Rainich-Type Conditions for Perfect Fluid Spacetimes
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Rainich-type Conditions for Perfect Fluid Spacetimes Synopsis • We describe and illustrate a relatively simple set of new Rainich-type conditions on an n-dimensional spacetime which are necessary and sufficient for it to define a perfect fluid solution of the Einstein field equations [1]. • Procedures are provided which implement these Rainich-type conditions and which reconstruct the perfect fluid from the metric. • These results provide an example of the idea of geometrization of matter fields in general relativity, which is a purely geometrical characterization of matter fields via the Einstein field equations. Analogous commands pertaining to the geometrization of non-null electromagnetic fields are RainichConditions and RainichElectromagneticField. • For a relatively complicated alternative approach using scalar invariants and which is valid in four dimensions, see [2]. • For an alternative approach which treats thermodynamic fluids in four dimensions, see [3]. Theory a a b • Let M, g be an n-dimensional spacetime, n > 2. Let u be a timelike unit vector field, gabu u =K1, and let µ : M/R and p : M/R be functions. The Einstein equations for a perfect fluid take the form 1 R K g R = µ C p u u C p g , ab 2 ab a b ab where Rab is the Ricci tensor of gaband R is the Ricci scalar. Note that a cosmological constant L can be incorporated into these field equations with the substitution µ / µ C L p / p K L . • Geometrization of the perfect fluid is achieved by uncovering necessary and sufficient conditions on the metric g such that a there exists µ, p, u obeying the perfect fluid Einstein equations. Note that we are not imposing any a priori relations, e.g., equations of state, on the energy density and pressure. • Let Sab be the trace-free Ricci tensor. Define a real function a : M/R, 1 2 3 n b c a a = K S S S . n K 1 n K 2 a b c • The geometrization of perfect fluids can be achieved as follows [1]. Theorem: An n-dimensional spacetime is a perfect fluid spacetime if and only if the following three conditions hold. (1) a s 0 1 1 (2) Ka bKc d = 0, where Kab = Sab K gab a n a b a (3) Kabv v O 0, for some v . Moreover, if a metric satisfies these conditions then it defines a perfect fluid spacetime where the energy density µ, pressure p, and velocity ua are given by 1 1 µ = n K 1 a K G , p = a C G , u u = K , where G is the trace of the Einstein tensor. n n a b ab • Einstein spaces are excluded by Condition (1). Note that a = µ C p in a perfect fluid spacetime. Conditions (2) and (3) guarantee the existence of a suitable fluid velocity relative to which the energy density and pressure can be defined. Procedures • For convenience, we assemble here several simple procedures which implement the geometrization theorem above. • CalcAlpha calculates a from a given metric. CalcK calculates Kab from a given metric. Kcondition calculates the quadratic quantity Ka bKc d from a given metric. EnergyAndPressure computes the energy density and pressure from a metric satisfying the geometrization conditions. FluidVelocity computes the fluid velocity from a metric and tensor K satisfying the geometrization conditions. • Execute the following code edit region to load all these procedures. Fluid Geometrization Example 1: Gödel spacetime The Gödel spacetime is a perfect fluid solution to the Einstein equations. Here we use it to illustrate the geometrization conditions and related procedures. Retrieve the Gödel metric from the Library and adjust the notation of the free parameter a. > g1 := eval(op(Retrieve("Stephani", 1, [12, 26, 1], manifoldname = M1, output = ["Metric"]) ), _a = a); a2 e2 x g1 := K a2 dt 5 dt K a2 ex dt 5 dz C a2 dx 5 dx C a2 dy 5 dy K a2 ex dz 5 dt K dz 5 dz (3.1) 2 Compute the scalar a and verify condition (1) is satisfied. We need to specify a O 0 to get the simplification we want. M1 > alpha1 := CalcAlpha(g1); 1 1 / 3 a1 := (3.2) a6 M1 > simplify(alpha1) assuming a > 0; 1 (3.3) a2 Compute the tensor Kab. Condition (3) is true by inspection (e.g, Ktt O 0). M1 > K1 := CalcK(g1) assuming a > 0; K1 := a2 dt 5 dt C a2 ex dt 5 dz C a2 ex dz 5 dt C a2 e2 x dz 5 dz (3.4) Verify condition (2). M1 > Kcondition(g1) assuming a > 0; 0 dt 5 dt 5 dt 5 dt (3.5) It follows that the metric g defines a perfect fluid solution to the Einstein field equations. We compute the energy density µ, pressure p. M1 > mu1, p1 := EnergyAndPressure(g1) assuming a > 0; 1 1 µ1, p1 := , (3.6) 2 a2 2 a2 The equation of state is "stiff" (µ = p) since we have absorbed the cosmological constant into the definition of the fluid. Next we compute the 4-velocity u from the metric. The metric only determines the 4-velocity up to an overall sign – reflecting a discrete symmetry of the Einstein-fluid equations. The solution of the quadratic equations determining the 4- velocity requires assumptions about a and x. M1 > u1 := FluidVelocity(g1, K1) assuming a > 0, x :: real; 1 1 u1 := K v , v (3.7) a t a t We choose the future-oriented 4-velocity. M1 > (3.3) M1 > U1 := evalDG(1/a*D_t); 1 U1 := v (3.8) a t Finally, we verify that the Einstein equations are in fact satisfied. We compute the energy-momentum tensor of the fluid, Tab = µ C p uaub C p gab and the Einstein tensor of the metric, and we check that they are equal. M > T1 := EnergyMomentumTensor("PerfectFluid", g1, U1, mu1, p1); 3 eKx 1 1 eKx eK2 x T1 := v 5 v K v 5 v C v 5 v C v 5 v K v 5 v C v 5 v (3.9) 2 a4 t t a4 t z 2 a4 x x 2 a4 y y a4 z t a4 z z M > G1 := EinsteinTensor(g1); 3 eKx 1 1 eKx eK2 x G1 := v 5 v K v 5 v C v 5 v C v 5 v K v 5 v C v 5 v (3.10) 2 a4 t t a4 t z 2 a4 x x 2 a4 y y a4 z t a4 z z M > evalDG(G1 - T1); 0 (3.11) Example 2: A static, spherically symmetric fluid It is possible to use the geometrization theorem to search for solutions to the Einstein-perfect fluid equations. The following static, spherically symmetric metric ansatz leads to a relatively simple fluid solution via the geometrization conditions. Initialize the coordinate chart and define a simple class of static, spherically symmetric metrics parameterized by a function f r . > DGsetup([t, r, theta, phi], M2); frame name: M2 (4.1) M2 > g2 := evalDG(-r^2*dt &t dt + f(r)*dr &t dr + r^2*(dtheta &t dtheta + sin(theta)^2* dphi &t dphi)); 2 2 2 2 g2 := K r dt 5 dt C f r dr 5 dr C r dq 5 dq C r sin q df 5 df (4.2) M1 > (3.3) M1 > (3.8) Compute the quadratic condition (2) from the geometrization theorem . The output is too long to be displayed. M2 > KC2 := Kcondition(g2): M2 > length(KC2); 16525 (4.3) Despite the apparent complexity of the condition KC2 = 0, it is possible to solve it. Using the DGsolve command, the quadratic condition on K is solved for f r and the result is substituted into the metric. Maple is thus able to find a one parameter family of metrics; we label the parameter l and call the final form of the fluid metric g2fluid. M2 > gsol := DGsolve(KC2, g2, [f(r)]); 2 2 2 2 2 gsol := K r dt 5 dt C dr 5 dr C r dq 5 dq C r sin q df 5 df (4.4) 2 _C1 r2 C 1 M2 > g2fluid := eval(op(gsol), _C1 = lambda); 2 2 2 2 2 g2fluid := K r dt 5 dt C 2 dr 5 dr C r dq 5 dq C r sin q df 5 df (4.5) 2 l r C 1 Verify that condition (2) of the geometrization theorem is satisfied. M2 > KC22 := Kcondition(g2fluid): M2 > DGsimplify(KC22) assuming r > 0; 0 dt 5 dt 5 dt 5 dt (4.6) Compute a and verify condition (1). M2 > CalcAlpha(g2fluid); 1 1 / 3 (4.7) r6 Compute Kab and check condition (3). M2 > K2 := CalcK(g2fluid); (4.8) M1 > (3.3) M1 > (3.8) 1 1 3 1 1 3 1 1 3 r2 C 3 r2 K 1 r2 K 1 r6 r6 r6 K2 := 1 3 dt 5 dt K 1 3 dr 5 dr K 1 3 dq 5 dq (4.8) 1 1 2 2 1 4 2 r 2 l r C 1 4 r6 r6 r6 1 3 2 2 1 sin q r K 1 r6 K df 5 df 1 1 3 4 r6 M2 > K2 := DGsimplify(K2) assuming r > 0; K2 := r2 dt 5 dt (4.9) Since Ktt O 0 condition (3) is satisfied. Compute (and simplify) the energy density µ2, pressure p2, and 4-velocity U2. M2 > mu20, p20 := EnergyAndPressure(g2fluid); 1 / 3 2 1 / 3 2 3 1 1 12 l r C 1 1 1 1 12 l r C 1 µ20, p20 := K , C (4.10) 4 r6 4 r2 4 r6 4 r2 M2 > mu2 := simplify(mu20) assuming r > 0; 2 1 6 l r K 1 µ2 := K (4.11) 2 r2 M2 > p2 := simplify(p20) assuming r > 0; 2 1 6 l r C 1 p2 := (4.12) 2 r2 M2 > U20 := FluidVelocity(g2fluid, K2) assuming r > 0, a > 0; 1 1 U20 := K v , v (4.13) r t r t M2 > U2 := evalDG(1/r*D_t); (4.14) M1 > (3.3) M1 > (3.8) (4.8) 1 U2 := v (4.14) r t Finally, we verify that we have built a solution to the perfect fluid Einstein equations.