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Hubble constant 
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#1
Jun803, 02:46 PM

P: 915

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 


#2
Jun803, 03:32 PM

Astronomy
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PF Gold
P: 23,098

It can be written as a mass density equivalentkilograms per cubic meter Or, by including a c^2, it can be written as an energy densityjoules per cubic meter. The actual energy density of the universe is now believed to be equal to rho critor very nearlyso 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
Jun803, 03:42 PM

Astronomy
Sci Advisor
PF Gold
P: 23,098

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 percentagenowadays 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 offwhich is fine since there is plenty of uncertainty. 


#4
Jun803, 05:09 PM

P: 915

Hubble constant



#5
Jun803, 06:06 PM

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#6
Jun803, 09:03 PM

P: 402

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
Jun803, 09:09 PM

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#8
Jun803, 10:34 PM

P: 915

The Hubble parameter is a variable thing. The Hubble constant is the value of the Huble parameter at the present moment
H^2=((8*pi*G*rho)/3)+(lambda/3)(k/(a^2)) http://super.colorado.edu/~michaele/Lambda/evol.html http://www.blackwellsynergy.com/lin...2.43110.x/html What represents rho in this equation? 


#9
Jun803, 11:16 PM

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#10
Jun803, 11:50 PM

P: 915

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? 


#11
Jun903, 12:06 AM

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#12
Jun903, 12:09 AM

P: 915

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



#13
Jun903, 12:32 AM

P: 915

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
Jun903, 07:12 AM

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#15
Jun903, 02:50 PM

Astronomy
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PF Gold
P: 23,098

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
Jun903, 09:40 PM

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P: 660

The olive branch lives. 


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