On Wednesday, September 11, 2024 at 4:33:51 AM UTC-6 Quentin
Anciaux wrote:
Le mer. 11 sept. 2024, 11:49, Alan Grayson
<[email protected]> a écrit :
On Wednesday, September 11, 2024 at 3:26:01 AM UTC-6
Quentin Anciaux wrote:
Le mer. 11 sept. 2024, 11:23, Alan Grayson
<[email protected]> a écrit :
On Tuesday, September 10, 2024 at 3:50:08 PM
UTC-6 Quentin Anciaux wrote:
Le mar. 10 sept. 2024, 23:19, Alan Grayson
<[email protected]> a écrit :
On Tuesday, September 10, 2024 at
2:19:42 PM UTC-6 John Clark wrote:
On Tue, Sep 10, 2024 at 3:57 PM Alan
Grayson <[email protected]> wrote:
*>> Even if you ignore Dark
Energy and postulate that the
Hubble constant really is
constant, every object a
megaparsec away (3.26 million
light-years) is moving away
from us at about 70
kilometers per second. So if
you try to look at objects a
sufficiently large number of
megaparsec away you will fail
to find any because they are
moving away from us faster
than the speed of light.*
>/That was in the past. At
present, the universe is
expanding at about 70 km/sec./
*Galaxies are receding from the Earth
at 70 km/sec for EACH
megaparsecdistant from Earth they
are. The further from Earth they are,
the faster they are moving away from
us, so if they are far enough away
they will be moving faster than the
speed of light away from us. *
*
*
/> You're assuming the universe
today is infinite,/
*NO! I said IF the entire universe is
infinite today then it was always
infinite, and IF it was finite 10^-35
seconds after the Big Bang then it's
still finite today. I also said
nobody knows if the entire universe
is infinite or finite. *
*>*///Hubble's law applies to the
past, not to the future,/
*What the hell?!*
*How about an intelligent reply?
Obviously, if the universe is infinite
today, it was always infinite. But that's
what I am questioning. For galaxies to
fall out of view, they have to moving at
greater than c. Now they aren't receding
that fast. How will they start moving
that fast? You're applying Hubble's law
without thinking what it says. Just
because a galaxy is now receding at less
than c, how will continued expansion
increase that speed to greater than c? AG *
The farther they are the faster they are
receding from you, so as they continue to get
farther away they receed faster from you till
the point they receed faster than c and go
out of your horizon.
Quentin
Instead of preaching the Gospel, why don't you
try to justify Brent's equation to prove your
point, if you can. I see the distance separation
along the equator for two separated galaxies as
linear as the radius of the sphere expands. Brent
uses Hubble's law, but the proof of what you
claim shouldn't depend on Hubble, but just the
geometry. AG
I did multiple times with the balloon analogy which
is purely geometrical, see previous answers.
I don't think so. You just asserted it. AG
The equation that links distance and recession velocity in
both cases comes from the same geometric principles of
uniform expansion in space. The proportionality between
distance and velocity is a natural consequence of how
expansion works, whether it’s on a 2D surface like a balloon
or in 3D space like our universe.
The expansion of the balloon and the universe follow similar
dynamics because, in both cases, the expansion is homogeneous
(the same everywhere) and isotropic (the same in all directions).
If you mark two points close to each other on the balloon and
start inflating it, those two points will move apart slowly.
However, if you mark two points farther apart, they will move
away from each other much more quickly as the balloon expands.
This is what you keep claiming, but have yet to offer a
*_mathematica_l _proof_*. Try this; two galaxies on the equator
of a sphere, with a separation distance s, and the equator
expanding as a function of its radius r to simulate expansion.
The recessional velocity is ds/dt, which depends on dr/dt. If
dr/dt is constant, so will be ds/dt, and the recessional velocity
is constant and cannot reach c or greater. What is wrong with
this proof, falsifying Hubble's law and your model? AGHHubble's
law says the recession velocity is proportional to the distance
so ds/dt=Hs whose solution is s=c*exp(Ht) So s is not constant
and r is not constant. What is constant is H=(1/s)*ds/dt.
