# Solving expansion rate for a variant of the Friedmann equation

Tags:
1. Oct 11, 2014

### Kyrios

1. The problem statement, all variables and given/known data
For the equation $$H^2 = \frac{8 \pi G \rho_m}{3} + \frac{H}{r_c}$$ how do I find the value of H for scale factor $a \rightarrow \infty$, and show that H acts as though dominated by $\Lambda$ (cosmological constant) ?

2. Relevant equations
$$\rho_m \propto \frac{1}{a^3}$$
$$H > 0$$

3. The attempt at a solution
I'm not sure how to show that H is driven by $\Lambda$, but have tried to sub in the scale factor in place of matter density and make the scale factor go to infinity.
As in,
$$H^2 = \frac{8 \pi G }{3 a^3} + \frac{H}{r_c}$$
This gets rid of the $\frac{8 \pi G \rho_m}{3}$ leaving $H = \frac{1}{r_c}$

2. Oct 12, 2014

### George Jones

Staff Emeritus
What is $r_c$? Is it constant?

3. Oct 12, 2014

### Kyrios

Yes $r_c$ is a constant which is called the cross over scale. I don't think we need to know the value of it

4. Oct 12, 2014

### George Jones

Staff Emeritus
Okay. Now, solve

$$H = \frac{1}{r_c}$$

for the scale factor $a$.

5. Oct 12, 2014

### George Jones

Staff Emeritus
Alternatively, what is $H$ for a universe that has a non-zero cosmological constant $\Lambda$, and that is otherwise empty, i.e., that has no matter or radiation content?

6. Oct 12, 2014

### Kyrios

ok, so $H = \frac{1}{r_c}$
$$H = \frac{1}{a}\frac{da}{dt} = \frac{1}{r_c}$$
$$\int_{0}^{a} \frac{1}{a} da = \int_{0}^{t} \frac{1}{r_c} dt$$
$$ln(a) = \frac{1}{r_c} t$$
$$a = \exp(\frac{t}{r_c})$$

Which is akin to dark energy domination $a \propto \exp(Hr_c)$ ?