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\phantomsection

\chapter{\uppercase{The conformal case in D-brane scenarios}} \label{App2}


In this section we show how to recover the pure conformal case from the D-brane picture, that is, $b=0$, $C_1=C_2$, $D_1=D_2$. 
We start by  expanding the square root in the scalar part of the action \eqref{S1A}. Doing this we get
\bea\label{Sphiconf}
S_\phi& =& -\int{d^4x  \sqrt{-g} \left[  M^4C_1^2 \left( 1+ \frac{D_1}{2C_1} (\partial \phi)^2   + \dots \right) + V(\phi)\right]}\nonumber \\
&=& -\int{d^4x  \sqrt{-g} \left[  \frac{M^4C_1 D_1}{2} (\partial \phi)^2   + M^4C_1^2(\phi)+ V(\phi) + \dots \right]}\nonumber \\ 
&=& -\int{d^4x  \sqrt{-g} \left[  \frac{M^4C_1 D_1}{2} (\partial \phi)^2   + {\mathcal V}(\phi) + \dots \right]} \,.
\eea
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On the other hand, the matter Lagrangean  takes the form
\bea
S_{DM} &=& -\int{d^4x  \sqrt{-\tilde g} \,{\cal L}_{DM}(\tilde g_{\mu\nu})}\nonumber \\
 &=& -\int{d^4x  \sqrt{-g} \, C_1^2(\phi) \left(1+ \frac{D_1}{2C_1} (\partial \phi)^2   + \dots  \right){\cal L}_{DM}( \tilde g_{\mu\nu})} \nonumber \\  &=&  -\int{d^4x  \sqrt{- g} \, C_1^2 (\phi) {\cal L}_{DM}( \tilde g_{\mu\nu}) + \dots } =-\int{d^4x  \sqrt{- \tilde g} \, {\cal L}_{DM}( \tilde g_{\mu\nu}) + \dots } \nonumber \\ 
 \eea
where now $\tilde g_{\mu\nu} = C_1(\phi) g_{\mu\nu}$ (and we have used that $\det \tilde g_{\mu\nu} = C_1^4 (1+ D_1/C_1 (\partial\phi)^2)$). 

Finally, to compare the D-brane case with the pure conformal case, we need to canonically normalise $\phi$. Calling $\varphi$ the canonically normalised field, this is obtained from $\phi$ as
\be
\varphi = \int M^2\sqrt{D_1C_1} \, d\phi \,.
\ee

It is clear that when $D_1=1/(M^4 C_1)$, $\varphi =\phi$ and therefore the action for the scalar field \eqref{Sphiconf} is already in the required form. 
We can now take the limit $\gamma \to 1$ into the  equations of motion \eqref{friedmann1A}-\eqref{kgA} and make  the identification $D_1=1/(M^4C_1)$ to recover the conformal case equations of motion. Note that in this limit $Q_0 \to \rho C_1'/2C_1$, and  is independent of $D_1$ (see \eqref{Q0A} with $C_1=C_2, D_1=D_2$).
