semitransparent pistons

The Problem Zeta Function Semitransparent Pistons Questions

Semitransparent Pistons

Pedro Morales-Almazan

Department of MathematicsBaylor University

pedro [email protected]

April, 15th 2011

Pedro Morales-Almazan Math Department

mailto:[email protected]

Outline

1 The ProblemCasimir Effect

Mathematical Model

2 Zeta FunctionDefinition

3 Semitransparent PistonsEigenvalue Problem

Operator DeterminantCasimir Force

4 Questions

Outline

Mathematical Model

4 Questions

Outline

Mathematical Model

4 Questions

Outline

Mathematical Model

4 Questions

Casimir Effect

Is a quantum field effect that arises when considering vacuumfluctuations

Casimir Effect

History

• Predicted theoretically in 1948 by Hendrik B. G. Casimir andDirk Polder when Casimir was trying to compute van derWaals forces between polarizable molecules.

• Confirmed experimentally in 1997 by S. K. Lamoreaux.

Casimir Effect

History

• Predicted theoretically in 1948 by Hendrik B. G. Casimir andDirk Polder when Casimir was trying to compute van derWaals forces between polarizable molecules.

• Confirmed experimentally in 1997 by S. K. Lamoreaux.

Casimir Effect

Why is so important?

• Believed to explain the stability of an electron

• Very sensitive to the geometry of the space (Quantum andComsmological implications)

• Provides a better understanding of the zero-point energy

Casimir Effect

Casimir Effect

Casimir Effect

Mathematical Model

In order to calculate the Casimir Energy of a system, consider aRiemannian manifold M possibly with boundary and theeigenvalue problem

(∆ + V )φ = λφ

where ∆ is the Laplacian on M, V is a potential and φ ∈ L2(M).If ∂M 6= ∅ boundary conditions must be imposed.

Casimir Effect

Mathematical Model

(∆ + V )φ = λφ

Casimir Effect

Mathematical Model

(∆ + V )φ = λφ

Casimir Effect

Definition of Casimir Energy

The self energy of the system is defined to be

E =1

2

∑λ

√λ

Since the self-adjointness of ∆, the eigenvalues λ are unboundedand hence, E is not well defined. Regularization methods to avoidinfinities are required.

Casimir Effect

E =1

2

∑λ

√λ

Since the self-adjointness of ∆, the eigenvalues λ are unboundedand hence, E is not well defined.

Regularization methods to avoidinfinities are required.

Casimir Effect

E =1

2

∑λ

√λ

Since the self-adjointness of ∆, the eigenvalues λ are unboundedand hence, E is not well defined. Regularization methods to avoidinfinities are required.

Definition

Zeta Function

Given a self-adjoint operator P with eigenvalues λn∞n=1, the zetafunction is defined by

ζP(s) =∞∑n=1

λ−sn

which is convergent for <s large enough.

Definition

Zeta Function

Given a self-adjoint operator P with eigenvalues λn∞n=1, the zetafunction is defined by

ζP(s) =∞∑n=1

λ−sn

which is convergent for <s large enough.

Definition

• Values at s = −1/2, 0 provide information of the Casimirenergy and the operator determinant

• An analytic continuation of the zeta function is required

• Lack of explicit eigenvalues requires an indirect method forcalculations

Definition

Definition

Eigenvalue Problem

Consider the piston configuration modeled by

Pφ = λ2φ

where P is the Laplace-type differential operator defined on[0, L]×N

P = − ∂2

∂x2−∆N + σδ(x − a)

N is a compact Riemannian manifold and we have Dirichletboundary conditions φ(0) = φ(L) = 0.

Eigenvalue Problem

Pφ = λ2φ

P = − ∂2

∂x2−∆N + σδ(x − a)

Eigenvalue Problem

Pφ = λ2φ

P = − ∂2

∂x2−∆N + σδ(x − a)

Eigenvalue Problem

Configuration

Eigenvalue Problem

Separation of variables

Using separation of variables

λ2k` = ν2

k + η2`

where ν2k and η2

` are the eigenvalues for the Laplacian on [0, L] andN respectively

Eigenvalue Problem

Zeta Function

ζ(s) =∞∑k=1

∞∑`=1

λ−2sk` =

∞∑k=1

∞∑`=1

(ν2k + η2

` )−s

Remark The eigenvalues ν2k cannot be calculated explicitly, an

indirect way of finding the zeta function is required

Eigenvalue Problem

Zeta Function

ζ(s) =∞∑k=1

∞∑`=1

λ−2sk` =

∞∑k=1

∞∑`=1

(ν2k + η2

` )−s

Remark The eigenvalues ν2k cannot be calculated explicitly, an

indirect way of finding the zeta function is required

Eigenvalue Problem

Contour Integration

Cauchy’s residue Theorem

Let f be a meromorphic function defined on a simply connectedregion Ω of the complex plane and let aknk=1 be its poles on Ω.Let γ be a closed curve in Ω, then

1

2πı

∫γf (z)dz =

n∑k=1

I (ak , γ) Res(f (z))|z=ak

Eigenvalue Problem

Integral Representation

ζ(s) =1

2πı

∞∑`=1

∫γ`

dν (ν2 + η2` )−s

d

dνlog F (ν)

where

F (ν) =σ sin(ν(L− a) sin(νa))

ν2+

sin(νL)

ν

where γ` is a contour enclosing νk∞k=1

Eigenvalue Problem

Integral Representation

ζ(s) =1

2πı

∞∑`=1

∫γ`

dν (ν2 + η2` )−s

d

dνlog F (ν)

where

F (ν) =σ sin(ν(L− a) sin(νa))

ν2+

sin(νL)

ν

where γ` is a contour enclosing νk∞k=1

Eigenvalue Problem

Contour Deformation

After deforming the contours γ` to the imaginary axis, the zetafunction becomes

ζ(s) =sin(πs)

π

∞∑`=1

∫ ∞η`

dν (ν2 − η2` )−s

d

dνlog F (ıν)

which converges for <s big enough

Eigenvalue Problem

Contour Deformation

After deforming the contours γ` to the imaginary axis, the zetafunction becomes

ζ(s) =sin(πs)

π

∞∑`=1

∫ ∞η`

dν (ν2 − η2` )−s

d

dνlog F (ıν)

which converges for <s big enough

Eigenvalue Problem

Analytic Continuation

In order to extend analytically ζ(s) to the left in the complexplane, we subtract the asymptotic behavior of log F (ıν),

log F (ıν) ∼ Lν − 2ν +∞∑n=1

(−1)n+1

n

( σ2ν

)n

Eigenvalue Problem

Finite Part

Subtracting the asymptotic terms enlarges the convergence region

ζ(s) = ζ(f )(s) + ζ(as)(s)

where

ζ(f )(s) =

sin(πs)

π

∞∑`=1

∫ ∞η`

dν (ν2 − η2` )−s

d

dν[log F (ıν)− asymptotics]

Eigenvalue Problem

Finite Part

Subtracting the asymptotic terms enlarges the convergence region

ζ(s) = ζ(f )(s) + ζ(as)(s)

where

ζ(f )(s) =

sin(πs)

π

∞∑`=1

∫ ∞η`

dν (ν2 − η2` )−s

d

dν[log F (ıν)− asymptotics]

Eigenvalue Problem

Asymptotic Part

ζ(as)(s) =sin(πs)

π

∞∑`=1

∫ ∞η`

dν (ν2 − η2` )−s

d

dν[asymptotics]

Eigenvalue Problem

The operator determinant for the differential operator P is definedas

Operator Determinant

exp(ζ ′(0))

which after some algebra, is computed to be...

Eigenvalue Problem

The operator determinant for the differential operator P is definedas

Operator Determinant

exp(ζ ′(0))

which after some algebra, is computed to be...

Eigenvalue Problem

ζ ′(0) =∞∑`=1

(log F (ıη`)− Lη` + log(2η`) +

N∑n=1

(−1)n+1

n

(σ

2η`

)n)

−L(

FPζN (−1

2)− ResζN (−1

2)(−2 + log 4)

)− 1

2ζ ′N (0)

+2N∑

n=1

(−1)n

n

(σ2

)n [FPζN

(n2

)+ ResζN

(n2

(γ + ψ(n

2

)))]

Eigenvalue Problem

Casimir Force

The casimir force is defined to be

Casimir Force

−1

2

∂

∂aζ

(−1

2

)

which after some small algebra, is computed to be...

Eigenvalue Problem

Casimir Force

The casimir force is defined to be

Casimir Force

−1

2

∂

∂aζ

(−1

2

)which after some small algebra, is computed to be...

Eigenvalue Problem

Casimir Force

1

2π

∞∑`=1

∫ ∞η`

dν (ν2 − η2` )1/2 ∂

∂a

∂

∂νlog F (ıν)

which after a lot of algebra, is computed to be... negative for0 < a < L/2 and positive for L/2 < a < L.

Eigenvalue Problem

Casimir Force

1

2π

∞∑`=1

∫ ∞η`

dν (ν2 − η2` )1/2 ∂

∂a

∂

∂νlog F (ıν)

which after a lot of algebra, is computed to be...

negative for0 < a < L/2 and positive for L/2 < a < L.

Eigenvalue Problem

Casimir Force

1

2π

∞∑`=1

∫ ∞η`

dν (ν2 − η2` )1/2 ∂

∂a

∂

∂νlog F (ıν)

which after a lot of algebra, is computed to be... negative for0 < a < L/2

and positive for L/2 < a < L.

Eigenvalue Problem

Casimir Force

1

2π

∞∑`=1

∫ ∞η`

dν (ν2 − η2` )1/2 ∂

∂a

∂

∂νlog F (ıν)

which after a lot of algebra, is computed to be... negative for0 < a < L/2 and positive for L/2 < a < L.

Eigenvalue Problem

Piston Behavior

Given the second order differential operator

P = − ∂2

∂x2−∆N + σδ(x − a) defined on [0, L]×N with Dirichlet

boundary conditions, the piston is then attracted to the closestwall.

Questions

Thanks

semitransparent pistons

Education

casimir anddirk polder

quantum field effect

better understanding

van derwaals forces

polarizable molecules