# Hubble constant

1. Jun 8, 2003

### 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*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: Jun 8, 2003
2. Jun 8, 2003

### marcus

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.

3. Jun 8, 2003

### marcus

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: Jun 8, 2003
4. Jun 8, 2003

### meteor

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: Jun 8, 2003
5. Jun 8, 2003

### jeff

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. Jun 8, 2003

### Tyger

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. Jun 8, 2003

### jeff

Re: Slightly unrelated but interesting

Maybe hubble parameter.

8. Jun 8, 2003

### meteor

The Hubble parameter is a variable thing. The Hubble constant is the value of the Huble parameter at the present moment
a and R are the same thing, the scale factor

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

What represents rho in this equation?

Last edited by a moderator: May 1, 2017
9. Jun 8, 2003

### jeff

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.

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.

rho is the total density and includes all contributions from matter, radiation and the vacuum.

Last edited: Jun 8, 2003
10. Jun 8, 2003

### 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?

Last edited: Jun 9, 2003
11. Jun 9, 2003

### jeff

Yep, everything you just wrote is correct.

12. Jun 9, 2003

### meteor

Ok, then i will delete. Occam's razor.

13. Jun 9, 2003

### 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)
and given that omega=rho/rhocrit=1 I think that Marcus should erase c^2 of the formula

14. Jun 9, 2003

### jeff

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.

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: Jun 9, 2003
15. Jun 9, 2003

### 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.

***********

16. Jun 9, 2003

### jeff

\

The olive branch lives.

Last edited: Jun 9, 2003