neutron star are not only the baryon number $N$ but also the “external field”, that is, the asymptotic value of the scalar field $\varphi_0$. For static systems, the partial derivative of $m$ in terms of $\varphi_0$ with $N$ constant is given by [@de1992] $$\label{eq:dermderphi0}
  \left(\frac{\partial m}{\partial \varphi_0}\right)_{N}=-\omega.$$ The energy injection of the system by increasing baryons is described as $$\label{eq:injection}
  \int 4\pi \tilde{u}_0 \delta\tilde{\rho}
  A^3 r^2 \left(1-\frac{2\mu}{r}\right)^{-1/2}dr 
  =\int 4\pi e^{\nu/2} \tilde{\mu}
  \delta \tilde{n}
  A^4 r^2 \left(1-\frac{2\mu}{r}\right)^{-1/2}dr,$$ where $\tilde{\mu}\equiv d\tilde{\rho}/d\tilde{n}$ is the chemical potential. The first law of thermodynamics in an adiabatic process is $$\label{eq:1stlaw}
  d\left(\frac{\tilde{\rho}}{\tilde{n}}\right)
    =-\tilde{p}d\left(\frac{1}{\tilde{n}}\right).$$ From Eqs. (\[eq:beq2\]), (\[eq:beq3\]), (\[eq:beq5\]), and (\[eq:1stlaw\]), we find that the quantity $A e^{\nu/2}\tilde{\mu} $ is constant all over the star. Therefore this quantity can be estimated by its value at the stellar surface. Using this fact, the expression of the energy injection, Eq. (\[eq:injection\]), is rewritten as $$A_s e^{\nu_s/2}\tilde{\mu}_s\int 4\pi A^3 r^2
  \left(1-\frac{2\mu}{r}\right)^{-1/2}\delta \tilde{n}dr 
  = A_s e^{\nu_s/2}\tilde{\mu}_s \delta N,$$ where the suffix “[*s*]{}” indicates that the quantity is evaluated at the stellar surface $r=r_s$. Therefore, the effective chemical potential, $\mu_{eff}$, is given by $$\mu_{eff}\equiv \left( \frac{\partial m}{\partial N} 
    \right)_{\varphi_0}
  = A_s e^{\nu_{s}/2}\tilde{\mu}_{s}.$$ From the above discussions the variation of $m$ for static systems 