### setting constant of integration \\chi for initial conditions

Posted:

**April 17 2015**Hello, I have a question about how [tex]\chi[/tex] is determined in CAMB. I know that it is set to [tex]-1[/tex], but see below.

\begin{equation}

\label{1}

\mathcal{R} = \pm (\Delta_{\mathcal{R}})^{1/2} = \pm \sqrt{A_s}

\end{equation}

at Planck's pivot scale [tex]k_{\star} = 0.05 ~\mathrm{Mpc}^{-1}[/tex], and In the synchronous gauge, using the (+ - - -) signature, the comoving curvature perturbation is

\begin{equation}

\label{2}

\mathcal{R} = \eta + \frac{\mathcal{H} v}{ k}

\end{equation}

where [tex]v \equiv \theta/k[/tex] using the notation of Ma and Bertschinger ({\tt arXiv:astro-ph/9506072}). For [tex]k<<\mathcal{H}[/tex] in the radiation epoch,

\begin{equation}

\label{3}

\eta= 2C - \frac{5+4 R_{\nu}}{6(15+4R_{\nu})} C (k \tau)^2,

\end{equation}

and

\begin{equation}

\label{4}

v_{rad} \equiv (1-R_{\nu}) v_{\gamma}+ R_{\nu} v_{\nu} = - \frac{C}{18} (k \tau)^3 \biggl(1-R_{\nu}+R_{\nu} \frac{23+4R_{\nu}}{15+4R_{\nu}}\biggr).

\end{equation}

It follows from Eqs. (\ref{1}) and (\ref{2}) that, for values of [tex]\tau[/tex] early enough during radiation domination such that [tex]k=k_{\star}[/tex] is super-horizon,

\begin{equation}

\label{5}

C \approx \mp 2 \cdot 10^{-5}

\end{equation}

for [tex]\pm \sqrt{A_s}[/tex] evaluated at [tex]k=k_{\star}[/tex]. I used [tex]R_{\nu}=\rho_{\nu}/(\rho_{\gamma}+\rho_{\nu})[/tex],

[tex]\rho_{\nu}/\rho_{\gamma}=(7 N_{\nu}/8)(4/11)^{4/3}[/tex], [tex]N_{\nu}=3.046[/tex], and [tex]\ln(10^{10} A_s)= 3.064[/tex], from Planck 2015.

Comparing equations for initial conditions in CAMB notes, we see that [tex]C = \chi/2[/tex].

However, in CAMB, [tex]\chi[/tex] is set to [tex]-1[/tex].

Am I doing something wrong here? Why this discrepancy? I know that using [tex]\chi=-1[/tex] in CAMB

gives a CMB angular power spectrum that agrees with

Planck's 2015 results, and using [tex]\chi=2C[/tex] gives an angular power spectrum with amplitudes that are too small. And [tex]A_s[/tex] is obtained from the CMB, so it makes sense

to me that [tex]\chi[/tex] should be constrained observationally.

Thank you for any help.

\begin{equation}

\label{1}

\mathcal{R} = \pm (\Delta_{\mathcal{R}})^{1/2} = \pm \sqrt{A_s}

\end{equation}

at Planck's pivot scale [tex]k_{\star} = 0.05 ~\mathrm{Mpc}^{-1}[/tex], and In the synchronous gauge, using the (+ - - -) signature, the comoving curvature perturbation is

\begin{equation}

\label{2}

\mathcal{R} = \eta + \frac{\mathcal{H} v}{ k}

\end{equation}

where [tex]v \equiv \theta/k[/tex] using the notation of Ma and Bertschinger ({\tt arXiv:astro-ph/9506072}). For [tex]k<<\mathcal{H}[/tex] in the radiation epoch,

\begin{equation}

\label{3}

\eta= 2C - \frac{5+4 R_{\nu}}{6(15+4R_{\nu})} C (k \tau)^2,

\end{equation}

and

\begin{equation}

\label{4}

v_{rad} \equiv (1-R_{\nu}) v_{\gamma}+ R_{\nu} v_{\nu} = - \frac{C}{18} (k \tau)^3 \biggl(1-R_{\nu}+R_{\nu} \frac{23+4R_{\nu}}{15+4R_{\nu}}\biggr).

\end{equation}

It follows from Eqs. (\ref{1}) and (\ref{2}) that, for values of [tex]\tau[/tex] early enough during radiation domination such that [tex]k=k_{\star}[/tex] is super-horizon,

\begin{equation}

\label{5}

C \approx \mp 2 \cdot 10^{-5}

\end{equation}

for [tex]\pm \sqrt{A_s}[/tex] evaluated at [tex]k=k_{\star}[/tex]. I used [tex]R_{\nu}=\rho_{\nu}/(\rho_{\gamma}+\rho_{\nu})[/tex],

[tex]\rho_{\nu}/\rho_{\gamma}=(7 N_{\nu}/8)(4/11)^{4/3}[/tex], [tex]N_{\nu}=3.046[/tex], and [tex]\ln(10^{10} A_s)= 3.064[/tex], from Planck 2015.

Comparing equations for initial conditions in CAMB notes, we see that [tex]C = \chi/2[/tex].

However, in CAMB, [tex]\chi[/tex] is set to [tex]-1[/tex].

Am I doing something wrong here? Why this discrepancy? I know that using [tex]\chi=-1[/tex] in CAMB

gives a CMB angular power spectrum that agrees with

Planck's 2015 results, and using [tex]\chi=2C[/tex] gives an angular power spectrum with amplitudes that are too small. And [tex]A_s[/tex] is obtained from the CMB, so it makes sense

to me that [tex]\chi[/tex] should be constrained observationally.

Thank you for any help.