Black hole in Nielsen-Olesen vortex

We consider the Einstein-Hilbert action coupled to electromagnetism

$$S=S_{G}+S_{V},$$

(1)

with

$$S_{G}=\int d^{3}x\sqrt{-^{(3)}g}\left(R-2\Lambda\right)$$

(2)

is the pure gravity part and

$$S_{V}=\int d^{3}x\sqrt{-^{(3)}g}\left(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}D_{\mu}\phi^{a}D^{\mu}\phi^{a}-V[\phi]\right),$$

(3)

is the part where gravity coupled to the Maxwell field. Here ${}^{(3)}g=\text{det}~{}g_{\mu\nu}$ ($\mu,\nu=1,2,3$) is the three dimensional metric, $R$ is the Ricci scalar, $\Lambda=-1/l^{2}$ is the negative cosmological constant, and $\phi^{a}$ ($a=1,2$) is a charged scalar field. The covariant derivative $D_{\mu}$ in terms of the gauge field $A_{\mu}$ is defined as

$$D_{\mu}\phi^{a}=\partial_{\mu}\phi^{a}-e\epsilon^{ab}A_{\mu}\phi^{b},$$

(4)

and $V[\phi]$ is the symmetry breaking potential:

$$V[\phi]=\frac{\lambda}{4}\left(|\phi|^{2}-\eta^{2}\right)^{2},$$

(5)

with the symmetry breaking coupling constant $\lambda$.

We take the field configuration vortex like, that is $\phi$ and $A$ are time independent and the components of the scalar field has the following form:

$$\phi^{1}=\eta\chi(r)\cos(n\theta),$$

(6)

and

$$\phi^{2}=\eta\chi(r)\sin(n\theta).$$

(7)

The abelian gauge field

$$A_{\mu}=\frac{1}{e}\left(n-P(r)\right)\partial_{\mu}\theta,$$

(8)

where $n$ is a topological quantity called winding number, which takes only integer values.

Finite energy condition imply that

$$|\phi|^{2}\to\eta^{2}~{}\text{as}~{}r\to\infty,$$

(9)

from Eq. (5) and $A_{\mu}$ asymptotically be a pure gauge rotation.

The above conditions lead us to write down the simplest field configuration for vortex.

$$\chi(0)=0~{}~{}~{}~{}~{}\text{and}~{}\chi(r)\to 1;r\to\infty,$$

(10)

where $\eta$ is the vacuum value, and

$$P(0)=n~{}~{}~{}~{}~{}\text{and}~{}P(r)\to 0;r\to\infty$$

(11)

The action (1) can be expressed in the Hamiltonian form as follows,

$$S=\int dt\int_{\Sigma}d^{2}x\pi^{ij}\dot{g}_{ij}-N^{\perp}H_{\perp}-N^{i}H_{i}$$

(12)

by adding appropriate surface terms, where

$$H_{\perp}=\frac{1}{\sqrt{g}}\left[\pi^{ij}\pi_{ij}-(\pi^{i}_{i})^{2}\right]-\sqrt{g}\left[R-2\Lambda^{\prime}\right]+\frac{1}{2\sqrt{g}}B^{2}$$

(13)

is the Hamiltonian constraint with

$$2\Lambda^{\prime}=2\Lambda+\frac{1}{2}g^{ij}(D_{i}\phi^{a})(D_{j}\phi^{a})+V[\phi]$$

(14)

and

$$H_{i}=\pi^{j}_{i,j}$$

(15)

are the momentum constraints. $\pi^{ij}$ are the conjugate momenta corresponding to the canonical variables $g_{ij}$, the two dimensional spacial metric, and $g=\text{det}(g_{ij})$. The magnetic field density $B$ is given by

$$F_{ij}=\epsilon_{ij}B.$$

(16)

The lapse function $N^{\perp}$ and the shift vector $N^{i}$ appear in the action as Lagrange multipliers; their variation leads to the field equations $H_{\perp}=0$ and $H^{i}=0$.

The stationary, axisymmetric (2+1)-dimensional metric can be expressed as the following spacetime line element

$$ds^{2}=-N^{\perp}(r)^{2}dt^{2}+f^{2}(r)dr^{2}+r^{2}(d\theta-N^{\theta}(r)dt^{2}),$$

(17)

With $0\leqslant r<\infty$ and $0\leqslant\theta<2\pi$. The spacial metric $g_{ij}$ can be written in the following form

$$g_{ij}=\text{diag}\left(f^{2}(r),r^{2}\right).$$

(18)

We can calculate the components of the momentum variable as:

$$\pi^{\theta r}=-\frac{r}{2N^{\perp}f}(N^{\theta})^{\prime}.$$

(19)

We vary with respect to the shift vector $N^{i}$, which gives

$$\pi^{ij}_{,j}=0=g^{il}\partial_{k}\pi^{k}_{l}-\frac{1}{2}g^{il}(\partial_{l}g_{jk})\pi^{jk}.$$

(20)

Since $g_{jk}$ is diagonal and $\pi^{jk}$ is off-diagonal,

$$g^{il}(\partial_{l}g_{jk})\pi^{jk}=0$$

(21)

which gives rise to the following result

$$\pi^{\theta r}=\frac{\mathcal{A}}{r^{2}},$$

(22)

where $\mathcal{A}$ is a constant determined by the boundary conditions later. We can rewrite the Hamiltonian constant as

$$H_{\perp}=f\left[\frac{2\mathcal{A}^{2}}{r^{3}}+\frac{d}{dr}(\frac{1}{f^{2}})+2\tilde{\Lambda}r+\frac{\tilde{B}^{2}}{2rf^{2}}\right],$$

(23)

where

$$2\tilde{\Lambda}=2\Lambda+\frac{\eta^{2}\chi^{2}P^{2}}{r^{2}}+V[\phi]$$

(24)

and

$$\tilde{B}^{2}=B^{2}+\eta^{2}r^{2}{\chi^{\prime}}^{2}.$$

(25)

Varying with respect to $N^{\perp}$ gives $H_{\perp}=0$, which leads to the following equation

$$\frac{dy}{dr}+\frac{\tilde{B}^{2}}{2r}y=-2\tilde{\Lambda}r-2\frac{\mathcal{A}^{2}}{r^{3}},$$

(26)

with the variable $y=\frac{1}{f^{2}}$. The equations of motion for the scalar field $(\phi)$ and gauge fields $(A_{\mu})$ arising from (1) are given by

$$\frac{1}{\sqrt{-g}}\partial_{\mu}\sqrt{-g}F^{\mu\nu}=-e\epsilon^{ab}\phi^{a}D^{\nu}\phi^{b}$$

(27)

and

$$\frac{1}{\sqrt{-g}}D_{\mu}\sqrt{-g}D^{\mu}\phi^{a}=\lambda(\phi^{2}-\eta^{2})\phi^{a}.$$

(28)

For the vortex like solution these equations boil down to

$$\frac{1}{2}\chi^{\prime}y^{\prime}+y(\chi^{\prime\prime}+\frac{\chi^{\prime}}{r})=\frac{\chi}{r^{2}}P^{2}+\lambda\eta^{2}(\chi^{2}-1)\chi$$

(29)

and

$$\frac{1}{2}P^{\prime}y^{\prime}+y(P^{\prime\prime}-\frac{P^{\prime}}{r})=e\eta^{2}\chi^{2}P.$$

(30)

The above equations (26), (30), (29) are highly coupled non linear second order differential equations. This complicated spacetime can admit more than one solution. There could be some nonsingular vortex solutions depending upon the parameters. Instead of solving these differential equations numerically, we can look for simpler solution considering a particular case. we shall go back to the equation (23) and we can rewrite our Hamiltonian constraint as

$$H_{\perp}=f\left[(a^{2}+b^{2})+c\right]$$

(31)

where,

$$a=\frac{P^{\prime}}{\sqrt{2}\sqrt{r}fe}\pm\frac{\eta\chi P}{\sqrt{r}},$$

(32)

$$b=\frac{\eta\chi^{\prime}\sqrt{r}}{\sqrt{2}f}\pm\frac{\sqrt{\lambda}\sqrt{r}\eta^{2}}{2}(\chi^{2}-1),$$

(33)

and

$$c=\frac{2\mathcal{A}^{2}}{r^{3}}+\frac{d}{dr}\left(\frac{1}{f^{2}}\right)+2\Lambda r\mp\frac{\eta}{\sqrt{2}rf}\left(\sqrt{\lambda}\eta^{2}r^{2}(\chi^{2}-1)\chi^{\prime}+\frac{2PP^{\prime}\chi}{e}\right).$$

(34)

For $H_{\perp}=0$ we can choose individually $a=0$, $b=0$ and $c=0$ as a possible solution, which leads to the following equations

$$P^{\prime}=\mp\sqrt{2}\eta efP\chi,$$

(35)

$$\chi^{\prime}=\mp\frac{\sqrt{\lambda}}{\sqrt{2}}\eta f(\chi^{2}-1),$$

(36)

and

$$\frac{d}{dr}y+\mathcal{P}(r)y=\mathcal{Q}(r).$$

(37)

The quantities $\mathcal{P}$ and $\mathcal{Q}$ are given by

$$\mathcal{P}=\pm\frac{\eta}{\sqrt{2}r}\left(\eta r^{2}{\chi^{\prime}}^{2}+\frac{{P^{\prime}}^{2}}{\eta e}\right),$$

(38)

and

$$\mathcal{Q}=-\left(2\Lambda r+\frac{2\mathcal{A}^{2}}{r^{3}}\right).$$

(39)

Since the above three equations are coupled differential equations we can solve these in a perturbative approach. In equations (35) and(36) we can choose particular form of $f$, say $f_{BTZ}$, to calculate $\chi$ and $P$ and put the value of these variables in equation (37) to obtain the value of the desired metric. For example from a sufficiently large distance from the origin (outside the horizon in case of black hole solution) we can take the following form of ansatz for the variables $\chi$ and $P$.

$$\chi(r)=1+\frac{\chi_{1}}{r}+\frac{\chi_{2}}{r^{2}}+...$$

(40)

and

$$P(r)=\frac{P_{1}}{r}-\frac{P_{2}}{r^{2}}+...$$

(41)

and we can impose suitable boundary conditions to know the value of $\chi_{i}$’s and $P_{i}$’s (for $i=1,2,...$). But now we are mainly focused on the behaviour of the metric (more precisely $f(r)$) not the scalar and gauge fields.

This differential equation (37) has the solution of the following form

$$y=e^{\int\mathcal{P}(r)dr}\int\mathcal{Q}(r)dr+\mathcal{D}~{}e^{\int\mathcal{P}(r)dr}$$

(42)

where

$$y=\frac{1}{f^{2}}.$$

(43)

We have to vary the action with respect to $g_{ij}$ which is equivalent to vary with respect to ‘$f$’ which leads to the following equation

$$\frac{{N^{\perp}}^{\prime}}{N^{\perp}}=-\frac{f^{\prime}}{f},$$

(44)

which gives rise to the following relation

$$N^{\perp}=f^{-1}.$$

(45)

We substitute the value of $N^{\perp}$ in (19) and we can calculate the value of $N^{\theta}$ from equation (22) in the following way

$$(N^{\theta})^{\prime}=-\frac{2\mathcal{A}}{r^{3}}\Rightarrow N^{\theta}=C+\frac{\mathcal{A}}{r^{2}}.$$

(46)

We have to find three constants $\mathcal{A}$, $\mathcal{D}$ and $C$. We will take the standard results Banados et al. (1992) which is solved for no magnetic field and scalar field. Which gives $\mathcal{D}=-M$, $\mathcal{A}=-\frac{J}{2}$ and $C=0$, with mass $M$ and angular momentum $J$, which gives rise to the following equations

$$y=e^{\int_{0}^{r}\mathcal{P}(r)dr}\left(\int\mathcal{Q}(r)dr-M\right).$$

(47)

The metric under our consideration can be given by

$$ds^{2}=-{N^{\perp}}^{2}dt^{2}+{N^{\perp}}^{-2}dr^{2}+r^{2}\left(d\theta+\frac{J}{2r^{2}}dt\right)^{2}$$

(48)

with

$${N^{\perp}}^{2}=e^{\int\mathcal{P}(r)dr}\left(-M+\frac{r^{2}}{l^{2}}+\frac{J^{2}}{4r^{2}}\right)$$

(49)

and the negative cosmological constant $\Lambda=-\frac{1}{l^{2}}$.

From the functions $\mathcal{P}(r),\chi(r),P(r)$ described in equations (38, 40, 41) we discuss the asymptotic behavior of the metric (48). At large $r$ the quantities

$$\chi^{\prime}=-\frac{\chi_{1}}{r^{2}}-O(r^{-3}),\text{ and }P^{\prime}=-\frac{p_{1}}{r^{2}}+O(r^{-3}),$$

(50)

which gives

$$\mathcal{P}(r)=\pm\frac{\eta^{2}\chi_{1}^{2}}{\sqrt{2}r^{3}}+O(r^{-4}),$$

(51)

and

$${N^{\perp}}^{2}=e^{\mp\frac{\eta^{2}\chi_{1}^{2}}{4\sqrt{2}r^{4}}}\left(-M+\frac{r^{2}}{l^{2}}+\frac{J^{2}}{4r^{2}}\right).$$

(52)

For large value of $r$ the quantity $e^{\mp\frac{\eta^{2}\chi_{1}^{2}}{4\sqrt{2}r^{4}}}\to 1$, and Eq. (48) boils down to the usual BTZ black hole metric.

If there is sufficient energy and mass density inside the vortex have it will form a magnetically charged black hole. The positions of the horizons $r_{\pm}$ can be calculated by putting $N^{\perp}=0$, which gives the following expression

$$r_{\pm}^{2}=\frac{l^{2}}{2}\left(M\pm\sqrt{M^{2}-\frac{J^{2}}{l^{2}}}\right).$$

(53)

The angular velocity for this black hole can be given by

$$\Omega=\frac{J}{2r_{+}^{2}}.$$

(54)

The famous semiclassical Hawking temperature has been calculated in the semiclassical approximation in various methodes, for e.g. Hawking (1975); Bekenstein (1972, 1973); Bardeen et al. (1973); Bekenstein (1974); Hawking (1976). We would like to compute the Hawking temperature adopting the tunnelling formalism Parikh and Wilczek (2000); Jiang et al. (2006); Srinivasan and Padmanabhan (1999); Kerner and Mann (2006). A general procedure based on the Hamilton-Jacobi method for calculation of Hawking temperature is done in Banerjee and Majhi (2008). First the $r$-$t$ sector of the metric is isolated through the following transformation,

$$d\xi=d\theta-\Omega dt.$$

(55)

Then using the transformation (55), in the the near horizon approximation, the metric(48) can be written in the form,

$$ds^{2}=-{N^{\perp}}^{2}dt^{2}+{N^{\perp}}^{-2}dr^{2}+r_{+}^{2}d\xi^{2},$$

(56)

where the $r$-$t$ sector is isolated from the angular part ($d\xi^{2}$). The expression of semiclassical Hawking temperature for $(2+1)$ dimensional black hole is given by

$$T=\frac{\hbar}{4}\left(\text{Im}\int_{c}\frac{dr}{N_{\perp}^{2}}\right)^{-1}$$

(57)

For our case, the expression for $N^{\perp}$ is given by

$$N^{\perp}=h(r){N^{\perp}}_{BTZ}$$

(58)

with

$$h(r)=e^{\int_{0}^{r}\mathcal{P}(r)dr}$$

(59)

from equation (49) and $N_{BTZ}$ is the usual lapse function for BTZ black hole.

In our case semiclassical Hawking temperature can be calculated as

$$T=\frac{\hbar}{4}\left(\text{Im}\int_{c}l^{2}\frac{r^{2}dr}{h(r)^{2}(r^{2}-r+^{2})(r^{2}-r_{-}^{2})}\right)^{-1}$$

(60)

with $r_{\pm}$ (53) is the inner and outer horizon radii of the usual BTZ black hole.

Doing the contour integration we can write down the expression of the Hawking temperature

$$T=h(r_{+})T_{BTZ}$$

(61)

where

$$T_{BTZ}=\frac{\hbar}{2\pi l^{2}}\frac{r_{+}^{2}-r_{-}^{2}}{r_{+}}$$

(62)

is the Hawking temperature for the usual BTZ black hole and

$$h(r_{+})=e^{\int_{0}^{r_{+}}\mathcal{P}(r)dr}$$

(63)

with $\mathcal{P}(r)$ is described at (38). In principle the factor $\int_{0}^{r_{+}}\mathcal{P}(r)dr\neq 0$ that is $h(r_{+})\neq 1$ so that the Hawking temperature is not same as ordinary BTZ black hole.

We can also describe the first law of the Black hole thermodynamics

$$dE=TdS+\Omega~{}dJ+\phi~{}dQ=\frac{\kappa}{8\pi}dA+\Omega~{}dJ+\phi~{}dQ,$$

(64)

where $E$ is the energy, $T$ is the temperature of the black hole, $S$ is the entropy, $\kappa$ is the surface gravity, $A$ is the horizon area, $\Omega$ is the angular velocity, $J$ is the angular momentum, $\Phi$ is the electrostatic potential and $Q$ is the electric charge. If we put $dE=0$ in the above equation (64), we see that the additional work terms from the scalar and gauge fields that try to compensate the change in the temperature relative to its usual value for the BTZ solution.

We shall try to prove the above statement by calculating the thermodynamic quantities explicitly. First, we start with the Bekenstein-Hawking entropy, which is directly proportional to the area of the black hole horizon. From the fundamental postulate of black hole thermodynamics, the entropy ($S$) Ganai et al. (2019) is defined as

$$S=\frac{A}{4},$$

(65)

where $A$ is the area of the event horizon. For $(2+1)$-dimensional black hole the area $A=2\pi r_{+}$, so that the the corresponding entropy is calculated as

$$S=\frac{\pi r_{+}}{2}.$$

(66)

In our case, the size of event horizon ($r_{+}$) for BTZ black hole with vortex is equal to the BTZ black hole without the vortex that is

$$r_{+}^{2}=\frac{l^{2}}{2}\left(M+\sqrt{M^{2}-\frac{J^{2}}{l^{2}}}\right)=r_{+BTZ}^{2}.$$

(67)

Therefore, the area of the $(2+1)$-dimensional black hole with vortex solution is the same as the area of the usual BTZ black hole, and so as the entropy; that is $S_{vortex}=S_{BTZ}$.

Now, we calculate the other thermodynamic quantities from the solution obtained in the previous subsection (II.2). The metric under our consideration is written as

$$ds^{2}=-\Delta dt^{2}+\Delta^{-1}dr^{2}+r^{2}\left(d\theta+\frac{J}{2r^{2}}dt\right)^{2}$$

(68)

with the lapse function

$$\Delta=e^{\int\mathcal{P}(r)dr}\left(-M+\frac{r^{2}}{l^{2}}+\frac{J^{2}}{4r^{2}}\right).$$

(69)

In article Ganai et al. (2019); Larrañaga (2008), the authors calculated the temperature and other thermodynamic quantities for an usual BTZ-black hole directly from the solution. Here we shall use the same formula for calculating the same quantities for our solution.

The surface gravity is calculated as:

	$\displaystyle\kappa$	$\displaystyle=$	$\displaystyle\frac{1}{2}\frac{\partial\Delta}{\partial r}\mid_{r=r_{+}}$		(70)
		$\displaystyle=$	$\displaystyle\frac{1}{2}\mathcal{P}(r_{+})\Delta(r_{+})+\frac{1}{2}e^{\int_{0}^{r_{+}}\mathcal{P}(r)dr}\left[\frac{2r_{+}}{l^{2}}-\frac{J^{2}}{8r_{+}^{3}}\right]$
		$\displaystyle=$	$\displaystyle e^{\int_{0}^{r_{+}}\mathcal{P}(r)dr}\left[\frac{r_{+}}{l^{2}}-\frac{J^{2}}{16r_{+}^{3}}\right].$

The Hawking temperature $T$ as follows,

$$T=\frac{\kappa}{2\pi}=e^{\int_{0}^{r_{+}}\mathcal{P}(r)dr}\left[\frac{r}{2\pi l^{2}}-\frac{J^{2}}{32\pi r^{3}}\right].$$

(71)

At $r=r_{+}$

$$\mathcal{P}(r_{+})=\frac{\eta^{2}\chi_{1}^{2}}{\sqrt{2}r_{+}^{3}}+O\left(r_{+}^{-4}\right),$$

(72)

such that the value of the Hawking temperature becomes

$$T=\frac{\kappa}{2\pi}=e^{-\frac{\eta^{2}\chi_{1}^{2}}{4\sqrt{2}r_{+}^{4}}}\left[\frac{r_{+}}{2\pi l^{2}}-\frac{J^{2}}{32\pi r_{+}^{3}}\right].$$

(73)

The angular velocity $\Omega$ is given by

	$\displaystyle\Omega$	$\displaystyle=$	$\displaystyle\frac{\partial\Delta}{\partial J}\mid_{r=r_{+}}$		(74)
		$\displaystyle=$	$\displaystyle e^{-\frac{\eta^{2}\chi_{1}^{2}}{4\sqrt{2}r_{+}^{4}}}\left[\frac{J}{2r_{+}^{2}}\right].$

From the above calculation we show that the entropy $S$ is same as the usual BTZ black hole, i.e. $dS=0$. As a result, there will be changes to the standard thermodynamic parameters, for e.g. $T,~{}\Omega$ etc., to hold the first law of the black hole thermodynamics.