The Heat-Kernel in a Schwarzschild Geometry and the Casimir Energy

29 Pages·0.23 MB·English

by  Frank Antonsen

The Heat-Kernel in a Schwarzschild
Geometry and the Casimir Energy
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9
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1 Frank Antonsen

n University of Copenhagen
a
J Niels Bohr Institute

9
Blegdamsvej 17, DK-2100 Copenhagen Ø, Denmark
1

2
v
0 Abstract
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1 We obtain an hybrid expression for the heat-kernel, and from that
0
the density of the free energy, for a minimally coupled scalar field in
1
7 a Schwarzschild geometry at finite temperature. This gives us the
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zero-point energy density as a function of the distance from the mas-
/
c sive object generating the gravitational field. The contribution to the
q
zero-pointenergyduetothecurvatureisextracted too, inthisway ar-
-
r rivingatarenormalisedexpressionfortheenergydensity(theCasimir
g
: energy density). We use this to find an expression for other physical
v
i quantities: internal energy, pressure and entropy. It turns out that
X
the disturbance of the surrounding vacuum generates entropy. For β
r
a smalltheentropyispositiveforr > 2M. Wealsofindthattheinternal
energy can be negative outside the horizon pointing to the existence
of bound states. The total internal energy inside the horizon turns
out to be finite but complex, the imaginary part to be interpreted as
responsible for particle creation.
1 Introduction
One of the most important quantum physical quantities is probably the zero-
pointenergy. Itisneeded, forinstance, when wewant tostudy back-reaction,
i.e., the influence the matter fields moving in a curved back-ground assert
1
on the back-ground geometry itself. This would be done by solving the Ein-
stein equations with the expectation value of the energy-momentum tensor
as source:
1
G = R g R = T ,
µν µν µν µν
− 2 h i
where we have chosen units such that κ = ~ = c = 1. It is also important for
the study of renormalisation properties of a quantum field theory in a curved
space-time.
In order to find this quantity, we evaluate the heat-kernel corresponding
to the equations of motion for a scalar field. This calculation is carried
out at finite temperature. The integral of this heat-kernel with respect to
some fictitious fifth coordinate, σ, will give the density of Helmholtz’ free
energy (its integral over the entire space-time will give the zeta-function,
which is essentially just Helmholtz’ free energy). The resulting expression is
regularised and renormalised in the subsequent sections.
We do this by using the spherical symmetry of the spacetime in order to
collect all the unknown bits of the heat kernel into a function g (r,r′;σ)
nl
depending only upon the radial coordinates and σ. A recursive relation for
an asympototic expansion of this unknow function can be found and solved,
thereby allowing us to find the heat kernel.
From the freen energy one can derive expressions for the entropy and the
pressure by using standard thermodynamic relations. We find these and
comment on their meaning. We also show that, to lowest order in the mass
generating the Schwarzschild geometry, the free energy is that of an infinite
family of particles moving in one dimension with an r-dependent mass.
2 Set-Up
We consider a minimally coupled scalar field φ moving in a Schwarzschild
back-ground, hence the action is
1
S = (∂ φ∂ φgµν µ2φ2) g d4x, (1)
µ ν
2 − | |
Z
p
where the metric is given by the standard expression
−1
2M 2M
ds2 = 1 dt2 1 dr2 r2dθ2 r2sin2θdφ2,
− r − − r − −
(cid:18) (cid:19) (cid:18) (cid:19)
2
whereM isthemassofthe(classical)massiveobjectgeneratingtheSchwarzschild
geometry, a black hole or a star, say. We will put the mass, µ, of the scalar
field equal to zero. We can later reinsert it if we find it desirable.
The d’Alembertian becomes
1 ∂2 1 ∂ ∂ L2
2 = r2h ,
h∂t2 − r2∂r ∂r − r2
where h(r) = (1 2M/r) and L2 is the square of the angular momentum
−
operator. A finite temperature can be included by complexifying the time
coordinate t, t τ. As φ describes a Bose field, it becomes periodic in τ,
7→
φ(τ+β) = φ(τ). This implies that the time direction is not only complexified
but also compactified – by appropriate scaling we can take τ to lie on the
unit circle S1 (see e.g. Ramond [1] or Itzykson and Zuber, [3], for further
details on this). The action, furthermore, becomes “euclideanised”.
Now, the partition function
Z = e−S φ = e− 21φ2φ√|g|d3xdτ φ,
D D
Z Z R
is simply a Gaussian and hence the functional integral can be carried out,
the result being
Z = (det2)−1/2. (2)
Seeforinstance[1]. Thequantityweareparticularlyinterestedin,isHelmholtz’
free energy, F, which is defined as
1 1
F = lnZ = lndet2. (3)
−β 2β
The internal energy and pressure, which appear in the energy momentum
tensor, are related to F by the usual thermodynamic relations, as is the
entropy.
3 Functional Determinants
2
The major problem is apparently the calculation of det . For consistency,
andin order to establish notation, Iwill give a short introduction to the topic
here. Descriptions can be found in e.g. Hawking [2] or Ramond [1].
3
A priori, the determinant of an operator A must be given by the product of
its eigenvalues λ
detA = λ. (4)
Now, obviously this is not an easy thingYto calculate. This is where the zeta
function, ζ (s), and the heat-kernel, G (x,x′;σ), comes into play. Define
A A
ζ (s) = λ−s, (5)
A
X
then
detA = e−ζA′ (0). (6)
The heat-kernel is defined through the differential equation
∂
A G (x,x′;σ) = G (x,x′;σ), (7)
x A A
−∂σ
subject to the boundary condition
limG (x,x′;σ) = δ(x x′). (8)
A
σ→0 −
The reason for the terminology is transparent: when A is d2 , the Laplacean
dθ2
on S1, then ζ (s) ζ(s), where ζ(s) = ∞ n−s, is the Riemann zeta-
A ∝ n=1
function, and when A = 2 and σ is the temperature, then the heat kernel
∇ P
satisfies the usual heat equation.
Denoting the eigenfunctions of A by ψ we have
λ
G (x,x′;σ) = ψ (x)ψ∗(x′)e−λσ, (9)
A λ λ
λ
X
and one easily proves the important relationship
1 ∞
ζ (s) = dσσs−1 d4xG (x,x;σ). (10)
A A
Γ(s)
Z0 Z
Notice that the integral is only along the diagonal x = x′. We don’t have
to know the eigenfunctions in order to solve the heat equation, in a d-
dimensional Euclidean space, for instance, one can show [1]
GA(x,x′;σ) = (4πσ)−d/2e−(x−4σx′)2,
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in Cartesian coordinates. We will later use a mixture of techniques to arrive
at a workable expression for the heat kernel. The angular and thermal part
will be handled by a mode sum, whereas we will have to make do with an
asymptotic expransion for the radial part in order to find the Casimir energy
density.
We can arrive at an important interpretation of the value of the heat kernel
along the diagonal by noting
1 ∂ 1 ∞
f(x) dσσs−1G (x,x;σ), (11)
A
≡ −2β ∂s Γ(s)
(cid:12)s=0 Z0
(cid:12)
(cid:12)
is the density of Helmholtz’ free energy
(cid:12)
F = f(x) g d4x, (12)
| |
Z
p
and is thus exactly the quantity we are interested in calculating.
4 Finding the Heat-Kernel
We now return to our particular problem, the minimally coupled scalar field
in a Schwarzschild back-ground, i.e. the calculation of the determinant of the
d’Alembertian. For this particular operator we know that the eigenfunctions
can be written
ψ (τ,r,Ω) = e−iωnτY (Ω)g (r), (13)
λ lm λ
where Ω denotes the angles, and where ω is the Matsubara frequency
n
2πn
ω = n = 0, 1, 2,.... (14)
n
β ± ±
Wehavenointentionoffindingtheeigenvaluesλ, noroffindingg (r),instead
λ
we will rewrite the d’Alembertian as
ω2 1 ∂ ∂ l(l+1)
2 = n r2h . (15)
− h − r2∂r ∂r − r2
And the heat-kernel will be written as
G(x,x′;σ) = g (r,r′;σ)Y (Ω)Y∗ (Ω′)e−λnl(r)σ−iωn(τ−τ′), (16)
nl lm lm
nlm
X
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where we have defined
l(l+1) ω2
λ (r) + n. (17)
nl ≡ r2 h
The reasoning behind this formula is as follows. The eigenfunctions can be
written according to (13) as a product of a spherical harmonic, a plane wave
invloving the Matsubara frequency and the complexified time, and finally
some unknown radial function. Clearly, we would expect the heat kernel,
being a sum of products of eigenfunctions, to have a similar form, but since
we cannot find the radial eigenvalues λ we cannot simply use (13) to find
G. On the other hand, it must be possible to expand it on the functions
Y (Ω)Y∗ (Ω′) and e−iωn(τ−τ′), hence the expression (16).
lm lm
A mass, µ, of the scalar field can be reinserted simply by adding µ2 to λ (r).
nl
The heat equation determines the unknown functions g . We will Taylor
nl
expand these in σ, writing (asymptotic expansion)1
∞
gnl(r,r′;σ) = 1 e−(r−4σr′)2 ak(r,r′)σk, (18)
√4πσ
k=0
X
where the first term is the form g would have in a flat space-time. The
nl
boundary condition implies a 1. This asymptotic expansion is similar
0
≡
to the standard Schwinger-DeWitt expansion [5, 6]. The difference here
lies in the fact that we only make an asymptotic expansion of the radial
part of the heat kernel, whereas we use a mode sum for the remaining part.
Our expansion is thus a hybrid, combining the asymptotic Schwinger-DeWitt
expansion with the exact mode sum.
Inserting this expansion into the heat equation yields a recursion relation
for the a ’s. As we only need to know the value of the heat-kernel along
k
the diagonal x = x′, we will put r = r′ to arrive at the following recursion
relation by straightforward computation
M
1+k + a = L a L a , (19)
k+1 1 k 2 k−1
r − −
(cid:18) (cid:19)
where we have defined the operators
2M d2 2 8M d
L = 1 +
1
− r dr2 r − r2 dr
(cid:18) (cid:19) (cid:18) (cid:19)
1For simplicity we will follow the common abuse of terminology and refer to this ex-
pression as “the heat kernel” although it strictly speaking is only an asymptotic formula
valid for σ not too large.
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d2 2 d
= h + (2h 1) , (20)
dr2 r − dr
d
L = hλ′ +2hλ′ r−1 +h′λ′ +hλ′′ . (21)
2 nldr nl nl nl
Putting
H = L a = 2hλ′ r−1 +h′λ′ +hλ′′ , (22)
2 0 nl nl nl
the first few functions a becomes
k
a = 1, (23)
0
a = 0, (24)
1
2H
a = − , (25)
2
5 h
−
( 1)222 H
a = − L , (26)
3 1
7 h 5 h
− −
( 1)323 1 H ( 1)222 H
a = − L L + − L , (27)
4 1 1 2
9 h 7 h 5 h 9 h 5 h
− − − − −
( 1)424 1 1 H ( 1)323 1 H
a = − L L L + − L L +
5 1 1 1 1 2
11 h 9 h 7 h 5 h 11 h 9 h 5 h
− − − − − − −
( 1)323 1 H
− L L . (28)
2 1
11 h 7 h 5 h
− − −
There are singularities in these at r = 2M, as one would expect. Noting that
both H and L are linear in ω2, we see that a can contain all even powers
2 n k
2[k/2]
of the Matsubara frequencies up to and including ω .
n
The method of summation of modes when calculating zeta functions have
been applied in a number of other spacetimes by various authors, [10], and
a number of explicit results exist for some rather simple cases.
5 The Free Energy Density
We have now found an asymptotic expansion for the heat-kernel and are
ready to integrate it. The σ integration is trivial (it usually is), and will just
give a sum of Γ-functions. Explicitly
1 ∞ 1 ∞
f˜(x;s) dσ Y (Ω) 2 a e−λnl(r)σσs+k−3/2
lm k
≡ Γ(s) | | √4π
Z0 nlm k=0
X X
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= Y 2 1 ∞ a (r)Γ(s+k − 12)λ21−s−k. (29)
| lm| √4π k Γ(s) nl
nlm k=0
X X
One should note that even though the asymptotic series is only valid for σ
small, this integral still makes sense since the factor λ acts as a mass term
nl
regularisation. If we hadn’t used our hybrid method we would have had to
assume a mass for the scalar field in order to obtain a meaningfull integral.
We can still do this, if we wish, by simply adding µ2 to λ . The addition of
nl
˜
such a finite mass, however, will not modify the convergence properties of f,
hence the succesive regularisations to be performed below are still needed.
This is so because only the sum over n need to be regularised, and for fixed
l, λ acts like a finite mass irrespective of whether µ2 = 0 or not.
nl
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For k = 0, differentiating this with respect to s at s = 0 simply amounts
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to removing the Γ-function in the denominator and putting s = 0 in the
remaining terms. Thek = 0termissingular andhastobetreatedseparately.
We have
f˜ (x;s) = Y 2 1 Γ(s− 21)λ21−s,
k=0 | lm| √4π Γ(s) nl
nlm
X
sowehaveadivergentsum(fors = 0)inthenumerator,namely √z2 +4π2n2,
n
where z2 = l(l+1)h(r)β2. We can carry out the sum for an arbitrary s, see e.g.
r2 P
Ramond [1]:
∞
1 z2 +4π2n2 21−s = 1 z1−2s + 2ζ(2s+1)+
Γ(s) Γ(s) " (4π2)s−21
n=−∞
X (cid:0) (cid:1)
2 ∞ x2ν ( 1)ν 1
− Γ(s+ν )ζ(2s+2ν 1) .
Γ(s+ 21) ν=1 ν! (4π2)s+ν−21 − 2 − #
X
Differentiating and putting s = 0 we obtain
1 z2 2 ∞ z2ν ( 1)ν 1
z + (γ 2ln2π)(4π )+ − Γ(ν )ζ(2ν 1),
2π3/2 − − π √π ν! (4π2)ν−1 − 2 −
2
ν=2
X
which is finite. This expression is to be multiplied by β−1h−1/2.
The free energy density f(x) is then
˜
1 ∂f
f(x) =
−2β ∂s
(cid:12)
(cid:12)s=0
(cid:12)
(cid:12)
(cid:12) 8
1 l(l+1)
= Y 2(4π)−1/2 2√π
−2β | lm| − r2 −
" r
lm
X
1 4π l(l+1)√hβ
2√π (γ 2ln2π) +
2π3/2 − √hβ − πr2 !
∞ ( 1)ν l(l+1)hβ2 ν 1
4√πβ−1h−1/2 − Γ(ν )ζ(2ν 1)
ν! 4π2r2 − 2 − −
#
ν=2 (cid:18) (cid:19)
X
∞
1 1
Ylm 2(4π)−1/2 ak(r)λnl(r)21−kΓ(k ). (30)
−2β | | − 2
nlm k=2
X X
On the face of it, this function depends on all the coordinates, but looking
closer we see that there is no explicit time-dependence, and using the sum
rule
2l+1
Y (Ω)Y∗ (Ω′) = P (u u′),
lm lm 4π l ·
m
X
where u,u′ are unit vectors given by the solid angles Ω,Ω′, we see that [4]
2l+1
Y (Ω) 2 = P (1)
lm l
| | 4π
m
X
[1l]
2l+1 2 ( 1)ν(2l 2ν)!
= − − .
2l+2π ν!(l ν)!(l 2ν)!
ν=0 − −
X
So the angular dependency also disappears. In accordance with what we
would expect, the energy density depends only on the radial coordinate. We
will write the density as a mode sum
f(r) = f (r), (31)
l
l
X
where
1 l(l+1)β2
f (r) π−1β−2 (4π)−3/2(2l+1)2−l +
l
≡ 2β r2
( r
1 4π l(l+1)√hβ2
(γ 2ln2π) +
2π3/2 − √hβ2 − πr2 !
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2 ∞ ( 1)ν l(l+1)hβ2 ν 1
− Γ(ν )ζ(2ν 1)
β√hπ ν! 4π2r2 − 2 − !−
ν=2 (cid:18) (cid:19)
X
∞ ∞
1 1
(4π)−3/2(2l+1)2−l Γ(k )a (r)λ (r)1/2−k
k nl
2β − 2 ×
)
n=−∞k=2
X X
[l/2]
( 1)ν(2l 2ν)!
− − . (32)
ν!(l ν)!(l 2ν)!
ν=0 − −
X
The major problem here is that the coefficients a (r) depend on n through
k
the Matsubara frequencies.
This expression needs regularisation. For instance, when l = 0 we encounter
a singularity. Remembering that a can include all even powers of n up to
k
and including n2[k/2] we write
[k]
2
a (r) = b(t)(r)n2t,
k k
t=0
X
where b(s) is independent of n. Defining λ = 4π2/hβ2 we then have
k
∞ ∞ ∞ [k/2] ∞
ak|l=0Γ(k − 21)λn120−k = 2 Γ(k − 21)λ21−k b(kt)(r) n2t+1−2k
n=−∞k=2 k=2 t=0 n=1
X X X X X
∞ [k/2]
1
= 2 Γ(k )λ21−k b(t)(r)ζ(2k 2t 1),
− 2 k − −
k=2 t=0
X X
which is divergent whenever 2k 2s 1 = 1, as s 1k this can only happen
− − ≤ 2
when k = 2,s = 1.
6 The Minkowski Space Contribution and the
Casimir Energy Density
To get a physical quantity we must subtract the Minkowski space contri-
bution [5, 6], because the physical energy must be such that it vanishes in
flat spacetime. Doing this we thereby get the “Casimir energy density”. The
relationship between this Casimir energy density, ζ-functions and cut-off reg-
ularisations has been studied in [9]. This subtracting off the flat spacetime
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