhere, then that changes how you estimate the other parameters. And the calculations you are doing do not help you at all in comparing the likelihood of those two different models, the one I propose and the one you refer to in what I quoted from you above. If all you have is the parameter estimation data, the best you can do is say that neither model is ruled out by the data (since $\Omega_k = 0$ is within the error bars of our current measurements). You can't give any relative likelihood.

 

[PeterDonis]

However, there is a related question that does have a calculable answer:

"Is the universe finite or infinite?"​

I just cannot understand why something that is what is denies the possibility that among the possible things it might be, each possibility has a probability.

The question is how you estimate that probability. The point is that you can't estimate it just from the $\Omega_k$ calculation.

I have a standard shuffled deck of 52 cards.

Yes, which means you have fixed the model, and that model already has known probabilities for all possible cards.

But the position we are in in trying to answer the title question of this post is not like that. We do not know what the correct model for the universe is. We can pick a model that looks like it might be right, and estimate parameters from that, but that does not answer the title question of this thread, because it does not answer the question of how likely that particular model is to be right among all the possible models that are consistent with the data.

It's as if, as @PeroK said, there are many different possible decks of cards and we don't know which one the card whose properties we are trying to calculate the probability of was drawn from. Calculating the probability of drawing a Spade from a standard deck does not help answer that question, because it doesn't tell you the likelihood that the deck is a standard deck to begin with.

 

[PeterDonis]

Are you implying that the value of Ω_k and its +/- standard deviation value are meaningless numbers? If so, what makes them meaningless? Also are the correspond numbers for H₀, Ω_r, Ω_m, and Ω_Λ also meaningless?

Of course they're not meaningless; they are estimates of those parameters given a particular assumption about the underlying model. But that in itself, as has been pointed out, is not enough to answer the title question of this thread.

 

[Buzz Bloom]

In general, probabilities only make sense given some specific assumption(s).

Hi @PeroK:

I would veru much appreciate your listing a few examples of probabilities making sense based on given some specific assumptions.

Regards,
Buzz

 

[PeroK]

Hi @PeroK:

I would veru much appreciate your listing a few examples of probabilities making sense based on given some specific assumptions.

Regards,
Buzz

If we stick to cosmology:

1) The probability that Betegeuse will go supernova in the next 100 years, as observed from Earth.

2) The probability that Dark Matter particles will be discovered in the next 20 years.

3) The probability that extraterrestrial life is discovered in the next 50 years.

4) The probability that universe is proved to be finite (or that a finite model is adopted) in the next 500 years.

Notice that if we remove the timescale, it makes things a lot less clear. If you bet that extraterrestrial life will never be discovered, then you can only ever lose that bet and never successfully win.

Likewise, if I bet that the universe is infinite, it's a bet that I (probably!) can't ever win. What would constitute proof of an infinite universe? I could only ever lose that bet.

 

[Buzz Bloom]

If all you have is the parameter estimation data, the best you can do is say that neither model is ruled out by the data (since Ω_k is within the error bars of our current measurements). You can't give any relative likelihood.

Hi @PeterDonis:

I am having a bit of confusion. The two models you describe seem to me to be plausibly (but not necessarily) in competition. The competition is based on the differences between the two with respect to the total sum of the differences between the input data (z and distance) with each corresponding model's fit of

da/dt to H₀ D (D= distance).​

It seems plausible that the value of this fit measurement for Ω_k=0 will be larger than the one for the other model's non-zero value for Ω_k.
This difference will enable a calculation of

(1) the fit measurement result M₁ based on the assumption that Ω_k=0, and​

(2) the fit measurement result based M₂ on the assumption that Ω_k has a Gaussian distribution.​

I do not know if the following would be useful, but I would find it to be interesting. A series of fit measurement M values could be calculated for a range of Ω_k values. From this result data, the average M_A of the M fit measurement result in a range R between

Ω_k-Q and Ω_k+Q​

which equals the value M₁. Then, I suggest that the integral of the Gaussian distribution for Ω_k over the R range is an approximate probability value that Ω_k=0.

The above described calculation is just a guess that something like that might possibly produce a probability for Ω_k=0.

Regards,
Buzz

 

[PeterDonis]

The two models you describe seem to me to be plausibly (but not necessarily) in competition.

They're in competition in the sense that they give different answers to the title question of this thread, yes.

each corresponding model's fit of
da/dt to H0 D (D= distance).

Neither model makes a single prediction for this. As I said before, each model estimates all of its free parameters based on the data. The difference is that the alternate model I proposed has one fewer free parameter, since it fixes $\Omega_k = 0$. But that still leaves multiple free parameters for both models, and it has to estimate all of them based on the data. The method of estimation, as I understand it, is basically a least squares fit to the data, but the data is not da/dt vs. H0 D.

(2) ... the assumption that Ωk has a Gaussian distribution.

The model under discussion for this item does not predict that $\Omega_k$ has a Gaussian distribution. To the extent it makes a prediction about $\Omega_k$, it says that any value that is not ruled out by our current evidence is about equally probable. (A Bayesian would call this a uniform prior over the interval of values not ruled out by evidence.)

The above described calculation is just a guess that something like that might possibly produce a probability for Ωk=0.

If the two models under discussion were known to be the only two possible models, then we could estimate the probability for $\Omega_k = 0$ based on the relative likelihood of the alternate model I proposed, which fixes $\Omega_k = 0$, vs. the standard inflation-based model, given the data. But that's still not the same as either the "fit" estimate you've described, or the parameter estimation I described.

If there are more possible models besides those two, then we would have to be able to estimate their relative likelihoods as well given the data.

In any case, the key point is that you are talking about estimating the probability of the data given a model, while what we actually need is the probability of each model given the data.

 

[Hornbein]

The standard deviation you are talking about is an estimate of the variation in the measurement of the curvature of the universe. One takes a number of measurements and finds how much they vary. Everyone thinks the curvature is practically constant, so the variation is due to imperfections in the measuring procedure. We then say that this standard deviation is an estimate of the amount of "error" in the measurement.

In other words, everyone thinks that during these measurements the curvature is practically a constant. This is the distribution we are hypothesizing about. If the measurements were perfect then they would always measure the same number. The variation in our sample is entirely due to measurement error.

If zero were more than 3 std deviations from the mean of the measurement then the no curvature hypothesis would be doubted.

Statistics can't be used to prove anything. One makes a hypothesis, takes some measurements, and calculates how consistent they are with the hypothesis. In this case the measurements are consistent with the zero hypothesis, while high curvature hypotheses are rejected. That's all you can get from these measurements.

Both zero and almost zero are consistent with the measurement. The measured statistics don't indicate one over the other.

 

[Buzz Bloom]

Of course they're not meaningless; they are estimates of those parameters given a particular assumption about the underlying model.

Hi @PeterDonis:

I remain confused. Please explain what the particular assumption about the underlying model is. My understanding is that the underlying model is the value of the five variables (and the assumption that Friedmann equation is the framework for making models). But these five values are not assumed. They are calculated as a best fit to a database of data. That is, they are not inputs, they are outputs.

By the way, I apologize for the card example. I now see that it was inappropriate.

Regards,
Buzz

 

[Buzz Bloom]

If there are more possible models besides those two, then we would have to be able to estimate their relative likelihoods as well given the data.

Hi @PeterDonis:

Again I get the impression that my confusion about your explanations is that I see a model as the result of calculations to find the model that is the best fit to the raw data. You discuss two models, (M1) with the assumed value that Ω_k=0, and the other (M2) with Ω_k as one of the model variables with a value to calculate from the relevant database of data. This comparison allows for a comparison between the results of the database fit to the two model results with respect to the database variables. For this pair of models it is likely that the fit by M1 is sufficiently close to the fit by M2 that it is plausible (perhaps with a numerical degree of confidence) that Ω_k=0 is a correct value.

This method is also a reasonable way to confirm that Ω_r=0 is a convenient assumption for calculations of the model's values for H(t) when Ω_r<<Ω_m.

I still do not see why the mean and standard deviation values for Ω_k do not represent a probability distribution of Ω_k that can be used to calculate the probability that Ω_k>0 (and also its expected value when assuming Ω_k>0). Perhaps the fact that there are five variables in the models creates a too complicated example of the reasons against the probability function interpretation. I will post later what I hope will be an acceptable context based on a one variable model: H₀.

Regards,
Buzz

 

[Buzz Bloom]

The standard deviation you are talking about is an estimate of the variation in the measurement of the curvature of the universe.

Hi @Hornbein:

I apologize for not understanding what the quote above is intended to communicate. You seem to be saying that measurements are made to obtain values for Ω_k. I am unable to guess what kinds of measurements are made. My guess about calculating a value for Ω_k (as well as for the other variables) is that the Friedmann equation is used with assumed values for the five variables to calculate values for H(t) for a list of t (time) values which correspond to some astronomical values. That is, a comparison is made between some database values and calculated values and the particular values for the Friedmann variables with the least sum of squares of differences with the database values are the means of the variable values The sum of differences can be used to calculate standard deviations.

So, my guess is that there is no measurement of curvature. The measurements are the sums of differences squared between calculated variables and database variables related to H(t).

Regards,
Buzz

 

[PeterDonis]

Please explain what the particular assumption about the underlying model is.

Your underlying model is that the universe is described by an FRW spacetime and that all five of the parameters you listed are free parameters; none of their values are fixed by the model, they are all to be estimated from the data.

The alternate model that I proposed is that the universe is described by an FRW spacetime and the four of the five parameters you listed are free parameters, but one, $\Omega_k$, is not: it is fixed at the value $0$ by constraints imposed by the model (I have given no details about what those constraints are because I am just proposing this model for comparison). The other four parameters are to be estimated from the data.

 

[PeterDonis]

I apologize for the card example. I now see that it was inappropriate.

No, it wasn't; it was a good illustration of the difference between calculating probabilities given a fixed underlying model, and trying to compare different underlying models.

 

[PeterDonis]

I see a model as the result of calculations to find the model that is the best fit to the raw data.

That is not how I am using the term "model". See post #42.

You can't even do "calculations to find the model that is the best fit to the raw data", using "model" in your sense, until you have decided on a "model" in my sense--an underlying theoretical model that tells you what the relevant equations are and what parameters in those equations are to be estimated from the data. And you cannot use "calculations to find the model that is the best fit to the raw data" to compare "models" in my sense. But comparing "models" in my sense is what you have to do to answer the title question of this thread.

In other words, the title question in this thread cannot be answered by doing calculations to find the best fit of the five parameters you listed, of which $\Omega_k$ is one, to the data, under the assumption that all five of them are free parameters to be estimated from the data. Those calculations can only answer a different question, namely: if we assume that $\Omega_k$ is a free parameter to be estimated from the data, what is the probability that the spatial curvature is positive (i.e., that the universe is spatially closed)? That is the question your calculations in the OP answer. But it is not the same question as the title question of this thread; the if clause in front makes a huge difference.

 

[Buzz Bloom]

Hi @PeterDonis:

If I am understanding your post #42 (further explained in post #44), you are making a distinction between two categories of models, both assuming and based on the Friedmann equation. Category 1 is a model where all five parameters are to be determined by what you describe as "estimated from the data", and I describe as making the best fit to a database of data. Category 2 is a model in which one or more (but not all five) of the five parameters are assumed to have a fixed value, and the other parameters are "estimated from the data" (or calculated by finding the best fit to a database). I also think you intend (and I agree) that for a Category 2 model to have a result that is useful, it needs to be compared with the Category 1 model. If the Category 2 model has a much worse fit to the database than Category 1, it will not be evaluated as being a reasonably reliable model.

Assuming that you and I agree about the above paragraph, I would very much appreciate your explaining why you do not accept that the Category 1 model has values for each variable which represent the mean and standard deviation of probabilities for the variables' values.

Regards,
Buzz

 

[PeterDonis]

you are making a distinction between two categories of models

Not two categories, two specific models, each of which makes a specific designation of which parameters are free parameters.

I also think you intend (and I agree) that for a Category 2 model to have a result that is useful, it needs to be compared with the Category 1 model.

Not at all. I can use my alternate model to estimate the fit of its free parameters to the data without even knowing of the existence of the other model, the "standard" one. And of course you can use the "standard" model to estimate the fit of its free parameters to the data without knowing of the existence of my alternate model (since that is what you did in the OP of this thread, before I had even proposed the alternate model).

What you can't do is answer the title question of this thread by using estimated parameter fits from only one model. But there are lots of other uses for particular models besides trying to answer the title question of this thread.

If the Category 2 model has a much worse fit to the database than Category 1, it will not be evaluated as being a reasonably reliable model.

If the best fit of any model's parameters to the data has huge variances (i.e., the model is unable to reproduce the data very closely at all, no matter what values you plug in for its parameters), then of course it's not going to be considered a viable model if there is another model whose fits are much better (i.e., which can reproduce the data much more closely with appropriate values for parameters). But that doesn't mean the other model is necessarily the best possible one. There could be still another model that could fit the data even better.

There is one particular sense in which the "standard" model, the one with all five parameters you listed as free parameters, is "more general" than any other model that only uses those parameters: it allows for the possibility that all five parameters could have values that we don't know for sure. Whereas any other model using only those parameters (such as my alternate model) must claim that we know for sure the value of at least one parameter (in the case of my alternate model, that is $\Omega_k$). That might be what you are trying to get at here. But it's still a fairly weak claim; in particular, it's not sufficient to ground any claim that considering only the "standard" model is sufficient to answer the title question of this thread.

I would very much appreciate your explaining why you do not accept that the Category 1 model has values for each variable which represent the mean and standard deviation of probabilities for the variables' values.

If by this you mean the claim about the "standard" being "more general" that I described above, then it is correct, but limited (as I said above).

If you mean anything stronger, such as the claim that the distribution of $\Omega_k$ in this model says anything useful about the value of $\Omega_k$ in my alternate model, then the statement is incorrect.

Also, we have so far not even discussed the possibility of another alternate model that had more free parameters than the standard one does (for example, a model in which the dark energy density is not treated as constant but is allowed to vary with time). In such a model, the estimate for $\Omega_k$ might be different from the one you calculate (because the presence of additional free parameters can change the overall best fit). You can't make any claims about that sort of model either from your calculations in the OP.

You can of course try to argue that models with additional free parameters are ruled out by Occam's razor since the fit to the data of the "standard" model is "good enough". But such arguments have nothing to do with the calculations you made in the OP.

 

[Buzz Bloom]

Not two categories, two specific models, each of which makes a specific designation of which parameters are free parameters.

Hi @PeterDonis:

Every time I think I am making progress, I am instead becoming more confused. The key issue I am trying to understand is the following.

Why is it not reasonable that calculating the values of the Friedman equation variables (five of them) by finding the Friedman variable values which have the best fit to database values (which are related to the Friedman equation, for example H(t) values) produce a mean and standard deviation for a probability distribution for each of the five Friedmann variables?​

A secondary issue is why isn't the two categories I defined meaningful? I get that Category 2 could be recognized as having several sub-categories, such as each sub-category having a specific subset (non-empty) of the five variables, and each such subset having preset fixed values.

My understanding of the term "Friedman model" is that such a model has a specific value for each of the five variables. You seem to be using this term with a different sense.

Regards,
Buzz

 

[PeterDonis]

Why is it not reasonable that calculating the values of the Friedman equation variables (five of them) by finding the Friedman variable values which have the best fit to database values (which are related to the Friedman equation, for example H(t) values) produce a mean and standard deviation for a probability distribution for each of the five Friedmann variables?

I have said no such thing. You aren't reading what I'm actually saying.

The calculations you refer to do produce means and standard deviations for the five variables, under the assumption that you are using a model in which all five of those variables are free parameters. Of course it is reasonable to make those calculations under that assumption. I have simply been pointing out, several times now, that the assumption I put in italics just now is an essential part of the calculations. The calculations only apply to the particular model that satisfies the assumption. They don't apply to other models that don't satisfy the assumption.

A secondary issue is why isn't the two categories I defined meaningful?

Again you aren't reading what I'm saying. I said they're not categories, they're individual models (with the proper usage of the term "model"--see below). Of course the difference between them is meaningful, and I never said otherwise.

My understanding of the term "Friedman model" is that such a model has a specific value for each of the five variables. You seem to be using this term with a different sense.

Of course I am, and I already explained the different sense in which I am using it, in post #44. Please read that post again, carefully.

Basically, you are using "category" to mean what I mean by "model", and you are using "model" to mean what I mean by "model with a particular set of values for its free parameters". That is a matter of words, not physics. My usage is, I believe, the standard usage in physics. Your usage is not. But either way, the concepts are clear, and it's the concepts that we are discussing, not the choice of words.

 

[Buzz Bloom]

That is a matter of words, not physics. My usage is, I believe, the standard usage in physics. Your usage is not.

Hi @PeterDonis:

It seems to be that a lot of my confusions are vocabulary issues. I try to avoid having a single word used to have more than a single meaning. Over the years my experience has been that when such usage is used my understanding of what a word means is the wrong definition, although it is one of several usages of the word. I have tried to make a distinction between (1) a "model" with five specific values, and (2) a kind of "model" where the variables have not been given specific values. I apologize for my confusion. In my readings I have not become familiar with the nuances of these two different meanings for "model".

Post #48

The calculations you refer to do produce means and standard deviations for the five variables, under the assumption that you are using a model in which all five of those variables are free parameters.

Post #4

That is for the assumed distribution of errors in the measurements underlying the calculations. It is not in any way a claim that there is a meaningful "probability distribution" for the spatial curvature of the universe, or that the results given in the paper express the parameters of such a distribution.

My interpretation of Post #4 led me to believe that your view is that it is unreasonable for me to use the

0.0007+/- 0.0019 values for Ω_k​

to calculate the probability that Ω_k>0, and that this result is a reasonable probability value that the universe is finite. A good bit of this thread was about my trying to understand why this was your view. Now, I interpret Post #48 as a correction to my misunderstanding of Post #4, and that you did not mean that my calculation of the probability that the universe is finite is an unreasonable conclusion. Do I now have a correct interpretation of your view?

Regards,
Buzz

 

[PeterDonis]

I try to avoid having a single word used to have more than a single meaning.

That's fine, but in order to have a useful discussion about a scientific topic, you need to use words with the single meaning that scientists who work on that topic use. You can't just pick your own meaning without regard to how the word is used by scientists working in the field.

I have tried to make a distinction between (1) a "model" with five specific values, and (2) a kind of "model" where the variables have not been given specific values.

The distinction itself is perfectly valid. The issue is that you have been using the word "model" to mean #1, whereas scientists use the word "model" to mean #2, so that's what I have been using the word "model" to mean as well. If you want to be consistent with standard scientific terminology, you need to find a different word for #1, since "model" is used in standard scientific terminology to mean #2.

In my readings I have not become familiar with the nuances of these two different meanings for "model".

There aren't two different meanings for "model" in standard terminology, which is why you have not seen anything in your readings about any such thing. There is only one. It is just not the one you have been using "model" to mean. See above.

 

[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.