# I What is the acceleration of expanding space?

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1. Dec 13, 2017

### DAirey

If me and my pet duck, who weighs next to nothing, were placed far away from any gravity source and in such a way that our mutual attraction was negligible, would I see my duck accelerate away from me? If so, how fast?

2. Dec 13, 2017

### newjerseyrunner

3. Dec 13, 2017

### PeroK

4. Dec 13, 2017

### newjerseyrunner

Ohh, you're right. Silly me. I was reading it like an acceleration and it's not, nothing is accelerating away, it's simply moving with the universe.

5. Dec 13, 2017

### kimbyd

You can certainly describe it as an acceleration, provided you use a particular definition for distance and therefore velocity. The second Friedmann equation provides a notion of acceleration that can be used:

$${\ddot{a} \over a} = -{4\pi G \over 3} \left( \rho + {3p \over c^2}\right)$$

Going through this equation is a little complicated, so I'll just roughly go over how you can calculate the answer then provide said answer (hopefully without making an error). Basically, you have to know two things: 1) How to convert each type of matter/energy density to a density fraction, as the density fractions are the values reported by cosmological experiments. 2) How the energy density of each type of matter relates to its pressure.

Once you have these, assuming a flat universe and only matter and a cosmological constant, with some understanding of how the scale factor relates to all of these, you get:

$${\ddot{a} \over a} = H_0^2 \left(-{\Omega_m \over 2a^3} + \Omega_\Lambda\right)$$

For an object at a distance of 3 million light years today (with $a=1$ today by convention), $\Omega_m = 0.32$, $\Omega_\Lambda = 0.68$, and $H_0 = 67$km/s/Mpc, the acceleration of distance today is roughly $7 \times 10^{-14} m/s^2$, or 2 meters per second per million years, if I did my math correctly.

Over time, as the matter dilutes and the object gets further away, this acceleration will increase.

6. Dec 13, 2017

### PeroK

A simpler and more naive calculation is that an object $1 Mpc$ away is receding at $67km/s$. One second later it is receding at this speed plus the expansion for the extra $67km$.

I get $(67km/s) (67km/1Mpc) = 1.6 \times 10^{-13}m/s$.

That gives an acceleration of $1.6 \times 10^{-13}m/s^2$

7. Dec 13, 2017

### Staff: Mentor

You have your units wrong; $\ddot{a} / a$ does not have units of acceleration. It has units of inverse time squared. It has no length dimension because of the $a$ in the denominator. That's what makes it useful--it "factors out" the scale, so it gives us a scale-invariant way of describing how the expansion of the universe is accelerating.

You are calculating something different from @kimbyd. Roughly speaking, you are multiplying $H$ by $a$ to get a change in velocity over a given time, then dividing by that time to get an acceleration. This does not give you $\ddot{a} / a$.

An easy way to see the difference is to note that your number will be positive if the universe is expanding; but @kimbyd's number is only positive if the expansion is accelerating. If there were no dark energy, we would have $\Omega_{\Lambda} = 0$ in @kimbyd's equation and his number would be negative; but your number would still be the same.

8. Dec 13, 2017

### rootone

If you and your duck have zero mass. then the distance between you and the duck will increase over time.
You don't have zero mass though, and even if you did you would have to live for millions of years to see anything changed.

9. Dec 13, 2017

### newjerseyrunner

Even if they had mass, they may still increase distance apart. It would depend on whether the influence of dark energy can overpower the influence of their mutual gravity. At the 3 million light year mark we’ve been using my assumption is that gravity would be even more neglicable than dark energy. The effects of gravity decrease with distance but the effects of dark energy increase with distance.

10. Dec 13, 2017

### kimbyd

Yes, I multiplied by the 3 million light years suggested by newjerseyrunner to get acceleration, based upon the idea that $d(t) = d_0 a(t)$, so that $\ddot{d} = d_0 \ddot{a}$ for two objects separated by some distance, both at rest with respect to the expansion. I didn't clarify that part.

Also, the scale factor $a$ is usually defined to be dimensionless. This is the convention I was assuming above, where $a=1$ at the current time. A dimensionful scale factor is possible, but seems to be less common. Of course, $\ddot{a}/a$ is dimensionless no matter the convention.

Her number.

11. Dec 14, 2017

### Staff: Mentor

Sorry about that, thanks for the correction.

12. Dec 14, 2017

### Arman777

Well we can apply the Hubble's Law. But we can use it only in large scales. Approximately larger then 100Mpc. Since we need to apply cosmological principle so that we can use comoving coordinate system.

If we look the basic equation we will see that

$D=a(t)Δx$ here $D$ is the proper distance and $Δx$ is the comoving distance. If we take second derivative we get accceleration and thats equal to.

$A=\frac {dD^2} {d^2t}=\frac {da^2(t)} {d^2t}Δx$.
Hence,σ $A=\frac {\frac {da^2(t)} {d^2t}} {a(t)}D$

In matter dominant universe a goes like;
$a(t)≅t^{\frac {2} {3}}$

In a distance about $100Mpc$,
$A≅-t^{-2}D$
$A≅\frac {-3.10^{21}km} {2.10^{35}s^2}$. Its as expected, really low and decelerating.

I did this case for matter dominant.In dark enegy dominant case which, $a(t)≈e^{√Λt}$
we would get a different answer.

In any case Hubble's Law is not valid on small scales. In my case I used matter dominant model but in other models it would be a bit complex.

Last edited: Dec 14, 2017
13. Dec 14, 2017

### Arman777

What happens If we put our duck near us like a 10pc. Then nothing happens for a long time. Since in a such a small scale it would take billions of years to see an real expansion.

First reason is we cant use friedmann equation or Hubble's Law to calculate the motion of the duck, cause distance or our scale is really low respect to umiverse and we cant talk about cosmological principle.

Second thing is, If such thing was possible then our galaxy would also feel such expansion. But we dont, If you think that our milky way has a radius of 15kpc. And we dont see that stars are getting apart or etc. If we place the duck in example between andromeda and milky way or between two superclusters, I think the result would be the same.

14. Dec 14, 2017

### newjerseyrunner

The galaxies are large enough and close enough together to be gravitationally bound. The universe still expands under them, gravity is just so much more dominate. We’re talking about a human and a duck which have essentially no mass one an astronomical scale.

15. Dec 14, 2017

### Arman777

Yes I said the same thing didnt I ? Or you mean duck and human will feel the expansion If we put them close?

I mean of course they will but it will be really slow and undetectable.

Here matter density is important I think or its related to gravity. Even we put them close (less then 50 Mpc or so) the region of the universe is the same, the physics rules are the same. My point is we ll not see a different type (value) of acceleration for duck respect to the galaxies.

For example, lets suppose theres a galaxy 20 Mpc away and our duck.

Does the acceleration due to expansion, will be the same for both objects ? I think they will be the same.

16. Dec 14, 2017

### DAirey

I'd be lying if I said I followed your derivation, but I don't see the lambda constant in the above equation. The discovery that the expansion of the universe is accelerating is due to the negative pressure of the vacuum that produces a repulsive gravity. Intuitively, it seems to me that this anti-gravity would be ubiquitous and constant (for a given epoch), hence, the question about the duck (would it appear like gravity, but in reverse). Have you considered the effects of vacuum energy in your calculation?

Last edited: Dec 14, 2017
17. Dec 14, 2017

### Staff: Mentor

Careful. The ordinary expansion of the universe is not a "force", and doesn't change anything "under" gravitationally bound systems. The ordinary expansion of the universe is just a statement about the relative motion of comoving objects, i.e., objects that are not gravitationally bound to each other. It doesn't tell you anything about gravitationally bound systems.

The presence of dark energy does create a kind of "force" (more precisely, tidal gravity) that, in principle, is present on all scales and within gravitationally bound systems; however, its magnitude on the scale of gravitationally bound systems is so tiny that it can't be measured and has no observable effect on the properties of such systems. It is only observable on very large distance scales and very long time scales (billions of light years and billions of years).

18. Dec 14, 2017

### Staff: Mentor

$\Lambda$ provides part of the $\rho$ and $p$ that appear in that equation.

Yes, she did. That's why there is an $\Omega_{\Lambda}$ on the RHS of @kimbyd's equation, and why that term is positive, whereas the matter term involving $\Omega_m$ is negative.

19. Dec 14, 2017

### Staff: Mentor

See my post #17 just now.

20. Dec 14, 2017

### DAirey

Ah, yes, I see that now. Thank you.
It would be very helpful if you could edit the above posts down to a single answer. Are you saying that that there's a single, constant acceleration that's too tiny to measure is correct or are you saying that @kimbyd's answer of $7 \times 10^{-13} m$ $s^{-2}$ per every 3 MLy years is correct? Does @kimbyd 's answer imply that you'd have an acceleration of $1.4 \times 10^{-12} m$ $s^{-2}$ every 6 MLy? I would think the negative pressure would be constant and, thus, the acceleration would be constant (for any given epoch).

Last edited: Dec 14, 2017
21. Dec 14, 2017

### DAirey

Are you sure that's $m$ $s^{-2}$ and not $km$ $s^{-2}$?

22. Dec 14, 2017

### Staff: Mentor

As I pointed out earlier in response to @kimbyd , the "acceleration" caused by dark energy does not have units of distance per time squared; it just has units of inverse time squared. In other words, the ordinary acceleration--distance per time squared--caused by dark energy does not have a single constant value; its value depends on how far apart the two objects are whose relative acceleration we are considering.

The negative pressure is constant at a given epoch, but negative pressure corresponds to the scale-invariant "acceleration" $\ddot{a} / a$, i.e., to a thing that has units of inverse time squared. That thing is constant at a given epoch. But, as above, to convert it into an ordinary acceleration, with units of distance per time squared, you need to multiply by the distance between two objects.

If you want to use @kimbyd 's equation to calculate $\ddot{a} / a$ and hence the relative acceleration of you and your duck, you need to obtain a value for $H_0$ that has units of inverse time; that means taking the usually quoted value of 67 km/s/Mpc and converting it to a value that has units of inverse seconds. You do that by dividing 67 by the number of km in a Mpc, which is about 31 trillion. Then you can plug in the result (squared, since the equation for $\ddot{a} / a$ has $H_0^2$) and the other numbers @kimbyd gave, to obtain a value for $\ddot{a} / a$; this is the number that is the same everywhere in the universe at the given epoch (our current epoch has $a = 1$ in the equation, as @kimbyd said). Then you multiply that number by the distance between you and your duck to get your relative acceleration.

23. Dec 14, 2017

### kimbyd

I used shorthand to include the cosmological constant in the energy density, to keep things simple. You can always include $\Lambda$ as an energy density which has $p = -\rho$, which also implies it doesn't change over time.

24. Dec 14, 2017

### kimbyd

It's only constant if you are in a universe that only has a cosmological constant, and no matter density. In that special case, the acceleration formula is much simpler:

${\ddot{a} \over a} = H_0^2$

Implying that $\dot{d} = H_0 d$ and $\ddot{d} = H_0^2 d$.

25. Dec 14, 2017

### kimbyd

I may have made a mistake, but I definitely didn't make that mistake.

If you want to check my math, what I did was I plugged the equation I wrote above (multiplied by the distance in post #2) into Google and let it handle the unit conversions. The precise search I used was:

(67 km/s/(1e6 parsecs))^2 * (-0.5*0.32 + 0.68) * 3 million light years

I then multiplied that by seconds per year to get the acceleration per year, which came in at $2 \times 10^{-6}$, giving 2m/s per million years.