Adventures Near a Sphere
helped by Bruce Liu, Mai Truong, Kevin Resil
Goals:
• Study vacuum stress near a curved surface by classical-path analysis.
• Resolve discrepancy between energy and pressure
calculations by partial-wave analysis (for scalar
field, with exponential ultraviolet cutoff).
All constants and signs are preliminary and subject to
change. (So are many conclusions.)
1
“The whole semiclassical approximation is only of
pedagogical value in the case of a sphere, since the
exact spectral representation of the Green function of
a massless particle excluded from a spherical region is
known.”
Schaden & Spruch, Ann. Phys. 313 (’04) 37
2
The paradox
Divergent surface energy E ∝ (4πa2 )t−3 (same inside
and out).
∂E
Generalized force F = −
∝ −2a(4π)t−3 .
∂a
F
2 −3
Radial pressure pr =
t .
∝
−
2
4πa
a
1
The constant is (probably)
× 2, certainly not 0.
8π
3
But
• Milton [PRD 68 (’03) 065020] following Bender and
Milton [PRD 50 (’94) 6547] finds no divergent pressure. No obvious a−1 term even at mode sum level!
• Fulling et al. [0806.2468] in rectangle find no divergent pressure with divergent energy next to a
side; divergence goes with energy near perpendicular side! In sphere, there is no perpendicular side.
α
β
4
Integral kernels in billiards
Heat: K(t, r, r′ ) = hr|e−tH |r′ i.
−3/2 −|r−r′ |2 /4t
K0 = (4πt)
e
.
−3/2 −|r−r′ |2 /4t
−itH
Quantum: U = he
i. U0 = (4πit)
e
.
* √ +
1
e−t H
1
√
Cylinder: T =
.
. T0 = − 2 2
′
2
2π t + |r − r |
− H
Resolvent: G(k, r, r′ ) = hr|(H − k 2 )−1 |r′ i.
G0 =
(also wave kernels and zeta function)
5
ik|r−r′ |
e
.
′
4π|r − r |
Eigenvalue density σ(k) dk =
1
Im G(r, r′ , k) d(k 2 ).
π
1 ∂2
1
Energy density ρ = −
).
T
(if
ξ
=
4
2
2 ∂t
Laplace transform in k 2 : σ → K, K → G.
Laplace transform in k: σ → T .
Z
∞
0
−1/2 −τ 2 /4t
t
e
√
K(t) dt = − π T (τ )
because T is G (with k = 0) in one higher dimension,
or
√
Z ∞
2
2
π −τ k
−1/2 −τ /4t −k t
e
.
t
e
e
dt =
k
0
6
Synge–DeWitt formalism: [Christensen, PRD 14
(’76) 2490; Molzahn et al., Ann. Phys. 204 (’90) 64]
l(r, r′ , y) ≡ distance from r′ to r along (say) a straight
path with specular reflections.
σ(r, r′ ) ≡ 12 l(r, r′ )2 . Then (∇ ≡ ∇r )
(1) ∇σ = l∇l = ln̂, n̂ ≡ unit vector at r in the
direction of the path;
(2) (∇σ)2 = l2 = 2σ;
(3) ∇2 σ = 1 + l∇ · n̂ = d + O(l2 ) (d = 3 today);
(4) For the direct path,
σ = 21 |r − r′ |2 ,
∇σ = r − r′ ,
7
∇2 σ = d.
Optical approximation: We attempt to construct
(approximately) the kernel for the billiard problem in
the form of a sum over specularly reflecting paths of
terms
Gj = (−1)j Dj (r, r′ )F (σ(r, r′ )),
where (−1)j means the parity of the number reflections (so that the Dirichlet condition is satisfied by
the sum), and D does not depend on the parameter t
or k (though F does).
Claim: D comes out the same for all the kernels. (for billiards!)
8
Proof and construction
• Plug ansatz into PDE.
• Group terms by order of singularity (for T , powers
of (t2 + 2σ)−1 ).
• Leading term vanishes if F ( 21 |r − r′ |) = G0 (·, r, r′ ).
(F depends on t or k.)
• Next term vanishes if D ≡ | det M |1/2 ,
Mjk
∂2σ
≡
.
∂rj rk′
det M = ld−1 ∆, ∆(r, r′ ) ≡ enlargement factor d(angle)
d(area)
[Scardicchio & Jaffe, NPB 704 (’05) 552].
9
Energy and pressure in spherical symmetry
2
2
2
1 ∂ T
∂ T
∂ T
2 ∂T
+
β
+
+
.
2
′
2
2 ∂t
∂r ∂r
∂r
r ∂r
2
2
∂ T
2β ∂T
1 ∂ T
r
.
−
−
pr = hTr i = −
4 ∂r ∂r ′
∂r 2
r ∂r
ρ = −hT00 i = −
p⊥ = hTθθ i = hTϕϕ i =
2
[β ≡ ξ − 14 ]
2
2
∂ T
∂ T
1 ∂T
1 ∂T
1 ∂ T
−
β
+
+
.
+
′
2
′
2
2r ∂θ ∂θ
4r ∂r
∂r ∂r
∂r
r ∂r
[Cf. Schwartz-Perlov & Olum, PRD 72 (’05) 065013;
Cavero-Peláez et al., PRD 73 (’06) 085004.]
−
10
The geometrical quantities
Sometimes write r1 and r2 instead of r and r ′ , etc. Ignore factor −1/2π 2 . T will mean the single-reflection
term to be subtracted from T 0 in Dirichlet case. From
S&J (± ≡ (out / in), ω ≡ reflection angle)
−1 −1
l ± 2l1 l2 cos ω
l ± 2l1 l2
.
∆=
a cos ω
a
l = l1 + l2 . When r′ = r,
∆ = l−2 1 ±
l = 2|r − a| = ±2(r − a),
11
l
2a
−2
.
ω = αj
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α
α
•
β
l
r
γ γ
r
β
•
θ
12
•
β
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l
α
r
α
ω = π − α2
l
β
γ γ
r
θ
13
•
Let ∆θ ≡ θ 1 − θ 2 > 0, u ≡ sin ω. Solve the triangles:
Interior:
q
p
lj = a 1 − u2 − rj2 − a2 u2 ,
au
au
−1
−1
−1
∆θ = −2 sin u + sin
+ sin
.
r1
r2
Exterior:
q
p
lj = −a 1 − u2 + rj2 − a2 u2 ,
au
au
∆θ = 2 sin−1 u − sin−1
− sin−1
.
r1
r2
14
Proceed by implicit differentiation. (Signs checked for
interior case only.)
∂∆θ
a
a
2
B≡
+p 2
.
=−√
+p 2
2
2
2
2
2
∂u
1−u
r1 − a u
r2 − a u
∂u
∂u
1
=−
= ,
∂θ 1
∂θ2
B
∂u
au
q
.
=
∂rj
Brj rj2 − a2 u2
Mathematica from then on!
15
Derivatives of σ: If u = 0 (path perpendicular to
sphere),
∂σ
= 0,
∂θ j
∂σ
= r1 + r2 − 2a = −l,
∂rj
∂2σ
∂2σ
=1=
,
∂r1 ∂r2
∂rj2
∂2σ
= 0,
∂r1 ∂θ2
etc.,
∂2σ
∂2σ
r1 + r2 − 2a
= − 2 = r1 r2 a
.
∂θ 1 ∂θ2
a(r1 + r2 ) − 2r1 r2
∂θ j
′
When r = r ,
r1 r2 a
∂2σ
= −ra.
=−
∂θ1 ∂θ 2
r
16
Derivatives of D: If u = 0,
2a2 − ar1 − ar2
D=
.
ar1 + ar2 − 2r1 r2
(“→” indicates setting both rj = r)
2a(a − r2 )2
a
∂D
=−
→− 2,
∂r1
(−2r1 r2 + a(r1 + r2 ))2
2r
4a(a − 2r2 )(a − r2 )2
a(a − 2r)
∂2D
=
→
,
2
3
3
∂r1
(−2r1 r2 + a(r1 + r2 ))
2(a − r)r
17
∂2D
4a2 (a − r1 )(a − r2 )
a2
=
→
.
3
3
4
∂r1 ∂r2
(−2r1 r2 + a(r1 + r2 ))
2ar − 2r
Note especially that
∂2D
a
∂2D
−
→
−
∂r12
∂r1 ∂r2
r 2 (a − r)
diverges at the boundary, r → a.
When r ′ = r, 1st and 2nd derivatives w.r.to θj are 0
when u = 0 (very complicated otherwise).
18
Derivatives of T
For any smooth curved surface,
∂T
= −2tD(t2 + 2σ)−2 ,
∂t
∂2T
2
2
−3
2
−2
=
8t
D(t
+
2σ)
−
2D(t
+
2σ)
,
2
∂t
∂D 2
∂T
2
−2 ∂σ
= −2D(t + 2σ)
+
(t + 2σ)−1 ,
∂r
∂r
∂r
19
2
2
2
∂ T
∂σ
2
−3
2
−2 ∂ σ
= 8D(t + 2σ)
− 2D(t + 2σ)
2
∂r
∂r
∂r 2
2
∂
D
∂σ
∂D
2
−1
2
−2
+ (t + 2σ)
,
− 4(t + 2σ)
∂r ∂r
∂r 2
2
∂2T
2
−3 ∂σ ∂σ
2
−2 ∂ σ
= 8D(t + 2σ)
−2D(t + 2σ)
′
′
′ ∂r
∂r ∂r
∂r
∂r
∂r
∂D ∂σ
∂D ∂σ
− 2(t2 + 2σ)−2
+
∂r ∂r ′
∂r ′ ∂r
2
∂
D
2
−1
,
+ (t + 2σ)
∂r ′ ∂r
and similarly for θ derivatives.
20
For sphere, with r′ = r, ξ = 14 ,
D=
l
1±
2a
−1
a
= ,
r
2σ = 4(a − r)2 .
Energy:
∂2T
2a 2
8t2 a 2
−3
−2
(t
+
2σ)
−
(t
+
2σ)
=
∂t2
r
r
2a 3t2 − 4(a − r)2
.
=
r [t2 + 4(a − r)2 ]3
21
Tangential pressure:
1 ∂T
a
a
= 4(a − r) 2 (t2 + 2σ)−2 − 3 (t2 + 2σ)−1 .
r ∂r
r
2r
First term = 0 on sphere. Sign changes in exterior(?).
a 2
1 ∂T
2
−2
(t
+
2σ)
.
=
2
′
r 2 ∂θ ∂θ
r
a
a
p⊥ = −
(t2 + 2σ)−2 − 3 (t2 + 2σ)−1 .
r
8r
So far, ρ and p⊥ are consistent with flat plate.
22
Radial pressure:
2
∂ T
2a 2al
16aσ 2
−3
(t + 2σ) −
+ 2
=
∂r 2
r
r
r
a(a − 2r) 2
−1
(t
+
2σ)
;
+
3
2(a − r)r
(t2 + 2σ)−2
∂2T
= (same top line)
′
∂r ∂r
a2
2
−1
(t
+
2σ)
.
+
3
2(a − r)r
All but last terms cancel in pr ; last terms diverge!
23
Why? Optical approximation and even higher-order
DeWitt–Christensen expansions break down at boundary [McAvity & Osborn, CQG 8 (’91) 603].
Possible responses:
1. Get pr from conservation law.
2. Use full multiple-reflection formula for T .
Conservation law
∂pr
2pr
2
+
− p⊥ = 0.
∂r
r
r
24
C
pr = 2
r
(homogeneous soln.)
a(a − r)
1
+
t2 r 2 t2 + 2σ
2
2
3
t
2a 4a + 3t
−1
tan
− 3 2
t r (4a2 + t2 )2
l
2
a 2
t + 2σ
1
ln
+
r
(4a2 + t2 )2
r2
1
a
.
+ 3 2
4r 4a + t2
25
π
(r = a, t ≪ a);
3
4at
π
correct pressure for surface energy, E =
, and
3
8t
should change sign on outside.
Third term → −
But how do I know that C = 0? Can’t look at r = 0
because optical approx. blows up there (caustic).
Also, E is independent of ξ, so pr (a) must be. How
does that turn out? (ξ = 16 ⇒ leading terms in ρ and
p⊥ vanish.)
26
Multiple-reflection expansion
[Balian & Bloch, Ann. Phys. . . . ]
T (r, r′ , t) =
Z
∞
Im G(r, r′ , k)e−kt dk.
0
Change notation: r1 and r2 ≡ distances of r and r′
from integration variable r∗ on the sphere; l = r1 + r2 .
ikr1 ikr2
∂
e
e
Im G(1 refl.) ∝ Im
∂r1
r1
r2
sin k(r1 + r2 ) k cos k(r1 + r2 )
+
.
=−
r12 r2
r1 r2
27
So single-reflection term is
Z
2
2
1
l
l −t
1
T ∝ dS cos φ
+ 2
,
r1 r2 (l2 + t2 )2
r1 r2 l2 + t2
φ = angle between r − r∗ and n̂.
When r′ = r, r ≡ |r1 |, z ≡ l = 2r1 , after geometry
2
Z
2
2
3z
+
t
z
1
+ a2 − r 2
T ∝ −4 dz 2
rz
4
(z 2 + t2 )2
lmax =2(a+r)
2
2
2
z + 4(a − r )
2
−1 z =
−
tan
.
r(t2 + z 2 )
rt
t lmin =2|a−r|
28
•
•
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..
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a
a
r
•
•
a
•
r
•
29
•
?
Lower limit’s contribution:
(a − r)2 + (a2 − r 2 )
2
−1
T ∝ −2
+
tan
r|a − r|(t2 + 4(a − r)2 ) rt
lmin
t
First term is the optical approximation:
1
D
a
= 2
.
T ∝
r t2 + 2σ
t + 2σ
But
• Sign discrepancy between interior and exterior?
• Upper limit’s term should have caustic sign?
• Regularity at r = 0?
• More reflections irrelevant?
30
.