Le sam. 14 sept. 2024, 11:17, Alan Grayson <[email protected]> a écrit :
> > > On Saturday, September 14, 2024 at 2:57:43 AM UTC-6 Quentin Anciaux wrote: > > > > Le sam. 14 sept. 2024, 09:32, Alan Grayson <[email protected]> a écrit : > > > > On Saturday, September 14, 2024 at 12:40:01 AM UTC-6 Quentin Anciaux wrote: > > > > Le sam. 14 sept. 2024, 07:09, Brent Meeker <[email protected]> a écrit : > > > > > On 9/13/2024 7:58 PM, Alan Grayson wrote: > > > > On Friday, September 13, 2024 at 8:48:25 PM UTC-6 Brent Meeker wrote: > > > > > On 9/13/2024 7:10 PM, Alan Grayson wrote: > > > > On Friday, September 13, 2024 at 4:03:45 PM UTC-6 Brent Meeker wrote: > > > > > On 9/13/2024 4:56 AM, Alan Grayson wrote: > > > > On Friday, September 13, 2024 at 4:06:49 AM UTC-6 Alan Grayson wrote: > > On Thursday, September 12, 2024 at 11:07:49 PM UTC-6 Alan Grayson wrote: > > On Thursday, September 12, 2024 at 11:00:21 PM UTC-6 Brent Meeker wrote: > > > > > On 9/12/2024 9:21 PM, Alan Grayson wrote: > > > > On Thursday, September 12, 2024 at 3:55:45 AM UTC-6 Quentin Anciaux wrote: > > > > Le jeu. 12 sept. 2024, 11:53, Alan Grayson <[email protected]> a écrit : > > > > On Thursday, September 12, 2024 at 2:40:56 AM UTC-6 Quentin Anciaux wrote: > > I just gave you a full proof that as long as the expansion is uniform and > expansion rate > 0, then it follows objects will sooner or later recess > from each other at speed > c. > > > What was the justification for the geometric progression? I made no such > assumption in my "proof". > > > As explained multiple times and in the quote you made, expansion is > uniform and happens at every point in space. > > > What bothers me about your method is that you* assume* a geometric > increase in the separation distance, when, IMO, that's the variable that > must be calculated (which I did). So no matter how many times you affirm > your proof as valid, I can't agree. AG > > > You didn't calculate the expansion parameter, which is the Hubble > constant. It's an observed value. > > Brent > > > Why must I do that, when I just want to show that eventually the > recessional velocity exceeds c? Also, I don't see why theta is fixed, when > the end of the arc defines the position of the receding galaxy. AG > > > Now I am not sure I proved the recessional velocity is greater than c, > after some time has passed. If the sphere is expanding, then the distance > between any two fixed points on the sphere will increase as time passes. > But that was obvious due to the expansion. What's wrong, if anything? AG > > > Now I see the light. We've been struggling to prove that a receding galaxy > will fall out of view if the universe is expanding, but all the so-called > "proofs" fail, but for different reasons. What Quentin offers is not a > proof. He's just repeating a result done by someone else,* using > mathematics*, which he believes (and might be true). Brent is mistaken in > his apparent belief that the proof of concept requires appeal to Hubble's > law. This is also mistaken IMO since the result to be proven depends > *exclusively* on the *geometry *of an expanding sphere. Finally, my proof > also fails, since it's obvious that the arclength, s, between two galaxies > on an expanding sphere, will obviously increase as the sphere expands. > That is, ds/dt will obviously be positive since the arclength is > increasing. IOW, a constantly increasing arclength s, assuming a uniformly > expanding sphere, necessarily yields ds/dt > 0, but it does NOT demonstrate > that the velocity of the receding galaxy eventually increases to be greater > than c. When I have the energy, I will calculate the *second time > derivative* of the arclength, s, hopefully to demonstrate, that for a > uniformly expanding sphere, the *four* terms of the second derivative of > s, imply a* positive acceleration*. This will establish that eventually > the receding galaxy will pass out of view for the observer on the assumed > stationary galaxy. Comments welcome. AG > > It's already proven that ds/dt=ks => s=j*exp(kt) where k with dimensions > of 1/time and j is an arbitrary constant of integration with the same > dimensions as s. Going to second derivatives won't gain any more. > > Brent > > > We have different objectives. Your equation represents Hubble's law, but > what I want to show is that Hubble's law is the inherent result of the > geometry of an expanding sphere. So I believe going to the second > derivative will demonstrate this. BTW, what's your argument that theta is a > constant? AG > > Notice I didn't mention Hubble's law and avoided using H. I deliberately > used j and k for undetermined constants. > > > But you didn't pull that equation out of the proverbial hat. You assumed > the recessional velocity ds/dt depends linearly on s, which is what Hubble > measured. AG > > > Theta's constant under the assumption that it's the angle between two > points that at FIXED on the expanding surface. > > > Yes, the two points are fixed but the distance between them, s. increases > due to the increase in the spatial separation, so I contend that rather > than assuming Hubble's law, I am trying to show that it is a property of an > expanding sphere. This, as I recall, was your claim ages ago when we > discussed this issue. AG > > > But it's not a property of an expanding sphere without the condition that > the expansion has a constant proportional rate; so the relative distances > keep the same proportions. The further away something is the faster it is > moving away. That's why your first assumption ds/dt=const gives a result > inconsistent with Hubble's law, it doesn't keep theta constant for every > point. > > Brent > > > > See my previous answer, the expansion is uniform hence exponential. > > > Your "hence" demands some mathematics! AG > > > Why uniform expansion implies exponential growth > > Uniform expansion does not necessarily mean that the sphere grows > linearly. In fact, uniform expansion implies that the proportion of growth > remains constant at every moment, which is the definition of exponential > growth. If the distance between the points increases proportionally to the > current distance, then we obtain exponential expansion. > > > You keep doing the same thing; asserting a result without proving it. > Please start by *defining *"uniform expansion", mathematically. AG > Defining "Uniform Expansion" Mathematically, uniform expansion means: At each moment, the distance between two points increases by a rate proportional to the current distance. We express this as the following differential equation: d/dt [d(t)] = k * d(t) where: d(t) is the distance between two points at time t, d/dt [d(t)] is the rate of change of the distance, k is a constant proportionality factor (the expansion rate). This equation states that the rate of change of the distance is proportional to the current distance. Solving the Differential Equation Rewriting the equation: d/dt [d(t)] = k * d(t) We can solve this by separating variables: 1 / d(t) * d(d(t)) = k * dt Integrating both sides: ln(d(t)) = k * t + C Exponentiating both sides: d(t) = e^(k * t + C) This simplifies to: d(t) = e^C * e^(k * t) Let d_0 = e^C represent the initial distance between the points at t = 0. The solution becomes: d(t) = d_0 * e^(k * t) Exponential Growth from Uniform Expansion The solution shows that uniform expansion — where the rate of change is proportional to the current distance — leads to exponential growth. The distance between the points increases exponentially over time: d(t) = d_0 * e^(k * t) Summary: "Uniform expansion" means distances grow proportionally to their current size. This leads to the exponential formula d(t) = d_0 * e^(k * t), where k is a constant. As a result, distances increase exponentially over time. > -- > You received this message because you are subscribed to the Google Groups > "Everything List" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected]. > > To view this discussion on the web visit > https://groups.google.com/d/msgid/everything-list/6d97ee23-4ddf-4022-9b99-b8fb0574643a%40gmail.com > <https://groups.google.com/d/msgid/everything-list/6d97ee23-4ddf-4022-9b99-b8fb0574643a%40gmail.com?utm_medium=email&utm_source=footer> > . > > -- > You received this message because you are subscribed to the Google Groups > "Everything List" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected]. > > To view this discussion on the web visit > https://groups.google.com/d/msgid/everything-list/e78e9b3b-9b7f-48e2-a84b-acf41c199b86n%40googlegroups.com > <https://groups.google.com/d/msgid/everything-list/e78e9b3b-9b7f-48e2-a84b-acf41c199b86n%40googlegroups.com?utm_medium=email&utm_source=footer> > . > > -- > You received this message because you are subscribed to the Google Groups > "Everything List" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected]. > To view this discussion on the web visit > https://groups.google.com/d/msgid/everything-list/54814637-3437-4e28-b8af-24aed1916e35n%40googlegroups.com > <https://groups.google.com/d/msgid/everything-list/54814637-3437-4e28-b8af-24aed1916e35n%40googlegroups.com?utm_medium=email&utm_source=footer> > . > -- You received this message because you are subscribed to the Google Groups "Everything List" group. 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