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My question refers to page 38 of Tong's Notes on Cosmology. Here we consider a matter dominated universe and derive the equation $$2h^\prime + h^2 + k = 0$$ for $h = a^\prime/a$, where $^\prime$ denotes differentiation with respect to the conformal time given by $dt = a d\tau$. The solutions are given as (after a suitable rescaling of $\tau$)

$$ h(\tau) = \begin{cases} \cot(\tau/2) & k=1\\ 2/\tau & k=0\\ \coth(\tau/2) & k=-1 \end{cases} $$

and then we can find that

$$ a(\tau) = A \times \begin{cases} \sin^2(\tau/2) & k=1\\ \tau^2 & k=0\\ \sinh^2(\tau/2) & k=-1 \end{cases}, $$

where $A$ is a constant of integration. I'm having trouble deriving these results. Firstly, what condition do we impose in the first set of equations to get rid of any constants of integration? The conditions in the second equation are not obvious either. For example, why is the integration constant $A$ the same across all three equations?

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1 Answer 1

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I suggest you differentiate the three h(tau) equations to get

h'(tau) = ...

The k= -1 and k=+1 equations are a bit messy, but for k=0 the derivative is easily seen as

h' = -2/r^2.

Combining this with the first equation yields

-4/tau^2 + h^2 = 0,

and therefore

a'/a = h = 2/tau.

Integrating yields (ignoring constant of integration)

ln(a) = 2 ln(tau) = ln(tau^2)

and therefore

a(tau) = tau^2.

Regarding "A": The value of a is by convention always 1 for t = now. The value of "A" depends on the value of now.

I hope you find this to be this helpful.

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