Solving Friedmann Equation w/ Friedmann Eqn Hmwk Stmt

[Logarythmic]

Homework Statement

By substituting in

$$\left( \frac{\dot{a}}{a_0} \right)^2 = H^2_0 \left(\Omega_0 \frac{a_0}{a} + 1 - \Omega_0 \right)$$

show that the parametric open solution given by

$$a(\psi)=a_0 \frac{\Omega_0}{2(1-\Omega_0)}(\cosh{\psi} - 1)$$

and

$$t(\psi)=\frac{1}{2H_0} \frac{\Omega_0}{(1 - \Omega_0)^{3/2}}(\sinh{\psi} - \psi)$$

solve the Friedmann equation.2. The attempt at a solution
I get

$$\dot{a} = a_0 \frac{\Omega_0}{2(1 - \Omega_0)}(\dot{\psi}\sinh{\psi})$$

and

$$\dot{\psi}=\frac{2H_0(1-\Omega_0)^{3/2}}{\Omega_0(\cosh{\psi}-1)}$$

but I can't get to the first equality. Is this the correct approach?

 

[Physics Monkey]

You may find it easier to write $$\dot{a} = \frac{\frac{da}{d\psi}}{\frac{dt}{d\psi}}$$. Dividing by $$a_0$$ and squaring should leave you with some expression involving only $$\psi$$. A little algebra is then all you need to re-express things in terms of $$a$$.

 

[Logarythmic]

I get the right hand side to equal

$$H^2_0 \frac{2(1-\Omega_0}{\cosh{\psi}-1}$$

and the LHS

$$H^2_0 \frac{(1-\Omega_0)\sinh^2{\psi}}{(\cosh{\psi}-1)^2}$$

but I can't get them equal...

 

[Physics Monkey]

I agree with your second expression from computing da/dt, but I think you've made a mistake with your first expression.