How does Casimir energy fall? Renormalization of
Gravitational and Inertial Masses
K. V. Shajesh
Oklahoma Center for High Energy Physics and Homer L. Dodge Department of Physics and
Astronomy, University of Oklahoma, Norman, OK 73019, USA
Collaborators: S. Fulling, K. A. Milton, P. Parashar, A. Romeo, J. Wagner
Date: Aug 06 - 08, 2007
Event: Quantum Vacuum meeting - 2007
Venue: Texas A & M University, Texas, USA.
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
1 / 32
Outline
1
Formalism in zero gravity
Single plate in zero gravity
Parallel plates in zero gravity
2
Rindler Coordinates
3
Formalism for parallel plates falling in a constant gravitational field
Single plate falling in a constant gravitational field
Parallel plates falling in a constant gravitational field
4
Renormalization
5
Report on current work in progress
Definition of force
Renormalization revisited
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
2 / 32
In the following the superscript (0) will be used to signify that it is the
zeroth order effect of gravity.
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
3 / 32
Formalism in zero gravity
Let us consider a scalar field φ(x) in the presence of a background V (0) (x)
described by the action
Z
(0)
W = d 4 x L(0) (φ(x))
written in terms of the Lagrangian density
1
1
L(0) (φ(x)) = − ∂µ φ(x)∂ µ φ(x) − V (0) (x)φ(x)2 .
2
2
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
4 / 32
Formalism in zero gravity
Let us consider a scalar field φ(x) in the presence of a background V (0) (x)
described by the action
Z
(0)
W = d 4 x L(0) (φ(x))
written in terms of the Lagrangian density
1
1
L(0) (φ(x)) = − ∂µ φ(x)∂ µ φ(x) − V (0) (x)φ(x)2 .
2
2
The background field V (0) (x) is used as a mathematical device to design
the plates. This is achieved by a judicious choice of the background field
which provides the appropriate boundary conditions to φ(x).
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
4 / 32
Formalism in zero gravity
Let us consider a scalar field φ(x) in the presence of a background V (0) (x)
described by the action
Z
(0)
W = d 4 x L(0) (φ(x))
written in terms of the Lagrangian density
1
1
L(0) (φ(x)) = − ∂µ φ(x)∂ µ φ(x) − V (0) (x)φ(x)2 .
2
2
The background field V (0) (x) is used as a mathematical device to design
the plates. This is achieved by a judicious choice of the background field
which provides the appropriate boundary conditions to φ(x).
Example: So called semitransparent parallel plates are represented by
V (0) (x) = λa δ(x) + λb δ(x − a)
Limit λa,b → ∞ leads to Dirichlet boundary conditions.
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
4 / 32
Zero gravity: Energy momentum tensor
Equation of motion: (Variation in φ(x))
h
i
∂ 2 − V (0) (x) φ(x) = 0
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
5 / 32
Zero gravity: Energy momentum tensor
Equation of motion: (Variation in φ(x))
h
i
∂ 2 − V (0) (x) φ(x) = 0
Energy momentum tensor (stress tensor): A class of arbitrary variation in
the field when generated due to infinitesimal general coordinate
transformation in space time lets us identify the energy-momentum tensor
(0)
tαβ (x) = ∂α φ(x)∂β φ(x) + gαβ L(0) (φ(x))
which is symmetric, tαβ = tβα .
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
5 / 32
Zero gravity: Energy momentum tensor
Equation of motion: (Variation in φ(x))
h
i
∂ 2 − V (0) (x) φ(x) = 0
Energy momentum tensor (stress tensor): A class of arbitrary variation in
the field when generated due to infinitesimal general coordinate
transformation in space time lets us identify the energy-momentum tensor
(0)
tαβ (x) = ∂α φ(x)∂β φ(x) + gαβ L(0) (φ(x))
which is symmetric, tαβ = tβα .
Conformal term: Identification of the stress tensor using this prescription is
(0)
unique only up to an additive term that satisfies ∂ β tαβ = 0.
(0)
tαβ (x) = ∂α φ(x)∂β φ(x) + gαβ L(0) (φ(x)) − θ ∂α ∂β − gαβ ∂ 2 φ(x)2
θ is called the conformal coupling constant. The particular choice, θ = 61 ,
which renders the stress tensor traceless, is called the conformal choice.
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
5 / 32
Zero gravity: Force
Force density: The field experiences a force due to the presence of the
background field, (the plate, or the source.)
(0)
fα(0) (x) = ∂ β tαβ (x)
1
= − φ(x)2 ∂α V (0) (x)
2
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
6 / 32
Zero gravity: Force
Force density: The field experiences a force due to the presence of the
background field, (the plate, or the source.)
(0)
fα(0) (x) = ∂ β tαβ (x)
1
= − φ(x)2 ∂α V (0) (x)
2
Force on the apparatus: The force density integrated over the complete
space time is interpreted to give the total change in four momentum
associated with the field
Z
Z
(0)
∆Pα = dt d 3 x fα(0) (x)
R
(0)
(0)
which is compatible with the definition ∆Pα = dσ β tβα (x), where σ β
is a time-like four vector, if we use Green’s theorem.
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
6 / 32
Zero gravity: Force
Force density: The field experiences a force due to the presence of the
background field, (the plate, or the source.)
(0)
fα(0) (x) = ∂ β tαβ (x)
1
= − φ(x)2 ∂α V (0) (x)
2
Force on the apparatus: The force density integrated over the complete
space time is interpreted to give the total change in four momentum
associated with the field
Z
Z
(0)
∆Pα = dt d 3 x fα(0) (x)
R
(0)
(0)
which is compatible with the definition ∆Pα = dσ β tβα (x), where σ β
is a time-like four vector, if we use Green’s theorem.
Casimir Force: Integration restricted to one plate.
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
6 / 32
Zero gravity: Green’s function
At the one loop level the Green’s function G (0) (x, x 0 ) satisfying
h
i
− ∂ 2 − V (0) (x) G (0) (x, x 0 ) = δ 4 (x − x 0 )
is related to the fields by the correspondence
1
hT φ(x)φ(x 0 )i = G (0) (x, x 0 )
i
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
7 / 32
Zero gravity: Green’s function
At the one loop level the Green’s function G (0) (x, x 0 ) satisfying
h
i
− ∂ 2 − V (0) (x) G (0) (x, x 0 ) = δ 4 (x − x 0 )
is related to the fields by the correspondence
1
hT φ(x)φ(x 0 )i = G (0) (x, x 0 )
i
(0)
For the case when V (x) has dependence only on z
Z +∞
Z 2
dω
d k⊥ −iω(t−t 0 ) ik⊥ ·(x⊥ −x0 ) (0)
⊥ g
G (0) (x, x 0 ) =
e
e
(z, z 0 )
2
2π
(2π)
−∞
where κ2 = k2⊥ − ω 2 and g (0) (z, z 0 ) satisfies
2
∂
2
(0)
−
− κ − V (z) g (0) (z, z 0 ) = δ(z − z 0 ).
∂z 2
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
7 / 32
Zero gravity: Green’s function
At the one loop level the Green’s function G (0) (x, x 0 ) satisfying
h
i
− ∂ 2 − V (0) (x) G (0) (x, x 0 ) = δ 4 (x − x 0 )
is related to the fields by the correspondence
1
hT φ(x)φ(x 0 )i = G (0) (x, x 0 )
i
(0)
For the case when V (x) has dependence only on z
Z +∞
Z 2
dω
d k⊥ −iω(t−t 0 ) ik⊥ ·(x⊥ −x0 ) (0)
⊥ g
G (0) (x, x 0 ) =
e
e
(z, z 0 )
2
2π
(2π)
−∞
where κ2 = k2⊥ − ω 2 and g (0) (z, z 0 ) satisfies
2
∂
2
(0)
−
− κ − V (z) g (0) (z, z 0 ) = δ(z − z 0 ).
∂z 2
Force using Green’s function:
hfα(0) (x)i = −
K. V. Shajesh (University of Oklahoma)
1 (0)
G (x, x) ∂α V (0) (x).
2i
How does Casimir energy fall?
QV-2007
7 / 32
Zero gravity: Energy density
The energy density of the field is defined to be the t00 (x) component of
the stress tensor
(0)
t00 (x) =
1
1 ∂φ ∂φ 1 ∂ 2
− φ 2 φ + (1 − 4θ) ∇ · ∇ φ2
2 ∂t ∂t
2 ∂t
4
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
8 / 32
Zero gravity: Energy density
The energy density of the field is defined to be the t00 (x) component of
the stress tensor
(0)
t00 (x) =
1
1 ∂φ ∂φ 1 ∂ 2
− φ 2 φ + (1 − 4θ) ∇ · ∇ φ2
2 ∂t ∂t
2 ∂t
4
Energy density in terms of Green’s function:
(0)
(0)
(0)
t00 (x) = Evol (x) + (1 − 4θ) Esur (x)
where we have introduced the definitions
1
∂2
∂ ∂
(0)
(0)
0
Evol (x) =
−
G (x, x )
2i
∂t ∂t 0 ∂t 2
x=x 0
1
(0)
∇ · ∇ G (0) (x, x).
Esur (x) =
4i
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
8 / 32
Zero gravity: Single plate
A single plate positioned at z = za is described by
V (0) (z) = λa δ(z − za ).
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
9 / 32
Zero gravity: Single plate
A single plate positioned at z = za is described by
V (0) (z) = λa δ(z − za ).
Total force on the plate:
(0)
Fa
λa
=−
A
2
+∞
Z
−∞
dω
2π
Z
d 2 k⊥
(2π)2
dζ
2π
Z
d 2 k⊥ 2
2ζ
(2π)2
∂ (0)
g (z, z)
∂z
z=za
Total energy of the plate:
(0)
Etot
A
= −
1
2
Z
+∞
−∞
Z
+∞
dz g (0) (z, z)
−∞
where
g (0) (z, z 0 ) =
1 −κ|z−z 0 |
λa
1 −κ|z−za | −κ|z 0 −za |
e
e
e
−
.
2κ
λa + 2κ 2κ
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
9 / 32
Zero gravity: Single plate (contd.)
Total force on the plate:
(0)
Fa
A
λa
= −
2
= 0
Z
+∞
−∞
dω
2π
Z
d 2 k⊥
(2π)2
∂ (0)
g (z, z)
∂z
z=za
Total energy of the plate:
Z
Z 2
Z
(0)
1 +∞ dζ
d k⊥ 2 +∞
Etot
= −
2ζ
dz g (0) (z, z)
A
2 −∞ 2π
(2π)2
−∞
(0)
=
Ebulk Ea
+
A
A
where
Ebulk
A
(0)
Ea
A
Z +∞
Z ∞
1
3
dz
= −
κ dκ
12π 2 0
−∞
Z ∞
Z ∞
1
λa
1
dy y 3
2
.
=
κ
dκ
=
12π 2 0
λa + 2κ
96π 2 0 y 1 + λya
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
10 / 32
Zero gravity: Parallel plates
Parallel plates at z = za and z = zb , (zb − za = a), is described by
V (0) (z) = λa δ(z − za ) + λb δ(z − zb ).
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
11 / 32
Zero gravity: Parallel plates
Parallel plates at z = za and z = zb , (zb − za = a), is described by
V (0) (z) = λa δ(z − za ) + λb δ(z − zb ).
Green’s function: g (0) (z, z 0 ) can be calculated explicitly.
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
11 / 32
Zero gravity: Parallel plates
Parallel plates at z = za and z = zb , (zb − za = a), is described by
V (0) (z) = λa δ(z − za ) + λb δ(z − zb ).
Green’s function: g (0) (z, z 0 ) can be calculated explicitly.
Force on apparatus:
(0)
(0)
(0)
Fa+b
Fb
Fa
=
+
=0
A
A
A
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
11 / 32
Zero gravity: Parallel plates
Parallel plates at z = za and z = zb , (zb − za = a), is described by
V (0) (z) = λa δ(z − za ) + λb δ(z − zb ).
Green’s function: g (0) (z, z 0 ) can be calculated explicitly.
Force on apparatus:
(0)
(0)
(0)
Fa+b
Fb
Fa
=
+
=0
A
A
A
Casimir force:
Z ∞
(0)
(0)
Fb
Fa
1
1 ∂
κ3 dκ
=−
= − 2
ln ∆(0) .
A
A
2π 0
2κ ∂a
∆(0) = 1 +
K. V. Shajesh (University of Oklahoma)
λa
λb
λa λb +
+
1 − e −2κa
2κ 2κ 2κ 2κ
How does Casimir energy fall?
QV-2007
11 / 32
Zero gravity: Parallel plates
Parallel plates at z = za and z = zb , (zb − za = a), is described by
V (0) (z) = λa δ(z − za ) + λb δ(z − zb ).
Green’s function: g (0) (z, z 0 ) can be calculated explicitly.
Force on apparatus:
(0)
(0)
(0)
Fa+b
Fb
Fa
=
+
=0
A
A
A
Casimir force:
Z ∞
(0)
(0)
Fb
Fa
1
1 ∂
κ3 dκ
=−
= − 2
ln ∆(0) .
A
A
2π 0
2κ ∂a
∆(0) = 1 +
λa
λb
λa λb +
+
1 − e −2κa
2κ 2κ 2κ 2κ
Dirichlet limit (λa,b → ∞):
(0)
(0)
Fb,D
F
π2
= − a,D = −
A
A
480 a4
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
11 / 32
Zero gravity: Total energy of parallel plates:
Total energy:
(0)
(0)
(0)
(0)
E
Ebulk
Ea
Ecas
Etot
−
=
+ b +
A
A
A
A
A
where
(0)
1
Ecas
=−
A
12π 2
Z
∞
0
K. V. Shajesh (University of Oklahoma)
"
1
1
κ dκ 2κa +
+
λa
λb
1 + 2κ
1 + 2κ
2
How does Casimir energy fall?
#
1 ∂
ln ∆(0) .
2κ ∂a
QV-2007
12 / 32
Zero gravity: Total energy of parallel plates:
Total energy:
(0)
(0)
(0)
(0)
E
Ebulk
Ea
Ecas
Etot
−
=
+ b +
A
A
A
A
A
where
(0)
1
Ecas
=−
A
12π 2
Z
∞
0
"
1
1
κ dκ 2κa +
+
λa
λb
1 + 2κ
1 + 2κ
2
#
1 ∂
ln ∆(0) .
2κ ∂a
Dirichlet limit (λa,b → ∞):
(0)
Ecas,D
π2
.
=−
A
1440a3
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
12 / 32
Zero gravity: Total energy of parallel plates:
Total energy:
(0)
(0)
(0)
(0)
E
Ebulk
Ea
Ecas
Etot
−
=
+ b +
A
A
A
A
A
where
(0)
1
Ecas
=−
A
12π 2
Z
∞
0
"
1
1
κ dκ 2κa +
+
λa
λb
1 + 2κ
1 + 2κ
2
#
1 ∂
ln ∆(0) .
2κ ∂a
Dirichlet limit (λa,b → ∞):
(0)
Ecas,D
π2
.
=−
A
1440a3
Force and Energy:
(0)
(0)
Fb
∂ Etot
=−
.
A
∂a A
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
12 / 32
Hyperbolic motion
Constant acceleration:
vµ =
aµ =
K. V. Shajesh (University of Oklahoma)
dx µ
= γ(1, v)
dτ
dv µ
= γ 4 (v · a)(1, v) + γ 2 (0, a)
dτ
How does Casimir energy fall?
QV-2007
13 / 32
Hyperbolic motion
Constant acceleration:
vµ =
aµ =
dx µ
= γ(1, v)
dτ
dv µ
= γ 4 (v · a)(1, v) + γ 2 (0, a)
dτ
If v is parallel to a
aµ aµ = γ 6 |a|2 =
K. V. Shajesh (University of Oklahoma)
1
ξ2
How does Casimir energy fall?
QV-2007
13 / 32
Hyperbolic motion
Constant acceleration:
vµ =
aµ =
dx µ
= γ(1, v)
dτ
dv µ
= γ 4 (v · a)(1, v) + γ 2 (0, a)
dτ
If v is parallel to a
aµ aµ = γ 6 |a|2 =
1
ξ2
Equation of motion: for z(0) = ξ and v(0) = 0 is a hyperbola
z(t)2 − t 2 = ξ 2
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
13 / 32
Rindler coordinates
Hyperbolic motion:
z(t)2 − t 2 = ξ 2
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
14 / 32
Rindler coordinates
Hyperbolic motion:
z(t)2 − t 2 = ξ 2
Rindler coordinates (ξ, τ ):
z = ξ cosh τ
and
t = ξ sinh τ
and the metric is
− ds 2 = − dt 2 + dz 2 + dx 2 + dy 2 = − ξ 2 dτ 2 + dξ 2 + dx 2 + dy 2 .
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
14 / 32
Rindler coordinates
Hyperbolic motion:
z(t)2 − t 2 = ξ 2
Rindler coordinates (ξ, τ ):
z = ξ cosh τ
and
t = ξ sinh τ
and the metric is
− ds 2 = − dt 2 + dz 2 + dx 2 + dy 2 = − ξ 2 dτ 2 + dξ 2 + dx 2 + dy 2 .
Example: Proper distance between two plates
(zb − za )2 − (tb − ta )2 = (ξb − ξa )2 cosh2 τ − sinh2 τ = (ξb − ξa )2
is preserved. Moves rigidly.
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
14 / 32
Rindler coordinates
Hyperbolic motion:
z(t)2 − t 2 = ξ 2
Rindler coordinates (ξ, τ ):
z = ξ cosh τ
and
t = ξ sinh τ
and the metric is
− ds 2 = − dt 2 + dz 2 + dx 2 + dy 2 = − ξ 2 dτ 2 + dξ 2 + dx 2 + dy 2 .
Example: Proper distance between two plates
(zb − za )2 − (tb − ta )2 = (ξb − ξa )2 cosh2 τ − sinh2 τ = (ξb − ξa )2
is preserved. Moves rigidly.
Weak gravity: ξ → ∞ and τ → 0, such that constructions, ξτ , and,
differences in ξ, remain finite.
In weak gravity we will have ∆z = ∆ξ and t = ξτ .
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
14 / 32
Formalism including gravity
In terms of the Rindler coordinates we can thus write the action describing
the motion of parallel plates moving with constant proper accelerations
ξa−1 and ξb−1 as
Z
p
W = d 4 x −g (x)L(φ(x))
where x ≡ (τ, x, y , ξ) represents the coordinates, d 4 x = dτ dξ dx dy is the
coordinate volume element, g (x) = det gµν (x) is the determinant of the
metric, gµν (x) = diag(−ξ 2 , +1, +1, +1) defines the metric, and the
Lagrangian density is
1
1
L(φ(x)) = − gµν (x)(∂ µ φ(x))(∂ ν φ(x)) − V (x)φ(x)2
2
2
where
V (x) = λa δ(ξ − ξa ) + λb δ(ξ − ξb ).
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
15 / 32
Gravity: Energy momentum tensor
Equation of motion:(Variation in φ(x))
1 ∂2
1 ∂ ∂
2
− 2 2+
ξ
+ ∇⊥ − V (x) φ(x) = 0
ξ ∂τ
ξ ∂ξ ∂ξ
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
16 / 32
Gravity: Energy momentum tensor
Equation of motion:(Variation in φ(x))
1 ∂2
1 ∂ ∂
2
− 2 2+
ξ
+ ∇⊥ − V (x) φ(x) = 0
ξ ∂τ
ξ ∂ξ ∂ξ
Energy momentum tensor:
tαβ (x) = ∂α φ(x)∂β φ(x) + gαβ (x)L(φ(x))
where have chosen the conformal term to be zero. A couple of
components relevant in the present calculation are
1 1
t33 (x) = + 2
2ξ
1
1 1
t00 (x) = + 2
2
ξ
2ξ
K. V. Shajesh (University of Oklahoma)
∂φ
∂τ
2
∂φ
∂τ
2
1
+
2
1
+
2
∂φ
∂ξ
2
∂φ
∂ξ
2
How does Casimir energy fall?
−
1
1
(∇⊥ φ)2 − V (x)φ2
2
2
+
1
1
(∇⊥ φ)2 + V (x)φ2 .
2
2
QV-2007
16 / 32
Gravity: Force density
The force density in a non Minkowskian metric is given as the covariant
derivative of the stress tensor and explicitly reads
fα (x) = tα β ;β
np
o 1
1
∂β
−g (x) tαβ (x) − t µν (x)∂α gµν (x)
= p
2
−g (x)
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
17 / 32
Gravity: Force density
The force density in a non Minkowskian metric is given as the covariant
derivative of the stress tensor and explicitly reads
fα (x) = tα β ;β
np
o 1
1
∂β
−g (x) tαβ (x) − t µν (x)∂α gµν (x)
= p
2
−g (x)
Force density in the ξ-direction
1 1
∂
f3 (ξ) =
t
(x)
+
t
(x)
+
t33 (x)
00
33
ξ ξ2
∂ξ
1 ∂ ∂φ ∂φ
∂φ
1
2 ∂
= − φ(x)
V (ξ) + 2
− ∇⊥ · (∇⊥ φ)
2
∂ξ
ξ ∂τ ∂τ ∂ξ
∂ξ
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
17 / 32
Gravity: Green’s function
Green’s function equation:
1 ∂2
1 ∂ ∂
− − 2 2+
ξ
+ ∇2⊥ − V (x) G (x, x 0 )
ξ ∂τ
ξ ∂ξ ∂ξ
δ(ξ − ξ 0 )
δ(τ − τ 0 )δ(x⊥ − x0⊥ )
=
ξ
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
18 / 32
Gravity: Green’s function
Green’s function equation:
1 ∂2
1 ∂ ∂
− − 2 2+
ξ
+ ∇2⊥ − V (x) G (x, x 0 )
ξ ∂τ
ξ ∂ξ ∂ξ
δ(ξ − ξ 0 )
δ(τ − τ 0 )δ(x⊥ − x0⊥ )
=
ξ
When V (x) has only ξ dependence
0
Z
+∞
G (x, x ) =
−∞
dω
2π
Z
d 2 k⊥ −iω(τ −τ 0 ) ik⊥ ·(x⊥ −x0 )
⊥ g (ξ, ξ 0 )
e
e
(2π)2
where g (ξ, ξ 0 ) satisfies
ω2
δ(ξ − ξ 0 )
1 ∂ ∂
2
ξ
+ 2 − k⊥ − V (x) g (ξ, ξ 0 ) =
−
ξ ∂ξ ∂ξ
ξ
ξ
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
18 / 32
Gravitational force
Force density in terms of the Green’s function
Z
Z 2
1 +∞ dζ
d k⊥
∂
f3 (ξ) = −
g (ξ, ξ) V (ξ)
2
2 −∞ 2π
(2π)
∂ξ
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
19 / 32
Gravitational force
Force density in terms of the Green’s function
Z
Z 2
1 +∞ dζ
d k⊥
∂
f3 (ξ) = −
g (ξ, ξ) V (ξ)
2
2 −∞ 2π
(2π)
∂ξ
Coordinate force:
1
∆P3
E3
= R
=−
A
2
A dτ
Z
+∞
−∞
dζ
2π
Z
d 2 k⊥
(2π)2
Z
∞
dξ ξ g (ξ, ξ)
0
∂
V (ξ)
∂ξ
where at this point E is introduced to be the coordinate force defined as
the change in momentum per unit coordinate time τ . The coordinate force
is intentionally, with foresight, denoted as E (for energy).
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
19 / 32
Gravitational force
Force density in terms of the Green’s function
Z
Z 2
1 +∞ dζ
d k⊥
∂
f3 (ξ) = −
g (ξ, ξ) V (ξ)
2
2 −∞ 2π
(2π)
∂ξ
Coordinate force:
1
∆P3
E3
= R
=−
A
2
A dτ
Z
+∞
−∞
dζ
2π
Z
d 2 k⊥
(2π)2
Z
∞
dξ ξ g (ξ, ξ)
0
∂
V (ξ)
∂ξ
where at this point E is introduced to be the coordinate force defined as
the change in momentum per unit coordinate time τ . The coordinate force
is intentionally, with foresight, denoted as E (for energy).
Physical force: The physical force on a body (or field) situated at the
coordinate ξ0 as measured by an observer while at the coordinate ξ0 is
F =
K. V. Shajesh (University of Oklahoma)
E
= Eg .
ξ0
How does Casimir energy fall?
QV-2007
19 / 32
Single plate falling in a constant gravitational field
A single plate moving with acceleration ξa−1 is described by
V (ξ) = λa δ(ξ − ξa )
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
20 / 32
Single plate falling in a constant gravitational field
A single plate moving with acceleration ξa−1 is described by
V (ξ) = λa δ(ξ − ξa )
Green’s function: Ia ≡ Iζ (k⊥ ξa )
g (ξa , ξa ) =
K. V. Shajesh (University of Oklahoma)
Ia Ka
.
1 + λa ξa Ia Ka
How does Casimir energy fall?
QV-2007
20 / 32
Single plate falling in a constant gravitational field
A single plate moving with acceleration ξa−1 is described by
V (ξ) = λa δ(ξ − ξa )
Green’s function: Ia ≡ Iζ (k⊥ ξa )
g (ξa , ξa ) =
Ia Ka
.
1 + λa ξa Ia Ka
Coordinate force:
E3
1
=
A
2
Z
+∞
−∞
K. V. Shajesh (University of Oklahoma)
dζ
2π
Z
d 2 k⊥ ∂
ln(1 + λa ξa Ia Ka ).
(2π)2 ∂ξa
How does Casimir energy fall?
QV-2007
20 / 32
Single plate falling in a constant gravitational field
A single plate moving with acceleration ξa−1 is described by
V (ξ) = λa δ(ξ − ξa )
Green’s function: Ia ≡ Iζ (k⊥ ξa )
g (ξa , ξa ) =
Ia Ka
.
1 + λa ξa Ia Ka
Coordinate force:
E3
1
=
A
2
Z
+∞
−∞
dζ
2π
Z
d 2 k⊥ ∂
ln(1 + λa ξa Ia Ka ).
(2π)2 ∂ξa
Weak gravity:
(0)
E3
Ea
=
+O
A
A
1
ξ0
(0)
F 3 = Ea g + O(g )2 .
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
20 / 32
Parallel plates falling in a constant gravitational field
Two plates with accelerations ξa−1 and ξb−1 are described by
V (ξ) = λa δ(ξ − ξa ) + λb δ(ξ − ξb )
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
21 / 32
Parallel plates falling in a constant gravitational field
Two plates with accelerations ξa−1 and ξb−1 are described by
V (ξ) = λa δ(ξ − ξa ) + λb δ(ξ − ξb )
Coordinate force:
E3 E3
E3
= a + b
A
A
A
where
3
Ea,b
λa,b
=
A
2
Z
K. V. Shajesh (University of Oklahoma)
+∞
−∞
dζ
2π
Z
d 2 k⊥
(2π)2
i
∂ h
ξ g (ξ, ξ)
∂ξ
ξ=ξa,b
How does Casimir energy fall?
QV-2007
21 / 32
Parallel plates falling in a constant gravitational field
Two plates with accelerations ξa−1 and ξb−1 are described by
V (ξ) = λa δ(ξ − ξa ) + λb δ(ξ − ξb )
Coordinate force:
E3 E3
E3
= a + b
A
A
A
where
3
Ea,b
λa,b
=
A
2
Z
+∞
−∞
dζ
2π
Z
d 2 k⊥
(2π)2
i
∂ h
ξ g (ξ, ξ)
∂ξ
ξ=ξa,b
Green’s function:


Iζ (k⊥ ξ< )Kζ (k⊥ ξ> ) −





g (ξ, ξ 0 ) = Iζ (k⊥ ξ< )Kζ (k⊥ ξ> ) −





I (k ξ )K (k ξ ) −
ζ ⊥ <
ζ ⊥ >
∆1 I (k ξ)I (k ξ 0 )
∆ ζ ⊥ ζ ⊥
∆ij i
I (k ξ)I j (k ξ 0 )
∆ ζ ⊥ ζ ⊥
∆2 K (k ξ)K (k ξ 0 )
ζ ⊥
∆ ζ ⊥
where ∆’s are functions of ξa and ξb in terms of Ia,b and
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
if {ξ, ξ 0 } < ξa < ξb
if ξa < {ξ, ξ 0 } < ξb
if ξa < ξb < {ξ, ξ 0 }
Ka,b .
QV-2007
21 / 32
Parallel plates falling in a constant gravitational field
Coordinate force:
K. V. Shajesh (University of Oklahoma)
E3 E3
E3
= a + b
A
A
A
How does Casimir energy fall?
QV-2007
22 / 32
Parallel plates falling in a constant gravitational field
Coordinate force:
E3 E3
E3
= a + b
A
A
A
which using the Green’s function evaluates to
Z
Z 2
3
Ea,b
1 +∞ dζ
d k⊥ ∂
=
ln ∆
A
2 −∞ 2π
(2π)2 ∂ξa,b
where ∆ = 1 + λa ξa Ia Ka + λb ξb Ib Kb + λa ξa λb ξb (Ia Ka )(Ib Kb ) − (Ia Kb )2
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
22 / 32
Parallel plates falling in a constant gravitational field
Coordinate force:
E3 E3
E3
= a + b
A
A
A
which using the Green’s function evaluates to
Z
Z 2
3
Ea,b
1 +∞ dζ
d k⊥ ∂
=
ln ∆
A
2 −∞ 2π
(2π)2 ∂ξa,b
where ∆ = 1 + λa ξa Ia Ka + λb ξb Ib Kb + λa ξa λb ξb (Ia Ka )(Ib Kb ) − (Ia Kb )2
Weak gravity: (ξ → ∞)
2
1
1
∆(ξa , ξb ) = ∆(0) (ξb − ξa ) + ∆(1) (ξa , ξb ) + O
ξ0
ξ0
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
22 / 32
Parallel plates falling in a constant gravitational field
Coordinate force:
E3 E3
E3
= a + b
A
A
A
which using the Green’s function evaluates to
Z
Z 2
3
Ea,b
1 +∞ dζ
d k⊥ ∂
=
ln ∆
A
2 −∞ 2π
(2π)2 ∂ξa,b
where ∆ = 1 + λa ξa Ia Ka + λb ξb Ib Kb + λa ξa λb ξb (Ia Ka )(Ib Kb ) − (Ia Kb )2
Weak gravity: (ξ → ∞)
2
1
1
∆(ξa , ξb ) = ∆(0) (ξb − ξa ) + ∆(1) (ξa , ξb ) + O
ξ0
ξ0
Coordinate force in weak gravity:
(0)
(0)
(0)
E
E3
Ea
Ecas
1
=
+ b +
+O
A
A
A
A
ξ0
h
i
(0)
(0)
(0)
F 3 = Ea + Eb + Ecas g + O(g )2 .
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
22 / 32
Renormalization
Thus, the vacuum energy, including the divergent contributions, falls under
a weak gravitational field exactly like conventional mass or energy. Thus
we can identify the gravitational force on the Casimir apparatus consisting
of two plates to be
i
h
(0)
(0)
(0)
F = ma + mb + Ea + Eb + Ecas g + O(g )2 .
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
23 / 32
Renormalization
Thus, the vacuum energy, including the divergent contributions, falls under
a weak gravitational field exactly like conventional mass or energy. Thus
we can identify the gravitational force on the Casimir apparatus consisting
of two plates to be
i
h
(0)
(0)
(0)
F = ma + mb + Ea + Eb + Ecas g + O(g )2 .
We identify ma,b as the bare mass of the plates, and interpret the
divergent Casimir energy contribution from each plate to renormalize the
bare mass. Thus we renormalize the masses by the prescription
(0)
ma,b + Ea,b → Ma,b
and define Ma,b to be the renormalized mass of the plates.
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
23 / 32
Renormalization
Thus, the vacuum energy, including the divergent contributions, falls under
a weak gravitational field exactly like conventional mass or energy. Thus
we can identify the gravitational force on the Casimir apparatus consisting
of two plates to be
i
h
(0)
(0)
(0)
F = ma + mb + Ea + Eb + Ecas g + O(g )2 .
We identify ma,b as the bare mass of the plates, and interpret the
divergent Casimir energy contribution from each plate to renormalize the
bare mass. Thus we renormalize the masses by the prescription
(0)
ma,b + Ea,b → Ma,b
and define Ma,b to be the renormalized mass of the plates. Thus we have
h
i
(0)
F = − Ma + Mb + Ecas g + O(g )2 .
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
23 / 32
Status on generalizations: Work in progress
1
Question: How will the result change if the plate is tilted?
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
24 / 32
Status on generalizations: Work in progress
1
Question: How will the result change if the plate is tilted?
=⇒ We expect the result to remain the same because the Casimir
energy (which is the analog of mass here) should be a scalar under
rotations.
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
24 / 32
Status on generalizations: Work in progress
1
Question: How will the result change if the plate is tilted?
=⇒ We expect the result to remain the same because the Casimir
energy (which is the analog of mass here) should be a scalar under
rotations.
=⇒ This led us to investigate the definition of force on a distributed
mass in general relativistic context.
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
24 / 32
Status on generalizations: Work in progress
1
Question: How will the result change if the plate is tilted?
=⇒ We expect the result to remain the same because the Casimir
energy (which is the analog of mass here) should be a scalar under
rotations.
=⇒ This led us to investigate the definition of force on a distributed
mass in general relativistic context.
2
A related question: What will be the pseudo force exerted on the
Casimir apparatus if we rotate it in a circle? Will the Casimir energy
experience a centrifugal force as we expect? Will
(0)
F = ECas ω 2 r + O(ωr )2
where ω is the angular speed of rotation?
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
24 / 32
Status on generalizations: Work in progress
1
Question: How will the result change if the plate is tilted?
=⇒ We expect the result to remain the same because the Casimir
energy (which is the analog of mass here) should be a scalar under
rotations.
=⇒ This led us to investigate the definition of force on a distributed
mass in general relativistic context.
2
A related question: What will be the pseudo force exerted on the
Casimir apparatus if we rotate it in a circle? Will the Casimir energy
experience a centrifugal force as we expect? Will
(0)
F = ECas ω 2 r + O(ωr )2
where ω is the angular speed of rotation?
=⇒ This exercise stemmed out as means of having a cross check for
our definition of force.
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
24 / 32
Local inertial frame associated with Rindler space
Rindler space time:
Coordinates ξ µ ≡ (τ, x, y , ξ)
Connections Γ003 = 1ξ , Γ300 = ξ,
K. V. Shajesh (University of Oklahoma)
Metric gµν (ξ) = (−ξ 2 , +1, +1, +1)
Curvature Rµναβ = 0
How does Casimir energy fall?
QV-2007
25 / 32
Local inertial frame associated with Rindler space
Rindler space time:
Coordinates ξ µ ≡ (τ, x, y , ξ)
Metric gµν (ξ) = (−ξ 2 , +1, +1, +1)
1
0
3
Connections Γ03 = ξ , Γ00 = ξ, Curvature Rµναβ = 0
Local inertial frame: A freely falling coordinate system (local inertial
frame) will be described by the coordinates x a ≡ (t, x, y , z) and metric
ηab = (−1, +1, +1, +1) with connections Γabc (x) = 0.
x a = x a (ξ)
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
25 / 32
Local inertial frame associated with Rindler space
Rindler space time:
Coordinates ξ µ ≡ (τ, x, y , ξ)
Metric gµν (ξ) = (−ξ 2 , +1, +1, +1)
1
0
3
Connections Γ03 = ξ , Γ00 = ξ, Curvature Rµναβ = 0
Local inertial frame: A freely falling coordinate system (local inertial
frame) will be described by the coordinates x a ≡ (t, x, y , z) and metric
ηab = (−1, +1, +1, +1) with connections Γabc (x) = 0.
x a = x a (ξ)
LIF corresponding to Rindler space: x a ’s are determined by solving the
differential equations
∂
∂ ∂ a
x (ξ) = Γλµν (ξ) λ x a (ξ)
µ
ν
∂ξ ∂ξ
∂ξ
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
25 / 32
Local inertial frame associated with Rindler space
Rindler space time:
Coordinates ξ µ ≡ (τ, x, y , ξ)
Metric gµν (ξ) = (−ξ 2 , +1, +1, +1)
1
0
3
Connections Γ03 = ξ , Γ00 = ξ, Curvature Rµναβ = 0
Local inertial frame: A freely falling coordinate system (local inertial
frame) will be described by the coordinates x a ≡ (t, x, y , z) and metric
ηab = (−1, +1, +1, +1) with connections Γabc (x) = 0.
x a = x a (ξ)
LIF corresponding to Rindler space: x a ’s are determined by solving the
differential equations
∂
∂ ∂ a
x (ξ) = Γλµν (ξ) λ x a (ξ)
µ
ν
∂ξ ∂ξ
∂ξ
which has solutions
t(ξ, τ ) = ξ sinh τ
z(ξ, τ ) = ξ cosh τ
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
25 / 32
Definition of Force
Force in LIF:
Fa (t) =
d
Pa (t)
dt
where
Z
Pa (t) =
K. V. Shajesh (University of Oklahoma)
d 3 x ta0 (x, t)
How does Casimir energy fall?
QV-2007
26 / 32
Definition of Force
Force in LIF:
Fa (t) =
d
Pa (t)
dt
where
Z
Pa (t) =
P3 (t)
A
=
d 3 x ta0 (x, t)
Z ∂z
∂z
∂ξ µ ∂ξ ν dξ
+ dτ
tµν (ξ, τ )
∂ξ
∂τ
∂z ∂t t
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
fixed
QV-2007
26 / 32
Definition of Force
Force in LIF:
Fa (t) =
d
Pa (t)
dt
where
Z
Pa (t) =
P3 (t)
A
=
=
d 3 x ta0 (x, t)
Z ∂z
∂z
∂ξ µ ∂ξ ν dξ
+ dτ
tµν (ξ, τ )
∂ξ
∂τ
∂z ∂t t
Z ∞
µ
ν
dξ
∂ξ ∂ξ q
tµν (ξ, τ )
2
∂z ∂t 0
1 + t2
ξ
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
fixed
t fixed
QV-2007
26 / 32
Definition of Force
Force in LIF:
Fa (t) =
d
Pa (t)
dt
where
Z
d 3 x ta0 (x, t)
Z ∂z
∂z
∂ξ µ ∂ξ ν =
dξ
+ dτ
tµν (ξ, τ )
∂ξ
∂τ
∂z ∂t t fixed
Z ∞
µ
ν
dξ
∂ξ ∂ξ q
=
tµν (ξ, τ )
2
∂z ∂t 0
1 + ξt 2
t fixed
2
Z ∞
1 1
t
dξ
= −t
t00 (ξ) + t33 (ξ) + O
2
ξ ξ
ξ
0
Pa (t) =
P3 (t)
A
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
26 / 32
Definition of Force
Force in LIF:
Fa (t) =
d
Pa (t)
dt
where
Z
d 3 x ta0 (x, t)
Z P3 (t)
∂z
∂z
∂ξ µ ∂ξ ν =
dξ
+ dτ
tµν (ξ, τ )
A
∂ξ
∂τ
∂z ∂t t fixed
Z ∞
µ
ν
dξ
∂ξ ∂ξ q
=
tµν (ξ, τ )
2
∂z ∂t 0
1 + ξt 2
t fixed
2
Z ∞
1 1
t
dξ
= −t
t00 (ξ) + t33 (ξ) + O
2
ξ ξ
ξ
0
Z ∞
F3 (t)
1 1
=−
dξ
t
(ξ)
+
t
(ξ)
+ O(g )2
33
2 00
A
ξ
ξ
0
Pa (t) =
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
26 / 32
Force on a single particle in Rindler space
As a first check let us determine the force on a single particle in the
Rindler space using
Z ∞
F3 (t)
1 1
dξ
=−
t00 (ξ) + t33 (ξ) + O(g )2
A
ξ ξ2
0
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
27 / 32
Force on a single particle in Rindler space
As a first check let us determine the force on a single particle in the
Rindler space using
Z ∞
F3 (t)
1 1
dξ
=−
t00 (ξ) + t33 (ξ) + O(g )2
A
ξ ξ2
0
Energy momentum for a single particle in an arbitrary metric space is
Z +∞
1
dx µ dx ν
µν
t (x) = m p
ds δ (4) (x − xa (s))
ds ds
−g (x) −∞
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
27 / 32
Force on a single particle in Rindler space
As a first check let us determine the force on a single particle in the
Rindler space using
Z ∞
F3 (t)
1 1
dξ
=−
t00 (ξ) + t33 (ξ) + O(g )2
A
ξ ξ2
0
Energy momentum for a single particle in an arbitrary metric space is
Z +∞
1
dx µ dx ν
µν
t (x) = m p
ds δ (4) (x − xa (s))
ds ds
−g (x) −∞
For a single particle only t 00 (x) is non zero. Using this we get the correct
result
F3
1
= − m + O(g )2
A
ξ
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
27 / 32
Centrifugal force on a single particle moving in a
circle
As a second check let us determine the centrifugal force on a single particle
moving in a circle. A particle moving in a circle is described by the metric
−ds 2 = −dt 2 (1 − ω 2 r 2 ) + dr 2 + 2ωr 2 dtdθ + r 2 dθ2 + dz 2
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
28 / 32
Centrifugal force on a single particle moving in a
circle
As a second check let us determine the centrifugal force on a single particle
moving in a circle. A particle moving in a circle is described by the metric
−ds 2 = −dt 2 (1 − ω 2 r 2 ) + dr 2 + 2ωr 2 dtdθ + r 2 dθ2 + dz 2
Local inertial coordinates for this space are
x(r , θ, t) = r cos(θ + ωt)
y (r , θ, t) = r sin(θ + ωt)
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
28 / 32
Centrifugal force on a single particle moving in a
circle
As a second check let us determine the centrifugal force on a single particle
moving in a circle. A particle moving in a circle is described by the metric
−ds 2 = −dt 2 (1 − ω 2 r 2 ) + dr 2 + 2ωr 2 dtdθ + r 2 dθ2 + dz 2
Local inertial coordinates for this space are
x(r , θ, t) = r cos(θ + ωt)
y (r , θ, t) = r sin(θ + ωt)
Momentum:
Z
Pa (t) =
K. V. Shajesh (University of Oklahoma)
d 3 x ta0 (x)
How does Casimir energy fall?
QV-2007
28 / 32
Centrifugal force on a single particle moving in a
circle
As a second check let us determine the centrifugal force on a single particle
moving in a circle. A particle moving in a circle is described by the metric
−ds 2 = −dt 2 (1 − ω 2 r 2 ) + dr 2 + 2ωr 2 dtdθ + r 2 dθ2 + dz 2
Local inertial coordinates for this space are
x(r , θ, t) = r cos(θ + ωt)
y (r , θ, t) = r sin(θ + ωt)
Momentum:
Z
d 3 x ta0 (x)
Z
d 3r
Pa (t) =
=
K. V. Shajesh (University of Oklahoma)
p
−g (r ) tµν (r , θ, t)
How does Casimir energy fall?
∂r µ ∂r ν
∂x a ∂t
QV-2007
28 / 32
Centrifugal force on a single particle moving in a
circle
As a second check let us determine the centrifugal force on a single particle
moving in a circle. A particle moving in a circle is described by the metric
−ds 2 = −dt 2 (1 − ω 2 r 2 ) + dr 2 + 2ωr 2 dtdθ + r 2 dθ2 + dz 2
Local inertial coordinates for this space are
x(r , θ, t) = r cos(θ + ωt)
y (r , θ, t) = r sin(θ + ωt)
Momentum:
Z
d 3 x ta0 (x)
Z
d 3r
Pa (t) =
=
p
−g (r ) tµν (r , θ, t)
∂r µ ∂r ν
∂x a ∂t
Centrifugal force:
F(t) = m √
K. V. Shajesh (University of Oklahoma)
ω2r
r̂
1 − ω2r 2
How does Casimir energy fall?
QV-2007
28 / 32
Force on a Casimir apparatus
Let us now calculate the gravitational force on a Casimir apparatus using
Z ∞
1 1
F3 (t)
dξ
=−
t00 (ξ) + t33 (ξ) + O(g )2
A
ξ ξ2
0
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
29 / 32
Force on a Casimir apparatus
Let us now calculate the gravitational force on a Casimir apparatus using
Z ∞
1 1
F3 (t)
dξ
=−
t00 (ξ) + t33 (ξ) + O(g )2
A
ξ ξ2
0
Let us remind ourselves of the zero gravity case
(0)
(0)
(0)
t00 (x) = Evol + (1 − 4θ) Esur
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
29 / 32
Force on a Casimir apparatus
Let us now calculate the gravitational force on a Casimir apparatus using
Z ∞
1 1
F3 (t)
dξ
=−
t00 (ξ) + t33 (ξ) + O(g )2
A
ξ ξ2
0
Let us remind ourselves of the zero gravity case
(0)
(0)
(0)
t00 (x) = Evol + (1 − 4θ) Esur
Gravitational force:
(0)
(0)
F = [ma + mb + Evol + Eextra ]g + O(g )2
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
29 / 32
Force on a Casimir apparatus
Let us now calculate the gravitational force on a Casimir apparatus using
Z ∞
1 1
F3 (t)
dξ
=−
t00 (ξ) + t33 (ξ) + O(g )2
A
ξ ξ2
0
Let us remind ourselves of the zero gravity case
(0)
(0)
(0)
t00 (x) = Evol + (1 − 4θ) Esur
Gravitational force:
(0)
(0)
F = [ma + mb + Evol + Eextra ]g + O(g )2
where
(0)
Evol
Z
=
K. V. Shajesh (University of Oklahoma)
(0)
(0)
(0)
(0)
d 3 xEvol = Evol,a + Evol,b + Evol,Cas
How does Casimir energy fall?
QV-2007
29 / 32
Force on a Casimir apparatus
Let us now calculate the gravitational force on a Casimir apparatus using
Z ∞
1 1
F3 (t)
dξ
=−
t00 (ξ) + t33 (ξ) + O(g )2
A
ξ ξ2
0
Let us remind ourselves of the zero gravity case
(0)
(0)
(0)
t00 (x) = Evol + (1 − 4θ) Esur
Gravitational force:
(0)
(0)
F = [ma + mb + Evol + Eextra ]g + O(g )2
where
(0)
Evol
(0)
Eextra
Z
d 3 xEvol = Evol,a + Evol,b + Evol,Cas
Z
d 3 xEsur = Esur,a + Esur,b + Esur,Cas
=
=
S
(0)
(0)
(0)
(0)
(0)
(0)
(0)
(0)
(0)
(0)
where all terms except Evol,Cas and Esur,Cas are divergent.
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
29 / 32
Explicit expressions for terms in force
(0)
(0)
(0)
(0)
(0)
(0)
F = [ma + mb + Evol,a + Evol,b + Evol,Cas + Esur,a + Esur,b + Esur,Cas ]g + O(g )2
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
30 / 32
Explicit expressions for terms in force
(0)
(0)
(0)
(0)
(0)
(0)
F = [ma + mb + Evol,a + Evol,b + Evol,Cas + Esur,a + Esur,b + Esur,Cas ]g + O(g )2
(0)
Evol,a/b = −
1 (0)
1 1
Esur,a/b =
3
3 32π 2
K. V. Shajesh (University of Oklahoma)
Z
0
∞
dy
y3
y 1 + λy
How does Casimir energy fall?
a/b
QV-2007
30 / 32
Explicit expressions for terms in force
(0)
(0)
(0)
(0)
(0)
(0)
F = [ma + mb + Evol,a + Evol,b + Evol,Cas + Esur,a + Esur,b + Esur,Cas ]g + O(g )2
(0)
Evol,a/b = −
(0)
Evol,Cas
1 1
= −
3 32π 2 a3
K. V. Shajesh (University of Oklahoma)
Z
0
∞
∞
dy
y3
y 1 + λy
0
a/b
h
i
1
1
1
+
+
y +aλa
y +aλb
dy 4
i
y h
y
y
y
1 + aλb e y − 1
1 + aλa
1 (0)
1 1
Esur,a/b =
3
3 32π 2
Z
How does Casimir energy fall?
QV-2007
30 / 32
Explicit expressions for terms in force
(0)
(0)
(0)
(0)
(0)
(0)
F = [ma + mb + Evol,a + Evol,b + Evol,Cas + Esur,a + Esur,b + Esur,Cas ]g + O(g )2
(0)
(0)
Evol,Cas
(0)
Esur,Cas
∞
dy
y3
y 1 + λy
0
a/b
h
i
1
1
Z ∞
1
+
+
y +aλa
y +aλb
1 1
dy 4
i
= −
y h
2
3
y
y
3 32π a 0 y
1 + aλb e y − 1
1 + aλa
h
i
1
1
Z ∞
+
y +aλa
y +aλb
1
dy 4
i
=
y h
y
32π 2 a3 0 y
1 + aλa
1 + aλy b e y − 1
Evol,a/b = −
1 (0)
1 1
Esur,a/b =
3
3 32π 2
K. V. Shajesh (University of Oklahoma)
Z
How does Casimir energy fall?
QV-2007
30 / 32
Explicit expressions for terms in force
(0)
(0)
(0)
(0)
(0)
(0)
F = [ma + mb + Evol,a + Evol,b + Evol,Cas + Esur,a + Esur,b + Esur,Cas ]g + O(g )2
(0)
(0)
Evol,Cas
(0)
Esur,Cas
∞
dy
y3
y 1 + λy
0
a/b
h
i
1
1
Z ∞
1
+
+
y +aλa
y +aλb
1 1
dy 4
i
= −
y h
2
3
y
y
3 32π a 0 y
1 + aλb e y − 1
1 + aλa
h
i
1
1
Z ∞
+
y +aλa
y +aλb
1
dy 4
i
=
y h
y
32π 2 a3 0 y
1 + aλa
1 + aλy b e y − 1
Evol,a/b = −
1 (0)
1 1
Esur,a/b =
3
3 32π 2
Z
Dirichlet limit (λa,b → ∞)
(0)
Evol,Cas = −
K. V. Shajesh (University of Oklahoma)
π2
1440 a3
and
How does Casimir energy fall?
(0)
Esur,Cas = 0
QV-2007
30 / 32
Conclusions and Things to do
1
We have exposed the appearance of extra terms in the gravitational
force, getting contributions from surfaces of the plates only.
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
31 / 32
Conclusions and Things to do
1
We have exposed the appearance of extra terms in the gravitational
force, getting contributions from surfaces of the plates only.
2
The divergent terms can be interpreted to renormalize bare mass.
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
31 / 32
Conclusions and Things to do
1
We have exposed the appearance of extra terms in the gravitational
force, getting contributions from surfaces of the plates only.
2
The divergent terms can be interpreted to renormalize bare mass.
How does one interpret the finite contribution from the extra surface
term, that goes to 0 in the Dirichlet limit?
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
31 / 32
Conclusions and Things to do
1
We have exposed the appearance of extra terms in the gravitational
force, getting contributions from surfaces of the plates only.
2
The divergent terms can be interpreted to renormalize bare mass.
How does one interpret the finite contribution from the extra surface
term, that goes to 0 in the Dirichlet limit?
3
To do: We have consistently ignored the conformal term in the energy
momentum tensor while dealing with the gravity situation.
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
31 / 32
Conclusions and Things to do
1
We have exposed the appearance of extra terms in the gravitational
force, getting contributions from surfaces of the plates only.
2
The divergent terms can be interpreted to renormalize bare mass.
How does one interpret the finite contribution from the extra surface
term, that goes to 0 in the Dirichlet limit?
3
To do: We have consistently ignored the conformal term in the energy
momentum tensor while dealing with the gravity situation.
4
It will be possible to address the case of tilted Casimir plates
unambiguously with the refined definition of force.
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
31 / 32
Conclusions and Things to do
1
We have exposed the appearance of extra terms in the gravitational
force, getting contributions from surfaces of the plates only.
2
The divergent terms can be interpreted to renormalize bare mass.
How does one interpret the finite contribution from the extra surface
term, that goes to 0 in the Dirichlet limit?
3
To do: We have consistently ignored the conformal term in the energy
momentum tensor while dealing with the gravity situation.
4
It will be possible to address the case of tilted Casimir plates
unambiguously with the refined definition of force.
5
To do: It would be illuminating to see the renormalization for a
sphere.
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
31 / 32
Conclusions and Things to do
1
We have exposed the appearance of extra terms in the gravitational
force, getting contributions from surfaces of the plates only.
2
The divergent terms can be interpreted to renormalize bare mass.
How does one interpret the finite contribution from the extra surface
term, that goes to 0 in the Dirichlet limit?
3
To do: We have consistently ignored the conformal term in the energy
momentum tensor while dealing with the gravity situation.
4
It will be possible to address the case of tilted Casimir plates
unambiguously with the refined definition of force.
5
To do: It would be illuminating to see the renormalization for a
sphere.
6
To do: Centrifugal force on the Casimir apparatus moving in a circle.
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
31 / 32
The talk was based on:
S. A. Fulling, K. A. Milton, P. Parashar, A. Romeo, K. V. Shajesh, J.
Wagner, Phys. Rev. D 76, 025004 (2007), arXiv:hep-th/0702091.
K. A. Milton, P. Parashar, K. V. Shajesh, J. Wagner, J. Phys. A:
Math. Theor. 40, 1 (2007), arXiv:0705.2611.
K. V. Shajesh (University of Oklahoma)
How does Casimir energy fall?
QV-2007
32 / 32