The Other ADM Result
with Pedro Mora (U. Florida)
and Nick Tsamis (U. Crete)
PRL 4 (1960) 375-377
1 Loop ∞’s of
“Sub-Gravity” + Matter
+φ ‘t Hooft & Veltman (1974)
+Aµ Deser & van N. (1974)
+Ψ Deser & van N. (1974)
+Aaµ Deser, Tseng & van N. (1974)
∂4 counterterms would renormalize . .
. . . but unstable Stelle (1977)
Conspiracy of Four Principles
Continuum Field Theory ∞ Modes
Q. Mechanics Can’t have q0=p0=0
1.
2.
3.
4.
Each mode has ½ω + interactions
General Relativity Energy gravitates
Pert. Theory ω’s add at 1st order
Maybe Perturbation Theory
Gives Wrong Asymptotic Exp.
What we want:
[Tree] {1 + # (GE2/c5) + . . .}
What perturbation theory gives:
[Tree] {1 + ln(∞) (GE2/c5) + . . .}
Same as if correct series were:
[Tree] {1 + # ln(GE2/c5) (GE2/c5) + . . .}
Eg. Statistical Mechanics of
Noninteracting Bosons
Z(T,V,N) for K = [m2c4 + p2c2]½ – mc2
Z
=
=
d3 xd3p −K/kB T
e
,
(2πh̄)3
∞
V
dK e−βK (K +mc2) K 2 +2Kmc2 .
3
2π 2 h̄ c3 0
Ln[Ξ(T,V,µ)] for K = p2/2m
ln(Ξ)
∞
d3 xd3p
−nβ(K−µ)
=
ln
e
,
3
(2πh̄)
n=0
∞
1
2V
m
1
3
= √ (
dK K 2 ln
,
)
2
−β(K−µ)
π 2πh̄
0
1−e
∞
mkB T 3 −5
2 ekβµ .
= V(
k
)
2πh̄2
k=1
Expanding Z
for x = mc2/kBT << 1
t = K/kBT
∞
V
2x
kB T 3
−t t2 1+ x
dt
e
1+
(
(
)
)
2π 2
h̄c
t
t
0
3
4
Wrong: 0 at x + ∞ at x
V kB T 3 ∞ −t 2
x 1 x2
x3 1 x4
Z=
dt e t 1+2· + · 2 +0· 3 + · 4 . . .
(
)
2
2π
h̄c
t 2 t
t
8 t
0
3
4
Right: ≠ 0 at x + x ln(x)
Z =
V kB T 3
1 2 1 3 1 4
Z=
(
) 2+2x+ x − x − x ln(x) + . . .
2
2π
h̄c
2
6
48
Expanding ln(Ξ)
for -βµ = x << 1
ln(Ξ) = V nQ f (x) ≡ V nQ
∞
−5
k 2 e−kx
k=1
Wrong: f(x) = ζ(5/2) - ζ(3/2) x + ∞ x2
∞ 3
1 −1 2
−5
−
f (x) =
k 2 − k 2x + k 2x + . . .
2
k=1
Right: f(x) = (Same) + 4π½/3 x3/2 +. . .
f ′′ (x) =
∞
k=1
−1
k 2 e−kx
≈
∞
0
−1
dk k 2 e−kx =
2nd order IS small, just not ~x2
π 1
2
x
Charged shell of radius R 0
(ADM 1960)
Without GR: mc2 = m0c2 + q2/8πǫR
“renormalize” with m0c2 = mobsc2 – q2/8πǫR
With GR:
mc2=m0c2+q2/8πǫR–Gm2/2R


2
Rc 
2Gm0
Gq 2 
m=
+
−1 + 1 +
 −→
2
2
2
G
Rc
4πǫ0R c
Perturbative Result:
Oscillating series of ever-higher ∞’s
q2
4πǫ0G
Lessons from the ADM Result
Gravity might well cancel ∞’s
But not perturbatively, eg m = α½mPl
Not analytic in α
Diverges for G 0
“Perturbative Conundrum”: Grav. response
always an order behind
Hopeless to compute exactly Seek new
expansion in which “gravity can keep up”
Past Efforts
Bryce DeWitt
Isham, Salam & Strathdee
PRL 13 (1964) 114-118
PRD3 (1971) 1805-1817
PRD5 (1972) 2548-2565
Mike Duff
PRD4 (1971) 1851-1855 (+ Huskins & Rothery)
PRD7 (1973) 2317-2326
PRD9 (1974) 1837-1839
Mass from the Propagator
|k,α> has ki & ωα = [k2 + α]½
Kallen-like Rep. for <Ω|φ(x)φ*(y)|Ω>
d3 k
1
∗ (y)|Ω
Ω|ϕ(x)|αα|ϕ
(2π)3 2ωα
α
d3 k Z(α)
−iωα (x0 −y 0 ) i!
k·(!
x−!
y)
=
e
e
(2π)3 2ωα
α
M = lim[x0+∞,y0-∞] i/(x0–y0) ×
3
∗
× ln d x Ω|ϕ(x)ϕ (y)|Ω
Integrate Matter Out
S[g,A] = ∫d4x L
√
√
1
1
µρ
νσ
L=
−g
R −g − Fµν Fρσ g g
16πG
4
S[g,A,φ*,φ] = ∫d4x φ*D[g,A]φ
√
√
µν
2
D[g, A] = (∂µ +ieAµ)[ −gg (∂ν +ieAν )] − m −g
<Ω|T[φ(x)φ*(y)]|Ω>
=
−1[g, A]|y
x|iD
[dg][dA] eiS[g,A] det(D[g, A])
A Different Expansion
Stationary phase, but include
<x|iD-1[g,A]|y> with S[g,A]
Doesn’t sum classes of loops
Adds 0 = ± (New Diagrams)
Cf. perturbative comparison for
2me
im(x0−y 0 )
d3x x|iD −1[g, A]|y = 1 + . . .
Comparing Diagrams for
ln[2meim∆t∫d3x<φ(x)φ*(y)>]
Physical Interpretation: A QM
Particle Causing Its Own Fields
Recall free result for x0 > y0
Generally u[g,A](x) ∋ Du = 0
d3k
1
−iωmx0+i!k·!x × √ 1
iωm y 0−i!k·!
y
√
i∆(x; y) =
e
e
(2π)3 2ωm
2ωm
∫dx3 selects for ki = 0
x|iD −1[g, A]|y =
0th Order
u[g, A](x) × u∗ [g, A](y)
u[g,A](x) moves in gµν & Aµ
u[g,A](x) sources gµν & Aµ
Looking for Bound States
Two Cases:
Simplifications
No bound states hard scat. problem
Bound states lowest one dominates
gµνdxµdxν = -A(r)dt2 + B(r)dr2 + r2dΩ
Aµdxµ = Φ(r)dt
Use variational techniques to bound
What if there is more than one?
What about Fermions?
Same representation works for fermions
Don’t need to drop (φ*φ)2 term
NB fermion kinetic operators give
bosonic QM problem
And fermions have spin
Potential Significance of Spin
Parts of Ψ(x) seen
thru BIG γ factors
Spinning disk model
½ = ½mR2ω
Rω = /mR
R = /mc Rω = c
Cons. Parallel Strips
Top strip rest frame
RS = γ(1+β2) Gδm/c2
Conclusions
QGR ∞‘s may be perturbative artifacts
This isn’t crazy
Physics is reasonable
Many examples from simple physics
But it IS hard to check
Hopeless to compute exactly
Need alternate expansion
One example would suffice!
Our Program: Generalize the
Other ADM Result to QFT
0th order: QM particle in fields it causes
Gauge invariant
Adds 0 = (New – New) to loop expansion
Breaks perturbative conundrum
Interesting case is bound states
Normal QGR response always an order behind
What are they if more than one?
Should be solvable numerically