Modified Metrics of Acoustic Black Holes: A Review

**Relativistic Acoustic Metric**

In order to determine the relativistic acoustic metric, we start by considering the following Lagrangian density:

$$\mathcal{L} = \partial_\mu \phi^* \partial^\mu \phi + m^2 |\phi|^2 - b |\phi|^4$$

Now, we decompose the scalar field as

$$\phi = \sqrt{\rho(x,t)} \exp(iS(x,t))$$ such that

$$\mathcal{L} = \rho \partial_\mu S \partial^\mu S + m^2 \rho - b \rho^2 + \frac{\rho}{\sqrt{\rho}} \left( \partial_\mu \partial^\mu \right) \sqrt{\rho}$$

Moreover, from the above Lagrangian, we find the equations of motion for $S$ and $\rho$ given respectively by

$$\partial_\mu (\rho \partial^\mu S) = 0$$

and

$$\frac{1}{\sqrt{\rho}} \partial_\mu \partial^\mu \sqrt{\rho} + \partial_\mu S \partial^\mu S + m^2 - 2b \rho = 0$$

Where the Eq. (3) is the continuity equation and Eq. (4) is an equation describing a hydro dynamical fluid, and the term,

$$\left( \sqrt{\rho} \right)^{-1} \partial_\mu \partial^\mu \sqrt{\rho}$$

called the quantum potential can be neglected in the hydrodynamic region. Now, by performing the following perturbations on equations of motion (3) and (4):

$$\rho = \rho_0 + \varepsilon \rho_1 + O\left(\varepsilon^2\right)$$

$$S = S_0 + \varepsilon \psi + O\left(\varepsilon^2\right)$$

We obtain

$$\partial_\mu \left( \rho_1 u_0^\mu + \rho_0 \partial^\mu \psi \right) = 0,$$

and

$$u_0^\mu \partial_\mu \psi - b \rho_1 = 0,$$

where we have defined $u_0^\mu = \partial^\mu S_0$, Hence, solving (8) for $\rho_1$ and substituting into (7), we have

$$\partial_\mu \left[ u_0^\mu u_0^\nu + b \rho_0 g^{\mu\nu} \right] \partial_\nu \psi = 0.$$

We can also write the above equation as follows:

$$\partial_i \left\{ \omega_0^2 \left[ -1 - \frac{b \rho_0}{2 \omega_0^2} \right] \partial_i \psi - \frac{\omega_0^2 v_0^2}{\omega_0} \partial_i \psi + \omega_0^2 \left[ -\frac{v_0^2 v_0^2}{\omega_0^2} \right] \partial_j \psi \right\} = 0,$$

(10)

$$ \text {W h e r e} \omega_ {0} = \partial^ {i} S _ {0} \text {a n d} v _ {0} ^ {i} = \partial_ {i} S _ {0} $$

(the local velocity field). In

addition, we define

2 2 0 0 / 2 sc bρ ω =

to be the speed of sound

$$ \text {a n d} v ^ {i} = v _ {0} ^ {i} / \omega_ {0}. $$

. However, the equation (10) becomes $$ \partial_ {t} \left\{\frac {b \rho_ {0}}{2 c _ {s} ^ {2}} \left[ \left(- 1 - c _ {s} ^ {2}\right) \partial_ {t} \psi - v ^ {i} \partial_ {i} \psi \right] \right\} + \partial_ {i} \left\{\frac {b \rho_ {0}}{2 c _ {s} ^ {2}} \left[ - v ^ {i} \partial_ {t} \psi \left(- v ^ {i} v ^ {j} + c _ {s} ^ {2} \delta^ {i j}\right) \partial_ {j} \psi \right] \right\} = 0 $$ ( ) ( ) 2 2 0 0 2 2 1 – – – 0. 2 2 i i i j ij t s t i i t s j s s (11) In this way, the above equation can be written as a Klein- Gordon equation in (3+1) dimensional curved space as follows:

( ) 1 0, g g g

$$ \frac {1}{\sqrt {- g}} \partial_ {\mu} \left(\sqrt {- g} g ^ {\mu \nu} \partial_ {\nu}\right) \psi = 0 $$ (12) Where 2 2 i s – 

0 ΛΛ ΛΛ (13)

δ  Hence, by determining the inverse of g µν , we find the relativistic acoustic metric given by

2 2 i s – 

0 ΛΛ ΛΛ (14)

δ  The metric depends on the density 0 ρ , the local sound speed in the fluid cs, the velocity of flow vρ. This is the acoustic black hole metric for high sc and  vρspeeds Note that, in the non-relativistic limit, up to an overall factor, the metric found by Unruh is obtained.

$$ = \frac {b \rho_ {0}}{2 c _ {s} ^ {2}} \left( \begin{array}{c c c} - c _ {s} ^ {2} + v ^ {2} & \vdots & - v ^ {i} \\ \dots \dots & \cdot & \dots \dots \\ - v ^ {j} & \vdots & \delta^ {i j} \end{array} \right). $$

2 2 i s – c v v b g c v

 µν ρ . 2

0 2 ΛΛ ΛΛ (15) s j ij

δ  The relativistic acoustic metric (14) has also been obtained from the Abelian Higgs model [20].

The Dispersion Relation

Here we aim to examine the dispersion relation. Hence, we will adopt the notation written below ρρ ρ $$ \psi \sim R e \left[ e ^ {i \omega t - i \vec {k} \cdot \vec {x}} \right], \omega = \frac {\partial \psi}{\partial t}, \vec {k} = \nabla \psi . $$ , , . i t i k x Re e k t (16) So we can write the Klein-Gordon equation (11) in terms of momentum and frequency as follows:

( ) ( ) ( ) 2 2 2 2 2 1 2 0. s s c v k c v k ω ω + + ⋅ − − = ρ ρ (17) Now, by making 1, i i k δ = we have

s s (18) In the limit of small 1_v_ , we find the modified dispersion relation $$ \omega \approx E \left(1 + \frac {v _ {1}}{2}\right), $$ (19) $$ \text {W h e r e} E = c _ {s} k $$

is the linear dispersion relation.

The Lorentz-Violating Model

At this point, we consider the Abelian Higgs model with Lorentz symmetry breaking that has been introduced as a change in the scalar sector of the Lagrangian [61]. Moreover, the relativistic acoustic metric violating Lorentz has been found in Anacleto MA, et al. [21]. Then, the corresponding Lagrangian for the Abelian Higgs model in the Lorentz- violating background is written as follows:

$$ \mathcal {L} = - \frac {1}{4} F _ {\mu \nu} F ^ {\mu \nu} + \left| D _ {\mu} \phi \right| ^ {2} + m ^ {2} \left| \phi \right| ^ {2} - b \left| \phi \right| ^ {4} + k ^ {\mu \nu} D _ {\mu} \phi^ {*} D _ {\nu} \phi , \tag {20} $$ being – F A A µν µ ν ν µ =∂ ∂ the field intensity tensor, – D ieA µ µ µ φ φ φ = ∂ the covariant derivative and kµν a constant tensor implementing the Lorentz symmetry breaking, given by Anacleto MA, et al. [21].

$$ = \left[ \begin{array}{c c c c} \beta & \alpha & \alpha & \alpha \\ \alpha & \beta & \alpha & \alpha \\ \alpha & \alpha & \beta & \alpha \\ \alpha & \alpha & \alpha & \beta \end{array} \right], $$ β α α α α β α α α α β α α α α β , kµν (21) where α and β are real parameters.

Next, following the steps taken in the previous section

to derive the relativistic acoustic metric from quantum

field theory, we consider

$$ \phi = \sqrt {\rho (x , t)} \exp \left(i S (x, t)\right) \mathrm {i n} \mathrm {t h e} $$

Lagrangian above. Thus, we have

1 2 4

2 2 2 µ µν µ µ µν µ µ µ F F S S e A S e A A m b

ρ ρ ρ ρ ρ = − + ∂ ∂ − ∂ + + −

Λ ρ ρ ρ ρ (22)

+ ∂ ∂ − ∂ + + ∂∂

2 ( 2 ) ( )

µν µ µ ν µ ν µ ν µ k S S eA S e A A

$$ \text {W h e r e} \tilde {\partial} ^ {\mu} = \partial^ {\mu} + k ^ {\mu \nu} \partial_ {\nu}. $$

. Then, the equations of motion

for S and ρ are:

$$ \partial \mu \left[ \rho u ^ {\mu} + \rho k ^ {\mu \nu} u _ {\nu} \right] = 0, \tag {23} $$ and (24)

where we have defined

$$ u _ {\mu} = \partial_ {\mu} S - e A _ {\mu}. $$

Now, by linearizing

the equations above around the background (

) 0 0 , S ρ , with $$ \rho = \rho_ {0} + \varepsilon \rho_ {1} + O \left(\varepsilon^ {2}\right) \tag {25} $$ $$ S = S _ {0} + \varepsilon \psi + O \left(\varepsilon^ {2}\right), \tag {26} $$ and keeping the vector field Aµ unchanged, we have $$ \partial_ {\mu} \left[ \rho_ {1} \left(u _ {0} ^ {\mu} + k ^ {\mu \nu} u _ {0 \nu}\right) + \rho_ {0} \left(g ^ {\mu \nu} + k ^ {\mu \nu}\right) \partial_ {\nu} \psi \right] = 0, \tag {27} $$ and $$ \left(u _ {0} ^ {\mu} + k ^ {\mu \nu} u _ {0 \nu}\right) \partial_ {\mu} \psi - b \rho_ {1} = 0, \tag {28} $$ by solving (28) for 1 ρ and replacing into equation (27), we obtain $$ \partial_ {\mu} \left[ u _ {0} ^ {\mu} u _ {0} ^ {\nu} + k ^ {\mu \lambda} u _ {0 \lambda} u _ {0} ^ {\nu} + u _ {0} ^ {\mu} k ^ {\nu \lambda} u _ {0 \lambda} + b \rho_ {0} \left(g ^ {\mu \nu} + k ^ {\mu \nu}\right) \right] \partial_ {\nu} \psi = 0. \tag {29} $$ Hence, we find the equation of motion for a linear acoustic disturbance ψ given by a Klein-Gordon equation curved space ( ) 1 0, g g g $$ \frac {1}{\sqrt {- g}} \partial_ {\mu} \left(\sqrt {- g} g ^ {\mu \nu} \partial_ {\nu}\right) \psi = 0 $$ (30) where µν g is the relativistic acoustic metrics. For β ≠0 and α=0 , we have Anacleto MA, et al. [21]

     − + −       = ⋅     − +    

2 2 c v v β β β ρ β i s

−   $$ \underline {{\tilde {\beta} _ {-}}} \quad \begin{array}{c c} \beta_ {+} & \beta_ {+} \\ \dots \dots \end{array} $$ b g c v f v v

0 ΛΛ Λ Λ (31) µν

2 2 Θ   s j ij i j

β δ β −  β +

+ + = + −    Θ and − − + = + −      .

2 2 1 sc v β β β

2 2 sc v fβ β β β β β −

+ − where The acoustic line element in the Lorentz-violating background can be written as follows    ρ ρ ρ ρ ρ    Θ $$ = \frac {b \rho_ {0} \sqrt {\tilde {\beta} _ {-}}}{2 c _ {s} \sqrt {\mathcal {Q}}} \left[ - \left(\frac {c _ {s} ^ {2}}{\tilde {\beta} _ {+}} - \frac {\tilde {\beta} _ {-} v ^ {2}}{\tilde {\beta} _ {+}}\right) d t ^ {2} - 2 \vec {v} \cdot d \vec {x} d t + \frac {\tilde {\beta} _ {-}}{\tilde {\beta} _ {+}} \left(\vec {v} \cdot d \vec {x}\right) ^ {2} + f _ {\beta} d \vec {x} ^ {2} \right] $$

2 2 2 0 2 2 2 2 2

b c v ds dt v dx dt v d x f dx c β ρ β β β β β β ( )

$$ \left| \frac {\beta_ {-}}{z} \right| - \left(\frac {c _ {s} ^ {2}}{z} - \frac {\beta_ {-} v ^ {2}}{z}\right) d t ^ {2} - 2 \vec {v} \cdot d \vec {x} d t + \frac {\beta_ {-}}{z} ( $$ s + + + s (32)

Now changing the time coordinate as

$$ s d \tau = d t + \tilde {\beta} _ {+} \vec {v} \cdot \vec {d} x, $$ , we find the acoustic metric in the stationary form       Φ Φ Θ $$ = \frac {b \rho_ {0} \sqrt {\tilde {\beta} _ {-}}}{2 c _ {s} \sqrt {\mathcal {Q}}} \left[ - \left(\frac {c _ {s} ^ {2}}{\tilde {\beta} _ {+}} - \frac {\tilde {\beta} _ {-} v ^ {2}}{\tilde {\beta} _ {+}}\right) d \tau^ {2} + \mathcal {F} \left(\frac {\tilde {\beta} _ {-} v ^ {i} v ^ {j}}{c ^ {2} - \tilde {\beta} _ {-} v ^ {2}} + \frac {f _ {\beta}}{\mathcal {F}} \delta^ {i j}\right) d x ^ {i} d x ^ {j} \right] $$ f b c v v v ds d dx dx c v c

2 2 0 2 2 2 2 2

β ρ β β β τ δ β β β i j ij i j s

$$ \left| \frac {\beta_ {-}}{-} \right| - \left(\frac {c _ {s} ^ {2}}{z} - \frac {\beta_ {-} v ^ {2}}{z}\right) d \tau^ {2} + \mathcal {F} \left(\frac {\beta_ {-} v}{2}\right) $$ $$ \beta_ {+} \quad \beta_ {+}) \quad \left(c ^ {2} - \beta_ {-} v\right) $$ s (33) $$ \mathrm {e} \mathcal {F} = \frac {\tilde {\beta} _ {+}}{\tilde {\beta} _ {-}} + \frac {c _ {s} ^ {2}}{\tilde {\beta} _ {+}} - \frac {\tilde {\beta} _ {-} v ^ {2}}{\tilde {\beta} _ {+}} $$

2 2 sc v β β β β β $$ - \mathrm {F o r} \tilde {\beta} = 1, $$

, we recover the result + − where found in Ref Ge XH, et al. [20]. Next, for 0 β = and 0 α ≠ , we have Anacleto MA, et al. [21]

 tt tj ΛΛ ΛΛ (34)

 where $$ g _ {t t} = - \left[ \left(1 + \alpha\right) c _ {s} ^ {2} - v ^ {2} + \alpha^ {2} \left(1 - v\right) ^ {2} \right] \tag {35} $$ $$ g _ {t j} = - \left(1 - \vec {\alpha}. \vec {\nu}\right) v ^ {j}, \tag {36} $$ $$ g _ {i t} = - \left(1 - \vec {\alpha} \cdot \vec {v}\right) v ^ {i}, \tag {37} $$ $$ g _ {i j} = \left[ \left(1 - \vec {\alpha} \cdot \vec {v}\right) ^ {2} + c _ {s} ^ {2} - v ^ {2} \right] \delta^ {i j} + v ^ {i} v ^ {j}, \tag {38} $$ $$ f = \left(1 + \alpha\right) \left[ \left(1 - \vec {\alpha} \cdot \vec {v}\right) ^ {2} + c _ {s} ^ {2} \right] - v ^ {2} + \alpha^ {2} \left(1 - v\right) ^ {2} \left[ 1 + \left(1 - \vec {\alpha} \cdot \vec {v}\right) ^ {2} c _ {s} ^ {- 2} \right] $$ (39) Thus, the acoustic line element in the Lorentz-violating background can be written as

(40) where ( ) 2 2 2 1 · s f v c v α α = − + − ρρ Now changing the time coordinate as ρρ ρ ρ ( )( )

2 2 2 2 1 · ·

v v d x d dt c v v

α τ α α − = +   + + −  

(41) ( ) ( )

1 – 1 s

We find the acoustic metric in the stationary form (42)

$$ \text {w h e r e} \Lambda = \left(1 - \vec {\alpha} \cdot \vec {v}\right) ^ {2} - g _ {t t}. \text {F o r} \alpha = 0, $$

, the result found in Ge XH, et al. [20] is recovered.

Noncommutative Acoustic Black Hole

The metric of a noncommutative canonical acoustic black hole has been found by us in Anacleto MA, et al. [22]. Here, starting from the noncommutative Abelian Higgs model, we briefly review the steps to generate the relativistic acoustic metric in the noncommutative background. Thus, the Lagrangian of the Abelian Higgs model in the noncommutative background is given by Ghosh S [62].

( ) κ κ φ φ φ

2 2 4 2 µν µν µ F F D m b

$$ C = - \frac {\kappa_ {+}}{4} F _ {\mu \nu} F ^ {\mu \nu} + \kappa_ {-} \left(\left| D _ {\mu} \phi \right| ^ {2} + m ^ {2} | \phi | ^ {2} - b\right) $$ Λ † † 4 1 2 ( ) ( )   + +     αβ µ µ αµ β β F D D D D θ φ φ φ φ (43) $$ \mathrm {b e i n g} \kappa_ {\pm} = 1 \pm \theta^ {\mu \nu} F _ {\mu \nu} / 2, F _ {\mu \nu} = \partial_ {\mu} A _ {\nu} - \partial_ {\nu} A _ {\mu} t $$

the field intensity tensor and – D ieA µ µ µ φ φ φ = ∂ the covariant derivative. The parameter αβ θ is a constant, real-valued antisymmetric D D × - matrix in D -dimensional spacetime with dimensions of length squared. Now, we use

$$ \phi = \sqrt {\rho (x , t)} \exp \left(i S (x, t)\right) \mathrm {i n} $$ the above Lagrangian, such that Anacleto MA, et al. [22].

2 2 , 4 F F g S S m b g µν µν µν µν µ ν µ ν κ ρ ρ θ ρ θ ρ ρ ρ $$ \mathcal {L} = - \frac {\kappa_ {+}}{4} F ^ {\mu v} F _ {\mu v} + \rho \bar {g} ^ {\mu v} \mathcal {D} _ {\mu} S \mathcal {D} _ {\nu} S + \tilde {\theta} m ^ {2} \rho - \tilde {\theta} b \rho^ {2} + \frac {\rho}{\sqrt {\rho}} \bar {g} ^ {\mu v} \partial_ {\mu} \partial_ {\nu} $$ (44) $$ \text {w h e r e} \mathcal {D} _ {\mu} = \partial_ {\mu} - e A _ {\mu} / S, \quad \bar {g} ^ {\mu \nu} = \tilde {\theta} g ^ {\mu \nu} + \Theta^ {\mu \nu}, \quad \tilde {\theta} = 1 + \vec {\theta} \cdot \vec {B}, $$ $$ \vec {B} = \nabla \times \vec {A} $$ $$ 1 \mathrm {a n d} \Theta^ {\mu \nu} = \theta^ {\alpha \mu} F _ {\alpha} ^ {\nu}. $$

. In our analysis we consider the case where there is no noncommutativity between space and time, that is $$ s \theta^ {0 i} = 0 \mathrm {a n d} \mathrm {u s e} \theta^ {i j} = \varepsilon^ {i j k} \theta_ {k}, F ^ {i 0} = E ^ {i} \mathrm {a n d} $$ ij ijk k F B ε = . In the sequence we obtain the equations of motion for S and ρ as follows:

$$ \partial_ {\mu} \left[ \tilde {\theta} \rho u ^ {\mu} + \rho \tilde {\Theta} ^ {\mu \nu} u _ {\nu} \right] = 0, \tag {45} $$ and $$ \frac {1}{\sqrt {\rho}} \bar {g} ^ {\mu \nu} \partial_ {\mu} \partial_ {\nu} \sqrt {\rho} + \bar {g} ^ {\mu \nu} u _ {\mu} u _ {\nu} + \tilde {\theta} m ^ {2} - 2 \tilde {\theta} b \rho = 0, \tag {46} $$ $$ \mathrm {w h e r e} \tilde {\Theta} ^ {\mu \nu} = \left(\Theta^ {\mu \nu} + \Theta^ {\nu \mu}\right) / 2. $$

. Hence, by linearizing the equations of motion around the background ( ) ρ0 0 , S , with $$ \rho = \rho_ {0} + \rho_ {1}, S = S _ {0} + \psi $$

and keeping the vector potential

unchanged, such that $$ \partial_ {\mu} \left[ \rho_ {1} \bar {g} ^ {\mu \nu} u _ {0 \nu} + \rho_ {0} \left(g ^ {\mu \nu} + \tilde {\Theta} ^ {\mu \nu}\right) \partial_ {\nu} \psi \right] = 0 \tag {47} $$ $$ \left(\tilde {\theta} u ^ {\mu} + \tilde {\Theta} ^ {\mu \nu} u _ {\nu}\right) \partial_ {\mu} \psi - b \tilde {\theta} \rho = 0. \tag {48} $$ Then, by manipulating the above equations, we obtain the equation of motion for a linear acoustic disturbance ψ in the form ( ) 1 0, g g g $$ \frac {1}{\sqrt {- g}} \partial_ {\mu} \left(\sqrt {- g} g ^ {\mu \nu} \partial_ {\nu}\right) \psi = 0 $$ (49) where

is the relativistic acoustic metric with noncommutative corrections in (3+1) dimensions and with g µν  given in the form Anacleto MA, et al. [22].

$$ \tilde {g} t t = - \left[ \left(1 - 3 \vec {\theta} \cdot \vec {B}\right) c _ {s} ^ {2} - \left(1 + 3 \vec {\theta} \cdot \vec {B}\right) v ^ {2} + 2 \left(\vec {\theta} \cdot \vec {v}\right) \left(\vec {B} \cdot \vec {v}\right) - \left(\vec {\theta} \times \vec {E}\right) \cdot \vec {v} \right], $$ (51) $$ \tilde {g} ^ {t j} = - \frac {1}{2} \left(\vec {\theta} \times \vec {E}\right) ^ {j} \left(c _ {s} ^ {2} + 1\right) - \left[ 2 \left(1 + 2 \vec {\theta} \cdot \vec {B}\right) - \left(\vec {\theta} \times \vec {E}\right) \vec {v} \right] \frac {v ^ {j}}{2} + \frac {B ^ {j}}{2} \left(\theta \cdot \vec {v}\right) + \frac {\theta^ {j}}{2} \left(B \cdot \vec {v}\right) $$ ( ) ( ) ( ) ( ) ( ) ( ) 2 1 1 2 1 2

• , 2 2 2 2 (52) $$ \tilde {g} ^ {i t} = - \frac {1}{2} \left(\vec {\theta} \times \vec {E}\right) ^ {i} \left(c _ {s} ^ {2} + 1\right) - \left[ 2 \left(1 + 2 \vec {\theta} \cdot \vec {B}\right) - \left(\vec {\theta} \times \vec {E}\right) \cdot \vec {v} \right] \frac {v ^ {i}}{2} + \frac {B ^ {i}}{2} \left(\theta \cdot \vec {v}\right) + \frac {\theta^ {i}}{2} \left(B \cdot \vec {v}\right) $$ ( ) ( ) ( ) ( ) ( ) ( ) 2 1 1 2 1 2
• , 2 2 2 2 (53) $$ \tilde {g} ^ {i j} = \left[ \left(1 + \vec {\theta} \cdot \vec {B}\right) \left(1 + c _ {s} ^ {2}\right) - \left(1 + \vec {\theta} \cdot \vec {B}\right) v ^ {2} - \left(\vec {\theta} \times \vec {E}\right) \vec {v} \right] \delta^ {i j} + \left(1 + \vec {\theta} \cdot \vec {B}\right) v ^ {i} v ^ {j}, $$ (54) ( )( ) ( ) ( ) ( )( ) 2 2 1 2
• 1 1 4
• 3
• 2
• . s f B c B v E v B v v θ θ θ θ   = − + − + − × +   ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ (55)

Setting $\theta = 0$, the acoustic metric above reduces to the acoustic metric obtained in Ref Ge XH, et al. [20].

**Canonical Acoustic Metric**

In this case the line element of the acoustic black hole is given by

$$ds^2 = -f(v_r)d\tau^2 + \frac{c_s^2}{f(v_r)}dr^2 + r^2(d\theta^2 + \sin^2\theta d\phi^2),$$

where the metric function, $f(v_r)$ takes de form

$$f(v_r) = c_s^2 - v_r^2 \rightarrow f(r) = c_s^2 \left(1 - \frac{r_h^4}{r^4}\right),$$

Here we have defined $v_r = c_s r_h^2/r^2$ being $r_h$ the radius of the event horizon. In this case we compute the Hawking temperature using the following formula:

$$T_H = \frac{f'(r_h)}{4\pi} = \frac{c_s^2}{\pi r_h},$$

By considering the above result for the Hawking temperature and applying the first law of thermodynamics, we can obtain the entropy (entanglement entropy [38]) of the acoustic black hole as follows

$$S = \int \frac{dE}{T} = \int \frac{dA}{4\pi r_h T_H} = \frac{A}{4c_s^2}$$

being $A = 4\pi r_h^2$ the horizon area of the canonical acoustic black hole.

**Canonical Acoustic Metric with Lorentz Violation**

In the limit $c_s^2 \ll 1$ and $v^2 \ll 1$ can be written as a Schwarzschild metric type. Thus, for $\beta \neq 0$ and $\alpha = 0$ and up to an irrelevant position-independent factor,

$$ds^2 = -f(v_r)d\tau^2 + \frac{c_s^2}{\sqrt{\beta_- \beta_+} f(v_r)}dr^2 + \sqrt{\beta_- \beta_+} r^2\left(d\theta^2 + \sin^2\theta d\phi^2\right),$$

where

$$f(v_r) = \sqrt{\beta_- \beta_+} \left[ \frac{c_s^2 - \beta_- v_r^2}{\beta_+} \right] \rightarrow f(r) = \sqrt{\beta_- \beta_+} \left[ \frac{c_s^2}{1 - \beta_- \frac{r_h^4}{r^4}} \right],$$

The Hawking temperature is given by

$$T_H = \frac{f'(r_h)}{4\pi} = \frac{c_s^2 \left(1 - \beta_- \frac{r_h^4}{r^4}\right)}{\left(1 + \beta_- \right)^3/2 \pi r_h} = \frac{c_s^2 \left(1 - 3\beta_- \right)}{\pi r_h},$$

Therefore, the temperature is decreased when we vary the parameter $\beta$. For $\beta = 0$ the usual result is obtained. Hence, from the above temperature, we have the following result for the entropy of the acoustic black hole in the background violating Lorentz.

$$S = \frac{(1 + 3\beta_-) A}{4c_s^2},$$

Now, for $\beta = 0$ and $\alpha \neq 0$, we find for $\alpha$ sufficiently small we have up to first order

$$f(v_r) = \frac{\tilde{\alpha} c_s^2 - v_r^2}{\sqrt{\tilde{\alpha} \left(1 - 2\alpha v_r\right)}}$$

where $\tilde{\alpha} = 1 + \alpha$. For $v_r = c_s r_h^2/r^2$ with $c_s = 1$, the metric function becomes

$$f(r) = \tilde{\alpha}^{-1/2} \left[ \tilde{\alpha} - \frac{r_h^4}{r^4} \left(1 + \alpha \frac{r_h^2}{r^2}\right) + \alpha \frac{r_h^2}{r^2} \right],$$

In the present case there is a richer structure such as charged and rotating black holes. The event horizon of the modified canonical acoustic black hole is obtained from the following equation:

$$\tilde{\alpha} - \frac{r_h^4}{r^4} \left(1 + \alpha \frac{r_h^2}{r^2}\right) + \alpha \frac{r_h^2}{r^2} = 0$$

which can also be rewritten in the form

$$r^6 + \alpha r_h^2 r^4 - \tilde{\alpha}^{-1} r_h^4 r^2 - \alpha r_h^6 = 0$$

we can also write

$$r^2 \left(r^2 - r_s^2\right) \left(r^2 - r_h^2\right) - \alpha r_h^6 = 0,$$

where

2 2 1 2 h r r α $$ r _ {\pm} ^ {2} = r _ {h} ^ {2} \left(- \frac {\alpha}{2} \pm \frac {1}{\sqrt {\tilde {\alpha}}}\right) $$ (69) Now, arranging the above equation (68), we have

6 2 2

2 2 2 hr r r r r r

α $$ = r _ {+} ^ {2} + \frac {\alpha r _ {h} ^ {6}}{r ^ {2} \left(r ^ {2} - r _ {-} ^ {2}\right)} $$ (70) ( ) + Therefore, we can find the event horizon by solving the above equation perturbatively. So, up to the first order in α we obtain

6 2 2 2 h h r r r r r r r

α α $$ \tilde {r} _ {+} ^ {2} \approx r _ {+} ^ {2} + \frac {\alpha r _ {h} ^ {6}}{r _ {+} ^ {2} \left(r _ {+} ^ {2} - r _ {-} ^ {2}\right)} = \left(1 - \frac {\alpha}{2}\right) r _ {h} ^ {2} \dots [ 7 1 ] $$

2 2 2 1 2 ( )

+ + $$ r _ {+} ^ {2} \left(r _ {+} ^ {2} - r _ {-} ^ {2}\right) $$

Then, we have

1 2 h r r α $$ \tilde {r} _ {+} = r _ {h} \sqrt {1 - \frac {\alpha}{2}} + \dots \tag {72} $$ For the Hawking temperature, we obtain

1 3 1 2 H T r α π +

$$ = \frac {1}{\pi \tilde {r} _ {+}} \left(1 + \frac {3 \alpha}{2}\right) [ 7 3 ] $$ In terms of hr , we have $$ \left| T _ {H} = \frac {1}{\pi r _ {h}} \left(1 + \frac {7 \alpha}{4}\right) \right| $$ (74) In this situation the temperature is increased when we vary the parameter α . For 0 α = one recovers the usual result.

In this case for entropy, we find $$ S = \left(1 - \frac {7 \alpha}{4}\right) \frac {A}{4} $$ (75)

Noncommutative Canonical Acoustic Metric

The noncommutative acoustic metric can be written as a Schwarzschild metric type, up to an irrelevant position- independent factor, in the nonrelativistic limit as follows Anacleto MA, et al. [22], ( ) 2 2 2 2 2 ( ) ( ) Φ Φ Φ r

2 2 2 ( ) r r

(76) where ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 1 3 1 3 2 r r s r r r r r θ θ θ θ   = = − ⋅ − + ⋅ − −   ρ ρ ρ ρ  Φ Φ  v v B c B B v v f v f v r r or (77) $$ f \left(v _ {r}\right) = 1 - 2 \vec {\theta} \cdot \vec {B} - 3 \theta \mathcal {E} _ {r} v _ {r} $$  (78) $$ \Lambda \left(v _ {r}\right) = 1 + \vec {\theta} \cdot \vec {B} - \theta \mathcal {E} _ {r} v _ {r} \tag {79} $$ $$ \Gamma \left(v _ {r}\right) = 1 + 4 \vec {\theta} \cdot \vec {B} - 2 \theta \mathcal {E} _ {r} v _ {r} \tag {80} $$ $$ \Sigma \left(v _ {r}\right) = \left[ \theta \mathcal {E} _ {r} - \left(B _ {r} v _ {r}\right) \theta_ {r} - \left(\theta_ {r} v _ {r}\right) B _ {r} \right] v _ {r} \tag {81} $$ $$ \begin{array}{l} \mathrm {B e i n g} \theta \mathcal {E} _ {r} = \theta \left(\vec {n} \times \vec {E}\right) _ {r}. \\ v = c r ^ {2} / r ^ {2}. \mathrm {w h e r e} \\ \end{array} $$

. Now, by applying the relation

$$ \begin{array}{l} \mathrm {B e i n g} \sigma c _ {r} = \sigma \left(n \times L\right) _ {r}. \\ v _ {r} = c _ {s} r _ {h} ^ {2} / r ^ {2}, \text {w h e r e} r \\ \end{array} $$

hr is the radius of the event horizon

and making 1

sc = and so, the metric function of the

noncommutative canonical acoustic black hole becomes

1/2 4 2 2 $$ (\theta) = \left[ 1 - 3 \vec {\theta} \cdot \vec {B} - \left(1 + 3 \vec {\theta} \cdot \vec {B} - 2 \theta_ {r} B _ {r}\right) \frac {r _ {h} ^ {4}}{r ^ {4}} - \theta \mathcal {E} _ {r} \frac {r _ {h} ^ {2}}{r ^ {2}} \right] \left[ 1 - 2 \vec {\theta} \cdot \vec {B} - 3 \theta \mathcal {E} _ {r} \frac {r _ {h} ^ {2}}{r ^ {2}} \right] ^ {- 1 /} $$ ρ ρ ρ ρ ρ ρ Φ  

4 2 2 1 3 1 3 2 1 2 3 h h h r r r r r r r r B B B B r r r θ θ θ θ θ θ ( ) ( ) (82) Next, we will do our analysis considering the pure magnetic sector first and then we will investigate the pure electric sector.

Hence, for 3 3 0, · 0, 0 r r B B θ θ θ θ = = ≠ = ρρ  (or 0 E = ) with small 3 3 B θ , ( ) 3 3 3 3

1 3 1 4 1 1 2 H h h

B B T r r B

θ θ π π θ $$ = \frac {\left(1 + 3 \theta_ {3} B _ {3}\right)}{\sqrt {1 - 2 \theta_ {3} B _ {3}}} \frac {1}{\pi r _ {h}} = \frac {1 + 4}{\pi} $$ (83)

3 3 $$ \mathrm {F o r} \theta = 0 \mathrm {t} $$

the usual result is obtained. Here the temperature

has its value increased when we vary the parameter θ .

However, for the temperature in (83) we can find the entropy given by ( ) 3 3 1 4 4 4 h H B A dE dA S T r T θ $$ T = \int \frac {d E}{T} = \int \frac {d A}{4 \pi r T _ {u}} = \frac {\left(1 - 4 \theta_ {3} B _ {3}\right) A}{4} \tag {84} $$ π $$ \text {w h e r e} A = 4 \pi r _ {h} ^ {2} \mathrm {i} $$

is the horizon area of the canonical acoustic

black hole.

At this point, we will consider the situation where

$$ At this point, we will consider the situation where B = 0 and $ \theta\mathcal{E}_{r}\neq0 $ . So, from (82), we have $$
$$

. So, from (82), we have

1/2 4 2 2 $$ \tilde {\mathcal {F}} (r) = \left[ 1 - \frac {r _ {h} ^ {4}}{r ^ {4}} - \theta \mathcal {E} _ {r} \frac {r _ {h} ^ {2}}{r ^ {2}} \right] \left[ 1 - 3 \theta \mathcal {E} _ {r} \frac {r _ {h} ^ {2}}{r ^ {2}} \right] ^ {- 1 / 2} \tag {85} $$

4 2 2 1 1 3 h h h r r r r r r r r r θ θ ( )

For this metric the event horizon is obtained by solving the equation below

4 2

4 2 1 0 h h r r r r r θ − − =  (86)

$$ r ^ {4} - \theta \mathcal {E} _ {r} r _ {h} ^ {2} - r _ {h} ^ {4} = 0 \tag {87} $$ So, solving the above equation, we obtain r h r r θ  (88) $$ \left| r _ {+} = \left(1 + \frac {\theta \mathcal {E} _ {r}}{4}\right) r \right| $$

1 4 For the Hawking temperature, we find ( ) 1 / 2 1 1 3 / 4 1 1 3

θ θ θ π π π θ + +

$$ = \frac {\left(1 - \theta \mathcal {E} _ {r} / 2\right)}{\sqrt {1 - 3 \theta \mathcal {E} _ {r}}} \frac {1}{\pi r _ {+}} = \frac {1 + \theta \mathcal {E} _ {r}}{\pi r _ {+}} = \frac {1 + 3}{\pi r _ {+}} $$    r r r H h r T r r r (89)  We also noticed that the temperature is increased when we vary the θ parameter. For entropy we have ( ) 1 $$ S = \frac {\left(1 - \theta \mathcal {E} _ {r}\right) A}{4} (9 0) $$

4 $$ \mathrm {w h e r e} A = 4 \pi r _ {+} ^ {2}. $$

Result Using GUP

At this point, we introduce quantum corrections via the generalized uncertainty principle (GUP) to determine the Hawking temperature and entropy of the canonical acoustic black hole in the Lorentz-violating and noncommutative background. So, we will adopt the following GUP [63, 64, 65, 66, 67, 68, 69, 70].

$$ \Delta x \Delta p \geq \frac {\hbar}{2} \left(1 - \frac {\lambda l _ {p}}{\hbar} \Delta p + \frac {\lambda^ {2} l _ {p} ^ {2}}{\hbar^ {2}} \left(\Delta p\right) ^ {2}\right) $$

2 2 2 η η η (91)

( )

2 1 2 Where α is a dimensionless positive parameter and pl is the Planck length.

In sequence, without loss of generality, we will adopt the

$$ G = c = k _ {B} = \hbar = l _ {p} = 1 a $$

and by assuming that

p E ∆∼

and following the steps performed in Anacleto MA, et

al. [38] we can obtain the following relation for the corrected

energy of the black hole   ≥ − + +       Λ (92)

2 λ λ ∆ ∆

2 1 2 2 gup E E x x

( ) ( ) Thus, applying the tunneling formalism using the Hamilton-Jacobi method, we have the following result for the probability of tunneling with corrected energy gup E given by $$ \Gamma = e x p \left[ - \frac {4 \pi E _ {g u p}}{\kappa} \right] $$ (93) where κ is the surface gravity. Comparing with the Boltzmann factor, / E T e− we obtain the following result for the Hawking temperature with quantum corrections

1 2 $$ \prime \leq T _ {H} \left[ 1 - \frac {\lambda}{2 \left(\Delta x\right)} + \frac {\lambda^ {2}}{2 \left(\Delta x\right) ^ {2}} + \dots \right] ^ {- 1} \tag {94} $$ λ λ ∆ ∆

2 1 2 2 H T T x x

( ) ( ) So, by applying it to temperature (58), we have the following result

2 c T $$ = \frac {c _ {s} ^ {2}}{\pi \left[ r _ {h} - \frac {\lambda}{4} + \frac {\lambda^ {2}}{8 r _ {h}} \dots \right]} $$ s (95)

2 4 8 r r λ λ π h h

Therefore, when 0 hr = the singularity is removed and the temperature is now zero. Next, we analyze the effect of GUP in the Lorentz-violating and noncommutative cases.

For this case we can calculate the entropy, which is given by

2

So due to the GUP we get a logarithmic correction term for the entropy.

Violation-Lorentz Case: In the situation where 0 β ≠ and 0 α = , the corrected temperature due to GUP is

1 2 $$ T = T _ {H} \left[ 1 - \frac {\lambda}{4 r _ {h}} + \frac {\lambda^ {2}}{8 r _ {h} ^ {2}} + \dots \right] ^ {- 1} \tag {97} $$ where $$ T _ {H} = \frac {1 - 3 \beta}{\pi r _ {h}} \tag {98} $$ Thus, we have

2 1 3 β − =   − +     Λ

T (99) λ λ π

4 8 h h r r

Note that when 0 hr → the Hawking temperature tends to zero, 0 T → . In the absence of the GUP the temperature, H T diverges when 0 hr = . Therefore, we observe that the GUP

has the effect of removing the singularity at

$$ r _ {h} = 0 \mathrm {i n t h e} $$

Hawking temperature of the acoustic black hole.

Now computing the entropy, we find

( ) $$ \mathrm {F o r} \beta = 0 \mathrm {a n d} \alpha \neq 0, $$

, we have

2 1 3 / 2 α $$ r = \frac {1 + 3 \alpha / 2}{\pi \left[ \tilde {r} _ {+} - \frac {\lambda}{4} + \frac {\lambda^ {2}}{8 \tilde {r} _ {+}} \dots \right]} $$ T (101) λ λ π + +

4 8 r r In terms of hr , we obtain

2 1 3 / 2 α $$ r ^ {\prime} = \frac {1 + 3 \alpha / 2}{\pi \left[ r _ {h} \left(1 - \frac {\alpha}{4}\right) - \frac {\lambda}{4} + \frac {\lambda^ {2}}{8 r _ {h}} \left(1 + \frac {\alpha}{4}\right) \dots \right]} $$ T (102) α λ λ α π

1 1 4 4 8 4 h h

r r In this situation, we can also verify the effect of the GUP

on the temperature that goes to zero when

$$ r _ {h} \rightarrow 0 \left(\tilde {r} _ {+} \rightarrow 0\right). \mathrm {I n} $$

addition, we note that in both cases the Hawking temperature

reaches a maximum value before going to zero, as we can see

in Figure 1. Therefore, presenting a behavior analogous to

what happens with the corrected Hawking temperature of

the Schwarzschild black hole.

For entropy, we obtain

(103) Again we find a logarithmic correction term and also the contribution of the parameter to the entropy.

Click to enlarge

Noncommutative Case: For the magnetic sector, the GUP- corrected Hawking temperature is given by

3 3 2 1 4 B T

θ $$ r ^ {\prime} = \frac {1 + 4 \theta_ {3} B _ {3}}{\pi \left[ r _ {h} - \frac {\lambda}{4} + \frac {\lambda^ {2}}{8 r _ {h}} \dots \right]} $$ (104) λ λ π

4 8 h h r r

Note that, the GUP acts as a temperature regulator

by removing the singularity when

0 hr =

. In addition, the

temperature goes through a maximum value point before

going to zero for

$$ r _ {h} = 0. $$

In this case entropy is given by

2

( ) Next, for the electrical sector, we find the following GUP- corrected Hawking temperature

2 1 θ $$ r = \frac {1 + \theta \mathcal {E} _ {r}}{\pi \left[ r _ {+} - \frac {\lambda}{4} + \frac {\lambda^ {2}}{8 r _ {+}} \dots \right]} $$  (106) r T λ λ π + +

4 8 r r In terms of hr , the temperature becomes

2 1 θ $$ r = \frac {1 + \theta \mathcal {E} _ {r}}{\pi \left[ r _ {h} \left(1 + \frac {\theta \mathcal {E} _ {r}}{4}\right) - \frac {\lambda}{4} + \frac {\lambda^ {2}}{8 r _ {h}} \left(1 - \frac {\theta \mathcal {E} _ {r}}{4}\right) \dots \right]} $$  T r (107) θ θ λ λ π  

1 1 4 4 8 4

r r r r h h

Hence, as has been verified in the violating-Lorentz case, here in both cases the temperature-corrected magnetic and electric sectors have the singularity removed when the horizon radius goes to zero. Also, in this case we can observe that the temperature reaches a maximum value and then goes to zero when the horizon radius is zero.

At this point, when determining the entropy, we have

( ) (108)

Result Using Modified Dispersion Relation

Near the event horizon the dispersion relation (19) becomes a E r ω   = +    

2 0 2 1 2 h

(109) where 0_a_ is a parameter with length dimension. By assuming $$ k \sim \Delta k \sim 1 / \Delta x = 1 / r _ {h}, $$

, we can write $$ \omega = E \left(1 + \frac {a _ {0} ^ {2} k ^ {2}}{2}\right) $$ (110) Thus, in terms of the energy difference, we have $$ \frac {\Delta E}{E} = \frac {\omega - E}{E} = \frac {a _ {0} ^ {2} k ^ {2}}{2} (1 1 1) $$ Next, by using the Rayleigh’s formula that relates the phase and group velocities $$ v _ {g} = v _ {p} + k \frac {d v _ {p}}{d k} (1 1 2) $$ where the phase velocity ( ) pv and the group velocity ( ) gv are given by $$ v _ {p} = \frac {\omega}{k} = 1 + \frac {a _ {0} ^ {2} k ^ {2}}{2} (1 1 3) $$ and $$ v _ {g} = \frac {d \omega}{d k} = 1 + \frac {3 a _ {0} ^ {2} k ^ {2}}{2} (1 1 4) $$ However, we find an expression for the velocity difference as following

2 2 0 – g p

$$ \frac {v _ {g} - v _ {p}}{v _ {p}} = a _ {0} ^ {2} k ^ {2} \tag {115} $$

p which corresponds to the supersonic case( ) g p v v > .

Furthermore, the Hawking temperature (58) can be corrected by applying the dispersion ratio (109), i.e

2 c T a r r π $$ = \frac {c _ {s} ^ {2}}{\pi \left(r _ {n} + \frac {a _ {0} ^ {2}}{2 r _ {h}}\right)} $$ s H (116)

2 0 2 n h Note that the singularity is removed, when hr and the temperature vanishes. In addition, the temperature reaches a maximum value before going to zero, as we can see in Figure 2. Now, by calculating the entropy, we find

2 0 2 2 2 ln 4 4 s s

$$ S = \frac {A}{4 c _ {s} ^ {2}} + \frac {2 \pi a _ {0} ^ {2} \ln A}{4 c _ {s} ^ {2}} \tag {117} $$ Here a logarithmic correction term arises in entropy on account of the modified dispersion relation. In order to correct the Hawking temperature and entropy for the Lorentz-violating and non-commutative cases, we will apply the modified dispersion relations obtained in Refs Anacleto MA, et al. [21, 22].

Click to enlarge

$$ \text{Figure 2: The Hawking temperature } T _ {H} \rightarrow \pi a _ {0} T _ {H} \text{ in function of } r _ {h} / a _ {0}. \text{ Note that the temperatures T2 (116), T2 (119) and T4 (122) reach maximum values and then} $$

a . Note that the temperatures T2 (116),

T3 (119) and T4 (123) reach maximum values, and then

decreases to zero as

$$ s r _ {h} / a _ {0} \rightarrow 0. $$ Violation-Lorentz Case: In the situation where

$$ \text{Violation-Lorentz Case: In the situation where } \beta = 0 \text{ and } \alpha \neq 0, \text{ we have the following dispersion relation:} $$

, we have the following dispersion relation:

$$ p = E \left(1 + \frac {\alpha}{2} + \frac {\alpha a _ {0} ^ {2}}{\tilde {r} _ {+} ^ {2}}\right) \tag {118} $$

2 0 2 1 2 a E r α α ω +

So for temperature (73), we get

1 3 / 2 α + =

T a r r (119)   + +      

2 0 α α π + +

2 Furthermore, the result shows that the temperature reaches a maximum point and then goes to zero when the horizon radius is zero. Moreover, entropy is given by $$ S = \left(1 - \frac {3 \alpha}{2}\right) \left[ \left(1 - \frac {\alpha}{4}\right) \frac {A}{4} + \frac {2 \sqrt {\pi} \alpha \sqrt {A}}{4} + \frac {4 \pi \alpha a _ {0} ^ {2} \ln A}{8} + \dots \right] (1 2 0) $$ Again due to the contribution of the modified dispersion relation, a logarithmic correction term arises in the entropy.

Noncommutative Case: At this point we consider the dispersion ratio for the pure electrical sector. So we have   = +    

2 1 0 2 1 4 a E r θ ω  (121)

+

For the temperature (89), we find

$$T = \frac{(1 + \theta E_r)}{\pi \left( r_y + \frac{\theta E_1 a_0^2}{4 r_y} \right)} (122)$$

which in terms of $r_h$ becomes

$$T = \frac{(1 + \theta E_r)}{\pi \left[ \left( 1 + \frac{\theta E_r}{4} \right) r_h + \frac{\theta E_1 a_0^2}{4 r_h} \right]} (123)$$

Here, we can see that the temperature goes through a maximum value before going to zero for $r_h = 0$. Hence, the result for entropy is

$$S = \left( 1 - \theta E_r \right) \left[ \left( 1 + \frac{\theta E_r}{4} \right) \frac{A}{4} + \frac{\pi \theta E_1 a_0^2 \ln A}{4} \right] (124)$$

In the above equation a logarithmic correction term arises in entropy as a consequence of the noncommutativity effect on the dispersion relation.