shear. Hence π and πβ² will both satisfy second order decoupled equations if, and only if, πΏ = πΏβ² = π = πβ² = 0. A natural name for a spacetime admitting such null vector ο¬elds seems to be:
Deο¬nition 4.2 A spacetime is doubly Kundt if and only if it admits a pair of non- expanding, non-shearing, non-twisting geodesic null vector ο¬elds β and π with β.πβ= 0.
4.3.2 The Schwarzchild Solution
Consider the special case of the higher-dimensional Schwarzschild solution, which is not Kundt. This solution has π = πβ2π 1 and π =0 (a consequence of spherical symmetry).
The latter implies that the terms inFandπ drop out of equation (4.20), leaving us with an equation of the form
(β‘π)π+ (πβ4)
(πβ2)2π2πβ²π = 0, (4.22) whereβ‘is a second order diο¬erential operator. The second term remains an obstruction to decoupling. For the Schwarzschild solution, the two multiple WANDs have identical properties so we can take the prime of the equation to obtain
(β‘β²πβ²)π+ (πβ4)
(πβ2)2πβ²2ππ = 0, (4.23)
and hence [
β‘β² ( 1
π2β‘π )]
πβ (πβ4)2
(πβ2)4πβ²2ππ = 0. (4.24) So in factπdoes satisfy a decoupled equation but it is fourth order in derivatives. Note that we had to make use of several special properties of the Schwarzschild solution to obtain this result. It would be interesting to investigate more generally the circumstances under which one can obtain a decoupled equation of higher order forπ.
4.4. DECOUPLING OF GRAVITATIONAL PERTURBATIONS 93 The ο¬nal result will be similar to that of the electromagnetic perturbations; we will ο¬nd that we can only achieve decoupling when the spacetime is Kundt. We will show that gravitational perturbations of such a Kundt spacetime are described by
(2ΓΎβ²ΓΎ+kπkπ+πβ²ΓΎβ6ππkπ+ 4Ξ¦β πβ12π Ξ) Ξ©ππ + 4(
ππk(πβ£βπ(πβ£kπ+ Ξ¦S(πβ£π+ 4Ξ¦A(πβ£π)
Ξ©πβ£π)+ 2Ξ¦ππππΞ©ππ= 0, (4.25) where all quantities exceptΞ©are evaluated in the background geometry (e.g. Ξ¦ denotes Ξ¦(0) etc.).
In a doubly Kundt spacetime, Ξ©β² also will satisfy a decoupled equation, which is given by taking the prime of the above equation.
4.4.2 Derivation of main result
We follow as closely as possible the 4D approach of Stewart & Walker [31]. Many of the equations in this section were checked using the computer algebra package Cadabra [162, 163]. We start by obtaining an equation in which second derivatives act only on Ξ©ππ. Consider the equations
0 = βkπΞ©ππ βπΏππkπΞ©ππ+kπΞ©ππβΓΎ(Ξ¨ππΏππ+ Ξ¨πππ) +πΏππ(Ξ¦ππβ2Ξ¦ππβΞ¦πΏππ)π π + (β2Ξ¦π[πβ£πΏπ]π+ 2πΏππΞ¦Aππ + Ξ¦ππππ)π π+πΏππ[βΞ¨ππβπππΞ¨πβ(Ξ¨πππ+ Ξ¨πππ)πππ]
+ 2(Ξ¨[πβ£πΏππ+ Ξ¨ππΏ[πβ£π+ Ξ¨π[πβ£π+ Ξ¨[πβ£ππ)ππβ£π]+ (Ξ©ππππβ²πΏππβΞ©ππππβ²+ Ξ©ππππβ²) (4.26) and
0 = β2ΓΎβ²Ξ©ππ +kπ(Ξ¨ππΏππ+ Ξ¨πππ) +(
βΞ©πππβ²+ 2Ξ©πππβ²[ππ])
β4(Ξ¨(ππΏπ)π+ Ξ¨(ππ)π)ππ
+ Ξ¦πππππβΞ¦πππππ+ Ξ¦πππππβΞ¦πππππ+ 2Ξ¦πππππβΞ¦πππ+ Ξ¦πππππππ+ Ξ¦πππ. (4.27) Equation (4.26) is obtained by taking various linear combinations and contractions of the Bianchi equations (B1), while equation (4.27) is constructed from the symmetric part of (B2) and a contraction of (B3). These equations are exact: no decomposition into background and perturbation has been performed at this stage.
Now we consider the linear combination kπ(4.26) +ΓΎ(4.27). This contains second derivatives acting onΞ©and onΞ¨. However, the point of taking this particular combina- tion is that the second derivatives ofΞ¨ occur in the combination β[ΓΎ,kπ](Ξ¨πππ+ Ξ¨ππΏππ) and therefore can be eliminated in favour of terms involving one or zero derivatives of Ξ¨using the formula (C2) for the commutator [ΓΎ,kπ].
We can also symmetrize the entire equation on ππ without losing any useful infor- mation, as the antisymmetric terms do not contain any second derivatives of Ξ©. This
reduces the equation to
0 = β(2ΓΎβ²ΓΎ+kπkπ)Ξ©ππ β2[ΓΎ,ΓΎβ²]Ξ©ππ β[k(πβ£,kπ]Ξ©πβ£π)+ΓΎ(πππππΞ© πβ²ππ)βkπ(πππππΞ© ππβ²) +[ΓΎ,kπ]ππππΞ¨ β4ΓΎ(ππππΞ¨ππ)β2k(πβ£(πβ£π)ππΞ¨ πππ) + 2kπ(π(πβ£ππΞ¨ ππβ£π))β2kπ(ππππΞ¨πππ)
+ΓΎ(πππππΞ¦ πππ)βkπ(πππππΞ¦ π π) (4.28)
where
πππππΞ¦ β‘ Ξ¦(πβ£πβ£π)π+ Ξ¦πΏ(πβ£ππΏβ£π)πβΞ¦SπππΏππ+ (2Ξ¦(πβ£πβΞ¦π(πβ£)πΏπβ£π)+ (2Ξ¦(πβ£πβΞ¦π(πβ£)πΏπβ£π), ππππΞ¨ β‘ Ξ¨(ππ)π+ Ξ¨(ππΏπ)π,
πππππΞ© β‘ βΞ©πππΏππ+ Ξ©(πβ£ππΏπβ£π)βΞ©(πβ£ππΏπβ£π). (4.29) Note that these quantities satisfy the following relations:
πππππΞ¦ =π(πβ£πβ£π)πΞ¦ =ππ(πβ£πβ£π)Ξ¦ , πππππΞ¦ = 0 and πππππΞ¦ =β(πβ2)Ξ¦Sππ+ Ξ¦πΏππ, (4.30) ππππΞ¨ =π(ππ)πΞ¨ , ππππΞ¨ = 0 and ππππΞ¨ = 12πΞ¨π (4.31) πππππΞ© =π(ππΞ©)ππ, πππ(ππ)Ξ© =βΞ©πππΏππ πππππΞ© = 0 and πππππΞ© =β(πβ2)Ξ©ππ. (4.32) In this notation, the parts of (4.26) and (4.27) symmetric on ππ become
ΓΎππππΞ¨ βkππππππΞ© =βπππππΞ© ππβ²+ 2π(πβ£ππΞ¨ ππβ£π)β2ππππΞ¨πππβ2ππ(πβ£πΞ¨ ππππΏπβ£π)βπππππΞ¦ π π (4.33) and
βkπππππΞ¨ + 2ΓΎβ²Ξ©ππ =πππππΞ¦ πππβ4ππππΞ¨ππ+πππππΞ© πβ²ππ. (4.34) Next we perform the following steps:
1. Use the commutator (C2) to eliminate the terms [ΓΎ,kπ]ππππΞ¨ from (4.28) (note that this introduces a new kind of term, of the schematic form πΏΓΎβ²Ξ¨).
2. Expand out the brackets using the Leibniz rule for GHP derivatives.
3. Eliminate the term ΓΎππππΞ¨ using equation (4.33).
4. Use the NP equations (NP1), (NP2) and (NP4)β² to eliminate terms in whichΓΎacts on π, π and πβ² respectively.
5. Take a linear combination of the Bianchi equations (B2,B3,B4) to get an equation ΓΎπππππΞ¦ = ΓΎβ²πππππΞ© +k(πβ£Ξ¨πβ£π)πβkπΞ¨(ππ)πβπΏ(πβ£ππΏβ£π)πkπΞ¨π+πΏππk(πΞ¨π)
+(β2kπΞ¨(πβ£+k(πβ£Ξ¨π)πΏπβ£π)+ (β2kπΞ¨(πβ£+k(πβ£Ξ¨π)πΏπβ£π)+. . . , (4.35) where the ellipsis indicates terms that involve no derivatives. Use this to eliminate ΓΎπππππΞ¦ from (4.28).
4.4. DECOUPLING OF GRAVITATIONAL PERTURBATIONS 95 6. Use a combination of (B5) and (B7) to show that
kππππππΞ¦ = 3ΓΎβ²ππππΞ¨ + 3πππππΞ¦ ππ+. . . (4.36) where the ellipsis denotes ο¬rst order terms not involving any derivatives. Use this to eliminate kππππππΞ¦ from (4.28).
The resulting equation is very long so we shall not write it out in full. It has the schematic form
(ΓΎβ²ΓΎ+k.k+ [ΓΎ,ΓΎβ²] + [k,k] +πβ²ΓΎ+πΓΎβ²+πk+πβ²k+π πβ² +ππβ²+Ξ¦) Ξ© +πΏΓΎβ²Ξ¨+πkΞ¨+ (π π+πβ²π+πΏπβ²+ΓΎβ²πΏ+kπ)Ξ¨
+ (π πΏ+πβ²πΏ+π2)Ξ¦+ (πΏπ)Ξ¨β² = 0 (4.37) Here, we neglect terms that are of quadratic order or higher when we decompose quan- tities into a background piece and a perturbation. Recall thatΞ© and Ξ¨ are ο¬rst order quantities. Note that the only terms containing derivatives of Weyl components other than Ξ©are of the schematic form πΏΓΎβ²Ξ¨ and πkΞ¨. For decoupling to occur, these must vanish for any possible perturbation. We shall now examine the circumstances under which we can eliminate these terms.
The detailed form of the πΏΓΎβ²Ξ¨ terms is
4π πΓΎβ²(Ξ¨(ππ)π+ Ξ¨(ππΏπ)π) (4.38)
IfπΏ(0) β=0 then there is nothing we can do to eliminate these terms. The only Bianchi equation containingΓΎβ²Ξ¨is (B5), and using this again would reintroduce the 1-derivative terms that we have eliminated above. Hence the only way for these terms to drop out is forπΏto vanish in the background. HenceπΏ(0) =0is a necessary condition for decoupling.
Henceforth we assumeπΏis a ο¬rst-order quantity, in which case the above terms become second order terms and can be neglected.
Recall that πΏ(0) = 0 is equivalent to the statement that β is geodesic in the back- ground. By Theorem 3.2, this places no further restrictions on the spacetimes that can be analysed.
Having set πΏ(0) =0, the only remaining terms involving derivatives of Weyl compo- nents other than Ξ©are of the form πkΞ¨. The detailed form of these terms is:
4π(πβ£πkπΞ¨β£π)+πππ[
2kπΞ¨(ππ)π+k(πβ£Ξ¨πβ£π)πβk(πΞ¨π)ππ+ 2kπΞ¨(ππ)πβk(πβ£Ξ¨πβ£π)π] +ππ(πβ£[
βkπΞ¨β£π)ππβkπΞ¨πβ£π)πβkβ£π)Ξ¨π+ 2kπΞ¨β£π)]
(4.39) For decoupling we need to eliminate these terms in favour of terms in which derivatives act only on Ξ©.
Certain combinations of terms of the formkΞ¨can be eliminated using Bianchi equa- tions. In order to understand precisely what kinds of terms can be so eliminated, we can decompose kπΞ¨πππ into parts that transform irreducibly under ππ(πβ2). If we do the same for the Bianchi equations at our disposal (or combinations of them such as (4.27)) then we will see which irreducible parts of kΞ¨can be eliminated from the above equation.
Decomposing into tracefree and trace parts gives, for π >4:
kπΞ¨πππ =πππππ+ 2πΏπ[πβ£ππβ£π]+πΏπππππ+ 2πΏπ[πβ£ππβ£π]+ 2πΏπ[πβ£πΏπβ£π]π, (4.40) whereπππππ is traceless and satisο¬es ππ[πππ]=πππ(ππ) = 0. The other terms are given by
π[ππ] = 1
2πππ = 1 π(πβ4)
(β(πβ3)kπΞ¨[ππ]π+k[πΞ¨π])
, (4.41)
π[ππ] = 1 π(πβ4)
(3kπΞ¨[ππ]πβ(πβ1)k[πΞ¨π])
, (4.42)
π(ππ)= 1 (πβ2)(πβ4)
(β(πβ3)kπΞ¨(ππ)π+k(πΞ¨π)βkπΞ¨ππΏππ)
, (4.43)
π(ππ)= 1 (πβ2)(πβ4)
(kπΞ¨(ππ)πβ(πβ3)k(πΞ¨π)+kπΞ¨ππΏππ)
, (4.44)
π = 1
(πβ2)(πβ3)kπΞ¨π. (4.45)
Note thatπ(ππ) and π(ππ) are traceless and π(ππ)= 0.
The traceless part πππππ can be decomposed further into parts that transform irre- ducibly underππ(πβ2). The relevant irreducible representations correspond to Young tableaux with 4 boxes. However, it turns out that we will not need to discuss these. As well as these quantities, we have two independent quantities transforming as , namely π[ππ] and π[ππ], two quantities transforming as , namely π(ππ) and π(ππ), and a singlet π.
Consider ο¬rst the singletπ. The contribution of this to equation (4.39) is 4(πβ3)ππππ = 4
πβ2πππkπΞ¨π, (4.46) where the shear π is the traceless symmetric part of π. In order to achieve decoupling, we would need to add to (4.39) a combination of Bianchi components containing a singlet term that cancelled this, and did not introduce any 1-derivative terms (e.g. ΓΎΞ¦ terms) that we have already eliminated. However, there is no such combination. For example, the singlet drops out of equation (4.27). Therefore, the only way to eliminate the singlet term from our equation, as required for decoupling, is to set π(0) = 0. This is the
4.4. DECOUPLING OF GRAVITATIONAL PERTURBATIONS 97 condition that, in the background geometry, the shear of the multiple WAND β must vanish. Henceforth we assume that this is the case.
Next consider the traceless symmetric tensors π(ππ) and π(ππ) that arise in the above decomposition ofkπΞ¨πππ. The contribution of these to (4.39) is:
β5π(π(ππ)+π(ππ))+12(πβ10)(
π(ππ)πππ+π(ππ)πππ)
β32(πβ2)(
π(ππ)πππ+π(ππ)πππ)
, (4.47) whereπππ β‘π[ππ].
Now consider again the Bianchi equations. The only combination of equations involv- ing π(ππ) and π(ππ) that does not introduce any 1-derivative terms that we have already eliminated is (4.34), which gives an expression for
kπππππΞ¨ β‘ β(πβ2)(
π(ππ)+π(ππ))
. (4.48)
We can use this to eliminate, say,π(ππ)from (4.47), via the expressionπ(ππ) =βπ(ππ)+. . ., where the ellipsis denotes terms in which derivatives act only onΞ©. Equation (4.47) then reduces to
2(πβ4)(
π(ππ)πππ+π(ππ)πππ)
+. . . . (4.49)
Since we have no independent equation that will allow us to eliminateπ(ππ), we conclude that in order for thekΞ¨terms to decouple we must haveπ =0in the background, i.e., the multiple WANDβ must be shearfreeand rotation free. Note the factor of πβ4: for π= 4, vanishing rotation isnot necessary for decoupling.4
Having set π(0) =π(0) =0, we ο¬nd that that the 1-derivative terms (4.39) reduce to 5π
πβ2
(kπΞ¨(ππ)π+k(πΞ¨π))
= 5π
πβ2kπππππΞ¨, (4.50) whereπβ‘πππ. These terms can be eliminated from (4.37) with equation (4.34).
In the resulting equation, we now use (NP3) to argue thatkππis a ο¬rst order quantity.
It appears only when multiplied byΨ, so such terms are second order and can be dropped.
The only Weyl components that are now acted on by derivatives areΞ©, and the equation has been reduced to the schematic form
(ΓΎβ²ΓΎ+k.k+[ΓΎ,ΓΎβ²]+[k,k]+πβ²ΓΎ+πΓΎβ²+πk+πβ²k+ππβ²+π πβ²+Ξ¦)Ξ©+π2Ξ¦+ππΞ¨= 0 (4.51) At this point, we can also simplify the form of the terms involving Ξ©, by using the commutators (C1,C3) to eliminate the terms of the form [ΓΎ,ΓΎβ²]Ξ©and [k,k]Ξ©respectively, in favour of terms that involve at most ο¬rst derivatives of Ξ©.
4Forπ= 4, Ξ¨πππ=β2πΏπ[πΞ¨π], so the irreducible parts ofkπΞ¨πππare just the trace, tracefree symmetric and antisymmetric parts of kπΞ¨π. Considering the trace gives π = 0 as for π > 4. The tracefree symmetric part can be eliminated with (4.27). The antisymmetric part simply drops out of (4.39), using the fact that all 2Γ2 antisymmetric matrices commute.
The terms of the formΞ¦πare simpliο¬ed by noting that equation (4.27), evaluated in the background geometry implies that
π(0)Ξ¦(0)ππ = 1
πβ2π(0)Ξ¦(0)πΏππ. (4.52) Equation (4.51) now reduces to something suο¬ciently simple to write out explicitly:
(2ΓΎβ²ΓΎ+kπkπ+πβ²ΓΎ+ π+6πβ2πΓΎβ² +πβ22 ππβ²β6ππkπ+ 4Ξ¦βπβ12π Ξ) Ξ©ππ
+(
4ππk(πβ£β4π(πβ£kπ+πβ22 π(πβ²π(πβ£βπβ²(πβ£π) + 4Ξ¦S(πβ£π+ 16Ξ¦A(πβ£π)
Ξ©πβ£π)+ 2Ξ¦ππππΞ©ππ
+ 2π2 πβ2
(Ξ¦Sππ β πβ21 Ξ¦πΏππ)
+ 2πππ(
Ξ¨(ππ)πβΞ¨(ππΏπ)π+πβ22 πΏππΞ¨π)
= 0. (4.53) This equation is the analogue of equation (4.20) for the Maxwell ο¬eld. Note that (4.52) implies thatπ(Ξ¦Sππ βπβ21 Ξ¦πΏππ) is a ο¬rst order quantity.
To achieve decoupling we have to eliminate the terms not involving Ξ©ππ, i.e., those on the ο¬nal line of this equation. For π = 4, this is automatic since the particular combination of Ξ¦ terms appearing in this equation vanishes identically (i.e. Ξ¦Sππ = 12Ξ¦πΏππ if π = 4), as does the particular combination of Ξ¨ terms. For π > 4, the only way of eliminating the Ξ¦terms above is to set π(0) = 0, i.e., take π to be ο¬rst order. All terms on the ο¬nal line above are then of higher order and can be neglected.
Hence we see that, for π >4, decoupling requires that
πΏ(0) =0 =π(0), (4.54)
that is the multiple WAND must be geodesic and free of expansion, rotation and shear.
The existence of such a vector ο¬eld implies, by deο¬nition, that the spacetime is Kundt.
This is a necessary condition for decoupling; it is also suο¬cient since we now have an equation in which the only perturbed Weyl components that appear areΞ©.
The resulting decoupled equation is:
(2ΓΎβ²ΓΎ+kπkπ+πβ²ΓΎβ6ππkπ+ 4Ξ¦β πβ12π Ξ) Ξ©ππ
+ 4(
ππk(πβ£βπ(πβ£kπ+ Ξ¦S(πβ£π+ 4Ξ¦A(πβ£π)
Ξ©πβ£π)+ 2Ξ¦ππππΞ©ππ= 0. (4.55) We remind the reader that Ξ©is a ο¬rst order quantity, so quantities multiplying Ξ©(e.g.
Ξ¦,π) must be evaluated in the background geometry.
4.4.3 Comment on the expanding case
Just as we did for Maxwell perturbations, it is interesting to consider what happens if β is geodesic with vanishing rotation and shear, but non-vanishing expansion (i.e. the spacetime is Robinson-Trautman). Under these circumstances, we have equation (4.53),
4.4. DECOUPLING OF GRAVITATIONAL PERTURBATIONS 99 a perturbation equation for a gauge invariant quantity Ξ©. However, it contains two terms that obstruct the decoupling of the equation. It is interesting to ask how these terms are consistent with gauge invariance. The answer is supplied by:
Lemma 4.3 Letβbe an expanding, non-twisting, non-shearing geodesic multiple WAND for an Einstein spacetime of dimension π >4. Then
(Ξ¦Sππ β πβ21 Ξ¦πΏππ)(1)
(4.56) is a gauge invariant quantity. If π(0) β=0, then
Ξ¨(1)π and ππ(0)Ξ¨(1)πππ (4.57) also are gauge invariant quantities.
The Schwarzschild black hole in arbitrary dimension is an example of a spacetime ad- mitting such a multiple WAND (although in this case, π(0) = 0). In four dimensions, (4.56) vanishes identically in all spacetimes, while the quantities (4.57) are not gauge invariant.
Proof: From equation (4.52) we have
Ξ¦(0)ππ = πβ21 Ξ¦(0)πΏππ (4.58) in any such spacetime. Hence we see immediately that (4.56) is invariant under inο¬nites- imal coordinate transformations, and also under inο¬nitesimal spins. Furthermore, Ref.
[158] showed that all such spacetimes are of algebraic Type D so we can choose our basis so that all Weyl tensor components with non-zero boost weight vanish. Under an inο¬nitesimal null rotation aboutβ, equation (2.58) implies that, to ο¬rst order inπ§π,
Ξ¦Sππ 7βΞ¦Sππ+π§(πΞ¨π)βπ§πΞ¨(ππ)π, (4.59) but Ξ¨ is a ο¬rst order quantity and hence Ξ¦S(1)ππ and Ξ¦(1) are both invariant in a Type D background. An identical argument applies to null rotations aboutπ, and hence (4.56) is a gauge invariant quantity.
For an algebraically special spacetime, Ξ¨πππ and Ξ¨π both vanish in the background, and so, to ο¬rst order, they are invariant under inο¬nitesimal spins and inο¬nitesimal coor- dinate transformations. They are also invariant under inο¬nitesimal null rotations about β, as these can only introduce terms involving Ξ©ππ which also vanishes in the background.
We now consider the eο¬ect of an inο¬nitesimal null rotation about π. Taking the prime of (2.61) implies that, to linear order,
Ξ¨πππ7βΞ¨πππ+πβ22 Ξ¦(0)πΏπ[ππ§π]+π§πΞ¦(0)ππππ (4.60)
and
Ξ¨π 7βΞ¨π β πβ1πβ2Ξ¦(0)π§π, (4.61) where we have used (4.58). We will show that the quantities (4.57) are invariant under this transformation ifππ(0) β= 0.
Take a double trace of the Bianchi equation (B7) for the background spacetimes.
This implies that (πβ4)kπΞ¦(0) = 0, and hence, forπ >4, kπΞ¦(0) = 0. The trace of (B5) gives
kπΞ¦(0)= πβ1πβ3ππ(0)Ξ¦(0), (4.62) and hence Ξ¦(0) = 0 if ππ(0) β= 0. From (4.58) we then have Ξ¦(0)ππ = 0. Putting these results back into (B5) implies that Ξ¦(0)ππππππ(0) = 0. Inserting these results into (4.60,4.61) implies that, although Ξ¨πππ is not invariant under inο¬nitesimal null rotations about π, both ππΞ¨πππ and Ξ¨π are invariant, and hence both of these are new gauge invariant quantities, provided that π >4 andππ(0) β= 0. β‘