Does the scale factor need to be normalized?

[hedgehug]

@Ibix Note that I only really need $a(t)/a(t_0)=1/(z+1)$, because $t_0=t_{rec}$ and $t=t_{emit}$.

 

[PeterDonis]

The ratio $a(t)/a_0$ is normalized, because $1/(z+1)\le 1$.

So by "normalized" you mean "less than or equal to 1"? Where are you getting that definition from?

in my opinion it's practically like normalizing the scale factor itself. This time even literally and explicitly.

I still don't see how this is "normalization". See above.

 

[PeterDonis]

Equal.

What is "equal" about two results that differ numerically by a factor $\left( 1100 + 1 \right)$?

 

[PeterDonis]

the ratio $a(t)/a(t_0)\le 1$

This is only true if $t \le t_0$. What if you're trying to compute something for a time $t$ that's later than now ($t_0$)?

 

[PeterDonis]

There has never been and there will never be other time for anyone

If this were true, no $t$ other than $t_0$ would have any meaning and this whole discussion would be pointless.

I think you have not thought through your position very carefully.

 

[hedgehug]

What is "equal" about two results that differ numerically by a factor $\left( 1100 + 1 \right)$?

Physical, proper distances are equal after substituting Ly/1101 for NLy.

 

[hedgehug]

If this were true, no $t$ other than $t_0$ would have any meaning and this whole discussion would be pointless.

I think you have not thought through your position very carefully.

I think you just need to have a last word. When I said that there has never been and there never will be other time for anyone, I meant the time when you are alive.

 

[PeterDonis]

When I said that there has never been and there never will be other time for anyone, I meant the time when you are alive.

Yes, but we still expect our physical models to cover other times besides those when we are alive. And that includes times to our future as well as to our past. For example, suppose we wanted to compute the radius of the observable universe five billion years from now, when the Sun is a red giant.

 

[hedgehug]

This is only true if $t \le t_0$. What if you're trying to compute something for a time $t$ that's later than now ($t_0$)?

$a(t_{emit})/a(t_{rec})=1/(z+1)\le 1$ for $t_{emit}\le t_0$ and $t_{rec}>t_0$ in the expanding universe.

 

[hedgehug]

So by "normalized" you mean "less than or equal to 1"? Where are you getting that definition from?

Bad habit from programming. Me and everyone I worked with have been calling a variable divided by its maximum value normalized.

 

[PeterDonis]

$a(t_{emit})/a(t_{rec})=1/(z+1)\le 1$

So now you're changing the definition of what variable you're using? Or did you really mean $t_{rec}$ before when you wrote $t_0$?

 

[hedgehug]

I'm changing nothing. When we calcuate the observed redshift of radiation emitted in the past, we can use $a(t_{emit})/a(t_{rec})=a(t)/a(t_0)=1/(z+1)\le 1$ formula. It's the same formula. The full, explicit form is needed for the redshift of light that will be received and observed in the future, $t_{rec}>t_0$. You asked about it, didn't you?

 

[PeterDonis]

I'm changing nothing.

You've been using $t_0$ all through this thread up until post #31, when you used $t_{rec}$, and that was in response to @Ibix using it, and then post #39. That's a change. (And for that matter, you didn't use $t_{emit}$ until post #31 either; your OP in this thread gave an integral that starts at the Big Bang, $t = 0$, which isn't the "time of emission" of anything.)

I would suggest that you read what @Ibix said at the start of post #30.

 

[hedgehug]

I started to use $t_{rec}$ because of Ibix, and I also used it, because I couldn't correctly answer your question about the future without it. That's the most important reason.

 

[PeterDonis]

Me and everyone I worked with have been calling a variable divided by its maximum value normalized.

Even if I accept this, the variable here is not the scale factor but the ratio of scale factors, and it only has a "maximum value" of $1$ because we put the earlier time in the numerator and because we insist on moving the time in the denominator to be the latest time we're considering. If we flip the ratio around (which makes more sense since then we're basically just looking at the redshift), it has no maximum value.

 

[hedgehug]

Even if I accept this, the variable here is not the scale factor but the ratio of scale factors, and it only has a "maximum value" of
because we put the earlier time in the numerator and because we insist on moving the time in the denominator to be the latest time we're considering. If we flip the ratio around (which makes more sense since then we're basically just looking at the redshift), it has no maximum value.

Correct, for the expanding universe.

What really asserts the scale factor normalization, is the proper distance calculation, $d(t_0)=a(t_0)\int_{0}^{t_0}cdt/a(t)$. Just like Ibis said, however I define $a(t_0)$, it cancels out after the integration.

 

[PeterDonis]

What really asserts the scale factor normalization, is the proper distance calculation

But the scale factor does not have to be normalized to the proper distance; that's the point. It doesn't matter how the scale factor is defined.

it cancels out after the integration

That doesn't mean it's normalized; it means it's irrelevant to the proper distance calculation because it cancels out.

 

[hedgehug]

But the scale factor does not have to be normalized to the proper distance; that's the point. It doesn't matter how the scale factor is defined.

That doesn't mean it's normalized; it means it's irrelevant to the proper distance calculation because it cancels out.

That's the point. The formula for the proper distance always gives the same result as for the normalized scale factor, because it cancels out its not normalized value $a(t_0)$.

In context of the proper distance, the scale factor is just like it was normalized.

Scale factor function is totally relevant to the proper distance calculation, which totally depends on it.

 

[PeterDonis]

the normalized scale factor,

I don't agree with you calling the definition $a(t_0) = 1$ "normalized", which is what you're doing here. It's a convenient convention, that's all. It's not normalized to anything.

I think we agree on the physics, we just disagree on use of language. Such disagreements are in the end a matter of opinion, but I don't think the use of the term "normalized" the way you are using it is at all common in the relevant literature.

 

[hedgehug]

Ok.

I just hope you can agree that $a(t_0)\ne 1$ and $a(t_0)/a(t_0)=1$ is very much like the normalization in the proper distance calculation with the upper limit of integration equal to $t_0$.

Still, scale factor function is totally relevant to the proper distance calculation, which totally depends on it.

 

[PeterDonis]

I just hope you can agree that $a(t_0)\ne 1$ and $a(t_0)/a(t_0)=1$ is very much like the normalization in the proper distance calculation with the upper limit of integration equal to $t_0$.

As I said, we agree on the physics. We just disagree on the usage of the term "normalization". I don't agree with your usage of it here, just as I don't agree with your other usages of it in this thread. If you are going to insist on such usage, I think you need to provide some references to the literature (textbooks or peer-reviewed papers) that support it.

scale factor function is totally relevant to the proper distance calculation, which totally depends on it.

I agree that you need to know the scale factor as a function of time, in whatever coordinates and with whatever conventions you've adopted, in order to evaluate the proper distance integral, yes.

 

[hedgehug]

Like I said, it's rather bad habit from programming, and my usage of this term was careless in this thread.

Thank you.

 

[hedgehug]

... Now I know exactly why the proper distance doesn't depend on $a(t_0)$ for a mathematical reason, but my common sense is asking the question Why shouldn't it depend on it?

The first answer that comes to my mind is that its initial value is 0, and scaling it to any other value is impossible, so we simply choose 1, because whatever we choose, it will make no difference.

Should it really make no difference for the proper size of the universe, how many times the universe has expanded since the BB, even if we can't really answer this question, because we can't divide by 0? What if we assume the Planck length?

 

[PeterDonis]

Should it really make no difference for the proper size of the universe, how many times the universe has expanded since the BB

The value of "how many times the universe has expanded since the BB", which is basically what your integral is computing, does not depend on how you define the scale factor, because, as you have already shown, the scale factor definition drops out of the integral. It depends on the ratio of scale factors, but that ratio is independent of how you define the scale factor (what time you pick for $a = 1$).

 

[hedgehug]

Because that's how the formula for the proper distance is constructed, and my common sense somehow can't accept it. You repeated the mathematical reason which I had already understood.

Also, the ratio $a(t_0)/a(0)$ is quite problematic to me. Why don't you have a problem with it? This ratio (not the proper distance calculation) should tell us how many times the universe has expanded since the BB, but at the same time it can't, because we can't divide by 0.

 

[PeterDonis]

the ratio $a(t_0)/a(0)$ is quite problematic to me.

You don't have to explicitly take such a ratio to evaluate the integral. You can take a limit as $t \to 0$, and find that you get a finite answer for the integral. Similar remarks would apply to any other calculation that on its face involves $t = 0$. Note that the "initial singularity" $t = 0$ is not actually part of the spacetime manifold.

This ratio (not the proper distance calculation) should tell us how many times the universe has expanded since the BB

Which of course is meaningless if you look at $t = 0$ itself. But as above, $t = 0$ is not actually part of the manifold. At any time $t > 0$, the ratio is finite and has the physical meaning you describe.

 

[hedgehug]

For the second time you didn't quote the second part of my sentence about the division by 0.

 

[PeterDonis]

For the second time you didn't quote the second part of my sentence about the division by 0.

Because it's irrelevant. You don't have to divide by zero to take the limit as I described. Or to calculate any other actual physical quantity from the FRW model.

 

[hedgehug]

Can you answer the question How many times the universe has expanded using your limit $t\rightarrow 0$?

 

[PeterDonis]

Can you answer the question How many times the universe has expanded using your limit $t\rightarrow 0$?

No, because, as I've already said, that particular quantity does not have a finite limit as $t \to 0$. But you don't need an answer to that question to evaluate any actual physical quantity. (Note that, as I've already said, $t = 0$ is not part of the spacetime manifold, so asking by what factor the universe has expanded since $t = 0$ has no physical meaning. Such a question only has physical meaning for $t \gt 0$.) For example, you don't need it to evaluate the integral that gives you the size of the observable universe.