Calculating the Hubble Constant at Different Redshifts

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SUMMARY

The discussion focuses on calculating the Hubble constant at different redshifts, specifically at z = 6, using the Friedmann equation. Participants emphasize that in a matter-dominated universe with Lambda = 0, the Hubble constant can be derived from the square root of the density, which increases as (z + 1)³. The significance of dark energy is noted, where 73% of the energy density remains constant while 27% increases with redshift. The Cosmos Calculator by Prof. Morgan is recommended as a practical tool for these calculations.

PREREQUISITES
  • Understanding of the Friedmann equation
  • Knowledge of cosmological parameters such as matter density and energy density
  • Familiarity with redshift concepts
  • Basic grasp of dark energy implications in cosmology
NEXT STEPS
  • Study the Friedmann equation in detail
  • Explore the implications of dark energy on cosmic expansion
  • Learn how to use the Cosmos Calculator for cosmological calculations
  • Investigate the effects of varying Lambda values on Hubble constant calculations
USEFUL FOR

Astronomers, astrophysicists, and students of cosmology who are interested in understanding the calculations behind the Hubble constant and the dynamics of the universe at different redshifts.

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How can you mathematically calculate the Hubble constant at different redshifts? I know it's possible, and I can get the number by using the Cosmos Calculator:

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?
 
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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 H2 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! :biggrin:

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:
Ahh ok, I think I understand. So basically I just solve the Friedmann equations for H with z = 6, 0.27 for the matter density, and 0.73 for the energy density?

I'm about to try it I'll let you know how it goes.
 
By the way, I got it, thanks!
 

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