Towards solving hierarchy problem with asymptotically safe gravity
Masatoshi Yamada
(Kanazawa Univ. → Kyoto Univ. → Heidelberg Univ.)
with Kin-ya Oda (Osaka Univ.) and Yuta Hamada (KEK & Wisconsin Univ.)
ERG2016@Trieste
Towards solving hierarchy problem with asymptotically safe gravity - - PowerPoint PPT Presentation
Towards solving hierarchy problem with asymptotically safe gravity Masatoshi Yamada (Kanazawa Univ. Kyoto Univ. Heidelberg Univ.) with Kin-ya Oda (Osaka Univ.) and Yuta Hamada (KEK & Wisconsin Univ.) ERG2016@Trieste LHC The
(Kanazawa Univ. → Kyoto Univ. → Heidelberg Univ.)
with Kin-ya Oda (Osaka Univ.) and Yuta Hamada (KEK & Wisconsin Univ.)
ERG2016@Trieste
The ATLAS and CMS collaborations, JHEP 08, 045
coupling to Higgs Particle mass
gi = 1 hhimi
taken from resonaances
What can we do at present? How to approach to them?
possible candidates.
to gravity
✖ ✖ = + +・・・
λ
Z Λ d4p 1 p2 ∼ Λ2
m2
Λ
m2
R = m2 Λ + Λ2
16π2 (λ + · · · )
R ⌧ m2 Λ
Γk = Z d4x 1 2(∂µφ)2 − m2
k
2 φ2 − λk 4 φ4
m2
k
k2 m2
k < 0
m2
k > 0
m2
k
k2 = − C 8π2 λk
λk m2
k
k2
m2
R = m2 Λ + CΛ2
8π2 λk=0
m2
R
(m2
Λ, λΛ)
λΛ ' λk=0
m2
k
k2 = − C 8π2 λk
λk m2
k
k2
m2
R = m2 Λ + CΛ2
8π2 λk=0
(m2
Λ, λΛ)
λΛ ' λk=0 = 0
m2
k
k2 = − C 8π2 λk
m2
Λ = −CΛ2
8π2 λΛ
m2
R
2 determines the position of phase boundary (critical line).
theory.
phase boundary.
λk m2
k
k2 m2
k < 0
m2
k > 0
m2
k
k2 = − C 8π2 λk
2
is spurious?
2
is always subtracted by the counter term or dimensional regularization.
C = 0
µdm2 dµ = m2 16π2 (12λ + · · · )
λk
m2
k
k2 m2
k
k2 = 0
λ0
k
m2
k
k2
m2
k
k2 = − C 8π2 λk
m2
k
k2 = − C 8π2 λk
m2
k
k2 = 0
boundary.
to gravity
by relevant operators.
k∂kΓk = 1 2Str[(Γ(2)
k
+ Rk)−1k∂kRk] g1 g2 gi Γk = Z d4x[g1O1 + g2O2 + · · · + giOi + · · · ] Γk ' Z d4x[g1O1 + g2O2]
S = ΓΛ
Γ = Γk=0
exact flow truncated flow
skippable!
釈迦に説法 (preaching to the experts)
gi(k) = g∗
i + N
X
j
ζi
j
✓Λ k ◆θj
θi > 0
θi < 0
relevant irrelevant
eigenvalue k → 0
∂t = −k∂k
∂tgi = βi(g∗) + ∂βi ∂gj
j ) + · · ·
f(R), αR + βR2 + γRµνRµν, etc.
Cf.
Cf.
etc.
Why is the universe critical?
Taken from Wiki M, Reuter, F. Saueressing, Phys. Rev. D65, 065016
Λcc << 10-120
to gravity
V (φ2) = Λcc + m2φ2 + λφ4 + · · · F(φ2) = M 2
pl + ξφ2 + · · ·
Γk = Z d4xpg 1 2gµν∂µφ∂νφ + V (φ2) F(φ2)R + ¯ ψ / rψ + yφ ¯ ψψ
gµν = ¯ gµν + hµν Rk(z) = (k2 − z)θ(k2 − z) Rk(z − R/4) = (k2 − (R/4))θ(k2 − (z − /R/4))
scalar and gravity fermion
{Mpl2, Λcc, m2, ξ, λ}
M 2
pl, Λcc
m2, ξ
λ
2.143 ± 2.879i
0.143 ± 2.879i
−2.627
θi =
¯ M 2
pl ∗ = 2.38 × 10−2
¯ Λ∗
cc = 8.82 × 10−3
¯ m2∗ = ¯ ξ∗ = ¯ λ∗ = 0
{Mpl2, Λcc, m2, ξ, λ, y}
M 2
pl, Λcc
m2, ξ
λ θi =
−0.4909 ± 2.461i 1.509 ± 2.4615i −2.6069 −1.464
y
¯ Λ∗
cc = 3.72 × 10−3
¯ M 2
pl ∗ = 1.63 × 10−2
Without non-minimal coupling:
¯ m2∗ = ¯ ξ∗ = ¯ λ∗ = ¯ y∗ = 0
2 is generated, m 2 grows up due to the canonical scaling
(2m
2).
2, Λcc is still needed.
∂t ¯ m2 = 2 ¯ m2 − 1 48π2 9¯ Λcc
ξ
¯ Mpl − ¯ Λcc 2 − 9
Λcc − ¯ Mpl 1 + 2¯ ξ 2 2 (1 + 2 ¯ m2) ¯ Mpl − ¯ Λcc 2 − 9
ξ 2 2 (1 + 2 ¯ m2)2 ¯ Mpl − ¯ Λcc − 18λ (1 + 2 ¯ m2)2
Mpl − 2 ¯ Mpl 96π2 ¯ Mpl − 2¯ ξ ¯ Mpl + 3 ¯ Mpl
ξ
¯ Mpl − ¯ Λcc 2 − 3 ¯ Mpl
ξ 2 2 (1 + 2 ¯ m2) ¯ Mpl − ¯ Λcc 2
1 96π2 ∂t ¯ ξ ¯ Mpl 2 − 3 ¯ Mpl ¯ Mpl − ¯ Λcc + 6 ¯ Mpl
ξ
m2) ¯ Mpl − ¯ Λcc
8π2 ,
∂t = −k∂k
m2
k → ∞, m2 → 0
k ∼ Mpl
m2 ∼ finite
Higgs mass
Gravitational couplings
{Mpl, Λcc} m2
Criticality of Higgs mass ⇄ Criticality of the universe
SJ = Z d4x√−g 1 + ξ h2 M 2
pl
+ · · · ! M 2
pl
2 R + 1 2(∂µh)2 − V
1 + ξ h2 M 2
pl
+ · · · ! gµν → gE
µν
V
→ V
⇣ 1 + ξ h2
M 2
pl + · · ·
⌘2
largeξ
∂tξ = − 1 576π2 1 + 2 ¯ m2 ¯ M 2
pl − ¯
Λcc 9 + 39 ¯ M 2
pl
¯ M 2
pl − ¯
Λcc + 60 ¯ M 2
pl 2
Λcc 2 ! + 3 (3 + 32ξ) ¯ M 2
pl − ¯
Λcc − 6 ¯ M 2
pl (11 + 2ξ)
⇣ ¯ M 2
pl − ¯
Λcc ⌘2 − 60 ¯ M 2
pl 2 (1 + 2ξ)
⇣ ¯ M 2
pl − ¯
Λcc ⌘3 + 216ξ (1 + 2ξ)2 (1 + 2 ¯ m2)3 ⇣ ¯ M 2
pl − ¯
Λcc ⌘ + 9 h ¯ Λcc (5 − 2ξ) − 2 ¯ M 2
pl (1 + 2ξ)
i (1 + 2ξ) (1 + 2 ¯ m2)
Λcc 2 + 27 (1 + 2ξ)
(1 + 2 ¯ m2)2 ⇣ ¯ M 2
pl − ¯
Λcc ⌘ + 108 ¯ M 2
plξ (1 + 2ξ)2
(1 + 2 ¯ m2)2 ⇣ ¯ M 2
pl − ¯
Λcc ⌘2 + 72λ4 (1 + 2 ¯ m2)2 1 + 12ξ + 2 ¯ m2 1 + 2 ¯ m2
∂t ¯ M 2
pl − 2 ¯
M 2
pl
1152π2 ¯ M 2
pl
1 + 2 ¯ m2 ¯ M 2
pl − ¯
Λcc ✓ 3 + 18 ¯ M 2
pl
¯ M 2
pl − ¯
Λcc + 20 ¯ M 2
pl 2
⇣ ¯ M 2
pl − ¯
Λ2
¯ mcc
⌘2 ◆ + 15ξ ¯ M 2
pl
− 6 (1 + ξ) ¯ M 2
pl − ¯
Λcc − 10 ¯ M 2
pl (3 + 4ξ)
⇣ ¯ M 2
pl − ¯
Λcc ⌘2 − 20 ¯ M 2
pl 2 (1 + 2ξ)
⇣ ¯ M 2
pl − ¯
Λcc ⌘3 − 3 h ¯ Λcc − ¯ M 2
pl (5 − 4ξ)
i (1 + 2ξ) (1 + 2 ¯ m2) ⇣ ¯ M 2
pl − ¯
Λcc ⌘2 + 36 ¯ M 2
plξ (1 + 2ξ)2
(1 + 2 ¯ m2)2 ⇣ ¯ M 2
pl − ¯
Λcc ⌘2
∂tξ 1152π2 ¯ M 2
pl
− 15 + 54 ¯ M 2
pl
¯ M 2
pl − ¯
Λcc + 20 ¯ M 2
pl 2
⇣ ¯ M 2
pl − ¯
Λcc ⌘2 − 6 ¯ M 2
pl (7 + 2ξ)
(1 + 2 ¯ m2) ⇣ ¯ M 2
pl − ¯
Λcc ⌘ − 144 ¯ M 2
plξ (1 + 2ξ)
(1 + 2 ¯ m2) ⇣ ¯ M 2
pl − ¯
Λcc ⌘
y2 48π2 ,
2, ξ becomes irrelevant.
symmetry: βy ∝ y
1604.02041
(in progress with Y. Hamada)
_
2 and Λcc prefer the critical?
2 and Λcc become
small?
energy.
Sgf = 1 α Z d4x√¯ gF
¯ gµνΣµΣν
Sgh = Z d4x√¯ g ¯ Cµ − δρ
µ ¯
D2 − ✓ 1 − 1 + β 2 ◆ ¯ Dµ ¯ Dρ + ¯ Rρ
µ
Σµ = ¯ Dνhνµ − β + 1 4 ¯ Dµh
¯ M 2
pl ∗ = 2.38 × 10−2
¯ Λ∗
cc = 8.82 × 10−3
¯ Λ∗
cc = 3.72 × 10−3
¯ M 2
pl ∗ = 1.63 × 10−2
∂βi ∂gj '
∂βξ ∂ξ ∂βξ ∂m2 ∂βm2 ∂ξ ∂βm2 ∂m2
✓2.85544 −6.51993 2.40051 −2.57031 ◆ ✓ 1.6814 −5.39674 1.99718 −2.66334 ◆
θi ' 1 2 ✓∂βξ ∂ξ + ∂βm2 ∂m2 ◆
θi ' 2.855442.57031 2 > 0
θi ' 1.68142.66334 2 < 0