Cont: Why James Webb Telescope rewrites/doesn't the laws of Physics/Redshifts (2)

[W.D.Clinger]

If that Hubble parameter diminishes to some constant H₁, then from that time onward we have

da/dt = H₁ a(t)​

from which it follows that

a(t) = e^{H₁ t} + C​

where C is some constant.

So from t = ∞ and onward?

As predicted:

(Given Mike Helland's struggles with calculus, he probably won't understand that either.)

 

[Steve]

So from t = ∞ and onward?

Sure thing, Buzz Lightyear.

 

[Mike Helland]

Sure thing, Buzz Lightyear.

That's why it's called "asymptotically de Sitter".

de Sitter means exponentially expanding, which means it has a constant expansion rate.

Hubble's parameter won't diminish to a value and stay there until t = ∞.

I'm not sure how Ethan got the value of 45 though.

If you do http://latex.codecogs.com/gif.latex?H = H_0 \sqrt{\Omega_\Lambda}, \Omega_\Lambda = 0.68, H_0 = 67.4 km/s/Mpc I get 55 km/s/Mpc.

If I forget to do the square root, which I usually do, I get 45. I wonder if that's what happened.

 

[W.D.Clinger]

If you do http://latex.codecogs.com/gif.latex?H = H_0 \sqrt{\Omega_\Lambda}, \Omega_\Lambda = 0.68, H_0 = 67.4 km/s/Mpc I get 55 km/s/Mpc.

That equation incorrectly assumes Ω_M,0 = 0.

You can correct that equation by replacing its H₀ by the value of the Hubble parameter H(t) for some future time t at which the mass-density term has become negligible compared to the dark energy term. Since H(t) is decreasing due to the gravitational effect Mike Helland has been at such pains to deny, that correction lowers the asymptotic value of the Hubble parameter below the value incorrectly computed by Mike Helland.

ETA: The paragraph above does not fully correct the equation, but I'm not striking it out because the paragraph does allude to the intuitive reason(s) Mike Helland's equation was incorrect. I'll try to get back to this sometime tomorrow.

 

[W.D.Clinger]

By way of apology to Mike Helland, I will derive his equation for the asymptotic value of the Hubble parameter H(t).

By definition, H(t) = (da/dt) / a(t), where a(t) is the scale factor. That gives us a differential equation

da/dt = H(t) a(t)​

If the Hubble parameter had a constant value H₁, then a(t) would have to be an exponential function of time:

a(t) = e^{H₁ t + C}​

where C is some constant. (In a previous post, I misplaced that constant C. It needs to be part of the exponent.)

According to mainstream cosmology, the Hubble parameter is not a constant, but it is expected to converge toward a constant H₁. To find the value of H₁ as predicted by mainstream cosmology, we need to start with the first Friedmann equation, which can be rearranged to obtain the following equation for the Hubble parameter H(t):

H(t) = sqrt( (8πGρ/3) − (kc²/(a²)) + (Λc²/3) )​

That equation can be rewritten in terms of density parameters such as Ω_M and Ω_Λ.

That rewritten equation is a bit messy. To derive Mike Helland's equation, we need to make several simplifying assumptions that Mike Helland neglected to mention. One thing we need to assume is an FLRW model for an expanding universe. We also need to assume the parameters of that model are such that the universe continues to expand forever. We will also need to make several more assumptions about the parameters of that model. According to mainstream cosmology, the curvature of our universe is very nearly zero and the values of density parameters other than Ω_M and Ω_Λ are also very small. If we assume the curvature is exactly zero and those other density parameters are also exactly zero, the equation for H(t) simplifies to

H(t) = H₀ sqrt( Ω_M (a(t))^-3 + Ω_Λ )​

We can then find the value of H₁ by taking the limit of that function as t increases without bound. With all of the assumptions we're making, the value of that limit will be the same as the limit of that function as the scale factor a(t) increases without bound. As a(t) increases, the Ω_M (a(t))^-3 term converges to zero, so the limit of H(t) is given by

H₁ = H₀ sqrt( Ω_Λ )​

which is Mike Helland's equation.

So Mike Helland's equation turns out to be correct after all, if we make all of the assumptions he failed to mention.

ETA:

The reason Mike Helland's equation looked wrong to me is that Ω_Λ = 0 would imply the Hubble parameter H(t) will eventually converge to zero.

But here's the catch: Mike Helland's equation depends upon the key assumptions of flatness and the assumption that the density parameters add up to 1. Under those assumptions, Ω_Λ = 0 implies Ω_M = 1, which is the critical density for a universe that converges toward a flat universe that is neither expanding nor contracting. If Ω_Λ = 0 and Ω_M < 1, then H(t) converges to a positive value (contrary to Mike Helland's equation). If Ω_Λ = 0 and Ω_M > 1, then H(t) eventually goes negative. But both of those possibilities are ruled out by the assumptions Mike Helland made but failed to state.

If we further assume H₀ = 67 km/s/Mpc and Ω_Λ = 0.675, we get

H₁ ≈ 55 km/s/Mpc​

If we instead assume H₀ = 74 km/s/Mpc and Ω_Λ = 0.72, we get

H₁ ≈ 62.8 km/s/Mpc​

The asymptotic value of the Hubble parameter H(t) probably lies somewhere between those numbers. The lower value corresponds to an older estimate for the age of the universe, while the higher value corresponds to a younger estimate.

Helland physics predicts H₀ = H₁ = 0 km/s/Mpc. I should emphasize that Helland physics rejects all FLRW models, so Helland physics rejects almost all of the assumptions that were necessary to derive Mike Helland's equation for H₁.

 

[Mike Helland]

So Mike Helland's equation turns out to be correct after all, if we make all of the assumptions he failed to mention.

If you include http://latex.codecogs.com/gif.latex?\Omega_r a(t)^{-4} + \Omega_k a(t)^{-2} you get the same equation as t goes to infinity.

The only assumption one needs to make is we're talking about an FLRW model.

Your "apologies" always seem to include the type of off-handed, back-handed remarks that lead to you having to "apologize" to begin with. You could save yourself the trouble and just be less rude to begin with.

 

[W.D.Clinger]

The only assumption one needs to make is we're talking about an FLRW model.

False.

That sentence is so false that it could not have been written by anyone who understood the family of FLRW models.

 

[Mike Helland]

False.

That sentence is so false that it could not have been written by anyone who understood the family of FLRW models.

I'll bite.

How is it that http://latex.codecogs.com/gif.latex?\Omega_m a(t)^{-3} = 0 when http://latex.codecogs.com/gif.latex?t = \infty but http://latex.codecogs.com/gif.latex?\Omega_r a(t)^{-4} + \Omega_k a(t)^{-2} doesn't?

 

[W.D.Clinger]

The only assumption one needs to make is we're talking about an FLRW model.

False.

That sentence is so false that it could not have been written by anyone who understood the family of FLRW models.

I'll bite.

How is it that http://latex.codecogs.com/gif.latex?\Omega_m a(t)^{-3} = 0 when http://latex.codecogs.com/gif.latex?t = \infty but http://latex.codecogs.com/gif.latex?\Omega_r a(t)^{-4} + \Omega_k a(t)^{-2} doesn't?

The following exercise will answer your question.

Exercise. Give an example of an FLRW model for which the scale factor a(t) is bounded, Ω_M = 0, and Ω_r is positive.​

Here are a couple of other exercises that would improve your understanding of FLRW models.

Exercise. Give an example of an FLRW model for which the scale factor a(t) is constant, Ω_M > 0, and Ω_k is nonzero.​

Exercise. Give an example of an FLRW model for which the scale factor a(t) converges to zero and Ω_M > 0.​

All of those examples will refute your claim that your equation would still hold when "The only assumption one needs to make is we're talking about an FLRW model."

 

[Mike Helland]

The following exercise will answer your question.

Exercise. Give an example of an FLRW model for which the scale factor a(t) is bounded, Ω_M = 0, and Ω_r is positive.​

Here are a couple of other exercises that would improve your understanding of FLRW models.

Exercise. Give an example of an FLRW model for which the scale factor a(t) is constant, Ω_M > 0, and Ω_k is nonzero.​

Exercise. Give an example of an FLRW model for which the scale factor a(t) converges to zero and Ω_M > 0.​

All of those examples will refute your claim that your equation would still hold when "The only assumption one needs to make is we're talking about an FLRW model."

I see what you're getting at.

1/a(t) doesn't go to all zero for all FLRW models. If 1/a(t) is not zero at t=infinity, obviously that equation doesn't simplify to just http://latex.codecogs.com/gif.latex?H_0 \sqrt{\Omega_\Lambda}.

The question was:

Assuming LCDM is asymptotically de Sitter, what expansion rate does it approach?

I came up with the right answer. Ethan appears to be wrong.

To get the right answer to the question you should plug LCDM's parameters into the Friedmann equation, and figure out what a(t) is at infinity and plug that in.

All I did is set 1/a(t) = 0 in the Friedmann equation (when written in terms of http://latex.codecogs.com/gif.latex?\Omega_m, \Omega_r, \Omega_k, \Omega_\Lambda).

The only assumptions made to get the answer are pretty much stated in the question.

 

[W.D.Clinger]

I see what you're getting at.

1/a(t) doesn't go to all zero for all FLRW models. If 1/a(t) is not zero at t=infinity, obviously that equation doesn't simplify to just http://latex.codecogs.com/gif.latex?H_0 \sqrt{\Omega_\Lambda}.

The question was:

Assuming LCDM is asymptotically de Sitter, what expansion rate does it approach?

So you were assuming an FLRW model that is asymptotically de Sitter.

You failed to mention that assumption until after you had made your ridiculous claim that

Our "accelerating universe" will never be expanding faster than a universe with a constant expansion rate of about 50 km/s/Mpc.

You got that completely backwards.

The only assumptions made to get the answer are pretty much stated in the question.

No one asked a question that implied those assumptions until you framed the question yourself less than half an hour ago, as I quoted you above.

 

[Mike Helland]

So you were assuming an FLRW model that is asymptotically de Sitter.

I was describing LCDM. So yes.

You got that completely backwards.

Probably.

If LCDM is accelerating, and is also approaching a de Sitter universe... wouldn't it be always slower (and thus never faster) than dS?

Hubble's parameter in LCDM will always be higher than the dS value it approaches. But it will also have some gravity pulling back on it, where dS does not.

 

[W.D.Clinger]

So you were assuming an FLRW model that is asymptotically de Sitter.

I was describing LCDM. So yes.

Most LCDM models are not asymptotically de Sitter.

Indeed, the Einstein-de Sitter universe, which is not asymptotically de Sitter, was quite popular among cosmologists until better data and theories ruled it out during the 1990s.

It is quite a remarkable coincidence that the universe in which we live appears to approximate an asymptotically de Sitter universe. Mainstream cosmology has no good answer to why that should be so.

You got that completely backwards.

Probably.

If LCDM is accelerating, and is also approaching a de Sitter universe... wouldn't it be always slower (and thus never faster) than dS?

No. Your unfamiliarity with differential equations is showing.

I suspect you are assuming similar initial conditions for both as of the early universe. You should instead be assuming similar initial conditions for both as of now.

Hubble's parameter in LCDM will always be higher than the dS value it approaches. But it will also have some gravity pulling back on it, where dS does not.

That is true for mainstream cosmology's current estimates of LCDM parameters.

But you don't appear to understand the mathematical and physical consequences of that.

 

[Mike Helland]

But you don't appear to understand the mathematical and physical consequences of that.

It just seems odd the universe expands faster and faster, but will never be as slow as dS.

 

[Mike Helland]

https://bigthink.com/starts-with-a-bang/universe-expansion-not-accelerating/

Even though the expansion rate — also known as the Hubble constant/parameter — still decreases, for the past ~6 billion years it’s been decreasing at a slow enough rate that as the volume of the Universe grows, these same distant objects now appear to recede away from us faster and faster; they’re now moving away from us in an accelerated fashion.

The Universe is expanding, the expansion rate is dropping, but it’s not dropping to zero; it’s in the process of asymptoting to a final value that’s only about 30% lower than its current value today. However, each individual object that’s receding from us will recede at faster and faster speeds as time goes on. Importantly, this implies that the recessional velocity of each galaxy is accelerating, but the expansion rate itself is not; it’s decreasing. It’s a challenging misconception to overcome, but hopefully now — armed with an in-depth explanation in plain English — you’ll understand that the objects within the Universe are accelerating, but the expansion rate of the Universe is not!

There seem to be two meanings of "accelerating" misconstrued here.

The farther an object gets, the faster its velocity gets, so the farther it gets.

That's true in an exponentially expanding universe, which is matterlesss, that has a constant expansion rate, which has a second derivative that equals zero.

Our universe, has matter, that pulls back on expansion.

The farther and object gets, the faster its velocity gets, so the farther it gets. But, thanks to gravity, not as fast as an exponentially expanding universe.

But thanks to dark energy, as the matter density drops over time, it gets closer and closer to the speed of an exponentially expanding universe. Its second derivative is positive.

I was under the impression that's what "accelerating" referred to. In any case, there doesn't seem to be much logical sense in a universe that is accelerating toward a slower model.

The most logical interpretation is that our universe will never be expanding as fast as an exponentially expanding universe.

A car moving at a constant 55 mph is moving faster than a car that's accelerating to its top speed of 55 mph at t = infinity.

 

[W.D.Clinger]

It just seems odd the universe expands faster and faster, but will never be as slow as dS.

These things will be much easier to explain to you if you ever get around to taking a course in first semester calculus.

There seem to be two meanings of "accelerating" misconstrued here.

Misconstruing is something you're very good at.

da/dt and (da/dt)/a do mean different things. If you understood first semester calculus, and weren't so determined to avoid giving any thought to the scale factor a(t), you'd understand what those two different things mean.

The farther an object gets, the faster its velocity gets, so the farther it gets.

That sentence expresses a lot of confusion.

The farther an object is, the faster its velocity. That fact is quantified by (da/dt)/a.

For any nonzero velocity, "the farther it gets" will be true in the direction of that velocity. That's true regardless of whether it is accelerating or decelerating or neither. Your confusion about this is highlighted by your very next sentence.

That's true in an exponentially expanding universe, which is matterlesss, that has a constant expansion rate, which has a second derivative that equals zero.

It's true in any universe that is either expanding or contracting, regardless of whether the universe contains any matter other than the object whose velocity we're discussing, regardless of whether the expansion rate is constant, and regardless of whether a second derivative equals zero.

In other words, everything you wrote in that sentence highlights your confusion.

Our universe, has matter, that pulls back on expansion.

That is true. You have denied that fact in the recent past, so it's good to see you are accepting that fact now.

The farther and object gets, the faster its velocity gets, so the farther it gets.

As I explained above, that sentence expresses a lot of confusion.

That confusion led you to the false conclusion you stated in your very next sentence.

But, thanks to gravity, not as fast as an exponentially expanding universe.

No. You got that wrong.

Had you said "thanks to gravity, not as fast as without gravity", you'd have been saying something true. But that's not what you wrote.

If you had a serious degree in computer science, you would understand the word "exponential", because you would have taken courses that discussed the difference between Θ(2^x) and other asymptotic complexity classes. But the fact that two functions f and g both belong to the class Θ(2^x) does not imply f(x) grows as fast as g(x). It means only that both f and g are asymptotically roughly proportional to the function 2^x (where the word "roughly" has a precise mathematical meaning I won't bother to state here).

Furthermore, the fact that a function f belongs to Θ(2^x) and some function g belongs to Θ(x³) (for example) does not mean f(x) > g(x) for all x.

In the situation you're trying to understand here, there are two functions f and g, both of which are asymptotically exponential. There is a particular time t₀ at which f(t₀) > g(t₀), but the second derivative of f is negative at t₀ while the second derivative of g is g itself times a constant.

ETA: That paragraph refers to scale factors rather than Hubble parameters. I should warn that the time t₀ is not the present time, but was back when the universe corresponding to f was dominated by matter.

You appear to be unable to understand the situation described within the previous paragraph. Your inability to understand that paragraph is due to your limited understanding of mathematics at the level of asymptotic complexity and calculus. Your inability to understand that paragraph has nothing to do with cosmology or general relativity or physics, other than the fact that your inability to understand cosmology and general relativity and physics can be traced to the same limited understanding of mathematics that prevents you from understanding the previous paragraph.

But thanks to dark energy, as the matter density drops over time, it gets closer and closer to the speed of an exponentially expanding universe. Its second derivative is positive.

No.

The expansion rate H(t) drops closer and closer to the expansion rate of an exponentially expanding universe because the first derivative of H(t) is negative, and the first derivative of H(t) is negative because the second derivative of a(t) is negative.

After all, the sign of the first derivative of H(t) will always be the same as negative when the sign of the second derivative of a(t) is negative. If you understood calculus and were willing to understand the definition of H(t), that fact would be obvious to you. But it isn't obvious to you, because you don't understand calculus, and because you have put serious effort into failing to understand the definition of the Hubble parameter H(t).

ETA: dH/dt = (d²a/dt²)/a − H²
I should clarify that the second derivative of a(t) was negative in the matter-dominated past, but is unlikely to be negative today even though the first derivative of H(t) remains negative.

I was under the impression that's what "accelerating" referred to. In any case, there doesn't seem to be much logical sense in a universe that is accelerating toward a slower model.

Your impressions mislead you because you don't understand calculus and you don't understand asymptotic complexity.

The most logical interpretation is that our universe will never be expanding as fast as an exponentially expanding universe.

No. You continue to get that completely backwards.

FLRW models of our universe are expanding faster than the de Sitter models to which those models' Hubble parameter is converging. That is true now and has been true since time began. It will continue to be true for all time, because although the Hubble parameter of each FLRW model converges to the constant Hubble parameter of the corresponding de Sitter model, it will never quite equal it.

(I spoke of models in that paragraph because the FLRW and de Sitter models are mathematically precise models of which I can state mathematically precise properties.)

A car moving at a constant 55 mph is moving faster than a car that's accelerating to its top speed of 55 mph at t = infinity.

But a car moving at a constant 55 mph is moving slower than a car that's decelerating toward its eventual constant speed of 55 mph.

The analogy you chose to use encapsulates your misunderstanding of the cosmological situation.

 

[Mike Helland]

FLRW models of our universe are expanding faster than the de Sitter models to which those models' Hubble parameter is converging.

Ok.

Just seems funny to say it's speeding up and slowing down.

 

[W.D.Clinger]

FLRW models of our universe are expanding faster than the de Sitter models to which those models' Hubble parameter is converging.

Ok.

Just seems funny to say it's speeding up and slowing down.

When discussing a situation in which some things are speeding up and others are slowing down, and some things were slowing down in the past but are now speeding up, or vice versa, using the same pronoun "it" to refer to all of those times and things does contribute to confusion.

 

[steenkh]

Ok.

Just seems funny to say it's speeding up and slowing down.

Isn’t it the case that the acceleration itself is slowing down, but it is still accelerating?

 

[Mike Helland]

Isn’t it the case that the acceleration itself is slowing down, but it is still accelerating?

I presume I'm supposed to be the one to answer that, so others can jump on perceived mistakes I've made in my answer.

I'll see what I can do.

In the meantime, here's my new video that explains the alternative model.

https://www.youtube.com/watch?v=lVoHmVafTns



Here's a question. Suppose we receive a light pulse from z=1, and suppose the source of that light pulse has emitted another one every year since the that one.

Let's assume a dS universe for simplicity, and 1/H = 14 billion years.

There are only 7 billion pulses in flight, between us and the source. Expansion has spread the ones closest to us by a factor of 2. If you count for how much expansion has spread out those 7 billion pulses, there would be 9.7 billion pulses evenly spaced by 1 year "in flight".

So we say z=1 has a lookback time of 9.7 billion years.

But there's really only 7 billion years of pulses that have been emitted. Right?

Does the "lookback time" actually refer to how far back in time we're looking? Or is it how far we back we're looking, plus the effects of time dilation?

It doesn't seem then, like saying these z>10 galaxies are within a billion years of the big bang would be right. If you consider how many things happened, sans time dilation, we're not actually back that far. Are we?