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 $a(t_0)=1$, because your unit rescaling is exactly like the assertion of the normalization $(1100+1)/(1100+1)=1$. It's no longer normalization of the scale factor itself, but the normalization of the product of the scale factor $a(t_0)$ and the scale of the distance unit, $1/a(t_0)$. In my opinion it's practically like normalizing the scale factor itself.

 

[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

This definition definitely requires $a(t_0)=1$.

 

[PeterDonis]

the result is exactly the same

What is "exactly the same"? Be specific.

 

[PeterDonis]

This definition definitely requires $a(t_0)=1$.

Yes, but it's easy to generalize it to not require that: just define the ratio of scale factors at the two times to be equal to $1 / \left( z + 1 \right)$.

 

[hedgehug]

Yes, but it's easy to generalize it to not require that: just define the ratio of scale factors at the two times to be equal to $1 / \left( z + 1 \right)$.

That's also my case with $a_0=a(t_0)=1100+1$ and $a(t)/a_0=1/(z+1)\le 1$.

The ratio $a(t)/a_0$ is normalized, because $1/(z+1)\le 1$. Again in my opinion it's practically like normalizing the scale factor itself. This time even literally and explicitly.

 

[hedgehug]

What is "exactly the same"? Be specific.

Equal.

 

[Ibix]

Is it worth stating carefully what's going on here? You're starting with the flat space FLRW metric, $ds^2=c^2dt^2-a^2(dr^2+d\Omega^2)$. Then you are setting $d\Omega=0$ for a radial path and $ds^2=0$ for a null one and solving to get $dr=\dfrac cadt$. Then you compute that for a light pulse starting at the origin, $r(t_0)=\int_0^{t_0}\dfrac c{a(t)}dt$. Then you are asking why you get a different value if you define $a=1$ at some time other than $t=t_0$, right?

To understand that, notice that $r$ is not a distance, it's a coordinate difference. The invariant physical quantity is $ar$. You can see this from the metric - if you set $t=\mathrm{const}$, so $dt=0$, to get a line in space at one cosmological time, the invariant quantity (the physical thing everyone can agree on) is $ds^2=a^2dr^2$. So when you replace $a$ with $a'=a/1101$ you do get $r'=1101r$, but the interesting quantity (i.e., how much string would you need to reach galaxy B from galaxy A at this time) is actually $a'r'$, and $a'r'=ar$.

What you've done is closely analogous to changing from using 1km×1km squares on a map to using 1m×1m squares. Sure, the number of squares between any two features on a map goes up by a factor of 1000, but the actual distance doesn't change. This is slightly obscured in the spacetime case because "moving one grid square" corresponds to different distances at different times, but the principle is the same.

 

[hedgehug]

Is it worth stating carefully what's going on here? You're starting with the flat space FLRW metric, $ds^2=c^2dt^2-a^2(dr^2+d\Omega^2)$. Then you are setting $d\Omega=0$ for a radial path and $ds^2=0$ for a null one and solving to get $dr=\dfrac cadt$. Then you compute that for a light pulse starting at the origin, $r(t_0)=\int_0^{t_0}\dfrac c{a(t)}dt$. Then you are asking why you get a different value if you define $a=1$ at some time other than $t=t_0$, right?

Right.

To understand that, notice that $r$ is not a distance, it's a coordinate difference. The invariant physical quantity is $ar$. You can see this from the metric - if you set $t=\mathrm{const}$, so $dt=0$, to get a line in space at one cosmological time, the invariant quantity (the physical thing everyone can agree on) is $ds^2=a^2dr^2$. So when you replace $a$ with $a'=a/1101$ you do get $r'=1101r$, but the interesting quantity (i.e., how much string would you need to reach galaxy B from galaxy A at this time) is actually $a'r'$, and $a'r'=ar$.

Proper distance is $d(t)=a(t)\chi$ where the comoving distance is $\chi=\int_{0}^{t_0}cdt/a(t)$, so the proper distance equal to the observable universe radius should be $d(t_0)=a(t_0)\int_{0}^{t_0}cdt/a(t)$, and in my case $a(t_0)=1100+1$. Is $d(t_0)=47\text{ GLy}$? If not, what should be changed to get this value?

 

[Ibix]

the proper distance equal to the observable universe radius should be $d(t_0)=a(t_0)\int_{0}^{t_0}cdt/a(t)$

So if you replace the conventional $a$ with your version that is $1101a$, the 1101 outside the integral cancels the one inside and you get the same proper distance as you do with the usual definition - as you must.

 

[hedgehug]

So if you replace the conventional $a$ with your version that is $1101a$, the 1101 outside the integral cancels the one inside and you get the same proper distance as you do with the usual definition - as you must.

Correct. Thank you. Again, for the third time, this cancellation looks exactly like normalization to me. Again it's dividing and multiplying 1 by the same factor, in my case 1101.

 

[Ibix]

Again, for the third time, this calcellation looks exactly like normalization to me.

I'm not sure what you mean by "normalisation" here. I'd just say it's part of your freedom to choose coordinates, just as you are free to pick meter grid squares or kilometre grid squares. Would you regard the maths required to handle the conversion from counted grid squares on a map to miles travelled as normalisation? I don't think I've heard the word used in that context, myself.

 

[hedgehug]

Would you regard the maths required to handle the conversion from counted grid squares on a map to miles travelled as normalisation? I don't think I've heard the word used in that context, myself.

No. I regard the ratio $a(t)/a(t_0)\le 1$ as normalized. That's what must be normalized and that answers my question. Thank you @PeterDonis and @Ibix.

 

[Ibix]

No. I regard the ratio $a(t)/a(t_0)$ as normalized.

I guess you could see that as normalisation, albeit normalisation to the value of $a$ at the arbitrary time we happen to be alive.

The usual way to think of it is that experiment can only determine $a$ up to a constant multiple. Cosmologists conventionally choose it so that $a=1$ today. I would see that as part of the process of specifying coordinates rather than a normalisation step, but I guess you can look at it as normalising $a$ to 1 at the present epoch if you wanted

 

[hedgehug]

I guess you could see that as normalisation, albeit normalisation to the value of $a$ at the arbitrary time we happen to be alive.

There has never been and there will never be other time for anyone, assuming atheism :) And whenever you live, you simply must have $a(t)/a(t_0)=1/(z+1)\le 1$ for your $t_0>t$.

If I define $A(t)=a(t)/a(t_0)$ and use it instead of $a(t)$ and $a(t_0)$, then it will be just like $a(t_0)=1$ and $a(t)\le 1$.

 

[Ibix]

There has never been and there will never be other time for anyone, assuming atheism :) And whenever you live, you simply must have $a(t)/a(t_0)=1/(z+1)\le 1$ for your $t_0>t$.

If I define $A(t)=a(t)/a(t_0)$ and use it instead of $a(t)$ and $a(t_0)$, then it will be just like $a(t_0)=1$ and $a(t)\le 1$.

As long as you clearly define the quantity (note that this thread would probably have been a lot shorter if you'd done so in the first post) and remember that coordinate-dependant quantities may change if you do so, I think that's fine.

Note that the usual approach of defining $a(t_0)=1$ means that $a(t)=1/(1+z)$ for all practical observations. You only really need to go to $a(t_{\mathrm{rec}})/a(t_\mathrm{emit})=1/(1+z)$ (the general form, which is basically what you are planning to define) if you expect to be interested in measurements made a long time away from now.