12.07.2015 Views

Contents

Contents

Contents

SHOW MORE
SHOW LESS
  • No tags were found...

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

230 15 Groups G 3 on non-null orbits V 2For vacuum, Einstein–Maxwell and pure radiation fields, and for dust, thegeneral solutions admitting a G 3 on V 2 are known (§§15.4, 15.5). Perfectfluid solutions in general are discussed in §15.6, and plane-symmetric perfectfluids in §15.7. Chapter 16 gives a survey of the spherically-symmetricperfect fluids.15.4 Vacuum, Einstein–Maxwell and pure radiation fieldsFor vacuum and Einstein–Maxwell fields (including the cosmological constantΛ), and pure radiation fields, the algebraic types of the energymomentumtensor are [(111, 1)], [(11) (1,1)], and [(11, 2)], see §5.2.The eigenvalues λ α of the Einstein tensor areλ 1 = λ 2 = G 1 1 = G2 2 ,λ 3,4 = 1 2 (G3 3 + G4 4 ) ± ∆1/2 , ∆ ≡ 1 4 (G3 3 − G4 4 )2 ∓ (G 4 3 )2 .(15.11)The symmetry implies that a double eigenvalue (λ 1 = λ 2 ) exists (seeTheorem 15.2). For the algebraic types under consideration, there aretwo double eigenvalues (which might coincide); λ 3 = λ 4 implies ∆ = 0.15.4.1 Timelike orbitsFor timelike orbits T 2 , Table 5.2 shows immediately that the algebraictype is [11(1, 1)]. Since the type [(11, 2)] is impossible we haveTheorem 15.4 Einstein–Maxwell fields with an electromagnetic nullfield, and pure radiation fields, cannot admit a G 3 on T 2 .For vacuum and non-null Einstein–Maxwell fields, we have to distinguishbetween Y ,a ̸= 0 andY ,a =0.IfY ,a ̸= 0, one can put Y = x 3 (canonicalcoordinates), because of Y ,a Y ,a > 0, and the evaluation of G 3 3 = G4 4and G 4 3 = 0 in the metric (15.3) (lower signs) gives λ′ + ν ′ = 0 and˙λ = 0. In ν an additive term dependent on x 4 can be made zero bytransforming x 4 .Thuswehaveν = −λ, ˙λ = 0. In the coordinates of(15.3), the general solution of the Einstein–Maxwell equations (includingΛ) reads (cp. (13.48))e 2ν = k − 2m/x 3 − e 2 /(x 3 ) 2 − Λ(x 3 ) 2 /3 > 0,λ = −ν, Y = x 3 , m,e = const, k =0, ±1,(15.12)the only non-vanishing tetrad component of the (non-null) electromagneticfield tensor being (up to a constant duality rotation, see §5.4)√ 2κ0 F 34 =2e/(x 3 ) 2 . (15.13)

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!