# I How much does the Universe accelerate?

1. Nov 9, 2018

### KurtLudwig

Please clarify. I have read that according to NASA the universe's expansion rate is 74.3 km/s +/- 2.1 at Mega parsec. At a distance of one Mega parsec, space is moving 74 km/s away from us. At 2 Mega parsec it is moving away at 148 km/s. Do I understand this correctly? But this is velocity, it is not acceleration. How fast was it moving away a billion years ago? The velocity must have been less that 74 km/s at 1 Mega parsec if the universe is accelerating. Are we able to measure past values with the red shift?

2. Nov 9, 2018

3. Nov 9, 2018

### rootone

Yes very approximately, by looking at the emmisions of long gone supernovea.

4. Nov 10, 2018

### Bandersnatch

Yes, this is correct.
It's perhaps worth noting, that when reading on the subject you will also encounter a different value of the Hubble constant, closer to 68 km/s/Mpc. This value is inferred from CMB data, whereas the one you cited is from observations of (relatively) nearby supernovae. The discrepancy is appreciated and a subject of investigations.

The velocity at 1 Mpc was higher than now. In other words, the Hubble constant was larger.
This is not a contradiction, because the magnitude you singled out here, i.e. recession velocity at the constant distance of 1 Mpc, is not the speed of expansion (rather, it's a rate). So, neither is how it changes with time the acceleration of the universe.
The particulars of how the rate of expansion changes does determine whether the universe is accelerating or not, though. It's just not that the time derivative of H must be >0 for acceleration to occur.

I think it'd be easiest to understand what is meant by acceleration if we focus on what happens to some particular test galaxy.
Imagine some random galaxy. It doesn't matter which we pick, since the same behaviour is true globally. To make things easier, let's say it's at 1 Mpc and is receding from us solely due to the expansion of space with ~70km/s. If it were coasting at that velocity, constantly, after some time it'd reach 2 Mpc. Since in this scenario it still recedes with 70 km/s, we could apply the Hubble's law to find out that the Hubble constant H is now 70 km/s/2 Mpc, or 35 km/s/Mpc.
What we notice here, is that while the initial distance has doubled from size 1 to size 2, and H went down as the distance grew (as 1/size), the recession velocity remained constant. In this scenario, there is expansion, but there is no acceleration.

If that recession velocity of this particular galaxy went down by the time the galaxy receded to 2 Mpc, that'd indicate deceleration. It'd correspond to H falling faster than 1/size. If it went up, that'd be acceleration - corresponding to H changing at least slower than 1/size (depending on by how much the velocity went up, this includes H being constant or growing).
In our universe, depending on the balance between the retarding matter and the accelerating dark energy, our test galaxy may either decelerate or accelerate. Since matter density decreases as the universe grows, while dark energy density remains constant (this is not certain), early universe should first be decelerated, while later on acceleration takes over.

Even if the universe is accelerating, Hubble constant can still go down with time. All it takes is for our test galaxy to accelerate to less than twice its initial recession velocity by the time it gets to twice the distance.

This graph shows the history of recession velocity of a generic galaxy currently at 14.4 Gly and receding with $V_{rec}=c$ (rather than 1 Mpc and 70 km/s we used earlier):

The next graph shows the changing Hubble parameter over the same period of time:

Both the monotonic decrease of H(t) and the switch from deceleration to acceleration at around 8 Gy can be seen.

There are two ways to go about it. One is looking at nearby supernovae and measuring how the redshift vs distance relationship deviates from linearity, which lets us see how H changed in the past. The other is by looking at features in the CMB radiation in order to obtain some parameters of the early universe, and using a model of expansion to find later values.
The graphs shown above were obtained using the second method.

(charts were generated using: http://www.einsteins-theory-of-relativity-4engineers.com/LightCone7-2017-02-08/LightCone_Ho7.html)

5. Nov 10, 2018

### KurtLudwig

Thank you all for your explanations and further reading, especially to Bandersnatch. The units are not km/s, but the rate is 1/s.

In Insights Blog I found Approximate LCDM Expansion in Simplified Math

Reference https://www.physicsforums.com/insights/approximate-lcdm-expansion-simplified-math/

I will visit Accelerating Expansion of the Universe in Wikipedia.

6. Nov 10, 2018

### Buzz Bloom

Hi @KurtLudwig:

You may find the following thread of some help.

Regards,
Buzz

7. Nov 10, 2018

### kimbyd

1) We measure how the expansion rate changed over time using a variety of methods to correlate redshift and distance. Redshift is the total amount of expansion between us and the source, so by looking at expansion vs. distance, we get an estimate of how fast the universe expanded at various times.
2) The acceleration of the universe does not refer to the change in the expansion rate. The expansion rate is actually decreasing over time (slowly).
3) Individual objects within the universe are accelerating away from one another. You can see this most easily by imagining the situation where the expansion rate is a constant. It isn't a constant today, but it is slow, and likely to become close to a constant in the far future. Similar to your example above but using larger distances (where the expansion is relevant: 1Mpc is too small scale), as an object moves from 100Mpc to 200Mpc, its recession velocity increases from 7,400km/s to 14,800km/s.

I won't explain the calculation here, but in our current universe the recession velocity once the object currently at around 100Mpc hits 200Mpc, it will be receding at closer to 10,700km/s which isn't all that close to twice the velocity. Once it hits 400Mpc it will be receding at 20,400km/s, which is very nearly twice the speed at twice the distance. Eventually the rate of change in recession velocity will be indistinguishable from an exponential (constant expansion rate). This is using your numbers of 74km/s for the Hubble expansion rate, and the current best-fit matter density fraction of 0.315.

Finally, a quick note on time scales: it will take roughly 10 billion years for an object currently at 100Mpc to get to 200Mpc, and roughly 10.9 billion years to get from 200Mpc to 400Mpc (since the expansion rate will have slowed a bit, the second doubling takes more time). In the far future, each additional doubling of distance will take just over 11 billion years.

8. Nov 10, 2018

### Buzz Bloom

Hi kimbyd:

Wikipedia
says
The accelerating expansion of the universe is the observation that the universe appears to be expanding at an increasing rate...​

Your quote and the Wikipedia quote seems to be saying opposing things. Is this correct? If not, would you please post your definition of "The acceleration of the universe"?

Regards,
Buzz

9. Nov 10, 2018

### Bandersnatch

I tried to explain this in post #4
The rate of expansion is expressed by the Hubble parameter. Its time derivative is and has always been negative.
The expansion itself (as opposed to contraction) means the time derivative of the scale factor is positive. Accelerated expansion means the second time derivative of the scale factor is positive.

I.e.:
rate: $H=\frac{\dot a}{a}$
and
rate decreases: $\dot H <0$
while
expansion: $\dot a>0$
and
accelerated expansion: $\ddot a >0$

10. Nov 10, 2018

### kimbyd

Yeah, that wording at Wikipedia is incorrect. I'm going to see if I can't fix that.

11. Nov 10, 2018

### Bandersnatch

Oh, bummer. I've only now noticed the issue @Buzz Bloom found. Never mind me.

12. Nov 10, 2018

### Buzz Bloom

Hi: Bandersnatch:

Would not the meaning be that the acceleration is the rate of change of the rate H0? That is:
acceleration = dH0/dt
= d ( (da/dt)/a ) dt
= (a (d2a /dt2) - (da /dt)2) / a2

I am very awkward using LaTeX, but I probably should have tried to use it here.

Regards,
Buzz

Last edited: Nov 10, 2018
13. Nov 10, 2018

### kimbyd

Removed this post for now, as it was incorrect. Working on a correct revision.

14. Nov 10, 2018

### kimbyd

An accelerated expansion occurs when $\ddot{a} > 0$. In terms of $H$, this condition can be represented as follows:

$$H = {\dot{a} \over a}$$
$$\dot{H} = {\ddot{a} \over a} - {\dot{a}^2 \over a^2}$$

If we then solve for $\ddot{a}/a$ and use the definition of $H$, we get:
$${\ddot{a} \over a} = \dot{H} + H^2$$

Typically, the rate of expansion $H$ decreases over time, such that $\dot{H} < 0$. In fact, it can only increase in exotic situations such as the Big Rip scenario. But in order to get an accelerated expansion ($\ddot{a} > 0$), you don't need $\dot{H} > 0$. You only need $\dot{H} > -H^2$. That is, if the expansion rate decreases slowly enough, objects within the universe accelerate away from one another.

15. Nov 13, 2018 at 1:55 PM

### jeremyfiennes

Summarizing: is the expansion rate increasing, as I have been led to believe, and which means no Big Crunch; or is it not? If it is, could the acceleration be due to a decreasing effect of gravity, rather than dark matter?

16. Nov 13, 2018 at 2:23 PM

### Staff: Mentor

The time derivative of the scale factor, $\dot{a}$, is increasing; or, to put it another way, the second time derivative of the scale factor, $\ddot{a}$, is positive. This is what is referred to when cosmologists say that the expansion of the universe is accelerating.

The Hubble constant $H$ is decreasing; but it will not decrease to zero, but to a finite positive value, because of the presence of dark energy.

The conditions for there not being a Big Crunch are much broader than either of the above; if the universe had no dark energy, it would still not end in a Big Crunch if the density were less than or equal to the critical density (which it is).

17. Nov 13, 2018 at 2:23 PM

### Staff: Mentor

What do you mean by "decreasing effect of gravity"?

18. Nov 13, 2018 at 3:13 PM

### kimbyd

It's proven very difficult to come up with a modified theory of gravity model which explains the accelerated expansion without a cosmological constant and also remains consistent with current observations of gravity's behavior (such models tend to make gravity behave very weirdly at short scales in ways that would be easy to measure and very clearly do not happen). So far, I'm not aware that anybody has come up with a working model for this.

Of course, our inability to come up with a model doesn't mean it's impossible. But at the very least it seems to indicate it's not all that likely that modified gravity is the answer. There remains the possibility that somebody clever will come up with an answer some day in the future. But there's no way to predict that.