# Google calculator impact on basic cosmo explanations

1. Dec 6, 2008

### marcus

If you like to calculate, it can help you learn cosmo basics. Partly just by making it fun. Partly because it gives a hands-on dimension to what you are learning. If you don't find calculating basic cosmo numbers fun, then this thread is not for you. There are other less number-crunchy approaches to understanding.

Google calculator changes the circumstances somewhat. You type something in the window and press return (or "search") and it evaluates it. It knows constants like G, hbar, k, pi. It knows units like lightyear and parsec and nanopascal. So it makes it easy. You don't have to look up constants and convert units and so on.

So suppose you want to try this out with a few basic cosmo quantities. Let's say you know that the present value of H is 71 km/s per megaparsec, and that the total energy density is very close to the critical density needed for spatial flatness, and that dark energy is estimated about 73 percent of total. Essentially you know two numbers 71 and 0.73. What can you calculate from that?

The Hubble radius is the distance at which objects at rest wrt background are receding at speed c. If an object's present distance from us is twice that radius, then it is receding from us at 2c. So put this into google:
"c/(71 km/s per megaparsec) in lightyears"
You get out:
c / (71 ((km / s) per megaParsec)) = 1.37720275 × 1010 lightyears
So 13.77 billion lightyears.

Total energy density (in nanopascal)
One joule per cubic meter is a metric unit of energy density called the Pascal. It is equivalent to one Newton per square meter, the corresponding unit of pressure. To find the energy density of the universe, in nanojoules per cubic meter, put this in:
"3 c^2 (71 km/s per megaparsec)^2/(8 pi G) in nanopascal"
You get out:
(3 * (c^2) * ((71 ((km / s) per megaParsec))^2)) / ((8 * pi) * G) = 0.85... nanopascals
We've used the critical density as an estimate of the real density, because of near flatness.

Dark energy density (in nanopascal)
Just take 73 percent of the above, put this in:
"0.73*3 c^2 (71 km/s per megaparsec)^2/(8 pi G) in nanopascal"
You get out 0.62... nanopascal.

So that's how much dark energy there is. 0.62 nanojoules per cubic meter, or if it seems easier to imagine, 0.62 joules per cubic kilometer. I like the latter. It is easy to imagine a joule of energy---raise a book a few centimeters up off the table, drop it, the satisfying thump is a joule of energy. About that much in a cubic kilometer.

If you are new and want to know where the formula for critical density comes from, that's good. This approach is meant to motivate you to find the underlying equations involved in a calculation. Just google Friedmann equations. You can read what critical density is from one of the two equations discussed there. Flat means k=0, so put it equal zero and solve for the density rho.

I'll get some other examples.

2. Dec 6, 2008

### marcus

If somebody has a favorite or neat cosmo calculation that demonstrates the virtues of the google calculator, it would be fine if they want to contribute it. If you do contribute please make it concise. Lengthier explanations can go in another thread, especially if someone asks to have something explained. I'm hoping to keep this thread reasonably compact. Here is another.

The cosmological event horizon
This is an interesting distance. If another galaxy is nearly that far, and at rest relative to background, then if you leave here today going the speed of light you can just barely reach the galaxy. On the other hand if it is beyond that distance, even if you leave at once and travel at c you will never reach it. Our universe is approximately deSitter and in that case there is a simple way to estimate the event horizon distance. It equals the socalled deSitter radius.

In a deSitter universe with cosmo constant Lambda the deSitter radius is L = sqrt(3/Lambda). This L tells us not only the distance to the event horizon but also a temperature associated with our universe's accelerating expansion.
A deSitter universe has a temperature--analogous to the Hawking BH temp, or the Unruh temp: kT = hbar*c/(2 pi L)

But first let's get a handle on L and Lambda.
We know from post #1 that dark energy density is 0.62 nanopascal.
Dark energy density is related to cosmological constant by
Lambda = 8 pi G (.62 nanopascal)/c^4
L = sqrt (3/Lambda) = sqrt (3 c^4/(8 pi G (.62 nanopascal))

We would like that in billions of lightyears. So put this into google:
"sqrt (3*c^4/(8 pi G (.62 nanopascal))) in lightyears"
You will get out 16 billion lightyears. If you have trouble with this, please let me know. The first time I tried it, it didn't work for some reason. I erased "in lightyears" and it worked. Then I put that phrase back and it still worked. So there may be some variation.

deSitter temperature
There is a temperature kT = hbar*c/(2 pi L) associated with the cosmological event horizon. Stuff that is at this moment beyond that horizon cannot ever get to us even if it travels at c. A bit like a black hole horizon in that sense---but we are not the ones in it, the stuff outside the event horizon is what is metaphorically in the black hole. It's not a real black hole, just faintly analogous. Well now we know the distance L = 16 billion lightyears, so let's calculate the energy kT and the temperature T.
Put in this:
"hbar*c/(k*2 pi (16 billion light years)) in kelvin"

You get out a ridiculously small temperature 2 x 10-30 kelvin.
But that is not too unfamiliar. Massive black holes have extremely low Hawking temperatures which is why they are so slow to evaporate. The Unruh temps associated with normal size accelerations are also very tiny.

Last edited: Dec 6, 2008
3. Dec 6, 2008

### marcus

Maybe I started with stuff that is too obscure. Here is something simpler.
Put this into the google window and press return:
G*solar mass/c^2

That will give the size of a black hole with the same mass as the sun. You probably know what it is---the schwarzschild radius for one solar mass. But you may not have calculated it.

4. Dec 7, 2008

### mysearch

Hi Marcus,
I don’t want to detract from the focus of your thread, but I think Hellfire’s calculator is an excellent reference: http://www.geocities.com/alschairn/cc_e.htm

Purely as a suggestion, maybe you might consider working through a specific example, e.g. z=1090, using this calculator and explaining the results given as it might help explain a number of key concepts within the LCDM model.