On Wednesday, September 11, 2024 at 6:01:29 PM UTC-6 Brent Meeker wrote:




On 9/11/2024 4:24 PM, Alan Grayson wrote:

On Wednesday, September 11, 2024 at 4:44:43 PM UTC-6 Brent Meeker wrote:




On 9/11/2024 9:04 AM, Alan Grayson wrote:

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 megaparsec 
distant 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 *mathematical 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.


*The phenomenon depends only on geometry, not on Hubble's law. Can you 
prove it without Hubble's law? AG*




*What the hell?  Do you think things are proven from nothing.  You've 
apparently never proven a theorem in you life.  "The phenomena" of  maximum 
observable distance is a consequence of Hubble's law, which  is an 
empirical observation...not an axiom of Euclid. Brent*


*The proof depends on the geometry of expanding circles or spheres (far 
from "nothing"). AG *

 


C'mon AG put some effort into understanding.


 
*Have you googled "chakra"? They're part of your body, but TOTALLY 
UNCONSCIOUS! AG *


Brent


In the same way, in the universe, the farther away a galaxy is, the more 
space there is between us and that galaxy. Since each portion of space is 
expanding, more distant galaxies experience the cumulative effect of the 
expansion over several portions of space. This means that for a galaxy at a 
great distance, the total expansion of space is larger, which results in a 
higher recession velocity.

* John* K Clark    See what's on my new list at  Extropolis 
<https://groups.google.com/g/extropolis>

hwt


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