Gravity and Strings

3.2 Gravity as a self-consistent massless spin-2 SRFT 89
3.2.7 Deser’s argument
In [299] Deser presented an argument that allows one to see GR as the self-consistent SRFT
of a spin-2 particle we were looking for in the sense that, in GR, the gravitational field
couples to its own energy–momentum tensor, at least for a certain choice of field variables,
Lagrangian, and energy–momentum tensor. The emphasis is on physical consistency rather
than on gauge-invariance, and, therefore, the choice of energy–momentum tensor is not
based on that criterion, as in our previous discussions about the Noether method. These
would be weak points if we wanted to take this work as proof of the uniqueness of GR as a
solution to our initial problem, but we should understand Deser’s work as a proof that GR
is a solution to our problem from the physical standpoint.
The starting point in Deser’s argument is a first-order version of the Fierz–Pauli action
that uses two (off-shell) independent fields ϕ µν and Ɣµν ρ (see [841] for a construction of
this action),
S (1)
FP [ϕµν ,Ɣµν ρ ] = 1
χ 2
d d x −χϕ µν 2∂[µƔρ]ν ρ + η µν 2Ɣλ[µ ρ Ɣρ]ν λ , (3.221)
which are Lorentz tensors symmetric in the pair of indices µν. This action is invariant up
to a total derivative under the gauge transformations
δɛϕµν =−2∂(µɛν) + ηµν∂ρɛ ρ , δɛƔµνρ =−χ∂µ∂νɛρ, (3.222)
and it is equivalent on-shell to the Fierz–Pauli action because it gives the same equations
of motion: the equations of motion of the fields ϕ µν and Ɣµν ρ are
χ δS(1)
δϕ µν =−∂(µƔν)ρ ρ + ∂ρƔµν ρ = 0,
2
δS(1)
χ
δƔµν ρ = 2Ɣρ (µν) − η µν Ɣρλ λ − η τσ Ɣτσ (µ η ν) ρ − χ∂ρϕ µν + χηρ (µ ∂σ ϕ ν)σ = 0.
The second equation is just a constraint for Ɣµν ρ .Oncontracting it with η ρσ ,weobtain
(3.223)
η ρσ Ɣρσ ν = χ∂λϕ λν , (3.224)
and, on contracting instead with ηµν and using the last result, we find
Ɣρλ λ =− 1
d − 2 χ∂ρϕ, ϕ = ϕµ µ . (3.225)
Using now these two last equations in the equation for Ɣµν ρ ,weobtain
Ɣρµν + Ɣνρµ = χ∂ρhµν, hµν = ϕµν − 1
d − 2 ηµνϕ. (3.226)
In order to solve for Ɣ, weadd to this equation (ρµν) the permutation µνρ and subtract
the permutation νρµ, obtaining, finally,
Ɣρµν = 1
2 χ{∂ρhµν + ∂µhνρ − ∂νhρµ}. (3.227)