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 $t_0$.