Solving Hubble Constant with rhocrit and Cosmological Constant

In summary: Originally posted by meteor Let's see if I can grasp it. Then the correct Freedmann equation is:H^2=((8*pi*G*rho)/3)+(lambda/3)-(k/(R^2))and here, rho represents the TRUE density of energy but not the CRITICAL density of energy. Ok?Yes, that equation is correct. And yes, rho represents the true density of energy.Originally posted by meteor I don't know what you mean with "unnecesary". Do you mean that if I erase (lambda/3) of the equation, then it all remains... the same?Yes, if you erase the lambda/3 term, then the equation is still valid.
  • #1
meteor
940
0
I see often that rhocrit(the critical density of matter), is expressed this way:
rhocrit=(3*(H^2))/(8*pi*G)
This is not correct because a cosmological constant is missed.
This is the Friedmann Equation:
H^2=((8*pi*G*rhocrit)/3)+(lambda/3)-(k/(a^2))
Since it is known that the curvature(k)of the universe is zero, the Friedmann equation can be reduced to this:
H^2=((8*pi*G*rhocrit)/3)+(lambda/3)
and rhocrit is:
rhocrit=(3*((H^2)-(lambda/3)))/(8*pi*G)
Now, i would like to solve this equation. I have a value for rhocrit of 10^(-26)kg/m^3, but i need the value of the cosmological constant to complete the equation. You know the value? (in SI units, please)

also given that H=((da/dt)/da), being a the scale factor and H=Hubble constant, it would be nice to know the current value of the scale factor
 
Last edited:
Astronomy news on Phys.org
  • #2
Originally posted by meteor
I see often that rhocrit(the critical density of matter), is r

You are mistaken at the outset. rho crit is not the critical density of matter but includes all forms of energy----light, dark matter, dark energy (the cosm. constant contribution).

It can be written as a mass density equivalent------kilograms per cubic meter

Or, by including a c^2, it can be written as an energy density----joules per cubic meter.

The actual energy density of the universe is now believed to be equal to rho crit---or very nearly---so that the universe is spatially flat.

The cosmological constant (sometimes referred to as dark energy term) is believed to represent around 73 percent of the total rho.


Originally posted by meteor
I see often that rhocrit(the critical density of matter), is expressed this way:
rhocrit=(3*(H^2))/(8*pi*G)
This is not correct because a cosmological constant is missed.
This is the Friedmann Equation:
H^2=((8*pi*G*rhocrit)/3)+(lambda/3)-(k/(a^2))
Since it is known that the curvature(k)of the universe is zero, the Friedmann equation can be reduced to this:
H^2=((8*pi*G*rho)/3)+(lambda/3)
and rhocrit is:
rhocrit=(3*((h^2)-(lambda/3)))/(8*pi*G)
Now, i would like to solve this equation. I have a value for rhocrit of 10^(-26)kg/m^3, but i need the value of the cosmological constant to complete the equation. You know the value? (in SI units, please)

also given that H=((da/dt)/da), being a the scale factor and H=Hubble constant, it would be nice to know the current value of the scale factor
 
  • #3
Last time I looked at rho crit in SI units it was, I believe,
0.83 joules per cubic kilometer.

the cosmological constant is usually given as a percentage----nowadays typically 73 percent of that.

So you just do 0.73 of 0.83 and get 0.61 joules per cubic kilometer.

Your figure is probably just 0.83 joules divided by the square of the speed of light (to give a kilogram equivalent) and rounded off---which is fine since there is plenty of uncertainty.
 
Last edited:
  • #4
Or, by including a c^2, it can be written as an energy density----joules per cubic meter.
I don't know if this is correct. I've seen in various texts that rhocrit is the critical mass density, but yes, you can always transform mass to energy with E=m*(c^2), and then the formula is like you say, rhocrit=((3*(H^2)-(lambda/3))*(c^2))/(8*pi*G), but doing this you have an energy density that doesn't includes the light, and remember that light has energy but not mass
 
Last edited:
  • #5
Originally posted by meteor
This is the Friedmann Equation:
H^2=((8*pi*G*rhocrit)/3)+(lambda/3)-(k/(a^2))

It's not rhocrit but rho, which does in fact - as marcus pointed out - include all forms of energy so there's no need for the lamda term (review the derivation of the FRW equation). Also, in the k-term, a^2 should be R^2.
 
  • #6
Slightly unrelated but interesting

if the "Hubble Constant" were in fact constant the Universe would expand in an exponential fashion. Which doesn't seem very likely! So I just call it the Hubble Value.
 
  • #7


Originally posted by Tyger
if the "Hubble Constant" were in fact constant the Universe would expand in an exponential fashion. Which doesn't seem very likely! So I just call it the Hubble Value.

Maybe Hubble parameter.
 
  • #8
The Hubble parameter is a variable thing. The Hubble constant is the value of the Huble parameter at the present moment
Also, in the k-term, a^2 should be R^2.
a and R are the same thing, the scale factor

It's not rhocrit but rho, which does in fact - as marcus pointed out - include all forms of energy so there's no need for the lamda term (review the derivation of the FRW equation).

I've seen the Friedmann Equation in this way:
H^2=((8*pi*G*rho)/3)+(lambda/3)-(k/(a^2))
http://super.colorado.edu/~michaele/Lambda/evol.html [Broken]
www.blackwell-synergy.com/links/doi/10.1046/j.1468-4004.2002.43110.x/html[/URL]

What represents rho in this equation?
 
Last edited by a moderator:
  • #9
Originally posted by meteor
a and R are the same thing, the scale factor

They're not the same: a(t) is dimensionless while R(t) has units of length. We define a(t):=R(t)/R(0) in which the present time occurs at t=0 so that a(0)=1.

Originally posted by meteor
I've seen the Friedmann Equation in this way:
H^2=((8*pi*G*rho)/3)+(lambda/3)-(k/(a^2))

Yes, so have I, it's very common. It's unnecessary though because isotropy (which by the way implies homogeneity) allows you - in this special case - to view rho as incorporating the vacuum contribution.

Originally posted by meteor
What represents rho in this equation?

rho is the total density and includes all contributions from matter, radiation and the vacuum.
 
Last edited:
  • #10
Yes, so have I, it's very common. It's unnecessary though because isotropy (which by the way implies homogeneity) allows you - in this special case - to view rho as incorporating the vacuum contribution.

Let's see if I can grasp it. Then the correct Freedmann equation is:
H^2=((8*pi*G*rho)/3)+(lambda/3)-(k/(R^2))
and here, rho represents the TRUE density of energy but not the CRITICAL density of energy. Ok?
I don't know what you mean with "unnecesary". Do you mean that if I erase (lambda/3) of the equation, then it all remains good?
 
Last edited:
  • #11
Originally posted by meteor
Let's see if I can grasp it. Then the correct Freedmann equation is:
H^2=((8*pi*G*rho)/3)+(lambda/3)-(k/(R^2))
and here, rho represents the TRUE density of energy but not the CRITICAL density of energy. Ok?
I don't know what you mean with "unnecesary". Do you mean that if I erase (lambda/3) of the equation, then it all remains good?

Yep, everything you just wrote is correct.
 
  • #12
Ok, then i will delete. Occam's razor. :wink:
 
  • #13
One thing, Marcus use to write the critical energy density in this way:
rhocrit=(3*(c^2)*(H^2))/(8*pi*g)
but if you look at the Friedmann equation, eliminating (lambda/3) and eliminating (k/(R^2)), because k is zero, then the Friedmann equation is simply:
rho=(3*(H^2))/(8*pi*g)
and given that omega=rho/rhocrit=1 I think that Marcus should erase c^2 of the formula
 
  • #14
Originally posted by meteor
One thing, Marcus use to write the critical energy density in this way:
rhocrit=(3*(c^2)*(H^2))/(8*pi*g)
but if you look at the Friedmann equation, eliminating (lambda/3) and eliminating (k/(R^2)), because k is zero, then the Friedmann equation is simply:
rho=(3*(H^2))/(8*pi*g)

rhocrit is simply the value of rho in the k=0 case, so the friedmann equation in this case should be rhocrit=(3*(H^2))/(8*pi*G), where I've made the correction g -> G.

Originally posted by meteor
and given that omega=rho/rhocrit=1 I think that Marcus should erase c^2 of the formula

Implicit in the way that rhocrit here has been written as rhocrit=(3*(H^2))/(8*pi*G) is the choice that has been made to work in units defined by requiring c=1. If we use units where c is it's usual value, c would appear in the above equation as rhocrit=3*(H^2)c^2/(8*pi*G) which has units of energy/volume. Although in units where c=1 rhocrit has units of mass//volume, this is just because we took c=1, really it's an energy density.
 
Last edited:
  • #15
Bravo A+ and thanks for responding to Meteor about my post. I had to be off the net for much of yesterday and was glad to see you were taking care of business.

***********

Originally posted by steinitz


--------------------------------------------------------------------------------
Originally posted by meteor
One thing, Marcus use to write the critical energy density in this way:
rhocrit=(3*(c^2)*(H^2))/(8*pi*g)
but if you look at the Friedmann equation, eliminating (lambda/3) and eliminating (k/(R^2)), because k is zero, then the Friedmann equation is simply:
rho=(3*(H^2))/(8*pi*g)
--------------------------------------------------------------------------------

rhocrit is simply the value of rho in the k=0 case, so the friedmann equation in this case should be rhocrit=(3*(H^2))/(8*pi*G), where I've made the correction g -> G.

quote:
--------------------------------------------------------------------------------
Originally posted by meteor
and given that omega=rho/rhocrit=1 I think that Marcus should erase c^2 of the formula
--------------------------------------------------------------------------------

Implicit in the way that rhocrit here has been written as rhocrit=(3*(H^2))/(8*pi*G) is the choice that has been made to work in units defined by requiring c=1. If we use units where c is it's usual value, c would appear in the above equation as rhocrit=3*(H^2)c^2/(8*pi*G) which has units of energy/volume. Although in units where c=1 rhocrit has units of mass//volume, this is just because we took c=1, really it's an energy density.

Last edited by steinitz on 06-09-2003 at 08:23 AM
 
  • #16
Originally posted by marcus
Bravo A+ and thanks for responding to Meteor about my post. I had to be off the net for much of yesterday and was glad to see you were taking care of business.

***********
\

The olive branch lives.
 
Last edited:

1. How is the Hubble Constant related to rhocrit and the Cosmological Constant?

The Hubble Constant, also known as H0, is a measure of the current expansion rate of the universe. It is related to the critical density of the universe, rhocrit, and the Cosmological Constant, Λ, through the equation H0 = √(8πG/3) * √(rhocrit + Λ/3), where G is the gravitational constant.

2. What is rhocrit and how is it calculated?

Rhocrit is the critical density of the universe, which is the density required for the universe to be flat. It is calculated using the equation rhocrit = 3H0^2/8πG, where H0 is the Hubble Constant and G is the gravitational constant.

3. How does solving for the Hubble Constant using rhocrit and the Cosmological Constant help us understand the universe?

By solving for the Hubble Constant, we are able to determine the expansion rate of the universe, which provides insight into the age and size of the universe. Additionally, the relationship between H0, rhocrit, and Λ can help us understand the overall structure and composition of the universe.

4. Are there any uncertainties or limitations in using rhocrit and the Cosmological Constant to solve for the Hubble Constant?

Yes, there are uncertainties and limitations in this method. The calculation of rhocrit and the Cosmological Constant relies on assumptions about the composition and evolution of the universe, which may not be entirely accurate. Additionally, there may be systematic errors in the measurements of H0, which can affect the accuracy of the final result.

5. How do scientists continue to refine and improve the accuracy of the Hubble Constant using rhocrit and the Cosmological Constant?

Scientists are constantly working to improve the accuracy of the Hubble Constant by refining measurements of rhocrit and the Cosmological Constant, as well as developing new methods and techniques for determining H0. They also use data from various astronomical observations, such as the cosmic microwave background radiation and the cosmic distance ladder, to cross-check and validate their results.

Similar threads

Replies
6
Views
1K
  • Astronomy and Astrophysics
Replies
5
Views
2K
Replies
18
Views
1K
  • Astronomy and Astrophysics
Replies
1
Views
4K
  • Cosmology
Replies
8
Views
2K
Replies
19
Views
378
  • Astronomy and Astrophysics
Replies
1
Views
3K
Replies
27
Views
4K
  • Astronomy and Astrophysics
Replies
7
Views
2K
Back
Top