Einstein-Aether Theory With and Without Einstein
aa r X i v : . [ g r- q c ] M a r Einstein-Aether Theory With and Without Einstein
Boris Hikin
E-mail: [email protected]: 310-922-4752 or 310-826-0209, USAThe exact static spherically symmetric solutions for pure-aether theory and Einstein-aether theory are presented. It is shown that both theories can deliver the Schwarzschildmetric, but only the Einstein-aether theory contains solutions with ”almost-Schwarzschild”metrics that satisfy Einstein’s experiments. Two specific solutions are of special interest:one in pure-aether theory that derives the attractive nature of gravitation as a result ofMinskowski signature of the metric, and one - the Jacobson solution- of Einstein-aethertheory with ”almost-Schwarzschild” metric and non-zero Ricci tensor.
IntroductionEinstein-aether theory was proposed by Jacobson and Mattingly al-most 10 years ago [1] (see latest review [2]) as a correction (extension)of Einstein’s GR. It postulates that gravitation, in addition to a curvedspace, is characterized by a unit vector G i ( G i G j g ij = 1), which to-gether with metric tensor g ij constitute the dynamic variables of theEinstein-aether theory. The equations of motion for g ij and for G i arederived by variational method from a Lagrangian that is a function ofthe metric tensor g ij and the vector field G i .With the requirement that the equations of motion are of the secondorder, the most general form of Lagrangian can be written as [1]: S = Z L ( g ij , G i ) √− gd xL = c R + c G i ; j G i ; j + c G i ; j G j ; i + c ( G i ; i ) + c G k ; i G i G k ; j G j + T ( G i G j g ij −
1) (1)where all c’s are constants, T is the Lagrange multiplier, and R isRicci scalar.The main question that Jacobson and Mattingly raised was: whatshould the value of the constants ”c”s be in order for the Einstein-aether heory to yield the same results as predicted by GR? In particular, incase of spherical symmetry the Einstein-aether theory must yield - inthe first order of approximation - the Schwarzschild’s metric.The Lagrangian of the aether theory, eq. (1), actually represensthree separate theories:a) General Relativity c = 0 all other c’s are zero ( c = c = c = c = 0)b) Einstein-aether theory c = 0 and at least one of others c’s (typ-ically c ) is not zeroc) Pure-aether theory with c = 0 and at least one of others c’s isnot zeroOne question that has not been addressed previously is the validity ofpure-aether theory. In other words, do we need Einstein’s term ( c R )in the Lagrangian? One might argue that the Einstein’s term ( c R )is not needed and the pure-aether theory is sufficient to explain theexperimental results - at least as far as the solar system is concerned.In this paper we would like to investigate this issue based on spher-ically symmetric solutions and we will show that on one hand the Ein-stein term is not needed - if one wants to get Schwarzschild’s metricsolution - and on the other hand, the Einstein term leads to uniquesolutions that are not present in pure-aether theory.Gravitation. Static, spherically symmetric solution.In a case of spherical symmetry the last term of Lagrangian (the c term) can be expressed thru the other terms (see [3]) and thus can bedropped out.The remaining four terms can be written in the following manner: S = Z L ( g ij , G i ) √− gd xL = − [ λ R + λ g ik g jl ( ∂ j G i − ∂ i G j )( ∂ l G k − ∂ k G l )+ λ R ij G i G j + λ ( G k ; k ) + T ( G i G j g ij − here the λ s are new constants that are linear combinations of c-constants.The equations of motion for the metric g ij and for the vector field G i have been obtained previously [3]. However, it is quite a laborioustask to derive the exact solution using these equations. Fortunately, incase of spherical symmetry, there is another approach that yields thedesired result with almost no hard labor involved.In a case of static (time independent) spherically symmetric solutionthe aether theory is governed by 5 functions: g (or g tt ), g (or g rr ), G (or Q t ), G (or Q r ) and T, which are all functions of radius only.We assume that g = − r By using the constraint G i G i = 1, one can eliminate two of thesefive functions - g and T. Our next step is to write the Lagrangianthru the three remaining functions ( G , g and G ). And then find theequations for these functions by means of variation of the Lagrangianwith respect to these functions.The problem is simplified if we choose new variables in this manner:ˆ g = − g g instead of g ¯ G = G √ g √− g instead of G x = 1 r ; ddx = () ′ (3)In these new variables g can be written as : g = ( G ) − ( ¯ G ) (4)and the Lagrangian (eq. 2) has this (amazingly) simple form (for detailssee appendix A): S = − Z d v √− gL = Z dtd Ω S r , whereS r = − Z r dr p ˆ gL ( g ij , G i ) = Z dxx p ˆ gL = (5) Z dx { λ [ 2 √ ˆ g ( G x ) ′ − √ ˆ gx ] − λ ( G ′ ) √ ˆ g − λ √ ˆ g ( ¯ G x ) ′ + λ √ ˆ g [ x ( ¯ G x ) ′ ] } here ¯ λ = λ + λ We now consider the solution of this problem with the followingboundary conditions:a) G i at infinity has only the time component: G i | x =0 or r = ∞ = (1 , , , orG | x =0 = 1; G | x =0 = ¯ G | x =0 = 0 (6)b) metric at infinity corresponds to a flat space: g ij | x =0 or r = ∞ = diag (1 , − , − r , − r sin θ ) (7)The requirement that metric g ij satisfies two Einstein experiments(bending of light and precession of Mercury) set these conditions onˆ g (bending of light) and g (precession of Mercury) as functions of x(1/r) with x ≈ → ∞ ) [4]:ˆ g = 1 + ¯ cx + ... (¯ c = constant ) g = 1 + c x + c x + ... ( c and c = constant ) (8)In other words, g has no quadratic terms in seria by x and ˆ g has nolinear terms.The case of General RelativityLet us - mostly to demonstrate the simplicity of this approach andas a sanity check - first consider the case of GR ( λ = λ = λ = 0).Variation of eq. (5) with respect to G and ˆ g yields: δS r δG = 0 − > ˆ g ′ = 0 or ˆ g = 1 δS r δ ˆ g = 0 − > ( G x ) ′ + 1 x = 0 or g = G = 1 + xC (9)The above solutions are exactly the expression of the Schwarzschildmetric. It is worth pointing out that the sign of the constant C in R is not set by the theory, but is taken as a seperate postulate thatgravitation is only attractive. The GR theory by itself does allow bothattraction ( C <
0) and repulsion ( C > λ = 0). Vari-ation of eq. (5) with respect to G ¯ G , and ˆ g yields the followingequations: a ) δS r δG = 0 → ( G ′ √ ˆ g ) ′ = 0 b ) δS r δ ¯ G = 0 → λ ( 1 √ ˆ g ) ′ ¯ G x − λ x [ x √ ˆ g ( ¯ G x ) ′ ] ′ = 0 (10) c ) δS r δ ˆ g = 0 → λ ( G ′ ) + λ ( ¯ G x ) ′ − λ x ( ¯ G x ) ′ ] = 0The first equation (eq. 10a) can be integrated and with the thirdequation (eq. 10c) it can be used to find the function ˆ g . a ) G ′ = C p ˆ g C = constantb ) λ [ ¯ G ′′ − G x ] = [ λ G ′ − G x ) − λ ¯ G x ] ˆ g ′ ˆ g (11) c ) ˆ g = − λ C λ ( ¯ G x ) ′ + λ C λ [ x ( ¯ G x ) ′ ] It is not difficult to see that if λ is not zero the system of eqs. (11)has no solutions that satisfy the conditions (8). Indeed, if ˆ g ≈ x ¯ c then ˆ g ′ ≈ x ¯ c and the rhs of eq. (11b) is about constant or zero. Inorder for the lhs of (11b) to be regular, ¯ G is about x . In this casethe ˆ g per eq. (11c) should be zero at x=0, which contradicts to therequirements (8).With λ = 0, the system of equations (11) can be easily integrated o give this result:ˆ g = 1; ¯ G = 1 + C x ; ¯ G = C x − λ λ C x and g = 1 + (2 C − C ) x + (1 + λ λ ) C x (12)In the above expressions C and C are constant and C > C = 0 or λ + λ = 0. In both cases the metric is the Schwarzschildone.If one sets C = 0 the time component of vector field is one ( Q = 1)and the radius component ¯ G is inverse to square root of the radius r : G = 1; g = − g = 1 − C r ; G = s ¯ C r (1 − C /r ) (13)In addition to Schwarzschild metric the pure-aether theory (with thecondition λ = 0) delivers the requirement that gravitation must be at-tractive, which is the consequence of Minkowski signature of the metrictensor.Einstein-aether theory ( λ = 0)We now can consider the Einstein-aether theory, or the case when λ is not zero. The presence of Einstein term ( λ R ) in Lagrangiansignificantly changes the number of possible solutions and the choice of λ parameters.The variation of action integral S r , eq. (5), leads to this set ofequations: ) δS r δG = 0 → λ ( G ′ √ ˆ g ) ′ − λ G x ( 1 √ ˆ g ) ′ = 0 b ) δS r δ ¯ G = 0 → λ ( 1 √ ˆ g ) ′ ¯ G x − λ x [ x √ ˆ g ( ¯ G x ) ′ ] ′ = 0 c ) δS r δ ˆ g = 0 → (14) λ p ˆ g ( G x ) ′ + λ x √ ˆ g − λ p ˆ g ( G ′ ) − ¯ λ p ˆ g ( ¯ G x ) ′ + λ p ˆ g [ x ( ¯ G x ) ′ ] = 0From the last equation (14c) one can express ˆ g as a function of G and ¯ G : ˆ g = x {− ( G x ) ′ + λ λ ( G ′ ) + ¯ λ λ ( G x ) ′ − λ λ [ x ( ¯ G x ) ′ ] } (15)Because of the x factor in front of the figure braket it is not diffi-cult to see that for any λ s the expression for ˆ g is always regular (nosingularities).Let us note that if λ is not zero from the equation (14b) followsthat for x → G ≈ x and the λ , λ terms of eq. (15) are about x .This means that if we are interested in the behavior of ˆ g near x = 0the λ , λ terms of eq. (15) could be dropped out. The remainingexpression for ˆ g ( function of G only) always has a right behavior thatsatisfies the condition of eq. (8) at x=0. Indeed for ˆ g we have:ˆ g = G − xG G ′ − λ x ( G ′ ) (16)If we write G near zero as a series by x , G = 1 + ax + bx + .. , (aand b are constants) and substitute it in eq. (16) above we will get thisapproximation for ˆ g :ˆ g = (1 + ax + bx + ... ) − x ((1 + ax + bx + ... )( a + 2 bx + ... ) − λ x A − λ x A = 1 + ( − a − b − λ a ) x + ... (17)which as we see has no linear term and thus satisfies the condition (8). n general the two equations, eq. (14), that describe variables G ,¯ G are coupled thru ˆ g , which depends on both functions. There arehowever four cases where the equations can be uncoupled and theresolutions can be presented in analytical forms:Case A: λ = 0 and ¯ G = 0Case B: ¯ G = 0 any λ , λ , λ Case C: ¯ λ = 0 (or λ = − λ )Case D: λ = 0Case AIf λ is zero (and G = 0), the equation (14b) yields that ˆ g = 1and from the equation (14a) follows that G is a linear function of x( G = 1 + C x ).The third equation (15) can be used to determine ¯ G :ˆ g = G − xG G ′ + λ λ x ( G ′ ) + ¯ λ λ x ( ¯ G x ) ′ or C x ) − x (1 + C x ) C + λ λ x C + ¯ λ λ x ( ¯ G x ) ′ ¯ G = C x + C x ( λ − λ ¯ λ ) (18)And for g we get: g = G − ¯ G org = 1 + x (2 C − C ) + C x (1 − λ − λ ¯ λ ) (19)This is practically (except for the value of the constants λ s) the sameresult as for pure-aether theory that we derived above - eq.(12).Case B, ( G = 0) and C, (¯ λ = 0)In both of these cases the system of equations (14) can be solvedanalytically.In the ”case B” ( G = 0 and thus ¯ G = 0) the eq. (14b) is satisfiedand in the remaining two equations λ and λ terms could be dropped,leaving these equations for G and ˆ g : λ ( G ′ √ ˆ g ) ′ − G x ( 1 √ ˆ g ) ′ = 0 or G ′′ = ( G ′ − G λ x ) ˆ g ′ ˆ g (20)ˆ g = G − xG G ′ + ˆ λ x ( G ′ ) where ˆ λ = λ λ (21)In the ”case C” (¯ λ = 0) the eq. (14b) has the solution ¯ G = C x ( C - constant), while G and ˆ g are defined by the same set of equations(20), (21) as in ”case B”.The equations (20), (21) - although in slightly different form - hadbeen obtained and investigated by Jacobson in his 2006 paper [1]. Theequations can be integrated analytically to yield a result in a form x = f ( G ) (for details see Appendix B): C x = G [( G − µ − G µ ] , where µ = r − λ λ ˆ g = 4 µ G [(1 − µ ) G − µ − (1 + µ ) G µ ] (22)where C is a constant equivalent to Schwarzschild radius.By direct calculation it is not difficult to show that for small x ( x ≈
0) ˆ g has no linear terms (ˆ g = 1 + ax + ... ) and metric has no quadraticterms ( g ≡ G ≈ − C x + bx + ... ) thus satisfying requirements ofthe Einstein experiments, eq. (8), for any parameter µ .The behavior of G vs. x outside x = 0 (small distance r) signifi-cantly depends on a sign of λ (we assume - as in GR - λ > λ =0 (the case of GR) the eqs.(22) and (4) yield: C x = 1 − G ˆ g = 1 G = 0 g = G = 1 − C x C = const (23)with the horizon point at x = 1 /C .If λ <
0, x as a function of G monotonically increases as G de-creases from 1 to 0. This means that G as a function of x monotonicallydecreases from 1 to 0 as x changes from x=0 to x = ∞ . or the ”case B” the metric, defined as g = G , also decreases from1 to 0 as x changes from 0 to infinity (the horizon point is x = ∞ orr=0). For small G the second term in eq. (22) could be dropped givingthis expression for G as a function of x: C x ≈ G − √ (1 − ˆ λ )0 G ≈ ( 1 C x ) √ (1 − ˆ λ − = ( rC ) √ (1 − ˆ λ − ˆ λ ≡ λ λ < g = G − ( C ) x , which always - due to Minkowski signature -leads toexistence of a horizon point a some point x, the value of which dependson the value of the constant C .If λ positive (1 > ˆ λ >
0) - analogues to Maxwell theory, x as afunction of G has a bell shape between two points G = 1 and G = 0with its maximum at some point in between. This means that G ( x )exists only from x = 0 to a certain point -”dead point”. It can beexplicitly illustrated for the case of ˆ λ = 8 / C x = G [( G − − G ] or G = ( 1 + √ − C x (25)with x = 1 / (4 C ) being a ”dead point”.For the metric again we have two possibilities:”Case B”: g = G and the metric exists up to a ”dead point”,which is not a horizon point, since g is not zero.”Case C”: g = G − ( C ) x and for sufficiently large C metric( g ) reaches zero - horizon point - at some point before the ”deadpoint”. One can choose C in such a way that at ”dead point” ( x =1 / C ) the time component of the metric ( g ) become zero. Thatwould represent the case when the ”dead point” is the horizon.Case D, λ = 0We add this case mostly for the sake of completeness. The conditionthat λ = 0 is probably non-physical, due to the fact that equations of otion for the vector field G i become of the first - instead of second -order: aR kj G k + b ( G k ; k ) ,j = 0 with a and b being constants. On the otherhand we must remember that the parameter λ in our considerations isa combination of two parameters - see eq. (1) - c and c , which wouldcanceled each other only in the case of spherical symmetry.If λ = 0, the eq. (14a) yields ˆ g = 1 and the equation for ¯ G can besolved to yield ¯ G = C x . Knowing ˆ g and ¯ G , we can determine thefunction G from eq. (15):1 = x {− ( G x ) ′ + ¯ λ λ C x } or G = 1 + C x + ¯ λ λ C x g = 1 + C x + ( ¯ λ λ − C x (26)General Case, λ ′ s = 0In the general case (both G and λ ′ s are not zero) the solutions havebehavior somewhat in between ”case C” and ”case D”. For the small x(large distance r) G linearly decreases, while ¯ G increases (in absolutevalue) as x . As x moves toward large numbers ( r →
0) the G startsdeviate from Jacobson’s solution while ¯ G deviates from x .The same is true for the time component of metric g . In addition,if radial component of the vector field ¯ G is present (not zero), themetric has horizon point, which is due to the Minkowski signature ofthe metric.Discussion and ConclusionA we saw above the general solution for aether theory is character-ized by two parameters C and C . The first one, C , typically sets thelinear dependence of g as a function of x with x → → ∞ ) and thuscan be identified with a Schwarzschild radius. A much more difficult uestion is the meaning of the other parameter C , which defines themagnitude of radial dependence of aether vector G .We have shown that if one requires from the aether theory to getSchwarzschild solution for the metric, one can choose both pure-aethertheory (no Einstein λ R term in Lagrangian) and Einstein-aether the-ory (with λ R term) with λ = 0 parameter.It also must be pointed out that the Ricci tensor in both pure-aetherand Einstein-aether (with exception for the Jacobson solution) theoriesis always proportional to the constant C - the radial component ( G or G r ) of vector field G i .There are two particular solutions of the aether theory that deservespecial attention.The first one is the solution of pure-aether or Einstein-aether theorywith λ = 0 and C = 0: G = 1; ¯ G = p C x ( C >
0) ˆ g = 1 and g = 1 − C x. (27)The presence of ”hard” matter does not change the time component ofthe aether field, but only adds the radial component.In this solution the atractive nature of gravitation is derived fromthe aether theory and is due to the Minkowski signature of the spacemetric.The second one is the Jacobson’s solution given by eq. (22):¯ G = 0 C x = G [( G − µ − G µ ] , where µ = r (1 − λ λ ) > g = 4 µ G [(1 − µ ) G − µ − (1 + µ ) G µ ] and g = G (28)As r changes from infinity toward zero, G declines from 1 to zero.Here we have that the hard matter ”replaces” the aether. This is op-posite to the situation in Maxwell electrodynamics where the vectorpotential increases toward the center of the charge. he Jacobson’s metric, eq. (28), has no horizon (or to be moreprecise its horizon point is r = 0) and it has singularity ( g = 0) at r = 0, which of course is an artifact of point-mass consideration.It is also worthwhile to mention that the distance from the point-mass (r=0) to any point along radius is finite. R = Z r √− g dr = − Z G ( r )0 x s ˆ gg dxdG dG = Z G µG µ − (1 − G µ ) dG < ∞ if µ > ds = ¯ g ( y ) dt − g c ( dρ + ρ d Ω). For theJacobson solution this can be done using these formula (Appendix C): G = ( 1 − µ y µ y ) µ where y = 1 /ρx = y (1 − µ y ) − µ (1 + µ y ) µ (30)¯ g ( y ) ≡ g ( x ( y )) = G = ( 1 − µ y µ y ) µ As we mentioned above, the Jacobson metric has no singularities.However, when presented in the conformly-Euclidean system coordi-nates it does have singularity at y = µ/
4. The reason for that is clearlyseen from the formula x vs. y in eq (30, line2). The x(y) transfers x = ∞ (r=0) to y = µ/ ρ = 4 /µ ). The singularity of conformly-Euclidean system coordinates is due to our ”bad choice” of systemcoordinates. Perhaps, the system coordinates with unity coefficient infront of dr ( ds = ¯ g ( y ) dt − dρ − g Ω ( ρ ) d Ω ) is a better choice. Aswe showed earlier in eq. (29), the function g Ω ≡ r ( ρ ) is regular for all ρ and the parameter ρ is the true distance betwee two points along theradius. ne puzzling issue of Einstein’s GR, that still has not been resolved,is the definition of the energy-momentum tensor of gravitation. It seemslogical to identify the tensor E ij ≡ − ( R ij + 1 / Rg ij ) (where R ij is theRicci tensor) as an energy-momentum tensor of the curved space. TheEinstein equations R ij − / Rg ij = T ij or Eij + T ij = 0 (31)then can be read in this manner: the total energy-momentum tensor ofthe system (matter and space) is zero.The difficulty here comes from consideration of the vacuum: T ij = 0(no matter) and thus E ij = 0, which leads to the unconventional (tosay the least) statement that in vacuum gravitation has no energy. Aswe saw in this paper, all the solutions of the pure-aether theory yieldSchwarzschild metric, which in its turns sets to zero Ricci (and thus E ij )tensor. In the Einstein-aether theory, on the other hand, this problemis resolved. Most of the solutions - and Jacobson’s metric (with G = 0)in particular - yield ”almost-Schwarzschild” (up to x terms) metric forwhich R ij (and thus E ij ) is not zero.This seems to be a key factor in resolving the competition betweenthe pure-aether ( λ = 0) and the Einstein-aether ( λ = 0) theories infavor of Einstein-aether theory.One more note on a physical nature of space. In the Einstein-aethertheory (as in Einstein’s GR) space (metric) is taken as independentphysical entity with some energy attached to it. It is expressed in exist-ing ”space only” Lagrangian term ( λ R - the Einstein term). However,in Einstein-aether theory there is another interpretation of space. Wecan write the Einstein term in this form: L R = λ RG k G k ≡ λ R due to G k G k = 1 (32)In this form the Einstein term ( λ R ) does not represent space asequal to matter entity, but rather a part of the aether ( G i ). The met-ric, that represents the curved space, is now only auxiliary entity thatties together all forms of matter including the aether as gravitationalmatter. ppendix AIn this appendix we derive the expression for the Lagrangian ofEinstein-Aether theory (see 1) thru variable g , ˆ g := − g g and G as a function of x = 1 /r .In the case of sphirical symmetry the differential of 4-volume dv canbe written as: √− gdv = p ˆ g r drd Ω dt = − p ˆ g x dxd Ω dt (33)The action integral can be written as: S = − Z dvL p ˆ g = Z dtd Ω S r where S r = − Z r drL p ˆ g = Z dxL ( x ) p ˆ g ( x ) x = 1 /r (34)Since Lagrangian is only a function of radius r (or x=1/r), to shortenthe formula everywhere below in writing action integral S we will dropthe term d Ω dt .For components of tensor Ricci we have: R = R i i = Γ i ,i − Γ i i, + Γ iim Γ m − Γ i m Γ m i = Γ , + [Γ + Γ + 2Γ ]Γ − [Γ Γ + Γ Γ ]= Γ , + [ − g , g + g , g + g , g ]Γ = ( Γ √− g g √ g ) , √ g √− g g = g gr ( g , r √ ˆ g ) , = x g √ ˆ g ( g ′ √ ˆ g ) ′ (35) R = R i i = Γ i ,i − Γ i i, + Γ iim Γ m − Γ i m Γ m i = Γ , − [Γ , + Γ , + 2Γ , ]+[Γ + Γ + 2Γ ]Γ − [Γ Γ + Γ Γ + 2Γ Γ ]= − Γ , − [Γ − Γ + 2Γ ]Γ + 2Γ (Γ + Γ ) − , + Γ Γ ] ouble underlined terms can be expressed through new variable ˆ g andthe single underlined terms cancel each other.= − Γ , − [Γ − Γ + 2Γ ]Γ + 2Γ ( ˆ g , g )= − ( Γ √ g g √− g ) , √− g √ g g + 2Γ ( ˆ g , g )= g gr ( g , r √ ˆ g ) , + ( ˆ g , r ˆ g ) = x g √ ˆ g [( g ′ √ ˆ g ) ′ + g ˆ g ′ x ˆ g √ ˆ g ]= x g √ ˆ g [( g ′ √ ˆ g − g x √ ˆ g ) ′ + ( g x ) ′ √ ˆ g ] (36) R = R i i = Γ i ,i − Γ i i, + Γ iim Γ m − Γ i m Γ m i = Γ , − Γ , + [Γ + Γ + 2Γ ]Γ − [2Γ Γ + Γ Γ ]= Γ , + [ g , g + g , g ]Γ + [ − Γ , − Γ Γ ]= (Γ √− g g ) , √− g g − [( g , g ) , + ( g , g ) ]= 1 √ ˆ g ( g r √ ˆ g ) , − [ ( sin θ ) ,θ sin θ ] ,θ − [ ( sin θ ) ,θ sin θ ] = 1 √ ˆ g ( g r √ ˆ g ) , + 1= x g √ ˆ g [( g x √ ˆ g ) ′ − √ ˆ gx ] (37)Combining expressions (35), (36) and (37) we get this expression forthe first term ( λ term) of the action integral: S rλ = − Z p ˆ g r dr λ R = Z p ˆ g dx ( x ) λ [ R g + R g + 2 R g ]= Z dx λ [( g ′ √ ˆ g ) ′ − ( 2 g x √ ˆ g ) ′ + 2( g x ) ′ √ ˆ g − √ ˆ gx ] (38) he first two terms in (38) are full differentials and could be droppedfrom the expression yielding this: S rλ = Z dx λ [2( G − ¯ G x ) ′ √ ˆ g − √ ˆ gx ] (39)where per eq. (4) we replace g with G − ¯ G .The λ -term can be straight forward written as: S rλ = Z dx λ [ − G ′ ) √ ˆ g ] (40)For λ -term we get the following expression: S rλ = − Z r dr λ R ij G i G j = Z x dx λ R ij G i G j = Z d x λ [ R g ( G ) g + R g ( G ) g ]= Z dx λ { ( g ′ √ ˆ g ) ′ [( G ) g + ( G ) g ] + ( g ˆ g ′ x ˆ g √ ˆ g )( G ) g } = Z dx λ { ( g ′ √ ˆ g ) ′ + ( g ˆ g ′ x ˆ g √ ˆ g )( G ) g } In the expression above the underlined term can be integrated out ofthis expression. In the second term we switch to the variable ¯ G = G √ g √− g and do a partial integration: S rλ = Z dx λ ( ˆ g ′ x ˆ g √ ˆ g )( ¯ G ) = Z dx ( − λ √ ˆ g )[ ( ¯ G ) x ] ′ (41)The λ term can be has this form: S rλ = − Z r √ gdr λ ( G k ; k ) = − Z r √ gdr λ [ ( G g √ g ) , √ g ] = Z dx λ √ ˆ g [( ¯ G x ) ′ x ] (42) ombining the expressions (39) for S rλ , (40) for S rλ , (41) for S rλ and (42) for S rλ , and introducing ¯ λ = λ + λ we get this final expres-sion for the action integral S r : S r = (43) Z dx { λ [ 2 √ ˆ g ( G x ) ′ − √ ˆ gx ] − λ ( G ′ ) √ ˆ g − λ √ ˆ g ( ¯ G x ) ′ + λ √ ˆ g [ x ( ¯ G x ) ′ ] } ppendix BIn this appendix we derive the result of Einstein-Aether theory forthe case when ¯ G = 0.Variation of Lagrangian eq.(4) yields this set of equations:a) with respect to ¯ G ( δS r /δ ¯ G = 0)2¯ λ ¯ G x ( 1 √ ˆ g ) ′ + λ [ x √ ˆ g ( ¯ G x ) ′ ] ′ x = 0 (44)which is satisfied if ¯ G = 0b) with respect to G ( δS r /δG = 0) λ ( G ′ √ ˆ g ) ′ − λ G x ( 1 √ ˆ g ) ′ = 0 orG ′′ = ( G ′ − G λ x ) ˆ g ′ ˆ g ; where ˆ λ = λ λ (45)c) with respect to ˆ g ( δS r /δ ˆ g = 0)ˆ g = x {− ( G x ) ′ + λ λ ( G ′ ) + λ λ ( G x ) ′ − λ λ [ x ( ¯ G x ) ′ ] } or with ¯ G = 0 → ˆ g = G − G G ′ x + ˆ λ ( G ′ ) x (46)We now introduce a new variable x = ln ( y ) and write equations(45), (46) as: y = ln ( x ) G ′ = ˙ G x G ′′ = ( G )˙˙ 1 x − ˙ G x a ) ˆ g = G − G ˙ G + ˆ λ ( ˙ G ) b ) ( G )˙˙ − ˙ G = ( ˙ G − G λ ) ˙ˆ g ˆ g (47)Substituting eq. (47a) in eq. (47b) we will get:[( G )˙˙ − ˙ G ][ G − G ˙ G + λ ( ˙ G ) ] =( ˙ G − G λ )[2 G ˙ G −
2( ˙ G ) − G ( G )˙˙ + 2 λ ˙ G ( G )˙˙] (48) nd after some algeraic manipulations we will get:( G )˙˙ G = ˆ λ ˙ G − G ˙ G + G ˙ G (49)The equation above has no explicit y-variable and thus can be reducedto the equation of first order by switching to the new varable V ( G ) =˙ G ( y ): dVdG = ˆ λ ( VG ) − VG + 1 (50)And after introducing new variable ¯ V = V /G : d ¯ VdG G = ˆ λ ¯ V − V + 1 (51)which can be integrated: Z d ¯ V ˆ λ ¯ V − V + 1 = ln ( G ) + C orln ( ¯ V − ¯ V )( ¯ V − ¯ V ) = µ ln ( G ) + C → ¯ V = ¯ V − C ¯ V G µ − C G µ (52) , where µ = ˆ λ ( ¯ V − ¯ V ); ¯ V , = 1+ p − ˆ λ λ In the formula (52) above C is an integration constant and ¯ V and ¯ V arethe roots of quadratic polymon on rhs of eq.(51). Taking into accountthe expression for ¯ V thru G ( x ) we get this equation: Z dG (1 − C G µ ) G ( ¯ V − C ¯ V G µ ) = Z dxx (53)The constant C above must be chosen as C = ¯ V / ¯ V for the reason thatlhs of equation above has logafifmic behavior at G near 1 as rhs atx=0. Z dG (1 − ¯ V ¯ V G µ ) G ¯ V (1 − G µ ) = Z dxx (54) ubstituting U = G µ the eq. can be written as: Z dU (1 − ¯ V ¯ V U ) U (1 − U ) = µ ¯ V ln ( x ) + C or Z dU [ 1 U + ( ¯ V ¯ V − U − ln ( C X V µ ) or U ( U − ¯ V V − = C x µV or G V ( G ¯ V − ¯ V V V −
1) = C x (55)And substituting in above the values for ¯ V and ¯ V thru λ we get: G ( G √ − ˆ λ − G − √ − ˆ λ ) = C x (56)The sign of C should be chosen to satisfy the condition of ”attractivegravity” - G decreases as x increases from zero on. ppendix CThe goal of this appendix is to derive the expression for the Jacobsonmetric in conformly-Euclidean coordinates. The Jacobson metric is aspherically symmetrical metric given by this expression - see eq. (11). g ≡ g tt = G ; C x = G ( G − µ − G µ ) x = 1 rg ≡ g rr = − ˆ gg ˆ g = 4 µ G [(1 − µ ) G − µ − (1 + µ ) G µ ] (57) C = const. µ = √ − λ λ < → ρ ) is done ac-cording to this equation: ds = ¯ g dt − ( rρ ) dl l = Euclidean lengthwhere ¯ g ( ρ ) = g ( r ( ρ )) and r ( ρ ) satisf ies √− g drdρ = rρ (58)The eq. 58 can be first written in x and y coordinates (x=1/r;y=1/ ρ ) and then in G , y coordinates: s ˆ gg dxx = dyy s ˆ gg dxxdG dG = dyy (59)Substituting expressions eq.(57) in to eq.(59) we will get:2 µG [ G µ − G − µ ] dG = dyy duu − dyy where u = G µ < r after integration: ln ( 1 − u u ) = ln ( y ) + Cor y = 4 µ ( 1 − G µ G µ ) G = ( 1 − µ y µ y ) µ and g = G = ( 1 − µ y µ y ) µ (61)where the constant C is taken as C = µ so g at small y (large ρ ) hasapproximation g = 1 − y . From here we can find the transformationcoordinates x → y : x = y (1 − µ y ) − µ (1 + µ y ) µ (62)IF µ = 1 ( the case of GR) the expression (62) becomes: x = y y ) or r = ρ (1 + 14 ρ ) (63)which is a well know expression from the theory of GR [5]. eferences [1] T. Jacobson and D. Mattingly, Gravity with a dynamical preferredframe, Phys. Rev. D 64, 024028 (2001) [arXiv:gr-qc/0007031].D. Mattingly and T. Jacobson, Relativistic gravity with dynamicalpreferred frame, in CPT and Lorentz Symmetry II, ed. V.A. Kost-elecky (World Scientific, Singapore, 2002) [arXiv:gr-qc/0112012].[2] T. Jacobson, Einstein-aether gravity: status report., [gr-cg/arXiv:0801.1547][3] C. Eling and T. Jacobson, Spherical Solutions in Einstein-AetherTheory: Static Aether and Stars, [arXiv:gr-qc/0603058v3][4] J.L. Synge, Relativity: The General Theory, North-Holland Publ.,Amsterdam (1960)C. W. Misner, K. S. Thorne, J. A. Wheeler. Gravitation. p1066.Publ. by Freeman & Co., San Francisco 1973.[5] L. D. Landau and E. M. Lifshitz, The Classical Theory of Fields,Adison-Wesley Publ., Reading, Mass (1962)[1] T. Jacobson and D. Mattingly, Gravity with a dynamical preferredframe, Phys. Rev. D 64, 024028 (2001) [arXiv:gr-qc/0007031].D. Mattingly and T. Jacobson, Relativistic gravity with dynamicalpreferred frame, in CPT and Lorentz Symmetry II, ed. V.A. Kost-elecky (World Scientific, Singapore, 2002) [arXiv:gr-qc/0112012].[2] T. Jacobson, Einstein-aether gravity: status report., [gr-cg/arXiv:0801.1547][3] C. Eling and T. Jacobson, Spherical Solutions in Einstein-AetherTheory: Static Aether and Stars, [arXiv:gr-qc/0603058v3][4] J.L. Synge, Relativity: The General Theory, North-Holland Publ.,Amsterdam (1960)C. W. Misner, K. S. Thorne, J. A. Wheeler. Gravitation. p1066.Publ. by Freeman & Co., San Francisco 1973.[5] L. D. Landau and E. M. Lifshitz, The Classical Theory of Fields,Adison-Wesley Publ., Reading, Mass (1962)