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\title{Hiding gravitationally coupled scalars through local symmetry restoration (after arxiv:1001.4525)}

\author{Andrei Vlasov}

\begin{document}

\maketitle
%\tableofcontents
%\section{General idea}
In the work is considered the scalar field, which is coupled with gravitation and the matter. The coupling is such that the field evolves in effective potential
\be
V_{eff}(\phi)=-\mu^2\phi^2+\lambda \phi^4+\frac{\rho \phi^2}{M^2}
\ee
where $\rho$ is the density of matter and $\mu,\lambda,M$ are the parameters of the theory. When $\rho\le \mu^2M^2$, the symmetry is broken, but in dense environments, $\rho > \mu^2 M^2$, the symmetry is restored. The interaction between matter and the field is proportional to the local average value of the field, and then in dense environments there is no interaction between matter and the field. \\
The action of the theory states:
\be
S=\int d^4x\sqrt(-g)(\frac{\mpl^2}{2}R-\frac12(\p \phi)^2-V(\phi))+\int d^4x L_m[\tilde{g}]
\ee
where
\be
\tilde{g}_{\mu\nu}=A^2(\phi)g_{\mu\nu},\quad A(\phi)=1+\frac{\phi^2}{2M^2}\\
V(\phi)=-\frac{1}{2}\mu^2\phi^2+\frac{1}{4}\lambda\phi^4 \\
\ee
%We consider the case when the "critical" density of matter coincides with current cosmological density (I don't know why?), so $H_0^2\mpl^2\approx \mu^2M^2$. Further, we will see that local tests of gravity require $M \lesssim 10^{-3}\mpl$, so the mass of small fluctuations equals $m_0 \approx \frac{\mpl}{M}H_0 \lesssim {\rm Mpc}$.
The existing limits:
\be
H_0^2\mpl^2 \approx \mu^2M^2 \\
M\lesssim 10^{-3}\mpl \\
\phi_0\equiv \frac{\mu}{\sqrt{\lambda}}\approx \frac{M^2}{\mpl}
\ee
Static equation on the field $\phi$:
\be
\phi''+\frac{2}{r}\phi'=V_{,\phi}+A_{,\phi}\rho
\ee
The prime denotes the derivative wrt $r$. we consider the case of homogeneous object of radius $R$ and density $\rho$. Approximate solution can be obtained by linearising the potential around the corresponding minima. The solution:
\be
\phi_{in}=A\frac{R}{r}\sinh\left(r\sqrt{\frac{\rho}{M^2}-\mu^2}\right) \\
\phi_{out}=B\frac{R}{r}e^{-\sqrt{2}\mu r}+\phi_0
\ee
There are three dimensionless parameters in this solutions: $\mu R, \frac{\rho}{\mu^2M^2},\rho R^2/M^2$. In the situation under consideration $\mu R \ll 1$, $\frac{\rho}{\mu^2M^2}\gg 1$. The third parameter, $\alpha$ can be either great or small. \\
The equations of gravitational field in the cases:
\be
R_{\mu\nu}-\frac12 g_{\mu\nu}R=8\pi G A(\phi) T_{\mu\nu}+8\pi G T^{(\phi)}_{\mu\nu}
\ee
where $T^{(\phi)}_{\mu\nu}$ is stress-energy tensor of the field $\phi$, and $T_{\mu\nu}$ is stress-energy tensor of the rest matter.
% The presence of field $\phi$ changes two things: first, the matter density (and entire stress-energy tensor) is multiplied by $A(\phi)=1+\frac{\phi^2}{2M^2}\approx 1+10^{-12}$, and the second
%The field itself adds to the mass of the object the "mass" $\de m=4\pi\int_0^R T_{00}$
The linearised Einstein equations state in this case:
\be
R_{\mu\nu}=\frac12 \de h_{\mu} \\
\De h_{00}=\frac{8\pi G}{c^2}(T_{00}-T_{11}) \\
\De h_{11}=\frac{8\pi G}{c^2}(T_{00}+3T_{11})
\ee
We assumed, that $h_{11}=h_{22}=h_{33}, h_{0i}=0, h_{ij}=0$ if $i\ne j$ - this is indeed the case in Cartesian coordinates. Gauge-fixing condition for this case states: $\p^\mu(h_{\mu\nu}-\frac12 h g^{\mu\nu})$  For light deflection and time delay experiments matters only one PPN parameter,
\be
 \gamma-1=\frac{h_{11}-h_{00}}{h_{11}+h_{00}}=\frac{4\int_0^RT_{11}r^2dr}{\int_0^RT_{00}r^2dr}=\frac{1}{\rho R^3}
\ee


\end{document}

