# Lay-ish approach to *both* Friedman equations (2nd try)

1. Jul 8, 2014

### marcus

I'll try to develop some notation and explanations that might make the Friedman equations more intuitive to some beginners. They might look a bit odd in their new get-up--if that bothers you please simply ignore this thread.

The scale-factor function a(t) keeps track of the size of a generic distance, as it grows over time. It is a real-number-valued function of time, normalized to equal one at present: a(now) = 1.

The derivative or slope a'(t) at a particular time t is the amount a(t) grows per unit time. For example it could be a "per second" or "per year" quantity. You need to specify a timescale.

The fractional growth rate H(t) = a'(t)/a(t) is what fraction or percentage of itself a(t) grows by, per unit time. Again you have to specify a timescale. For example the present distance growth rate is 1/144% per million years, or 1/14400 per million years.

A generic distance (between objects that are at rest relative to the ancient light Background) grows by 1/14400 of itself per million years.

According to standard cosmic model, this fractional growth rate of distances is expected to continue declining (more and more slowly) and level out at around H = 1/17300 per million years.

Or if you like to think in percentages think of the eventual steady growth rate as H = 1/173% per million years.

You can see these numbers (in reciprocal form) in the R column of Jorrie's Lightcone tables, and get an idea of how they change over time. Just click on the Lightcone link that some of us have in our PF signatures at the end of our posts. You will get a table, where you can increase the number of rows or steps in the table (and, if you wish, shorten the span of time covered) to get finer resolution.

Obviously if you have a way of finding the history of what H(t) the fractional growth rate of a(t) has been and will be, from early times on into the future, then you have a way to plot the history of a(t) itself. Remember that a(now) = 1. So you or your computer can start at the present value and (using the growth rate) work forwards and backwards in time, constructing the entire history.

What I want to do in this thread is introduce the Friedman equations which give cosmologists a handle on the distance growth rate H(t) and the scalefactor a(t).

2. Jul 8, 2014

### marcus

The expansion history a(t) and the growth rate H(t) interact with the average density which is conventionally written with a greek letter rho ρ . It is recognizably different from the letter p. If you are writing a post and want to use the symbol for density, it is in the third row of the table of Quick Symbols off to the right. Just clicking on it makes it appear in the text you are writing.

In this thread the convention is that ρ(t) is the energy density at time t, including radiation energy and the energy equivalent of both ordinary and dark matter. That is all. It does not include an energy equivalent of the cosmological constant Lambda.

Just ordinary and dark matter, and radiation (eg. light) expressed as a total energy density. The metric unit is joules per cubic meter, which the google calculator insists on calling "pascals". I'm inclined not to fight google.* At present the average density in the universe is 0.239 or say 0.24 nanojoule per cubic meter, or as google would put it: 0.24 nanopascal.

For most of history, matter and light have exerted negligible pressure so we study a simplified "pressureless" case. Space has just been too empty to have a significant amount of pressure. Also the overall average spatial CURVATURE has been negligible. So we focus on the zero curvature case. This makes the equations, and the job of introducing them, simple. Eventually when we are through with this easy case, if there is interest, we could see how to add terms allowing for pressure and spatial curvature.

*after all joule=Newton meter. So in the expression for energy density we have Newton meter per cubic meter, and a term cancels to give Newton per square meter, which is the official definition of the Pascal unit. So in a sense the google calculator usage is justified and we shouldn't argue.

3. Jul 8, 2014

### marcus

Here's how H(t) and a(t) evolve.

H2(t) - H2 = 8πG/(3c2) ρ

a"(t)/a(t) - H2 = -4πG/(3c2) (ρ + 3p)

And in the second one here we are going to consider pressureless case and set 3p=0. The RHS of the second equation is just going to be minus half (either side of) the first. In any case we seldom need to use the second. I think these are potentially fairly easy to understand, at least the first one is (which is the one we mostly always use.) If there is anything off-putting about the first equation it is the somewhat bizarre-looking constant out in front of the rho.

To have an equation we must have the same kind of quantity on the right as on the left. So we have to have some constant to multiply rho by, that will turn an energy density into "the growth of a growth rate". What we have on the left is a quantity like "some number per second squared".
Or "some fractional growth per year per year". This is a bit weird. We aren't accustomed to encountering and dealing with such quantities. It's the reciprocal of the square of a time. Analogous to an acceleration but without involving a distance or length unit. What would you call it? A "fractional acceleration"?

Let's give the constant a name W = 8πG/(3c2) and try to understand it a bit better. a curious thing about W is that its reciprocal is something very much like a FORCE. IN FACT if you take the reciprocal 1/W, and multiply it by c2 you get a force.
Maybe this force is important---and multiplying by c2 is how we relate mass to energy, and square time to area, and number per area (a common measure of spatial curvature) to number per square time (which is our "growth rate of the fractional growth rate" type of measure).

So there could be something to understand about this quantity W. Lets take the reciprocal, multiply it by c2 to get a force c2/W = 3c4/(8pi G) and ask google calculator to tell us how big the force is.

Paste this into google: 3c^4/(8pi G)

You should get something in Newtons (the metric unit of force, about a quarter pound of force, 9.8 of them make the weight of a kilogram mass in standard gravity etc etc.)
Have to run some errands, back later.

Last edited: Jul 8, 2014
4. Jul 8, 2014

### marcus

So we are looking at the case where overall spatial curvature and average pressure are negligible and the first Friedman equation boils down to:
H2(t) - H2 = W ρ(t)

Using that, a little algebra shows that the second is equivalent to:
a"(t)/a(t) = H2 - 0.5(H2(t) - H2)
which simplifies to:
a"(t)/a(t) = 1.5 H2 - 0.5 H2(t)

I'm not sure why I bothered to include the second equation. H(t) and a(t) could be calculated in this case just from the first. Starting at a(now) = 1, and putting in the observationally determined parameters such as H(now) = 1/144 % per million years and H = 1/173% per million years, one can determine values of a(t) forward one step into the future or back one step into the past. Then with the new a(t) calculate the new density ρ(t) = ρ(now)/a(t)3. Solve the Friedman equation to find the new H(t), and repeat.

The first equation, and the two parameters suffice to tell us the presentday density ρ(now).
It is ρ(t) = (H2(t) - H2 )/W, for t=now.
If you want to use google calculator to find it, simply paste in
((1/14400^2 - 1/17300^2) per (million years)^2)*3c^2/(8pi G)
since dividing by W is the same as multiplying by 3c^2/(8pi G).

Google calculator will give the answer in nanopascals, so remember to interpret that as the energy density in nanojoules per cubic meter.

The first equation gives a bird's eye view of the future of the universe: it says that as distances expand and the density rho falls off as the cube of a(t), the difference on the left hand side of the equation has to go to zero. So H(t) will gradually decline and settle down at H
So the current 1/144% growth rate will continue to decline (as it has throughout past history) and will level off at 1/173%.

Last edited: Jul 8, 2014
5. Jul 9, 2014

### marcus

In the preceding post I used the symbol W to stand for the reciprocal of a force-like quantity 3c2/(8πG). This is what converts the energy density rho, on the righthand side, to the square of a growth rate, on the left. An alternative notation might be to define Phi as the force-like quantity itself: Φ = 3c2/(8πG). So then the reciprocal 1/Φ = 8πG/3c2, the central constant in the Friedman equation, and we can write the Friedman equation like this:
$$H(t)^2 - H_\infty^2 = {1 \over \Phi} \rho (t)$$This is what the equation boils down to in the case where overall spatial curvature and average pressure are negligible.

You can see the analogous parts in the Einstein GR equation:
$$G_{\mu\nu} + \Lambda g_{\mu\nu} = {8\pi G \over c^4} T_{\mu\nu}$$In fact the central constant 8πG/c4 is the reciprocal of an actual intrinsic quantity of force that is, so to speak, built in to nature. This universal constant force is what relates MATTER (expressed by the T** tensor on the right) to the dynamically responding GEOMETRY, expressed by G** the so called "Einstein tensor". If you want to know the size (in quarter pound metric force units) of that innate force woven into the fabric of existence just paste this into google:
c^4/(8pi G)
You should get some large number of Newtons.

The Friedman equation is a gross simplification of this Einstein GR equation. Instead of the full T** tensor (essentially a 4x4 matrix varying over space and time, describing the concentrations of energy and momentum that matter represents) we just have the one density quantity rho of t varying over time: the average energy density ρ(t).
Instead of the Einstein G tensor (basically a 4x4 matrix expressing the changes in geometry felt by probing in various spacetime directions) all we have is the square of a simple expansion rate H(t).
And the Lambda term in the original GR equation is reflected in what is written here as H squared. In fact H2 = Λc2/3, so it is almost a verbatim transcription of the Lambda term in the original.

The version of the GR equation I've put here is iconic. It is what you see up in the righthand corner if you go to any Wikipedia article on General Relativity or related topics such as the Friedman equations. Lambda, in this iconic form of GR, is a reciprocal area. The metric unit for Lambda would be "per square meter" or "m-2". It's a common measure of spatial curvature. But when studying expansion cosmology we are mainly interested in fractional growth rates and growth rates squared. These are fractional or percentage numbers "per unit time"--reciprocal times--and reciprocal times squared (e.g. "per second per second" or "per year per year".) In a sense we are shifting our focus from curvature in space to "curvature in time"--distance expansion rates, and their rates of change. So it was convenient to change from Lambda, a reciprocal area, to the reciprocal square time quantity Λc2. That is all multiplying by c2 really does here--convert from square length to square time.

Last edited: Jul 9, 2014
6. Jul 10, 2014

### marcus

I decided to continue this in the "getting us all on the same page" sticky thread
https://www.physicsforums.com/showthread.php?p=4793648#post4793648
That link takes you to post #487 and the discussion continues a few posts on from there. Comments are welcome.
What I'm looking for is a kind of introductory springboard to help a newcomer to cosmology gain familiarity with the equation model and not need to rely totally on verbal concepts, analogies, and imagery like the balloon model.
It would be a help to hear other people's ideas about how to do this, or critique, or related questions.

Reply to next post (#7)
DRAKKITH, looking forward to it. Basically time for me to turn in too. See you then!

Last edited: Jul 10, 2014
7. Jul 10, 2014

### Drakkith

Staff Emeritus
Ugh, I'm WAY too tired to get into this now, but I'll try tomorrow.

8. Jul 10, 2014

### marcus

Jorrie made a comment up in the "all on the same page" sticky thread, where I started developing these ideas. It basically is an "all on the same page" introductory pedagogical project. I want a way of introducing the Friedman equation (mainly the first one, since more often used) that won't put non-mathy newcomers off too much and is intuitive and accessible. So our collective understanding doesn't have to be all VERBAL and analogistic.

Jorrie's comment suggested a way that calculations could be made trimmer…

For example to find the current energy density ρ(now) in the universe, given the two Hubble times 14.4 and 17.3 billion years that are the main input parameters for the universe model we are using all the time (in Lightcone calculator).

I habitually translate those into distant growth rates 1/144% per million years and 1/173% per million years.
But they could be expressed as fractional growth rates 1/14.4 per billion years and 1/17.3 per billion years.

The Friedman equation says you multiply (H(t)2 - H2) by that force-like constant Φ = 3c2/8πG and it gives you the overall average density ρ(t).

So here goes, paste this into google:
3c^2/(8 pi G)(1/14.4^2 - 1/17.3^2) per (billion years)^2

Bingo! It comes out 0.239 nanopascals, which is right! That is, 0.239 nanojoules of energy per cubic meter.

It is more compact to write the calculation that way, which you are going to feed into the google calculator.

9. Jul 10, 2014

### marcus

And then suppose, for example, you want to know what the Hubble rate, i.e. the distance growth rate, was back in the day when distances were ONE THIRD what they are today. Well, volumes back then were 1/27 what they are now. So the density ρ(t) was 27 times ρ(now).
ρ(then) = 27ρ(now) = 27*0.239 = 6.453 nanopascal

Now we can plug in the density (of that past era) and get the distance growth rate back then H(then).

H(then)2 = (1/Φ)ρ(then) + H2
= 8 pi G/(3c^2)*6.453 nanopascal + 1/17.3^2 per (billion years)^2

Now I need to take the square root of that to get the growth rate H(then) and I have to take the RECIPROCAL (one over it) to get the Hubble time. So why not combine those two steps? Raise it to the minus 1/2 power:

Hubble time (then) = 1/H(then) = (8 pi G/(3c^2)*6.453 nanopascal + 1/17.3^2 per (billion years)^2)^(- .5)
That is what I paste in, and bingo! I get the Hubble time 4.8 billion years.
that means the growth rate back then was 1/48 percent per million years.

We could express the fractional rate of growth a different way, as 1/4.8 per billion years. But I still prefer to think of it as smaller (percentage) growth per the proportionally smaller time interval: 1/48% per My.

Last edited: Jul 10, 2014
10. Jul 10, 2014

### marcus

To summarize, I've been continuing this kind of pedagogical development of the basic cosmic model equation up in "all on the same page" thread:
https://www.physicsforums.com/showthread.php?p=4793648#post4793648
I hope it turns out to be right for some newcomers to PF cosmology.

Here's the iconic (e.g. wikipedia) form of the GR equation, and a form of the Friedman equation that shows the analogous parts clearly.
$$G_{\mu\nu} + \Lambda g_{\mu\nu} = {8\pi G \over c^4} T_{\mu\nu}$$
$$H(t)^2 - H_\infty^2 = {1 \over \Phi} \rho (t)$$

In the GR equation the central constant 8πG/c4 is the reciprocal of an actual constant quantity of force that is built in to nature. This universal constant force relates features of matter (such as energy density) to features of geometry (such as curvature, expansion rate is a kind of spreading-out "curvature in time").

In this form of the Friedman equation the corresponding central constant (1/Φ) is the reciprocal of a force-like quantity Φ = 3c2/(8πG). This is what converts the energy density rho on the righthand side, to the square of growth rate quantity on the left. Also, for simplicity, in this version the overall average spatial curvature and average pressure are assumed negligible.

Last edited: Jul 10, 2014
11. Jul 11, 2014

### Jorrie

It's neither here nor there, but while we are commenting on simplifying things for "Lay-ish readers", I think a statement like:

"That means it took 48 million years (then) for distances to grow by 1%"

would be slightly easier than yours to wrap one's head around.

 PS: I'm surely biased, because I have used this "time for 1% growth" in my original one-shot (single redshift) cosmo-calculator.

Still a useful tool in the sense that it gives more calculated values per redshift. I have changed it lately to also use Planck 2013 values as defaults.

Last edited: Jul 11, 2014
12. Jul 11, 2014

### marcus

I see your point. I'll have to take a look at the one-shot redshift calculator.
One thing that keeps me anchored to the "per million years" way of speaking is that I don't imagine the growth rate changing significantly in such a short time as a million years.

But over longer spans of time the growth rate might change somewhat and throw one off.

As it is, suppose that the growth rate is constant and such that distance grow exactly 1/48 % in a million years. Then I think it takes 47,766,563 years for them to grow 1% :^)
Close enough, just a quibble.

13. Jul 11, 2014

### Jorrie

Yes, I see your point too, but maybe the thing that bothers more is the arbitrariness of the 1% that we are both using.
It now strikes me that the most physically natural characteristic may be that the instantaneous growth rate at a distance equal to the the Hubble radius (R) is one light year per year. Hence all distances grow by a fraction 1/R per year. Why don't we simply say (for you example at a=1/3):

"That means all distances then grew by a fraction 1/(4.8 billion) per year"?

Does the 1% really add anything to the understanding?

Bed time in my valley; will think again tomorrow...

14. Jul 11, 2014

### marcus

One of the issues we are thinking about here is whether to use a million year or a billion year time interval as a kind of step or unit in talking about distance growth.
I agree that, when doing a calculation with google, it's trimmer to switch to something like
(1/14.4^2 - 1/17.3^2) per (billion years)^2

The user can learn to recognize there the same numbers he sees in Lightcone.

The alternative is not as good, it would make some users eyes glaze over:
(1/14400^2 - 1/17300^2) per (million years)^2

Either one gives the right density of the universe ρ(t). They give the same answer.

You have suggested very reasonably that therefore we should TALK about the growth rate as
"1/14.4 per billion years", which makes sense for consistency with the model calculation as well as other reasons.

I still want to have it both ways. I want to sometimes use a compact (Gy) notation when calculating, but to TALK about the growth rate as 1/144 percent per million years. Partly because people are used to the idea of PERCENTAGE GROWTH. That's the handle most people have on exponential growth, using the word "percentage" activates familiar routines in the mind.

Also partly for the sake of precision. If something grows by 1/144 percent per My, then what fraction does it grow by in a billion years?
(1+1/14400)^1000 = 1.0719099243 = 1 + 1/13.906…

So the actual fractional growth at the present rate we are talking about is 1/13.9 in a billion years. With compounding. The fractional increment is not really 1/14.4 , it is more than that.

So it can confuse people if we say "1/14.4 per billion years".

It is actually less confusing, I think, to say "1/144 percent per million years".

Also we do not really expect the fractional growth rate to stay the same for a billion years. So using that time step could be misleading people by making them think of it as constant over such a time scale.

Anyway I'm glad you raised the issue and I think fractional OR percentage growth rate is the way to talk about Hubble parameter, hopefully in some kind of notation that shows the Hubble radius or Hubble time numbers explicitly, as we do in all these notation styles. So beginners can immediately get the reciprocal relation. And be learning about Hubble radius at the same time they learn about Hubble rate.

For now I think I'll continue on with describing H(now) as 1/144 % per My, but make sure readers know that as LINEAR RATES that's the same as 1/14.4 per Gy. That should get hammered in and become very familiar so that when one does a google calculation it can be second nature to write it compactly the way I did at start of post
H(now)^2 = 1/14.4^2 per (billion years)^2, and so on.

Last edited: Jul 12, 2014
15. Jul 12, 2014

### Jorrie

The idea I had my prior post was actually a fraction per year basis, i.e. at a=1/3, all distances between large scale structures increased at a fractional 1/(4.8 billion) per year (not 1/4.8 per billion years), perhaps emphasizing the 'instantaneousness' of the rate a little more.

I also feel that you can (newbie-) simplify your H(then) Google calculation a little, to:

(8 pi G/(3c^2)*6.453 nanopascal + 1/(17.3 billion years)^2)^(- .5)

I find the "1/17.3^2 per (billion years)^2" a little confusing, but that's perhaps just me. However, if we are talking to newcomers, we should set our own preferences aside and strive for simplicity - the subject is complex enough!

16. Jul 12, 2014

### marcus

What I'm hoping people might have ideas about, is how one could explain what the Friedman equation says in words that could give a newcomer some intuition. Ill recall the short summary here:
To me what this equation
$$H(t)^2 - H_\infty^2 = {1 \over \Phi} \rho (t)$$
says is there must be a balance between the amount of (spreading-type) curvature on the left and the amount (of converging-type) caused by matter on the right.

But that we only count spreading curvature on the left that is above a certain constant amount. Somehow nature knows there is this zero point curvature a spreading intrinsic to space. Matter, which causes the opposite type of curvature (where lines that start parallel tend to converge) only has to balance the part on the LHS of the equation that exceeds this zero point or "vacuum curvature".

Another way of talking about this balance (which would be more familiar to mathematicians) would be where you assign spreading, distance growth, curvature a MINUS SIGN and call it negative and call the type on the right side caused by matter positive and then say they have to add to zero. But let's not get distracted by elegance we don't need. I hope this idea of balancing raw unsigned amounts of curvature (presumably to maintain overall spatial flatness) is intuitive enough and will suffice to help a beginner make some sense of the equation. If anyone has a different way to account for it, please tell us.

What I'm hoping further is that since we have written Friedman so it LOOKS LIKE the full Einstein GR, if a beginner can make sense of the Friedman and get familiar with it then that will carry over and give intuition about the GR equation. It won't seem as mysterious or unapproachable as it otherwise might.

17. Jul 12, 2014

### marcus

My clock says we posted #15 and #16 at the exact same minute! Thanks for continuing to help wrestle with this. I think you may be right about combining and only doing one squaring operation. I will see if google calculator accepts that. But only after sleep and coffee next morning :^D.

18. Jul 12, 2014

### Jorrie

I agree with the essence of how you explained the balancing in the simplified Friedman equation, but I'm a little worried about the term "zero-point curvature" - it may create impressions that it is related to zero-point energy. If you want retain the equation in the 'iconic form', what about a line of discussion like:

The RHS ${1 \over \Phi} \rho (t)$, represents positive spatial curvature and $H_\infty^2$ represents intrinsic negative spatial curvature (not because of the negative sign, but built into the constant), so that $-H_\infty^2$ is positive. When Einstein originally postulated his 'static universe', he essentially equated $-H_\infty^2 = {1 \over \Phi} \rho (t)$, so that $H(t)$ was identically zero and this represented a static, spatially flat* universe (no expansion or contraction).

Presently $H(t) > 0$ (by observation), but Marcus' Friedman equation $H(t)^2 - H_\infty^2 = {1 \over \Phi} \rho (t)$ must still balance. Together with a variety of other observations, this fact limits the values of $H_\infty^2$ (essentially the cosmological constant) and $\rho$ (matter+dark matter density) to certain narrow ranges. These ranges point towards a universe that will expand forever at an accelerating rate.

Critique welcome...

* PS: I'm not sure if the "spatially flat" necessarily follows from Einstein's "balancing act". Maybe someone with more knowledge care to say?

Last edited: Jul 12, 2014
19. Jul 13, 2014

### Jorrie

I noticed the Ned Wright's web page on the cosmological constant starts with:
I interpret 'spherical geometry' as meaning a positively curved, spatially closed model. (?)

20. Jul 13, 2014

### marcus

Thanks for post #18. I couldn't say it better, so will just bring your explanation forward (since we just turned a page.)
You are right I shouldn't say "zero point curvature" but it might be OK to say "vacuum curvature" for Einstein's cosmological curvature constant. Bianchi and Rovelli use that term to refer to Lambda, in their 2010 essay on how to think about Lambda (google "why all these prejudices against a constant?" :^)

I think the sign could have to do with the Lorentz signature (-+++) which is an accessory which we don't have when writing the Friedman equation

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