# Accelerated expansion and its rate

1. Apr 22, 2004

### kristobal hunta

I am graduate student working now at the project in cosmology. I would appreciate any links or references to the papers or publications casting a light on the problem of accelerated expansion of the universe.

Did they found the rate of accelerated expansion?

2. Apr 22, 2004

### marcus

please look at this easy survey paper
"Making Sense of the New Cosmology"
by Michael Turner

http://arxiv.org/astro-ph/0202008 [Broken]

he is at Chicago and is one of the most eminent of today's cosmologists
you can rely on what he says
it is representative of the main consensus
(but the numbers are sharper in more recent papers)

the acceleration of expansion is measured by the dimensionless number q

see the definition of q on page 7 near the bottom

see equation (1) on page 7
where you see the quantity R"/R which measures acceleration
is given by the second Friedmann equation

R is the scale factor in the RW metric (the standard metric used in cosmology),
R is usually normalized so it is equal to 1 at present time
and the increase in R tracks the expansion

R' is the increase rate (the time derivative) of R
R'' is the increase in the increase in R

to get rid of units and get a more pure number we can divide first by R

R''/R has dimension of reciprocal time2
that is it is a "per second per second" or a "per year per year" type of quantity

Now one can further normalize this if one divides by the square of the H parameter, because the H parameter is itself a reciprocal time!

therefore one can have a pure dimensionless number that expresses the amount of acceleration if one writes

$$\frac{R''}{R H^2}$$

this is the q which he mentions

it would be good to know this number since it expresses how much acceleration there is

(from the 1998 supernova observations)

http://arxiv.org/astro-ph/0305179 [Broken]

Last edited by a moderator: May 1, 2017
3. Apr 26, 2004

### meteor

Ok, marcus, but you forgot the signus minus in the formula for q
I have problems understanding the expansion parameter. The expansion parameter a is

$$a= \frac {R} {R_0}$$
R is the scale factor at time t, and R0 the scale factor at time t0. What's the exact value of the expansion parameter at this moment?

4. Apr 27, 2004

### Orion1

Ultimate Acceleration...

The Universe is far more dimentionally vast now at this instant in time than what is currently optically viewable through primordial radiation.

Integrating Hubble's Law into the average acceleration theorem yields the following solution:

WMAP Ho = 71 +/- 4 km/sec/Mpc

$$\Delta d_t = (d_f - d_i)$$ (31.5 BLy - 13.18 BLy)
$$a_u = 2 H_o^2 \left( \Delta d_t - \frac{nc}{H_o} \right)$$

au = 4.679*10^-10 m*s^-2

n - fractional luminous galaxy velocity
au - Universe Expansion Acceleration Rate
Ho - Hubble 'Constant'

$$H_o = \frac{ (V_i + V_f)}{ 2(d_f - d_i)} = \frac{ c(n_i + n_f)}{ 2(d_f - d_i)}$$

$$H_o = \frac{a}{(V_f - V_i)} = \frac{a}{c(n_f - n_i)}$$

According to these solutions, Hubble's 'Constant' is not actually a constant, but a function of the Universe Expansion Acceleration Rate and the differential galaxy velocities which compose it.

Reference:
http://www.astro.ucla.edu/~wright/cosmolog.htm
http://www.sltrib.com/2004/mar/03022004/nation_w/144063.asp

5. Apr 27, 2004

### marcus

they usually use subscript zero to mean the present
so time t0 is the present
and H0 is the hubble parameter at present

by your definition, a, at present is equal to 1.

You know the common estimate is that the universe is flat.
the energy density (all forms) is estimated to be about equal to one times the critical density. Actually they estimate 1.04 plus or minus some error bar.
But let's assume it is simply flat, to make things easy.

Then the "deceleration" parameter q (with the minus sign you mentioned)
comes out to be about -0.6
or maybe we should be focusing on -q, the "acceleration" parameter, which comes out to be about 0.6

expansion is accelerating so this dimensionless number 0.6 is positive.

I actually got 0.595, when I calculated it. But that is close to 0.6.

Assuming flat, which although not sure is within the uncertainty range, then to calculate this thing all you need is the 0.73 dark energy fraction of total energy density.

The formula is

$$-q = \frac{a''}{a H^2} = \frac{a'' a}{a'^2} = - \frac{1}{2}(1 - 3 (0.73))$$

the formula comes from the two Friedmann equations

Last edited: Apr 27, 2004
6. Apr 27, 2004

### marcus

for me, this is the first time that I have calculated the acceleration parameter (or minus the deceleration)
anyone want to confirm or point out mistakes?

Meteor does this seem OK?

by definition H = a'/a
so dividing by H^2 is the same as dividing by (a'/a)^2
which is why the two alternative forms of writing -q

then one looks at the two friedmann equations and one equation
gives a''/a
and the other equation gives (a'/a)^2
and one divides first equation by second equation
I think it is not very hard or complicated to see
if you want friedmann equantions written out, please tell me
(or maybe someone else will)

7. Apr 27, 2004

### meteor

marcus, I don't see how the expansion parameter can be 1. That means that R and R0 have the same value. But aren't they the value of the scale factor at different stages of the evolution?

8. Apr 27, 2004

### marcus

you asked how it is at this moment
this moment is t = t0
(they often use this subscript zero for the present)

at this moment R = R(t) = R(t0) = R0 by definition

(the cosmologists often use the subscript zero to indicate the current value of some parameter, at this moment)

so dividing R by R0 is a way of normalizing the scale parameter so that it will be forced to equal 1 at the present moment

at present R/R0 = 1

they really have a bit too much redundant notation---and some cosmologists are using R instead of a, or a instead of R---they dont have completely consistent conventions
as a general rule either a or R will stand for a scale parameter that is very often normalized to be equal to one at the present moment.

then the Hubble parameter H(t) is defined to be a'/a
or equivalently R'/R

and the Hubble parameter at the present moment H0 is defined to be H(t0 ) = present value of a'/a
or equivalently R'/R

it is like having to learn French, or some even less efficient language, where they have many ways to say the same thing---or perhaps this is good?