CosmoCoffee

Hi Antony,

I have a very simple question, but I want to be sure. You define dgpi in /camb/equations.f90 as

[tex]\text{dgpi} = \sum 8 \pi G a^2 \rho_i \Pi_i[/tex]

So, in Newtonian gauge, I should write the Einstein equation as

[tex]k^2 (\phi - \psi) = 12 \pi G a^2 \sum (\rho + P )\sigma = \frac{3}{2} \text{dgpi}[/tex].

Am I wrong?

Thanks,
Alex

Not sure about the factor of 3/2, e.g. see the second equation on p12 of the CAMB notes.

Great, thanks!

Alex