# What's the universe radius function

1. May 10, 2015

### Quarlep

We know that universe is expanding .I am curious about how universe radius change with time or simply R(t)=? (Observable universe radius)

2. May 10, 2015

### marcus

Hi Q,
we do not know that the universe has a radius. the radius of the OBSERVABLE universe does not grow according to the same curve as the ordinary expansion of distances.

this is because as time passes light from more and more distant matter comes in and so what they call the observable universe includes more and more matter and extends farther and farther out even if you discount the geometric expansion of distances.

the radius of the observable region (as cosmologists normally understand that term) is NOT a good thing to talk about if you want to track ordinary expansion.

If you want to track expansion, according to Hubble law and friedmann equation, then you should plot the SCALE FACTOR a(x) as a function of time x
that is the size of any large-scale distance, normalized to equal 1 at present.
a(x) = sinh2/3(1.5x)/1.311

This is exactly the answer to your question. It is how the "radius" of the universe would change with time if the universe was known to have radius. Since it is not known that a radius exists and we have no idea what it might be if it did, this function a(x) can be thought of as the expansion history of a typical or generic large-scale distance. Say large-scale so it can be distance between galaxies or clouds of matter with are far enough apart not to be gravity-bound and also which are not moving significantly in their surrounding space.

Last edited: May 10, 2015
3. May 10, 2015

### Quarlep

In that equation If I put 13.7/17.3 what we get ? An expansion radius ?

4. May 10, 2015

### marcus

The model parameters I normally use are 14.4 and 17.3 which lead to the age being 13.787 billion years.
Let's use that age, if OK with you.

So if you put 13.787/17.3 you get 0.797 for the age. (measured in what I think of as a natural cosmic time unit.)

So if you put 0.797 into that formula you should get a(.797) = 1
the formula is normalized to equal one at the present.

It makes a(x) extra useful to have it normalized to equal one at the present. It means we can interpret 1/a as a stretch factor.

If for some x we have a(x) = 0 .5 we can say "back then at that time distances were 1/2 present size" and light coming to us today from a galaxy back then will have wavelengths stretched by a factor of 2.

And if you find by measuring standard candle that a certain galaxy is now 3 billion LY from us, you can have the whole history of the distance to that galaxy simply by multiplying 3 billion LY by a(x). You get that convenience because a(x) is NORMALIZED to equal 1 at present.
At a time when a(x) = 0.25, the distance to that galaxy was 0.75 billion LY and so on.

Astronomers call the a(...) function the scale factor.
You just now were suggesting calling it "expansion radius". I think that is a bad idea. Why not call it scale factor, like everybody else. We don't know that the universe has a radius. And the radius of the observable does not behave like other distances so it would just give nonsense to multiply the current radius of observable by a(x). It would confuse other people to call it "expansion radius"

Scale factor is a very important function in cosmology. It is the size of a generic or typical largescale distance normalized to equal one at present. I advise calling it that.

What formula you use for the scale factor depends on what time scale you use. If you use billions of years as time unit, then you get a different formula. If you use 17.3 billion years as your time scale you get the formula I wrote, which I like because it is simple.

Last edited: May 10, 2015
5. May 10, 2015

### ChrisVer

The "expansion" is given by the 1st Friedmann equation, and is defined by the Hubble parameter.
It's not like a "real" radius that is expanding, since that wouldn't make much sense...Some people would try to sketch the scale factor $R(t)$ as the radius of the expanding balloon (in the expanding balloon analogy).But the balloon analogy is not a good choice to speak for an expanding "radius" since the universe doesn't have a curvature (it's rather flat).
It only shows you how things scale, and that's why it's called "Scale factor". Distances scale with the scale factor, but how they do depends on what distances you are talking about (in cosmology they have like 5 different types of distances).

Last edited: May 10, 2015
6. May 10, 2015

### marcus

Here are three curves describing the universe where the time unit is 17.3 billion years. The present is at x≈0.8
View attachment 83266
The three curves are the Hubble distance, c/H(x), the scale factor a(x), and a curve that tracks expansion speed of a sample distance.
You can tell which is the scale factor a(x) because it equals zero at x=0, and 1 at present x≈0.8

The Hubble distance, defined as c/H(x) is the one which rises from zero and levels out at one for large x.
Its current size is 0.83.

The third curve is the expansion speed history of the distance which at present has size 0.83.

Last edited: May 11, 2015
7. May 10, 2015

### Quarlep

R(t)=a(t)D (D=Observable universe radius which marcus mention it 49 billion light years)I thought like this so it make me sense.But in this condition we use R(t)=a(t)/1.311D and we get R(t)=D.Differantiation respect to t will gave 0.So Observable radius will not depend time If we normalized the scale factor.If all things are correct until here I want to ask something why we normalized it.Normalized scale factor is a(t)/1.311 or always 1.It seems to me always one but I want to be sure.

8. May 10, 2015

### Quarlep

Your first post marcus you mention normalized scale factor and universe radius.Lets forget universe radius and say a distance 0.25 light year .

a(t)/1.311 equals 0.5 when t is something. So we multiply them and we get 0.125 light year.It means In that t, 0.25 light year was 0.125 light year.Normalization makes the distance correct observable distances.I saw a galaxy 0.25 light year away but thats true only normalized scale factor.Cause t=0.8 when I calculate 0.25.So every time I need to normalize to scale factor to track objects distance from us that means Observable radius will be the same all the time.I am soo confused

Last edited: May 10, 2015
9. May 11, 2015

### ChrisVer

Of course something will not depend on time when you go and write it at a given point. When you normalize the scale factor, you commonly do it for "today", that means you write: $a(t_0) \equiv a_0 =1$ with $t_0$ the current moment. So how do you expect to find the observable radius that depends on time? It's like : <<I tell you that a body has some velocity, and you look at its position say $x_0$. Because that position is a fixed number, you turn back and say that there is no velocity because the derivative of a constant wrt time is zero.>>
It doesn't matter what normalization you will choose, you can derive everything just by dragging with you some factor $a_0$ in all your calculations. Normalizing it at a given value is like setting a particular point of reference.

10. May 11, 2015

### Quarlep

Yeah I noticed it latrr

11. May 11, 2015

### Quarlep

So R(t) is 49 billion light year and it doesnt change with time ?

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook