Beyond IDCS J1426.5+3508

In article , "Richard D. Saam"
writes:

On 9/10/12 7:18 PM, Phillip Helbig---undress to reply wrote:
Right. However, what does it mean that it is 2.5 percent of the present
universe? What is the luminosity distance to z=infinity? (Hint: much
larger than the speed of light times the age of the universe.) Yes, one
can express it as a percentage, but the question is whether this is
useful.

Yes in the conventional view, luminosity distance to z=infinity is much
larger than speed of light times the age of the universe.
This points to an error in concept.

What is the error?

What is a non linear distance function of redshift
such that distance approaches
speed of light times the age of the universe
as redshift approaches infinity?

If that is what you want, then you can just define your distance as the
speed of light multiplied by the light-travel time.

In article , "Richard D. Saam"
writes:

It is understood that universe radius is subject to the particular model
used and can be expressed as a function of z
A model can be used wherein the Hubble sphere expands at c

This is not relevant to the main thread, but perhaps interesting
nonetheless:

The Hubble sphere is, by definition, at the distance at which the
current velocity (proper distance per proper time) of recession due to
cosmological expansion is the speed of light. If the Hubble constant is
constant in time (which is true in the case of exponential expansion,
e.g. in the de Sitter model), then the Hubble sphere has a fixed size.
If the Hubble constant decreases, the Hubble sphere expands. At a given
time, one could have a model in which the Hubble sphere expands with c.
I don't think there is a conventional cosmological model (i.e. one
described by the Friedmann-Lemaître equation) in which the Hubble sphere
expands with c at all times.

The radius of the universe is generally understood to be the radius of
curvature of the universe. It could be equal to the Hubble radius, but
not in general. (If the radius of curvature is infinite, i.e. in a flat
universe, then the scale factor is often DEFINED as the radius of the
Hubble sphere, rather than the curvature radius.) The scale factor
changes with time. Since the ratio of the scale factor now to that at a
previous time is just 1+z, then the relationship between the scale
factor and redshift is trivial (though the relationship between the
scale factor and time, or the redshift and time, is not).

The particle horizon expands (locally) at the speed of light in all
cosmological models, always increases with time (at least in comoving
coordinates) and corresponds to what is generally thought of as the
observable universe. It doesn't have much connection with the Hubble
sphere.

On 9/11/12 10:03 AM, Phillip Helbig---undress to reply wrote:
In article , "Richard D. Saam"
writes:

It is understood that universe radius is subject to the particular model
used and can be expressed as a function of z
A model can be used wherein the Hubble sphere expands at c

This is not relevant to the main thread, but perhaps interesting
nonetheless:

The Hubble sphere is, by definition, at the distance at which the
current velocity (proper distance per proper time) of recession due to
cosmological expansion is the speed of light. If the Hubble constant is
constant in time (which is true in the case of exponential expansion,
e.g. in the de Sitter model), then the Hubble sphere has a fixed size.
If the Hubble constant decreases, the Hubble sphere expands. At a given
time, one could have a model in which the Hubble sphere expands with c.
I don't think there is a conventional cosmological model (i.e. one
described by the Friedmann-Lemaître equation) in which the Hubble sphere
expands with c at all times.

referencing Harrison equation 16.4
for universal expanding sphe

(dR/dt)^2 = (8/3)*pi*G*rho*R^2

dR/dt could equal constant c at all times if

G*(1+z)^n gravitational parameter
rho*(1+z)^n universe density parameter
R*(1+z)^-n universe radius parameter

perhaps n=1 for Hubble sphere and n=3 for particle horizon

which would not negate the Friedmann-Lemaître equation.

In article , "Richard D. Saam"
writes:

The Hubble sphere is, by definition, at the distance at which the
current velocity (proper distance per proper time) of recession due to
cosmological expansion is the speed of light.

referencing Harrison equation 16.4

A good reference. Harrison wrote a couple of papers distinguishing
concepts which are often confused, including horizons and Hubble
spheres.

for universal expanding sphe

(dR/dt)^2 = (8/3)*pi*G*rho*R^2

Right, but that is not necessarily Hubble sphere.

dR/dt could equal constant c at all times if

G*(1+z)^n gravitational parameter
rho*(1+z)^n universe density parameter
R*(1+z)^-n universe radius parameter

Presumably you mean replacing G with G*(1+z)^n etc?

On 9/14/12 4:52 AM, Phillip Helbig---undress to reply wrote:
Right, but that is not necessarily Hubble sphere.
not necessarily but let R be constrained to Hubble length c/H


Presumably you mean replacing G with G*(1+z)^n etc?

Yes and for present values
G = G
rho = (3*H^2/(8*pi*G)
R = c*H

(dR/dt)^2 = (8/3)*pi*G*(1+z)^n*rho*(1+z)^n*(R*(1+z)^-n)^2 = c^2

The expression collapses to c^2 for all z
but is there any meaning in the uncollapsed form?

It is interesting to note that the expression with n=3
G*(1+z)^3
rho*(1+z)^3
R*(1+z)^-3
is dimensionally and numerically consistent
with CMBR temperature scaling
with (1+z)^1 as observed.
The Hubble sphere may contain another coordinate system
(represented by CMBR temperature scaling)
that expands differently
but in a well defined manner relative to the Hubble sphere R^3.

A big question results:
Does Newtonian G scale with z
G*(1+z)^3 ?

[Mod. note: quoted text trimmed. In the local universe (z1), we can
write lookback time t = z/H_0, thus G = G_0(1+z)^3 = (binomial
expansion) G = G_0(1+3z) = G = G_0(1+3H_0t) = dG/dt = 3G_0 H_0 =
(dG/dt)/G_0 = 3H_0 where 3H_0 = 2.1e-10 yr^-1. Current experimental
evidence puts an upper limit on (dG/dt)/G of a few times 10^-12 yr^-1
(http://arxiv.org/abs/gr-qc/0411113). So this strong a scaling is
inconsistent with experiment -- mjh]

In article , "Richard D. Saam"
writes:

On 9/14/12 4:52 AM, Phillip Helbig---undress to reply wrote:
Right, but that is not necessarily Hubble sphere.
not necessarily but let R be constrained to Hubble length c/H

Presumably you mean replacing G with G*(1+z)^n etc?

Yes and for present values
G = G
rho = (3*H^2/(8*pi*G)
R = c*H

(dR/dt)^2 = (8/3)*pi*G*(1+z)^n*rho*(1+z)^n*(R*(1+z)^-n)^2 = c^2

You need to specify a) what you are postulating and what you are
deriving and b) the physics (e.g. general relativity or something else)
which is used in your derivation.

What is R? Is it some scale factor, is it the radius of curvature or is
it both? (Convention is to choose the scale factor to be equal to the
radius of curvature except when the latter is infinite in which the
scale factor is set to 1---all in units of c/H_0.)

On 9/17/12 9:34 AM, Richard D. Saam wrote:

A big question results:
Does Newtonian G scale with z
G*(1+z)^3 ?

[Mod. note: quoted text trimmed. In the local universe (z1), we can
write lookback time t = z/H_0, thus G = G_0(1+z)^3 = (binomial
expansion) G = G_0(1+3z) = G = G_0(1+3H_0t) = dG/dt = 3G_0 H_0 =
(dG/dt)/G_0 = 3H_0 where 3H_0 = 2.1e-10 yr^-1. Current experimental
evidence puts an upper limit on (dG/dt)/G of a few times 10^-12 yr^-1
(http://arxiv.org/abs/gr-qc/0411113). So this strong a scaling is
inconsistent with experiment -- mjh]


but is the observed dG/dt)/G
of a few times 10^-12 yr^-1 (essentially constant)
a consequence of
G*(1+z)^3*R*(1+z)^-3 = constant
and also
rho*(1+z)^3*R*(1+z)^-3 = constant

because speed of GR gravity
is the same as Newtonian gravity
that is infinite
http://math.ucr.edu/home/baez/physic...rav_speed.html
?

In article , "Richard D. Saam"
writes:

but is the observed dG/dt)/G
of a few times 10^-12 yr^-1 (essentially constant)
a consequence of
G*(1+z)^3*R*(1+z)^-3 = constant

This reduces to G*R = constant.

and also
rho*(1+z)^3*R*(1+z)^-3 = constant

This reduces to rho*R = constant

Due to conservation of matter, rho should be proportional to (1+z)^3.

Again, since these equations imply a very different universe than the
one we live in, and even a very different one than the possible ones
which exist given the known laws of physics, such a discussion makes
sense only if it is clear what is postulated (and why) and what is
derived (and how).

because speed of GR gravity
is the same as Newtonian gravity
that is infinite
http://math.ucr.edu/home/baez/physic...rav_speed.html
?

No. In GR gravity propagates at the speed of light. What you might be
thinking of is that people often assume it propagates at the speed of
light. This is wrong, but something else is wrong as well: it should
propagate from where the object was then, not where it is now. It turns
out that the two effects cancel.

Richard D. Saam wrote:
but is the observed dG/dt)/G
of a few times 10^-12 yr^-1 (essentially constant)

Actually the limits are considerably tighter than that. Quoting
section 3 of
Stephen M. Merkowitz
"Tests of Gravity Using Lunar Laser Ranging",
Living Reviews in Relativity 13 (2010), 7
http://www.livingreviews.org/lrr-2010-7
(n.b. this is online open-access!),

| Recent analysis of LLR data by the JPL group sets a limit on
| (dG/dt)/G = (6 +/- 7) * 10^{-13}/year [69]. Similarly, Mueller
| et al find (dG/dt)/G = (2 +/- 7) * 10^{-13}/year and
| and (d^2G/dt^2)/G = (4 +/- 5) * 10^{-15}/year^2 [35].
| These limits translate to less than a 1% variation of G over
| the 13.7 billion year age of the universe.

ciao,

--
-- "Jonathan Thornburg [remove -animal to reply]"
Dept of Astronomy & IUCSS, Indiana University, Bloomington, Indiana, USA
currently on sabbatical in Canada
"Washing one's hands of the conflict between the powerful and the
powerless means to side with the powerful, not to be neutral."
-- quote by Freire / poster by Oxfam

On 9/18/12 4:35 AM, Phillip Helbig---undress to reply wrote:
because speed of GR gravity
is the same as Newtonian gravity
that is infinite
http://math.ucr.edu/home/baez/physic...rav_speed.html
?

No. In GR gravity propagates at the speed of light. What you might be
thinking of is that people often assume it propagates at the speed of
light. This is wrong, but something else is wrong as well: it should
propagate from where the object was then, not where it is now. It turns
out that the two effects cancel.

What Baez is saying is different:

"In that case, one finds that the "force" in GR is not quite central—it
does not point directly towards the source of the gravitational
field—and that it depends on velocity as well as position. The net
result is that the effect of propagation delay is almost exactly
cancelled, and general relativity very nearly reproduces the newtonian
result."

Such an explanation could explain in part the world we live in

wherein the observed dG/dt)/G
of a few times 10^-12 yr^-1 (essentially constant)
no matter where you look.

Such an explanation appears to outside of current universe expansion models.