# Why is the Hubble sphere expanding?

1. Jun 25, 2012

### andrewkirk

Hubble’s law, according to Wikipedia, says that the recession velocity of a distant object P comoving with Earth is vrec=HD where H is Hubble’s constant and D is the proper distance from Earth to P.
The Hubble sphere is the area of space in which all objects comoving with Earth have vrec less than c. From Hubble’s law, one would expect the radius of the Hubble sphere to be c/H, a constant.
Yet Davis and Lineweaver (2003) say the Hubble sphere is expanding, by which I understand that the Hubble radius – the proper distance from Earth to the boundary of the Hubble sphere - is increasing.
How can the Hubble radius be increasing if both Hubble’s constant and the speed of light are constant?

2. Jun 26, 2012

### marcus

Andrew, H(t) is not constant. It changes over time. Shouldn't be called by the name "constant".

The basic equation of cosmology is called the Friedmann equation. (The whole field is based on it, that is how the universe is modeled.) The F. equation is an equation describing how H *changes* over time.

H(t) has been 10X larger in the past, even 100X larger if you go back far enough.

One way to get a feel for this is to use any one of the online cosmological calculators which convert redshift to distance now, and also tell what the distance was when the light was emitted, and how long ago that was etc. It's good to get used to working with a calculator---they're easy.

http://www.einsteins-theory-of-relativity-4engineers.com/cosmocalc_2010.htm
Go there and press "calculate".
It will tell you what Hubble rate H(t) was at the time the light was emitted that we now see as
the CMB (cosmic microwave background).

then put a different number in the "redshift of the source" box.
Like put redshift 9 in the box. The most distant galaxies we have seen so far have redshifts up in the range 9 to 10
Press "calculate" and you will find out what the Hubble rate H(t) was back then when those galaxies emitted the light that we are now getting.

"constant" is just a misnomer that got stuck in the language, once people start using a certain term and writing it in books and get used to calling whatever it is by that name it's *very hard to change*.
We call it that "for historical reasons"

Last edited: Jun 26, 2012
3. Jun 26, 2012

### andrewkirk

Thank you Marcus. Is H(t) currently decreasing? That would cause an expansion of the Hubble sphere if I've understood it correctly.
If so, is that the sole reason for the expansion of the Hubble sphere, or are there other factors as well?

4. Jun 26, 2012

### Chalnoth

H is not a constant, but a function of time. The term, "Hubble constant," is a misnomer.

5. Jun 26, 2012

### marcus

If you took the hint and went to the calculator
http://www.einsteins-theory-of-relativity-4engineers.com/cosmocalc_2010.htm
and pressed "calculate"
then you know that Hubble rate now is about 70 or 71 in the usual conventional units
call it 70.4 since that's the most recent best estimate.
and BACK THEN when the ancient light was emitted the rate is estimated to have been
1,518,900
that is about one and a half MILLION. (in the same units)
So let's see by what ratio it is less now 1518900/70.4 = 21575

So H(t) was roughly 21 thousand times bigger back then

and therefore the Hubble radius was 21 thousand times SMALLER back then.

Obviously the Hubble radius has grown a lot. So what Lineweaver and Davis say is not so surprising.

Check it out for some other redshifts, using the calculator, like for z = 2, 3, 4, 5 etc.
It's a way of getting some hands-on experience of the standard cosmic model (based on the Friedmann equation)

6. Jun 26, 2012

### marcus

Yes it is currently decreasing but at an ever slower rate.
You know it is now about 70.4 (earlier the most common figure used was 71 but let's say 70.4).

It is expected to kind of level out at around 60 in the same units.

That is the only reason for the increase in the Hubble radius (as you suggest) because the by definition the Hubble radius is simply c/H(t)
So if H(t) decreases, smaller denominator, c/H(t) has to increase. No other cause.

Last edited: Jun 26, 2012
7. Jun 26, 2012

### andrewkirk

Thank you Marcus. Sorry for missing the second part of your first post, which is why I didn't get the hint. I got distracted by after reading the first para and when I returned to the PC I had forgotten that there was more.

How can we reconcile the fact that the Hubble "constant" is decreasing, with the recent discovery (Perlmutter, Schmidt & Riess) that the expansion is accelerating? Does this discovery imply the need for modifications to the Friedman equations on which the calculator is based? Does the discovery imply that, despite the Hubble constant being now well below its value in the distant past, it is now increasing ?

Thank you.

8. Jun 26, 2012

### jobigoud

Acceleration of the expansion means that the Hubble parameter is decreasing less rapidly than before. But it's still decreasing.

9. Jun 26, 2012

### clamtrox

Indeed; the expansion would need to be faster than exponential for the Hubble parameter to increase. That's what it means to be a Hubble parameter. (it's defined to be $H = \dot{a}/a$)

10. Jun 26, 2012

### marcus

I agree with the concise clear answers that Clamtrox and Jobigoud just gave.

You can track expansion of distances either as an absolute rate or as a "percentage" rate.
To see the absolute rate you look at a quantity called the SCALEFACTOR a(t).
This is normalized so that it equals one at present a(present) = 1
and back when distances were only 1/10 as big it was a(back then) = 0.1

So you can get an idea of the absolute rate distances are growing by looking at the time-derivative a'(t) = da/dt. This is just the slope of the scalefactor a(t) curve.

it is a'(t) that is increasing---the absolute distance growth rate is increasing, the scalefactor curve is getting steeper. But very very slowly. You wouldn't believe---it's hard to comprehend how slow.

The percentage or fractional growth rate puts that absolute rate OVER the whole.
It is a'(t)/a(t).
You can think of it as the amount the distance increased DIVIDED by the distance itself.
It turns out that this fractional rate of increase is DECREASING because the DENOMINATOR is already big and getting bigger.

So yes, a'(t) is increasing, the slope is increasing---that is the "acceleration" people talk about. But a(t) the scalefactor itself is putting on size faster than a'(t) is! So the denominator is growing faster than the numerator. So a'(t)/a(t) is actually decreasing and is expect to continue doing so.

11. Jun 26, 2012

### andrewkirk

OK I think I get it. Under Hubble's law, the proper distance between any two distant objects would increase exponentially over time if the Hubble 'constant' were actually constant. So if the Hubble 'constant' is only slightly decreasing, and is expected to asymptotically approach 60 from above (per Marcus's post), the universe will continue to expand approximately exponentially.

Is that right?

Thank you very much for the answers provided. They really help me understand this, which I didn't before.

Here is a follow-up question, if I may:

Can anybody recommend a good freely-available derivation of the FLRW metric and the Friedman equation, assuming an understanding of Einstein's equation and the associated differential geometry as a starting point? I have Schutz's 'A first course in GR', which deals with this in Chapter 12, but his presentations are hard to follow as they skip important steps, use inconsistent and sometimes undefined terms and sometimes mis-state things.

Last edited: Jun 26, 2012
12. Jun 26, 2012

### nicksauce

I think Carroll's derivation is good: http://preposterousuniverse.com/grnotes/grnotes-eight.pdf [Broken]

Last edited by a moderator: May 6, 2017
13. Jun 27, 2012

### clamtrox

That is correct. You can see this easily by looking at the Friedmann equation:

$$H^2 = \frac{8\pi G}{3} \rho = H_0^2 (\Omega_\Lambda + \Omega_m a^{-3} + \Omega_r a^{-4})$$

The energy density of dust, radiation and all other normal stuff dilutes away as the universe expands. Dark energy does not dilute, and it's energy density remains constant. As time passes and the universe expands, you have

$$\lim_{a \rightarrow \infty} H^2 = H_0^2 \Omega_\Lambda,$$
which is a constant. You can also see right away where the 60 comes from. Currently we have $H_0 \simeq 70 km/s/Mpc$ and $\Omega_\Lambda \simeq 0.72$. From this you can easily calculate $H \rightarrow 60 km/s/Mpc$

14. Jun 27, 2012

### jobigoud

I think the reasoning is sound, but the Hubble constant is not simply "slightly decreasing". For most of the time, the expansion has been much closer to linear than to exponential.

If I understood correctly, what is observed is a slight acceleration from linear. With time, recession velocities build up and the expansion eventually catch up with exponential growth.

15. Jun 28, 2012

### Naty1

Chalnoth
yes, that had me confused for a while when I first read it...."Hubble parameter" is a more appropriate term.
H0 is a designation meaning 'Hubble now', but I do not know how universal that is.

16. Jun 28, 2012

### marcus

AFAIK completely universal in the sense you mean, widely used and recognized.

17. Jun 28, 2012

### andrewkirk

In fact, it seems so universal that, despite coming across the term several times over the last week, in none of the places where it was used did the writer even bother to define it. I surmised that it probably meant "Hubble now" but it's nice to see that explicitly confirmed!