marcus said:
Wikipedia has an article on "Friedmann equations" that includes the acceleration one.
I checked the article
http://en.wikipedia.org/wiki/Friedmann_equations
and it only had the one constant Lambda. The presentation was simpler than some others I remember seeing, that had a "dark energy equation of state w". But that's OK. Maybe w is an unnecessary complication and the Wikipedia treatment gets the basic idea across.
Well, maybe you already know all of this, but the Friedmann equation can just be written as
\left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3}\rho_{\textrm{tot}}
where a is the scale factor and \rho_{\textrm{tot}} is the total energy density of the universe. Now, assuming that \rho_{\textrm{tot}} = \rho_m + \rho_{\textrm{de}}, i.e. matter and dark energy are the two important components, then we have:
\left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3}\rho_{m} + \frac{8 \pi G}{3}\rho_{\textrm{de}}
Now, the interesting part is the relation that determines how the density of a component varies with scale factor. It's just given by:
\frac{d\rho}{da} +\frac{3}{a}(\rho + P) = 0
I think that the proper way to derive this is from the condition in general relativity that the covariant derivative of the stress-energy tensor vanishes. However, you can derive it from simple thermodynamic arguments. Just consider that any given volume in the universe expands adiabatically (no heat transfer in or out) so that the 1st law of thermodynamics says dU = -PdV where U is the total internal energy in that volume. Then you just consider that V ~ a
3 and you can get the equation above. Now, consider the equation of state (the relationship between pressure and density) for a given constituent of the universe:
P = w\rho
Non-relativistic matter is essentially pressure-less so that w
m = 0. Plugging that into the equation above gives you:
\frac{d}{da}(\rho_m a^3) = 0~~~~\Rightarrow \rho_m a^3 = \textrm{const.}~~~~\Rightarrow \rho_m \propto a^{-3}
Now, for a generic component where we may not be entirely sure what w is, solving the equation gives you:
\rho a^{3(1+w)} = \textrm{const.}
So, the evolution of the energy density of dark energy with scale factor depends on the equation of state of dark energy i.e. it depends on w
de, which I'll just call w from now on. You can see that if w = -1 (which we think it is close to), we just end up with the result that:
\rho_{\textrm{de}} = \textrm{const.}
If that's true, then I can write my Friedmann equation above as:
\left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3}\rho_{m} + \frac{\Lambda}{3}
where
\Lambda \equiv 8 \pi G \rho_{\textrm{de}}
is a
constant. That leads me to the whole point of my post, which was to address marcus' quoted statement above by explaining that
for dark energy, w = -1 is consistent with a cosmological constant \Lambda.
If w is not exactly equal to -1, then you have some other more exotic form of dark energy that cannot be expressed as a cosmological constant term in the Friedmann equation. But you could still express that equation in terms of the parameter w and other constants. To do this, we need two facts. One is that the density parameter \Omega_i for the i
th constituent of the universe is defined as \rho_{i,0}/\rho_{\textrm{cr}} where the zero in the subscript denotes the value of the density today, and the reciprocal of the critical density is just 8 \pi G / 3H_0^2. Using these relations, the Friedmann equation becomes:
\left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3}(\rho_{m,0})a^{-3} + \frac{8 \pi G}{3}(\rho_{\textrm{de,0}})a^{-3(1+w)}
$$\left(\frac{\dot{a}}{a}\right)^2 = H_0^2\left(\frac{\rho_{m,0}}{\rho_{\textrm{cr}}} \right)a^{-3}+ H_0^2\left(\frac{\rho_{\textrm{de,0}}}{\rho_{\text{cr}}}\right)a^{-3(1+w)} $$
\left(\frac{\dot{a}}{a}\right)^2 = H_0^2\Omega_m a^{-3} + H_0^2\Omega_{\textrm{de}} a^{-3(1+w)}
Since the left-hand side is just the the square of Hubble parameter H, it's nice to express the final result as
H^2 = H_0^2[\Omega_m a^{-3} + \Omega_{\textrm{de}} a^{-3(1+w)}]
So you do need to explicitly include w in the case where it is not -1 and the effect of dark energy cannot be expressed as a cosmological constant term in the equations. I'm not sure whether any of this adds to the discussion, but I thought it was good to include for completeness.
I've omitted the curvature term in the Friedmann equations throughout this discussion, so I guess I've implicitly assumed a flat universe.
