This page intentionally left blank Introduction to General Relativity A student-friendly style, over 100 illustrations, and numerous exercises are brought together in this textbook for advanced undergraduate and beginning graduate students in physics and mathematics Lewis Ryder develops the theory of General Relativity in detail Covering the core topics of black holes, gravitational radiation and cosmology, he provides an overview of General Relativity and its modern ramifications The book contains a chapter on the connections between General Relativity and the fundamental physics of the microworld, explains the geometry of curved spaces and contains key solutions of Einstein’s equations – the Schwarzschild and Kerr solutions Mathematical calculations are worked out in detail, so students can develop an intuitive understanding of the subject, as well as learn how to perform calculations Passwordprotected solutions for instructors are available at www.cambridge.org/Ryder Lewis Ryder is an Honorary Senior Lecturer in Physics at the University of Kent, UK His research interests are in geometrical aspects of particle theory and its parallels with General Relativity Introduction to General Relativity Lewis Ryder University of Kent, UK CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo, Delhi, Dubai, Tokyo Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521845632 © L Ryder 2009 This publication is in copyright Subject to statutory exception and to the provision of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press First published in print format 2009 ISBN-13 978-0-511-58004-8 eBook (EBL) ISBN-13 978-0-521-84563-2 Hardback Cambridge University Press has no responsibility for the persistence or accuracy of urls for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate For Mildred Elizabeth Ryder It is always a source of pleasure when a great and beautiful idea proves to be correct in actual fact Albert Einstein [letter to Sigmund Freud] The answer to all these questions may not be simple I know there are some scientists who go about preaching that Nature always takes on the simplest solutions Yet the simplest by far would be nothing, that there would be nothing at all in the universe Nature is far more interesting than that, so I refuse to go along thinking it always has to be simple Richard Feynman 323 9.2 Radiation from a rotating binary source Nkl ¼ Pik Pjl Mij (9:51) Nkl n k ¼ Nkl nl ¼ 0; (9:52) is transverse: since Nkl nl ẳ ik ni nk ịjl nj nl ịMij nl ẳ ik ni nk Þðnj À nj ðn Á nÞÞMij ¼ 0; but it is not traceless: Nk k ¼ Pik Pjk Mij ¼ Pij Mij ẳ Pji Mij ẳ trPM ị; where we have used (9.50) and the fact that Pij = Pji However, the tensor (matrix) (9:53) Rkl ¼ Pik Pjl À Pkl Pij Mij is traceless: Rkk ẳ Pik Pkj 1=2Pkk Pij ịMij ¼ À Pkk Pij Mij ; but Pkk = δkk nk n k = = 2, so Rkk ¼ 0: (9:54) R is then the transverse and traceless version of M: R ¼ M TT ¼ PMP À Pðtr PM Þ (9:55) or (9:56) M TT i j ¼ Pik Pjl À Pij Pkl Mkl : This completes the mathematics we need to find hTTij at any point Equation (9.47) will be amended to τ 00 ¼ G €_TT €_TT I ij I ij ; 8πc6 r2 (9:57) with I TT i j ¼ Pik Pjl À Pij Pkl Ikl : To evaluate (9.57) observe that 1 I TT ij I TT ij ¼ Pik Pjl À Pij Pkl Pip Pjq À Pij Ppq Ikl Ipq 2 ¼ ðPik Pjl Ikl ÞðPip Pjq Ipq Þ À ðPik Pjl Ikl ÞðPij Ppq Ipq ị ỵ Pij Pij Pkl Ikl ịPpq Ipq Þ: (9:58) (9:59) There are three terms on the right hand side of this equation After some algebra it is seen that the first one is equal to Iij Iij np np Iip Iiq ỵ ni nj np nq Ii j Ipq ; 324 Gravitational radiation and the second and third terms are, respectively À nk nl np nq Ikl Ipq and ỵ nk nl np nq Ikl Ipq ; giving, finally, I TTij I TTij ¼ Iij Iij À 2np nq Ii p Iiq þ nk nl np nq Ikl Ipq : (9:60) We have, then, that from (9.57) the gravitational energy density is τ 00 ¼ G €_TT €_TT I ij I ij 8πc6 r2 (9:61) where the ‘averaging’ sign is put in to deal with possible oscillating terms; and the energy passing through a spherical surface of radius r in unit time is ð ð TT dE G €_ ¼ cτ 00 r dΩ ¼ I TT ij dΩ: I ij €_ (9:62) dt 8πc The integral above is, from (9.60) ð ð TT €_TT €_ €_ I ij €_ I ij À np nq €_ I ip €_ I iq þ nk nl np nq €_ I kl €_ I pq dΩ: I ij I ij dΩ ¼ (9:63) The factors €_ I i j and so on come outside the integral sign The remaining angular integrals are ð ð 4π δi j ; dΩ ¼ 4π ; ni nj dΩ ¼ ð (9:64) 4π ni nj nk nl d ẳ ij kl ỵ ik jl ỵ il δjk Þ: 15 These are easily seen: for putting n1 ¼ ^x ¼ sin θ cos , n2 ¼ ^y ¼ sin θ sin , n3 ¼ ^z ¼ cos θ and dΩ = sinθ dθ d , it is simple to check that ð ð ð 4π 4π n1 n2 dΩ ¼ 0; n21 dΩ ¼ ; n21 n22 dΩ ¼ ; 15 and so on Hence (9.63) gives (noting (9.48)) ! ðD E 24π 4π 4π TT TT €_ €_ €_ €_ I ij I ij dΩ ¼ I ij I ij 4π ỵ ẳ _ I ij €_ ; I ij 15 15 and, finally, the rate of energy loss is dE G €_ ¼ €_ I ij I ij : dt 5c Returning to our orbiting binary, from (9.38) sin 2ωt €_ I ij ¼ 8MR ω @ À cos 2ωt I i j = 128M2R4ω6 and hence €_ I i j €_ À cos 2ωt À sin 2ωt (9:65) 0 A; 325 9.2 Radiation from a rotating binary source dE 128G M R ω : ¼ dt 5c5 (9:66) Substituting further from (9.37) for ω, the frequency of rotation, gives dE 2G4 M ¼ 5: 5c R dt (9:67) This is an expression for the gravitational luminosity of the source: we may write it as GM c5 : (9:68) L¼ Rc2 G The quantity (GM/Rc2) is dimensionless and (c5/G) has the dimensions of luminosity Note that L depends on 1/R5, so most energy is emitted from compact binaries (R is the radius of the binary system) 9.2.3 Spin-up and the binary pulsar PSR 1913+16 The total kinetic energy of the binary system is T ¼2 GM ; M ν2 ¼ 4R GM (since the distance between the bodies 2R is 2R) so the total energy of the binary system, including rest masses, is from (9.37) The potential energy is V ¼ À E¼ GM GM GM ỵ 2Mc2 ẳ 2Mc2 : 4R 2R 4R (9:69) As the system emits energy E must decrease, so R must also decrease (assuming M is unaffected) and the orbit will shrink; the system ‘spins up’ and eventually the stars coalesce The frequency of the emitted radiation is f = 2ω/2π, which from (9.37) is f ¼ GM 1=2 : π 4R3 (9:70) As R decreases the pitch of the radiation increases, and the signal is known as a chirp, the song of a sparrow Let us make a crude estimate of the characteristic time for the radius of the orbit to shrink from R to R/2 From above it is clear that an amount of energy GM2/4R will have been emitted, and if we assume that the luminosity L is constant during this time, the characteristic time tch would be Ltch ¼ giving, from (9.68), GM ; 4R 326 Gravitational radiation tch 5R GM ¼ : 8c R c2 (9:71) This is the chirp time for the binary The factor 5R/8c is the time taken for light to travel a distance 5R/8 and the second factor is dimensionless It must be admitted that this is a rather crude estimate, but note that because of the 1/R5 dependence of the luminosity, the time taken for a further decrease in radius, say from R/2 to R/4, is very much less, so the chirp time is actually not a bad order-of-magnitude estimate for the time taken for the stars to coalesce Now let us be more specific, and consider a system where both stars are neutron stars, M ¼ 1:4MS ; Then MG/Rc2 = 2.06 ì 10 R ẳ 106 km: (9:72) and tch ¼ 2:2  10 17 s % 7:1  109 years; comparable to the age of the Universe! This is hardly measurable, but an interesting and measurable quantity is the change in the period of an orbit, as a result of its shrinking The period is, from (9.37), R3 τ ¼ 4π GM 1=2 / R3=2 : As R changes τ will also change, and from above dτ dR ¼ : τ R From (9.69) we have dE/E = (9:73) dR/R, so dτ dE ¼À : τ E We want to find dτ/dt, the rate of change of the period From the above equations we find dτ=dt dE=dt 3L 12 G3 M ¼À ¼À ¼À ; τ E 2E R c5 hence dτ 48π GM 5=2 ¼À : τ Rc2 With MG/Rc2 = 2.06 × 10 (9:74) this gives dτ ¼ À1:84  10 dt 13 ; the period of the orbit decreases by 1.84 × 10 13 seconds every second This is a dramatic prediction of General Relativity, coming about as a consequence of the emission of gravitational radiation from the binary system 327 9.2 Radiation from a rotating binary source This prediction was verified in 1974 by Russell Hulse and Joseph Taylor who observed the decrease in the period of the pulsar PSR 1913+16 in a binary system like the one considered above: both the pulsar and its companion have masses close to 1.4MS.3 The orbits are tight but eccentric: the semi-major axis is about × 108 m and the eccentricity of the ellipse is 0.62; the formula (9.74) has to be corrected for this Hulse and Taylor took measurements over many years and concluded that d ẳ 2:422 ặ 0:006Þ Â 10 dt 12 ; in agreement with the general relativistic prediction to an accuracy of better than 0.3% We may thus conclude that we have definite – though indirect – evidence for the existence of gravitational radiation 9.2.4 Search for gravitational waves The search for gravitational waves actually has a long history, going back to the pioneering experiments of Weber in the 1960s These experiments, which in a modified form are still being pursued by many research groups today, consist of a large metal cylinder which, when hit by a gravitational wave, will oscillate, and the oscillations are converted by transducers into electrical signals Modern versions of this experiment may achieve sensitivities of the order of 10 18 To put this figure into context, let us calculate the amplitude of the signal from PSR 1913+16 Equations (9.37) and (9.39) give h¼ 2G2 M c4 r R and with M = 1.4MS, R = 109 m and r, the distance away of the source, equal to kpc = 2.5× 1020 m, we have h = 3.4 × 10 23, many orders of magnitude down on 10 18! An apparatus of this type would only be able to register signals from much more powerful emitters than this binary pulsar – and of course there are more powerful sources Even with a sensitivity of 10 18, however, in a bar of length (say) 10 m, we would be looking for a change in length of 10 17 m – one hundredth the size of a nucleus! Gravitational wave detection certainly offers an experimental challenge More recently a lot of effort is being put into interferometric methods of finding gravitational waves These experimeents are basically of the same design as a Michelson interferometer Mirrors are placed along perpendicular arms and when a gravitational wave impinges one of the arms contracts while the other expands, followed a short time later by the opposite motion Thus what is looked for is an oscillation in the path difference and therefore the interference pattern Some of these experiments are earthbound but the most interesting of them are planned to be set up in space The LISA (Large Interferometric Space Antenna) project will consist of three spacecraft about × 106 km apart The expectation is that sensitivities of about 10 22 might be achieved in this type of experiment and there is Hulse & Taylor (1975); see also Weissberg & Taylor (1984) 328 Gravitational radiation considerable optimism that, in the words of Bradaschia and Desalvo,4 the gravitational wave community is poised to prove Einstein right and wrong: right in his prediction that gravitational waves exist, wrong in his prediction that we will never be able to detect them 9.3 Parallels between electrodynamics and General Relativity: Petrov classification In considering the question whether gravitational radiation exists, much of our thinking has been guided by the case of electromagnetism Maxwell’s equations give rise to a wave equation whose solutions describe waves carrying energy Correspondingly, in the linear (weak field) approximation of General Relativity, where the metric tensor differs from its Minkowski value by the small quantity hμν, we saw that this field also obeys a wave equation; and that these waves also carry energy In the last section we considered convincing, though indirect, evidence that gravitational waves exist, so it might be thought that all that remains is to discover them directly, but this is not true Even if gravitational waves are found conclusively to exist experimentally, we still need to have a proper theoretical understanding of them – and the weak field approximation does not provide this In the first place it is linear, whereas General Relativity in its complete version is non-linear In this complete version there is a graviton–graviton coupling, but there is no analogous photon– photon coupling in quantum electrodynamics The weak field approximation, however, is linear so the obvious question must be asked: what is the status of this approximation? On iteration, does it yield General Relativity, and if it does, how does the non-linearity enter? Is the iteration even convergent? To my knowledge, these questions are not yet satisfactorily answered 9.3.1 A geometric approach to electrodynamics Reflections such as these encourage us to look at the question of gravitational radiation in a different way, but still bearing in mind the model of electrodynamics Consider, for example, two types of electromagnetic field, the Coulomb field and the radiation field, in which the electric and magnetic components have the following behaviour: ; B ¼ 0; r2 Plane wave radiationị : E $ ; jEj ẳ jBj; r Coulomb field : E $ E : B ¼ 0: (9:75) The Coulomb field is quasi-static and carries no energy but the radiation field carries energy – it has a non-vanishing Poynting vector What we want to show is that these Bradaschia & Desalvo (2007) 329 9.3 Electrodynamics and General Relativity cases correspond to a classification of the field tensor Fμν into distinct categories – and then to show that a similar situation exists for the gravitational field and the Riemann tensor ~μν we may form the Lorentz invariant quantities (see From the tensor Fμν and its dual F Section 2.6) Fμ ν F μ ν ¼ ÀjEj2 þjBj2 ; ~ μ ν ¼ ÀE:B: Q ¼ Fμ ν F P¼ (9:76) Electromagnetic fields with P = Q = are called null fields The field of a Coulomb (electric) charge has P < 0, Q = 0, so is not null The field of a single magnetic charge is also Coulomblike, but in this case it is purely magnetic and P > 0, Q = In summary Field of a point charge : P 6¼ 0; Q ¼ 0; purely electric; P50; purely magnetic; P 0: (9:77) On the other hand, a plane wave travelling in the direction k has c c E ẳ E0 expẵik : x tị; B ẳ k E ẳ k E0 expẵik : x À ωtÞ: ω ω In a vacuum ∇ · E = so k · E = and jk  E0 j2 ¼ k E0 À ðk : E0 ị2 ẳ k E0 so jBj2 ¼ c2 k 2 E0 ¼ E0 ¼ jEj2 ) Fμν F μν ¼ 0; ω2 and in addition ẾB ¼ c : ~ μν ¼ 0: E k Eị ẳ ) F F We therefore have P = Q = 0, a null field: Plane wave : P ¼ 0; Q ¼ 0: null field: (9:78) We have thus characterised the fields of a point charge and radiation algebraically The next step is to characterise them in the form of an eigenvalue problem This will enable us to make the comparison with gravitational fields.5 The eigenvalue problem takes the form Fμ ν k ν ¼ λ kμ ; (9:79) kμ is the eigenvector and λ the eigenvalue of Fμν For the next part of the argument we give Fμν – an antisymmetric rank tensor – a purely algebraic and geometric interpretation.6 We describe it as a bivector – an object constructed from two vectors A bivector is simple if it can be written as Much of the following mirrors the work of Frolov (1979) For a more complete version of the argument that follows see Frolov (1979) 330 Gravitational radiation F ẳ aẵ b ẳ a b aν bμ Þ: (9:80) In general Fμν is not simple, but it can always be decomposed into a pair of simple bivectors: F ẳ aẵ b ỵ cẵ d : (9:81) It can be shown7 that a bivector is simple if and only if Fẵ F ẳ 0: (9:82) From the definition ~ μν ¼ εμνλ Fμν Fλ ; Fμν F it is clear that the right hand side is totally antisymmetric in all its indices so (9.82) implies that ~ μν ¼ 0: Fμν F (9:83) We now introduce an important transformation, the dual rotation: ~ : F ! F ị ẳ cos F ỵ sin θ F (9:84) It is actually a transformation which mixes electric and magnetic fields; for example with ~~ ¼ ÀF we have (μν) = (10), Ex → Ex cos θ + Bx sin θ Since F μv μv ~μν sin F ; ~ ị ẳ cos F F so ~ ị ẳ cos F F ~ μν À sin 2θ Fμν F μν Fμν ðθÞ F (9:85a) ~ μ ν: Fμ ν ðθÞ F ị ẳ cos F F ỵ sin F F (9:85b) and ~ μv (θ) = 0, From (9.85a) it follows that, given an arbitary Fμν there exists a θ such that Fμν (θ) F so Fμν is simple (by (9.82) and (9.83)) Then by a duality rotation any Fμν may be converted into the form (9.80) F ị ẳ aẵ b ; (9:86) ~ μν(θ) = 0, F ~ μν(θ) is also simple, and since F() F ~ ị ẳ c½μ dν : F (9:87) ~μv (θ) simply as F ~μv Equation (9.86) tells us From here on we write Fμν(θ) simply as Fμν and F ~ that the bivector Fμν defines a plane Π(a, b) and similarly Fμv defines a plane Π(c, d) And ~μv ¼ εμvλ a½ bλ , the plane Π(a, b) is orthogonal to the plane Π(c, d) since F Schouten (1954) 331 9.3 Electrodynamics and General Relativity With these geometric considerations in mind we now examine more closely the cases Fμν Fμν = and Fμν F μν ≠ (corresponding to radiation and Coulomb-type fields) Consider the following: Proposition: If Fμν is a simple bivector there exists a pair of vectors p, q such that F ẳ pẵ qν ; pμ qμ ¼ 0; (9:88) in other words, pμ and qμ are orthogonal Proof: If Fμν is simple then we can write Fμν = a[μbν], from (9.86), so we have only to find pμ, qμ such that pμqμ = Case (i): if aμaμ ≠ 0, choose p μ ¼ aμ ; qμ ¼ bμ À ða a Þ aμ ðaλ aλ Þ so that pμ qμ ¼ a b a a ị a b ẳ 0; a a ị and clearly pẵ q ẳ aẵ b ; (since a[ μaν] = 0) Case (ii): if bμb μ ≠ 0, interchange a and b above Case (iii): if aμ and bμ are both null, aμa μ = bμb μ = 0, then choose pμ ¼ p a b ị; q ẳ p a ỵ b ị; so that 1 pẵ q ẳ a bịẵ a ỵ bị ẳ faẵ b b a g ẳ aẵ b 2 and 1 pμ q μ ¼ ðaμ À bμ ịa ỵ b ị ẳ a a b b ị ẳ 0: 2 If the simple bivector is represented in the canonical form (9.88), then 1 Fμν F μν ¼ ðpμ qν À pν qμ Þðp μ qν À pν q μ Þ ¼ ðpμ p μ Þðqμ qμ Þ: (9.89) We may then consider the cases in which Fμν F μν < and Fμν F μν = 0: in which Fμν is said to be timelike or null From the mathematical point of view we should also consider the case Fμν F μν > 0, but for our present purposes that is of no physical interest 332 Gravitational radiation Case I: Fμν F μν < Then from (9.89) one of the vectors (say pμ) must be timelike and the other spacelike In the plane Π(p, q), however, there must exist a pair of vectors, lμ and nμ, with Fμν ¼ lμ nν À lν nμ ; (9:90) such that lμ and nμ are both null, but they are not orthogonal: lμ l μ ¼ nμ nμ ¼ 0; lμ nμ ¼ α 6¼ 0; (9:91) since then Fμν F μν ¼ ðlμ nν À lν nμ Þ ðl μ nν À l ν nμ Þ ¼ À2α2 50: Case II: Fμν F μν = Then one of the vectors (say pμ) must be null and the other spacelike (timelike and null, or two independent null vectors, are not orthogonal) In this case we may write Fμν ¼ lμ aν À lν aμ (9:92) with lμ null and aμ spacelike, and the two vectors orthogonal, lμ l μ ¼ 0; aμ a μ 40; lμ a μ ¼ 0: (9:93) We now turn to the eigenvalue problem (9.79) Fμν k ν ¼ λkμ : (9:79) In the case that Fμν is timelike we have, from (9.90), l n l n ịk ẳ k ; l n : kị n l : kị ẳ λkμ ; so either kμ = lμ and (n · k) = λ = (n · l), i.e λ = α, or kμ = nμ and (l · k) = (l · n) = λ = α In other words, the equation (9:94) λ, i.e Fμν k ν ¼ λkμ (9:79) kμ ¼ lμ or nμ ; with λ ¼ Æðn : lÞ: (9:95) has two solutions It clearly follows from (9.79) that kρ Fμν k ν ¼ λ kρ kμ : The right hand side of this equation is symmetric under ρ ↔ μ so the antisymmetric part under this interchange vanishes: kẵ F k ẳ 0; and we recall that this equations has two solutions in the case where Fμν is timelike (9:96) 333 9.3 Electrodynamics and General Relativity When Fμν is null (9.79) becomes, with (9.92) l a l a ịk ẳ k or lμ ða : kÞ À αμ ðl : kÞ ¼ λ kμ : (9:97) It might be thought that this equation has two solutions, the first being kμ = lμ, in which case the left hand side of (9.97) is l a l a ịl ẳ l a : lị l : lị ẳ by (9.93), implying that λ = 0; and the second being kμ = aμ, in which case the left hand side of (9.97) is ðlμ aν À lν aμ Þaν ¼ lμ ða : aÞ À aμ ðl : aÞ ¼ ða : aÞ lμ ; but this is incompatible with the right hand side, so this second case is not a solution There is therefore only one solution to the eigenvalue problem F μ ν k ν ¼ λ kμ when Fμν is null, which is kμ ¼ Alμ A ẳ constị; and ẳ 0: (9:98) We conclude that in the case P ≠ 0, Q = 0, describing a Coulomb-type field, the eigenvalue equation is (9.96) k½ρ Fμ ν k ν ¼ (9:99) and this has two solutions This is called the non degenerate case and the field tensor is denoted Fμν[1,1] On the other hand, in the case P = 0, Q = 0, the null case corresponding to a pure radiation field, the eigenvalue equation is Fμ ν k ν ¼ 0: (9:100) There is only one eigenvalue, λ = 0; this is the degenerate case and the field tensor is denoted Fμν[2] The retarded field from an isolated extended source has the asymptotic (r → ∞) behaviour 1 F μ ν ¼ F ẵ2 ỵ F ẵ1;1 ỵ Or ị: r r (9:101) 9.3.2 Petrov classification We now move on to the gravitational field, described by the Riemann tensor The relevant classification is the Petrov classification, which applies in the first instance to the Weyl tensor Cλμνρ This is closely related to the Riemann tensor and is defined as follows: 334 Gravitational radiation Rλ μ ν ρ ¼ Cλ g B ỵ g ν Bλ ρ À gλν Bμρ À gμ ρ Bλ ν Þ Rðgλ ρ gμ ν À gλ ν gμ ρ Þ; À 12 (9:102) with Bμ ν ¼ Rμ ν ¼ À gμ ν R: (9:103) Clearly, in a vacuum, where Rμν = R = 0, in vacuoị R ẳ C ν ρ : (9:104) Since gμνBμν = R – ¼δμμR = 0, it is clear, multiplying (9.102) by gλν that R ẳ g C ỵ R ρ ; hence gλ v Cλ μ v ρ ¼ 0: (9:105) The Weyl tensor has the same symmetries as the Riemann tensor: from (9.102) Cλ μ ν ρ ¼ À Cμ λ ν ρ ¼ À Cλ μ ρν ẳ ỵ Cv ; C ỵ C ỵ C ν ¼ 0; (9:106) so Cλμνρ, like Rλμνρ, has 10 independent components; but gλνCλμνρ is always zero, even in the presence of matter The classification of a tensor depends only on its symmetry properties, so the classification of the Riemann tensor in vacuo is the same as the classification of the Weyl tensor anywhere – even in matter We shall consider the matter of classification by the method of eigenvectors and eigenvalues, as we did above for the electromagnetic field tensor.8 If we have, then, a rank tensor Tij we look for a vector V j with the property Ti j V j ¼ λVi ¼ λ gik V k ; (9:107) or À Á Ti j À λgi j V j ¼ 0; and the eigenvalues λ are the solutions to the equation jTi j À λ gi j j ¼ 0: (9:108) It is most important to note that the classification which we are about to discuss depends crucially on the fact that we choose a locally inertial frame at any point P of space-time, at which, therefore, the metric tensor takes on its Minkowski values gμ v ẳ diag 1; 1; 1; 1ị: (9:109) Much of the material below follows the treatments of Landau & Lifshitz (1971) and Papapetrou (1974) 335 9.3 Electrodynamics and General Relativity It follows from this that the Petrov classification is a local one – in practice, a physical gravitational field may change from one class to another, or a mixture of classes, as we move from one point to another in space In the spirit of (9.107) we should define a rank tensor related to the Weyl tensor, and we this by proceeding, as in Section 4.4, by making the association Cλ μ v ρ $ C A B ; A $ ðλ μÞ; B $ ðv ρÞ: (9:110) The indices A, B take on the values 1, 2, …, and in view of (9.106) CAB = CBA is a symmetric tensor in a dimensional space with metric tensor γAB This must be symmetric, and is defined by γ A B $ g λ μ v ρ gλ v g μ ρ À gλ ρ g μ v : (9:111) It has the same symmetries as the Weyl tensor, so in the bivector space is symmetric The eigenvalue equation is of the form ðCA B AB ị W B ẳ 0: (9:112) It actually corresponds to the equation ðCλ μ v ρ À g v ị W v ẳ 0; where W νρ = – W ρν, but we shall find it easier to work with Equation (9.112) The correspondence between the indices A, B and μν, etc., is A μν 01 02 03 23 31 12 so, for example, γ11 ¼ g0101 ¼ g00 g11 À g01 g01 ¼ À1; γ44 ¼ g2323 ¼ g22 g 33 g23 ị2 ẳ 1; 12 ẳ g0102 ¼ g00 g12 À g02 g01 ¼ 0; and so on, giving γAB ¼ diag ðÀ1; À1; À1; 1; 1; 1Þ: (9:113) 336 Gravitational radiation An analogous ‘bivector’ relabelling of the Riemann and Weyl tensors may be carried out, so, for example, Rμ v λ $ RA B ẳ RB A : (9:114) R0123 ỵ R0231 ỵ R0312 ẳ (9:115) R14 ỵ R25 ỵ R36 ẳ 0; (9:116) Then the relation (see (4.34iv)) becomes and similarly, since the Weyl tensors has the same symmetries, C14 ỵ C25 þ C36 ¼ 0: (9:117) To proceed further (noting all the while that the symmetries of the Riemann and Weyl tensors are the same) we now write the 20 independent components of the Riemann tensor as a collection of 3-dimensional tensors9 Mik, Nik and Pik: Mi k ¼ R0 i k ; Ni k ¼ εi m n R0 k m n ; Pi k ¼ εi m n εk p q Rm n p q ; (9:118) noting that in this 3-dimensional locally Minkowski space there is no need to distinguish upper and lower indices By virtue of the symmetry properties of the Riemann tensor we have Mi k ẳ Mk i ;6 componentsị; (9:119) and 4 Pk i ¼ εk m n εi p q Rm n p q ¼ εi p q εk m n Rp q m n ¼ Pi k ð;6 componentsÞ: (9:120) In addition 2 N11 ¼ ε1m n R01m n ¼ ðR0123 R0132 ị ẳ R0123 ; N22 ẳ R0231 ; N33 ¼ R0312 ; hence, from (9.115), Ni i ¼ N11 ỵ N22 ỵ N33 ẳ 0: (9:121) Further, N12 ¼ R0131 ; N13 ¼ R0112 ; N23 ¼ R0331 ; N32 ¼ R0212 ; (9:122) and so on We now make use of the vacuum field equations Rμν = We have R00 ¼ g ik R0i0k ¼ R0i0i ¼ Mii ; hence Mii ¼ 0: That is, tensors whose indices take on only the values 1, 2, (9:123) 337 9.3 Electrodynamics and General Relativity We also have P12 ¼ ε1mn ε2 pq Rmnpq ¼ R2331 ¼ ÀR3132 : On the other hand, R12 ¼ gρσ Rρlσ2 ¼ R0l02 À Rili2 ¼ R0l02 À R3132 ¼ 0; hence, from (9.118) M12 = P12, and in general Mi k ¼ À P i k ði 6¼ kÞ: (9:124) In a similar way it follows from R01 = that N32 = N23, and in general Ni k ¼ Nk i : (9:125) Finally, the equations R11 = 0, R22 = and R33 = give, respectively M11 ẳ P22 ỵ P33 ; M22 ẳ P33 þ P11 ; M33 ¼ P11 þ P22 ; (9:126) which are, on rearrangement, P11 ẳ M22 ỵ M33 M11 ị; P22 ẳ M11 ỵ M33 M22 ị; P33 ẳ M11 ỵ M22 À M33 Þ; from which, using (9.123) Pi i ¼ Mi i ¼ 0: It then follows from (9.126) that M11 = – P11 and so on, so (9.124) becomes Mi k ¼ ÀPi k for all i; k: (9:127) We can now enumerate the number of independent components of these tensors Because of (9.127) the independent tensors are Mik and Nik Equations (9.119) and (9.123) imply that Mik has = independent components, and (9.121) and (9.125) that Nik has offdiagonal and diagonal independent components, making in all This gives a total of 10 independent components for the Riemann tensor, as expected when Rμν = We can now represent the components of the Riemann tensor as a × matrix, which in terms of Mik and Nik is M N CA B ¼ ; (9:128) N ÀM with M and N both symmetric traceless × matrices The eigenvalues λ of CAB are the roots of det CA B AB ị ẳ 0; (9:129) i.e Mi k À λδi k Ni k Ni k ¼ 0: ÀMi k À λδi k (9:130) ... parallels with General Relativity Introduction to General Relativity Lewis Ryder University of Kent, UK CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore,... Delhi, Dubai, Tokyo Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www .cambridge. org... written permission of Cambridge University Press First published in print format 2009 ISBN-13 978-0-511-58004-8 eBook (EBL) ISBN-13 978-0-521-84563-2 Hardback Cambridge University Press has no responsibility