of stable equilibrium solutions changes as 1, 2 (degenerate in Fig. 8), 3, 2, 0 as $N$ is increased continuously, as is seen in Fig. 8. For $\beta\alt-4.6$ the maximum ADM energy with $\varphi_0=0$ is greater than that in general relativity, while, for $\beta\agt -4.6$, it is the same as that in general relativity.

The Kepler mass, which governs the Newtonian orbital motion of a test body, is not the ADM energy $m$ in general, but [@de1992] $$\tilde{\mu}=\frac{1+\alpha_0\alpha_A}{1+\alpha_0^2}m,$$ where $$\alpha_A\equiv\frac{\partial\ln m}{\partial\varphi}=
  -\frac{\omega}{m}.$$ When we consider the case of $\alpha_0=\beta\varphi_0=0$, the Kepler mass is identical to the ADM energy. Therefore the argument above for $\varphi_0=0$ is also valid for the Kepler mass.

Here we present the physical interpretation as to why spontaneous scalarization occurs. In spite of the absence of the potential in the Lagrangian, the scalar field $\varphi$ obtains an effective potential term $W(\varphi)$ which satisfies $$\frac{\partial W}{\partial \varphi}=-4\pi \alpha(\varphi) T_*$$ because of the coupling with matter. Note that $T_*$ depends on $\varphi$. Then, if we consider $A(\varphi)$ of the form (\[eq:quadratic\]), $$\frac{\partial V}{\partial\varphi}=-4\pi \beta\varphi T_*,$$ and if $T_*= A^4(-\tilde{\rho}+3\tilde{p})$ is negative, $\varphi=0$ is an unstable stationary point of the effective potential, if $\beta<0$. On the other hand, the term 