- #1

- 10

- 0

http://www.uni.edu/morgans/ajjar/Cosmology/cosmos.html

But I want to know the equation they used to calculate the Hubble Constant at z = 6? So, given the matter density, energy density, and redshift, how do you calculate the Hubble constant?

- Thread starter Epistimonas
- Start date

- #1

- 10

- 0

http://www.uni.edu/morgans/ajjar/Cosmology/cosmos.html

But I want to know the equation they used to calculate the Hubble Constant at z = 6? So, given the matter density, energy density, and redshift, how do you calculate the Hubble constant?

- #2

marcus

Science Advisor

Gold Member

Dearly Missed

- 24,738

- 787

You probably know the Friedmann equation, if not just google it.

Take a simple case with k=0 (spatial flat) and Lambda=0, just to see how it goes.

The Friedmann equation says that H^{2} goes as the density. So H goes as the square root of the density. But assuming matter is the main content, density increases as (z+1)^{3}.

So in this Lambda=0 case it is really easy.

The radiation content of the U at present is such a small fraction that one can neglect it.

Radiation density goes as the fourth power of (z+1) because the number of photons per volume increases as the cube, and the energy of each one increases linearly and distances and wavelengths shrink.

But the universe is "matter dominated" i.e. we can neglect radiation, so we don't use the 4th power we just use the cube.

The other complication is if Lambda > 0. Think of it as an equivalent dark energy. Dark energy density does not change. So part of the energy density of the U (73% at present) is not changing as z+1 increases, and part of the density (27%) is going up as the cube of z+1.

Maybe someone will step in with solved equations. But it's easy in any case, even with Lambda >0.

I'm glad you found Prof. Morgan's calculator! It's simple and good for learning with and I like her for putting it online. I understand she collects classic superhero comicbooks, another appealing trait. Dont forget to put .27, .73, and 71 in the first three boxes on the left margin. HAVE FUN!

I'm guessing that you already know that the factor by which distances and wavelengths are shorter is z+1. Light from a redshift z = 1 galaxy has wavelengths TWICE as long when the light gets here. Light from a z=2 galaxy has wavelenghts THREE times what they were when the light was emitted. And distances were 1/3 then what they are today. and volumes 1/27. And matter densities 27 times what they are today etc etc.

You are always having to add one to the z in order to get useful factor.

Take a simple case with k=0 (spatial flat) and Lambda=0, just to see how it goes.

The Friedmann equation says that H

So in this Lambda=0 case it is really easy.

The radiation content of the U at present is such a small fraction that one can neglect it.

Radiation density goes as the fourth power of (z+1) because the number of photons per volume increases as the cube, and the energy of each one increases linearly and distances and wavelengths shrink.

But the universe is "matter dominated" i.e. we can neglect radiation, so we don't use the 4th power we just use the cube.

The other complication is if Lambda > 0. Think of it as an equivalent dark energy. Dark energy density does not change. So part of the energy density of the U (73% at present) is not changing as z+1 increases, and part of the density (27%) is going up as the cube of z+1.

Maybe someone will step in with solved equations. But it's easy in any case, even with Lambda >0.

I'm glad you found Prof. Morgan's calculator! It's simple and good for learning with and I like her for putting it online. I understand she collects classic superhero comicbooks, another appealing trait. Dont forget to put .27, .73, and 71 in the first three boxes on the left margin. HAVE FUN!

I'm guessing that you already know that the factor by which distances and wavelengths are shorter is z+1. Light from a redshift z = 1 galaxy has wavelengths TWICE as long when the light gets here. Light from a z=2 galaxy has wavelenghts THREE times what they were when the light was emitted. And distances were 1/3 then what they are today. and volumes 1/27. And matter densities 27 times what they are today etc etc.

You are always having to add one to the z in order to get useful factor.

Last edited:

- #3

- 10

- 0

I'm about to try it I'll let you know how it goes.

- #4

- 10

- 0

By the way, I got it, thanks!

- Last Post

- Replies
- 7

- Views
- 3K

- Last Post

- Replies
- 7

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 17

- Views
- 3K

- Replies
- 8

- Views
- 2K

- Replies
- 3

- Views
- 1K

- Replies
- 5

- Views
- 997

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 5

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 1K