PDA

View Full Version : Friedmann paradox


meteor
May8-04, 02:25 AM
Please show me how to solve this paradox
cos the Friedmann equation is:

H^{2}= \frac{8*pi*G*rho}{3}-\frac{k*c^{2}}{R^{2}}

but the concordance model says that k=0, so we can eliminate the last term of the equation
then we isolate rho, the density of energy:

rho=\frac{3*H^{2}}{8*pi*G}

but however, the formula for the critical energy density is

rho_{crit}=\frac{3*H^2*c^2}{8*pi*G}

but the concordance model says that rho=rhocrit
but you see that the 2 formulae are not equal, there's an extra c2 in the formula for rhocrit

I can't figure where is the mistake

Stingray
May8-04, 04:00 AM
rho is mass density. rho_crit is energy density (as you wrote it). Mass and energy have different units, and you use c^2 to convert between them.

Most people like to use units such that c=1, and then energy and mass are interchangable. This convention is so common that books are sometimes careless about distinguishing the two.

meteor
May8-04, 05:45 AM
please, take a look to this thread
http://www.physicsforums.com/showthread.php?t=2864
Here's jeff, and I think that he is a knowledgeable person, and says that rho includes all kinds of energy, not only mass


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)

Stingray
May8-04, 01:46 PM
please, take a look to this thread
http://www.physicsforums.com/showthread.php?t=2864
Here's jeff, and I think that he is a knowledgeable person, and says that rho includes all kinds of energy, not only mass

That's true. I was being a little sloppy myself. rho includes everything, but the way you wrote it, it has units of mass/volume, whereas rho_crit has units of energy/volume. Wherever you're quoting rho_crit from has a slightly different form for Friedmann's equation than you do, so their answer is different by c^2. This is just a convention. You can choose either mass units or energy units for rho as long as H works out as 1/time in the end.

Chronos
May19-04, 03:48 AM
use the conversion c^2=m/e