Rotating Casimir systems: magnetic-field-enhanced perpetual motion, possible realization in doped nanotubes, and laws of thermodynamics

2.3 MB·English

by  M. N. Chernodub

Rotating Casimir systems: magnetic–field–enhanced perpetual motion, possible
realization in doped nanotubes, and laws of thermodynamics
M. N. Chernodub∗
CNRS, Laboratoire de Math´ematiques et Physique Th´eorique, Universit´e Fran¸cois-Rabelais Tours,
F´ed´eration Denis Poisson, Parc de Grandmont, 37200 Tours, France and
Department of Physics and Astronomy, University of Gent, Krijgslaan 281, S9, B-9000 Gent, Belgium
(Dated: July 12, 2012)
Recently, we have demonstrated that for a certain class of Casimir–type systems (“devices”)
the energy of zero–point vacuum fluctuations reaches its global minimum when the device rotates
aboutcertainaxisratherthanremainsstatic. Thisrotationalvacuumeffectmayleadtoemergence
of permanently rotating objects – philosophically similar to “time crystals” proposed recently by
ShapereandWilczekinclassicalandquantummechanicalsystems–providedthenegativerotational
energy of zero–point fluctuations cancels the positive rotational energy of the device itself. In this
2 paper we show that for massless electrically charged particles the rotational vacuum effect should
1 be drastically (astronomically) enhanced in the presence of magnetic field. As an illustration, we
0 show that in a background of experimentally available magnetic fields the zero–point energy of
2 massless excitations in rotating torus–shaped doped carbon nanotubes may indeed overwhelm the
classicalenergyofrotationforcertainangularfrequenciessothatthepermanentlyrotatingstateis
l
u energeticallyfavored. Thesuggested“zero–pointdriven”devices–whichhavenointernallymoving
J parts – correspond to a perpetuum mobile of a new, fourth kind: they do not produce any work
2 despitetheirequilibrium(ground)statecorrespondstoapermanentrotationeveninthepresenceof
1 anexternalenvironment. Weshowthatourproposalisconsistentwiththelawsofthermodynamics.

] PACSnumbers: 03.70.+k,42.50.Lc,42.50.Pq
h
p
- I. INTRODUCTION tationalfieldtheCasimirenergygravitatesasrequiredby
t
n the equivalence principle, so that the gravitational and
ua Themostdirectmanifestationoftheexistenceofzero– inertial masses, associated with the Casimir energy EC
q pointfluctuationsofthevacuumistheCasimireffect[1]. are both mC = EC/c2 [6]. This statement is valid re-
[ The essence of the Casimir effect, in its original formu- gardless of the sign of the Casimir energy and therefore
lation, is the appearance of an attracting force between the negative Casimir energy corresponds to a negative
1 two parallel infinitely large plates in the vacuum. The inertial mass.
v
plates are taken to be perfectly conducting and electri- A negative mass should have a negative moment of
2
5 callyneutralsothattheattractingforceappearsonlydue inertia. As a result, if a system possessing a negative
0 to the vacuum fluctuations of the electromagnetic field. zero–point energy is rotated, then the rotational energy
3 Themechanismoftheeffectisasfollows: thepresenceof corresponding to zero–point fluctuations should decrease
7. the plates affects the zero–point fluctuations of the elec- as the angular frequency is increased. This effect, called
0 tromagnetic field thus modifying their energy spectrum. therotationalvacuumeffect,wasindeedfoundinRef.[7].
2 Due to the change in the energy spectrum the total en- Moreover, if the (positive–valued) classical moment of
1 ergy of the vacuum fluctuations in the presence of the inertia of the system is small enough then at certain
v: plates is different from the total energy of the fluctua- nonzero frequency Ω0 (cid:54)=0 the negative rotational energy
i tions in the absence of the plates. Both energies are in- of the zero–point fluctuations may make the total rota-
X finitelylargebuttheirdifferenceisafinitequantitywhich tional energy E of the system smaller compared to the
ar iscalledtheCasimirenergyorzero–pointenergy. Inthis totalrotationalenergyinthestaticstate,E(Ω0)<E(0).
particularexampletheCasimirenergyisanegativequan- In this case the ground state of the system should corre-
tity, the absolute value of which increases rapidly as the spond to Ω (cid:54)= 0 and the system should prefer to rotate
distance between the plates gets smaller. As a result the forever in its ground state even in the presence of ex-
plates are attracting to each other. The existence of the ternal environment such as a thermal bath. This “zero–
attracting force was confirmed experimentally [2]. The pointdriven”perpetualmotionisphilosophicallysimilar
current progress in this rapidly evolving field is reviewed to“timecrystals”proposedrecentlybothinspecificclas-
in various books [3–5]. sical, semiclassical as well as quantum–mechanical sys-
The interesting property of the zero–point energy is tems[8]andtosuggested“space–timecrystal”systemof
that it has a mass. More precisely, in the external gravi- a permanently rotating cold ions [9].
However, the rotational vacuum effect as small as the
usual Casimir effect [7]. The first goal of this paper is
to show that the rotational vacuum effect for electrically
∗ maxim.chernodub(at)lmpt.univ-tours.fr; chargedmasslessparticlesshouldbestronglyenhancedin
OnleavefromITEP,Moscow,Russia. thepresenceofthemagneticfieldsothatitcanprobably
2
be tested in metallic carbon nanotubes. The second goal estimate the rotational energy of the zero–point fluctu-
is to demonstrate that the very existence of the zero– ations in the device and demonstrate that the magnetic
point driven perpetual motion of a macroscopic system enhancementoftherotationalvacuumeffectmayberea-
does not violate the laws of thermodynamics. sonably large for experimental detection even in a very
Our strategy is as follows. In Section II we provide conservativeestimation. Asanexample,weestimatethe
detailsofthecalculationsoutlinedinRef.[7]anddemon- period τ of perpetual rotation (about one second) for
stratethat(i)therotationalzero-pointenergyisnegative the minimal radius R of the torus (about a width of a
and (ii) its absolute value increases with increase of an- humanhair,0.1mm)atthehighestexperimentallyavail-
gular frequency Ω. To this end we consider a simplest able magnetic field (B =50 Tesla). At weaker magnetic
system exhibiting the rotational vacuum effect: a mass- field the radius of the device and the time period should
lessscalarfielddefinedonathincirclewithapointwhere both increase.
the field vanishes (i.e. the point represents an infinites- In Section V we argue that the existence of such a de-
imally thin “Dirichlet cut”), Fig. 1. For shortness, we vice – which can be considered as a “perpetuum mobile
often call the infinitely thin circle with the Dirichlet cut of the fourth1 kind driven by zero–point fluctuations” –
as “the device”. isconsistentwiththelawsofthermodynamicsduetoab-
sence of the energy transfer and due to a discontinuous
nature of the rotational energy of the zero–point fluctu-
ations.
The last Section is devoted to our conclusions.
II. EXAMPLE OF SPONTANEOUS ROTATION:
REAL-VALUED SCALAR FIELD ON A CIRCLE
WITH A DIRICHLET CUT
FIG. 1. A simplest device which demonstrates the existence A. Nonrotating circle with the Dirichlet cut
of the rotational vacuum effect (i.e., a negative moment of
inertia of zero–point fluctuations): A massless scalar field is
1. Lagrangian, boundary conditions and Casimir energy
defined on a thin circle with an infinitesimally thin cut. The
cut imposes a Dirichlet boundary condition on the field.
Considerareal-valuedmasslessscalarfieldφ=φ(t,ϕ)
definedonacirclewithafixedradiusR. Thecorrespond-
In Section IIA we consider a static device with a neu-
ing Lagrangian is:
tralscalarfield. Formethodologicalreasons,weoverview
the computation of the corresponding zero–point energy 1 1(cid:34)(cid:18)∂φ(cid:19)2 1 (cid:18)∂φ(cid:19)2(cid:35)
in a Green’s function approach using both the time– L= 2∂µφ∂µφ≡ 2 ∂t − R2 ∂ϕ , (1)
splitting regularization and the ζ–function regulariza-
tion. For completeness, we also discuss a formal compu- where ϕ∈[0,2π) is the angular coordinate.
tation of the same zero-point energy via the ζ–function In order to make the circle sensitive to rotations (this
regularization of a sum over the individual energies of case will be considered in Section IIB), we make an in-
all vacuum modes. Having reviewed these methods we finitesimally small cut at ϕ = 0, Fig. 1. The role of the
findthatitisveryconvenienttocalculatethezero–point cut is to impose the simplest, Dirichlet boundary condi-
energy using an explicit form of the Green’s function tions on the field φ at the position of the cut:
and the time–splitting regularization. We will use this (cid:12) (cid:12)
method henceforth. φ(t,ϕ)(cid:12)(cid:12) ≡φ(t,ϕ)(cid:12)(cid:12) =0, (2)
In Section IIB we repeat the calculation for a uni- (cid:12)ϕ=0 (cid:12)ϕ=2π
formly rotating circle with a cut. We demonstrate that
where the points ϕ = 0 and ϕ = 2π are identified. For
the negative zero–point energy of the circle with the
shortness, we call this cut as the “Dirichlet cut”.
Dirichlet cut decreases further as the angular frequency
The field φ experiences quantum (zero–point) fluctua-
of the rotation increases thus confirming our earlier cal-
tions. The mean energy density of the field fluctuations
culation [7]. E¯is given by a sum over energies ε (R) of all individual
m
In Section III we consider the same device but with
fluctuation modes m:
an electrically charged massless scalar field. We show
that the background magnetic field drastically enhances E¯(R)= 1 (cid:88) ε (R). (3)
2πR m
the rotational vacuum effect. The enhancement depends
modesm
quadraticallyonthenumberofelementaryintegerfluxes
(quanta) of the magnetic field piercing the circle.
InSectionIVweproposetoconstructthedevicefroma 1 We call it the “perpetuum mobile of the fourth kind” because
metalliccarbonnanotubeinaformofaclosedtoruswith firstthreekindsdonotexistastheyviolatethelawsofthermo-
a cut made by a suitable chemical doping. We roughly dynamics.
3
TheinfinitesuminEq.(3)isadivergentquantityboth The Green function G(t,t(cid:48);ϕ,ϕ(cid:48)) in Eqs. (7) and (12)
for the circle with a finite radius R and in a free space satisfies the following equation:
(R → ∞). However, a difference between these vacuum (cid:18)∂2 1 ∂2 (cid:19) 1
energy densities, − G(t,ϕ;t(cid:48),ϕ(cid:48))= δ(ϕ−ϕ(cid:48))δ(t−t(cid:48)),(13)
∂t2 R2∂ϕ2 R
E¯phys(R)=E¯(R)−E¯(∞), (4) which is valid in the region 0 < ϕ,ϕ(cid:48) < 2π. Due to the
Dirichlet boundary condition (2) the Green’s function G
is a finite physical quantity since the divergency in the
should vanish at the Dirichlet cut at ϕ = 0,2π and at
energy density (3) is independent of the radius R. The
ϕ(cid:48) =0,2π.
total energy of the quantum fluctuations,
It is convenient to express the Green function G via
Ephys(R)≡2πRE¯phys(R), (5) the eigenvalues
m2
is an experimentally measurable finite observable called λω,m = 4R2 −ω2, (14)
the Casimir energy [1].
and the eigenfunctions
φ (t,ϕ)≡φ(C)(t,ϕ)−iφ(S) (t,ϕ)=e−iωtχ (ϕ), (15)
ω,m ω,m ω,m m
2. Energy of nonrotating circle: time–splitting and 1 mϕ
ζ–regularizations in Green’s function approach χm(ϕ)= √πRsin 2 , (16)
ofthesecond–orderdifferentialoperatorintherighthand
The local energy density of the quantum fluctuations
side of Eq. (13):
E(x) is given by the quantum expectation value of a cer-
tain component of the stress–energy tensor, (cid:104)Tµν(x)(cid:105): (cid:18)∂2 1 ∂2 (cid:19)
− φ (t,ϕ)=λ φ (t,ϕ), (17)
∂t2 R2∂ϕ2 ω,m ω,m ω,m
E(x)=(cid:10)T00(x)(cid:11) . (6)
where m = 1,2,3,... and ω ∈ R. For convenience, the
The expectation value of the stress–energy tensor can real-valued eigenfunctions φ(C) and φ(S) – which have
ω,m ω,m
be computed using a Feynman–type Green function, the same eigenvalues – were combined into one complex
eigenfunction (15). All eigenfunctions satisfy the Dirich-
G(x,x(cid:48))=i(cid:104)Tφ(x)φ(x(cid:48))(cid:105) , (7) let boundary condition (2) at the cut ϕ=0.
The system of the eigenfunctions (15) is orthonormal,
via the following relation [3, 4]:
ˆ+∞ ˆ2π
(cid:18) (cid:19) (cid:12)
(cid:104)Tµν(x)(cid:105)= ∂µ∂(cid:48)ν − 12gµν∂λ∂λ(cid:48) 1iG(x,x(cid:48))(cid:12)(cid:12)(cid:12) . (8) R dt dϕφ†ω1,m1(t,ϕ)φω2,m2(t,ϕ)
x→x(cid:48) −∞ 0
=2πδ(ω −ω )δ , (18)
In Eq. (7) the symbol “T” stands for the time–ordering 1 2 m1,m2
operator. and complete:
In our case, the coordinates are x≡(x0,x1)=(t,Rϕ) ˆ+∞
∞
and the corresponding derivatives are as follows: dω (cid:88) 1
φ (t ,ϕ )φ† (t ,ϕ )= δ(t −t )
2π m,ω 1 1 m,ω 2 2 R 1 2
∂ = ∂ , ∂ = 1 ∂ , ∂(cid:48) = ∂ , ∂(cid:48) = 1 ∂ . (9) −∞ m=1
0 ∂t 1 R∂ϕ 0 ∂t(cid:48) 1 R∂ϕ(cid:48) (cid:88)(cid:2) (cid:3)
· δ(ϕ −ϕ +4πn)−δ(ϕ +ϕ +4πn) .(19)
1 2 1 2
The line element is: n∈Z
Outside the boundary, ϕ (cid:54)= 0,2π, the right hand
ds2 ≡g dxµdxν =(dx0)2−(dx1)2 1,2
µν side of Eq. (19) is proportional to the product of the δ
≡dt2−R2dϕ2, (10) functions, δ(ϕ − ϕ )δ(t − t ). At the boundary, the
1 2 1 2
right hand side of Eq. (19) vanishes, as expected.
so that gµν =diag(1,−1) and TheGreen’sfunction(7)isgivenbyageneralequation,
∂λ∂λ(cid:48) ≡ ∂∂t∂∂t(cid:48) − R12∂∂ϕ∂∂ϕ(cid:48) . (11) G(t,t(cid:48);ϕ,ϕ(cid:48))= ˆ+∞dω (cid:88)∞ φω,m(t,ϕ)φ†ω,m(t(cid:48),ϕ(cid:48)), (20)
2π λ −i(cid:15)
ω,m
−∞ m=1
Then the energy density (6) is given by the following
which in our case takes the following form:
formula:
(cid:10)T00(t,ϕ)(cid:11)=(cid:18)∂ ∂ + 1 ∂ ∂ (cid:19) G(t,t(cid:48);ϕ,ϕ(cid:48))= 1 ˆ+∞dω (cid:88)∞ (cid:18) m2 −ω2−i(cid:15)(cid:19)−1
∂t∂t(cid:48) R2∂ϕ∂ϕ(cid:48) πR 2π 4R2
(cid:12) −∞ m=1
21iG(t,t(cid:48);ϕ,ϕ(cid:48))(cid:12)(cid:12)(cid:12)tϕ(cid:48)(cid:48)→→tϕ. (12) ·eiω(t(cid:48)−t)sinm2ϕsinm2ϕ(cid:48) , (21)
4
where the infinitesimally small imaginary term i(cid:15) in the Eq. (26) gives us a finite Casimir energy density of the
denominatorguaranteesthattheGreen’sfunction(20)is vacuumfluctuationsduetothevacuumfluctuations,and
oftheFeynmantype[3],Eq.(7): theintegrationcontour other terms in Eq. (26) vanish in the limit of vanishing
passesbelow(above)thepolesonthenegative(positive) time splitting, δt→0.
part of the real axis, Fig. 2. Finally, we get the following expression for the zero–
pointenergydensityofthescalarfieldatthenonrotating
circle with the Dirichlet cut:
1
E¯phys =− . (27)
96πR2
Thisenergydensityturnszerointhelimitoftheinfinitely
large circle, R → ∞. The corresponding total vacuum
FIG. 2. The integration contour in the propagator (20). energy (5) is given by the following expression:
1
Next, we substitute the integral representation of the Ephys =2πRE¯phys =− . (28)
48R
Green’s function (12) into the expression for the energy
density(20)andthenwecalculatethemeanenergyden- This result is known as the Casimir energy of the string
sity (3): of the length l=2πR [3, 10].
ˆ
E¯= 2π dϕ (cid:10)T00(t,ϕ)(cid:11) . (22) Time–splitting regularization: explicit Green’s
function. It is very convenient to compute the local en-
2π
0 ergy density via explicit calculation of the Green’s func-
There are various ways to perform this calculation. tion (21):
regTuilmariez–espthliettienngergryeg(u2l2a)riwzaetiuosne [t3h,e4]G. rIeneno’srdfeurntco- G(t,t(cid:48);ϕ,ϕ(cid:48))= (cid:88)∞ i sinmϕsinmϕ(cid:48)e−im2Ω0|(t−t(cid:48))|.
πm 2 2
tion (20) with m=1
i (cid:18)ϕ ϕ(cid:48) |(t−t(cid:48))|(cid:19)
t−t(cid:48) =δt, (23) = G , , , (29)
π 2 2 2R
where δt is a small but finite parameter (to be later sent
where the contour of integration, Fig. 2, is closed by a
tozero). Theregularizedzero-pointenergythenreadsas large semicircle in the upper (lower) half plane if t(cid:48) > t
follows: (t(cid:48) <t)sothatthecontourenclosesonlythepoleslocated
ˆ+∞ ∞ (cid:16)m2 +ω2(cid:17)e−iωδt at the negative (positive) part of the real axis.
E¯= Ω0 dω (cid:88) 4R2 The function G in Eq. (29) is defined as follows:
4π 2πi m2 −ω2−i(cid:15)
−∞ m=1 4R2 (cid:88)∞ sin(mx)sin(my)
∞ G(x,y,z)= e−imz (30)
1 (cid:88) m
= me−imδt/(2R). (24) m=1
8πR2 m=1 1 (cid:2)1−ei(x+y−z)(cid:3)(cid:2)1−e−i(x+y+z)(cid:3)
= ln
(cid:2) (cid:3)(cid:2) (cid:3)
Wehavetakenthetime–splittingparameterδttobepos- 4 1−ei(x−y−z) 1−ei(−x+y−z)
itive, δt > 0 (one could equivalently use δt < 0 as well), (cid:12) (cid:12)
1 (cid:12)cos(x+y)−cosz(cid:12)
and closed the contour in Fig. 2 at the infinitely large ≡ ln(cid:12) (cid:12)
4 (cid:12)cos(x−y)−cosz(cid:12)
semicircle in the lower half plane. Thus, only positive–
valued poles at the real axis enter the expression for the −i(cid:0)[z−x−y] +[z+x+y]
energy density (24), and the corresponding residues are 8 2π 2π
(cid:1)
as follows: −[z−x+y] −[z+x−y] ,
2π 2π
ω2e−iωδt m where we have used the following formula2:
res =− e−imδt/(2R). (25)
ω=m m2 −ω2−i(cid:15) 4R
2R 4R2 − (cid:88)∞ eimx =ln(cid:0)1−eix(cid:1)≡ln(cid:16)2(cid:12)(cid:12)sinx(cid:12)(cid:12)(cid:17)+ i ([x] −π).
The sum in Eq. (24) may be taken explicitly, and it m (cid:12) 2(cid:12) 2 2π
gives in the limit δt→0: m=1
Using the explicit representation of the Green’s func-
E¯=− 1 − 1 +O(cid:0)δt2/R4(cid:1) . (26) tion (29) the local energy density can be computed
2πδt2 96πR2
The first term in this expression is a divergent quantity
which is independent on the circle’s radius R. Therefore
the first term corresponds for the divergent energy den- 2 The cut of the logarithmic function lnz is located on the real
sity of the zero–point fluctuations. The second term in axisatRez<1[11].
5
straightforwardly: The spatial eigenmodes (16) are orthonormal,
ˆ
(cid:10)T00(t,ϕ)(cid:11)= 1 lim lim (cid:18)∂ ∂(cid:48)+ 1 ∂ ∂(cid:48)(cid:19) (31) R 2πdϕχ(0)†(t,ϕ)χ(0)(t,ϕ)=δ , (39)
2it(cid:48)→tϕ(cid:48)→ϕ t t R2 ϕ ϕ m m(cid:48) mm(cid:48)
0
1 1 1
G(t,t(cid:48);ϕ,ϕ(cid:48))=− lim − . and their basis is complete:
2π t(cid:48)→t(t(cid:48)−t)2 96πR2
∞
(cid:88) 1 (cid:88)
We again have arrived to Eq. (27). Notice that the local χ(0)(ϕ)χ(0)†(ϕ(cid:48))= [δ(ϕ−ϕ(cid:48)+4πk)
m1 m2 R
energy density (31) turns out to be independent on the m=1 k∈Z
angular variable ϕ. −δ(ϕ+ϕ(cid:48)+4πk)] . (40)
Formal ζ–function regularization. One can also The second δ–function in the right hand side of Eq. (40)
formally start from Eq. (24) at δt=0 and regularize the guarantees that the sum in the left hand side is zero at
divergent sum over the modes m, the position of the Dirichlet cut, ϕ = 0,2π and at ϕ(cid:48) =
0,2π. This property is expected because of the Dirichlet
1
E¯(s)= ζ(s), (32) boundary condition (2).
8πR2
Secondly, we should sum all energies of the individual
via the Riemann’s ζ function: fluctuations (38) and get the zero–point energy of the
scalar field at the circle with the Dirichlet cut:
∞
ζ(s)= (cid:88) m−s. (33) E = 1 (cid:88)∞ ε(0) ≡ Ω0 (cid:88)∞ m. (41)
m=1 2 m 4
m=1 m=1
The regularization of the sum in Eq. (32) may be done
Since this sum a divergent quantity, we following the
via an analytical continuation of the ζ function,
standard approach [3, 4, 12–14], and we regularize the
∞ sum (41) using the ζ regularization:
(cid:88) reg 1
m = lim ζ(s)=ζ(−1)=− , (34)
m=1 s→−1 12 E(s)= 1 (cid:88)∞ (cid:2)ω(0)(cid:3)−s= 2s−1 (cid:88)∞ m−s ≡ 2s−1ζ(s), (42)
2 m Ωs Ωs
which – after substitution to Eq. (32) – again gives the m=1 0 s=1 0
expression for the known vacuum energy density (27).
where ζ is the Riemann’s ζ function (33).
In order to calculate the physical part of the regular-
ized energy (41) we use, as usual, the analytical con-
3. ζ–function regularization of explicit sum over the modes
tinuation of the ζ function to the point s = −1, which
gives us ζ(−1) = −1/12, Eq. (34). Then the physical
A quick derivation of the vacuum energy can also be part of the energy density (41) becomes a finite quan-
doneviaadirectζ–functionregularizationofthesum(3) tity which coincides with the result obtained in all other
over the energies of individual field’s fluctuations. approaches (28).
Firstly,weshouldfindtheenergyspectrumofthecircle
withthecut. Theenergyeigenmodessatisfytheclassical
equation of motion of the Lagrangian (1), B. Rotating circle with the Dirichlet cut
(cid:18)∂2 1 ∂2 (cid:19)
− φ(0)(t,ϕ)=0, (35) In the previous section we have overviewed various
∂t2 R2∂ϕ2 m
well-known approaches to standard calculation of the
zero–point energy of a scalar field in the circle with the
and the boundary condition (2). As usual, the corre-
Dirichlet cut, and we have highlighted the equivalence
sponding real-valued eigenfunctions can conveniently be
of these approaches. Although there are various ways to
combined into the complex valued function,
calculate the zero–point energy for the case of the uni-
φ(0)(t,ϕ)≡φ(0),C(t,ϕ)−iφ(0),S(t,ϕ). (36) formly rotating circle, below we chose the simplest and
m m,ω m
most straightforward approach based on the explicit cal-
The eigenfunctions φ(0) and eigenenergies (cid:15)(0) are la- culation of the Green’s function and the time–splitting
m m
belled by a positive integer number m, regularization. We recover the result of Ref. [7] where a
zeta–function method [12–14] was used.
φ(m0)(t,ϕ)=e−iε(m0)tχ(m0)(ϕ), (37)
m
(cid:15)(m0) = 2R, (38) 1. Eigenfunctions in the rotating circle
m=1,2,3,... ,
Let us consider the circle with the Dirichlet cut which
where the spatial eigenfunction χ(0) is given in Eq. (16). rotates uniformly about its central point with a uniform
m
6
angular velocity Ω, Fig. 1. The rotation leads to the In the rotating frame the Dirichlet boundary condi-
following time–dependent boundary condition: tion (43) takes a simpler form,
(cid:12)
(cid:12)
φ(t,ϕ)(cid:12) =0, (43) (cid:12)
(cid:12)ϕ=[Ωt] φ(cid:101)(ϑ,θ)(cid:12)(cid:12) =0, (48)
2π (cid:12)
ϑ=0
where
[x] =x+2πn, n∈Z, 0(cid:54)[x] <2π, (44) where φ(cid:101)(ϑ,θ) ≡ φ(ϑ,θ + Ωt). The eigenvalue equa-
2π 2π
tion (17) takes a new form:
denotesthemodulooperationwiththebaseof2π. Inthe
nonrotating limit, Ω=0, Eq. (43) reduces to Eq. (2).
As in the nonrotating case, the energy density is given (cid:34)(cid:18) ∂ ∂ (cid:19)2 1 ∂2 (cid:35)
byEq.(12),wheretheGreen’sfunction(20)isexpressed ∂ϑ −Ω∂θ − R2∂θ2 φ(cid:101)ω,m(ϑ,θ)
viathesolutionsofEq.(17)butnowthetime–dependent
boundary conditions (43) should be used. In the labora- =λω,mφ(cid:101)ω,m(ϑ,θ), (49)
tory frame the corresponding wavefunctions are
(cid:114) Solving Eqs. (48) and (49) and coming back to the labo-
1 (cid:104)m (cid:105)
φ (t,ϕ)= sin [ϕ−tΩ] ratory frame one gets the wavefunctions (45).
m,ω πR 2 2π
(cid:26) (cid:18) ΩR2[ϕ−tΩ] (cid:19)(cid:27)
·exp −iω t− 2π ,(45)
1−Ω2R2
2. Time–splitting: explicit Green’s function calculation
withm=1,2,3,.... Onecanshowthatthesewavefunc-
tionssatisfyboththeorthonormality(18)andcomplete-
ness (19) conditions. We substitute the wavefunction (45) into the general
The corresponding eigenvalues are as follows: expression(20)andrepeatallthestepswhichledusfrom
Eq. (20) to the derivation of the Green’s function in the
1−Ω2R2 ω2
λ = m2− . (46) nonrotating case (29). To this end we notice that the
ω,m 4R2 1−Ω2R2 wavefunctionoftherotatingcircle(45)isverysimilar(up
to a redefinition of the time and angular coordinates) to
Technically, a simple derivation of the wavefunc-
the wavefunction (15) and (16) of the nonrotating circle.
tions (45) can be done, for example, by changing the
Thesameistrue(uptoarecallingofωandofaprefactor
coordinates from the laboratory frame, (t,ϕ), to the ro-
intheGreen’sfunction)fortheeigenvalues(46)and(14).
tating frame, (ϑ,θ), in which our object is static:
Asaresult, wegetthefollowingGreen’sfunctionforthe
θ =ϕ−Ωt, ϑ=t. (47) scalar field on a circle with the Dirichlet cut:
(cid:32) (cid:12) (cid:12)(cid:33)
i [ϕ−Ωt] [ϕ(cid:48)−Ωt(cid:48)] (cid:12)(1−Ω2R2)(t−t(cid:48))−ΩR2([ϕ−Ωt] −[ϕ(cid:48)−Ωt(cid:48)] )(cid:12)
G (t,t(cid:48);ϕ,ϕ(cid:48))= G 2π, 2π, 2π 2π , (50)
Ω π 2 2 2R
where the function G(x,y,z) is given in Eq. (30). zero–point (ZP) fluctuations:
Then the local zero–point energy density is: EZP(t,ϕ)≡(cid:10)T00(t,ϕ)(cid:11)phys =−1+Ω2R2 . (52)
(cid:10)T00(t,ϕ)(cid:11)= 1 lim lim (51) The phyΩsical energy density of the rota9t6iπngR2device de-
2it(cid:48)→tϕ(cid:48)→ϕ
(cid:18) (cid:19) pends neither on time t nor on the angular variable ϕ.
1
∂ ∂(cid:48)+ ∂ ∂(cid:48) G(t,t(cid:48);ϕ,ϕ(cid:48)) Forastaticdevice,Ω=0,Eq.(52)reducestotheknown
t t R2 ϕ ϕ
result (27).
1 1 1+Ω2R2 Finally, we get the following expression for the total
=− lim − .
2π t(cid:48)→t(t(cid:48)−t)2 96πR2 zero–point energy of the rotating circle [7]:
ˆ2π
The divergent part of this expression is equivalent to the 1+R2Ω2
EZP ≡R dϕEZP(t,ϕ)=− . (53)
one in the static case (31). This divergence depends nei- Ω Ω 48R
ther on the radius of the circle nor on the angular fre- 0
quency of rotation and therefore it does not contribute Notice that Eq. (53) provides us with the exact rela-
to the physical part of the rotational energy density of tivistic expression for an infinitely thin circle with an
7
infinitely thin Dirichlet cut. Thirdly, it is important to stress that no transition
from a nonrotating state (Ω=0) to a rotating one (Ω(cid:54)=
0) may occur for an isolated device because the angular
3. Physical features of the device momentumisaconservedquantity. Atransitiontowards
the rotating ground state may only be realized in the
There are three important general physical features of presence of environment.
our device. In order to support this statement we calculate the
Firstly, the rotational energy is a negative-values forceFZPwhichisexperiencedbytheDirichletcutdueto
quantity, the absolute value of which increases quadrati- the zero–point fluctuations. Following line of arguments
cally with the angular frequency Ω. One can define the of Ref. [15], the force can be expressed as follows:
“moment of inertia” of the zero–point fluctuations3: (cid:18) (cid:12)
FZP = 1 (cid:104)Tϕϕ(t,ϕ)(cid:105)(cid:12)(cid:12)
∂2 (cid:126)R R2 (cid:12)
IZP ≡ EZP =− , (54) ϕ=[Ωt−0]2π
∂Ω2 Ω 24c (cid:12) (cid:19)
where we have restored the Planck constant (cid:126) and the −(cid:104)Tϕϕ(t,ϕ)(cid:105)(cid:12)(cid:12)(cid:12) , (56)
ϕ=[Ωt+0]
speed of light c. 2π
wherethedifferenceoftheexpectationvalueofthecom-
The negative value of the moment of inertia of the
ponentTϕϕ(t,ϕ)istakenattheleftandattherightsides
zero–point fluctuations is a natural fact. Indeed, the in-
of the cut (43).
ertial mass corresponding to the Casimir energy E is
c
alwaysE /c2 regardlessofthesignoftheCasimirenergy According to the general expression (8),
c
itself[6],sothatthenegativemassshouldhaveanegative (cid:104)Tϕϕ(t,ϕ)(cid:105)≡−(cid:10)T00(t,ϕ)(cid:11) . (57)
momentofinertia. Ifthedeviceitselfweremasslessthen
This relation is a natural fact because the theory (1) is
the ground state of the device would correspond to the
conformal both at classical and quantum levels (there
permanently rotating state due to the negative moment
is no conformal anomaly), so that the expectation value
of inertia of the zero–point fluctuations.
of the trace of the stress–energy tensor should be zero.
In our simplest example, the zero–point moment of
Taking into account the fact that the expectation value
inertia (54) is tiny,
of the T00 component (12) does not depend on angular
(cid:18) (cid:19)
R variable (31), we obtain that the zero–point fluctuations
IZP =−2.2×10−42· ·kgm2. (55)
m produce no force on the Dirichlet cut:
It is too small (55) to overcome a classical moment of FZP =0. (58)
inertia of a real device. However, in the next Section we Thus, the isolated device is not self–accelerating be-
show that the negative moment of inertia can be drasti- causeoftheconservationoftheangularmomentum. The
callyenhancedbyanexternalmagneticfieldifthemass- absence of the force on the Dirichlet cut is one of the
lessparticlesareelectricallycharged,sothatafabrication majordifferencesoftherotationalvacuumeffectandthe
of a permanently rotating device may become closer to conventional Casimir effect: despite the ground state of
reality. the device corresponds to a rotating state, the device –
Secondly,thezero–pointrotationalenergy(53)isun- even if it is not residing in its ground state – will not
bounded from below at the relativistically large angular self–accelerate unless it exchanges the angular momen-
frequencies |Ω|→1/R. This feature is an artifact which tumwithenvironmentor,equivalently,unlessitemitsthe
emergesduetoourmathematicalsimplificationwhichas- extraangularmomentumviaradiationof,e.g.,aphoton.
sumes that the thickness of our circle is infinitely small.
One can show that for spatially extended systems – such
as a cylinder – the rotational zero–point energy has its III. CONDUCTING CIRCLE WITH
minimum at finite values of the angular frequency. In THE DIRICHLET CUT IN MAGNETIC FIELD
this case the dependence of the zero–point energy on the
rotational frequency has a form of a double–well poten- A. The device
tial [7] with nontrivial minima at Ω(cid:54)=0.
Consider an electrically charged massless scalar field
Φ = Φ(t,ϕ) which is defined, as in the previous section,
on a circle with a fixed radius R, Fig. 3. The field Φ
3 We put the term “moment of inertia” in quotation marks be-
is electrically charged and it is interacting with a back-
cause we have identified it as if the system is nonrelativistic.
Indeed,therotationalpartoftheenergyinEq.(53)resemblesa ground electromagnetic field Aµ. The corresponding La-
nonrelativistic behavior while it is computed for the zero–point grangian is as follows:
fluctuationscorrespondingtorelativisticmasslessparticles. The
system is neither fully relativistic nor fully nonrelativistic. In L=[DµΦ]∗DµΦ
this unusual case Eq. (53) should be considered as a definition 1
ofthemomentofinertiaofthezero–pointfluctuations. ≡[DtΦ]∗DtΦ− R2 [DϕΦ]∗DϕΦ, (59)
8
Below we calculate the energy of zero–point fluctua-
tions for this device following the line of previous sec-
tions.
B. Energy density of zero-point fluctuations
1. The eigensystem
TheeigensystemproblemfortheLagrangian(59)with
the gauge field (62) is the following:
(cid:34)∂2 1 (cid:18) ∂ (cid:19)2(cid:35)
− −iγ Φ (t,ϕ)
∂t2 R2 ∂ϕ B ω,m
FIG. 3. A simplest device which demonstrates an enhance- =Λω,mΦω,m(t,ϕ). (65)
ment of the negative moment of inertia of zero–point fluctu-
The eigenvalues and eigenfunctions of the equa-
ations due to the magnetic field B. The circle supports elec-
tion (65) with the boundary condition (60) are, respec-
trically charged and massless scalar excitations while the cut
imposes a Dirichlet boundary condition on these excitations. tively, as follows:
1−Ω2R2 (ω+γ Ω)2
Λ = m2− B . (66)
where D =∂ −ieA is the covariant derivative. ω,m 4R2 1−Ω2R2
µ µ µ
(cid:114)
Asusual,weconsiderthesimplest,Dirichletboundary 1 (cid:16)m (cid:17)
Φ (t,ϕ)= sin [ϕ−tΩ] (67)
condition at the position of the cut. The circle rotates ω,m πR 2 2π
uniformly about its central point with an angular veloc- (cid:26) γ +ωΩR2 (cid:27)
ity Ω, so that the rotation leads to the following time– ·exp −iωt+i B [ϕ−tΩ] ,
1−Ω2R2 2π
dependentboundaryconditionatthepositionofthecut:
(cid:12) where m = 1,2,3,.... The wavefunctions are orthonor-
(cid:12)
Φ(t,ϕ)(cid:12) =0, (60) mal and they form a complete basis.
(cid:12)
ϕ=[Ωt]
2π
wherethemodulooperation[...] isdefinedinEq.(44).
2π
We consider our circle in a background of a uniform 2. The energy density
(i.e., space and time independent) magnetic field B.
Since the model (59) is invariant under Maxwellian U(1) Thelocalenergydensityofthezero–pointfluctuations
gauge transformations, E(x) is given by the vacuum expectation value (6) of
the T00 component of the stress–energy tensor. This ex-
U(1): Φ→eieωΦ, A →A +∂ ω, (61)
µ µ µ pectation value can be computed using a Feynman–type
it is convenient to choose a gauge where the gauge field Green function
has the following form4
G(x,x(cid:48))=i(cid:104)TΦ(x)Φ∗(x(cid:48))(cid:105) , (68)
γ
A = B , A =0, A =0, A =0, (62)
ϕ e t ρ z via the following familiar relation:
where (cid:104)Tµν(x)(cid:105)=(DµD(cid:48)ν∗+DνD(cid:48)µ∗
eF (cid:12)
γB = 2πB , (63) −gµνDλDλ(cid:48)∗(cid:1)1iG(x,x(cid:48))(cid:12)(cid:12)(cid:12) , (69)
x→x(cid:48)
is a constant and F is the flux of the magnetic field B
B
so that
whichpiercesthesurfaceSspannedonthecircleC ≡∂S:
(cid:20) (cid:18) (cid:19)(cid:18) (cid:19)(cid:21)
‹ ˛ ˆ2π (cid:10)T00(t,ϕ)(cid:11)= ∂ ∂ + 1 ∂ −iγ ∂ +iγ
F = d2s·B ≡ dx·A=R dϕA . (64) ∂t∂t(cid:48) R2 ∂ϕ B ∂ϕ(cid:48) B
B ϕ (cid:12)
S C 0 1G(t,t(cid:48);ϕ,ϕ(cid:48))(cid:12)(cid:12) . (70)
i (cid:12)t→t(cid:48)
ϕ→ϕ(cid:48)
The Green’s function can be expressed via the eigen-
4 Weworkwiththecylindricalcoordinates,A=Aρρˆ+Aϕϕˆ+Azzˆ, functions (67) and eigenvalues (66) similarly to Eq. (7):
where ρˆ, ϕˆ, and zˆ are unit orthogonal vectors. We consider
the vacuum of the scalar particles at the circle and not in the ˆ+∞
ainrteercioomrpolreteexltyeriirorreleovfatnhtefcoirrcoluersoprtohbaltemA,ϕwahnildeAthzecboemhpavoinoernotsf GΩ,B(t,t(cid:48);ϕ,ϕ(cid:48))= d2ωπ (cid:88)∞ Φω,m(tΛ,ϕ)Φ−∗ω,mi(cid:15)(t(cid:48),ϕ(cid:48)),(71)
theAϕ=Aϕ(ρ)componentisrelevantonlyatρ=R. −∞ m=1 ω,m
9
The positions of poles ω = ω are determined by the both positive and negative and it depends both on the
m
following equation: Λ =0. According to Eq. (66) the angularfrequencyΩofthecircleandonthenetmagnetic
ω,m
poles are located at the real axis: fluxF whichpiercesthecircle. ThelastlineofEq.(74)
B
is written for a uniform magnetic field B, so that the
1−Ω2R2 magnetic flux going through the circle is
ω = m−γ Ω≡µ −γ Ω, (72)
m 2R B m B
F =πBR2. (75)
B
where m∈Z is an integer number.
We have restored the dependence on (cid:126) and c in the last
An important novelty of Eq. (72) is that the positions
line of Eq. (74).
of the poles are no more symmetric with respect to the
In order to evaluate the Green’s function (71) we use
reflections ω →−ω due tothe presence ofthe flux ofthe
the following relations, valid for an arbitrary parameter
magnetic field (63),
αandanevenfunctionf (withf =f andf =0):
m m −m 0
2πγ
FB = eB . (73) ˆ∞ dω (cid:88)∞ e−iαωfm = ˆ∞ dω (cid:88)(cid:48) e−iαωfm
2π Λ −i(cid:15) 4π Λ −i(cid:15)
As the magnetic flux increases at nonzero angular fre- −∞ m=1 ω,m −∞ m∈Z ω,m
quency Ω, some of the poles (72) may cross the origin,
ieiαΩγB (cid:88)(cid:48) e−iαµm
ω = 0, coming from a negative part of the real axis to = f sign(α)Θ(αω )
2Ω m m m
the positive part and vice versa, as illustrated in Fig. 4. 0 m∈Z
∞
ieiαΩγB (cid:88)(cid:48) e−i|α|µm
= f , (76)
2Ω m m
0
m=NΩ,B(α)
where prime in the sum indicates that the term with
m = 0 is omitted. In Eq. (76) Θ(x) is the Heaviside
function, and we have also defined the following integer
number:
(cid:20) (cid:21)
1 1
N (α)= + M + sign(α)
Ω,B 2 Ω,B 2
(cid:26)
M +1, α>0
≡ Ω,B , (77)
−M , α<0
Ω,B
is an integer number.
FIG. 4. Schematic illustration of the positions of the
Then we notice that for an arbitrary integer number
poles(72)intheGreen’sfunction(71)vsthemagneticfieldB
(arbitraryunits)foraclockwiserotation,Ω<0. Foracoun- N and an arbitrary function Km the following relation
terclockwise rotation (Ω<0) the slopes are negative. holds:
∞ ∞
(cid:88)(cid:48) (cid:88)
The number of poles which have crossed (due to the Km = Km+S[Km,N] , (78)
presence of the magnetic field B) the origin in the nega- m=N m=1
tive direction is
where we have defined the following finite sum
(cid:22) (cid:23) (cid:22) (cid:23)
MΩ,B =(cid:22)12e−γBBΩΩΩ2RRR32 c=(cid:23) 1−ΩΩR2R2eFπB (74) S[K ,N]=−Nm(cid:80)−=11Km0,, NN =>10,1 . (79)
≡ . m
c2−Ω2R2 (cid:126)  (cid:80)−1 Km, N <0
m=N
Here the floor operator (cid:98)x(cid:99) defines the largest integer
which is smaller than the real number x (with e.g., ApplyingEq.(76)totheEqs.(71)and(67)wegetthe
(cid:98)0.1(cid:99) = 0, (cid:98)1.9(cid:99) = 1 etc). The number M can be following explicit representation of the Green’s function:
Ω,B
∞
GΩ,B(t,t(cid:48);ϕ,ϕ(cid:48))= πiei(Ω(t−t(cid:48))+[ϕ−tΩ]2π−[ϕ(cid:48)−t(cid:48)Ω]2π)γB (cid:88)(cid:48) hm(t,t(cid:48);ϕ,ϕ(cid:48)) (80)
m=NΩ,B(α)
(cid:26) (cid:27)
=ei(Ω(t−t(cid:48))+[ϕ−tΩ]2π−[ϕ(cid:48)−t(cid:48)Ω]2π)γB GΩ(t,t(cid:48);ϕ,ϕ(cid:48))+ πiS(cid:2)hm(t,t(cid:48);ϕ,ϕ(cid:48)),NΩ,B(cid:0)α(t,t(cid:48);ϕ,ϕ(cid:48))(cid:1)(cid:3) ,
10
where G is the Green’s function (50) for the rotating
Ω
circle with the Dirichlet cut in the absence of the back-
ground of the magnetic flux F . In Eq. (80) we have
B
defined the following functions:
ΩR2([ϕ−tΩ] −[ϕ(cid:48)−t(cid:48)Ω] )
α(t,t(cid:48);ϕ,ϕ(cid:48))=t−t(cid:48)− 2π 2π ,
1−Ω2R2
hm(t,t(cid:48);ϕ,ϕ(cid:48))=e−i(1−2ΩR2R2)|α(t,t(cid:48);ϕ,ϕ(cid:48))|m
1 (cid:16)m (cid:17) (cid:16)m (cid:17)
· sin [ϕ−tΩ] sin [ϕ(cid:48)−t(cid:48)Ω] . (81)
m 2 2π 2 2π
Finally,wecalculatetheenergydensityusingEq.(70):
EZP ≡(cid:10)T00(cid:11)phys
Ω,B Ω,B
FIG. 5. First ten “enhancement” bands for a circle of the
(cid:104) (cid:105)1+Ω2R2 radius R=1 cm in the plane “angular frequency – magnetic
=− 1+6M (M +1) . (82)
Ω,B Ω,B 48πR2 field”. Theboundariesofthebandsaredefinedbytherelation
M =n with n∈Z and M is given in Eq. (74) or (86).
The strength of the background magnetic field B enters Ω,B Ω,B
this expression via the integer number M , Eq. (74),
Ω,B
which depends on the angular frequency Ω as well. The
TheintegernumberM ,thecharacteristicfrequency
Ω,B
energy density (82) does not depend on the angular co-
Ω and the corresponding characteristic time period τ
ch ch
ordinate ϕ, so that the total energy of the zero–point
of the device can be written in physical units as follows:
fluctuations (5) is:
 
 5.1×106·(cid:0) Ω (cid:1)(cid:0)B(cid:1)(cid:0)R(cid:1)3 
(cid:104) (cid:105)1+Ω2R2  1/s T m 
EΩZP,B =− 1+6MΩ,B(MΩ,B +1) 24R . (83) MΩ,B (cid:39)1−1.1×10−17·(cid:0) Ω (cid:1)2(cid:0)R(cid:1)2 , (86)
1/s m
Notice that in the absence of the magnetic field, B=0, (cid:18) B (cid:19)−1(cid:18) R (cid:19)−3
Eq. (83) equals to Eq. (53) multiplied by a factor of two Ωch(B)(cid:39)2×10−7· 1T 1m s−1, (87)
because a charged field contains two degrees of freedom
compared to one degree of freedom of a neutral field. 2π (cid:18) B (cid:19)(cid:18) R (cid:19)3
τ (B)(cid:39) =5.1×106· s, (88)
ch Ω (B) 1T 1cm
ch
respectively. Values of the characteristic frequencies and
C. Magnetic-field-enhanced zero-point energy
time periods for a certain set of B and R are shown in
Table I.
Itisclearlyseenthatthepresenceofthemagneticfield
InFig.5weillustratethestructureoftheenhancement
enhances the negative energy of the zero–point fluctua-
bands for a circle of the radius R = 1 cm. In zeroth
tions (83) because the integer number M is a rising
Ω,B (M = 0) band – filled by the reddish color in Fig. 5
stepwise function of the magnetic field, Eq. (74). In or- Ω,B
– the enhancement is absent and the zero–point energy
der to characterize the quantity M it is convenient to
Ω,B is given by the B =0 expression, Eq. (53), multiplied by
introduce a characteristic frequency Ω of the vacuum
ch two due to presence of two scalar degrees of freedom in
fluctuations,
thecomplexscalarfield. InthenextbandwithM =1
Ω,B
π (cid:126)c the enhancement of the rotational zero–point energy is
Ω (B)= ≡ , (84)
ch eFBR eBR3 equalto13,whilethehighestshownbandwithMΩ,B =9
givestheenhancementfactorof433. Thus,themagnetic
where the magnetic flux is given in Eq. (75). Then the
flux enhances the negative–valued moment of inertia of
integer number M can be rewritten as follows:
Ω,B the zero–point fluctuations.
(cid:22) (cid:23) It is important to mention that the enhancement pref-
Ω 1
MΩ,B = Ω (B)1−Ω2R2 . (85) actor in the expressionfor thezero-point energy(83) de-
ch pends on the quantity M while this integer quantity
Ω,B
Notice that the characteristic frequency Ω is a positive isastepwisefunctionoftheproductBΩ(multipliedbya
ch
number which is not limited from above. nonzerorelativisticfactor),Eq.(74). Thus,theenhance-
AtsmallangularfrequenciesΩ∼Ω (or,equivalently, menteffectworksonlyforarotatingdeviceandonlyina
ch
at weak magnetic fields) the quantity M is of the or- background of the magnetic field (otherwise the product
Ω,B
der of unity and the mentioned enhancement factor for BΩ is zero). For a static device and/or in the absence of
the zero–point energy (83) is always of the order of 10. the magnetic field the enhancement effect is absent.
However, as the angular frequency and/or magnetic flux According to Eqs. (83) and (85), one can distinguish
through the circle increase, the enhancement factor rises threedifferentlimitsintermsofthestrengthofthemag-
drastically as we will see below. netic field B and the angular frequency Ω: