In 1998 Hubble volume meant a cube 12 billion LY on a side

marcus

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In 1998 "Hubble volume" meant a cube 12 billion LY on a side

Majin asked what is hubble volume.
I did a google search and found that in 1998 there was
a computer simulation study at one of the Max Planck institutes
which simulated a cube
12 billion LY on a side (comoving distance)

other sites indicated the simulation cube was 3/0.71 billion parsecs on a side

which is slightly different---about 13.8 billion LY on a side

they use the symbol h = H0/(100 km/s per Mpc)
which has now been determined to be 0.71
and they make the cube 3/h billion pc on a side

Back in 1998 and before there was wide uncertainty about the hubble parameter like 50-100 km/sec per megaparsec
corresponding to widely varying ideas of the Hubble length.

Now the estimate for the Hubble length is reasonably stable
at around 13.7 or 13.8 billion LY

According to today's notion of the Hubble length, the volume should be a cube 13.7 billion LY on a side.

They indicated that this cube was NOT the volume of the observable universe but, instead, was "a significant fraction" of it.

In fact the present volume of the observable universe is several times the Hubble volume and the two should not be confused.

(But it appears that the two are being confused in a recent
scientific american article!)

There is a lot of google stuff on hubble volume but it mostly
involves this simulation project.

I found a journal article---very technical---with hubble volume in a technical sense of L^3, where L is the hubble length. But that
was kind of drowned out by links to the simulation projects

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marcus

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Majin referred to the Max Tegmark article in sci am "Infinite earths in parallel universes etc"
and i checked and indeed Max does use the term
Hubble volume to mean the volume of the observable universe.

this is confusing and a waste of a good term, but human language is full of stuff like that

the present distance to the farthest objects whose light has reached us is around 40 billion LY (eg. see Wright cosmology tutorial) . Usually one takes the estimated age of U and multiplies by 3 to get the LY----so 13 billon age implies 39 billion LY radius,
or 14 age implies 42 billion LY radius

I will write Max Tegmark and ask for clarification.

The Hubble volume simulation project (Harching Planck Inst)
used a cube which (it turns out) is exactly the Hubble length
13.8 billion LY on one side-----i.e. NOT the volume of the observable universe.

So there is a discrepancy by a factor of (4pi/3)x27----a factor of 113----between what is meant technically and what Max Tegmark says in the Scientific American.

much is excused in writing for general public but this is trashy
and blurs the meaning of a potentially useful term.

Alexander

Would not 40 bly be the circumference of closed universe with the curvature 13 bly-1? (Distance you have to travel to return back from back?)

My humble opinion, for what it's worth = they shouldn't have used a cube.

marcus

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Originally posted by schwarzchildradius
My humble opinion, for what it's worth = they shouldn't have used a cube.
oh excellent Schwarzschildradius with whom I humbly desire
amicable disputation, why in heaven's name should they not have used a cube?

they didnt say they were trying to model a volume equal to the
observable
they just said it was a "substantial part" of the observable

it turns out, after a careful look, that they made it
13.7 by 13.7 by 13.7 billion LY. A cube is good for computer
simulations.

The whole observable is way bigger than that and it is spherical.
the accepted radius is around 3 times the hubble length
(this is a "cosmology 101" type of fact) and hence generally assumed to be about 40 billion LY. NedW cosmology tutorial
is good at explaining that---he teaches cosmology at UCLA and
is a nice guy and a good explainer

did you mean it as a joke that they should not have used a cube
or if serious why?

Alexander

Why do you think that the whole universe is spherical? Do you know for sure that it has overall positive curvature (overcritical density?).

My understanding is that there is way not enough matter to make it positive, it is either negative or barely flat.

marcus

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Originally posted by Alexander
Why do you think that the whole universe is spherical? Do you know for sure that it has overall positive curvature (overcritical density?).

My understanding is that there is way not enough matter to make it positive, it is either negative or barely flat.
the universe is flat
I do not imagine it as the surface of a 3-sphere!
the observable universe is a spherical volume, a solid ball,
containing all those objects whose light has reached us
I cannot figure out how you could have gotten the mistaken
idea that I was postulating positive curvature---the observations
strongly imply zero curvature

BTW it is a standard "Astro 101" fact that the radius of
the observable universe---that ball or spherical volume around us---is about 40 billion LY. the usual estimate is gotten
by multiplying whatever is assumed for the age of the universe by 3. So if the current age is 14 billion Y, then the radius of the observable is currently about 42 billion LY. Do you agree?

OK so what model are you using to multiply by 3? With the standard Friedman model, you can analyze H=dr/rdt for dt (with dr=r) to get t=1/H, (alexander points out that H/c is just the curvature) which is around 13-15 e9 years. Of course this just assumes that space has been expanding at the same rate since the beginning, which is apparently untrue.

marcus

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Originally posted by schwarzchildradius
OK so what model are you using to multiply by 3? With the standard Friedman model, you can analyze H=dr/rdt for dt (with dr=r) to get t=1/H, (alexander points out that H/c is just the curvature) which is around 13-15 e9 years. Of course this just assumes that space has been expanding at the same rate since the beginning, which is apparently untrue.
One of the FAQ in cosmology is:

"If the Universe is only 13 billion years old, how can we see objects that are now 39 billion light years away?"

It's just what one would expect bewildered undergrads to ask. And indeed it is in the FAQ of Ned Wright who teaches cosmology at UCLA. The factor of three is approximate but gives the right idea. To keep the numbers nice, for a simple explanation, one can rephrase it with an age of 10 billion years so the radius of the observable is 30.

Ned Wright's cosmology FAQ has it this way and I quote:

"If the Universe is only 10 billion years old, how can we see objects that are now 30 billion light years away?"

"When talking about the distance of a moving object, we mean the spatial separation NOW, with the positions of both objects specified at the current time. In an expanding Universe this distance NOW is larger than the speed of light times the light travel time due to the increase of separations between objects as the Universe expands. This is not due to any change in the units of space and time, but just caused by things being farther apart now than they used to be.

What is the distance NOW to the most distant thing we can see? Let's take the age of the Universe to be 10 billion years. In that time light travels 10 billion light years, and some people stop here. But the distance has grown since the light traveled. The average time when the light was traveling was 5 billion years ago. For the critical density case, the scale factor for the Universe goes like the 2/3 power of the time since the Big Bang, so the Universe has grown by a factor of .5^-2/3 = 1.59 since the midpoint of the light's trip. But the size of the Universe changes continuously, so we should divide the light's trip into short intervals. First take two intervals: 5 billion years at an average time 7.5 billion years after the Big Bang, which gives 5 billion light years that have grown by a factor of 0.75^-2/3 = 1.21, plus another 5 billion light years at an average time 2.5 billion years after the Big Bang, which has grown by a factor of 0.25^-2/3 = 2.52. Thus with 1 interval we got 1.59*10 = 15.9 billion light years, while with two intervals we get 5*(1.21+2.52) = 18.7 billion light years. With 8192 intervals we get 29.3 billion light years. In the limit of very many time intervals we get 30 billion light years. With calculus this whole paragraph reduces to this..."

If you don't mind some easy integral calculus it reduces to one line, but he is trying to make the FAQ understandable also to people who arent used to integrals and might just turn off if he said something like

"integral zero to one of t-2/3 dt"

but indeed that integral is just equal to 3, which is what he gets
by breaking the interval down and riemann-summing.

http://www.astro.ucla.edu/~wright/cosmology_faq.html#DN

Alexander

Interesting that universe does not have certain "shape" (distribution of galaxies), as well as certain time and certain position of objects in it - all these values vary with observer. So, now and here only refers to us, and in no way to someone else far away. His "now" and "here" is completely different than what we "see" from Earth.

marcus

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Originally posted by Alexander
Interesting that universe does not have certain "shape" (distribution of galaxies), as well as certain time and certain position of objects in it - all these values vary with observer. So, now and here only refers to us, and in no way to someone else far away. His "now" and "here" is completely different than what we "see" from Earth.
I used to think that too!!! But then I discovered that for cosmologists there is a preferred frame with simultaneity well-defined for all comoving observers. They use this a lot in their work.

what you say sounds like you are thinking from a SR point of view. No preferred frame. No idea of simultaneity.

In cosmology they often say things like "for all observers at rest with respect to the CMB (or with respect to the Hubble flow)"

there is an absolute universal idea of being at rest
(you are at rest if there is no dipole in the CMB, basically)

There is a single calendar for all observers everywhere, as long as they are at rest.

Read Ned Wright Tutorial Part 2 (the comoving distance definition, for example, requires this preferred frame)

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