UFIFT-QG-10-08
Scalar Contribution to the Graviton Self-Energy during Inflation
Sohyun Park† and R. P. Woodard‡
Department of Physics
University of Florida
Gainesville, FL 32611
ABSTRACT
We use dimensional regularization to evaluate the one loop contribution to
the graviton self-energy from a massless, minimally coupled scalar on a locally de Sitter background. For noncoincident points our result agrees with
the stress tensor correlators obtained recently by Perez-Nadal, Roura and
Verdaguer. We absorb the ultraviolet divergences using the R2 and C 2 counterterms first derived by ’t Hooft and Veltman, and we take the D = 4 limit
of the finite remainder. The renormalized result is expressed as the sum of
two transverse, 4th order differential operators acting on nonlocal, de Sitter
invariant structure functions. In this form it can be used to quantum-correct
the linearized Einstein equations so that one can study how the inflationary
production of infrared scalars affects the propagation of dynamical gravitons
and the force of gravity.
PACS numbers: 04.62.+v, 98.80.Cq, 04.60.-m
†
‡
e-mail: spark@phys.ufl.edu
e-mail: woodard@phys.ufl.edu
1
Introduction
The linearized equations for all known force fields do two things:
• They give the linearized force fields induced by sources; and
• They describe the propagation of dynamical particles which carry the
force but are, in principle, independent of any source.
This is the classic distinction between the constrained and unconstrained
parts of a force field. In electromagnetism it amounts to the Coulomb potential versus photons. In gravity there is the Newtonian potential, plus its
three relativistic partners, versus gravitons.
Quantum corrections to the linearized field equations derive from how
the 0-point fluctuations of various fields in whatever background is assumed,
respond to the linearized force fields. These quantum corrections do not
change the dichotomy between constrained and unconstrained fields but they
can, of course, modify classical results. Around flat space background there
is no effect, after renormalization, on the propagation of dynamical photons
or gravitons but there are small corrections to the Coulomb and Newtonian
potentials. As might be expected, the long distance effects are greatest for
the 0-point fluctuations of massless particles and they take the form required
by perturbation theory and dimensional analysis [1, 2],
∆Φ Φ
Coul.
∼−
e2 r ln
h̄c
r0
,
∆Φ Φ
Newt.
∼−
h̄G
,
c3 r 2
(1)
where r is the distance to the source, r0 is the point at which the renormalized
charge is defined, and the other constants have their usual meanings.
Schrödinger was the first to suggest that the expansion of spacetime can
lead to particle production by ripping the virtual particles (which are implicit
in 0-point fluctuations) out of the vacuum [3]. Following early work by
Imamura [4], the first quantitative results were obtained by Parker [5]. He
found that the effect is maximized during accelerated expansion, and for
massless particles which are not conformally invariant [6], such as massless,
minimally coupled (MMC) scalars and (as noted by Grishchuk [7]) gravitons.
The de Sitter geometry is the most highly accelerated expansion consistent with classical stability. For de Sitter background with Hubble constant
H and scale factor a(t) = eHt it is simple to show that the number of MMC
1
scalars, or either polarization of graviton, created with wave vector ~k is [8],
Ha(t) 2
.
N (t, ~k) =
2ck~kk
(2)
It is these particles which comprise the scalar and tensor perturbations produced by inflation [9], the scalar contribution of which has been imaged [10].
Of course the same particles also enter loop diagrams to cause an enormous
strengthening of the quantum effects caused by MMC scalars and gravitons.
A number of analytic results have been obtained for one loop corrections to
the way various particles propagate on de Sitter background and also to how
long range forces act:
• In MMC scalar quantum electrodynamics, infrared photons behave as if
they had an increasing mass [11], and the charge screening very quickly
becomes nonperturbatively strong [12], but there is no big effect on
scalars [13];
• For a MMC scalar which is Yukawa-coupled to a massless fermion,
infrared fermions behave as if they had an increasing mass [14] but the
associated scalars experience no large correction [15];
• For a MMC scalar with a quartic self-interaction, infrared scalars behave as if they had an increasing mass (which persists to two loop
order) [16];
• For quantum gravity minimally coupled to a massless fermion, the
fermion field strength grows without bound [17]; and
• For quantum gravity plus a MMC scalar, the scalar shows no secular effect but its field strength may acquire a momentum-dependent
enhancement [18].
The great omission from this list is how inflationary scalars and gravitons
affect gravity, both as regards the propagation of dynamical gravitons and as
regards the force of gravity. This paper represents a first step in completing
the list.
One includes quantum corrections to the linearized field equation by subtracting the integral of the appropriate one-particle-irreducible (1PI) 2-point
function up against the linearized field. For example, a MMC scalar ϕ(x)
2
in a background metric gµν (x) whose 1PI 2-point function is −iM 2 (x; x′ ),
would have the linearized effective field equation,
∂µ
h√
i
µν
Z
gg ∂ν ϕ(x) − d4 x′ M 2 (x; x′ )ϕ(x′ ) = 0 .
(3)
To include gravity on the list we must therefore compute the graviton selfenergy, either from MMC scalars or from gravitons, and then use it to correct
the linearized Einstein equation. In this paper we shall evaluate the contribution from MMC scalars; a subsequent paper will solve the linearized effective
field equations to determine quantum corrections to the propagation of gravitons and the gravitational response to a point mass.
It should be noted that the vastly more complicated contribution from
gravitons was derived some time ago [19]. However, that result is not renormalized, and is therefore only valid for noncoincident points. To use the
graviton self-energy in an effective field equation such as (3), where the integration carries x′µ over xµ , one must extract differential operators until the
remaining structure functions are integrable. That is the sort of form we
will derive, using dimensional regularization to control the divergences and
BPHZ counterterms to subtract them.
This paper contains five sections. In section 2 we give those of the Feynman rules which are needed for this computation, and we describe the geometry of our D-dimensional, locally de Sitter background. Section 3 derives
the relatively simple form for the D-dimensional graviton self-energy with
noncoincident points. We show that this version of the result agrees with the
stress tensor correlators recently derived by Perez-Nadal, Roura and Verdaguer [20]. Section 4 undertakes the vastly more difficult reorganization which
must be done to isolate the local divergences for renormalization. At the end
we subtract off the divergences with the same counterterms originally computed for this model in 1974 by ’t Hooft and Veltman [21], and we take the
unregulated limit of D = 4. Our discussion comprises section 5.
2
Feynman Rules
In this section we derive Feynman rules for the computation. We start with
expressing the full metric as
gµν = g µν + κhµν ,
3
(4)
where g µν is the background metric, hµν is the graviton field whose indices
are raised and lowered with the background metric, and κ2 ≡ 16πG is the
loop counting parameter of quantum gravity. Expanding the MMC scalar
Lagrangian around the background metric we get interaction vertices between
the scalar and dynamical gravitons. We take the D-dimensional locally de
Sitter space as our background and introduce de Sitter invariant bi-tensors
which will be used throughout the calculation. We close this section by
providing the MMC scalar propagator on the de Sitter background.
2.1
Interaction Vertices
The Lagrangian which describes pure gravity and the interaction between
gravitons and the MMC scalar is,
L=
i
√
1 h
1
R − (D−1)(D−2)H 2 − ∂µ ϕ∂ν ϕg µν −g .
16πG
2
(5)
where R is Ricci scalar, G is Newton’s constant and H is the Hubble constant.
Computing the one loop scalar contributions to the graviton self-energy
consists of summing the 3 Feynman diagrams depicted in Figure 1.
x
x′
+
x
+
×
x
Figure 1: Graviton self-energy coupled to the MMC scalars at one loop order.
The sum of these three diagrams has the following analytic form:
−i[µν Σρσ ](x; x′ )
2
2
X
1X
µναβ
=
TJρσγδ (x′ ) × ∂α ∂γ′ i△(x; x′ ) × ∂β ∂δ′ i△(x; x′ )
TI
(x)
2 I=1
J=1
+
4
1X
FIµνρσαβ (x) × ∂α ∂β′ i△(x; x′ ) × δ D (x − x′ )
2 I=1
+2
2
X
I=1
CIµνρσ (x) × δ D (x − x′ ) .
4
(6)
The 3-point and 4-point vertex factors TIµναβ and FIµνρσαβ derive from expanding the MMC scalar Lagrangian using (4),
√
1
− ∂µ ϕ∂ν ϕg µν −g
(7)
2
√
1
√
1
κ
= − ∂µ ϕ∂ν ϕg µν −g − ∂µ ϕ∂ν ϕ hg µν − hµν −g
2
2
2
)
(
i
h1
√
1 µν
κ2
1
µν
2
ρσ
µ ρν
−g + O(κ3 ) . (8)
− ∂µ ϕ∂ν ϕ
h − h hρσ g − hh +h ρ h
2
8
4
2
The resulting 3-point and 4-point vertex factors are given in the Tables 1
and 2, respectively. The procedure to get the counterterm vertex operators
CIµνρσ (x) is given in section 4.
TIµναβ
√
− iκ
−g g µν g αβ
2
√
+iκ −g g µ(α g β)ν
I
1
2
Table 1: 3-point vertices TIµναβ where g µν is the de Sitter background metric
and κ2 ≡ 16πG
I
1
2
3
4
2
+ iκ2
√
FIµνρσαβ
2√
− iκ4 −g g µν g ρσ g αβ
2√
+ iκ2 −g g µ(ρ g σ)ν g αβ
−g g µ(α g β)ν g ρσ + g µν g ρ(α g β)σ
√
−2iκ2 −g g α(µ g ν)(ρ g σ)β
Table 2: 4-point vertices FIµνρσαβ where g µν is the de Sitter background metric
and κ2 ≡ 16πG
These expressions for interaction vertices are valid for any background
metric g µν . In the next subsections we introduce the locally de Sitter space
as our background geometry and give the scalar propagtor i△(x; x′ ) on it.
5
2.2
Working on de Sitter Space
We specify our background geometry as the open conformal coordinate submanifold of D-dimensional de Sitter space. A spacetime point xµ = (η, xi )
takes values in the ranges
−∞ < η < 0
− ∞ < xi < +∞ .
and
(9)
In these coordinates the invariant element is,
ds2 ≡ g µν dxµ dxν = a2 ηµν dxµ dxν ,
(10)
where ηµν is the Lorentz metric and a = −1/Hη is the scale factor. The
Hubble parameter H is constant for the de Sitter space. So in terms of ηµν
and a our background metric is
g µν ≡ a2 ηµν .
(11)
De Sitter space has the maximam number of space-time symmetries in a
given dimension. For our D-dimensional conformal coordinates the 21 D(D+1)
de Sitter transformations can be decomposed as follows:
• Spatial transformations - (D − 1) transformations.
η ′ = η , x′i = xi + ǫi .
(12)
• Rotations - 21 (D − 1)(D − 2) transformations.
η ′ = η , x′i = Rij xj .
(13)
• Dilation - 1 transformation.
η ′ = kη , x′i = kxj .
(14)
• Spatial special conformal transformations - (D − 1) transformations.
xi − θ i x · x
η =
, x =
.
1 − 2θ~ · ~x+ k θ~ k2 x · x
1 − 2θ~ · ~x+ k θ~ k2 x · x
′
η
′
6
(15)
It turns out that the MMC scalar contribution to the graviton self-energy
is de Sitter invariant. This suggests to express it in terms of the de Sitter
length function y(x; x′ ),
′
′
y(x; x ) ≡ aa H
2
"
#
2
2
′
~
x − x~′ − |η−η |−iǫ
.
(16)
Except for the factor of iǫ (whose purpose is to enforce Feynman boundary
conditions) the function y(x; x′ ) is closely related to the invariant length
ℓ(x; x′ ) from xµ to x′µ ,
y(x; x′ ) = 4 sin2
1
2
Hℓ(x; x′ ) .
(17)
With this de Sitter invariant quantity y(x; x′ ), we can form a convenient
basis of de Sitter invariant bi-tensors. Note that because y(x; x′ ) is de Sitter
invariant, so too are covariant derivatives of it. With the metrics g µν (x) and
g µν (x′ ), the first three derivatives of y(x; x′ ) furnish a convenient basis of de
Sitter invariant bi-tensors [13],
∂y(x; x′ )
0
′
=
Ha
+2a
yδ
H∆x
,
µ
µ
∂xµ
∂y(x; x′ )
0
′
yδ
−2aH∆x
=
Ha
ν ,
ν
∂x′ν
∂ 2 y(x; x′ )
2
′
0 0
′
0
0
=
H
aa
yδ
δ
+2a
H∆x
δ
−2aδ
H∆x
−2η
.
µ
ν
µν
µ
ν
ν
µ
∂xµ ∂x′ν
(18)
(19)
(20)
Here and subsequently ∆xµ ≡ ηµν (x−x′ )ν .
Acting covariant derivatives generates more basis tensors, for example
[13],
D2 y(x; x′ )
= H 2 (2−y)g µν (x) ,
µ
ν
Dx Dx
D2 y(x; x′ )
= H 2 (2−y)g µν (x′ ) .
Dx′µ Dx′ν
(21)
(22)
The contraction of any pair of the basis tensors also produces more basis
tensors [13],
g µν (x)
∂y ∂y
∂xµ ∂xν
= H 2 4y − y 2 = g µν (x′ )
7
∂y ∂y
,
∂x′µ ∂x′ν
(23)
∂y ∂ 2 y
∂xν ∂xµ ∂x′σ
∂ 2y
∂y
g ρσ (x′ ) ′σ µ ′ρ
∂x ∂x ∂x
2
∂
y
∂ 2y
g µν (x) µ ′ρ ν ′σ
∂x ∂x ∂x ∂x
∂ 2y
∂ 2y
g ρσ (x′ ) µ ′ρ ν ′σ
∂x ∂x ∂x ∂x
∂y
,
∂x′σ
∂y
= H 2 (2 − y) µ ,
∂x
= H 2 (2 − y)
g µν (x)
2.3
(24)
(25)
∂y ∂y
,
∂x′ρ ∂x′σ
∂y ∂y
= 4H 4 g µν (x) − H 2 µ ν .
∂x ∂x
= 4H 4 g ρσ (x′ ) − H 2
(26)
(27)
Scalar Propagator on de Sitter
From the MMC scalar Lagrangian of (5) we see that the propagator obeys
i
h√
√
(28)
∂µ −g g µν ∂ν i△(x; x′ ) = −g i△(x; x′ ) = iδ D (x − x′ )
The scalar propagator on de Sitter is expressed in terms of the de Sitter
length function y(x; x′ ) [22],
i△(x; x′ ) = A y(x; x′ ) + k ln(aa′ ) .
Here the constant k is given as,
k≡
(29)
H D−2 Γ(D−1)
,
D
D
(4π) 2 Γ( 2 )
(30)
and the function A(y) has the expansion,
πD Γ(D−1)
H D−2 Γ( D2 ) 4 D2 −1 Γ( D2 +1) 4 D2 −2
+
−π
cot
A(y) ≡
D
D
D
2
−2 y
Γ( D2 )
(4π) 2 2 −1 y
2
(
+
Γ(n+ D2 +1) y n− D2 +2
−
4
4
n− D2 +2 Γ(n+2)
"
∞
X
1 Γ(n+D−1) y n
n=1
n Γ(n+ D2 )
1
#)
. (31)
The infinite series terms of A(y) vanish for D = 4, so they only need to be
retained when multiplying a potentially divergent quantity, and even then
one only needs to include a handful of them. This makes loop computations
manageable.
We note that the MMC scalar propagator (29) has the de Sitter breaking
term, k ln(aa′ ). However the graviton self-energy only involves the terms
like ∂α ∂β′ i△(x; x′ ). Being differentiated twice, once with respect to x and
then with respect to x′ , the de Sitter breaking term drops out. So the scalar
contribution to the graviton self-energy remains de Sitter invariant.
8
3
One Loop Graviton Self-energy
In this section we calculate the primitive diagrams. It turns out the the
graviton self-energy with coincident point vanishs in D = 4 dimensions. We
reach the relatively simple form for the D-dimensional graviton self-energy
with noncoincident points. We check that this result agrees with the stress
tensor correlator recently derived by Perez-Nadal, Roura and Verdaguer [20].
3.1
No Contributions from the 4-point Vertices
The 4-point contributions,
4
1X
FIµνρσαβ (x) × ∂α ∂β′ i△(x; x′ ) × δ D (x − x′ ) ,
2 I=1
(32)
vanish on de Sitter space. Because the sum contains a factor of (D − 4)
and there is no divergence from the coincidence limit of the propagator in
dimensional regularization, the 4-point diagrams make no net contribution
to the final, renormalizaed result.
To check this, we calculate the 4-point contributions explicitly. In the
coincidence limit x′ = x we have
∂y(x; x′ )
0
′
lim
=
lim
Ha
yδ
+2a
H∆x
=0,
(33)
µ
µ
x′ →x
x′ →x
∂xµ
∂y(x; x′ )
′
0
lim
=
lim
Ha
yδ
−2aH∆x
(34)
ν = 0 ,
ν
x′ →x
x′ →x
∂x′ν
∂ 2 y(x; x′ )
2
′
0 0
′
0
0
=
lim
H
aa
yδ
δ
+2a
H∆x
δ
−2aδ
H∆x
−2η
lim
µ
ν
µν
µ
ν
ν
µ
x′ →x
x′ →x ∂xµ ∂x′ν
(35)
= −2H 2 g µν .
Therefore,
lim
x′ →x
∂µ ∂ν′ i△(x; x′ )
∂y ∂y
∂ 2y
′
A
= lim
+
A
x′ →x
∂xµ ∂x′νh
∂xµi∂x′ν
= 0 + A′ (y = 0) × −2H 2 gµν
′′
(36)
From the definition of A(y), A′ (y) is
4 D −1
D
1 H D−2
D 4 D2
2
−Γ(
A (y) =
−Γ(
)
+1)
4 (4π) D2
2 y
2
y
′
(
)
4 D −2
Γ(D)
1 D
2
+ · · ·−0+ D
− Γ( +2)
,
2 2
y
Γ( 2 +1)
9
(37)
Now we recall that in dimensional regularization any D-dependent power of
vanishes. Then we see that when y = 0, the first three terms of this (37)
vanish in the dimensional regularization scheme. Thus,
A′ (y = 0) =
1 H D−2 Γ(D)
,
4 (4π) D2 Γ( D2 + 1)
(38)
and therefore we have
∂µ ∂ν′ i△(x; x′ )δ D (x − x′ ) = lim
∂µ ∂ν′ i△(x; x′ ) = −
′
x →x
1 HD
Γ(D)
g .(39)
D
2 (4π) 2 Γ( D2 + 1) µν
So the 4-point contributions become
4
X
I=1
FIµνρσαβ (x) × ∂α ∂β′ i△(x; x′ ) × δ D (x − x′ )
√ 1 µν ρσ αβ 1 µ(ρ σ)ν αβ
1 HD
Γ(D)
2
=−
g × iκ −g − g g g + g g g
2 (4π) D2 Γ( D2 + 1) αβ
4
2
i
h
1
+ g µ(α g β)ν g ρσ + g µν g ρ(α g β)σ − 2g α(µ g ν)(ρ g σ)β
2
√ 1 µν ρσ 1 µ(ρ σ)ν 1 HD
Γ(D)
2
=−
(D − 4) . (40)
iκ −g − g g + g g
2 (4π) D2 Γ( D2 + 1)
4
4
We explicitly see that with no divergence factors multiplied by (D − 4), this
becomes zero in D = 4.
3.2
Contributions from the 3 point vertices
In this section we divide our discussion into two subsections: first for the
graviton self-energy in terms of A(y) and second for the decomposition of it
into two parts.
3-point contributions in terms of A(y)
3.2.1
In this section we calcalate the graviton self-energy with noncoincident points,
−i[µν Σρσ ]3-point (x; x′ )
=
2
2
X
1X
TJρσγδ (x′ ) × ∂α ∂γ′ i△(x; x′ ) × ∂β ∂δ′ i△(x; x′ ) .
TIµναβ (x)
2 I=1
J=1
10
(41)
Using the de Sitter background MMC scalar propagator (29), we can express
the 3-point contributions in terms of the de Sitter invariant function A(y)
given in equation (31). The key computation is,
∂y
+ kHa′ δγ0 ,
(42)
∂x′γ
∂y ∂y
∂ 2y
∂α ∂γ′ i△(x; x′ ) = A′′ α ′γ + A′ α ′γ + 0 .
(43)
∂x ∂x
∂x ∂x
We note that the de Sitter breaking term drops out being differenciated twice
and therefore the graviton self-energy turns out de Sitter invariant.
Multiplying by one more propagator with derivatives and contracting
with the vertex factors we can arrange the 3-point contribution in terms of
the five basis tensors given in section 2.2:
∂γ′ i△(x; x′ ) = A′
−i[µν Σρσ ]3-point (x; x′ )
(
i
√ q ′
∂ 2y h
∂ 2y
α
(A)
= −g −g
A
∂xµ ∂x′(ρ ∂x′σ) ∂xν
i
∂y
∂y h
∂ 2y
β
(A)
A
∂x(µ ∂xν) ∂x′(ρ ∂x′σ)
i
∂y ∂y ∂y ∂y h
γ
(A)
+
A
∂xµ ∂xν ∂x′ρ ∂x′σ
+
h
+g µν g ′ρσ H 4 δA (A)
i
i
∂y ∂y ′ρσ i 2 h
∂y ∂y
+
g
+g
H
ǫ
(A)
A
∂x′ρ ∂x′σ ∂xµ ∂xν
h
µν
)
.
(44)
Here the coefficients are funtions of A(y):
i
h 1
αA (A) = κ2 − (A′ )2 ,
h 2
i
2
βA (A) = κ −A′ A′′ ,
h 1
i
γA (A) = κ2 − (A′′ )2 ,
2 1 h ′′ 2
(A ) (4y − y 2 )2 + 2A′ A′′ (2 − y)(4y − y 2 )
δA (A) = κ2 −
8
i
+(A′ )2 [4(D − 4) − (4y − y 2 ) ,
h
i
2 1
ǫA (A) = κ
(4y − y 2 )(A′′ )2 + 2(2 − y)A′ A′′ − (A′ )2 .
4
11
(45)
(46)
(47)
(48)
(49)
We can check this result agrees with the stress tensor correlators recently
derived by Perez-Nadal, Roura and Verdaguer [20]. We compare our result
for the MMC scalar contribution to the graviton self-energy, (44) with the
stress tensor correlators Fµνρσ in the massless limit which is from [20],
Fµνρσ = ∂µ ∂ρ′ G+ ∂ν ∂σ′ G+ + ∂µ ∂σ′ G+ ∂ν ∂ρ′ G+
′
−g µν ∂α ∂ρ′ G+ ∂ α ∂σ′ G+ − g ′ρσ ∂µ ∂α′ ′ G+ ∂ν ∂ ′α G+
1
′
+ g µν g ′ρσ ∂α ∂α′ ′ G+ ∂ α ∂ ′α G+ .
2
(50)
For comparison we first note the conversion of the essential variables. Our
biscalar function y(x; x′ ) corresponds to Z(x; x′ ) with the relation,
Z =1−
y
.
2
(51)
In the massless limit, the function G+ in equation (50) becomes
∞
X
Γ(D − 1 + n)Γ(n) 1 1 + Z n
1
.
G = G(Z) =
(4π)D/2 n=0
n!
2
Γ( D2 + n)
+
(52)
Recalling the hypergeometric function,
2 F1 (α, β; γ; z) =
∞
X
Γ(α + n) Γ(β + n)
n=0
Γ(α)
Γ(β)
Γ(γ) z n
,
Γ(γ + n) n!
(53)
we see that G(Z) can be written as
G(Z) = G(y) =
D
Γ(D − 1)Γ(0)
1
y
;1 − ) .
2 F1 (D − 1, 0;
D
D/2
(4π)
2
4
Γ( 2 )
(54)
Now we use one of the transformation formulas for the hypergeometric functions (See for example, 9.131 of Gradshteyn and Ryzhik [32]) to write G as
the series expansion of y/4:
H D−2 Γ( D2 ) 4 D2 −1 Γ( D2 +1) 4 D2 −2
Γ(D−1)
G(y) =
+ D
−Γ(0)
D
D
Γ( D2 )
−2 y
(4π) 2 2 −1 y
2
(
+
Γ(n+ D2 +1) y n− D2 +2
−
4
4
n− D2 +2 Γ(n+2)
"
∞
X
1 Γ(n+D−1) y n
n=1
n Γ(n+ D2 )
1
12
#)
. (55)
Here we recover H for which they used H = 1 unit. We see that this is
almost the same as the function A(y) except the singular term
−Γ(0)
Γ(D−1)
,
Γ( D2 )
(56)
which we regulated using the de Sitter breaking term as
−π cot
πD Γ(D−1)
2
Γ( D2 )
.
(57)
Noting that the stress tensor correlator (50) only involves derivatives of G(y)
this constant term does not effect the answer. Then we can check that
the stress tensor correlator (50) agrees with our result for the MMC scalar
contributions to the graviton self-energy:
Fµνρσ = ∂µ ∂ρ′ A(y)∂ν ∂σ′ A(y) + ∂µ ∂σ′ A(y)∂ν ∂ρ′ A(y)
′
−g µν ∂α ∂ρ′ A(y)∂ α ∂σ′ A(y) − g ′ρσ ∂µ ∂α′ ′ A(y)∂ν ∂ ′α A(y)
1
′
+ g µν g ′ρσ ∂α ∂α′ ′ A(y)∂ α ∂ ′α A(y) ,
2
1
4
= − 2 × √ √ ′ × −i[µν Σρσ ]3-point (x; x′ ) .
κ
−g −g
3.2.2
(58)
(59)
Separation into spin 0 and spin 2 parts
Now provided by the gauge symmetry of our Lagrangian (5), it is convenient
to decompose the graviton self-energy into spin 0 and spin 2 parts:
√ q ′n
′
−i[µν Σρσ ]3-point (x; x ) = −g −g [transverse, spin 0 operator]F1 (y)
o
+[transverse, traceless, spin 2 operator]F2 (y) .
(60)
Here the structure functions F1 (y) and F2 (y) are scalar functions of y. Actually they are arbitrary scalar functions, but because the result for the graviton
self-energy is de Sitter invariant we can assume the scalar is a function of the
de Sitter invariant length, y.
We can guess, from the flat space case which is summarized in section 4,
the de Sitter generalization of the spin 0 operator as
[spin 0 operator]
= [ g µν − Dµ Dν + (D − 1)H 2 g µν ][
13
′ ′
g ρσ
− Dρ′ Dσ′ + (D − 1)H 2 g ′ρσ ] .(61)
The question arises as to what the de Sitter generalization of the spin 2
operator is. It is hard to guess because it involves mixed index groups: µν
is an index group associated to x and ρσ is another index group associated
to x′ , and unlike ηµν which is the same everywhere, g µν (x) 6= g µν (x′ ). So our
strategy is to construct a tranverse, traceless spin 2 operator using the Weyl
tensor. We recall that the Weyl tensor is traceless. It is not transverse in
general but in the linearized order which we actually consider, it is transverse.
Varying the linearized Weyl tensor with respect to the graviton field hµν we
can form a transverse, traceless differential operator,
[spin 2 operator]
′ ′ ′ ′
∂δC αβγδ (x) ∂δC α β γ δ (x′ ) h ∂ 2 y
∂ 2y
∂ 2y
∂ 2y i
=
. (62)
′
′
′
∂hµν
∂h′ρσ
∂xα ∂x′α ∂xβ ∂x′β ∂xγ ∂x′γ ∂xδ ∂x′δ′
Here the linearized Weyl tensor, δC αβγδ (x) is given as
δCαβγδ (x)
1
= + [Dγ Dβ hαδ − Dγ Dα hδβ − Dδ Dβ hαγ + Dδ Dα hγβ ]
2
(
1
+g αγ [Dλ Dβ hλδ + Dδ Dλ hλβ − hβδ − Dδ Dβ h]
−
2(D − 2)
−g γβ [Dλ Dδ hλα + Dα Dλ hλδ −
λ
λ
λ
λ
+g βδ [D Dα hλγ + Dγ D hλα −
−g δα [D Dγ hλβ + Dβ D hλγ −
−
hδα − Dα Dδ h]
hαγ − Dγ Dα h]
hγβ − Dβ Dγ h]
)
1
[g g − g αδ g βγ ][ h − Dτ Dω hτ ω ] ,
(D − 2)(D − 1) αγ βδ
(63)
and we get δCα′ β ′ γ ′ δ′ (x′ ) by replacing α, β, γ, δ → α′ β ′ γ ′ δ ′ respectively and
calculating all h’s and g’s at x′ . Further we can act ∂h∂µν on δCαβγδ (x) to get
∂
δCαβγδ (x)
∂hµν
i
1h
ν)
(µ ν)
(µ
= 1) δα(µ δδ Dγ Dβ − δβ δδ Dγ Dα − δα(µ δγν) Dδ Dβ + δβ δγν) Dδ Dα
2
(
1
(µ
(µ
−g αδ [δγ(µ Dν) Dβ − δβ δγν) − g µν Dγ Dβ + δβ Dγ Dν) ]
2) −
2(D − 2)
14
+g βδ [δγ(µ Dν) Dα − δα(µ δγν)
(µ
(µ ν)
(µ
−g βγ [δδ Dµ) Dα
ν)
δα(µ δδ
+g αγ [δδ Dν) Dβ − δβ δδ
3) +
−
− g µν Dγ Dα + δα(µ Dγ Dν) ]
(µ
− g µν Dδ Dβ + δβ Dδ Dν) ]
µν
− g Dδ Dα +
δα(µ Dδ Dν) ]
)
1
[g αγ g βδ − g αδ g βγ ][D(µ Dν) − g µν ] .
(D − 2)(D − 1)
(64)
µν
To write this simpler let us define Dαβγδ
(x) as
µν
Dαβγδ
(x) ≡
i
1 h (µ ν)
(µ ν)
(µ
δα δδ Dγ Dβ − δβ δδ Dγ Dα − δα(µ δγν) Dδ Dβ + δβ δγν) Dδ Dα . (65)
2
Then the terms of the first line in 2) become
µν
µν
Dβδ
≡ g κλ Dκβλδ
=
1 h (µ ν)
(µ ν)
δ D Dβ − δ β δ δ
2 δ
(µ
i
− g µν Dδ Dβ + δβ Dδ Dν) . (66)
and the terms in 3) become
µν
Dµν ≡ g κλ g θφ Dκθλφ
(x) = D(µ Dν) − g µν
.
(67)
We also write
∂ 2y
′
≡ T αα .
′
∂xα ∂xα′
(68)
Then we can write the spin 2 part a little simpler:
[spin 2 operator]F2 (y)
(
h
i
1
µν
µν
µν
µν
µν
−g αδ Dβγ
= Dαβγδ
−
+ g βδ Dαγ
+ g βγ Dαδ
− g αγ Dβδ
(D − 2)
)
1
µν
+
[g g − g αδ g βγ ]D
(D − 2)(D − 1) αγ βδ
(
i
h
1
Dαρσ′ β ′ γ ′ δ′ −
−g α′ δ′ Dβρσ′ γ ′ + g β ′ δ′ Dαρσ′ γ ′ − g α′ γ ′ Dβρσ′ δ′ + g β ′ γ ′ Dαρσ′ δ′
(D − 2)
)
1
ρσ
[g ′ ′ g ′ ′ − g α′ δ′ g β ′ γ ′ ]D
+
(D − 2)(D − 1) α γ β δ
T
αα′
T
ββ ′
T
γγ ′
T
δδ ′
F2 (y) .
(69)
15
With the graviton self-energy separated into two pieces
−i[µν Σρσ ]3-point (x; x′ )
√ q ′
= −g −g [Spin 0 Operator]F1 (y) + [Spin 2 Operator]F2 (y) ,
(70)
the next step is to find the structure functions F1 (y) and F2 (y) by comparing
(70) with the original expression, (44). In order to compare them, we need to
arrange (70) in terms of the five basis tensors. To accomplish that we have
to act the spin 0 and spin 2 operators on F1 and F2 , respectively. Acting the
spin 0 operator on F1 gives,
[spin 2 operator]F1 (y) =
i
∂ 2y
∂ 2y h
α
(F
)
1
1
∂xµ ∂x′(ρ ∂x′σ) ∂xν
i
∂y
∂ 2y
∂y h
+ (µ ν) ′(ρ ′σ) β1 (F1 )
∂x ∂x ∂x ∂x
i
∂y ∂y ∂y ∂y h
+ µ ν ′ρ ′σ γ1 (F1 )
∂x ∂x ∂x
h ∂x i
′
4
+g µν g ρσ H δ1 (F1 )
h
+ g µν
i
∂y ∂y ′ i 2 h
∂y ∂y
+
g
H
ǫ
(F
)
. (71)
1
1
∂x′ρ ∂x′σ ∂xµ ∂xν ρσ
Here the coefficients are
2F1′′ ,
(72)
′′′
4F1 ,
(73)
′′′′
F1 ,
(74)
2 2 ′′′′
2
′′′
2
′′
(4y−y ) F1 + 2(D+1)(2−y)(4y−y )F1 − 4(4y−y )F1
+(D2 −3)(2−y)2 F1′′ + (D−1)2 (2−y)F1′ + (D−1)2 F1 ,
(75)
2
′′′′
′′′
′′
ǫ1 (F1 ) = −(4y−y )F1 − (D+3)(2−y)F1 + (D+1)F1 .
(76)
α1 (F1 )
β1 (F1 )
γ1 (F1 )
δ1 (F1 )
=
=
=
=
Acting the spin 2 operator on F2 gives,
[spin 2 operator]F2 (y) =
i
∂ 2y h
∂ 2y
α
(F
)
2
2
∂xµ ∂x′(ρ ∂x′σ) ∂xν
i
∂y
∂ 2y
∂y h
+ (µ ν) ′(ρ ′σ) β2 (F2 )
∂x ∂x ∂x ∂x
i
∂y ∂y ∂y ∂y h
+ µ ν ′ρ ′σ γ2 (F2 )
∂x ∂x ∂x ∂x
16
h
+g µν g ′ρσ H 4 δ2 (F2 )
h
+ g µν
i
i
∂y ∂y ′ i 2 h
∂y ∂y
+
g
H
ǫ
(F
)
. (77)
2
2
ρσ
∂x′ρ ∂x′σ ∂xµ ∂xν
Now we remind ourselves that the construction of the spin 2 operator using
Weyl tensor is not unique. It can differ by constants. So the coefficients
α2 , β2 , γ2 , δ2 , ǫ2 multiplied by a constant can also be a good set of the coefficients. One way to determine the constant is to compare with the flat space
D−2
case which is given in section 4. It turns out that if we multiply by 16(D−3)
the leading term of the F1 and F2 are exactly agree with the corresponding
D−2
we get,
ones in flat space. So by multiplying by 16(D−3)
α2 (F2 ) = α21 F2 + α22 F2′ + α23 F2′′ + α24 F2′′′ + α25 F2′′′′ ,
(78)
and β2 (F2 ), · · · , ǫ2 (F2 ). Here the coefficients α21 , · · · are quite lengthy so we
give them in the Tables 3 - 7.
Now in order to solve for F1 and F2 what we have to do is just to equate
the correspoding coefficients:
αA (A)
βA (A)
γA (A)
δA (A)
ǫA (A)
=
=
=
=
=
α1 (F1 ) + α2 (F2 ) ,
β1 (F1 ) + β2 (F2 ) ,
γ1 (F1 ) + γ2 (F2 ) ,
δ1 (F1 ) + δ2 (F2 ) ,
ǫ1 (F1 ) + ǫ2 (F2 ) .
(79)
(80)
(81)
(82)
(83)
We recall that αA , · · · ǫA are given as the functions of A(y) defined in (31).
Because the infinite series which vanishs for D = 4, we only need to keep
potentially divergent terms,
Γ( D2 +1) 4 D2 −2
Γ( D2 +2) 4 D2 −3
H D−2 Γ( D2 ) 4 D2 −1
A(y) =
+
+
D
D
D
−2 y
2( D2 −3) y
(4π) 2 2 −1 y
2
(
)
πD Γ(D−1)
Γ(D) 4 −1
+ D
− π cot
+ ··· .
2
Γ( 2 +1) y
Γ( D2 )
(84)
We see this function A(y) is basically power series of ( y4 . Keeping only the
potentially divergent powers, we have
αA
"
#
4 D−1 D(D+1) 4 D−2
K 4 D
= − 5
+D
+
+· · · ,
2
y
y
2
y
17
(85)
βA
γA
δA
ǫA
"
#
4 D (D−2)D(D+1) 4 D−1
K 4 D+1
= 7 D
+(D−1)D
+
+· · · ,(86)
2
y
y
2
y
"
4 D+2
4 D+1
K
+(D−2)D2
= − 11 D2
2
y
y
#
2
2 D
(D −3D−2)D 4
+
+· · · ,
(87)
2
y
"
4 D
4 D−1
K
+ (D3 −5D2 +4D − 4)
= − 5 (D2 −D − 4)
2
y
y
#
4
3
2
D −8D +19D −28D + 8 4 D−2
+
+· · · ,
(88)
2
y
"
4 D+1
4 D
K
+ (D3 −5D2 +6D−4)
= 8 (D−2)D
2
y
y
#
3
2
D(D −7D +12D−12) 4 D−1
+
+· · · .
(89)
2
y
where the constant K is,
κ2 H 2D−4 Γ2 ( D2 )
.
K≡
(4π)D
(90)
Now to solve for F1 and F2 we can choose any two of the coefficient equations
(79 - 83). For example, we can pick the “α-” and “β-” coefficient equations,
αA (A) = α1 (F1 ) + α2 (F2 ) ,
βA (A) = β1 (F1 ) + β2 (F2 ) .
These equations are not easy to solve in general, but cosidering αA and βA
given in terms of ( y4 ) we can assume that F1 (y) and F2 (y) have the series
expansions of ( y4 ) as well. So let us set
4 D−2
+ b1
4 D−3
+ c1
4 D−4
+ ··· ,
y
y
y
4 D−3
4 D−4
4 D−2
+ b2
+ c2
+ ··· .
F2 (y) ≡ a2
y
y
y
F1 (y) ≡ a1
(91)
(92)
After arranging the two “α-” and “β-” equations in powers of ( y4 ) we can
determine the coefficients of F1 and F2 as,
a1 = −K
1
,
8(D − 1)2
(93)
18
1
,
4(D −
− 1)(D + 1)
D(D2 − 5D + 2)
−K
,
8(D − 4)(D − 3)(D − 1)2
D4 − 3D3 − 8D2 + 60D − 96
,
−K
4(D − 4)(D − 3)(D − 2)3 (D − 1)(D + 1)
D2 (D4 − 12D3 + 39D2 − 16D − 36)
−K
,
16(D − 6)(D − 4)2 (D − 3)(D − 1)2
1
−K
2
8(D − 6)(D − 4) (D − 3)(D − 2)4 (D − 1)(D + 1)
×[D8 − 8D7 − 13D6 + 348D5 − 1136D4
−1024D3 + 15056D2 − 38208D + 34560] .
a2 = −K
b1 =
b2 =
c1 =
c2 =
2)2 (D
(94)
(95)
(96)
(97)
(98)
First we note that the leading coefficients a1 and a2 agree with the flat space
case which is summarized in section 4 for the correspondence limit. Second,
we note that the subleading terms are ultraviolet finite in D = 4 but have
1
1
divergent factors either D−4
or (D−4)
2 . Third, the divergent factors for the
1
1
4
same powers of ( y ) are the same: D−4 for b2 and c2 and (D−4)
2 for b3 and c3 .
In the next section we introduce the appropriate counterterms to absorb the
divergences and fully renormalize this result.
4
Renormalization
In this section we fully renormalize the graviton self-energy. The ultraviolet
divergent terms are all absorbed by BPHZ counterterms and the remained
terms are completely finite. We begin with reviewing the BPHZ procedure
of obtaining counterterms for our case. Next we summarize the flat space
case to provide the correspondence limit as well as to illustrate the procedure
of our computation briefly. By taking trace we renormalize the spin 0 part
1
first. This enables us to subtract all the divergent factors D−4
for subleading
terms of the spin 2 parts. Finally we completely renormalize the spin 2 part.
4.1
BPHZ counterterms
The theorem of Bogoliubov, Parasiuk, Hepp and Zimmerman (BPHZ) tells us
how to construct the local counterterms to absorb the ultraviolet divergences
19
of any quantum field theory at some fixed order in the loop expansion [23].
For our case, the 1PI 2-point function at one loop order, it implies we need
two counterterms
√
∆LC1 = αR2 −g ,
(99)
√
ρσµν
∆LC2 = βC
Cρσµν −g .
(100)
for D = 4 spacetime dimensions. The divergent parts of which were originally
obtained by ’t Hooft and Veltman in 1974[21].
The procedure is summarized as follows: First we check the superficial
degree of divergence (SDD) of the 1PI function we consider at a certain order
for a particular dimension; in our case we consider the graviton self-energy
at one loop order for D = 4 dimensions. Its SDD is four and this means
it is quartically divergent in D = 4 dimensions. Then the BPHZ theorem
says that we need four or fewer derivatives of the field, in our case, gµν .
So the counterterm vertex functions have to be in the form of four or fewer
derivatives acting on a delta function. Now recalling that the Lagrangian has
to be generally coordinate invariant, the only possibilities for the counterterm
Lagrangian are,
√
−g × [constant, R, Rαβγδ Rαβγδ , Rµν Rµν , R2 ] .
(101)
It turns out we do not need the constant and R terms [21]. Also we note that
there exists a scalar called Gauss-Bonnet Scalar, which is a total derivative
h
i√
(102)
Rαβγδ Rαβγδ − 4Rµν Rµν + R2 −g = ∂µ S µ
in D = 4 dimensions. Because the total derivative makes no contribution
to the action the remained possibilities are two of the three 4th-order curvature scalars. We choose the two linearly independent combinations, (99)
and (100). Further we make a slight change to (99) for computational convenience,
h
R2 → R − D(D−1)H 2
i2
(103)
With this change it is left with only quadratic dependence on h:
h
R 2 ∼ 1 + h + h2 ,
R − D(D−1)H 2
i2
20
∼ h2 .
(104)
i2
h
For the price of writing R2 of the form R − D(D − 1)H 2 , it turns out
1
we need two more counterterms to absorb the divergent factor D−4
in the
subdominant terms:
h
i
√
∆LC3 = γ R − (D − 1)(D − 2)H 2 H 2 −g ,
(105)
√
∆LC4 = δH 4 −g .
(106)
In fact we can show that these two extra counterterms and the subtracted
terms from R2 add making only finite contributions:
−α × 2D(D − 1) + γ = finte at D = 4 ,
αD (D − 1) − γ(D − 1)(D − 2) + δ = finte at D = 4 .
2
2
(107)
(108)
This means we only need the two counterterms (99) and (100) to absorb all
the divergences.
One problem for quantum general relativity concerns us is that these
counterterms are not present in the original Lagrangian. We could include
(99); it would add a massive, positive energy scalar particle which poses no
essential problem for the theory. However, incorporating (100) on a nonperturbative level would add a negative energy, spin two particle whose presence
would cause the universe to decay instantly. We must therefore treat the one
loop counterterms perturbatively, and regard them as proxies for the still
unknown ultraviolet completion of the theory.
4.2
Flat space limit
In this section we consider the flat space case to illustrates the procedure
as well as to give the correspondence limit. In D dimensional flat space the
MMC scalar propagator is,
i△(x; x′ ) =
Γ( D2 − 1) h
D
4π 2
i D −1
1
2
,
∆x2 (x; x′ )
2
(109)
where ∆x2 (x; x′ ) = k~x − x~′ k2 − |t−t′ |−iǫ . All the vertices in flat space
follow from those of de Sitter given in Tables 1 and 2 by taking H → 0 and
g µν → ηµν .
21
Recalling that dimensional regularization works by choosing D so that
any D-dependent power of zero vanishes, we note that all the 4-point contributions,
4
1X
FIµνρσαβ (x) × ∂α ∂β′ i△(x; x′ ) × δ D (x − x′ ) ,
(110)
2 I=1
vanish in flat space as well as on de Sitter.
The 3-point one loop scalar contribution to the graviton self-energy in
flat space is,
−i[µν Σρσ ]3-point (x; x′ )
=
2
2
X
1X
J
I
Tρσγδ
(x′ ) × ∂ α ∂ ′γ i△(x; x′ ) × ∂ β ∂ ′δ i△(x; x′ )
Tµναβ
(x)
2 I=1
J=1
(
h 4D i
2 i
+
∆x
η
∆x
(µ ν)(ρ
σ)
∆x2D
∆x2D+2
h
h 1 (D 2 − D − 4) i
2D2 i
+∆xµ ∆xν ∆xρ ∆xσ − 2D+4 + ηµν ηρσ −
∆x
∆x2D
)2
h D(D − 2) i
.
(111)
+[ηµν ∆xρ ∆xσ + ∆xµ ∆xν ηρσ ]
∆x2D+2
h
= C ηµ(ρ ησ)ν −
κ2 Γ 2 ( D )
Here C ≡ 16πD2 for simplicity. We decompose this into the spin 0 part and
spin 2 part as well as on de Sitter:
−i[µν Σρσ ]3-point (x; x′ )
= [spin 0 operator]F1 (∆x2 ) + [spin 2 operator]F2 (∆x2 )
= [ηµν ∂ 2 − ∂µ ∂ν ][ηρσ ∂ 2 − ∂ρ ∂σ ]F1 (∆x2 )
n
+ ηµ(ρ ησ)ν ∂ 4 − 2∂(µ ην)(ρ ∂σ) ∂ 2 + ∂µ ∂ν ∂ρ ∂σ
o
1
[ηµν ∂ 2 − ∂µ ∂ν ][ηρσ ∂ 2 − ∂ρ ∂σ ] F2 (∆x2 ) .
−
D−1
(112)
Next we determine the structure functions F1 and F2 by comparing this eqn
(112) with eqn (111). The procedure is the same as on de Sitter: First we
act all the derivatives on F1 and F2 to arrange the spin 0 and spin 2 parts
in terms of the five basis tensors. Then we compare the coefficient of each
tensor to determine F1 and F2 . Here the flat space version of the five basis
tensors are from (111),
α : ηµ(ρ ησ)ν ,
22
β:
γ:
δ:
ǫ:
∆x(µ ην)(ρ ∆xσ) ,
∆xµ ∆xν ∆xρ ∆xσ ,
ηµν ηρσ ,
ηµν ∆xρ ∆xσ + ∆xµ ∆xν ηρσ .
Acting all the derivatives on F1 and F2 eqn (112) becomes
−i[µν Σρσ ]3-point (x; x′ )
= [spin 0 operator]F1 (∆x2 ) + [spin 2 operator]F2 (∆x2 )
n
= ηµ(ρ ησ)ν α1 (F1 ) + α2 (F2 )
n
o
+∆x(µ ην)(ρ ∆xσ) β1 (F1 ) + β2 (F2 )
n
o
+∆xµ ∆xν ∆xρ ∆xσ γ1 (F1 ) + γ2 (F2 )
n
+ηµν ηρσ δ1 (F1 ) + δ2 (F2 )
o
n
o
o
+[ηµν ∆xρ ∆xσ + ∆xµ ∆xν ηρσ ] ǫ1 (F1 ) + ǫ2 (F2 ) .
(113)
Now we can get the equations for F1 and F2 by equating the coefficients of
the five basis tensors from (113) and (111):
α
β
γ
δ
ǫ
=
=
=
=
=
α1 (F1 ) + α2 (F2 ) ,
β1 (F1 ) + β2 (F2 ) ,
γ1 (F1 ) + γ2 (F2 ) ,
δ1 (F1 ) + δ2 (F2 ) ,
ǫ1 (F1 ) + ǫ2 (F2 ) .
(114)
Here the coefficients are different combinations of derivatives, up to four four
derivatives, of F1 and F2 . Now considering the origianl coefficients α, β, γ, δ
and ǫ are functions of ∆x1 2 we can assume
K1
,
[∆x2 ]D−2
K2
=
.
[∆x2 ]D−2
F1 =
(115)
F2
(116)
To determine the two constants K1 and K2 , we pick any two of the five
equations (114). For example, if we choose β and γ coefficient equations, we
23
have
i
16(D − 2)D h
4D
=
−2(D
−
1)K
−
(D
−
2)(D
+
1)K
,
1
2
∆x2D+2
∆x2D+2
i
2D2
16(D − 2)D(D + 1) h
γ : −C
=
(D
−
1)K
+
(D
−
2)K
1
2 .
∆x2D+4
∆x2D+4
β: C
(117)
(118)
Solving for K1 and K2 gives,
K1
K2
κ2 Γ2 ( D2 )
1
,
= −
D
64π
2(D − 1)2
κ2 Γ2 ( D2 )
1
= −
.
D
2
64π
(D − 2) (D − 1)(D + 1)
(119)
(120)
So the flat space 3-point contributions become,
−i[µν Σρσ ]3-point (x; x′ )
(
κ2 Γ2 ( D2 )
1
1
[spin 0 operator]
=−
64π D
2(D − 1)2 ∆x2D−4
)
1
1
. (121)
+[spin 2 operator]
(D − 2)2 (D − 1)(D + 1) ∆x2D−4
We can check that these results for F1 and F2 in flat space agree with the
leading terms of F1 and F2 on de Sitter. We can compare them with the
following correspondences:
y
∂y
∂xµ
∂y
∂x′ρ
2
∂ y
∂xµ ∂x′ρ
∂
∂y
→
H 2 ∆x2 ,
(122)
→
2H 2 ∆xµ ,
(123)
→
− 2H 2 ∆xρ ,
(124)
→
− 2H 2 ηµρ ,
(125)
→
1 ∂
.
H ∂∆x2
(126)
This means we can read off the leading term of F1 and F2 on de Sitter by
the correspondences,
(F1 )flat = −
1 D−2
κ2 Γ2 ( D2 )
1
,
64π D 2(D − 1)2 ∆x2
24
(127)
→ (F1 )de Sitter
(F2 )flat
→ (F2 )de Sitter
4 D−2
κ2 Γ2 ( D2 )H 2D−4
1
= −
+ ··· ,
(128)
(4π)D
8(D − 1)2 y
1 D−2
κ2 Γ2 ( D2 )
1
,
(129)
= −
64π D (D − 2)2 (D − 1)(D + 1) ∆x2
4 D−2
κ2 Γ2 ( D2 )H 2D−4
1
= −
(4π)D
4(D − 2)2 (D − 1)(D + 1) y
+··· .
(130)
Now let us recall that we can use the graviton self-energy to get the
quantum corrections to the linearized field equation,
−
Z
dD x′ [µν Σρσ ](x; x′ )hρσ (x′ ) .
(131)
Basically we need to do the following integral,
Z
d D x′
1
∆x2D−4
hρσ (x′ ) .
(132)
We note that this is logarithmically divergent in D = 4 dimensions. After
one partial integration we can get a finite integral,
Z
1
∂2
hρσ (x′ ) ,
d D x′
2D−6
(D − 4)(2D − 6)
∆x
(133)
1
however this integral is multiplied by a divergent factor, D−4
for D = 4
dimension. Thus we need to cancel this divergence using appropriate counterterms. Recalling that conterterms should be local, we need to isolate the
divergence in the form of delta function. Our strategy for doing that is adding
zero in the form,
∂2
1
i4π D/2
−
δ D (x − x′ ) = 0 .
∆xD−2 Γ(D/2 − 1)
(134)
Using this equation (134) we add zero to 1/∆x2D−4 in the form of a delta
function,
∂2
1
2D−4
2D−6
∆x
(D − 4)(2D − 6) ∆x
1
µD−4
i4π D/2
D
′
∂2
−
δ
(x
−
x
)
. (135)
−
(D − 4)(2D − 6)
∆xD−2 Γ(D/2 − 1)
1
=
25
Here a dimensional factor µD−4 is added to make the dimensions consistent
and µ is called the dimensional regularization mass scale. Now we can write
1/∆x2D−4 as a nonlocal finite term plus a local divergent term,
1
∆x2D−4
=−
∂ 2 ln(µ2 ∆x2 ) 4π D/2 µD−4
iδ D (x − x′ )
.
+
4
∆x2
Γ(D/2 − 1) 2(D − 4)(D − 3)
(136)
Here for the finite part the D = 4 limit is taken. Then the 3-point contribution (121) can be written as
−i[µν Σρσ ]3-point (x; x′ )
(
κ2 Γ2 ( D2 )
1
′
=−
Πµν (x)Πρσ (x )
64π D
2(D − 1)2
)
1
1
1
Πµν Πρσ ]
,
+[Πµ(ρ Πσ)ν −
D−1
(D − 2)2 (D − 1)(D + 1) ∆x2D−4
io ln(µ2 ∆x2 )
κ2 n 10
1h
1
= + 10
Π
Π
+
Π
Π
Π
Π
−
∂2
µν
ρσ
µν
ρσ
µ(ρ
σ)ν
2 · 5π 4 9
3
3
∆x2
D
D−4 n
2
κ Γ( 2 )µ
D−2
Πµν Πρσ
−
D
2(D − 4)(D − 3)(D − 1)2
64π 2
1
1
+
[Πµ(ρ Πσ)ν −
Πµν Πρσ ]
(D − 4)(D − 3)(D − 2)(D − 1)(D + 1)
D−1
o
iδ D (x − x′ ) .
(137)
Here we symbolically write the spin 0 and spin 2 operators as
spin 0 operator = Πµν Πρσ ,
(138)
1
Πµν Πρσ .
(139)
D−1
To absorb the divergences we add the two invariant one loop counterterms
(99) and (100) with H → 0 and g µν → ηµν to the original Lagrangian,
spin 2 operator = Πµ(ρ Πσ)ν −
∆L1 = αR2 ,
∆L2 = βC ρσµν Cρσµν .
(140)
(141)
By expanding them, we get two counterterms
iδ 2 ∆S1 = 2ακ2 Πµν Πρσ iδ D (x − x′ ) ,
δhµν δhρσ hαβ =0
(142)
D−3 n
o
iδ 2 ∆S2 1
= 2βκ2
[Πµ(ρ Πσ)ν −
Πµν Πρσ ] iδ D (x − x′ ) . (143)
δhµν δhρσ hαβ =0
D−2
D−1
26
To cancel the divergences requires
Γ( D2 )µD−4
D−2
+ finite ,
4(D − 4)(D − 3)(D − 1)2
64π
Γ( D2 )µD−4
1
+ finite .
β=
D
64π 2 2(D − 4)(D − 3)2 (D − 1)(D + 1)
α=
D
2
(144)
(145)
These results were originally obtained in momemtum space by ’t Hooft and
Veltman in 1974[21]. In the next sections we compare these results with the
de Sitter case.
Note the agreement of the flat space limit of our final result in section
four with [24].
4.3
Spin 0 Part: Fully Renormalize F1
In the previous section we separated the graviton self-energy in two pieces:
We could find the F1 (y) and F2 (y) by comparing (44) with (70). Let us recall
that the subdominant terms of F1 and F2 have the same divergent structure:
1/(D − 4) and 1/(D − 4)2 otherwise finite powers of y4 . Thus we expect the
counterterms from LC3 and LC4 will take care of the divergent factors for
both F1 and F2 . (Better reasoning?) Thus recalling that the spin 2 part is
traceless, taking trace on (44) and (70) gives an equation of F1 . To solve this
equation for F1 and fully renormalize it is the task of this section.
4.3.1
Determine F1 alternatively by taking trace
Because the spin 2 part is traceless, taking trace on both sides of this equation
(70) gives an equation including only F1 :
[
+ DH 2 ][ ′ + DH 2 ]F1 (y)
n
o
4
1
(A′ )2 .
= − (D − 2)2 κ2 H 4 [(2 − y)A′ − k]2 +
8
D−1
(146)
Noting
F (y) = H 2 [(4y − y 2 )F ′′ + D(2 − y)F ′ ] =
′
F (y) ,
(147)
we can write the equation (146) as
n
o
4
1
(A′ )2 . (148)
[ +DH 2 ]2 F1 (y) = − (D −2)2 κ2 H 4 [(2−y)A′ −k]2 +
8
D−1
27
Again we note that A(y) is essentially a series expansion of y4 , and we only
need to keep the powers which become divergent in D = 4 dimensions because
the infinite series of A(y) vanishes for D = 4. Thus keeping only potentially
divergent terms, the right hand side of the equation (148) can be written as
n
o
1
4
− (D − 2)2 κ2 H 4 [(2 − y)A′ − k]2 +
(A′ )2
8
D−1
2D−4 2 D n
4 D
H
Γ (2)
1
(D − 2)2 4 D−1
D
= − (D − 2)2 κ2 H 4
+
8
(4π)D
4(D − 1) y
4(D − 1) y
3
2
o
D − 7D + 16D − 8 4 D−2
.
(149)
+
8(D − 1)
y
Now we have the differential equation
4 D−1
4 D−2
4 D
[ + DH 2 ]2 F1 (y) = A1
+ B1
+ C1
+ ··· .
(150)
y
y
y
where the constants A1 , B1 and C1 are
κ2 Γ2 ( D2 )H 2D−4 D(D − 2)2
×
,
(151)
(4π)D
32(D − 1)
κ2 Γ2 ( D2 )H 2D−4
(D − 2)4
B1 = −H 4 ×
,
(152)
×
(4π)D
32(D − 1)
κ2 Γ2 ( D2 )H 2D−4 (D3 − 7D2 + 16D − 8)(D − 2)2
4
×
. (153)
C1 = −H ×
(4π)D
64(D − 1)
A1 = −H 4 ×
and want to solve for F1 (y). Again we set F1 (y) as a series expansion of y4 .
F1 (y) ≡ a1
4 D−2
+ b1
4 D−3
+ c1
4 D−4
+ ··· .
(154)
y
y
y
Plugging (154) into (150) and comparing both sides we can get the coefficients
a1 , b1 and c1 :
κ2 Γ2 ( D2 )H 2D−4
1
×
,
(155)
(4π)D
8(D − 1)2
κ2 Γ2 ( D2 )H 2D−4
D(D2 − 5D + 2)
b1 = −
×
,
(156)
(4π)D
8(D − 4)(D − 3)(D − 1)2
κ2 Γ2 ( D2 )H 2D−4 D2 (D4 − 12D3 + 39D2 − 16D − 36)
c1 = −
×
. (157)
(4π)D
16(D − 6)(D − 4)2 (D − 3)(D − 1)2
a1 = −
Of course these results agree with the coefficients for F1 (y) we have found in
the previous section. In the subsequent section we fully renormalize F1 .
28
4.3.2
Renormalization of Spin 0 part
Here we add the counterterms to absorb the divergences occuring in F1 (y).
The structure function F1 (y) is given as,
F1 (y) = −K a1 (D)
4 D−2
y
b1 (D) 4 D−3 c1 (D) 4 D−4
+
+
+· · ·
(D−4) y
(D−4)2 y
.(158)
Here we factored the divergent coefficient out and redefined b1 and c1 as
follows:
1
a1 (D) =
,
(159)
8(D − 1)2
D(D2 − 5D + 2)
b1 (D) =
,
(160)
8(D − 3)(D − 1)2
D2 (D4 − 12D3 + 39D2 − 16D − 36)
.
(161)
c1 (D) =
16(D − 6)(D − 3)(D − 1)2
We see that the first term is logarithmically divergent in D = 4 dimensions
as well as the flat space case. The new problem on de Sitter is that the
second and third terms which will give finite integrals have the divergent
1
1
factors, D−4
and (D−4)
2 , respectively. Our procedure is again to segregate
the divergence on the local term (in the form of a delta function). For doing
that the key identity we can use is,
4 D/2−1
H2 y
=
(4π)D/2 iδ D (x − x′ ) D(D − 2) 4 D/2−1
√
.
+
4
y
Γ( D2 − 1) H D −g
Let us first note that in D = 4 dimensions
4 D−3
4 D/2−1
=
,
y
y
4 D/2−2
4 D−4
=
=1.
y
y
(162)
(163)
(164)
For simplicity, let’s change the variable X ≡ y/4. For the second and third
term of eqn (158) we may add
1 D/2−1
b1 (D)
−
(D − 4)
X
1 D/2−2
c1 (D)
−
2
+
1
(D − 4)2
X
29
to
to
b1 (D) 1 D−3
,
(D − 4) X
c1 (D) 1 D−4
.
(D − 4)2 X
(165)
(166)
Then by taking D = 4 limit, we have
b1 (4) 1
b1 (D) 1 D−3 1 D/2−1
→ −
−
ln(X) ,
(D − 4) X
X
2 X
1 D/2−2
c1 (4) 2
c1 (D) 1 D−4
−2
+1 →
ln (X) ,
2
(D − 4)
X
X
4
(167)
(168)
which are finite in D = 4. We expect that theh divergences
from these, which
i2
will occur acted by the differential operator H 2 + D , will be canceled by
the γ- and δ- counterterms.
Now let us review what the possible counterterms are. From the counterterm Lagrangians we discussed in the previous section, we get the corresponding counterterms, by varying the Lagragians with respect to the graviton field
hµν :
h
∆LC1 = α R − D(D − 1)H 2
iδ 2 ∆S1 ⇒
δhµν δhρσ hαβ =0
q
i2 √
−g ,
√
= 2ακ2 −g[ g µν − Dµ Dν + (D − 1)H 2 g µν ]
− D′ρ D′σ + (D − 1)H 2 g ′ρσ ]
√
∆LC2 = βC ρσµν Cρσµν −g ,
q
√
iδ 2 ∆S2 µν
ρσ
2
′
=
2βκ
⇒
−g(
)
αβγδ −g ( )α′ β ′ γ ′ δ ′
δhµν δhρσ hαβ =0
× −g ′ [
′ ′ρσ
g
iδ D (x − x′ )
√
, (169)
−g
iδ D (x − x′ ) i
√
,
−g
h
i
√
∆L3 = γ R − (D − 1)(D − 2)H 2 H 2 −g ,
h
× Tαα′ Tββ ′ Tγγ ′ Tδδ′
(170)
√
γ
iδ 2 ∆S3 = κ2 H 2 −g 2g µ(ρ Dσ) Dν − g µ(ρ g σ)ν
⇒
δhµν δhρσ hαβ =0 2
−(g µν Dρ Dσ + g ρσ Dµ Dν ) + g µν g ρσ
− 2(D − 1)H 2 g µ(ρ g σ)ν
+(D − 1)H 2 g µν g ρσ iδ D (x − x′ ) ,
√
∆L4 = δH 4 −g ,
δ 2 4√
iδ 2 ∆S4 µν ρσ
µ(ρ σ)ν
= κ H −g g g − 2g g
iδ D (x − x′ ) .
⇒
δhµν δhρσ hαβ =0 4
30
(171)
(172)
Taking trace on the primitive terms (70) and the counterterms (169 - 172)
gives,
√ q
(D − 1)2 −g −g ′ [ + DH 2 ][ ′ + DH 2 ]F1 (y) + 0
√
+2ακ2 (D − 1)2 −g[ + DH 2 ][ ′ + DH 2 ]iδ D (x − x′ ) + 0
√
γ
+ κ2 H 2 (D − 1)(D − 2) −g[ + DH 2 ]iδ D (x − x′ )
2
√
δ 2 4
+ κ H D(D − 2) −giδ D (x − x′ ) .
(173)
4
We note that the added terms on the subleading terms of F1 produce the
delta function and therefore can be absorbed by the γ- and δ- counterterms:
Note that
h
H2
+D
i2 1 D/2−1
X
=
h
H2
+D
i (4π)D/2 iδ D (x − x′ )
√
Γ( D2 − 1) H D −g
D(D + 2) (4π)D/2 iδ D (x − x′ )
√
4
Γ( D2 − 1) H D −g
D2 (D + 2)2 1 D/2−1
+
,
16
X
D − 4 (4π)D/2 iδ D (x − x′ )
√
= −
2 Γ( D2 − 1) H D −g
+
h
H2
+D
i2 1 D/2−2
X
(174)
(D − 4)(D2 + 2D − 4) 1 D/2−1
−
4
X
(D − 2)2 (D + 4)2 1 D/2−2
+
(175)
16
X
We write F1 = F 1 + ∆F1 where F 1 is,
(
1 D−2
b1 (D) 1 D−3 1 D/2−1
F 1 ≡ −K a1 (D)
+
−
X
(D − 4) X
X
)
1 D/2−2
1 D−4
c1 (D)
−
2
+1
(176)
+
(D − 4)2 X
X
(
)
1 D−2
b1 (4) 1
c1 (4) 2
→ −K a1 (D)
−
ln(X) +
ln (X) . (177)
X
2 X
4
Then we have,
h
H2
i2
+ D F1
31
1 D−2
b1 (4) 1
c1 (4) 2
ln(X) +
ln (X) ,
2
H(
X
2 X
4
1 D−1
1 D−2
1 D
+ n2 (D)
+ n3 (D)
= −K n1 (D)
X
X
X
=
h
+D
i2 a1 (D)
−
(178)
i (4π)D/2 iδ D (x − x′ )
b1 (D) h
√
−
+D
D − 4 H2
Γ( D2 − 1) H D −g
+ −
c1 (D)
D − 4 (4π)D/2 iδ D (x − x′ )
b1 (D) D(D + 2)
√
−2
×
−
D−4
4
(D − 4)2
2
Γ( D2 − 1) H D −g
)
8
4 1
ln(X) − ln2 (X) .
−
3X
3
(179)
Here the constants n1 (D), n2 (D) and n3 (D) are
D(D − 2)2
n1 (D) =
,
32(D − 1)
(D − 2)4
,
n2 (D) =
32(D − 1)
(D3 − 7D2 + 16D − 8)(D − 2)2
n3 (D) =
.
64(D − 1)
(180)
(181)
(182)
These n1 (D), n2 (D) and n3 (D) agree with the original expressions (151, 152,
153) as they should be. We see that the localized divergent terms of the eqn
(179) in the form of the γ- and δ- counterterms so that can be canceled by
them. The left-over term,
−
8
4 1
ln(X) − ln2 (X) ,
3X
3
(183)
can be canceled by adding ∆F1 which satisfies
h
H2
i2
+ D ∆F1 = +
4 1
8
ln(X) + ln2 (X) .
3X
3
(184)
The explicit function of ∆F1 is given in the next subsection.
Now we require that the localized divergences of (179) cancel with γ- and
δ- counterterms
(D − 1)2
b1 (D)
γ
(4π)D/2
+ κ2 (D − 1)(D − 2) = 0 ,
×K × D
D
D−4
2
Γ( 2 − 1)H
32
(185)
(D − 1)
2
b1 (D) D(D + 2) c1 (D)
(4π)D/2
−
× (−K) × D
+
D−4
4
D−4
Γ( 2 − 1)H D
δ
+ κ2 D(D − 2) = 0 .
4
Thus we determine γ- and δ as,
(186)
Γ( D2 )H D−4
D(D2 − 5D + 2)
×
,
8(D − 4)(D − 3)(D − 1)
(4π)D/2
D(D3 − 11D2 + 24D + 12) Γ( D2 )H D−4
δ =
×
.
16(D − 6)(D − 3)
(4π)D/2
(187)
γ = −
(188)
Now to take care of the leading divergence ( X1 )D−2 we extract a covariant
d’Alembertian from it as we did for the flat space case:
1 D−2
1 D−3
2
4 1 D−3
=
−
.
(189)
X
(D − 3)(D − 4) H 2 X
D−4 X
Using the key identity (162), we add zero to
1 D−2
X
1
X
D−2
:
1 D/2−1
1 D−3
2
−
(D − 3)(D − 4) H 2 X
X
4 1 D−3
2
D(D − 2) 1 D/2−1
−
+
D−4 X
(D − 3)(D − 4)
4
X
(4π)D/2 iδ D (x − x′ )
2
√
+
.
(190)
(D − 3)(D − 4) Γ( D2 − 1) H D −g
=
Now taking D = 4 it becomes,
1 D−2
ln(X)
ln(X)
1
+2
→ 2 −
−
X
H
X
X
X
+
2
(4π)D/2 iδ D (x − x′ )
√
.
(D − 3)(D − 4) Γ( D2 − 1) H D −g
(191)
Then F 1 becomes
2
(4π)D/2 iδ D (x − x′ )
√
F 1 → −K a1 (D)
(D − 3)(D − 4) Γ( D2 − 1) H D −g
ln(X)
ln(X)
1
+2
−
+a1 (4) 2 −
H
X
X)
X
c1 (4) 2
b1 (4) 1
ln(X) +
ln (X) .
−
2 X
4
(
33
(192)
Requiring that this leading divergence canceled by the α- counterterm we
can determine the coefficient α:
(D − 1)2 a1 (D)
=0.
2
(4π)D/2
+ 2ακ2 (D − 1)2
× (−K) × D
(D − 4)(D − 3)
Γ( 2 − 1)H D
(193)
This gives,
α=+
Γ( D2 )H D−4
(D − 2)
×
.
16(D − 4)(D − 3)(D − 1)2
(4π)D/2
(194)
We note that this agrees with the flat space case (144).
Spin 0 part: Determine ∆F1
4.3.3
To completely determine F1 we need to find ∆F1 which satisfies in D=4
dimension,
h
i2
4 1
8
∆F1 = +
+
4
ln(X) + ln2 (X)
(195)
2
H
3X
3
We can invert one differential operator first:
+ 4]∆F1 ≡ ∆G
(196)
H2
By omiting the numerical factor 34 which we can recover later easily, we want
to solve the differential equation for G(x):
[
[
H2
+ 4]∆G(X) = [X(1 − X)(
d 2
d
) + (2 − 4X)
+ 4]∆G(X) (197)
dX
dX
1
ln(X) + 2 ln2 (X) .
X
First we can find the Green’s function for the operator,
=
X(1 − X)(
d 2
d
) + (2 − 4X)
+4,
dX
dX
(198)
(199)
as
G(X; Y ) = (2 − 4X)(2 − 4Y )Y (1 − Y )
(
1
4
3
3
1
+
+
+ ln(X) − ln(1 − X)
× −
4X 4(1 − X) 2 − 4X 2
2
)
1
4
3
3
1
+
+
+ ln(Y ) − ln(1 − Y ) . (200)
− −
4Y
4(1 − Y ) 2 − 4Y
2
2
34
Then basically we can find ∆F1 (X) using the Green’s function by integrating
twice over the Green’s function,
∆F1 (X) =
where f (X) =
∆F1 (X) as
1
X
Z
X
′
′
dX G(X; X )
X′
Z
dX ′′ G(X ′ ; X ′′ )f (X ′′ ) ,
(201)
ln(X) + 2 ln2 (X). This gives after some simplification,
(
1
62468X 3 − 95547X 2 + 28829X + 4340
∆F1 (X) = −
54000(1 − X)X
−60(1 − X)(1418X 2 − 1129X + 70) ln(1 − X)
−120X(884X 2 − 1041X + 217) ln(X)
−43200(1 − X)X(1 − 2X) ln(1 − X) ln(X)
+18000(1 − X)X 2 ln2 (X)
)
−68400(1 − X)X(1 − 2X)Li2 (X) .
(202)
Here the Li2 (X) is the dilogarithm function.
4.4
Spin 2 Part: Renormalize F2
In the previous section we have found the F1 = F 1 + ∆F1 and fully renormalized it. Here we can use F1 to find F2 and renormalize it. The way
we determine the coefficients of F2 is again to compare the powers of (4/y)
using the five coefficient equations (79 - 83). Now with the F1 complitely
determined, we need only one of the five equations (79 - 83). Recalling the
form of F1 ,
(
F1 = F 1 + ∆F1 = −K a1 (D)
c1 (D)
+
(D − 4)2
1 D−4
X
1 D−2
X
−2
b1 (D) 1 D−3 1 D/2−1
+
−
(D − 4) X
X
1 D/2−2
X
+1
)
+ ∆F1 ,
(203)
we suppose that F2 has the same form,
(
b2 (D) 1 D−3 1 D/2−1
+
−
F2 = −K a2 (D)
X
(D − 4) X
X
)
1 D/2−2
1 D−4
c2 (D)
−2
+ 1 + ··· .
+
2
(D − 4)
X
X
1 D−2
35
(204)
Then we just need to determine the coefficients a2 (D), b2 (D) and c2 (D). The
procedure is the same as before. We compare the coefficient of each power
of (4/y) then we get
1
,
(205)
4(D −
− 1)(D + 1)
D4 − 3D3 − 8D2 + 60D − 96
,
(206)
b2 (D) = −K
4(D − 3)(D − 2)3 (D − 1)(D + 1)
1
c2 (D) = −K
8(D − 6)(D − 3)(D − 2)4 (D − 1)(D + 1)
×[D8 − 8D7 − 13D6 + 348D5 − 1136D4 − 1024D3
+15056D2 − 38208D + 34560] .
(207)
a2 (D) = −K
2)2 (D
Of course this is the same as the solutions we have in the Section 3. But now
with this form we can take D = 4 limit. Then we have
(
F2 → −K a2 (D)
1 D−2
X
)
b2 (4) 1
c2 (4) 2
−
ln(X) +
ln (X) .
2 X
4
(208)
Now to make F2 integrable in D = 4 we do one more partial integration
using the identity (191),
1 D−2
X
ln(X)
ln(X)
1
+2
→ 2 −
−
H
X
X
X
D
D/2
iδ (x − x′ )
(4π)
2
√
+
(D − 3)(D − 4) Γ( D2 − 1) H D −g
as D → 4 − ǫ
Then F2 becomes
(
(4π)D/2 iδ D (x − x′ )
2
√
(D − 3)(D − 4) Γ( D2 − 1) H D −g
ln(X)
ln(X)
1
+2
−
+a2 (4) 2 −
H
X
X) X
b2 (4) 1
c2 (4) 2
−
ln(X) +
ln (X) .
2 X
4
F2 → −K a2 (D)
(209)
Now we can cancel the local divergent part with the β-coundterterm by
requiring
2βκ2 + a2 (D)
2
(4π)D/2
=0.
(D − 3)(D − 4) Γ( D2 − 1)H D
36
(210)
And we can determine β as
β=
Γ( D2 )H D−4
1
.
8(D − 4)(D − 3)(D − 2)(D − 1)(D + 1) (4π)D/2
(211)
This agrees with the flat space case (145).
Finally the finite part of the graviton self-energy becomes
−i[µν Σρσ ]3-point (x; x′ )
(
(
√ q ′
= −g −g [Spin 0 Operator] a1 (4)
H2
−
ln(X)
ln(X)
1
+2
−
X
X
X
)
c1 (4) 2
b1 (4) 1
ln(X) +
ln (X) + ∆F1
−
2 X
4
(
ln(X)
ln(X)
1
+[Spin 2 Operator] a2 (4) 2 −
+2
−
H
X
X
X
))
c2 (4) 2
b2 (4) 1
ln(X) +
ln (X) + ∆F1
.
(212)
−
2 X
4
1
1
Here a1 (4) = −K× 72
, b1 (4) = −K×− 19 , c1 (4) = −K×− 32 , a2 (4) = −K× 240
,
1
b2 (4) = −K × 6 , c2 (4) = −K × −1.
Note the agreement of the flat space limit of our final result in section
four with [24].
5
Discussion
• Summarize results.
• Note that the result is de Sitter invariant.
• Note agreement of the form reached in section 3 with the stress tensor
correlator of Perez-Nadal, Roura and Verdaguer [20].
• Note agreement of our counterterms with those found for this model
by ‘t Hooft and Veltman [21].
• Note the agreement of the flat space limit of our final result in section
four with [24].
37
• Discuss how one uses our result to quantum correct the linearized Einstein equations. In particular, give the prescription for getting the
Schwinger-Keldysh effective field equations [25, 26, 27, 28, 29, 30, 31].
• Discuss what we expect from solving the equation: nothing much for
dynamical gravitons (because scalars have no spin) but a potential
fractional enhancement of the gravitational potential of the form,
∆Φ Φ
h̄GH 2
∼ − 5 ln[a(t)] .
c
(213)
We have computed the one loop contribution to the graviton self-energy
−i[ Σρσ ](x; x′ ) from a massless, minimally coupled (MMC) scalar on a locally de Sitter background. We used dimensional regularization and renormalized by absorbing the divergences with the BPHZ counterterms. The
finite result is given in (212).
This result can be used to quantum correct to quantum correct the linearized Einstein equations.
µν
Dµνρσ hρσ (x) −
Z
1
dD x′ [µν Σρσ ](x; x′ )hρσ (x′ ) = κT µν .
2
(214)
√
Here Dµνρσ hρσ (x′ ) = 12 κT µν is ( κ−g times) the linearized classical Einstein
equation. In de Sitter background the differential operator Dµνρσ is
D
µνρσ
1√
=
−g g µρ Dσ Dν +g νρ Dσ Dµ −g µρ g νσ D2 −g ρσ Dµ Dν
2
(
µν
ρ
σ
µν ρσ
2
2 µρ νσ
2 µν ρσ
−g D D +g g D −2(D−1)H g g +(D−1)H g g
)
. (215)
Here Dµ is the covariant derivative operator with respect to the de Sitter
background. For a stationary point mass the source is,
Tµν
mδ 3 (~x)g µ0 g ν0
=
.
√
−g 00
(216)
The usual, in-out formalism has two features which make it undesirable
for applications in cosmology: quantum corrections can derive from the future of the observation point and quantum corrections to Hermitian operators need not be real. Neither of these features means the in-out formalism
38
is wrong. In fact it is the right answer to questions about matrix elements
between scattering states if the system begins in free vacuum in the asymptotic past and ends up the same way in the far future. However, that question isn’t typically very relevant for cosmology in which the universe begins
with an initial singularity and no one knows how it ends. The question
of greater relevance for cosmology is, “what happens when the universe is
released at some finite time in a prepared state.” The Schwinger-Keldysh
formalism[25, 26, 27, 28, 29, 30, 31]. provides a way of answering this more
relevant question which is almost as simple to use as the Feynman diagrams
of the in-out formalism.
We can get the four propagators of the Schwinger-Keldysh formalism
from the Feynman propagator once that is known. The rule is just to replace
∆x2 (x; x′ ) in the Feynman propagator with ∆x2±± (x; x′ ) where
∆x2++ (x; x′ ) ≡ k~x − x~′ k2 − |η−η ′ |−iǫ
∆x2+− (x; x′ ) ≡ k~x − x~′ k2 − η−η ′ +iǫ
2
∆x2−+ (x; x′ ) ≡ k~x − x~′ k2 − η−η ′ −iǫ
2
(217)
,
(218)
,
(219)
2
∆x2−− (x; x′ ) ≡ k~x − x~′ k2 − |η−η ′ |+iǫ
,
2
.
(220)
The “graviton self-energy” which belongs in equation (214) in the S-K
formalism becomes
h
µν
i
Σρσ (x; x′ ) −→
h
µν
Σρσ
i
(x; x′ ) +
++
h
µν
Σρσ
i
(x; x′ ) .
+−
(221)
The linearized gravitational field equations tell us two things:
• How dynamical gravitons propagate through the background geometry;
and
• What the force of gravity is in the background geometry.
The quantum correction term in equation (214) describes how both of these
things are affected by virtual particles, in our case virtual scalars. If there
are not many virtual particles or they interact with gravitons only weakly
then there will not be much effect once the field is renormalized. That is
what happens in flat space.
39
In a inflationary background, for example, the de Sitter background we
consider we can show that the number of particles with wave number k grows
in time as the square of the scale factor,
Ha(t)
N (k, t) =
2k
2
(222)
Note from expression (222) that there is not much excitation above the
instantaneous 0-point energy of k/2a(t) for “ultraviolet” wavelengths with
k ≫ Ha(t). It is only for cosmological scale wavelengths with k ∼ Ha(t)
that the occupation number N (k, t) becomes of order unity.
At this point it is useful to compare my thesis problem with those of two
recent UF graduates, Shun-Pei Miao [34] and Emre Kahya [35]. Whereas they
studied the effect of inflationary gravtions (which are produced copiously for
the same reason as MMC scalars) on different types of matter particles, I
am studying the effect of inflationary scalars on gravitons. Dr. Miao studied
the effect of inflationary gravitons on fermions and she found that one loop
corrections increase the fermion field strength by a factor of GH 2 ln(a) [36].
Dr. Kahya studied the effect of infrared gravitons on scalars and he found
no similar growth [37]. The reason why they got different results is spin:
gravitons and fermions both have spin and this leads to an extra interaction,
in addition to the one due to their kinetic energies. This extra interaction is
highly significant because it persists even for highly infared particles, whose
kinetic energies redshift to zero. In a subsequent paper, Dr. Miao showed
that the secular growth factors of GH 2 ln(a) derive entirely from the vertices
of the spin-spin interaction [38]. The absence of a secular enhancement in
Dr. Kahya’s thesis computation derives from the fact that scalars have zero
spin.
Now recall that I am really making two studies:
1. How inflationary scalars affect dynamical gravitons; and
2. How inflationary scalars affect the force of gravity.
The first of these studies is a sort of mirror-image of Dr. Kahya’s; whereas
he studied the effect of inflationary gravitons on scalars, I am studying the
effect of inflationary scalars on dynamical gravitons. I expect the result of
this first study to be the absence of any secular enhancement for the same
reason that Dr. Kahya found none: dynamical gravitons have spin two, and
scalars have spin zero. On the other hand, the force of gravity is partly
40
carried by the non-dynamical (constrained) spin zero sector of the graviton
field. This might well experience a secular enhancement.
To find the force of gravity one first puts down a point mass and computes the classical response to that, then this classical potential is integrated
against the one loop graviton self-energy to serve as a source for the one loop
correction to the potential. When Donoghue and collaborators worked out
the same thing for flat space background they got [39],
(
)
GM
41 G
Φ=−
1+
+ O(G2 ) .
r
10π r2
(223)
There are two ways to view this: from the perspective of dimensional analysis,
and from the perspective of how virtual particles interact with the point mass.
In the context of dimensional analysis, any one loop correction in quantum
gravity must carry a factor of G, which goes like a length-squared. The
only other dimensionful parameters in this problem are the length r and the
particle mass m. If we focus on the linearized response to m then any one
loop effect must give a fractional correction of G/r2 .
The physical interpretation of (223) is that the classical potential GM/r
interacts with the energy density of virtual particles to reduce the measured
mass of the source, in the same way that the combined mass of the EarthMoon system is lower than the sum of their masses by their gravitational
interaction energy. The effect is infinite because the energy density of ultraviolet virtual particles diverges, but if we renormalize things so that M is the
measured mass as r goes to infinity, then the residual potential must grow
as one approaches the mass because the radius r encloses less and less of the
negative interaction energy between the mass and the virtual particles. The
reason it grows like 1/r2 is that the energy of a virtual particle of wavelength
r goes like 1/r.
Now consider my problem in de Sitter background. From the perspective
of dimensional analysis one sees first that there is a new parameter with the
dimensions of an inverse length: the Hubble constant H. That means one
can get a fractional correction of the form GH 2 . The physical interpretation
is the same as in Donoghue’s case: the one loop correction to the source’s
potential comes from the gravitational interaction between it and the density
of virtual particles. However, because inflation rips more and more virtual
scalars out of the vacuum, as is evident in expression (222), the effect ought
to get bigger with time. Of course the flat space result (223) must be present
41
as well, but I epxect an additional contribution of the form,
(
)
Gm
1 − const × GH 2 ln(a) + . . . .
Φ∼−
r
(224)
This intrinsic inflation effect would manifest as a secular decrease in the
effective Newton constant,
n
o
G −→ G 1 − const × GH 2 ln(a) + . . . .
(225)
That is all pure theory, but it is worth asking how big the effect would
be for the phase of primordial inflation which is believed to have actually
occurred. Based on the measured value of the scalar amplitude of cosmic
microwave anisotropies, and the current upper limit on the tensor-to-scalar
ratio [33], the Hubble parameter during primordial inflation can have been
no greater than about 1013 GeV. Because Newton’s constant is the inverse
square of about 1019 GeV, the dimensionless loop counting parameter is very
small,
1013 2
∼ 10−12 .
(226)
GH 2 ∼
1019
On the other hand, the secular factor of ln(a) that I expect (and that Dr.
Miao actually found for her problem) grows as long as inflation persists,
which is eternity for de Sitter space. So I expect that the secular reduction
of Newton’s constant could be significant for a very long period of inflation.
Acknowledgements
We are grateful to A. Roura for discussions on this topic. This work
was partially supported by NSF grant PHY-0855021 and by the Institute for
Fundamental Theory at the University of Florida.
References
[1] A. F. Radkowski, Ann. Phys. 56 (1970) 319; D. M. Capper, M. J. Duff
and L. Halperin, Phys. Rev. D10 (1974) 461; D. M. Capper and M. J.
Duff, Nucl. Phys. B84 (1974) 147; D. M. Capper, Nuovo Cimento A25
(1975) 29; M. J. Duff, Phys. Rev. D9 (1974) 1837; H. Hamber and S.
Liu, Phys. Lett. B357 (1995) 51, hep-th/9505182; M. J. Duff and J. T.
Liu, Phys. Rev. Lett. 85 (2000) 2052, hep-th/0003237; J. F. Donoghue,
42
Phys. Rev. Lett. 72 (1994) 2996, gr-qc/9310024; Phys. Rev. D50 (1994)
3874, gr-qc/9405057; I. J. Muzinich and S. Kokos, Phys. Rev. D52
(1995) 3472, hep-th/9501083; A. Akhundov, S. Belucci and A. Shiekh,
Phys. Lett. B395 (1997) 16, gr-qc/9611018; I. B. Khriplovich and G.
G. Kirilin, J. Exp. Theor. Phys. 95 (2002) 981, gr-qc/0207118; J. Exp.
Theor. Phys. 98 (2004) 1063, gr-qc/0402018. N. E. J. Bjerrum-Bohr,
J. F. Donoghue and B. R. Holstein, Phys. Rev. D67 (2003) 084033,
Erratum-ibid. D71 (2005) 069903, hep-th/0211072; N. E. J. BjerrumBohr, Phys. Rev. D66 (2002) 084023, hep-th/0206236; B. R. Holstein
and A. Ross, “Spin Effects in Long Range Gravitational Scattering,”
arXiv:0802.0716; A. Campos and E. Verdaguer, Phys. Rev. D49 (1994)
1816, gr-qc/9307027; F. C. Lombardo and F. D. Mazzitelli, Phys. Rev.
D55 (1997) 3889, gr-qc/9609073; R. Martin and E. Verdaguer, Phys.
Rev. D61 (2000) 124024, gr-qc/0001098; A. Satz, F. D. Mazzitelli and E.
Alvarez, Phys. Rev. D71 (2005) 064001, gr-qc/0411046; D. A. R. Dalvit
and F. D. Mazzitelli, Phys. Rev. D50 (1994) 1001, gr-qc/9402003; E.
D. Carlson, P. R. Anderson, A. Fabbri, S. Fagnocchi, W. H. Hirsch and
S. Klyap, arXiv:1008.1433.
[2] Sohyun Park and R. P. Woodard, Class. Quant. Grav. 27 (2010) 245008,
ArXiv:1007.2662.
[3] E. Schrödinger, Physica 6 (1939) 899.
[4] T. Imamura, Phys. Rev. 118 (1960) 1430.
[5] L. Parker, Phys. Rev. Lett. 21 (1968) 562; Phys. Rev. 183 (1969) 1057;
Phys. Rev. D3 (1971) 346.
[6] L. H. Ford and L. Parker, Phys. Rev. D16 (1977) 1601.
[7] L. P. Grishchuk, Sov. Phys. JETP 40 (1975) 409.
[8] T. Prokopec and R. P. Woodard, Am. J. Phys. 72 (2004) 60, astroph/0303358.
[9] A. A. Starobinsky, JET Lett. 30 (1979) 682; V. F. Mukhanov and G.
V. Chibisov, JETP Lett. 33 (1981) 532.
[10] E. Komatsu et al., arXiv:1001.4538.
43
[11] T. Prokopec, O. Tornkvist and R. P. Woodard, Phys. Rev. Lett. 89
(2002) 101301, astro-ph/0205331; Annals Phys. 303 (2003) 251, grqc/0205130; T. Prokopec and R. P. Woodard, Annals Phys. 312 (2004)
1, gr-qc/0310056.
[12] H. Delgueldre and R. P. Woodard, “One Loop Field Strengths of Charges
and Dipoles on a Locally de Sitter Background,” UF preprint in preparation.
[13] E. O. Kahya and R. P. Woodard, Phys. Rev. D72 (2005) 104001, grqc/0508015; Phys. Rev. D74 (2006) 084012, gr-qc/0608049.
[14] T. Prokopec and R. P. Woodard, JHEP 0310 (2003) 059, astroph/0309593; B. Garbrecht and T. Prokopec, Phys. Rev. D73 (2006)
064036, gr-qc/0602011.
[15] L. D. Duffy and R. P. Woodard, Phys. Rev. D72 (2005) 024023, hepph/0505156.
[16] T. Brunier, V. K. Onemli and R. P. Woodard, Class. Quant. Grav. 22
(2005) 59, gr-qc/0408080; E. O. Kahya and V. K. Onemli, Phys. Rev.
D76 (2007) 043512, gr-qc/0612026.
[17] S. P. Miao and R. P. Woodard, Class. Quant. Grav. 23 (2006) 1721,
gr-qc/0511140; Phys. Rev. D74 (2006) 044019, gr-qc/0602110.
[18] E. O. Kahya and R. P. Woodard, Phys. Rev. D76 (2007) 124005,
arXiv:0709.0536DA; Phys. Rev. D77 (2008) 084012, arXiv:0710.5282.
[19] N. C. Tsamis and R. P. Woodard, Phys. Rev. D54 (1996) 2621, hepph/9602317.
[20] G. Perez-Nadal, A. Roura and E. Verdaguer, JCAP 1005 (2010) 036,
arXiv:0911.4870.
[21] G. ’t Hooft and M. Veltman, Ann. Inst. Henri Poincaré XX (1974) 69.
[22] V. K. Onemli and R. P. Woodard, Class. Quant. Grav. 19 (2002) 4607,
qc-gr/0204065; Phys. Rev. D70 (2004) 107301, gr-qc/0406098.
44
[23] N. N. Bogoliubov and O. Parasiuk, Acta Math. 97 (1957) 227; K. Hepp,
Commun. Math. Phys. 2 (1966) 301; W. Zimmermann, Commun. Math.
Phys. 11 (1968) 1; 15 (1969) 208.
[24] L. H. Ford and R. P. Woodard, Class.and Quant. Grav. 22 (2005) 1637,
gr-qc/0411003.
[25] J. Schwinger, J. Math. Phys. 2 (1961) 407.
[26] K. T. Mahanthappa, Phys. Rev. 126 (1962) 329.
[27] P. M. Bakshi and K. T. Mahanthappa, J. Math. Phys. 4 (1963) 1; J.
Math. Phys. 4 (1963) 12.
[28] L. V. Keldysh, Sov. Phys. JETP 20 (1965) 1018.
[29] K. C. Chou, Z. B. Su, B. L. Hao and L. Yu, Phys. Rept. 118 (1985) 1.
[30] R. D. Jordan, Phys. Rev. D33 (1986) 444.
[31] E. Calzetta and B. L. Hu, Phys. Rev. D35 (1987) 495.
[32] Gradshteyn and Ryzhik
[33] E. Komatsu et al., Astrophys. J. Suppl. 180 (2009) 330,
arXiv:0803.0547.
[34] S. P. Miao, “The Fermion Self-Energy during Inflation, ” University of
Florida PhD Thesis 2007, hep-th/0705-0767.
[35] E. O. Kahya, “Quantum Gravitational Correction to Scalar Field Equations during Inflation, ” University of Florida PhD Thesis 2008.
[36] S. P. Miao and R. P. Woodard, Class. Quant.Grav. 23 (2006) 1721,
gr-qc/0511140; Phys. Rev. D74 (2006) 024021, gr-qc/0603135.
[37] E. O. Kahya and R. P. Woodard, Phys. Rev. D76 (2007) 124005,
arXiv:0709.0536; Phys. Rev. D77 (2008) 084012, arXiv:0710.5282.
[38] S. P. Miao and R. P. Woodard, Class. Quant. Grav. 25 (2008) 145009,
arXiv:0803.2377.
[39] N. E. J. Bjerrum-Bohr, J. F. Donoghue and B. R. Holstein, Phys. Rev.
D67 (2003) 084033, Erratum-ibid. D71 (2005) 069903, hep-th/0211072.
45
Coefficient of F2
α21
h
−(D−3)D2 (D+1)2 −4(D−2) + (D−1)(4y−y 2 )
β21
2(D−3)(D−1)D2 (D+1)2 (2−y)
γ21
(D−3)(D−1)D2 (D+1)2
δ21
4(D−3)D(D+1)2 −4(D−2) + D(4y−y 2 )
ǫ21
h
−4(D−3)D2 (D+1)2
Table 3: Coefficient of F2 : each term is multiplied by
i
i
1
16(D−2)(D−1)
Coefficient of F2′
h
α22
4(D−3)(D+1)2 (2−y) −2(D−2)D + (D−1)(D+1)(4y−y 2 )
β22
8(D−3)(D+1)2 −3D2 + (D−1)(D+1)(4y−y 2 )
γ22
δ22
ǫ22
h
−4(D−3)(D−1)(D+1)3 (2−y)
h
i
−16(D−3)(D+1)2 (2−y) −2(D−2) + (D+1)(4y−y 2 )
16(D−3)(D+1)3 (2−y)
Table 4: Coefficient of F2′ : each term is multiplied by
46
i
i
1
16(D−2)(D−1)
Coefficient of F2′′
α23
h
2 8(D−2)2 D(D+1) − 4(D+1)(3D3 −8D2 −6D+12)(4y−y 2 )
+(D−3)(D−1)(3D2 +9D+7)(4y − y 2 )2
h
−4(2−y) −2D(D+1)(3D2 −5D−10)
β23
+(D−3)(D−1)(3D2 +9D+7)(4y−y 2 )
h
−2 −12(D4 −D3 −7D2 +D+10)
γ23
i
+(D−3)(D−1)(3D2 +9D+72)(4y−y 2 )
h
i
−8 8(D−2)2 (D+1) − 2(D+1)(6D2 −11D−18)(4y−y 2 )
δ23
ǫ23
i
+(D−3)(3D2 +9D+7)(4y−y 2 )2
h
i
8 −2(D+1)(5D2 −6D−24) + (D−3)(3D2 +9D+7)(4y−y 2 )
Table 5: Coefficient of F2′′ : each term is multiplied by
i
1
16(D−2)(D−1)
Coefficient of F2′′′
h
−4(D−1)(2−y)(4y−y 2 ) −2(D−2)(D+1)
α24
+(D−3)(D+2)(4y−y 2 )
β24
h
i
−8 4(D−2)D(D+1) − (5D3 −8D2 −23D+22)(4y−y 2 )
+(D−3)(D−1)(D+2)(4y−y 2 )2
h
i
i
γ24
4(2−y) −4(D−2)(D2 −5) + (D−3)(D−1)(D+2)(4y−y 2 )
δ24
16(2−y)(4y−y 2 ) −2(D−2)(D+1) + (D−3)(D+2)(4y−y 2 )
ǫ24
−16(2−y) −2(D−2)(D+1) + (D−3)(D+2)(4y−y 2 )
h
h
Table 6: Coefficient of F2′′′ : each term is multiplied by
47
i
i
1
16(D−2)(D−1)
Coefficient of F2′′′′
h
i
α25
−(D−1)(4y−y 2 )2 −4(D−2) + (D−3)(4y−y 2 )
β25
2(D−1)(2−y)(4y−y 2 ) −4(D−2) + (D−3)(4y−y 2 )
γ25
δ25
ǫ25
h
h
ih
i
4(D−2) − (D−3)(4y−y 2 ) 4(D−2) − (D−1)(4y−y 2 )
h
4(4y−y 2 )2 −4(D−2) + (D−3)(4y−y 2 )
h
i
−4(4y−y 2 ) −4(D−2) + (D−3)(4y−y 2 )
Table 7: Coefficient of F2′′′′ : each term is multiplied by
48
i
i
1
16(D−2)(D−1)