N ov 2 01 9 Exact solutions of Einstein-æther gravity in Bianchi Type V Cosmology

We present the solution space of the field equations in the Einstein-æther theory for the case of a vacuum Bianchi Type V space-time. We also find that there are portions of the initial parameters space for which no solution is admitted by the reduced equations. Whenever solutions do exist, their physical interpretation is examined through the behavior of Ricci and/or Kretsmann scalar, as well as with the identification of the effective energy momentum tensor in terms of a perfect fluid. There are cases in which no singularities appear and others where the effective fluid is isotropic.


INTRODUCTION
Bianchi spacetimes contain various important cosmological models that have been used to describe the anisotropies of primordial universe and the evolution to an isotropic universe as observed in the present epoch. [1][2][3][4]. In Bianchi models the spacetime manifold is foliated along the time axis, with three dimensional homogeneous hypersurfaces [5]. The corresponding geometries are thus spatially homogeneous, which means that the physical variables depend on the time variable only, reducing the Einstein and other governing equations to coupled ordinary differential equations. In total there are nine Bianchi models which are defined from the possible nine different three-dimensional algebras constructed from the isometries of the metric tensor, see for example [5,6].
In this work we are interested in the exact solutions of field equations for Bianchi V spacetime in Einstein-aether gravity [28,29]. This is a Lorentz violating gravitational theory, because a preferred frame is introduced through the aether field present in the gravitational action integral [30]. Mathematically, the aether field is described by a unit time-like vector field, with its kinematic quantities coupled to gravity in the Einstein-Hilbert action. By definition Einstein-aether gravity is a second-order theory however, because of the existence of nonlinear terms in the aether field, there are very few exact solutions. For applications of Einstein-aether gravity in cosmological studies we refer the reader in [31][32][33][34][35].
Recently, in [36] exact solutions where found for the case of Friedmann-Lemaître-Robertson-Walker (FLRW) and LRS Bianchi III spacetimes in Einstein-aether theory. It was there proved that exact solutions exist only for specific ranges of values of the free coupling parameters for the aether field. These exact solutions have been applied in [37] for the determination of inhomogeneous cosmological solutions in Einstein-aether theory. Other studies which can be found in the literature on anisotropic Einstein-aether gravitational models are based on the method of the critical point analysis, where special solutions are determined at the critical points which describe asymptotic behavior for the evolution of the field equations [38][39][40][41][42]. The plan of the paper is as follows: In Section 2 we briefly present the field equations in Einstein-aether gravity. Section 3 includes the main material of our study where we present the solution of the field equations for the case where the aether field is either parallel to the comoving observer (Class A), or tilted (Class B). For these two classes we find the relations between the coupling constants c i of the theory in order that the field equations do admit solutions. For both Classes there exist anisotropic solutions, whereas in the first class, where the aether field is parallel to the comoving observer, there exist an isotropic solution which corresponds to a FLRW spacetime with nonzero spatial curvature. Furthermore, the physical description of the corresponding energy momentum tensor is calculated in all cases and then used to understand the physical properties of the resulting spacetimes. Finally, at Section 4 we draw our conclusions.

EINSTEIN-AETHER GRAVITY
The action integral in Einstein-aether gravity is where u α is the aether field i.e. a unit time-like vector field, which defines the preferred frame. The Lagrange multiplier λ ensures the unitarity of the aether field, u α u α = −1. Finally, the Lorentz violetion terms are introduced by the tensor K αβµν defined as where c 1 , c 2 , c 3 and c 4 are the dimensionless coupling parameters of the aether field with the gravity. In [43] the first two terms of action integral (2.1) have been written with the use of the kinematic quantities of the aether field in 1 + 3 analysis as follows with α µ the acceleration, θ the expansion rate, σ µν the shear tensor, ω µν the vorticity tensor and h µν the projective tensor defined as h µν = g µν − u µ u ν . The coefficient constants c θ , c σ , c ω and c a are related with c 1 , c 2 , c 3 and c 4 as follows The gravitational field equations of the Einstein-aether theory are obtained by variation of (2.1) with respect to the metric g µν , fielding where G µν is the Einstein tensor and T ae µν is the effective aether energy-momentum tensor which defined as T ae µν = 1 2 g µν K αβρσ u ρ;α u σ;β + 1 2 g µν λ(g αβ u α u β + 1)+ (2.8) − c 1 g αβ (u α;µ u β;ν + u µ;α u ν;β ) − c 2 g αβ (u µ;ν u α;β + u β;α u ν;µ ) − c 3 g αβ (u α;µ u ν;β + u µ;α u β;ν )+ − c 4 u α u β u µ;α u ν;β − c 4 (g λα u µ u β u λ;ν u α;β + g βλ u α u ν u β;α u λ;µ ) − λu µ u ν + The equation of motion for the aether field follows by varying u a c 4 g µν u α u ν;β u µ;α g κβ − c 4 g µκ g αλ u λ;β u β u µ;α − c 4 g µκ u α u β ;β u µ;α − K αβµκ u µ;α;β + λg ακ u α = 0 (2.9) while its constraining unit condition is obtained by varying λ u a u a + 1 = 0. (2.10) The above equations make it clear that Einstein-aether gravity is a second-order theory; in a four dimensional manifold this system comprises of fifteen equations. It is important to mention that in our consideration we have not assumed any matter content and the effective aether energy-momentum tensor is produced from the second term of the action (2.1).

Physical Interpretation
In order to assign a possible physical meaning to the solutions obtained we calculate, on mass-shell, the Einstein tensor G µν = R µν − 1 2 Rg µν and interpret it as an effective energy-momentum tensor of a perfect fluid by writing The energy momentum tensor in the 1+3 decomposition can be written as follows where ρ is the energy density of the fluid, u µ the 4-velosity, q µ the heat flux vector, p the pressure and π µν the anisotropic stress tensor. The relations that make the identification possible are in which h µν is the projection tensor orthogonal to velosity u µ defined by In the next section we begin our analysis by selecting the underlying geometry to be that of a Bianchi Type V space-time; in this case the field equations are reduced to ordinary, coupled differential equations with time as the independent variable.

SOLUTION SPACE
The general diagonal Bianchi Type V line element is If we apply the isometries of the Bianchi V spacetime to the aether vector field, also demand that the corresponding one form be curl-free and finally utilize (2.10) we arrive at We choose the time so that M (t) = a(t)b(t); thus, the second component of the equation (2.9) assumes the form implying that we have to separately investigate the two classes of solutions where: Class A when λ 1 = 0 and Class B when λ 1 = 0. The general approach we adopt in order to reveal the solution space is to algebraically solve two of the equations in terms of the accelerations a ′′ (t), b ′′ (t) and substitute the result into the rest. In doing so some particular branches appear when the denominators of the corresponding expressions vanish. This may happen either for specific value ranges of the constants or for particular relations among a(t), b(t). In what follows we present all the cases that emerge.

Class A solutions
In a way similar to [36] the existence of the solution of the field equations is directly related to the ranges of values of the coefficients c 1 , c 2 , c 3 and c 4 of the theory. In this Class there are three possible cases of study, corresponding to . The only non-zero component of (2.9) fixes the Lagrange multiplier to which is then substituted into (2.7), giving the final set of equations to be solved in this class.

Case A2
The assumptions of this case make the (0, 0) component of (2.7) indicating that c 2 = 1 must hold for solutions to exist. As we can easily see the above equation admits the scaling symmetry a → ω 1 a, b → ω 2 b; thus, if we make the replacement the aforementioned equation transforms to which, being a quadratic form in a 1 (t), b 1 (t), can be parametrized by If we substitute the above forms of a 1 (t), b 1 (t) into the rest of (2.7) we find that a solution exists if f (t) satisfies the differential equation , the solution is real if c 2 < 1 and then a(t), b(t) becomes yielding the line element The Riemann tensor is zero when c 2 = − 2 3 , so we obtain the Minkowski space-time in this case. The corresponding Ricci scalar is which has a curvature singularity at t → ±∞ depending on the value of ǫ. For this solution the physical components of the energy momentum tensor describe a perfect fluid with energy density and pressure component as follows (3.14) • If f (t) = −2 tanh −1 (tan (m (t − t 0 ))), the solution of the field equations is given by the functions Hence, the line element reads with c 2 > 1.
The above line element corresponds to a perfect fluid solution with with equation of state parameter It is interesting to note that as the argument τ ≡ (m (t − t 0 )) spans the prime interval [0, 2π] the role of the coordinates t, x as time-like and space-like respectevely is interchanged (i.e. when π 2 < τ < 3π 2 t is time-like while when 3π 2 < τ < π 2 x is time-like). The variables y, z remain space-like in this interval. The corresponding Ricci scalar is 3 m 2 + 1 cos(4m(t − to)) + 7m 2 + 3 sec 3 (2m(t − to)) which develops curvature singularities at finite intervals.
With q = 6 5 (1 − c 2 ) we substitute the above mentioned accelerations into the equation (2.7). The (0,0) component of the latter equation admits a scaling symmetry a → ω 1 a, b → ω 2 b; thus, if we make a replacement similar to (3.6), namely a(t) = exp (−a 1 (t) + b 1 (t)) dt , b(t) = exp (a 1 (t) + b 1 (t)) dt , the (0,0) component of the field equations is simplified as In order to parametrize the above equation, we have to check the cases when q equals 1 5 (9c 2 − 9) or not.
• If q = 9 5 (c 2 − 1) then the equation (3.20) gives Then we substitute the above form of a(t), b(t) into the equations of the accelerations and we arrive at a differential equation for a 1 (t) which has the solution a 1 (t) = m 1 exp (∓4nt) with m 1 an integration constant and n −1 = 6(1 − c 2 ).
Finally the scale factors a(t), b(t) are given by where ǫ 2 = 1, c is a redefinition of m 1 and c2 < 1. The corresponding Ricci scalar reads which has curvature singularities at t → ±∞ for ǫ = −1 and at t → −∞ for ǫ = 1.
The above solution describes a perfect fluid when ǫ = 1, and a non perfect one for ǫ = −1.
The energy density and the pressure component are given by while the nonzero components of the stress tensor π µν are • If (9c 2 − 5q − 9) = 0 equation (3.20) can be parametrized as Now if we substitute the above values of a 1 (t), b 1 (t) into the rest of (2.7) we find that a solution exists if f (t) satisfies the differential equation cosh(f (t)) √ 10(6c 2 + 5q − 6)f ′ (t) + 20 6c 2 + 5q − 6 cosh(f (t)) = 0, (3.29) -When cosh(f (t)) = 0 we have that For these functions f (t), the solution of the field equations is determined to be and the line element becomes In the special case when q = − 1 5 (6c 2 +4) the Riemann tensor is 0, and we obtain the Minkowskin space-time. For any other value of q the corresponding Ricci scalar is which has a curvature singularity at t → ±∞ depending on ǫ, being plus or minus 1. The energy momentum tensor describes a perfect fluid, similarly to the previous case.
-When cosh f (t) = 0 the solution of the differential equation (3.29) is which implies that, The corresponding line element is where we removed t 0 by a redefinition of the coordinate t → t + t 0 . It is easy to see that this line element describes a perfect fluid solution.
We proceed our analysis with the derivation of the second-class of solutions in which u a is not the comoving observer.

Class B solutions
For the second class of solutions where λ 1 = 0, the strategy is now to first solve the second component of (2.9) for the Lagrange multiplier λ(t) obtaining: Next we substitute this λ(t) into the first component of (2.9) and also into the components of (2.7). If, from the resulting equations, we solve the first component of (2.9) and the (2,2) component of (2.7) in terms of the accelerations a ′′ (t), b ′′ (t), the possible vanishing of the following expression appearing in the denominators of a ′′ (t), b ′′ (t) must be considered, giving rise to the following cases This case finally results into a, b being both constants functions. When we substitute these constant values in (2.7), we end up with (0, 0) + (1, 1) = −2(c 1 + c 3 + 1) = 0 which is impossible; thus there is no solution in this case.

Case B2
With the above assumption the constrain equation, i.e. the (0,0) component of (2.7), becomes As before, the existing scaling symmetry indicates that the replacement satisfies the above expression, while the rest of the equations (2.7) are solved if f (t) obeys the first order differential equation Its solutions read where m 1 is an integration constant.
• If f (t) = iκπ the solution is 43) and the line element becomes with the Riemann tensor equal to zero, indicating the Minkowski space time.
• If f (t) = 2 coth −1 e 2t−m1 the solution is: e m1 − e 2t e m1 + e 2t with the corresponding line-element where κ 2 = 2e −m1 and τ = t + ln( κ √ 2 ). The Ricci tensor is derived to be R µν = 0; thus the above line element describes a vacuum solution. The Kretsmann scalar equals indicating the existence of a singularity at τ = 0. Note that κ is a gravitational essential constant and that the time-like character between τ, x is interchanged as τ spans the interval (−∞, ∞).

Case B3
In this case we end up with a(t)b(t) = λ 1 √ −2c 2 (c 1 + 3c 2 + c 3 ) −1/2 and if we substitute this expression into the components of (2.7) we can see that a(t) must be constant, which entails the result (0, 0)+ (1, 1) = −2(c 1 + c 3 + 1) = 0. However, this relationship makes the rest of equation (2.7) incompatible with the assumptions of the given branch; we thus conclude that there is no solution.
The above analysis corresponds to the last case where the denominators vanish. Thus we can now use the forms of the accelerations a ′′ (t), b ′′ (t) and replace them into equation (2.7). This action results in the (0,1) component being Various cases will emerge from the above equation. When c 1 + c 3 = 0, the only independent components of (2.7) are the (0,0) and (1,1) which are quadratic in the velocities. If we differentiate these with respect to t, use again the accelerations and subtract appropriate linear combinations of the expressions themselves, we end up with the following two cases: • If c 2 = 0, c 1 + c 3 = 0, ab = c b then the combination (0,0)-(1,1) of the components of (2.7) reads − cast the line element into the simple final form (3.51) while the aether one-form transform into u µ = − µ 2 − λ 1 2 , λ 1 , 0, 0 which, in order to be real, implies the restriction λ1 2 < |µ| 2 .
The above line-element represents a singularity free geometry, on account of having vanishing covariant derivative of its Riemmann tensor and all fourteen curvature scalars monomialls of the Ricci scalar R = 8 µ 2 . Note that, since the space admits only one extra Killing field ξ 4 = (1, 0, ǫY, −ǫZ), we have the intersting case of a CSI space time [44] Despite the non physical character of a neutral singnature line element it is not unreasonable to investigate the resulting effecting T µν in terms of perfect fluid: inferring that it corresponds to an isotropic perfect fluid with an equation of state p + ρ = 0 and zero heat conduction, i.e. it mimics that of a cosmological constant.
• If c 2 = 0, c 1 + c 3 = 0, ab = c b then with the application of the same steps as before we end up with ab = const., which contradicts the initial assumption of this case. Thus there is no solution here although, of course, the previous solution hold true.
When this is substituted into (2.7) we get a differential equation for a(t) which leads to the final scale factors under the assumpion c 1 + 3c 2 + c 3 = 2.
The resulting line element is where we have used the transformation t → t + m 1 + ln λ 1 , x → x + m 2 1 + ln 2, y → 4m −2 2 λ 2 1 y, z → e −2m1 λ −2 1 m 2 2 z. Furthermore, it holds that R λµνκ = 1 λ 2 1 (g λν g µκ − g λκ g µν ), revealing the space-time as maximally symmetric with Ricci scalar R = 12 .. If we further investigate the physical properties of the energy-momentum tensor for the above solution, we find that , q κ = 0, π µν = 0 (3.56) inferring that it corresponds to an isotropic perfect fluid with an equation of state p + ρ = 0 and zero heat conduction, i.e. it mimics that of a cosmological constant.

CONCLUSION
In this work we studied the existence of exact solutions for the field equations in the Lorentz violating theory known as Einstein-aether gravity. For the background geometry we assumed that of a Bianchi Type V spacetime, with a diagonal scale factor matrix dictated by the vanishing of the G tx constraint equation. Note that this form of the line-element, although with two independent functions of time, is not Locally Rotationally Symmetric, i.e. it does not admit a fourth Killing field. We have also restricted the form of the aether field by imposing on it the symmetries of the geometry; thus arriving at two possible cases, that of tilted or non-tilted u µ . For both models of our study we determined the field equations and exhibited their solution space. Like in the previous similar work of ours concerning Bianchi Type III geometry [36], we also here find that exact solutions exist only for specific values of the coupling parameters of the Einstein-aether theory.
When the aether field is parallel to the comoving observer, we found that there exists an exact isotropic solution in case A 2 , when c 1 + 3c 2 + c 3 − 2 = 0, q = 0. In this case, the spacetime reduces to the Friedmann-Lemaître-Robertson-Walker spacetime with non-zero spatial curvature. On the other hand, in case A 3 , when c 1 + 3c 2 + c 3 − 2 = 0, q = 0 both anisotopic and isotropic solutions are found; the latter case being when (9c 2 − 5q − 9) = 0, an isotropic solution is found.
In a similar way when the aether field is not parallel to the comoving observer anisotropic and isotropic exact solutions are found for specific relations between the coupling constants of the Einstein-aether theory. There are two possible constraining relations for the coupling constants, when exact solutions exist. These occur in case B 2 where c 2 = 0, c 1 + c 3 = 0, and in case B 6 where c 1 + c 3 = 0.
Lastly, we would like to briefly comment about the LRS case: The scale factor matrix is now γ µν = diag{a 2 , a 2 , b 2 } which results in a non-zero G tx component of the Einstein tensor. Now the geometry admits the fourth Killing field ξ 4 = −z ∂ ∂y + y ∂ ∂z . The aether field remains the same, since it commutes with ξ 4 , and thus the classification according to λ 1 being zero or non-zero is also valid. There is however a substantial difference in the solution space. and b(t) = m 1 a(t) where ǫ = ±1, Q = c 1 + 3c 2 + c3 < 2 for the solution to be real, with the Minkowski space-time corresponding to the value Q = 0. Note also that, if a free electromagnetic field was considered instead of the aether field, the line element used in the present work would lead only to the vacuum solution whereas the LRS element would have non-trivial solutions. All these issues are under consideration and will be presented in a future publication.