Special Cases

k = 0

Here, as noted above, our definition of $\lambda$ becomes arbitrary. However, by a change of variables to

$$\rho_{\Lambda}^{} \rho_{m}^{} \lambda$$ (9)

the solution can be written:

$${R_\Lambda \over c} \int^{x}_{0} {dx \over \sqrt{3 (1 + x) \vert x\vert}}$$ (10)

Now we only have an overall scale factor, R_$\Lambda$; there are just two functional forms, depending on whether $\Lambda$ is positive or negative. If x < 0 (i.e. negative $\Lambda$), we have a turnaround at x = - 1.

Examination of the general solution (Eq. 4) shows that for very large $\lambda$, k becomes irrelevant: if r is small enough that the term in $\lambda$ becomes comparable to unity, then 3kr $\ll$ 1, whereas if 3kr > 1 then the term in $\lambda$ dominates. Thus these cases are actually equivalent to the cases where $\lambda$ = ±$\infty$.

$\lambda$ = 0

This is a similar situation to the previous one: there are two functional forms, depending on the sign of k. For k = + 1 there is the `cycloid' universe

$$\tau {1 \over 3} \eta \eta {1 \over 3} \eta$$ (11)

where the auxiliary parameter $\eta$ is the `conformal time', and ranges from 0 to 2$\pi$. For k = - 1 we have an equivalent form with hyperbolic functions:

$$\tau {1 \over 3} \eta \eta {1 \over 3} \eta$$ (12)

Here $\eta$ ranges from zero to infinity. Obviously these solutions are just the limit for $\lambda$ $\rightarrow$ 0.

k =0 and $\lambda$ =0

This corresponds to just one universe, the Einstein-de Sitter case. This is a scale-free solution, i.e. there is no characteristic time or size; instead the density is a unique function of time (or vice versa); our integral solution reduces to

$${2 \over 3} \sqrt{3 \over 8 \pi G \rho_m} {2 \over 3 H}$$ (13)

All universes where M_* > 0 initially follow the Einstein-de Sitter solution; only when the scale becomes comparable to R_m do they deviate from it. Hence this can be considered as the limiting case when R_m $\rightarrow$ $\infty$.

M_* = 0

There are also six special cases when the matter density is strictly zero. There is a trivial empty static universe when k = 0 and $\Lambda$ = 0, in which case H = 0. This can be thought of as the limit of the Einstein-de Sitter universe when t $\rightarrow$ $\infty$.

Next is the Milne solution, when $\Lambda$ = 0 and k = - 1 (to make H² positive) and hence

R = ct. (14)

Universes with $\lambda$ $\ll$ 1 and k = - 1 go through a Milne phase, as noted above and this is the limiting form as t $\rightarrow$ $\infty$ if $\Lambda$ is strictly zero and k = - 1.

In the de Sitter solution k = 0 and $\Lambda$ > 0. The Hubble parameter is then constant at

$$\Lambda$$ (15)

and we have exponential expansion:

$${R_\Lambda \over c} \sqrt{3}$$ (16)

Note that this case reaches zero size only infinitely far back in time, so we have to choose an arbitrary value of R₀. Any universe that expands to infinity tends to this solution at late times, unless $\Lambda$ is strictly zero.

The other three cases correspond to both $\Lambda$ and k non-zero: note that for M_* = 0, $\Lambda$$\le$ 0, k = + 1, is ruled out because it would give a negative H² (similarly $\Lambda$ < 0, k = 0). For all these R_$\Lambda$ sets the absolute scale, and there are no continuous parameters that affect the shape.

To summarize the M_* = 0 cases:

$\Lambda$	k	Rturn	Type
< 0	-1	(- 3$\Lambda$)1/2	Bang-crunch.
0	0	-	Static
0	-1	-	Milne: Linear Expansion
> 0	-1	-	Continuous expansion
> 0	0	0	de Sitter: Exponential expansion
> 0	+1	(3$\Lambda$)1/2	Bounce

H = 0

A final special case is the static Einstein solution, which in our notation has $\lambda$ = 1 and r = 1.