Energy density for a massive scalar field in (1+1)D
Fernando Daniel Mera
Northeastern University-Department of Physics
mera.f@husky.neu.edu
May 16, 2013
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
May 16, 2013
1 / 23
Introduction
Vacuum energy density for a massive scalar field
This paper is based on Patrick Hays’s paper on a confined massive field in
two dimensions. In the paper “Vacuum fluctuations of a confined massive
field in two dimensions,” the zero-point energy of a massive scalar field
confined to a two-dimensional M.I.T. bag model, is computed.
Motivation
We follow the mathematical style of Fulling’s paper “Vacuum energy as
spectral geometry.” The vacuum energy is treated as a purely
mathematical problem, an underdeveloped aspect of the spectral theory of
self-adjoint second-order differential operators. What I am basically doing
is to note the common generalization between the P. Hay’s paper and S.
Fulling’s paper.
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
May 16, 2013
2 / 23
Vacuum energy density
Boundary vacuum energy from closed and periodic orbits
We consider a finite interval with either a Dirichlet or a Neumann boundary condition at each end. Thus
d2
2
2
H = − dx
2 + m acts in L (0, L) on the domain defined by
u (1−l) (0) = 0,
u (1−r ) (L) = 0,
l, r ∈ {0, 1}
(1)
The Green function can be constructed from G∞ by the method of images. The Green function can be expressed as
G (ω 2 , x, y ) = G∞ (y ) + (−1)l G∞ (−y ) + (−1)r G∞ (2L − y ) + (−1)l+r G∞ (2L + y )
(2)
+ (−1)l+r G∞ (−2L + y ) + (−1)2l+r G∞ (−2L − y )
(3)
+ (−1)l+2r G∞ (4L − y ) + (−1)2l+2r G∞ (4L + y ) + · · ·
(4)
and the above Green function has to satisfy the equation:
δ(x − y ) = −
∂2G
+ (m2 − λ)G
∂x 2
(5)
This is the same as the equation satified by G∞ and G in [2] except that −λ √
has been replaced by m2 − λ . So, we
should be abe to use the same formulas as in [2] but we need to replace ω(≡ λ) by
p
κ ≡ ω 2 − m2 .
(6)
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
May 16, 2013
3 / 23
Green Function
Let’s check from first principles that the new G∞ satisfies the right Green equations. We want
−
∂2G
− κ2 G = δ(x − y ),
∂x 2
(7)
so for x − y we want
∂2G
= −κ2 G .
∂x 2
Thus
G (x, y ) =
(8)
(
Ae −iκ(x−y ) , x < y ,
Be iκ(x−y ) ,
x > y.
(9)
Therefore, our G∞ is given by
i −κ|x−y |
e
.
2κ
When we go to the variable κ the situation is slighly more complicated; κ is not just ω minus a constant
G∞ (ω 2 , x, y ) =
(10)
Remark:The Weyl and periodic terms will not be the same as in the massless case,
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
May 16, 2013
4 / 23
The Hamiltonian, now contains a potential term which comes from the massive scalar field. I will adopt the
convention that
(Hx − κ)G (ω 2 , x, y ) = δ(x − y ).
(11)
All we know is that G (ω 2 , x, y ) must satisfy the above equation. Let us not loose sight of our objective, to find the
local spectral density from closed and periodic orbits. So, we have
κ
π σ(ω) ≡ 2κ Im G (ω 2 , x, x)
ω
n=∞
∞
X
X
=
(−1)n(l+r ) cos(2κnL) +
(−1)l+n(l+r ) cos(2κ(nL + x))
+
n=0
∞
X
(13)
n=0
(−1)−l+n(l+r ) cos(2κ(nL − x)) +
n=1
=1+2
(12)
∞
X
(−1)n(l+r ) cos(2κnL) +
n=1
∞
X
(−1)n(l+r ) cos(2κnL)
(14)
n=1
∞
X
(−1)l+n(l+r ) cos(2κ(x + nL))
(15)
n=−∞
κ
κ
≡ π (σav + σper + σbdry ) ≡ π (σav + σosc )
ω
ω
(16)
√
where κ ≡ ω 2 − m2 .
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
May 16, 2013
5 / 23
In the case ξ = 14 , the contribution of the space derivatives is identical to that of the time derivatives, so we can
write
Z ∞
1 ∂
T00 (t, x) ≡ E (t, x) = −
σ(ω)e −ωt dω ≡ EWeyl (t, x) + Eper (t, x) + Ebdry (t, x).
(17)
2 ∂t 0
Using Equation 3.914.1 from [4], we obtain the following expression:
Z ∞
q
√
mt
2
2
e −t m +κ cos(2nLκ) dκ = p
K1 (m t 2 + (2nL)2 )
2
2
t + (2nL)
0
(18)
Let’s compute the EWeyl term for the massive case:
EWeyl (t) = −
1d
2 dt
Z
∞
σWeyl (ω)e −ωt dω
(19)
0
and doing the change of variables, ω 2 = κ2 + m2 , gives
Z ∞√ 2
√
1 d
κ + m2
κ
2
2
EWeyl (t) = −
·√
e −t κ +m dκ
2π dt 0
κ
κ2 + m 2
Z ∞
√
1 d
1 d
2
2
=−
e −t κ +m dκ = −
mK1 (mt).
2π dt 0
2π dt
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
(20)
(21)
May 16, 2013
6 / 23
When ν is fixed and z → 0,
−ν
1
1
Kν (z) ∼ Γ(ν) z
2
2
and in our case, ν = 1 and hence
(Re ν > 0)
(22)
−1
1
1
1
K1 (z = mt) ∼ Γ(1) mt
=
2
2
mt
(23)
Therefore for small mt, our expression becomes
EWeyl (t) ∼ −
1 d
1 d
1
1
mK1 (mt) = −
m
=
2π dt
2π dt
mt
2πt 2
(24)
To put equation 24 into the usual form for renormalization calcualtions, we need to expand it in power (Laurent)
series in t. The leading term will be O(t −2 ) and should match the massless case. The Laurent series in t for
equation can be expressed as
1 d 1 1
+ m(mt) 2 log(mt) + 2γ − 1 − 2 log(2) + O (mt)2
EWeyl (t) ∼ −
(25)
2π dt t
4
1
m2
1 2
∼
−
−
m (2 log(mt) + 2γ − 1 − 2 log(2)) + O (t)
(26)
2πt2
4π
8π 1 1
m2
mt
m2
∼
−
log
−
(1 + 2γ) + O (t)
(27)
2π t 2
2
2
4
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
May 16, 2013
7 / 23
The periodic term for the massive case is given by
Z ∞
∞
1 d X
ω
(−1)n(l+r )
σper (ω)e −ωt dω
π dt
κ
0
n=1
Z ∞
∞
√
1 d X
2
2
e −t m +κ cos(2nLκ) dκ
=−
(−1)n(l+r )
π dt
0
n=1
∞
q
mt
1 d X
=−
K1 (m (2nL)2 + t 2 )
(−1)n(l+r )p
2 + t2
π dt
(2nL)
n=1
Eper (t) = −
(28)
(29)
shri
(3
Therefore,
q
∞
∞
1 d X
t
1 d X m2 t
2 log m (2nL)2 + t 2
− lim
2
2
m→0
π dt
(2nL) + t
π dt
4
n=1
n=1
2
2 2
2
+ 2γ − 1 − 2 log(2) + O m (4L n + t )
!
∞
πt
1 d X
t
1 d 1 π coth 2L
2
=
−
−
∼−
π dt
(2nL)2 + t 2
π dt 4
L
t
n=1
πt π
1
−
∼ 2 csch2
8L
2L
2πt 2
lim Eper (t) ∼ −
m→0
(31)
(32)
(33)
(34)
and we can clearly see that the above result agrees with [2, p. 15].
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
May 16, 2013
8 / 23
The periodic term, Eper (t), will approach a contant value as t → 0:
∞
1 d X
t
π dt
(2nL)2 + t 2
n=1
q
∞
1 d X m2 t
− lim
2 log m (2nL)2 + t 2 + 2γ − 1 − 2 log(2)
t→0 π dt
4
n=1
2
2 2
+ O(m (4L n + t 2 ))
lim Eper (t) ∼ − lim
t→0
t→0
(35)
(36)
(37)
At this point, let’s split the periodic terms into the massless contribution and the massive contribution. The
m=0 (t) and the massive contribution will be
massless contribution to the periodic energy will be denoted by Eper
m (t).
denoted by Eper
So,
∞
∞
t
1X d
t
1 d X
m=0
=
lim
−
(38)
lim Eper
(t) = lim −
t→0
t→0
t→0
π dt
(2nL)2 + t 2
π
dt (2nL)2 + t 2
n=1
n=1
∞
πt
πcsch2 2L
1 X 4L2 n2 − t 2
1
= lim −
= lim
−
(39)
2 n 2 + t 2 )2
t→0
t→0
π
16L2
4πt 2
(4L
n=1
2 πt
πcsch 2L
1
π
= lim
−
=−
(40)
t→0
8L2
2πt 2
24L2
and this agrees with [2, pg. 16].
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
May 16, 2013
9 / 23
Calculation of the term Ebdry (t, x)
The interesting term is the boundary term, Ebdry (t, x), which is given by
Z ∞
∞
(−1)l ∂ X
ω
cos(2κ(x + nL))e −ωt dω
(−1)n(l+r )
2π ∂t n=−∞
κ
0
Z ∞
∞
√
(−1)l ∂ X
2
2
(−1)n(l+r )
cos(2κ(x + nL))e −t κ +m dκ
=−
2π ∂t n=−∞
0
Ebdry (t, x) = −
(41)
(42)
Now, we do a change of variables so that we can integrate with respect to κ instead of ω. After doing the change
of variables we obtain
Z ∞
∞
√
(−1)l ∂ X
2
2
(43)
Ebdry (t, x) = −
(−1)n(l+r )
cos(2κ(x + nL))e −t κ +m dκ
2π ∂t n=−∞
0
∞
q
mt
(−1)l ∂ X
(−1)n(l+r )p
=−
K1 (m (2(x + nL))2 + t 2 )
(44)
2
2
2π ∂t n=−∞
(2(x + nL)) + t
since ω dω = κ dκ.
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
May 16, 2013
10 / 23
In the boundary case, things get more complicated because we now have to deal with position, x. Then the
boundary term, Ebdry (x, t) can be expressed as
Ebdry (x, t) =
q
∞
(−1)l ∂ X
mt
(−1)n(l+r )p
K1 m 4(x + nL)2 + t 2
2
2
2π ∂t n=−∞
4(x + nL) + t
(45)
Let’s go back to computing the boundary term. The boundary term can be expressed as
Ebdry (x, t) = −
q
∞
mt
(−1)l ∂ X
(−1)n(l+r )p
K1 m 4(x + nL)2 + t 2
2
2
2π ∂t n=−∞
4(x + nL) + t
(46)
(47)
Using the asymptotic expansion of K1 (z) for small argument yields
Ebdry (x, t) ∼
∞
mt
(−1)l ∂ X
1
p
(−1)n(l+r )p
2π ∂t n=−∞
4(x + nL)2 + t 2 m 4(x + nL)2 + t 2
=−
∞
t
(−1)l ∂ X
(−1)n(l+r )
2π ∂t n=−∞
4(x + nL)2 + t 2
(48)
(49)
and this agrees with the result obtained in [2].
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
May 16, 2013
11 / 23
Let’s go back to the massive case. In the massive case, we quickly discover that we can’t obtain an explicit formula
for the infinite sum. Then
∞
(−1)l ∂ X
t
Ebdry (x, t) = −
(−1)n(l+r )
(50)
2π ∂t n=−∞
4(Ln + x)2 + t 2
q
2
m t
+
(51)
2 log m 4(Ln + x)2 + t 2 + 2γ − 1 − 2 log(2)
4
q
2
2
2
+ O (m 4(Ln + x) + t )
(52)
Let’s assume that l + r is an odd integer. For the odd case, we have
(−1)l π
π(t − 2ix)
π(t − 2ix)
Ebdry (x, t) = −
coth
csch
16L2
2L
2L
π(t + 2ix)
π(t + 2ix)
+ coth
csch
2L
2L
(−1)l m2 t
π(t − 2ix)
π(t + 2ix)
+
csch
+ csch
16L
2L
2L
q
∞ (−1)l X (−1)n 2
+
m 2 log m 4(Ln + x)2 + t 2 + 2γ − 1 − 2 log(2)
2π n=−∞
4
+ O(m4 (4(Ln + x)2 + t 2 ))
(53)
(54)
(55)
(56)
(57)
or,
πx
csc πx
t 2 π 2 2 cot3 πx
csc
L
L + cot L
−
8L2
64L4
2 2
t πm cot xL csc xL
+ −
+ ···
16L2
q
∞ (−1)l X (−1)n 2
+
m 2 log m 4(Ln + x)2 + t 2
2π n=−∞
4
+ 2γ − 1 − 2 log(2)) + O(m4 (4(Ln + x)2 + t 2 )) + O(t 4 )
Ebdry (x, t) ∼
cot
πx
L
πx
L
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
+ ···
(58)
(59)
(60)
(61)
May 16, 2013
12 / 23
and when t = 0, we obtain
Ebdry (x, 0) =
cot
πx
L
πx
L
csc
8L2
−
∞
(−1)l X (−1)n 2
m 2 log (mLn) + 2γ − 1
2π n=−∞ 4
(62)
For the even case we obtain,
πt
L
2x
L −1
πt 2
L
(−1)l πm2 t sinh πt
L πt
8L(cos 2x
4L2 cos L − cosh
L − cosh L )
q
∞ (−1)l X 1 2
+
m 2 log m 4(Ln + x)2 + t 2 + 2γ − 1 − 2 log(2)
2π n=−∞ 4
+ O m4 (4(Ln + x)2 + t 2 )
Ebdry (x, t) = −
(−1)l π cosh
2x
cos
−
(63)
(64)
(65)
Can we match this result against some results of [1] or Appendix B of the predecessor paper by Bender and Hays
[3]?
Ignoring the mass terms, we obtain the following expression:
2x
(−1)l π cosh πt
L cos L − 1
(66)
Ebdry (x, t) ∼ −
πt 2
4L2 cos 2x
L − cosh L
and now let’s assume that t is very small and that x is fixed. Assuming that l = 1, and using a power series
expansion we obtain the following expression:
4 x π csc2 xL
t 2 π 3 cos 2x
L + 2 csc
L
Ebdry (x, t) ∼
−
+ O(t 4 ).
2
4
8L
32L
(67)
When t = 0, we obtain
Ebdry (x, 0) ∼
π csc2
8L2
x
L
(68)
and the above result agrees with [2].
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
May 16, 2013
13 / 23
Then,
4 x π csc2 xL
t 2 π 3 cos 2x
π 2 m2 t 2 csc2 xL
L + 2 csc
L
−
+
2
4
2
8L
32L
16L
q
∞ (−1)l X 1
2
2
2
+
2m log 2m (Ln + x) + 2γm − m2 − 2m2 log(2)
2π n=−∞ 4
m2 t 2
+
+ O t4
16(Ln + x)2
Ebdry (x, t) ∼
(69)
(70)
(71)
and when t = 0, we have
Ebdry (x, 0) ∼
π csc2
8L2
x
L
+
q
∞
(−1)l X 1
2m2 log 2m (Ln + x)2 + 2γm2 − m2 − 2m2 log(2)
2π n=−∞ 4
(72)
When the mass is sufficiently small or equal to 0, the above analysis yields the correct answers time after time. So
far, Dr. Fulling and I haven’t spotted any serious errors with the above asymptotic analysis. It seems to me that
the above is valid when m → 0 because the above answers also seem to agree with Hay’s paper [1].
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
May 16, 2013
14 / 23
The boundary term can be expressed as
Ebdry (x, t) = −
∞
q
mt
(−1)l ∂ X
(−1)n(l+r )p
K1 (m 4(x + nL)2 + t 2 )
2π ∂t n=−∞
4(x + nL)2 + t 2
(73)
(74)
Let’s assume that l + r is an even integer. Then we integrate the local energy denisty and we obtain
Ebdry (t) = −
∞ Z
q
(−1)l ∂ X L
mt
p
K1 (m 4(x + nL)2 + t 2 ) dx
2 + t2
π ∂t
4(x
+
nL)
0
n=0
(75)
and hence,
(2Ln − t + 2x)(2Ln + t + 2x)K mpt 2 + 4(Ln + x)2
∞ Z
1
(−1)l X L
m
Ebdry (t) = −
π
(4(Ln + x)2 + t 2 )3/2
n=0 0
p
mt 2 K0 m t 2 + 4(Ln + x)2 −
dx
4(Ln + x)2 + t 2
(76)
(77)
and doing a change of variables x 0 = Ln + x, we obtain
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
May 16, 2013
15 / 23
√
√
∞ Z
(−1)l X L(n+1)
(2x 0 − t) (t + 2x 0 ) K1 (m t 2 + 4x 02 ) mt 2 K0 (m t 2 + 4x 02 )
m
dx 0
−
2
02
3/2
2
02
π
t + 4x
(t + 4x )
n=0 Ln
√
√
Z
(−1)l ∞
(2x 0 − t) (t + 2x 0 ) K1 (m t 2 + 4x 02 ) mt 2 K0 (m t 2 + 4x 02 )
=−
m
dx 0
−
2
02
3/2
2
02
π
t + 4x
(t + 4x )
0
Ebdry (t) = −
and doing another change of variables u = 4x 02 + t 2 , we have
!
Z
u 2 − 2t 2 K1 (mu) mt 2 K0 (mu)
u
(−1)l ∞
√
Ebdry (t) = −
m
−
du
π
u3
u2
2 u2 − t 2
t
!
Z
u 2 − 2t 2 K1 (mu) mt 2 K0 (mu)
(−1)l ∞
√
=−
m
− √
du
π
2u 2 u 2 − t 2
u u2 − t 2
t
1
1
Ebdry (t) = − πm2 tEi(−mt) + πm mtEi(−mt) + e −mt
4
4
(78)
(79)
(80)
(81)
(82)
and,
1
(−1)l
1
lim Ebdry (t) = −
lim
πm2 tEi(−mt) + πm mtEi(−mt) + e −mt
π t→0 4
4
(−1)l mπ
(−1)l m
=−
=−
π
4
4
t→0
And for the Dirichlet case (l = 0) we obtain
Ebdry (0) = −
m
4
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
(83)
(84)
(85)
May 16, 2013
16 / 23
Future Work
There are 4 calculations in Section 4 of [2]:
local spectral density (σ, p. 12),
“global” eigenvalue density (ρ or N, p. 13 and p. 14),
total energy (E , pp. 15-16), and
local energy density (T00 or E (t, x), pp. 18-20).
Question: Where do we stand on these four calculations?
Answer: And, my answer is simple, I have been focusing all of my attention on the local energy density calculations.
I ignored the other three calcualtions because I thought obtaining the local energy density was the top priority.
Hopefully, I will get around to improving the structure of the paper itself, but before I do that I would like to
receive more feedback.
The massive analogs of the formulas for ρWeyl , ρper , ρbdry will be computed along the following lines:
The massive analogs of the formulas for σ(ω) are related to the massless case by simply substituting κ for ω
πω
and also multiplying it by the factor
.
κ
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
May 16, 2013
17 / 23
Future Work & The Current State of Affairs
At this point, the research notes lack structure, but the notes don’t lack direction. The direction that I am taking
is now to compute the massive analog of the counting function N(ω).
Once again, there should be agreement between the massive counting function and the massless counting function
when m → 0.
Let’s examine the global situation first. The eigenvalue density is
Z
L
ρ(ω) =
σ(ω, x) dx = ρWeyl (ω) + ρper (ω) + ρbdry (ω),
(86)
0
where
Z
L
ρWeyl (ω) =
Z
L
σav dx =
0
0
Lω
ω
dx =
,
πκ
πκ
ρper =
∞
2Lω X
(−1)n(l+r ) cos(2κnL),
πκ
(87)
n=1
The eigenvalue density is
Z
L
ρ(ω) =
σ(ω, x) dx = ρWeyl (ω) + ρper (ω) + ρbdry (ω),
(88)
0
where
Z
ρWeyl (ω) =
L
Z
σav dx =
0
ρbdry =
0
L
Lω
ω
dx =
,
πκ
πκ
ρper =
∞
2Lω X
(−1)n(l+r ) cos(2κnL),
πκ
(89)
n=1
∞
(−1)l X (−1)n(l+r ) ω
[sin(2κL(n + 1)) − sin(2κLn)],
2π n=−∞
κ2
(90)
√
where κ = ω 2 − m2 .
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
May 16, 2013
18 / 23
Future Work and State of Affairs
The eigenvalue counting function N(ω) is zero for ω < m and
NWeyl (ω) =
L
π
κ
Z
0
=
=
2L X
(−1)n(l+r )
π
2L
π
1
π
n=1
∞
X
ρ for ω > m. Therefore (for ω > m),
Z κ√
0
(91)
κ
κ2 + m2
√
cos(2nL κ) dκ
κ
κ2 + m2
(92)
sin(2nL κ)
2Ln
(93)
(−1)n(l+r )
sin(2nL κ)
n
(94)
(−1)n(l+r )
n=1
∞
X
n=1
0
√
ω κ
Lκ
L ω 2 − m2
· dκ =
=
.
κ ω
π
π
∞
Nper (ω) =
Rω
The Fourier series in Nper can be evaluated to a sawtooth function. [See GR 1.441.1, GR 1.441.3, and [2] pp. 14
and pg. 9-10].
Z
ρbdry (ω) =
σbdry (ω) dx =
0
=
L
Z L
∞
ω
(−1)l X
(−1)n(l+r )
cos(2κ(x + nL)) dx
π n=−∞
0 κ
∞
(−1)l X (−1)n(l+r ) ω
[sin(2κL(n + 1)) − sin(2κnL)]
2π n=−∞
κ2
(95)
(96)
and,
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
May 16, 2013
19 / 23
Future Work & State of Affairs
Z
ω
Nbdry (ω) =
ρbdry (ω) dω =
m
Z κ√ 2
∞
κ + m2
(−1)l X
κ
dκ
sin(2κL(n + 1)) − sin(2κnL) √
(−1)n(l+r )
2
2π n=−∞
κ
κ2 + m 2
0
(97)
Z κ
∞
1
(−1)l X
(−1)n(l+r )
[sin(2κL(n + 1)) − sin(2κnL)] dκ
=
2π n=−∞
0 κ
(98)
In other words, we end up with the following expression:
(
R
(−1)l P∞
n(l+r ) κ sin(2κL(n+1))−sin(2κLn) dκ if l + r is even,
n=0 (−1)
π
κ
0
Nbdry (ω) =
0
if l + r is odd.
(99)
Consider the regularized vacuum energy
E (t) = −
d 1
dt 2
Z
∞
ρ(ω)e −ωt dω ≡ EWeyl (t) + Eper (t) + Ebdry (t)
(100)
0
where
Z ∞
∞
√
L d X
2
2
(−1)n(l+r )
cos(2κnL)e −t κ +m dκ
π dt
0
n=1
q
∞
L d X
mt
K1 m t 2 + (2nL)2
=−
(−1)n(l+r )p
2
2
π dt
t + (2nL)
n=1
Eper (t) = −
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
(101)
(102)
May 16, 2013
20 / 23
Taking the limit of m → 0 of Eper (t) yields
lim Eper (t) = lim −
m→0
m→0
∼ lim −
m→0
q
∞
L d X
mt
K1 m t + (2nL)2 )
(−1)n(l+r )p
2
2
π dt
t + (2nL)
(103)
L d
π dt
(104)
n=1
∞
X
n=1
mt
(−1)n(l+r )p
t 2 + (2nL)2
1 1
+ z(2 log(z) + 2γ − 1 − 2 log(2)) + O z 2
z
4
∞
t
L d X
(−1)n(l+r ) 2
,
=−
π dt
t + (2nL)2
×
(105)
(106)
n=1
p
where z = m t 2 + (2nL)2 . So, the massive analog of the periodic energy agrees with the massless case.
The End
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
May 16, 2013
21 / 23
References
P. Hays, Vacuum fluctuations of a confined massive field in two dimensions, Ann.
Phys. 121 (1979) 32-46.
S. A. Fulling, Vacuum energy as spectral geometry, SIGMA 3 (2007) 094;
arXiv:0706.2831.
C.M. Bender and P. Hays, Phys. Rev. D 14 (1976) 2622.
I.S. Gradshteyn and I.M. Ryzhik, Table of Integrals, Series and Product, Academic
Press Inc.; 2nd edition (June 1980).
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
May 16, 2013
22 / 23
The End
Fernando Daniel Mera (Northeastern University)Energy density for a massive scalar field
May 16, 2013
23 / 23