Gravity and Strings

64 A perturbative introduction to general relativity
By subtracting the trace of this equation we can simplify it without any loss of information:
ˆDµν(h) ≡ Dµν(h) − 1
d − 2 ηµνDρ ρ (h) = ∂ 2 hµν + ∂µ∂νh − 2∂ λ ∂(µhν)λ = 0. (3.86)
Sometimes the equation of motion (3.85) is written in terms of the convenient variable ¯hµν:
Dµν( ¯h) = ∂ 2 ¯hµν − 2∂ λ ∂(µ ¯hν)λ + ηµν∂λ∂σ ¯h λσ = 0, (3.87)
where
¯hµν ≡ hµν − 1
2ηµνh. (3.88)
Finally, we can write the Fierz–Pauli wave operator as the divergence of a tensor η µνρ ,
D νρ (h) = 2∂µη νρµ , (3.89)
but the tensor η µνρ is not uniquely defined. Some possible candidates are
where
K µνρσ = 1
2
H µσ νρ = 1
2
η νρµ
T
η νρµ
LL
η νρµ
AD
= η(νρ)µ
T
= ην[ρµ]
LL
= η[ν|ρ|µ]
AD
=−∂σ H µσ νρ ,
=−∂σ K νσρµ ,
=−∂σ K νµρσ ,
η µσ ¯h νρ + η νρ ¯h µσ − η µρ ¯h νσ − η νσ ¯h µρ ,
η σρ ¯h µν + η σν ¯h µρ − η νρ ¯h µσ − η µσ ¯h νρ ,
(3.90)
(3.91)
H is symmetric in the last two indices and K is antisymmetric. In fact, K has exactly the
same symmetries as the Riemann tensor (in the Levi-Cività case).
has the defining property
On the other hand, η µνρ
T
∂LFP
∂∂µhνρ
= η νρµ
T , (3.92)
for the Fierz–Pauli Lagrangian written in Eq. (3.84).
Using any of the last two ηνρµ s, the fact that the Fierz–Pauli wave operator D µν (h) is
divergenceless becomes manifest.
Let us now determine the gauge symmetry of the Fierz–Pauli Lagrangian. Under a general
variation of hµν, the variation of the action is, up to a total derivative,
δSFP =− 1
2
d d x D µν δhµν. (3.93)
If δhµν is a gauge transformation, we know that, up to total derivatives, the integrand of the
variation of the action has to be proportional to the gauge identity Eq. (3.61), i.e.
d d x D µν
δhµν ∼ d d x ∂µD µν ɛν, (3.94)
(the gauge parameter ɛµ(x) has to be a local Lorentz vector). On integrating the r.h.s. by