Does the scale factor need to be normalized?

[hedgehug]

Redefining the scale factor from its current value of 1 to 1 at the time of the CMB emission, and from 1/(1100+1) at the time of the CMB emission to 1100+1 at the present time, shouldn't change the result of the integral for calculating the radius of the observable universe, right?

What exactly would the current integral $\int_{0}^{t_0}cdt/a(t)$ look like after this change? Integration limits must remain the same: 0 and $t_0=13.8$ billion years, and the integral must obviously remain consistent with the FLRW metric for null geodesic.

 

[PeterDonis]

Redefining the scale factor from its current value of 1 to 1 at the time of the CMB emission, and from 1/(1100+1) at the time of the CMB emission to 1100+1 at the present time, shouldn't change the result of the integral for calculating the radius of the observable universe, right?

It will certainly change the numerical value of the result because redefining the scale factor means redefining the units of distance.

It won't change the physical meaning of the result.

 

[hedgehug]

So the integral would be exactly the same, right?

In what units would be $c$? In what units would be the integration limits?

 

[PeterDonis]

So the integral would be exactly the same, right?

The form of the integral would be the same. But the numerical value of the function $a(t)$ that appears in the integrand would be very different over the range of integration, because you rescaled it.

In what units would be $c$?

In whatever units are deterrmined by your units of distance and your units of time.

In what units would be the integration limits?

The integration limits are values of $t$, i.e., time, and you didn't say anything about changing the units of time. So presumably they would be whatever units of time you originally had in mind.

 

[hedgehug]

My unit of time is either a second, or a year. If my original distance unit is light year, is my new distance unit equal to (1100+1)Ly?

 

[PeterDonis]

My unit of time is either a second, or a year.

Ok.

If my original distance unit is light year, is my new distance unit equal to (1100+1)Ly?

No, it's equal to $1 / (1100 + 1)$ Ly, since that's the distance that a scale factor of $1$ corresponds to with your redefinition. The numerical value of the integral goes up by a factor of $1100 + 1$ because it takes that many more of the new (much smaller) distance units, with your redefined scale factor, to span the same physical distance.

 

[hedgehug]

The original scale factor of 1 corresponds to 1100+1 in my definition after the change, and you said that I have to take its inverse for my new distance unit.

Doesn't it mean that the scale factor is normalized anyway, since the distance unit must be inversely proportional to the observed redshift+1, so that it cancels out with non-normalized scale factor value?

 

[PeterDonis]

The original scale factor of 1 corresponds to 1100+1 in my definition after the change, and you said that I have to take its inverse for my new distance unit.

Yes.

Doesn't it mean that the scale factor is normalized anyway, since the distance unit must be inversely proportional to the observed redshift+1, so that it cancels out with non-normalized scale factor value?

I'm not sure what you think the word "normalized" means.

 

[hedgehug]

Normalized in a sense that the product of its value as it is defined today, and the distance unit, will always be equal to 1 light year.

 

[PeterDonis]

Normalized in a sense that the product of its value as it is defined today, and the distance unit, will always be equal to 1 light year.

That's only true if you define the scale factor "now" as $1$, and you're using light-years as your distance unit. But both of those choices are conventions, made for convenience, not required by the physics. No particular choice for either of these things is required by the physics.

In that sense, I think the answer to the title question of this thread is "no".

 

[hedgehug]

It's also true for the scale factor defined to be 1100+1 "now", because the distance unit is proportional to the inverse of 1100+1, so their product is equal to the original distance unit, whatever it is.

 

[PeterDonis]

It's also true

What is also true? Think carefully.

the distance unit is proportional to the inverse of 1100+1

The distance unit in terms of the scale factor is. In other words, the distance that a scale factor value of $1$ represents is.

But nothing says that has to be the distance unit you use anywhere else. Note that the distance unit actually doesn't appear at all in your integral except in the definition of $c$, and that can be anything you want. You just have to accept that the numerical value of the integral when you evaluate it will have a different relationship to the actual physical distance (which is what "the distance unit" really means). No single such relationship is dictated by the physics; it's a matter of convenience.

I probably should have gone into this in post #8, but I wasn't sure at that point quite what your question was.

 

[hedgehug]

Since the unit of time remains unchanged, I can only change the unit of $c$. The new distance unit called New Light Year (NLy) is NLy=Ly/(1100+1), so the new unit of $c$ is (1100+1)NLy/y. Otherwise it won't give Ly/y. As a result I multiply and divide Ly by the same factor and get the same original unit.

And that's for the scale factor defined to be 1100+1 "now".

 

[PeterDonis]

the new unit of $c$ is (1100+1)NLy/y.

That isn't the new "unit" of $c$, that's the new value of $c$ in your new chosen distance units (and the same time units as before).

And with that choice of units, the numerical value of the integral for the size "now" of the observable universe will be $1100 + 1$ times as large--it will be telling you that the size "now" of the observable universe is about $46 \times \left( 1100 + 1 \right)$ billion NLy (whereas with the original units of just Ly, the integral was telling you that the size "now" of the observable universe was about $46$ billion Ly).

 

[hedgehug]

Yup, my bad - the value of $c$, not its unit.

At the end of the day after the integration we substitute Ly/(1100+1) for NLy, and the result is exactly the same as for the normalized $a(t_0)=1$, so your unit redefinition is exactly like the assertion of its normalization. $(1100+1)/(1100+1)=1$.

 

[renormalize]

If my original distance unit is light year, is my new distance unit equal to (1100+1)Ly?

I'm just curious: is it somehow significant that you write $1100+1$ instead of $1101\,$?

 

[hedgehug]

I'm just curious: is it somehow significant that you write $1100+1$ instead of $1101\,$?

Yup

According to the FLRW metric which is used to model the expanding universe, if at present time we receive light from a distant object with a redshift of z, then the scale factor at the time the object originally emitted that light is $a(t)=1/(z+1)$.

https://en.wikipedia.org/wiki/Scale_factor_(cosmology)#Detail