Questions about the accelerating Hubble expansion

[Kairos]

I don't understand the discovery of "accelerating expansion" in 1998 (hence the postulate of dark energy, etc..), because Hubble's old law already described exponentially accelerating expansion in 1929, right?

 

[Orodruin]

Hubble's old law already described exponentially accelerating expansion in 1929, right?

No. All Hubble’s law says is that velocity is proportional to distance. This is true in any expanding FLRW universe.

 

[Ibix]

Hubble's old law already described exponentially accelerating expansion in 1929, right?

Under Hubble's law the recession speed of objects varies with distance, but it may or may not vary with time. Whether it varies with time and how depends on the mix of matter, radiation and dark energy in the universe. You cannot match the observed history of the expansion rate without some dark energy, and the observed mix leads to a Hubble constant that increases with time.

 

[Kairos]

Velocity dx(t)/dt is proportional to distance x(t), so dx(t)/dt=k*x(t), so x(t)=exp(k*t), which accelerates with time? without the need to increase k

 

[Orodruin]

the observed mix leads to a Hubble constant that increases with time.

This is incorrect. The Hubble constant decreases with time towards a constant value in a universe dominated by a cosmological constant.

An accelerated universe refers to a universe where the second derivative of the scale factor, $\ddot a$, is larger than zero. This is not the same thing as $dH/dt > 0$ as $H = \dot a/a$ implies
$$
\frac{dH}{dt} = \frac{\ddot a}{a} - \frac{\dot a^2}{a^2} = \frac{\ddot a}{a} - H^2
$$
Thus, the Hubble constant can indeed be negative even if $\ddot a > 0$.

 

[PeterDonis]

Velocity dx(t)/dt is proportional to distance x(t), so dx(t)/dt=k*x(t), so x(t)=exp(k*t)

Wrong, because $k$ is not constant in time--the proportionality between velocity and distance changes with time.

 

[Kairos]

of course, but that's not relevant to my question.

 

[Ibix]

An accelerated universe refers to a universe where the second derivative of the scale factor, $\ddot a$, is larger than zero.

Ah - that makes sense. Thanks.

 

[PeterDonis]

of course, but that's not relevant to my question.

Are you responding to my post #6? If so, you are wrong, the fact that $k$ is not a constant is highly relevant since it makes your claimed mathematical solution in terms of an exponential wrong. And that claimed mathematical solution is the basis of your claim in the OP.

 

[Ibix]

of course, but that's not relevant to my question.

See @Orodruin's reply to me - it's about the sign of $\ddot a$, not $\ddot x$.

 

[Kairos]

k is not a constant is highly relevant since it makes your claimed mathematical solution wrong

This is certainly very important, but I pointed out that even if k had been constant, the expansion would still have accelerated.

 

[PeterDonis]

Under Hubble's law the recession speed of objects varies with distance, but it may or may not vary with time. Whether it varies with time and how depends on the mix of matter, radiation and dark energy in the universe.

Actually, it does vary with time in every FRW solution except a pure de Sitter spacetime, i.e., pure dark energy, no matter or radiation at all.

 

[PeterDonis]

I pointed out that even if k had been constant, the expansion would still have accelerated.

You're looking at it backwards. The case with $k$ constant is a pure de Sitter spacetime, just dark energy (aka cosmological constant), no matter or radiation at all. This is the "pure" case of "accelerated expansion". Adding matter or radiation makes $k$ not constant, and may take away the acceleration, depending on the relative densities of matter/radiation as compared to dark energy. Until a few billion years ago, the density of matter in our universe was high enough that the expansion was decelerating, not accelerating.

Up until the late 1990s, cosmologists thought the expansion of our universe was decelerating now because they thought there was zero dark energy. That decelerating solution is the original one that Hubble's evidence was taken to support. What changed in the late 1990s was that evidence was discovered that the expansion of our universe is accelerating now, and had been until a few billion years ago. That required adding nonzero dark energy to our best current model in order to account for the data.

 

[Kairos]

You're looking at it backwards. The case with $k$ constant is a pure de Sitter spacetime, just dark energy (aka cosmological constant), no matter or radiation at all. This is the "pure" case of "accelerated expansion". Adding matter or radiation makes $k$ not constant, and may take away the acceleration, depending on the relative densities of matter/radiation as compared to dark energy. Until a few billion years ago, the density of matter in our universe was high enough that the expansion was decelerating, not accelerating.

Up until the late 1990s, cosmologists thought the expansion of our universe was decelerating now because they thought there was zero dark energy. That decelerating solution is the original one that Hubble's evidence was taken to support. What changed in the late 1990s was that evidence was discovered that the expansion of our universe is accelerating now, and had been until a few billion years ago. That required adding nonzero dark energy to our best current model in order to account for the data.

Hubble's law certainly describes well de Sitter's universe, but beyond the details, I'm just saying that this law describes an acceleration of expansion. You seem to agree?

 

[PeterDonis]

Hubble's law certainly describes well de Sitter's universe, but beyond the details, I'm just saying that this law describes an acceleration of expansion. You seem to agree?

No. Hubble's law, as @Orodruin has already told you, is true in any expanding FRW universe, not just de Sitter spacetime; de Sitter spacetime is the particular solution where the proportionality between velocity and distance, what you call $k$, is not a function of $t$. If $k$ is a function of $t$, you do not get exponential expansion. You might not even get accelerating expansion at all. But you will still get a solution in which Hubble's law is true.

 

[Orodruin]

Hubble's law certainly describes well de Sitter's universe, but beyond the details, I'm just saying that this law describes an acceleration of expansion. You seem to agree?

Hubble’s original law said nothing about the time dependence of the proportionality constant. Just that observed velocity now is proportional to distance now. As such, it would allow any type of expansion. Accelerated, constant, or decelerated.

 

[Kairos]

particular solution where the proportionality between velocity and distance, what you call k, is not a function of t.

OK, if k is a constant of proportionality, then the expansion is exponential and therefore accelerated.

 

[Orodruin]

OK, if k is a constant of proportionality, then the expansion is exponential and therefore accelerated.

You are completely missing the point.

While k could technically be a constant in time - Hubble's original law says nothing about this. It only states that, for relatively nearby galaxies, velocity is proportional to distance. The data Hubble had available only contained relatively nearby objects and so he could have concluded nothing about the time dependence of the constant that would later bear his name ... and not be a constant. In fact, in modern cosmology, it is often referred to as the Hubble parameter or Hubble rate - not the Hubble constant.

 

[Orodruin]

And to continue on that theme. People contemporary with Hubble would most likely have assumed it to be decreasing with time simply due to gravitational attraction and - in the absence of gravity - that the objects would end up further and further away, thereby decreasing the constant.

 

[PeterDonis]

OK, if k is a constant of proportionality, then the expansion is exponential and therefore accelerated.

You are not even responding to the point I made in post #15. Go back and read it again. Carefully.

 

[Kairos]

You are completely missing the point.

While k could technically be a constant in time - Hubble's original law says nothing about this. It only states that, for relatively nearby galaxies, velocity is proportional to distance. The data Hubble had available only contained relatively nearby objects and so he could have concluded nothing about the time dependence of the constant that would later bear his name ... and not be a constant. In fact, in modern cosmology, it is often referred to as the Hubble parameter or Hubble rate - not the Hubble constant.

If k is not constant, this changes everything (but can we still talk about proportionality?).
So how does the Hubble constant evolve over time according to the 1998 discovery?

 

[PeterDonis]

If k is not constant, this changes everything

More precisely, if k is not constant in time. Which it isn't in our universe, as you were already told in posts #3, #5, #6, #9, #13, #16, #18, and #19.

(but can we still talk about proportionality?).

Proportionality with respect to distance at an instant of time (more precisely an instant of FRW coordinate time).

So how does the Hubble constant evolve over time according to the 1998 discovery?

You were already told that in post #5. Go back and read it.

 

[PeterDonis]

the observed mix leads to a Hubble constant that increases with time.

No, decreases with time. See @Orodruin's post #5. What increases with time is $\dot{a}$. But $H$ is $\dot{a} / a$, and that decreases with time (asymptotically towards the constant de Sitter value associated with the dark energy density).

 

[Kairos]

And to continue on that theme. People contemporary with Hubble would most likely have assumed it to be decreasing with time simply due to gravitational attraction and - in the absence of gravity - that the objects would end up further and further away, thereby decreasing the constant.

Orodruin, Reading other threads on the same topic, I read in one of your posts, "In fact, in a dark energy dominated universe, H is constant.
I get it!

 

[PeterDonis]

I read in one of your posts, "In fact, in a dark energy dominated universe, H is constant.

You need to quote the specific post so we can see the context. The term "dark energy dominated" is ambiguous: many cosmologists describe our current universe as "dark energy dominated" because the expansion is accelerating, even though $H$ is not constant with time in our current universe, it is decreasing, as @Orodruin said in post #5 of this thread.

 

[Orodruin]

... and given enough time to evolve, it will approach a constant value given by the cosmological constant.

It should be noted that there are ways to achieve an increasing value of $H$, but it would require an energy component with an equation of state for which $w < -1$, i.e., pressure needs to be negative and larger in magnitude than the energy density.

 

[PeterDonis]

... and given enough time to evolve, it will approach a constant value given by the cosmological constant.

Yes, and I suspect that's what you were actually saying in whatever other thread the OP read and quoted from out of context in post #24.

 

[Orodruin]

Yes, and I suspect that's what you were actually saying in whatever other thread the OP read and quoted from out of context in post #24.

... or it was not worth making the distinction of being dominated by a cosmological constant or just having a cosmological constant ... I have probably said things like that a lot of times over the years.

 

[Kairos]

... and given enough time to evolve, it will approach a constant value given by the cosmological constant.

If H will approaches a constant value given by the cosmological constant, there will still be a continuous acceleration ?(exponential).

 

[Orodruin]

If H will approaches a constant value given by the cosmological constant, there will still be a continuous acceleration ?(exponential).

...

An accelerated universe refers to a universe where the second derivative of the scale factor, $\ddot a$, is larger than zero. This is not the same thing as $dH/dt > 0$ as $H = \dot a/a$ implies
$$
\frac{dH}{dt} = \frac{\ddot a}{a} - \frac{\dot a^2}{a^2} = \frac{\ddot a}{a} - H^2
$$
Thus, the Hubble constant can indeed be negative even if $\ddot a > 0$.

If $dH/dt = 0$ then clearly $\ddot a = aH^2 = \dot a^2/a$ and therefore the universe is accelerating.

This is also clear from the scale factor being on the form $a = e^{Ht}$ meaning that $\dot a = Ha$ and therefore $\ddot a = H^2 a$ for constant $H$.

 

[PeterDonis]

If H will approaches a constant value given by the cosmological constant, there will still be a continuous acceleration ?(exponential).

If $H$ is asymptotically approaching a constant value, the expansion of the universe is asymptotically approaching exponential. At what point during that process the expansion becomes accelerating depends on the details, but it will become accelerating at some point, and once it does, it will stay accelerating.

 

[DAH]

In a spatially flat universe the Hubble parameter is given by $$H^2=\frac{8\pi G(\rho_m+\rho_r)}{3}+\frac{\Lambda c^2}{3}$$
$\rho_m$ and $\rho_r$ decrease with time and $\Lambda$ stays constant, so the Hubble parameter decreases with time. In the remote future $\rho_m$ and $\rho_r$ will tend to zero and the Hubble parameter will become a constant given by $$H=\sqrt{\frac{\Lambda c^2}{3}}$$

 

[Kairos]

In a spatially flat universe the Hubble parameter is given by $$H^2=\frac{8\pi G(\rho_m+\rho_r)}{3}+\frac{\Lambda c^2}{3}$$
$\rho_m$ and $\rho_r$ decrease with time and $\Lambda$ stays constant, so the Hubble parameter decreases with time. In the remote future $\rho_m$ and $\rho_r$ will tend to zero and the Hubble parameter will become a constant given by $$H=\sqrt{\frac{\Lambda c^2}{3}}$$

Thank you. I understand that the densities decrease and lambda is left alone at the end. What I don't understand is why rho and lambda affect H in the same direction. I would have thought - and + signs since lambda increases expansion while rho slows it down.

 

[PeterDonis]

What I don't understand is why rho and lambda affect H in the same direction. I would have thought - and + signs since lambda increases expansion while rho slows it down.

First, what you mean by "expansion" here is not $H$, its $\dot{a}$. They're not the same.

Second, when you say "lambda increases expansion while rho slows it down", what you mean is that dark energy, by itself, causes expansion to accelerate, while rho, by itself, causes expansion to decelerate. But the equation for $H^2$, by itself, tells you nothing about acceleration or deceleration. What you need to look at is the equation for $\ddot{a}$.

 

[Kairos]

First, what you mean by "expansion" here is not $H$, its $\dot{a}$. They're not the same.

Second, when you say "lambda increases expansion while rho slows it down", what you mean is that dark energy, by itself, causes expansion to accelerate, while rho, by itself, causes expansion to decelerate. But the equation for $H^2$, by itself, tells you nothing about acceleration or deceleration. What you need to look at is the equation for $\ddot{a}$.

yes, that's why I spoke of expansion in this post, and not acceleration. dark energy favors expansion contrary to gravity