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.