# I What is the acceleration of the universe?

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1. Oct 9, 2018

### DLeuPel

When I think about it, I see galaxies getting farther away with a constant acceleration. An acceleration that must be caused by a force ( pinned Dark Energy ). I lack in knowledge of the subject but yet I wonder what is the value of the acceleration in the expansion of the universe. Just like in kinematic we calculate the acceleration of a moving point of mass, there should also be a value for the expansion of the universe.For example, x ms-2.

2. Oct 9, 2018

### Bandersnatch

I suspect you're looking at the Hubble law, and are assuming that it shows acceleration since V=HD should mean that once any particular galaxy gets farther it will recede faster. It doesn't show that.
The Hubble law, with any one value of the Hubble parameter, is a snapshot of the expansion at one given time. It doesn't track the motion of any single galaxy. It doesn't say that the galaxy today receding with velocity V will recede with 2xV once it gets twice as far. By the time that one galaxy gets to twice the initial distance, the constant of proportionality (Hubble parameter) will have changed (in our universe = decreased), giving different velocity.
In fact, if we were to imagine a universe without any energy whatsoever (no matter, radiation, or dark energy), but that still somehow had some galaxies in it, those galaxies would recede at constant velocities.
If we add matter and radiation, those velocities go down with time. If we add dark energy to the mix, those velocities can go down before going up.

Try to think of expansion not as a velocity, but as a rate, where current rate is equivalent to growth of all distances by approx. 1/144 % per million years. Then we can ask what is the rate of change of this rate of growth. This is expressed as the deceleration parameter q.

It is a dimensionless parameter. It is defined as $q=- \frac{\ddot a a}{\dot a^2}$ where $a$ is the scale factor of the universe (with dots of course indicating its time derivatives).
The intuitive meaning is that when q is negative, the expansion is accelerating. When it's equal to -1, the expansion is accelerating exponentially. When it's 0, the expansion is steady (i.e. each velocity obtained from the Hubble law at any given time remains constant, but farther galaxies still recede faster), and when it's positive, the universe is decelerating.
The present-day value of the deceleration parameter is $q_0 = - 0.55$.

Historically, the universe was thought to be decelerating at all times, hence the name, and hence the minus sign.
The parameter changes with time together with changing densities of various energy components in the universe. It used to be positive for good half of the history of the universe while matter and radiation were sufficiently dense to retard recession velocities, indicating decelerated expansion (hugely so in the earliest moments). Today it's negative, and it will asymptotically approach -1 as all energy densities other than dark energy dilute to negligible values.

3. Oct 9, 2018

### Orodruin

Staff Emeritus
It is however of interest to point out that, in a Universe dominated by a cosmological constant, the Hubble parameter is constant and this is exactly what the Hubble law says.

4. Oct 9, 2018

### Staff: Mentor

Be careful here. In GR only proper acceleration requires a force. The cosmological acceleration you are referring to here is not a proper acceleration and it does not require any force.

5. Oct 9, 2018

### kimbyd

Or to reword this a little bit with less jargon:
In GR, free-falling objects follow geodesic trajectories. The shapes of those trajectories are determined by the space-time geometry and the position/velocity of the object. These galaxies are free-falling (no force), and they're simply following the trajectory determined by the space-time geometry. The space-time geometry, in turn, is determined by the matter/energy density as well as the cosmological constant (if any).

One way to understand this is that by looking at how Newtonian gravity looks if you add a cosmological constant. Here is the force an object of mass $m$ experiences from an object of mass $M$ in this picture (here the vector $\vec{r}$ points from $M$ to $m$):

$$\vec{F}(\vec{r}) = -{G M m \over r^2}\hat{r} + {1 \over 3} \Lambda m \vec{r}$$

So if there's a cosmological constant (or some dark energy that behaves that way), the gravitational effect of that effectively creates a repulsive gravitational force between any two objects. You may notice that the force is proportional only to the mass of the object that the force is acting on (meaning this force breaks Newton's third law). This is because what is actually happening is it's creating a constant acceleration of any two objects away from one another proportional to their distance. It's not caused by the mass of the other object, but just by the amount of intervening space between them.

6. Nov 9, 2018 at 3:44 PM

### Buzz Bloom

Hi Bandersnatch:

I would much appreciate a citation of a reference to the calculation giving q0 = -0.55, implying acceleration..
I have unsuccessfully been trying to find a reference to any article about any recent calculation of q0. Wikipedia
gives a good presentation of the theory, but nothing about the calculation of a value.

The 1998 Reiss et all article
describes their model fitting based on a database of supernovas, and concludes that q0<0 with a 3.9 confidence level.

A 2016 nature paper
disagrees with the earlier results saying the earlier modeling uses of a smaller supernova database was flawed.
Cosmological parameters are then fitted, along with the parameters determining the light curves, by simple χ2 minimisation. This method has a number of pitfalls as has been emphasised earlier.
With ever increasing precision and size of SN Ia datasets, it is important to also improve the statistical analysis of the data. To accomodate model comparison, previous work has introduced likelihood maximisation. In this work we present an improved maximum likelihood analysis, finding rather different results.​
This paper however does not seem to make any calculation of q0. Instead it seems to compare correlations involving 10 other parameters, and I am not sufficiently knowledgeable about the subject to understand how this related to q0. Can yo help me? Also. do you know of any more recent papers about this topic?

Regards,
Buzz

7. Nov 9, 2018 at 4:08 PM

### Bandersnatch

It's from PLANCK mission data releases. Although now that I look at the 2018 data, it appears to be closer to -0.53.
Look for $\Omega_m$ and $\Omega_{\Lambda}$ parameters. The relationship is $q_0 = \frac{1}{2}\Omega_m - \Omega_{\Lambda}$
Wikipedia has an article on the deceleration parameter, showing how the above relation is obtained.

8. Nov 9, 2018 at 6:13 PM

### Buzz Bloom

I looked at the 2018 Planck report abstract for cosmology
and found the value
Ωm = 0.315±0.007.​
Using the equation in your post, and
ΩΛ = 1- Ωm,​
this gives
q0 = -0.53±0.01,​
matching the value you gave.

Regards,
Buzz

9. Nov 10, 2018 at 2:31 PM

### kimbyd

The force is actually just gravity. If we assume that the acceleration is caused by the cosmological constant, then one way to understand this acceleration is to consider the Newtonian limit. Here's the acceleration an object feels due to the gravity of an object of mass $M$, according to Newtonian gravity:

$$a_g = -{GM \over r^2}$$

I'm neglecting vector notation, but using the minus sign to indicate that the acceleration of the accelerating object is towards the object of mass $M$.

If we add the cosmological constant in the Newtonian approximation, the above is modified to become:

$$a_g = -{GM \over r^2} + {\Lambda c^2 \over 3} r$$

Based upon the current cosmological parameters, the $\Lambda c^2 / 3$ term is approximately $3.3 \times 10^{-36} s^{-2}$.

Thus the cosmological constant creates an acceleration between any two objects separated by some distance, but that acceleration is very small. Two objects separated by ten billion light years experience a mutual acceleration away from one another of about 3 billionths of a meter per second per second, or nearly 10 m/s over 1000 years.