Noether Symmetry Approach to the Non-Minimally Coupled Y(R) F 2 Gravity
NNoether Symmetry Approach to the Non-MinimallyCoupled Y ( R ) F Gravity ¨Ozcan SERT a , Fatma C¸ EL˙IKTAS¸ b a Department of Physics, Pamukkale University, 20017 Denizli, Turkey b Department of Mathematics, Pamukkale University, 20017 Denizli, Turkey
Abstract
We use Noether symmetry approach to find spherically symmetric static solutions of the non-minimally coupled electromagnetic fields to gravity. We construct the point-like Lagrangian underthe spherical symmetry assumption. Then we determine Noether symmetry and the correspondingconserved charge. We derive Euler-Lagrange equations from this point-like Lagrangian and showthat these equations turn out to be the differential equations derived from the field equations of themodel. Also we give two new exact asymptotically flat solutions to these equations and investigatesome thermodynamic properties of these black holes.
I. INTRODUCTION
The late time expansion of the universe and the missing matter at large astrophysical scalesstill remain the most important issues of modern cosmology. Although Einstein’s gravity with thecosmological constant known as ΛCDM is in agreement with the observations, it has some problemssuch as fine tuning and coincidence. Since dark matter particle has not yet been observed directly,the efforts to modify the Einstein’s theory of gravity have recently increased by studying the modelssuch as f ( R ) gravity, scalar-tensor gravity and vector-tensor gravity.Due to the modification of the Einstein’s gravity, it is possible to modify the Einstein-Maxwelltheory to the f ( R )-Maxwell theory in the presence of the electromagnetic field. The sphericallysymmetric static solution of this theory with constant Ricci scalar, which is similar to the Reissner-Nordstrom-AdS black hole solution, was given in [1]. However, in general cases with dynamicalnon-constant Ricci scalar, it is not easy to find more general solutions. Then one can take into a r X i v : . [ g r- q c ] A p r ccount the non-minimal couplings between electromagnetic fields and gravity like Y ( R ) F -type.The feature of the non-minimal theory is to have a large class of solutions such as the sphericallysymmetric [2–6] and cosmological solutions [7–9]. It is interesting to note that such non-minimalcouplings, which are first order in R , obtained from a five dimensional Lagrangian via dimensionalreduction [10–13] and they investigated in various aspects [14–17] such as the charge conservationand the relationship between electrical charge and geometry. The general couplings in R n F -formapplied to the generation of primordial magnetic fields in the inflation stage [7, 8, 18–22]. Therefore,it is possible to consider the more general couplings in Y ( R ) F form and their solutions can give usmore information about the relation between electric charge and space-time curvature. Especiallyin the presence of medium with very high density electromagnetic fields, these couplings may ariseand their effects can be significant even far from the source.The Noether symmetry approach is one of the effective techniques to find solutions of a La-grangian without using field equations. This symmetry approach allows us to find conservedquantities of a model by using the symmetry of the Lagrangian which is invariant along a vectorfield. Then the vector field can be determined by this symmetry and each symmetry of the La-grangian gives a conserved quantity. This symmetry approach has been applied to f ( R ) gravitysuccessfully to find out solutions and select the corresponding f ( R ) function which is compatiblewith the Noether symmetry [23–25].This study is organized as follows: In the second section, we find the first order point-like actionof the model for the spherically symmetric static metric and electric field. After we apply theNoether symmetry approach to the action, we obtain the system of partial differential equations.By solving the system, we find the Noether charge and the corresponding vector field for the non-minimal Y ( R ) F model. In the third section, after we obtain Euler-Lagrange equations we givetwo new solutions to the equations and investigate some thermodynamic properties of the blackhole solutions. Finally, we summarize the results in the last section. I. NOETHER SYMMETRY APPROACH FOR THE NON-MINIMAL MODEL
Let us start with the following action of the non-minimally coupled electromagnetic fields togravity [2, 3, 22] I = (cid:90) (cid:20) R κ ∗ − Y ( R ) F ∧ ∗ F (cid:21) . (1)By taking the variation of the action, and obtaining the field equations, we can find the solutions[2–6] for the spherically symmetric static metric ds = − A ( r ) dt + 1 A ( r ) dr + B ( r ) dθ + B ( r ) sin θdϕ . (2)Here the corresponding Ricci curvature scalar is R = − A (cid:48)(cid:48) − AB (cid:48)(cid:48) B − A (cid:48) B (cid:48) B + AB (cid:48) B + 2 B . (3)Alternatively, we can find the solutions also from Noether symmetry approach by taking thefollowing action of the non-minimally coupled model with the Lagrange multiplier λ I = (cid:90) (cid:20) R κ ∗ − Y ( R ) F ∧ ∗ F − λ ( R − ¯ R ) ∗ (cid:21) . (4)Here variation of the action with respect to λ gives us R = ¯ R and ¯ R is defined as¯ R = R ∗ − A (cid:48)(cid:48) − AB (cid:48)(cid:48) B (5)to eliminate the second order derivatives in the action via integration by parts. The variation withrespect to R gives λ = 12 κ ∗ − Y R ( R ) F ∧ ∗ F . (6)If we substitute (5) and (6) in the action (4), we obtain the following Lagrangian L = R κ ∗ − Y ( R ) F ∧ ∗ F − (cid:18) κ ∗ − Y R ( R ) F ∧ ∗ F (cid:19) (cid:18) R − R ∗ + A (cid:48)(cid:48) + 2 AB (cid:48)(cid:48) B (cid:19) . (7) e see that Y R ( R ) F ∧ ∗ F term in the Lagrangian has higher order derivatives which complicatesthe Noether approach. But, fortunately we have the following equation from the trace of the fieldequations Y R ( R ) F ∧ ∗ F = − κ (8)which corresponds to the conservation of the energy-momentum tensor [8] and eliminates the higherorder derivatives in the Lagrangian. By taking the electromagnetic tensor F , F = φ (cid:48) ( r ) e ∧ e , (9)which has only the electric potential φ ( r ), the Lagrangian of the model is obtained as L = Bκ (cid:20) − R κ Y ( R ) φ (cid:48) + R ∗ − A (cid:48)(cid:48) − f B (cid:48)(cid:48) B (cid:21) . (10)In the Lagrangian, the second order derivatives can be eliminated by integration by parts and itturns out to be the following point-like Lagrangian L = AB (cid:48) κ B + A (cid:48) B (cid:48) κ − BR κ + BY ( R ) φ (cid:48) + 2 κ . (11)By considering the the configuration space Q which has the generalized coordinates q i ≡ { A, B, φ, R } and its tangent space T Q ≡ { q i , q (cid:48) i } , we look for the symmetries of the Lagrangian. Noether’stheorem states that if the Lie derivative of a Lagrangian vanishes L X L ( q i , q (cid:48) i ) = XL ( q i , q (cid:48) i ) = 0 (12)along a vector field X X = α i ∂∂q i + α (cid:48) i ∂∂q (cid:48) i , (13)then X is a symmetry of the action and each symmetry of the action corresponds to a conservedquantity or first integral such as Σ = α i ∂L∂q (cid:48) i . (14) hen we take the Lie derivative of the point-like Lagrangian in the configuration space to find thefirst integral L X L = α ∂L∂A + α ∂L∂B + α ∂L∂φ + α ∂L∂R + α (cid:48) ∂L∂A (cid:48) + α (cid:48) ∂L∂B (cid:48) + α (cid:48) ∂L∂φ (cid:48) + α (cid:48) ∂L∂R (cid:48) (15)= α B (cid:48) κ B + α (cid:18) − AB (cid:48) κ B − R κ + Y ( R ) φ (cid:48) (cid:19) + α (cid:18) − B κ + BY R ( R ) φ (cid:48) (cid:19) + α (cid:48) B (cid:48) κ + α (cid:48) (cid:18) AB (cid:48) κ B + A (cid:48) κ (cid:19) + α (cid:48) BY ( R ) φ (cid:48) = 0 (16)where α i = α i ( A, B, φ, R ). Here the derivatives α (cid:48) i can be written by the chain rule α (cid:48) i = ∂α i ∂A A (cid:48) + ∂α i ∂B B (cid:48) + ∂α i ∂φ φ (cid:48) + ∂α i ∂R R (cid:48) (17)and the Lie derivative of the point-like Lagrangian gives us the following system of partial differ-ential equations α B − α A B + ∂α ∂B + ∂α ∂B AB = 0 , (18) α Y ( R ) + α BY R ( R ) + 2 ∂α ∂φ BY ( R ) = 0 , (19) ∂α ∂φ κ + ∂α ∂φ κ AB + 2 ∂α ∂B BY ( R ) = 0 , (20) ∂α ∂A + ∂α ∂A AB + ∂α ∂B = 0 , (21) ∂α ∂φ κ + 2 ∂α ∂A BY ( R ) = 0 , (22) ∂α ∂R + ∂α ∂R AB = 0 , (23) ∂α ∂A = 0 , ∂α ∂R = 0 , ∂α ∂R BY ( R ) = 0 , (24) α R + α B = 0 . (25)A solution to the system for an arbitrary Y ( R ) function can be found as α = c √ B , α = 0 , α = c , α = 0 . (26) ere c , c are arbitrary constants. Then the X vector field can be found as X = c √ B ∂∂A − c B / ∂∂A (cid:48) + c ∂∂φ (27)and the the constant of motion (14) becomesΣ = 2 c κ + 2 c BY ( R ) φ (cid:48) . (28) III. EULER-LAGRANGE EQUATIONS
In order to determine the non-minimal function and the metric functions, we calculate the Euler-Lagrange equations from ddr ( ∂L∂q (cid:48) i ) − ∂L∂q i = 0 (29)for the Lagrangian (11). Then we obtain the following differential equations for A, B, φ, R , respec-tively: B (cid:48)(cid:48) − B (cid:48) B = 0 , (30) A (cid:48) B (cid:48) B + AB (cid:48)(cid:48) B − AB (cid:48) B + A (cid:48)(cid:48) + R − Y ( R ) φ (cid:48) κ = 0 , (31)( Bφ (cid:48) Y ( R )) (cid:48) = 0 , (32) Y R ( R ) φ (cid:48) = 12 κ . (33)From (30) and (32) we find B ( r ) = b ( r + b ) , Y ( R ) φ (cid:48) = qB , (34)where q is an integration constant and it corresponds to the electric charge of the source. We notethat the condition (33) can be found by taking the derivative of equation (31) with respect to r asin [6]. Then we have only the following differential equation (31) to solve A (cid:48)(cid:48) − AB (cid:48) B + 1 B − κ Y ( R ) φ (cid:48) = 0 (35) hich is same with the differential equation obtained from the field equations of the model in [3–6]for B = r . Thus we show that these two different methods give the same differential equation(35). The conserved charge (28) of the model turns out to beΣ = 2 c κ + 2 qc (36)for the Noether symmetry. We see that the conserved quantity involves the gravitational couplingconstant κ and the electric charge of the system q . We also calculate the energy function from E L = q (cid:48) i ( ∂L∂q i ) − L (37)and find E L = BR κ + AB (cid:48) κ B + A (cid:48) B (cid:48) κ + BY ( R ) φ (cid:48) − κ . (38)By substituting the Ricci scalar (3) in the energy function (38), we find that the function is equal tozero, since equation (38) is nothing more than equation (35). Furthermore, we can choose B = r without loss of generality then (35) becomes A (cid:48)(cid:48) − Ar − κ Y ( R ) φ (cid:48) = 0 . (39) A. Some New Solutions
In order to obtain solutions of the differential equation (39), we can choose the non-minimalfunction Y ( R ) that determines the strength of the coupling and find the metric function A ( r ) as afirst method. Alternatively, we can choose possible geometries which are asymptotically flat andinvolve correction terms to the known Reissner-Nordstrom solution as a second method. Then wecan find the corresponding non-minimal function Y ( R ). Here we consider the second method andwe take the following metric function with the Yukawa-like correction term A ( r ) = 1 − Mr + q r − a e − r r (40) hich gives the solution φ ( r ) = qr − ae − r ( r + 4) qr , (41) Y ( r ) = 8 q /κ q − a ( r + 2) e − r . (42)Then the Ricci scalar becomes R = ae r r for the metric function and by taking the inverse function r = 2 W ( x ), we can re-express the non-minimal function (42) in terms of R as Y ( R ) = 8 q /κ q − W ( x ) + 1) e − W ( x ) (43)where W ( x ) is the Lambert function with x = (cid:112) a R .Secondly, we choose another metric function with the Yukawa-like correction term A ( r ) = 1 − Mr + q r − (1 + 4 r + 6 r ) ae − r (44)which is also asymptotically flat. Then we obtain the following solution φ ( r ) = qr − a ( r + 6 r + 18 r + 24) e − r qr , (45) Y ( r ) = 8 q /κ q − a ( r + 4 r + 12 r + 24 r + 24) e − r . (46)We calculate the Ricci scalar for the second metric as R = ae − r and the inverse function r = lnx with x = aR . Then the non-minimal function becomes Y ( R ) = 8 q /κ q − ( ln x + 4 ln x + 12 ln x + 24 lnx + 24) R . (47)
B. Some Thermodynamic Properties of the Solutions
The above metric functions (40) and (44) may describe a naked singularity without horizon ora black hole with one horizon or two horizons which are called event horizon and Cauchy horizondepending on the choice of the parameters. In the cases with event horizon r = r h , the Hawkingtemperature is defined by T = A (cid:48) ( r h )4 π (48) a) (b) FIG. 1:
The Hawking temperature versus the event horizon radius with q = 0 .
4, thedotted curve is the Reissner-Nordstrom case and the solid curve is the non-minimal model.and the temperatures can be found4 πT = a ( r h + 2) r h e r h + 2 Mr h − q r h , (49)4 πT = a ( r h + 4 r h + 10 r h + 12) r h e r h ) + 2 Mr h − q r h (50)for the above metric functions (40) and (44). We give the variation of the temperatures with theevent horizon radius r h in Fig. 1 for this non-minimal model and the Reissner-Nordstrom case. Byusing the entropy of black hole S = πr h , we calculate the heat capacity from C = T (cid:18) ∂S∂T (cid:19) q (51)for the above two metric functions and obtain C = 2 πr h (cid:0) ( r h − q ) e r h + a ( r h + 1) (cid:1) (3 q − r h ) e r h − a ( r h + 3 r h + 3) , (52) C = 2 πr h (cid:0) ( r h − q ) e r h + a ( r h + 3 r h + 6 r h + 6) (cid:1) (3 q − r h ) e r h − a ( r h + 3 r h + 9 r h + 18 r h + 18) . (53)Heat capacity gives us information about thermal stability intervals and phase transition pointsof a black hole. Heat capacity must be positive and finite for a stable black hole. The points where eat capacity is zero give us the Type-1 instability and the points where heat capacity diverges giveus Type-2 instability points which correspond to the second order phase transition for a black hole.We plot also the heat capacity versus the horizon radius r h for the Reissner-Nordstrom solution andthe non-minimal model in Figure 2 with different ranges to see these points clearly. Numerically, wecan find upper bounds for the non-minimal parameter a as a = 0 .
16 for A solution and a = 0 . A solution with q = 0 .
4, to have a stable black hole. Furthermore, the type-2 instability pointsdecrease from 0 . a increases from 0 to the upper bounds, respectively. Moreover, theseupper bounds of the parameter a can increase to higher values as the electric charge increases.On the other hand, the first law of thermodynamics is given by dM = T dS + φdq (54)for a non-rotating black hole with mass M and electromagnetic potential φ . It is interesting toshow that the mass M in the first law can be expressed by the Smarr formula [26] M = 2 T S + φQ (55)for the Reissner-Nordstrom black hole in the Einstein-Maxwell theory.Furthermore, the Smarr relation can be also obtained from the Komar integral [27–30] with acorrection term as M = 2 T S + qφ − (cid:90) τ dV (56)where τ is the trace of energy-momentum tensor obtaining from the gravitational field equation G a = κ τ a (57)and it can be related with the work density [31]. In the minimal Einstein-Maxwell case, thisrelation (56) is automatically satisfied, since the trace of Maxwell energy-momentum tensor iszero. In contrast to the Einstein-Maxwell theory, the non-minimally coupled Y ( R ) F theory hasa non-vanishing trace of energy-momentum tensor. By taking κ = 8 π , the trace is found a) C for 0 ≤ r h ≤ . C for 0 ≤ r h ≤ . C for 0 ≤ r h ≤ . C for 0 ≤ r h ≤ . FIG. 2:
The heat capacity versus the event horizon radius with q = 0 .
4, the dotted curvecorresponds to the Reissner-Nordstrom case and the solid curve to the non-minimal case. τ = 18 π ∗ ( G a ∧ e a ) = R π = − π [ A (cid:48)(cid:48) + 4 A (cid:48) r + 2 r ( A − . (58)In order to investigate whether these metric functions (40) and (44) satisfy the Smarr formula wecalculate the correction term as (cid:90) V τ dV = 116 π (cid:90) ∞ r h R ( r )4 πr dr = M r h A (cid:48) ( r h )4 − r h or the the metric functions. Then we firstly consider the electric potential (41) at the event horizon φ = (cid:90) ∞ r h Edr = qr h − a e − r h ( r h + 4)4 qr h . (60)Thus the mass obtained from the Smarr formula (56) turns out to be M = q + r h − ae − r h r h (61)and it is equal to the mass obtained from A ( r h ) = 0. The Smarr formula (56) is also satisfied forthe second solution (45) similarly, and the mass is found M = q + r h − a ( r h + 4 r h + 6) e − r h r h . (62) IV. CONCLUSION
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