*quantitative* fun with the cosmos (easy)

In summary: G))*(850 nanojoules per m^3)And what does that equal?It equals 107 kilometers per second per megaparsec!In summary, the Hubble expansion rate is about 71 kilometers per second per megaparsec. This corresponds to a fractional growth in distance of about 7 percent per billion years. If you want it in percent then paste this in the window: 71 kilometers per second per megaparsec*10^9 years in percent.
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marcus
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The basic number you hear all the time is that the Hubble expansion rate is

71 km/s per megaparsec

So get the google calculator to tell you want actual percentage growth rate that turns out to be. Say over a billion year span of time. Past this blue thing verbatim into the google window:

71 km/s per megaparsec*10^9 years

That's the approximate fractional growth in a billion years. If you want it in percent then paste this in the window:
71 km/s per megaparsec*10^9 years in percent

If you don't specify to the calculator in what terms you want the answer, it will say the answer in whatever terms it thinks are appropriate. If you know what terms you want it in, it can help to specify (as we did here.)

Now the expansion rate actually changes gradually over time, so it would be more precise to use a shorter time interval, like a year, or a thousand years, or a million years. But those more precise tactics lead to very small percentage numbers. What we're really talking about is an instantaneous rate, but for convenience we express it crudely with a billion-year time step.

And 71 is just a commonly used ballpark figure. There is a plus-minus errorbar for it.
You can see from google that it corresponds to about 7 percent increase in distance per billion years. (But I really mean the instantaneous rate 7 billionths of a percent per year.)

The actual figure the googleator gives is 7.26. Round it off to 7.3 or 7 if you want.
I will keep using 7.26, and not round off, because it is just an alias for that conventional number 71.

What I want to suggest doing is use this expansion rate of about 7 percent per billion years to estimate the density of the universe (including everything: ordinary and dark).

Corresponding to every density, there is an ideal perfect expansion rate which is just the rate needed so the universe is spatially flat and stays flat.

So conversely, given any particular expansion rate there is a perfect ideal density (the critical density) that is just right so she is spatially flat and stays that way.

What I'm suggesting is that anyone at cosmo forum who hasn't already calculated the critical density (that goes with 7.26 percent expansion) should try it.

And I'll show how. It's easy.
 
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there is a good quantity to know, a coefficient that you use

"ate pig over three see square" 8 pi G/(3 c^2)

The critical relation needed for a nice universe is:
square of expansion rate equals that coefficient times the density.

H^2 = 8 pi G/(3 c^2) * density

If you know H, like it's the observed 7.26 percent per 10^9 years, then you can solve for the critical or perfect density to go with that rate of expansion.

This involves turning over the coefficient and getting

"three see square over ate pig" 3 c^2/(8 pi G)

If we solve our density-expansion equation for the density we get:
density = (3 c^2/(8 pi G))*H^2

In other words, using what we know about the present expansion rate:

[critical] density = (3 c^2/(8 pi G))*(7.26 percent per 10^9 years)^2

Now that is ready to go into the google calculator except we need to specify what UNITS we want the density expressed in. I'm assuming we convert everything into the common currency of energy and want it expressed in the standard metric form of joules per cubic meter. The matter is all converted to energy so we can add it together with the dark energy and radiation energy etc.

Since the actual density in the universe is very sparse it will turn out nanojoules per m^3, so we can specify that in what we paste into the google window:

(3 c^2/(8 pi G))*(7.26 percent per 10^9 years)^2 in nanojoules per m^3

Let's see how that works. Does it give us some reasonable density figure in nanojoules per cubic meter?

YES! Check it out. Google says the presentday critical density is 0.85 nanojoule per m^3.
 
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  • #3
So now we can begin to have fun and play around with the universe.

And we can ask what was the percentage expansion rate back when the universe was 1000 times denser?

Later we can see what redshift that would correspond to. Roughly ballpark you can see (without worrying about the dark energy component) that distances would be roughly 10 times shorter, to make the density 1000 times bigger. We can be more exact about that later. But now let's see what the expansion rate would have needed to be.

We have to manipulate the basic equation we used earlier so that now it will give the expansion rate when you plug in the density.

And then we have to plug in a much larger density of 850 nanojoules per m^3, instead if today's 0.85 nanojoules per m^3.

So here is the basic equation we used earlier:
square of expansion rate equals the pig coefficient times the density.

H^2 = 8 pi G/(3 c^2) * density

And if we want it to give us the expansion rate we just need to take the square root:
expansion rate equals square root of (coefficient times density).

H = sqrt( 8 pi G/(3 c^2) * density)

So let's paste the following blue thing directly into google and see if it works:

sqrt( 8 pi G/(3 c^2) * 850 nanojoules per m^3) in percent per 10^9 years

Yes! Way cool. It gives a percentage addition to distance that happens per billion years. When the universe is 1000 times denser than today.

Actually as we discussed earlier, a billion years is too crude and we should be using a shorter time-step, and getting smaller percentage add-ons to distance. But very small numbers can be a pain so I'm trying it this way.
 
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1. What is quantitative fun with the cosmos?

Quantitative fun with the cosmos refers to using mathematical and scientific methods to study and understand the universe. It involves collecting and analyzing data to make predictions and draw conclusions about the cosmos.

2. How is quantitative fun with the cosmos different from other types of science?

Quantitative fun with the cosmos is unique in that it focuses specifically on using quantitative methods and data to study the universe. This often involves complex mathematical equations and statistical analyses, making it different from other types of science that may rely more on qualitative observations.

3. What are some examples of quantitative fun with the cosmos?

Examples of quantitative fun with the cosmos include measuring the distance between celestial objects, calculating the speed of light, and analyzing the chemical composition of stars. Other examples may include predicting the movement of planets and galaxies, and studying the behavior of black holes.

4. Why is quantitative fun with the cosmos important?

Quantitative fun with the cosmos is important because it allows us to gain a deeper understanding of the universe and its workings. By using quantitative methods, we can make more accurate predictions and discoveries about the cosmos, leading to advancements in technology and our understanding of the world around us.

5. Can anyone participate in quantitative fun with the cosmos?

Yes, anyone with an interest in science and mathematics can participate in quantitative fun with the cosmos. While it may require a strong understanding of these subjects, there are many resources available for individuals to learn and engage in this type of scientific exploration.

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