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Or has it been as long as six billion years? Over the last few years I keep seeing longer and longer numbers...

Have researchers noticed something different recently? Or found a mistake in their previous calculation(s) or theory?

Or has it been six billion years since a noticeable increase in the rate of deceleration due to (we think) dark energy, and four billion years since a change from deceleration to actual acceleration?

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  • $\begingroup$ See How does the Hubble parameter change with the age of the universe? for full details on how the expansion has changed over time. The transition from deceleration to acceleration is the point of inflection on the a(t) graph. $\endgroup$ Sep 19, 2023 at 6:10
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    $\begingroup$ Over the last few years I keep seeing longer and longer numbers... that is to be expected. Wait another billion years and the number will be even higher. $\endgroup$
    – Falco
    Sep 19, 2023 at 8:46

2 Answers 2

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It is likely that you or your sources are conflating two different events.

  • Dark energy begins to comprise the majority of the Universe's energy density about 4 billion years ago.
  • The expansion of the Universe transitions from deceleration to acceleration about 6 billion years ago.

Here I assume $\Lambda$CDM with Planck cosmological parameters.

Why do these events happen at different times? The simple answer is that dark energy's repulsive gravity is twice as efficient as matter's attractive gravity, so acceleration begins when the dark energy density is only half the matter density.

More precisely, we can examine the second Friedmann equation, which describes how the cosmic expansion factor $a$ accelerates or decelerates in response to the energy content of the universe. If the Universe contains matter with density $\rho_m$ and dark energy at density $\rho_\Lambda$, then the matter contributes 0 pressure while the dark energy contributes pressure $p=-\rho_\Lambda c^2$, and the equation becomes

$$\frac{\ddot{a}}{a} = -\frac{4 \pi G}{3}\left(\rho_m-2\rho_\Lambda\right).$$

If $\rho_\Lambda<\rho_m/2$, then $\ddot{a}<0$, i.e. the expansion decelerates. If $\rho_\Lambda>\rho_m/2$, then $\ddot{a}>0$, i.e. the expansion accelerates.

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If we consider that the universe is Euclidean and that the radiation density to be negligible compared to that of matter and dark energy, the instant when the expansion shifted from deceleration to acceleration occurred when $\rho_M \simeq 2\rho_{\Lambda}$ as our colleague @sten explains.

However, this still doesn't answer the question posed by our colleague @Kurt Hikes when he started the thread:

When did that happen? 4 billion years ago? 4.5? 5?

What follows, is the detailed calculation of that moment in time. According to the ΛCDM model, the evolution of the expansion of the universe is given by the time evolution of the scale factor $a=a(t)$

The expansion was decelerated as long as the second derivative of this function was negative (the function was concave) and became accelerated when the second derivative became positive (convex function).

Time evolution of scale factor

The transition point is an inflection point of the function (black dot) and we know from high school that at this point the second derivative is zero.

If we had the explicit expression of the scale factor "$a$" as a function of time "$t$"

$a=a(t)$

we would derive 2 times, equal zero and clear the value of the time "$t$" at which the acceleration was zero. There is no explicit function in the general case, but there is an alternative method to find the second derivative, that is using the generalised Hubble - Lemaître Law, a law deduced from the Friedmann Equations and which states:

$$\dot a=H \ a \tag 1$$

$H$ is the Hubble Parameter and is a function of time. We use the usual convention that one point on a variable is its first time derivative, and two points the second derivative with respect to time. We derive that product:

$\ddot a=\dot H \ a + H \ \dot a=\dot H \ a + H \ (H \ a)$

$$\ddot a=(\dot H+H^2) \ a \tag 2$$

There is also no explicit expression of the Hubble Parameter $H$ as a function of time, but there is an explicit expression as a function of the scale factor:

$$H=H_0 \sqrt{\Omega_{R_0} a^{-4}+\Omega_{M_0} a^{-3}+\Omega_{K_0} a^{-2}+\Omega_{\Lambda_0}}\tag 3$$

$H_0$ is the Hubble Constant = Hubble Parameter at the current time and the Omegas are the current density ratios of the energies that compose the Universe, respectively: Radiation R, Matter (dark + baryonic) M, and Dark Energy Λ.

$\Omega_{K_0}=1-\Omega_{R_0}-\Omega_{M_0}-\Omega_{\Lambda_0}$

Deriving $H$ with respect to time by applying the chain rule, (derivative of a function of function) we obtain:

$\dot H=\dfrac{H_0^2 (-4\Omega_{R_0} a^{-5} \ \dot a-3\Omega_{M_0} a^{-4} \ \dot a-2\Omega_{K_0} a^{-3} \ \dot a)}{2 H}$

Substituting $\dot a$ according to (1) and simplifying we obtain the derivative of the Hubble Parameter:

$$\dot H=H_0^2 \ \left (-2\Omega_{R_0} a^{-4}-\dfrac 3 2 \Omega_{M_0} a^{-3}-\Omega_{K_0} a^{-2}\right ) \tag 4$$

Substituting (4) in the expression (2) and simplifying we obtain that the value of the acceleration of the scale factor is:

$$\boxed{\ddot a=H_0^2 \ \left (\Omega_{\Lambda_0} a-\dfrac 1 2 \Omega_{M_0} a^{-2}-\Omega_{R_0} a^{-3}\right )}\tag 5$$

Making now $\ddot a=0$ we will obtain the equation that fulfils the scale factor of the inflection point:

$\Omega_{\Lambda_0} \ a_{ip}^4 - \dfrac{\Omega_{M_0}} 2 \ a_{ip} - \Omega_{R_0} = 0$

The values of the density ratios according to the best data currently available, which have been published by the Planck Collaboration in 2018, are:

$\Omega_{\Lambda_0}=0.6889$

$\Omega_{M_0} = 0.3111$

$\Omega_{R_0} < 0.0001$

At the level of precision we are working with, it is licit to consider this last density negligible, then the 4th degree equation is simplified and we obtain:

$a_{ip} \simeq \sqrt[3]{\dfrac{\Omega_{M_0}}{2 \ \Omega_{\Lambda_0}}}$

$a_{ip} \simeq 0.609$

That means that the inflection point took place when the distances in the Universe were 60.9% of today's distances.

OK, we now know at what scale factor the inflection point took place, now we only need to calculate at what cosmic time instant that scale factor occurred. As we have said previously, in the general case there are no explicit functions $a=a(t)$ or $t=t(a)$ but the Friedmann Equations allow us to relate both variables by means of an integral, which states that the cosmic time $t$ can be calculated from the scale factor $a$ by means of:

$$\displaystyle t=\dfrac 1{H_0} \int_0^{a} \dfrac{dx}{\sqrt{\Omega_{\Lambda_0} x^2+\Omega_{K_0}+\Omega_{M_0} x^{-1}+\Omega_{R_0} x^{-2}}}\tag 6$$

Again neglecting the radiation density, and now also considering that according to the best observations currently available the Universe is Euclidean $\Big (\Omega_{K_0} \simeq 0 \Big )$ the following approximation is obtained for the integral:

$$t \simeq \dfrac 2{3 H_0 \sqrt{\Omega_{\Lambda_0}}} \ \sinh^{-1} \sqrt{\dfrac{\Omega_{\Lambda_0}}{\Omega_{M_0}} \ a^3}$$

As we have seen above:

$a_{ip}^3 \simeq \dfrac{\Omega_{M_0}}{2 \ \Omega_{\Lambda_0}}$

And then:

$$t_{ip} \simeq \dfrac 2{3 H_0 \sqrt{\Omega_{\Lambda_0}}} \ \sinh^{-1} \left ( \dfrac 1 {\sqrt 2}\right )$$

Substituting values with $H_0=67.66 \ (km/s)/Mpc$ according to the Planck Satellite, we obtain that the universe started its accelerated expansion with an age of:

$t_{ip} \simeq 7642$ millions years.

As the current age of the Universe is 13787 million years, this happened $\boldsymbol{6145}$ million years ago.

Best regards.

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