The Initial value of the Hubble Constant

The initial value of the Hubble constant is used for setting the field derivatives (equation (6.72)) and as an initial condition for the second order evolution equation for the scale factor. The derivative of $a$ is determined by the equation

$$H^2 = \left({\dot{a} \over a}\right)^2 = {8 \pi \over 3} \rho.$$ (6.87)

Initially $a$ is set to $1$ and

$$H_{pr}^2 = a'^2 = {8 \pi \over 3 B^2} \rho.$$ (6.88)

In setting initial values we assume all inhomogeneities are small and thus use only the homogeneous values of the fields $<f>$ and $<\dot{f}>$. Typically the initial field values will be one for the inflaton and zero for all other fields but they can be set to any values by the user. In general, the initial energy density is thus

$$\rho \approx {1 \over 2} \sum_{fields}\dot{f}^2 + V.$$ (6.89)

Converting to program variables

$$\rho = {1 \over 2} {B^2 \over A^2} \sum_{fields}\left(a^{2s-2r} f_{pr}'^... ...{pr}' + r^2 a^{2s-2r-2} a'^2 f_{pr}^2\right) + {B^2 \over A^2} a^{2s-2r} V_{pr}$$ (6.90)

$$= {B^2 \over A^2} \left(\sum_{fields}\left({1 \over 2} f_{pr}'^2 - r a' f_{pr} f_{pr}' + {1 \over 2} r^2 a'^2 f_{pr}^2\right) + V_{pr}\right)$$

where the second step uses the fact that initially $a=1$. Since initially $H_{pr}=a'$ we can plug equation (6.90) into equation (6.88) to get an equation we can solve for $H_{pr}$. Solving this quadratic equation gives

$$H_{pr,0} = {1 \over {3 A^2 \over 4 \pi} - r^2 f_{pr}^2} \lef... ...V_{pr} \left({3 A^2 \over 4 \pi} - r^2 f_{pr}^2\right)}\right)$$ (6.91)

where $H_{pr,0}$ refers to the initial value of $H_{pr}$ and each term with field or field derivative values is understood to be summed over all fields.