**2. Model**

Here, we will discuss solutions to the following (3+1)-dimensional model :

$$\mathcal{S} = \int \mathrm{d}^4 \mathbf{x} \sqrt{-\mathbf{g}} \mathcal{L} \quad , \quad \mathcal{L} = \frac{\mathcal{R}}{16\pi G} - \left(D\_{\mu}\Psi\right)^\* D^\mu \Psi - \frac{1}{4} F\_{\mu\nu} F^{\mu\nu} - \mathcal{U}(|\Psi|) \,. \tag{1}$$

This is the Einstein–Hilbert action with R the Ricci scalar and *G* Newton's constant minimally coupled to a complex valued scalar field Ψ that is charged under a U(1) gauge field *<sup>A</sup>μ*. *<sup>D</sup>μ* = *∂μ* − *iqAμ* is the covariant derivative of the scalar field and *<sup>F</sup>μν* = *∂μ<sup>A</sup>ν* − *∂νAμ* the field strength tensor of the U(1) gauge field. The scalar field potential *U*(|Ψ|) will turn out to be a crucial ingredient in our construction in the following. We will choose the potential as follows [13,14] :

$$M(|\Psi|) = \mu^2 \eta^2 \left[1 - \exp\left(-\frac{|\Psi|^2}{\eta^2}\right)\right],\tag{2}$$

where *μ* corresponds to the mass of the scalar field and *η* is an energy scale.

In the following, we are interested in spherically symmetric and stationary solutions. The Ansatz for the metric is:

$$ds^2 = -N(r)(r(r))^2dt^2 + \frac{1}{N(r)}dr^2 + r^2(d\theta^2 + \sin^2\theta d\varphi^2) \; , \; N(r) = 1 - \frac{2m(r)}{r} \; , \tag{3}$$

while the matter fields are chosen according to :

$$\Psi(r,t) = \eta e^{i\hat{\omega}t} \Psi(r) \quad , \ A\_0 = \eta V(r) \tag{4}$$

with *ω*˜ a real constant. Note that, although the scalar field is time-dependent, the associated energy-momentum tensor is static and hence likewise the space-time. Defining the following dimensionless quantities

$$\infty = \mu r \quad , \ \omega = \frac{\tilde{\omega}}{\mu} \; , \ \mathfrak{e} = \frac{\eta \eta}{\mu} \; , \ \mathfrak{a} = 4\pi \text{G} \eta^2 \tag{5}$$

the equations resulting from the variation of the action (1) depend only on *e* and *α* and read (with the prime denoting derivative with respect to *x*) :

$$m' = -\alpha x^2 \left[ \frac{V'^2}{2\sigma^2} + N\psi'^2 + \mathcal{U}(\psi) + \frac{(\omega - \varepsilon V)^2}{N\sigma^2} \psi^2 \right] \tag{6}$$

$$\sigma^{\prime} \quad = 2a\text{ax}\sigma \left[ \psi^{\prime 2} + \frac{(\omega - \varepsilon V)^2}{N^2 \sigma^2} \psi^2 \right] \tag{7}$$

for the metric functions and

$$V'''' + \quad \left(\frac{2}{\pi} - \frac{\sigma'}{\sigma}\right)V' + \frac{2(\omega - \varepsilon V)\psi^2}{N} = 0\tag{8}$$

$$\left(\psi''\right)' + \left(\frac{2}{\mathfrak{x}} + \frac{N'}{N} + \frac{\sigma'}{\sigma}\right)\psi' + \frac{(\omega - eV)^2 \psi}{N^2 \sigma^2} - \frac{1}{2N} \frac{d\mathcal{U}}{d\psi} = 0 \,. \tag{9}$$

for the matter fields. As is obvious, these equations depend only on the combination *ω* − *eV*(*x*). Note that the Lagrangian density and with it the energy density are now given in units of *η*<sup>2</sup>*μ*2.

The asymptotic behaviour of the metric and matter field functions is:

$$N(\mathbf{x}\gg 1) = 1 - \frac{2M}{\mathbf{x}} + \frac{aQ^2}{\mathbf{x}^2} + \dots \quad , \text{ } \sigma(\mathbf{x}\gg 1) = 1 + \mathcal{O}(\mathbf{x}^{-4}) \text{ },\tag{10}$$

$$V(\mathbf{x}) = V\_{\infty} - \frac{\mathbf{Q}}{\mathbf{x}} + \dots \; \; \; \; \psi(\mathbf{x} \to \infty) \sim \frac{\exp(-\mu\_{\text{eff,os}}\mathbf{x})}{\mathbf{x}} + \dots \; \; \; \; \; \mu\_{\text{eff,os}} = \sqrt{1 - \Omega^2} \; \; \; \; \tag{11}$$

where Ω<sup>2</sup> := (*ω* − *eV*∞)2. *M* and *Q* denote the (dimensionless) mass and electric charge of the solution, respectively. *V*∞ can be understood to be a "chemical potential", i.e., the resistance of the system against the addition of extra charges *e* to the system. Note that the scalar field possesses an effective mass *μ*eff smaller than the bare mass of the scalar field, which—in our dimensionless units—is equal to unity.

*Symmetry* **2020**, *13*, 2

Next to the electric charge, the solutions possess a Noether charge which results from the unbroken U(1) symmetry of the system. This reads :

$$Q\_N = \int \mathrm{d}x \, \frac{2\pi^2 ev\psi^2}{N\sigma} \,. \tag{12}$$

This quantity can be interpreted as the number of scalar bosons that make up the solution. For globally regular solutions *eQN* ≡ *Q*, while, for black holes, *Q* = *eQN* − *Ex*(*xh*)*x*2*h*/*σ*(*xh*), where *Ex*(*x*) = −*<sup>V</sup>* (*x*) is the radial electric field of the solution. The second term in this equality is the horizon electric charge and is the consequences of the fact that the horizon corresponds to a surface on which boundary conditions have to be imposed. Finally, the temperature *TH* and entropy S of the black hole solutions are given by

$$T\_H = (4\pi)^{-1} \sigma(\mathbf{x}\_h) N'|\_{\mathbf{x} = \mathbf{x}\_h} \ \ \ \ \ \mathcal{S} = \pi \mathbf{x}\_h^2 \ . \tag{13}$$

The energy-density , radial pressure *pr*, and tangential pressure *pθ* = *pϕ*, respectively, read :

$$\boldsymbol{\epsilon} = -T\_t^t = \boldsymbol{\epsilon}\_1 + \boldsymbol{\epsilon}\_2 + \boldsymbol{\epsilon}\_3 + \boldsymbol{\mathcal{U}}(\boldsymbol{\psi}) \tag{14}$$

$$p\_I = T\_r' = -\varepsilon\_1 + \varepsilon\_2 + \varepsilon\_3 - lI(\psi) \tag{15}$$

$$p\_{\theta} = T\_{\theta}^{\theta} = \epsilon\_1 - \epsilon\_2 + \epsilon\_3 - lI(\psi) \,. \tag{16}$$

with

$$
\epsilon\_1 = \frac{V^{\prime 2}}{2\sigma^2} \; , \; \epsilon\_2 = N\psi^{\prime 2} \; , \; \epsilon\_3 = \frac{(\omega - \varepsilon V)^2 \psi^2}{N\sigma^2} \; . \tag{17}
$$
