The Power Wall
Steve Fulling and Jef Wagner
Quantum Vacuum Workshop, TAMU, July 2010
1
Motivation
• Physical understanding requires hT µν (r)i.
• Perfectly reflecting wall ⇒ divergence.
• Ultraviolet cutoff ⇒ pressure paradox:
∂E
6= −
∂h
∂E
=+
∂x
Z
p
Z
p.
Sh
in dimension 2, intersecting planes.
Sx
2
∂E
= +2
∂h
Z
p in dimension 3
Sh
(for both intersecting planes and sphere).
• This shouldn’t happen in an internally consistent
dynamical theory.
• A steeply rising smooth potential
∗ mocks up a reflecting wall.
∗ should define a nonsingular, internally consistent
theory.
• In principle this can be done for a spherical boundary, etc., but for now we do only a plane one.
3
Renormalization
Cylinder kernel divergences ⇐⇒ heat kernel.
In dimension 3, hT 00 i should contain
• universal t−4 divergence
• t−2 term ∝ V (r)
• ln t terms ∝ V 2 and ∇2 V .
After removing these we should be able to set t = 0.
Only the first occurs where V = 0.
Elsewhere hT µν i should include V ϕ2 interaction term.
4
The model
v(x, y, z) =
0,
z<0
λz α , z > 0. (1 ≤ α ∈ R)
α
z
.
Get dimensions right: v = λ0
z0
1
α α+2
−1
z0
α+2
=
.
Only one length scale: ẑ = λ
λ0
For any α, v(z0 ) = λ0 ; increasingly steep as α → ∞.
5
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................
α=2
α=1
λ0
•
z0
6
0
ϕ(r, x ) =
X
n
∂2
∂t2
0
0
1 √
an φn (r)e−iωn x + a†n φn (r)eiωn x ].
2ω n
2
∂
T
1
00
, etc.
T = lim −
t→0
2 ∂t2
X φ (r)φ (r′ )
n
n
′
T (t, r, r ) =
e−ωn t .
−ω n
n
+∇2 − v(z) T (t, x, y, z, z ′ ) = 2δ(t)δ(x)δ(y)δ(z−z ′ ).
7
Eigenfunctions
φn (r) = eik⊥ ·r⊥ φp (z)
(r⊥ , k⊥ ∈ R2 , z ∈ R, p ∈ (0, ∞).)
2
∂
− 2 + v(z) − p2 φp (z) = 0.
∂z
r
2
sin[pz − δ(p)] when z < 0.
φp (z) =
π
8
When z > 0,
2
−
φp (z) = C(p)Pα
z
ẑ
d
α
+
z
− E Pα (z, E) = 0,
2
dz
, (ẑp)
2
,
Pα (+∞, E) = 0.
√
P2 (z, E) ∝ D 21 (E−1) ( 2 z).
√
For hard wall at z0 , P∞ (z, E) ∝ sin[ E(z − z0 )].
√
(Henceforth usually ẑ = 1, E = p (z0 = 1 = λ0 ).)
P1 (z, E) ∝ Ai(z − E),
9
The solutions must match at z = 0:
Pα (0, p2 )
.
tan δ(p) = −p ′
2
Pα (0, p )
C(p)2 =
1
2
.
2
2
2
2
−2
′
π Pα (0, p ) + p Pα (0, p )
Even for P = Ai, these formulas are unpleasant.
10
Small p
When p = 0 the solution is known:
α+2
2
1/2
1
z 2
.
Pα (z, 0) = z K α+2
α+2
Perturbation expansion:
Pα (z, E) = Pα (z, 0) + EPα(1) (z) + · · · .
−1
2
α+2
α+1
α+3
Γ
+ O(p3 ).
Γ
δ(p) = p α + 2
α+2
α+2
11
Large p (WKB)
φp (z) ∼ [p2 − v(z)]
− 41
cos
Z
a
z
p
π
,
p2 − v(z̃) dz̃ −
4
turning point a = p2/α .
Z ap
π
2
p − v(z) dz +
δ(p) =
mod π
4
0
3 1
1 1+2/α
π
B
,
= p
+ .
α
2 α
4
12
α = 1 (Airy function):
 2/3 4
2
 p 3 Γ( 3 )/Γ( 3 ), p → 0,
δ(p) ∼ 2p3
π

+ ,
p → ∞.
3
4
α = 2 (parabolic cylinder function):

5
3
 2p Γ( 4 )/Γ( 4 ), p → 0,
δ(p) ∼ πp2
π

+ ,
p → ∞.
4
4
13
Cylinder kernel calculations
Recall
X φ (r)φ (r′ )
n
n
T (t, r, r′ ) =
e−ωn t .
−ω n
n
∂2
(2)
2
′
′
+∇
−
v(z)
T
(t,
r
,
z,
z
)
=
2δ(t)δ
(r
)δ(z
−
z
).
⊥
⊥
∂t2
φp (z ′ )
−2
.
T̂ (ω, k⊥ , p) =
2
2
2
3/2
ω + k⊥ + p
(2π)
14
Cartesian calculations
1
′
T (t, r⊥ , z, z ) = −
2π
Z
∞
0
dp Y (s, p)φp (z)φp (z ′ ),
p
e−sp
Y (s, p) ≡
,
s ≡ t2 + |r⊥ |2 .
s
In potential-free region, z < 0,
Z ∞
1
′
dp Y (s, p) sin pz − δ(p) sin pz − δ(p) .
T =− 2
π 0
15
1
1
2π 2 t2 + r⊥2 + (z − z ′ )2
Z ∞
1
′
dp Y (s, p) cos p(z + z ) − 2δ(p)
+ 2
2π 0
T =−
≡ T free + T ren .
Hard wall: δ(p) = z0 p ⇒ (correctly)
T ren
1
1
=
.
2π 2 t2 + r⊥2 + (z + z ′ − 2z0 )2
16
The bad news:
∞
e−sp
′
cos p(z + z ) − 2δ(p)
dp
T ren
s
0
√
is poorly convergent when s ≡ t2 + r⊥2 is small,
which is precisely where we want it. In fact, we should
be able to take s = 0 and get a finite answer when
z + z ′ > 0, but the integrand is pointwise infinite there!
There is a genuine divergence for δ(p) = Ap + B unless
B = 0. Asymptotics for small p as well as large is
critical. Presumably large k⊥ is at fault, so . . .
1
=
2π 2
Z
17
Polar calculations
Do the integral in polar coordinates in the Fourier
space: (Z ≡ z + z ′ , s = (t, r⊥ ), v = (ω, k⊥ ))
T ren
∞
1
eiv·s
= 4
dv 2
cos pZ − 2δ(p)
dp
2
4π 0
v +p
R3
Z ∞ Z 1
p
1
−1
= 3
dρ
du s sin(sρ 1 − u2 )
π 0
0
× cos Zρu − 2δ(ρu) .
Z
Z
18
With s = 0 and z = z ′ ,
Z ∞ Z 1
p
1
T ren (0, 0, z, z) = 3
dρ
du ρ 1 − u2
π 0
0
× cos 2zρu − 2δ(ρu) .
Integration by parts (appealing to δ(0) = 0 !) improves
the convergence but worsens the algebra:
#
" √
Z ∞ Z 1
1
d
1 − u2
T ren (0, 0, z, z) = − 3
dρ
du
2π 0
du z − δ ′ (ρu)
0
× sin 2zρu − 2δ(ρu) .
19
Numerics (barely started)
• Padé interpolation between small and large p asymptotics works fairly well.
• The oscillatory integrals appear intractable so far,
even with Riesz–Cesàro averaging.
20