Finite temperature Casimir pistons for electromagnetic field with mixed boundary conditions and its classical limit

0.22 MB·English

by  L. P. Teo

Finite temperature Casimir pistons for electromagnetic field with mixed boundary
conditions and its classical limit
L.P. Teo∗
Faculty of Information Technology, Multimedia University,
Jalan Multimedia, Cyberjaya, 63100,
Selangor Darul Ehsan, Malaysia.
Phone: [phone].
Inthispaper,thefinitetemperatureCasimirforceactingonatwo-dimensionalCasimirpistondue
to electromagnetic field is computed. It was found that if mixed boundary conditions are assumed
onthepistonanditsoppositewall,thentheCasimirforcealwaystendstorestorethepistontowards
the equilibrium position, regardless of the boundary conditions assumed on the walls transverse to
the piston. In contrary, if pure boundary conditions are assumed on the piston and the opposite
9
wall,thentheCasimirforcealwaystendtopullthepistontowardsthecloserwallandawayfromthe
0
equilibrium position. The nature of the force is not affected by temperature. However, in the high
0
temperatureregime,themagnitudeoftheCasimirforcegrowslinearlywithrespecttotemperature.
2
This shows that theCasimir effect has a classical limit as has been observed in otherliteratures.

n
a PACSnumbers: 11.10.Wx
J Keywords: Casimirforce,pistongeometry,finitetemperature,electromagneticfield,mixedboundarycondi-
3 tions.
2

]
h
t I. INTRODUCTION
-
p
e
Since the work of Cavalcanti [1], the Casimir effect of the piston geometry (see FiG. 1) has attracted considerable
h
interest for it was shown to be free of divergence problem. Some studies have been devoted to this subject [2, 3, 4,
[
5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]. It was found that for massless scalar field with periodic boundary conditions

1 (b.c.), Dirichlet b.c. and Neumann b.c., and for electromagnetic field with perfect electric conductor (PEC) b.c. and
v
perfectmagneticconductor(PMC)b.c. ina d-dimensionalspace,the Casimirforceactingonthe pistonalwaystends
4
to pull the piston to the closest wall. This might create undesirable effect known as stiction in the functionality of
8
5 nanodevices. In[5],Bartonshowedthatforathinpistonwithweaklyreflectingdielectrics,theCasimirforceatsmall
3 separationsis attractive,butturntorepulsiveasthe separationincreases. Anotherscenariowhichbringstorepulsive
. Casimirforcewasconsideredin[9,16],whereamasslessormassivescalarfieldisassumedtosatisfyNeumannb.c.on
1
the piston and Dirichlet b.c. on all other walls. In this case, the zero temperature Casimir force was shown to be
0
9 alwaysrepulsive. In[10],itwassuggestedthataperfectlyconductingpistoninsidearectangularcavitywithinfinitely
0 permeable walls will lead to repulsive Casimir force.
: In this paper, we consider the thermal correction to the repulsive Casimir force due to electromagnetic field with
v
mixed boundary conditions (PEC b.c. on one wall and PMC b.c. on the opposite wall) and determine whether
i
X temperature will change the nature of the force. We only consider the case where the space dimension d = 2. This
r will simplify the mathematical computation but it gives enough indications for the general case of higher dimensions
a
which will be considered in future. The two dimensional rectangular Casimir pistons for electromagnetic field with
purely PEC b.c. and purely PMC b.c. have been studied. The Casimir effect due to electromagnetic field with PMC
b.c. coincides with the Casimir effect due to a massless scalar field with Dirichlet b.c. whose zero temperature limit
is studied in the pioneering work [1]. The Casimir effect due to electromagnetic field with PEC b.c. coincides with
the Casimir effect due to a massless scalarfield with Neumann b.c. whose zero temperature limit is considered in [8].
ThefinitetemperatureCasimireffectwasrecentlyconsideredin[15]. Itwasfoundthatforpureboundaryconditions,
the Casimir force is alwaysattractive atany temperature. Thereforeit will be interesting to see whether the thermal
correction affects the repulsive nature of the Casimir force due to electromagnetic field with mixed b.c. This is the
issue addressed in this paper. We consider more general case of mixed b.c. where each pair of parallel plates can
either assume pure boundary conditions (both PEC b.c. or both PMC b.c.) or mixed boundary conditions.
In this paper, we work in the units where ~ (reduced Planck constant), c (speed of light) and k (Boltzmann
B
constant) are equal to unity.
∗Electronicaddress: [email protected].
2
Region I Region II
L
2
a L − a
1
FIG. 1: The two dimensional rectangular piston
II. CASIMIR ENERGY FOR ELECTROMAGNETIC FIELD WITH MIXED BOUNDARY
CONDITIONS INSIDE A RECTANGULAR CAVITY
Recall that the finite temperature Casimir energy is defined as the sum of the zero temperature Casimir energy
and the temperature correction, i.e.,
ECas =EC0as+∆ECas = 12 ωk+T log 1−e−ωTk ,
ωXk6=0 ωXk6=0 (cid:16) (cid:17)
where ωk runs through all zero point energies. The sum corresponding to the temperature correction ∆ECas is a
convergentsum. However,thezerotemperaturecontributionE0 isdivergent. Therearedifferentwaystoregularize
Cas
this sum. In the zeta regularizationscheme [17, 18, 19, 20], we define the zeta function
ζ (s)= ω−2s
0 k
ωXk6=0
and analytically continue it to a neighborhoodof s=−1/2. If ζ (s) is regularat s=−1/2,the zeta–regularizedzero
0
temperature Casimir energy is then defined as
1 1
E0,zetareg = ζ − .
Cas 2 0 2
(cid:18) (cid:19)
Correspondingly, the finite temperature Casimir energy can be computed by using the zeta function
∞
ζ(s)= ωk2 +(2πlT)2 −s.
ωXk6=0l=X−∞(cid:0) (cid:1)
It can be shown that (see [21, 22, 23, 24]) if ζ(s) has an analytic continuation to a neighborhood of s = 0 with
ζ(0)=0, then
ζ′(0)=−1ζ0 −1 −2 log 1−e−ωTk .
T 2
(cid:18) (cid:19) ωXk6=0 (cid:16) (cid:17)
Consequently, the zeta regularized finite temperature Casimir energy is equal to
T
E reg =− ζ′(0).
Cas 2
3
A disadvantage of applying the zeta regularizationscheme is that all the divergence terms in the Casimir energy has
been renormalized to zero. However, it can be shown as in [15] that in the piston scenario, the divergence terms of
the Casimir force acting on the piston due to Region I and Region II always cancel without renormalization due to
the fact that the divergence terms of the Casimir energies are linear in L .
1
For an electromagnetic field inside a d-dimensional space Ω, the field strength is represented by a totally anti-
symmetric rank two tensor Fµν, µ,ν =0,1,...,d, satisfying the equations
∂ F˜µν1...νd−2 =0, ∂ Fµν =jν, (1)
µ µ
whereF˜µ1...µd−1 =εµ1...µd−1,ν,λFνλ isthedualtensorofFµν andjν isthe current. Inthevacuumstate,jν =0. There
are two ideal boundary conditions that are of particular interest, i.e., the perfect electric conductor (PEC) boundary
conditions (b.c.) characterizedby nµ F˜µν1...νd−2 =0 andthe perfect magnetic conductor(PMC) b.c.characterized
∂Ω
by n Fµν| =0. Introducing the potentials A(cid:12)µ so that
µ ∂Ω (cid:12)
(cid:12)
Fµν =∂µAν −∂νAµ, ∂0 =∂ , ∂i =−∂ , 1≤i≤d;
0 i
and working in the radiation gauge
A0 =0, ∂ Ai =0,
i
eq. (1) is equivalent to
d
∆Ai =0, ∆:=∂2− ∂2,
0 j
j=1
X
when jµ = 0. When the space Ω is a rectangular cavity Ω = [0,L ]×...×[0,L ], the PEC b.c. on a wall x = 0 or
1 d i
x =L is equivalent to
i i
∂ A −∂ A | =0
µ ν ν µ xi=0orxi=Li
for all µ6=ν ∈{0,1,...,d}\{i}; whereas the PMC b.c. is equivalent to
∂ A −∂ A | =0
i µ µ i xi=0orxi=Li
forallµ∈{0,1,...,d}\{i}.Restrictedtothecased=2,weconsiderthefollowingdifferentcombinationsofboundary
conditions:
Case I Mixed boundary conditions (i.e., one wall PEC b.c. and one wall PMC b.c.) on both x and x directions.
1 2
Case II Mixed boundary conditions on one direction, say x , and purely PEC b.c. in the other direction.
1
Case III Mixed boundary conditions on one direction, say x , and purely PMC b.c. in the other direction.
1
Now we derive the finite temperature Casimir energy of the electromagnetic field for each of the above boundary
conditions:
Case I In this case, we are looking for solutions of A (x ,x ,t) and A (x ,x ,t) satisfying
1 1 2 2 1 2
(∂2−∂2 −∂2 )A =0, i=1,2; ∂ A +∂ A =0, (2)
t x1 x2 i x1 1 x2 2
and the boundary conditions
∂ A | =0, ∂ A | =0, (∂ A −∂ A )| =0.
t 1 x1=L1,x2=0 t 2 x1=0,x2=L2 x1 2 x2 1 x1=L1,x2=L2
It is easy to verify that a basis of solutions are given by
A1(x1,x2,t) = α1cosπ(k1L+112)x1 sinπ(k2L+212)x2 e−ωkt, k ,k ∈N˜ =N∪{0},
(cid:18)A2(x1,x2,t)(cid:19) α2sinπ(k1L+121)x1 cosπ(k2L+212)x2 1 2
 
subjected to the condition
α k + 1 α k + 1
1 1 2 + 2 2 2 =0.
L L
(cid:0) 1 (cid:1) (cid:0) 2 (cid:1)
4
Here
k + 1 2 k + 1 2
ωk =π 1 2 + 2 2 .
s L L
(cid:18) 1 (cid:19) (cid:18) 2 (cid:19)
The corresponding zeta function is
π−2s 1 1 1 1
ζ(s)= Z s; , ,2T −Z s; , ,2T
3 3
4 2L 2L 2L L
( (cid:18) 1 2 (cid:19) (cid:18) 1 2 (cid:19)
1 1 1 1
−Z s; , ,2T +Z s; , ,2T ,
3 3
L 2L L L
(cid:18) 1 2 (cid:19) (cid:18) 1 2 (cid:19))
where Z (s;c ,...,c ) is the homogeneous Epstein zeta function defined by
n 1 n
−s
n
Z (s;c ,...,c )= [c k ]2 , (3)
n 1 n j j
 
kX∈Zcn Xj=1
 
and Zn = Zn \{0}. Since Z (0;c ,...,c ) = −1, we find that the regularized Casimir energy for electromagnetic
n 1 n
field with mixed boundary conditions in both x and x directions of a rectangular cavity is given by
1 2
c
T 1 1 1 1
EI,reg(L ,L )=− Z′ 0; , ,2T −Z′ 0; , ,2T
Cas 1 2 8 3 2L 2L 3 2L L
( (cid:18) 1 2 (cid:19) (cid:18) 1 2 (cid:19)
(4)
1 1 1 1
−Z′ 0; , ,2T +Z′ 0; , ,2T .
3 L 2L 3 L L
(cid:18) 1 2 (cid:19) (cid:18) 1 2 (cid:19))
Explicit formulas for Z′(0;c ,...,c ) are given in the Appendix A.
n 1 n
Case II In this case, we are looking for solutions of A (x ,x ,t) and A (x ,x ,t) satisfying (2) and the boundary
1 1 2 2 1 2
conditions
∂ A | =0, ∂ A | =0, (∂ A −∂ A )| =0.
t 1 x1=L1,x2=0,x2=L2 t 2 x1=0 x1 2 x2 1 x1=L1
A basis of solutions are given by
A1(x1,x2,t) = α1cosπ(k1L+121)x1 sinπkL22x2 e−ωkt, k ,k ∈N˜,
(cid:18)A2(x1,x2,t)(cid:19) α2sinπ(k1L+121)x1 cosπkL22x2 1 2
 
k + 1 2 k 2
ωk =π 1 2 + 2 ,
s L L
(cid:18) 1 (cid:19) (cid:18) 2(cid:19)
subjected to the condition
α k + 1 α k
1 1 2 + 2 2 =0.
L L
(cid:0) 1 (cid:1) 2
The corresponding regularized Casimir energy is
T 1 1 1 1
EII,reg(L ,L )=− Z′ 0; , ,2T −Z′ 0; , ,2T
Cas 1 2 8 3 2L L 3 L L
( (cid:18) 1 2 (cid:19) (cid:18) 1 2 (cid:19)
(5)
1 1
+Z′ 0; ,2T −Z′ 0; ,2T .
2 2L 2 L
(cid:18) 1 (cid:19) (cid:18) 1 (cid:19))
5
Case III In this case, we are looking for solutions of A (x ,x ,t) and A (x ,x ,t) satisfying (2) and the boundary
1 1 2 2 1 2
conditions
∂ A | =0, ∂ A | =0, (∂ A −∂ A )| =0.
t 1 x1=L1 t 2 x1=0,x2=0,x2=L2 x1 2 x2 1 x1=L1,x2=0,x2=L2
A basis of solutions are given by
A1(x1,x2,t) = α1cosπ(k1L+121)x1 cosπkL22x2 e−ωkt, k ∈N˜,k ∈N,
(cid:18)A2(x1,x2,t)(cid:19) α2sinπ(k1L+112)x1 sinπkL22x2 1 2
 
k + 1 2 k 2
ωk =π 1 2 + 2 ,
s L L
(cid:18) 1 (cid:19) (cid:18) 2(cid:19)
subjected to the condition
α k + 1 α k
− 1 1 2 + 2 2 =0.
L L
(cid:0) 1 (cid:1) 2
The corresponding regularized Casimir energy is
T 1 1 1 1
EIII,reg(L ,L )=− Z′ 0; , ,2T −Z′ 0; , ,2T
Cas 1 2 8 3 2L L 3 L L
( (cid:18) 1 2 (cid:19) (cid:18) 1 2 (cid:19)
(6)
1 1
−Z′ 0; ,2T +Z′ 0; ,2T .
2 2L 2 L
(cid:18) 1 (cid:19) (cid:18) 1 (cid:19))
Notice that there is a slight difference between the set of eigenmodes in Case II and Case III. In Case II, we allow
k =0 which corresponds to solutions
2
(cid:18)AA12((xx11,,xx22,,tt))(cid:19)= α2sinπ0(k1L+112)x1!e−ωkt, k1 ∈N˜, ωk = π(cid:0)kL11+ 12(cid:1).
However, in Case III, k =0 implies that α =0 and A =A ≡0. Therefore there is no eigenmode with k =0.
2 1 1 2 2
III. CASIMIR FORCE ACTING ON THE PISTON FOR ELECTROMAGNETIC FIELD WITH MIXED
BOUNDARY CONDITIONS
Inthissection,weconsidertheCasimirforceactingonatwo-dimensionalrectangularpistonduetoelectromagnetic
field with mixed boundary conditions. The boundary conditions on the walls of Region I are the Cases I, II, III as
considered in the previous section. In Region II, we assume that the boundary condition on the wall x = L is the
1 1
same as on the wall x =0. We have the following cases.
1
A. Case MBC-A
We assume mixed boundary conditions on both directions. In this case, we find that the total regularized Casimir
energy of the piston system is
EA,reg(a;L ,L )=EI,reg(a,L )+EI,reg(L −a,L ).
Cas 1 2 Cas 2 Cas 1 2
Applying Chowla–Selberg formula (A2) to (4), we find that
T L L 3 1 L L 3 1
EI,reg(L ,L )=− 1 2Z ;2L , − 1 2Z ;L ,
Cas 1 2 8 2πT 2 2 2 2T 4πT 2 2 2 2T
( (cid:18) (cid:19) (cid:18) (cid:19)
∞ ∞ ∞ (−1)k1 k + 1 2
+4 exp −2πk L 2 2 +(2lT)2 .
1 1
kX1=1kX2=0l=X−∞ k1  s(cid:18) L2 (cid:19) )
 
6
Therefore, in the limit L →∞, the Casimir force acting on the piston is given by
1
FA,L1=∞(a;L )= lim FA (a;L ,L )
Cas 2 Cas 1 2
L1→∞
∂
=− lim EA,reg(a;L ,L )
L1→∞∂a Cas 1 2
∞ ∞ k2+12 2+(2lT)2 (7)
L2
=πT r .
(cid:16) (cid:17)
kX2=0l=X−∞exp 2πa k2+12 2+(2lT)2 +1
L2
r !
(cid:16) (cid:17)
Notice that this is a positive decreasing function in a. Consequently, when L is finite, the Casimir force acting on
1
the piston
FA (a;L ,L )=FA,L1=∞(a;L )−FA,L1=∞(L −a;L )
Cas 1 2 Cas 2 Cas 1 2
is positive if a<L −a, and is negative if a>L −a. In other words, at any temperature, the Casimir force always
1 1
tends to restore the piston to the equilibrium position x =L /2, which is the middle of the cavity.
1 1
The infinite summation in the expression (7) for the Casimir force converges very fast if a≫L . It shows that in
2
the limit L →∞, the magnitude of the Casimir force decays exponentially when the plate separation a is large. In
1
mostpracticalsituation,we areinterestedinthe oppositecasewhere a≪L . Inthis latter case,the Chowla–Selberg
2
formula (A2) gives
3ζ (3)L L (−1)k2
FA,L1=∞(a;L )= R 2 − 2
Cas 2 32π a3 32π (k3X,ℓ)∈Zc2 [k2L2]2+ 2lT 2 23
(8)
πL ∞ 1 2 2(cid:16)π k + 1 (cid:2) (cid:3) (cid:17) l 2
+ 2 (−1)k2 k + K 1 2 [k L ]2+ .
2a3 1 2 0 a s 2 2 2T 
kX1=0(k2X,ℓ)∈Zc2 (cid:18) (cid:19) (cid:0) (cid:1) (cid:20) (cid:21)
 
This shows that at any temperature, when the plate separation a is small, the leading behavior of the Casimir force
is given by
3ζ (3)L
FA,L1=∞(a;L )∼ R 2 +O(a0).
Cas 2 32π a3
It implies that when a → 0+, the magnitude of the Casimir force approaches ∞ and behaves as 1/a3. From this we
can conclude that at any temperature, the Casimir force acting on the piston, consideredas a function of a∈(0,L ),
1
decreases from ∞ to 0 when a∈(0,L /2) and increases from 0 to ∞ when a∈(L /2,L ).
1 1 1
The formula(7) canalsobe usedto study the hightemperaturebehaviorofthe Casimirforce. Itshowsthatin the
high temperature regime, the leading behavior of the Casimir force is
πT ∞ k + 1
FCAas(a;L1,L2)∼L2 exp 2πa 2k +2 1 +1 −(a←→L1−a), (9)
kX2=0 L2 2 2
whichislinearinT. Theremainingtermsdecaysexpon(cid:16)entia(cid:0)llyasT(cid:1)(cid:17)→∞. Ifwerestoretheunits~,candk intothe
B
expressionfor Casimir force, we find that a term with Tj will be accompaniedby ~j−1. Therefore (9) shows that the
Casimir force acting on the piston has a classical (~→0) limit, as has also been observed in other works in Casimir
effect (see e.g. [25, 26, 27, 28]). The right hand side of (9) is called the classical term of the Casimir force.
In the low temperature (T ≪ 1) regime, the Casimir force is dominated by the zero temperature Casimir force,
with correction term being the temperature correction:
FA (a;L ,L )=FA,T=0(a;L ,L )+∆ FA (a;L ,L ).
Cas 1 2 Cas 1 2 T Cas 1 2
Applying the Chowla–Selberg formula (A1), we have
L (−1)k2 L 3 1 3 1
2 2
− =− 2Z ;2L , −Z ;L ,
32π (k3X,ℓ)∈Zc2 [k2L2]2+ 2lT 2 32 32π (cid:18) 2(cid:18)2 2 2T(cid:19) 2(cid:18)2 2 2T(cid:19)(cid:19)
(cid:16) (cid:2) (cid:3) (cid:17) =3ζR(3) − T ∞ ∞ k2+ 21K π k2+ 21 l .
64πL2 L l 1 L T
2 2 kX2=0Xl=1 (cid:0) 2 (cid:1) !
7
With this, we can read from the formula (8) that the zero temperature Casimir force is given by
3ζ (3)L 3ζ (3) πL ∞ ∞ 1 2 2πk k + 1 L
FA,T=0(a;L ,L )= R 2 + R + 2 (−1)k2 k + K 2 1 2 2
Cas 1 2 32π a3 64πL2 a3 1 2 0 a
2 kX1=0kX2=1 (cid:18) (cid:19) (cid:0) (cid:1) !
−(a←→L −a);
1
and the thermal correction is
∆ FA (a;L ,L )=− T ∞ ∞ k2+ 21K π k2+ 21 l + πL2 ∞ ∞ ∞
T Cas 1 2 L l 1 L T a3
2 kX2=0Xl=1 (cid:0) 2 (cid:1) ! kX1=0k2X=−∞Xl=1
×(−1)k2 k + 1 2K 2π k1+ 21 [k L ]2+ l 2 −(a↔L −a).
1 0 2 2 1
2  a s 2T 
(cid:18) (cid:19) (cid:0) (cid:1) (cid:20) (cid:21)
 
Notice that if L →∞, the thermal correction to the Casimir force decays to zero exponentially fast when T →0+.
1
In the limit L ,L →∞, the geometric configurationbecomes that of a pair of infinite parallelplates separatedby
1 2
a distance a. In this case, since
L (−1)k2 L T3 π T ∞ ∞ (−1)k2
− 2 =− 2 ζ (3)+ −2T2 lK (4πk lL T),
32π (k3X,ℓ)∈Zc2 [k2L2]2+ 2lT 2 32 2π R 48L2 kX2=1Xl=1 k2 1 2 2
(cid:16) (cid:2) (cid:3) (cid:17)
eq. (8) then implies that in the infinite parallel plates limit, the Casimir force acting on a wall is given by
FCAa,s||(a)=L2 33ζ2Rπ(a33) − T2π3ζR(3)+ aπ3 ∞ ∞ k1+ 12 2K0 πl ka1T+ 21 . (10)
( kX1=0Xl=1(cid:18) (cid:19) (cid:0) (cid:1)!)
This shows that for infinite parallel plates, the zero temperature Casimir force is
3ζ (3)
FA,||,T=0(a)= R L .
Cas 32πa3 2
The temperature correction is of order T3 as T → 0+. The remaining terms decays to zero exponentially fast when
T →0+. In the high temperature regime,
π 2T2 ∞ ∞ l ∞ ∞
FA,||(a)=L T − (−1)k1 K (4πlk Ta)−8πT3 (−1)k1l2K (4πlk Ta) .
Cas 2 48a2 a k 1 1 0 1
( 1 )
kX1=1Xl=1 kX1=1Xl=1
ThisshowsthattheclassicallimitoftheCasimirforceactingonapairofinfiniteparallelplateswithmixedboundary
conditions is
πL
FA,||,classical(a)= 2T.
Cas 48a2
B. Case MBC-B
We assume mixed boundary conditions in the x direction and purely PEC b.c. in x direction. Using the same
1 2
method as the previous section, we find that the Casimir force acting on the piston is given by
FB (a;L ,L )=FB,L1=∞(a;L )−FB,L1=∞(L −a;L ),
Cas 1 2 Cas 2 Cas 1 2
where
2
k2 +(2lT)2
FCBa,Ls1=∞(a;L2)=πT r(cid:16)L2(cid:17) . (11)
2
(k2,l)∈XN˜×Z\{0}exp 2πa k2 +(2lT)2 +1
r L2 !
(cid:16) (cid:17)
8
As in the previous case, this shows that at any temperature, the Casimir force tends to pull the piston to the
equilibrium position x = L /2. Moreover, it shows that in the high temperature limit, the leading term of the
1 1
Casimir force is given by the classical term
∞
πT k
FCBas(a;L1,L2)∼L2 exp 2πk22a +1 −(a↔L1−a). (12)
kX2=1 L2
(cid:16) (cid:17)
An alternative expression for FB,L1=∞(a;L ) that can be used to study the small a and low T behavior of the
Cas 2
Casimir force is
3ζ (3)L π ζ (3) πT2 T ∞ ∞ k πk l
FB,L1=∞(a;L )= R 2 + − R − − 2K 2
Cas 2 32π a3 96a2 16πL2 6 L l 1 L T
2 2 kX2=1Xl=1 (cid:18) 2 (cid:19)
πL ∞ 1 2 2π k + 1 l 2
+ 2 k + K 1 2 [k L ]2+
2a3 1 2 0 a s 2 2 2T  (13)
kX1=0(k2X,ℓ)∈Zc2(cid:18) (cid:19) (cid:0) (cid:1) (cid:20) (cid:21)
π ∞ k + 1  
+ 1 2 .
2a2 kX1=0exp (cid:0)π(kT1+a12)(cid:1) −1
(cid:18) (cid:19)
It shows that when the plate separation a is small, the leading terms of the Casimir force is given by
3ζ (3)L π
FB (a;L ,L )∼ R 2 + +O(a0).
Cas 1 2 32π a3 96a2
Notice thatthe firsttermbehavesas1/a3 whena→0+. Onthe otherhand, (13)givesthe zerotemperatureCasimir
force as
3ζ (3)L π ζ (3) πL ∞ ∞ 1 2 2π k + 1 k L
FB,T=0(a;L ,L )= R 2 + − R + 2 k + K 1 2 2 2
Cas 1 2 32π a3 96a2 16πL22 a3 kX1=0kX2=1(cid:18) 1 2(cid:19) 0 (cid:0) a (cid:1) ! (14)
−(a←→L −a)
1
The thermal correction goes to zero exponentially fast when T →0+.
Intheparallelplatelimit,itcanbecheckedthatonewouldobtainthesameresultas(10). Thisshouldbeexpected
since in the limit L →∞, the boundary conditions assumed on the x direction become immaterial.
2 2
C. Case MBC-C
We assume mixed boundary conditions in the x direction and purely PMC b.c. in x direction. This case is very
1 2
similar to the MBC-B case. We find that the Casimir force acting on the piston is given by
FC (a;L ,L )=FC,L1=∞(a;L )−FC,L1=∞(L −a;L ),
Cas 1 2 Cas 2 Cas 1 2
where
2
∞ ∞ k2 +(2lT)2
FCCa,Ls1=∞(a;L2)=πT r(cid:16)L2(cid:17) (15)
2
kX2=1l=X−∞exp 2πa k2 +(2lT)2 +1
L2
r !
(cid:16) (cid:17)
The difference between this term and the corresponding term in the case of MBC-B lies in the summation over k ,
2
where now k starts from 1 instead of 0. As in the previous case, (15) shows that at any temperature, the Casimir
2
forcetends to pull the pistonto the equilibriumpositionx =L /2. Moreover,it showsthatin the high temperature
1 1
limit, the leading term of the Casimir force is given by the classical term
∞
πT k
FC (a;L ,L )∼ 2 −(a↔L −a).
Cas 1 2 L2 exp 2πk2a +1 1
kX2=1 L2
(cid:16) (cid:17)
9
One notice that this classicalterm is the same as in the case of MBC-B given by (12). In other words, the difference
between the Casimir forces for case MBC-B and case MBC-C is insignificant at high temperature.
An alternative expression for FC,L1=∞(a;L ) that can be used to study the small a and low T behavior of the
Cas 2
Casimir force is
∞ ∞
3ζ (3)L π ζ (3) T k πk l
FC,L1=∞(a;L )= R 2 − − R − 2K 2
Cas 2 32π a3 96a2 16πL2 L l 1 L T
2 2 kX2=1Xl=1 (cid:18) 2 (cid:19)
+ πL2 ∞ k + 1 2K 2π k1+ 21 [k L ]2+ l 2
2a3 1 2 0 a s 2 2 2T  (16)
kX1=0(k2X,ℓ)∈Zc2(cid:18) (cid:19) (cid:0) (cid:1) (cid:20) (cid:21)
π ∞ k + 1  
− 1 2 .
2a2 kX1=0exp (cid:0)π(kT1+a21)(cid:1) −1
(cid:18) (cid:19)
When the plate separation a is small, the leading terms of the Casimir force is given by
3ζ (3)L π
FC (a;L ,L )∼ R 2 − +O(a0),
Cas 1 2 32π a3 96a2
with leading order 1/a3 when a→0+. On the other hand, the zero temperature Casimir force is
3ζ (3)L π ζ (3) πL ∞ ∞ 1 2 2π k + 1 k L
FC,T=0(a;L ,L )= R 2 − − R + 2 k + K 1 2 2 2
Cas 1 2 32π a3 96a2 16πL22 a3 kX1=0kX2=1(cid:18) 1 2(cid:19) 0 (cid:0) a (cid:1) ! (17)
−(a←→L −a),
1
which only differs with the MBC-B case by the sign of the term π/(96a2). The thermal correction also goes to zero
exponentially fast when T →0+.
We would like to remark that the regularizedCasimir energy and Casimir force acting on the piston in this case is
the same as the correspondingquantities for massless scalar field which assume Neumann boundary condition on the
piston and Dirichlet boundary conditions on the other walls. In fact, the zero temperature Casimir force (17) agrees
with the corresponding result in [9].
D. Case MBC-D
We assume PEC b.c. in the x direction and mixed boundary conditions in x direction. In this case,
1 2
ED,reg(a;L ,L )=EII,reg(L ,a)+EII,reg(L ,L −a).
Cas 1 2 Cas 2 Cas 2 1
Similar computations give
FD (a;L ,L )=FD,L1=∞(a;L )−FD,L1=∞(L −a;L ),
Cas 1 2 Cas 2 Cas 1 2
where
∞ ∞ k2+12 2+(2lT)2
FCDa,sL1=∞(a;L2)=−πT r(cid:16) L2 (cid:17) . (18)
kX2=0l=X−∞exp 2πa k2+21 2+(2lT)2 −1
L2
r !
(cid:16) (cid:17)
Contrary to the previous cases, now we find that the Casimir force acting on the piston always tends to pull the
piston towards the closer wall, and awayfrom the equilibrium position. (18) also shows that in the high temperature
regime, the Casimir force is dominated by the classical term, i.e.
πT ∞ k + 1
FD (a;L ,L )∼− 2 2 −(a↔L −a)
Cas 1 2 L2 kX2=0exp 2πa(Lk22+21) −1 1
(cid:18) (cid:19)
10
as T →∞. The remaining terms decay exponentially.
An alternative expression for the Casimir force is given by
L 3ζ (3) T ∞ ∞ k + 1 π k + 1 l
FD (a;L ,L )=− 2 ζ (3)+ R − 2 2K 2 2
Cas 1 2 8πa3 R 64πL2 L l 1 L T
2 2 kX2=0Xl=1 (cid:0) 2 (cid:1) !
(19)
∞ 2
πL 2πk l
+ 2 × (−1)k2k2K 1 (k L )2+ −(a↔L −a).
2a3 1 0 a s 2 2 2T  1
kX1=1(k2X,l)∈Zc2 (cid:18) (cid:19)
 
This shows that when the plate separation is small, the leading term of the Casimir force is
L
FD (a;L ,L )∼− 2 ζ (3)+O(a0),
Cas 1 2 8πa3 R
which is of order 1/a3. (19) also shows that in the low temperature limit, the Casimir force is dominated by the zero
temperature Casimir force given by
∞ ∞
L 3ζ (3) πL 2πk k L
FD,T=0(a;L ,L )=− 2 ζ (3)+ R + 2 (−1)k2k2K 1 2 2
Cas 1 2 8πa3 R 64πL2 a3 1 0 a
2 kX1=1kX2=1 (cid:18) (cid:19)
−(a↔L −a).
1
The thermal correction terms tends to zero exponentially fast when T →0+.
In the infinite parallel plate limit, we find that
ζ (3) T3 π ∞ ∞ πlk
FCDa,s||(a)=L2 −8Rπa3 − 2πζR(3)+ a3 k12K0 aT1 , (20)
( kX1=1Xl=1 (cid:18) (cid:19))
which gives the zero temperature Casimir force as
ζ (3)L
FD,||(a)=− R 2,
Cas 8πa3
agreeing with well-known results (see e.g. [29]). An alternative expression for (20) is given by
πT 2T2 ∞ ∞ l ∞ ∞
FD,||(a)=L − − K (4πlk Ta)−8πT3 l2K (4πlk Ta) ,
Cas 2 24a2 a k 1 1 0 1
( 1 )
kX1=1Xl=1 kX1=1Xl=1
which shows that the classical limit of the Casimir force is given by
πL
FD,||,classical(a)=− 2T.
Cas 24a2
E. Case MBC-E
We assume PMC b.c. on x direction and mixed boundary conditions on x direction. In this case, although the
1 2
regularized Casimir energy is different with the regularized Casimir energy for Case MBC-D, one can verify that
their difference is a term independent of L . Consequently, the Casimir force acting on the piston for case MBC-E is
1
identical to that for case MBC-D.
We donotdiscussthecaseswherethe electromagneticfieldassumespurelyPECb.c.onbothdirectionsorassumes
purely PMC b.c. on both directions. This has been considered in [15]. Another case we do not consider here is the
casewherethefieldassumespurelyPECb.c.ononedirectionandpurelyPMCb.c.ontheotherdirection. Theresult
is not much different from the cases of purely PEC b.c. or purely PMC b.c. on all directions.