# Kindercosmos--basic Friedmann

1. Feb 21, 2015

### marcus

This thread is meant to give a basic introduction to the equation that governs distance growth, how it changes over time.
If you've worked with Jorrie's calculator ("Lightcone") you may know that the present distance growth rate is about 1/144 of a percent per million years, and the longterm rate that it is tending towards is 1/173 % per million years.
The calculator makes tables of universe history showing what the growth rate (and other descriptors) have been in the past and what they will be at various times in future. It's running the standard cosmic model and it's based on the Friedmann equation (derived from Einstein GR around 1923).

I'll denote the rate at time t by H(t) and the longterm rate, which is an important constant, by H.

Some people like to use metric units consistently. If you're not a metric purist and are more comfortable with familiar quantities like percents and years, then skip the next two lines. On the other hand if you like metric units you can find out the two key rates by pasting the percentage versions into google like this:
(1/144) percent per million years in attohertz
(1/173) percent per million years in attohertz

In this thread I want to be able to use the google calculator easily, with its automatic unit conversions, so I don't want to go against its conventions. It measures a key quantity, energy density, in pascals: a pascal can just as well be read as an energy density of one joule per cubic meter, as it can be as one newton of force per square meter. Algebraically they are the same: N/m2 = Nm/m3

Everything about the Friedmann equation, and the standard cosmic model that runs on it, is simpler if we assume large-scale spatial curvature is zero. Spatial curvature has been measured and the observed value is near zero (within less than one percent, with high confidence) so for the sake of a basic introduction we can assume spatial flatness. That doesn't mean spacetime curvature is zero. the 4D geometry can still have curvature, e.g. related to the the expansion or contraction of distances over time.

Last edited: Feb 21, 2015
2. Feb 21, 2015

### marcus

The most important function of time in this model is the scale factor a(t). This is the size of a generic distance, compared with its present size a(now)=1.
It is normalized to equal one at present.
I'll denote time-derivatives by prime (easier to type than putting a dot over the symbol).

The fractional rate of change of distances, that we were calling H(t), is therefore a'(t)/a(t). H(t) = a'(t)/a(t) is the change in a(t) as a fraction of the whole a(t), per unit time.

We want to follow how that rate changes over time. It will be connected with the energy density in space----radiation, ordinary matter, dark matter---the energy density and the distance growth rate evolve together. That is what the Friedmann equation tells us.

It is customary to use the letter rho ρ for density. Don't confuse with the letter p. The average energy density at present is about a quarter of a nanopascal
ρ(now) = 0.24 nanopascal = 0.24 nanojoule per cubic meter.

H(t)2 - H2 = (8πG/3c2

Last edited: Feb 21, 2015
3. Feb 21, 2015

### marcus

Don't be put off by this somewhat ungainly looking thing. It is just a way to make google tell us the average density of the universe:
(1/144^2 - 1/173^2) (percent per million years)^2 * 3c^2/(8 pi G)
If you copy and paste that into the google window what you get back is:
2.39109977 × 10-10 pascals
In other words you get ρ(now) = 0.24 nanopascal
the average energy density including dark matter, ordinary matter, and light.

So we know H(now) and the constant H and the present energy density ρ(now).
We can already use the Friedmann equation to get a little intuition.
But first let's make a convenient abbreviation Γ = 8πG/(3c2). We use that same combination of constants over and over again, so it will save typing if we take the capital Greek Gamma as a shorthand. Then the equation is:

H(t)2 - H2 = Γρ(t)

one thing we can see from that is that as distances (and therefore VOLUMES) keep growing, the energy density keeps getting less. So the Lefthand of the equation has to go to zero. that means the percentage distance growth rate H(t) has to tend towards H, which is its longterm value.

So the Friedmann is telling us exactly how the growth rate H is going to decline in future and tend towards its limiting value, the constant H.

Actually this constant H is a disguised version of the famous cosmological curvature constant Λ. The cc is given by Λ = 3H2.
the fact that there is a 3 tacked on is basically just a consequence of how Einstein first wrote the cosmological constant Λ when he included it in the GR equation in 1917.

For people who like taking derivatives, as one does in beginning calculus, with Leibniz rule, or product rule, and chain rule etc. I want to take the derivative of both sides of the Friedmann equation and show what happens. Anyone who wants to can proceed right away because we know that H(t) = a'(t)/a(t).
So finding H'(t) is just the quotient rule.

the physically interesting part is taking the derivative of ρ(t) when the energy is that of LIGHT on one hand, versus when it is in the form of slow-moving matter---think of a pressure-less cloud of dust. As distances expand--volumes grow--the dust cloud just thins out. But the swarm of light both thins out with increasing volume and redshifts. So it loses energy density as a4.
It loses energy density as a3 because of volume increase, but also loses as a(t) increases because wavelengths get stretched out.

Last edited: Feb 21, 2015
4. Feb 21, 2015

### marcus

I want to continue this intro to Friedmann equation by taking derivative of both sides.
H(t)2 - H2 = Γρ(t)

Obviously the H term is going to drop out---the derivative of a constant term is zero.
(H(t)2)' is the chain rule
(H(t)2)' = 2H H'
H' is the quotient rule
H' = (a'/a)' = a''/a - (a'/a2)a' = a''/a - H2

I'll pause here and pick up with taking derivative of the Righthand ρ(t) when I get back. We should split it into two cases, light (radiation) and pressureless matter.

If we take the latter case, "dust", and a sample volume that currently has density ρ0 = ρ(now), the density is clearly going to go as ρ(t) = ρ0/a(t)3
ρ'(t) = (-3ρ0/a(t)4)a'(t) = -3ρ(t)H(t)

Similarly if light is the dominant form of energy, or any sort of radiation,
the density is going to go as ρ(t) = ρ0/a(t)4
ρ'(t) = (-4ρ0/a(t)5)a'(t) = -4ρ(t)H(t)

These two cases can be combined as ρ'(t) = -3(1+w)ρ(t)H(t)
where the symbol w takes the value 0 for dust (so we just get -3ρ(t)H(t)
and takes the value 1/3 for light, so that 1+w = 4/3,
and we get the right answer for light, namely -4ρ(t)H(t)
Physicists already had the required w symbol, it was the ratio p/ρ of pressure to energy density, which happens to be 1/3 for light, and of course must be zero for a medium that exerts no pressure at all. Since the w parameter was already in use it provided a natural way to combine the two cases, which was moreover physically grounded. Mixtures of light and matter with intermediate ratios of pressure to energy density could be handled the same way. ρ'(t) = -3(1+w)ρ(t)H(t) works generally.

Now let's see what happens when we put the above elements together and differentiate both sides of the Friedmann equation:
H(t)2 - H2 = Γρ(t)

2H H' = Γρ' = -3(1+w)ρHΓ
H' = a''/a - H2 = -3(1+w)ρΓ/2
a''/a - H2 - ρΓ = -3(1+w)ρΓ/2
a''/a - H2 = (2 -3(1+w))ρΓ/2
a''/a - H2 = -(1+3w)ρΓ/2
Have to leave this unfinished, for the night.

Last edited: Feb 22, 2015
5. Feb 22, 2015

### marcus

So what have we got so far? We have the scale factor a(t), comparing the size of a generic distance at time t with its size now. a(now) = 1.
that is basically what all this is about: the expansion history (past and future).
And we have the energy density ρ(t)
and the growth rates H(t) = a'(t)/a(t) which have been measured for two special times H(now) = 1/144% per million years and H = 1/173% per million years.
And this relation (assuming zero spatial curvature) called Friedmann equation:
H(t)2 - H2 = Γρ(t)

And we derived from that a further relation giving a handle on the second derivative of a(t)
a''/a - H2 = -(1+3w)ρΓ/2
Here (1+3w) is a special number which is 1 for dust, and 2 for light.

Square growth rates are a way of expressing curvature of spacetime geometry. The Friedmann says that there is a kind of intrinsic spacetime curvature which will be there even after matter and radiation have thinned out so much that ρ is essentially zero and the Righthand vanishes. Then H(t) will equal H. Spacetime will have settled down to the amount of curvature it is comfortable with (it has been called the "vacuum curvature") and so H(t)2 will equal H2.

BTW back a few posts I mentioned that some people like to see stuff in metric units (instead of, for example, in terms of percentages and years) and I actually think its a good idea in general. Be amphibious, use both eyes, or something.
If you paste these into google:
(1/144) percent per million years in attohertz
and
(1/173) percent per million years in attohertz
it will give you:
2.20 attohertz (for the distance growth rate now)
and
1.83 attohertz (for the eventual longterm growth rate.)

And you can see that the Friedmann is really an equation about the difference between two curvatures.

If you paste this into google:
((1/144) percent per million years)^2 - ((1/173) percent per million years)^2 in square attohertz
it will give you:
1.49 square attohertz.

I swear this happens . Google actually knows the spacetime geometry curvature unit "square attohertz".

So what about that constant Γ, or that ever-present combination of constants 8πG/(3c2) we are abbreviating by the Greek letter?
Well that is on the righthand of the Friedmann turning an energy density like 0.24 nanopascal into a curvature like 1.49 square attohertz.

So in the Friedmann equation context, the right units for Γ should be "square attohertz per nanopascal".

Last edited: Feb 22, 2015
6. Feb 22, 2015

### marcus

8 pi G/(3c^2) in square attohertz per nanopascal
I swear to you by the planet Ceres and all the laws of physics, google comes back:
((8 * pi) * G) / (3 * (c^2)) =
6.2208967 (square attohertz) per nanopascal

So that temporary convenience constant Γ = 6.22 square aHz per nPa.

To put it in a nutshell, since the average energy density in our day and age is 0.24 nanopascal (and near spatial flatness abounds) we should multiply 0.24 by 6.22 and get 1.49 (six quarters is a dollar fifty)
1.49 is by how much the spacetime curvature has to decline before it is down to its intrinsic vacuum level.
Kinda. This is just a way of getting some intuition about what the Friedmann says, and there have to be a bunch of different ways to see it.

7. Feb 22, 2015

### Jorrie

Marcus, I find the playing around with units as interesting as you seem to, but are we not at risk of confusing beginners?

For instance: the normal units for curvature is m-1, but 8πG/(3c2) actually gives standard SI units of m/kg, which can be converted to "(square attohertz) per nanopascal", if you like. But is it not a little 'contrived' to express Γ that way? Is it correct to equate it to curvature? Is it not rather a radius of curvature per unit of mass?

Last edited: Feb 22, 2015
8. Feb 22, 2015

### marcus

Jorrie thanks for reading! In GR a common using of curvature is inverse area so it would be m-2, not m-1 as you suggest. This is how a saddle surface can have NEGATIVE curvature, you can think of it as having two radiuses of curvature coming from opposite directions.
It's a good undergrad 2D differential geometry lesson to think about it. Think of an ellipsoid balloon surface, or an egg shape. If you just measure the curvature along one curve in one direction you don't capture it. So you measure the curvature at a point along two orthogonal paths that cross there, and multiply the two together. So it is a m-1 quantity times another m-1 quantity. A number per unit area, aka a reciprocal area.

What I've abbreviated as Γ is a "curvature per energy density" quantity. This is basically what the Einstein GR equation is about. Curvature on the left side and Energy density on the right (multiplied by a central constant that converts from energy density to curvature).

The stiffness or flexibility of geometry---geometry's responsiveness to matter ---this is the feel or intuition we can get across here.

The nice thing about working with the Friedmann equation is it says the same thing as the GR equation but in much simpler terms. Distance growth rate---and the square of distance growth rate. I've tried to strip it down to the barest of bare essentials, as you can see. I hope it's not too naked

9. Feb 22, 2015

### marcus

Oh, to mention the obvious: in a spacetime context both time and distance are length quantities, so since curvature is (distance)-2
it is also (time)-2.
A fractional growth rate is time-1 so the square of a growth rate is the same type of quantity as inverse area--i.e. a curvature.
My guess is that for beginners, the idea of percentage growth rate is readily intuitive.
Which is one of the appealing things about your calculator. The direct intutive grasp of 1/144 percent per million years, and its longterm counterpart.
So we can start with that, and when we square it we have hold of a curvature (without further rigamarole) and can do equations with it etc etc.
It is a good doorway.

10. Feb 23, 2015

### Jorrie

But doesn't this give you curvature squared? Maybe I'm just confused between spatial curvature and spacetime curvature, but then we should make the difference in units between the two very clear. As it stands, I would have been much more comfortable with "Γ is a (squared curvature) per energy density".

11. Feb 23, 2015

### Jorrie

I see Ben has stated here that curvature units depend on one's choice of coordinates:
I take it then that inverse area fits the Friedmann equations best in terms of the energy densities, especially in the way Marcus has manipulated them for simplicity, i.e. H(t)2 - H2 = Γρ(t)

12. Mar 1, 2015

### marcus

Γ was a convenient shorthand for 8πG/(3c2)
If we want to calculate the time when a'' goes from negative to positive, we can do that with those two equations

13. Mar 1, 2015

### marcus

Assuming matter dominates over radiation (w=0),
It looks to me by inspection of those two equations that a''(t) is zero (the inflection point in the a(t) curve) when H(t)2 = 3H2.
That would be when H(t) = √3 H = 1/100 % per million years.
You just divide 173 by √3
So the inflection point should come when the Hubble radius is right around 10.0 billion LY.

Lightcone calculator tells me that Hubble radius 10.0 Gly comes right around year 7.60 billion or stretch factor 1.65

Last edited: Mar 1, 2015