Hubble Time

Op zondag 2 december 2012 14:43:07 UTC+1 schreef Jonathan Thornburg [remove -animal to reply] het volgende:
Nicolaas Vroom wrote:

The current accepted values for omega(m) and omega(Lambda) are 0.26 and 0.74
See for definition:
http://en.wikipedia.org/wiki/Friedma...#The_equations
Using my simulation program discussed at:
http://users.telenet.be/nicvroom/friedmann's%20equation.htm#Q6
you can calculate Lambda which is equal to 0.01129
This Lambda is a Universal constant.

I don't think we know whether Lambda is a universal constant or not.

Lambda, G and c are considered Universal constants.
H, Omega(L), omega(M), omega(k), R are varying in time.
C defined as rho*R^3 is also an Universal constant.
See: Ray A d'Inverno, Introducing Einstein's Relativity, ISBN 0-19-859686-3.

That raises the question: Why is the present age of the Universe
not larger than 13.7 ?

Why should it be? Or, if you prefer, why should the Hubble constant be
less than 72 km/sec/Mpc? To be more precise, could you explicate the
line of reasoning which suggests a larger age of the universe and/or
a smaller Hubble constant?

The question is IMO:
Why is 1/H0 at this moment equal to the age of the Universe.
The following table shows evolution with Lambda = 0,01129
Age Omega(M) Omega(L) rho 1/H0 H0 Mpc
10 0,473 0,527 0,00161 11,84 82,7
14 0,263 0,738 0,00064 14,00 70
16 0,190 0,810 0,00042 14,6 66,8
20 0,096 0,904 0,00019 15,5 63,2

As I said in the feature the people will find stars
which have an age which is older than the value of 1/H0
calculated/observed at that epoch.

Nicolaas Vroom

In article , Nicolaas Vroom
writes:

Op zondag 2 december 2012 14:43:07 UTC+1 schreef Jonathan Thornburg
[remove -animal to reply] het volgende:
Nicolaas Vroom wrote:

The current accepted values for omega(m) and omega(Lambda) are 0.26 and 0.74
See for definition:
http://en.wikipedia.org/wiki/Friedma...#The_equations
Using my simulation program discussed at:
http://users.telenet.be/nicvroom/friedmann's%20equation.htm#Q6
you can calculate Lambda which is equal to 0.01129
This Lambda is a Universal constant.

I don't think we know whether Lambda is a universal constant or not.

Lambda, G and c are considered Universal constants.
H, Omega(L), omega(M), omega(k), R are varying in time.
C defined as rho*R^3 is also an Universal constant.
See: Ray A d'Inverno, Introducing Einstein's Relativity, ISBN 0-19-859686-3.

I think Jonathan introduced a red herring here. Within standard
physics, Lambda is constant. However, people have considered models
with varying Lambda, with varying G, with varying c and so on. I think
it is fair to say that constraints on the variation of Lambda are not as
strong as for the other constants.

Why is 1/H0 at this moment equal to the age of the Universe.

No-one knows. As far as we know, it is a coincidence. Check out this
paper for some thoughts: http://arxiv.org/abs/1001.4795

Consider that the Moon is moving away from the Earth, so in the past was
closer and thus appeared larger. Why is it the same angular size as the
Sun just during a relatively small period of time during the lifetime of
the Earth? Is it just a coincidence?

As I said in the feature the people will find stars
which have an age which is older than the value of 1/H0
calculated/observed at that epoch.

feature -- future ?

Yes, but there is absolutely nothing at all puzzling about this.

20 years ago, it was somewhat puzzling since people assumed that the age
of the universe was LESS than the Hubble time, which is ALWAYS the case
in a non-empty universe with no cosmological constant. One didn't know
the values of lambda and Omega, but assumed lambda was 0 (and many
assumed Omega was 1), so WHATEVER the value of Omega, the age of the
universe was ALWAYS less than the Hubble time. Since no-one knew the
value of Omega for sure, people used the Hubble time as an upper limit
on the age of the universe. So, historically, this was interesting, but
in a universe with a positive cosmological constant which expands
forever, the age of the universe is always larger---for most of the
time, arbitrarily larger---than the Hubble time, except for a time near
the beginning when the universe was decelerating before acceleration
started. So, in such a universe---which we live in if the currently
accepted values for the cosmological parameters are correct---then the
generic case is that the age of the universe is greater than the Hubble
time. As the universe tends towards the de Sitter universe, the Hubble
constant and thus the Hubble time tend towards constant values but of
course the universe keeps getting older.

Op dinsdag 4 december 2012 08:46:14 UTC+1 schreef Phillip Helbig---undress to reply het volgende:
In article , Nicolaas Vroom
writes:

Lambda = lambda*3H^2 is the usual definition. Be aware that different
units are used; sometimes there is a factor of the square of the speed
of light.

When you perform the following calculation:
0.73743 * 3 /(14 * 14) you get 0.011287
(with lambda = 0.73743 and H = 1/14)
The program does not use the equation Lambda = omega(Lambda) * 3H^2

Unless you are using something equivalent, you are defining it
differently than everyone else does

What I wrote is maybe misleading.
If you know H you cannot perform this equation to calculate Lambda and omega(Lambda).
You have to do that by trial and error.
First you try 0.01 and 0.02 for Lambda which give 1/H0 is 14,5 and 11.48
Next you try 0.012 wich gives 13,74. Try 0,011 which gives 14,109 etc.
Internally the program uses the equation. First Omega(M) is calculated.

1/H0 at present is 13.7 (13.7 billion years after BB)

Yes, this is measured.
IMO this value is not measured but calculated.
There are two ways (?) to calculate H0:
1) using H0=c*z/d
2) using WMAP data.
See for details:
http://users.telenet.be/nicvroom/fri...ation.htm#Ref4


1/H0 20 billion years after BB should be 20 (?)

Why?

Note that, as far as we know, it is a COINCIDENCE that 1/H0 now is very
close to the age of the universe. There are some special models in
which this relation always holds, but it doesn't hold in our model. In
the future, the age of the universe will be more than 1/H0. In fact, H0
(and hence 1/H0) will converge on a constant value, while of course the
age will always increase.
It seems to me that a large part of the calculation of 1/H0 is
based on the fact that globular clusters are already 13.2 billion years
of age. This is more or less a yardstick for 1/H0

The interesting thing is that 6.3 billion years the globular clusters
will be roughly 20 billion years old while 1/H0 will be smaller.

In order to do such a calculation, you also have to specify Omega and
lambda. The age of the universe is a function of Omega and lambda; H0
is essentially a scalingactor.

IMO both the age of t0 and H0 are both a function of Omega(Lambda) and
Omega(M) (and omega(k)

Nicolaas Vroom

In article , Nicolaas Vroom
writes:

1/H0 at present is 13.7 (13.7 billion years after BB)

Yes, this is measured.
IMO this value is not measured but calculated.
There are two ways (?) to calculate H0:
1) using H0=c*z/d
2) using WMAP data.
See for details:
http://users.telenet.be/nicvroom/fri...ation.htm#Ref4

In this sense, NOTHING is measured in astronomy, except perhaps counts
of photons, and EVERYTHING is calculated.

The interesting thing is that 6.3 billion years the globular clusters
will be roughly 20 billion years old while 1/H0 will be smaller.

Why is that interesting?

IMO both the age of t0 and H0 are both a function of Omega(Lambda) and
Omega(M) (and omega(k)

Omega(k) follows from Omega(M) and Omega(Lambda), so it is not an
independent parameter. Given Omega(M) and Omega(Lambda), one can
calculate the age in units of the Hubble time, so to convert it to
something with the dimenstion time, you have to specify the Hubble
constant.

On 12/4/12 2:11 AM, Nicolaas Vroom wrote:
C defined as rho*R^3 is also an Universal constant.
See: Ray A d'Inverno, Introducing Einstein's Relativity, ISBN 0-19-859686-3.
page 323
dimensionality requires:

C = (8*pi/3)*G*rho*R^3

Does this mean that the represented Friedmann theory
allows for variation of
G, rho and R
such that C remains constant?

RDS

In article , "Richard D. Saam"
writes:

On 12/4/12 2:11 AM, Nicolaas Vroom wrote:
C defined as rho*R^3 is also an Universal constant.
See: Ray A d'Inverno, Introducing Einstein's Relativity, ISBN 0-19-859686-3.
page 323
dimensionality requires:

C = (8*pi/3)*G*rho*R^3

Does this mean that the represented Friedmann theory
allows for variation of
G, rho and R
such that C remains constant?

No. C is a constant above becasuse everything except rho and R are
constants and rho is proportional to R^{-3} due to the conservation of
mass, thus rho*R^3 is constant. Writing C instead of the longer term
just makes for a neater equation.

Rho of course changes; it drops as the universe expands. Similarly, R
increases as the universe expands. One could, of course, examine the
consequences of a varying G and people have done so, but in this case
"Friedmann theory" no longer holds: if one changed the dependence of R
and/or rho with time to match the change in G, then mass would no longer
be conserved.

On 12/17/12 2:14 AM, Phillip Helbig---undress to reply wrote:
In article , "Richard D. Saam"
writes:

On 12/4/12 2:11 AM, Nicolaas Vroom wrote:
C defined as rho*R^3 is also an Universal constant.
See: Ray A d'Inverno, Introducing Einstein's Relativity, ISBN 0-19-859686-3.
page 323
dimensionality requires:

C = (8*pi/3)*G*rho*R^3

Does this mean that the represented Friedmann theory
allows for variation of
G, rho and R
such that C remains constant?

No. C is a constant above becasuse everything except rho and R are
constants and rho is proportional to R^{-3} due to the conservation of
mass, thus rho*R^3 is constant. Writing C instead of the longer term
just makes for a neater equation.

Rho of course changes; it drops as the universe expands. Similarly, R
increases as the universe expands. One could, of course, examine the
consequences of a varying G and people have done so, but in this case
"Friedmann theory" no longer holds: if one changed the dependence of R
and/or rho with time to match the change in G, then mass would no longer
be conserved.

On review of Ray A d'Inverno page 323,
It is noted that C is a constant of integration
which implies function continuity over the range of investigation.
Are there functional discontinuities such that "Friedmann theory"
is inadequate?
Also
Has anyone investigated other mass conservation theory
such as:
C = (8*pi/3)*G*N*rho*R^3
where N is a dimensionless number
and N*rho*R^3 represents mass conservation.
?

In article , "Richard D. Saam"
writes:

Has anyone investigated other mass conservation theory
such as:
C = (8*pi/3)*G*N*rho*R^3
where N is a dimensionless number
and N*rho*R^3 represents mass conservation.
?

If N is a constant number, then this doesn't lead to mass conservation.

On 12/19/12 3:26 AM, Phillip Helbig---undress to reply wrote:

If N is a constant number, then this doesn't lead to mass conservation.

If N is constant then there is mass conservation if
C ~ N*rho*B^3
and
If N is variable then there is mass conservation if
C ~ N*rho*R^3
When BR and B^3 is 3D tessellated within R^3
and rho = m/B^3 (constant m)

What determines the constant m?