Doubt continues to linger over the reality of quantum vacuum energy. There is
some question whether fluctuating fields gravitate at all, or do so
anomalously. Here we show that for the simple case of parallel conducting
plates, the associated Casimir energy gravitates just as required by the
equivalence principle, and that therefore the inertial and gravitational masses
of a system possessing Casimir energy $E_c$ are both $E_c/c^2$. This simple
result disproves recent claims in the literature. We clarify some pitfalls in
the calculation that can lead to spurious dependences on coordinate system.Comment: 5 pages, 1 figure, REVTeX. Minor revisions, including changes in
  reference

August Romeo

H. B. G. Casimir

J. Schwinger

Jef Wagner

K. V. Shajesh

K. A. Milton

Kimball A. Milton

Prachi Parashar

S. Weinberg

Stephen A. Fulling

English

arXiv

Crossref

How does Casimir energy fall?

Fulling, Stephen A.

Milton, Kimball A.

Parashar, Prachi

Romeo, August

Shajesh, K. V.

Wagner, Jef

Texas A&amp;M Repository

arXiv:hep-th/0702091v2  22 Mar 2007How Does Casimir Energy Fall?Stephen A. Fulling,1, ∗ Kimball A. Milton,2, † Prachi Parashar,2 August Romeo,3 K.V. Shajesh,2 and Jef Wagner21Departments of Mathematics and Physics, Texas A&M University, College Station, TX 77843-3368 USA2Oklahoma Center for High-Energy Physics and Homer L. Dodge Department of Physics and Astronomy,University of Oklahoma, Norman, OK 73019-2061 USA3Societat Catalana de F´ısica, Laboratori de F´ısica Matema`tica (SCF-IEC), 08028 Barcelona, Catalonia, Spain(Dated: July 9, 2018)Doubt continues to linger over the reality of quantum vacuum energy. There is some questionwhether fluctuating fields gravitate at all, or do so anomalously. Here we show that for the simplecase of parallel conducting plates, the associated Casimir energy gravitates just as required by theequivalence principle, and that therefore the inertial and gravitational masses of a system possessingCasimir energy Ec are both Ec/c2. This simple result disproves recent claims in the literature. Weclarify some pitfalls in the calculation that can lead to spurious dependences on coordinate system.PACS numbers: 03.70.+k, 04.20.Cv, 04.25.Nx, 03.65.SqThe subject of quantum vacuum energy (the Casimireffect) dates from the same year as the discovery of renor-malized quantum electrodynamics, 1948 [1]. It puts thelie to the naive presumption that zero-point energy isnot observable. On the other hand, it continues to besurrounded by controversy, in large part because sharpboundaries give rise to divergences in the local energydensity near the surface (see Refs. [2, 3, 4]). The mosttroubling aspect of these divergences is in the couplingto gravity. Gravity has its source in the local energy-momentum tensor, and such surface divergences promiseserious difficulties. The gravitational implications ofzero-point energy are an outstanding problem in view ofour inability to understand the origin of the cosmologicalconstant or dark energy [5, 6, 7].As a prolegomenon to studying such issues, we here ad-dress a simpler question: How does the completely finiteCasimir energy of a pair of parallel conducting plates cou-ple to gravity? The question turns out to be surprisinglyless straightforward than one might suspect! Previousauthors [8, 9, 10, 11, 12] have given disparate answers,including gravitational forces, or gravitationally modi-fied Casimir forces, that depend on the orientation of theCasimir apparatus with respect to the gravitational fieldof the earth. We will here resolve some of this confusionwith a convincingly calculated result consistent with theequivalence principle. That is, the renormalized Casimirenergy couples to gravity just like any other energy. Inour opinion, this fact is evidence that vacuum energymust be taken seriously in gravitational theory and thatthe problem of boundary divergences must be resolved bya better understanding of the modeling and renormaliza-tion processes.We start by recalling the electromagnetic Casimirstress tensor between a pair of parallel perfectly con-ducting plates separated by a distance a, with transversedimensions L≫ a, as given by Brown and Maclay [13]:〈T µν〉 = Ecadiag(1,−1,−1, 3), (1)where the third spatial direction is the direction normalto the plates. This is given in terms of the Casimir en-ergy per unit area, Ec = −π2~c/(720a3). Outside theplates, 〈T µν〉 = 0. Omitted here is a constant divergentterm that is present both between and outside the plates,and also in the absence of plates, which cannot haveany physical significance. Because the electromagneticfield respects conformal symmetry, there is no surface di-vergent term such as is present for a minimally coupledscalar field subject to Dirichlet conditions on the plates,or more generally for curved surfaces [14]. (Henceforthwe will set ~ = c = 1.)Now we turn to the question of the gravitational in-teraction of this Casimir apparatus. It seems to us thatthis question can be most simply addressed through useof the gravitational definition of the energy-momentumtensor, as a variation of the matter part of the action,Wg ≡ δWm ≡ 12∫(dx)√−g δgµνT µν . (2)Following Schwinger (note the factor of 2 in the defini-tion), for a weak field we define gµν = ηµν+2hµν. To firstorder we can ignore√−g. The gravitational energy, fora static situation, is therefore given by (δW ≡ − ∫ dt δE)Eg ≡ δEm = −∫(dx)hµνTµν . (3)We then replace T µν here by the one-loop expectationvalue of the electromagnetic stress tensor (1).Calloni et al. [9] and Bimonte et al. [12] use the Fermimetricg00 = − (1 + 2gz) , gij = δij , (4)in terms of the gravitational acceleration g. This is evi-dently appropriate for a constant gravitational field. Wewill discuss its relation to the field due to the earth below.Let the apparatus be oriented at an angle α with respectto the direction of the gravitational field. The Cartesian2yzζ ηαFIG. 1: Relation between two Cartesian coordinate frames:One attached to the earth (x, y, z), where −z is the directionof gravity, and one attached to the parallel-plate Casimir ap-paratus (ζ, η, χ), where ζ is in the direction normal to theplates. The parallel plates are indicated by the heavy linesparallel to the η axis. The x = χ axis is perpendicular to thepage.coordinate system attached to the earth is denoted by(x, y, z), where, as noted above, z is the direction of −g.The Cartesian coordinates associated with the Casimirapparatus are (ζ, η, χ), where ζ is normal to the plates,and η and χ are parallel to the plates. (See Fig. 1.) Thecenter of the apparatus is located at (ζ0 , η = 0, χ = 0).Now we calculate the gravitational energy from Eq. (3):Eg =∫(dx) gzT 00 =gEcaL2a ζ0 cosα, (5)up to an additive constant, independent of ζ0 . (Anyconstant added to the gravitational potential gz simplyshifts the value of this constant.) Thus, the gravitationalforce per area A = L2 on the apparatus isFA= − 1A∂Eg∂z0= −gEc ≡ − ǫ2aEc , (6)because z0 = ζ0 cosα. Note that on the earth’s surface,the dimensionless number ǫ ≡ 2ga/c2 is very small. Fora plate separation of 1µm, ǫ = 2 × 10−22, so the effectsthat we are considering are very small compared to thegravitational forces on the plates. (However, Calloni etal. [9] discuss the possibility of experimentally observingthis force.)It is a bit simpler to use the energy formula (3) tocalculate the force by considering the variation in thegravitational energy directly, that is,δEg = −∫(dx) δhµνTµν. (7)It is easy to verify that this gives the correct force on amass point, F = mg. If we use this formula to calculatethe gravitational force on the rigid Casimir apparatus, byconsidering a virtual displacement upward by an amountδz0, we find the same α-independent result (6).Alternatively, we can start from the definition of thegravitational field [15],δWg =∫(dx) δT µνhµν , (8)which can again be checked to yield the correct force ona mass point. For the constant field (4) the force on aCasimir apparatus is obtained from the change in theenergy density T 00; that is, recalling that z0 = ζ0 cosα,δT 00 =Ec δz0a1cosα[δ(ζ−ζ0−a/2)−δ(ζ−ζ0+a/2)], (9)where the δ functions arise from the step functions at theboundaries. This yields from Eq. (8) the same result (6).Our answer is consistent with the principle of equiva-lence, and with the second analysis of Jaekel and Rey-naud [16], who state that the inertia of Casimir energy(at least in two dimensions) is Ec/c2. However, it is only14that found by Bimonte et al. [12], which is also thefirst force formula [Eqs. (7) and (8)] provided by Cal-loni et al. [9]. Our Eq. (6) does, however, reproduce thesecond formula [Eq. (9)] given in Ref. [9], which thoseauthors describe as the one that should be observable.We discuss this situation further below.We now digress to consider whether the constant-field approximation (4) is adequate for an apparatussuspended above the earth or some other pointlikemass. Should we instead use the perturbation of theSchwarzschild metric? One might expect that the result-ing curvature corrections are of order LR ≪ 1, at worst,relative to the main term, where R is the earth’s radius.The point, however, is that naive attempts to do the cal-culation in curved space change the answer by factors like2, and also differ among themselves, and the resolutionsof the fallacies are sufficiently instructive to justify ourbelaboring the point.The Schwarzschild metric in isotropic coordinates [17]and for weak fields (GM/r ≪ 1) isds2 = −(1− 2GMr)dt2 +(1 +2GMr)dr2. (10)If we expand a short distance z above the earth’s sur-face, we find the nonzero components of the gravitationalfield to be h00 = h11 = h22 = h33 = GM/R − gz, interms of the acceleration of gravity, g = GM/R2. (Itis important to recognize that the constant GM/R is ir-relevant in the following, and that correspondingly theresults do not depend on the origin of z.) The virtueof isotropic coordinates is that the spatial line element(apart from an overall factor) has the usual Cartesianform dr2 = dx2 + dy2 + dz2. Now when we compute thegravitational energy from Eq. (7) each component of theCasimir stress tensor contributes with equal weight:δEg = gA δz0∫ ζ0+a/2ζ0−a/2dζ (T 00+ T 11+ T 22+ T 33), (11)which gives the force− 1AδEgδz0=FA= −2gaT 00 = −2gEc , (12)3since T = T λλ = 0. This is twice the previous result (6).Note that again the result is independent of α. The sameresult is obtained if we start from Eq. (8).We should be able to obtain the same result usingthe original Schwarzchild coordinates, where h00 = −gz,hρρ = −gz, and all other components of hµν are zero.However, now if we use the first method (7), the resultis F/A = −4gEc cos2 α, so now the force depends on theorientation of the apparatus. Even if α = 0, the magni-tude differs from Eq. (12) by an additional factor of 2.(It thereby fortuitously agrees with the result in Ref. [12]for that angle.)What is going on here? The reason we get differ-ent answers in different coordinate systems is that ourstarting point (3) is not gauge-invariant. Under a co-ordinate redefinition, which for weak fields is a gaugetransformation of hµν [15], hµν → hµν + ∂µξν + ∂νξµ ,where ξµ is a vector field, Eq. (3) is invariant only ifthe stress tensor is conserved, ∂µTµν = 0 (in the weak-field context). Otherwise, there is a change in the action,∆Wg = −2∫(dx)ξν∂µTµν . Now in our case (where wemake the finite size of the plates explicit, but ignore edgeeffects on T µν because L≫ a)T µν =Ecadiag(1,−1,−1, 3)θ(ζ − ζ0 + a/2)θ(a/2− ζ + ζ0)θ(η + L/2)θ(L/2− η)θ(χ + L/2)θ(L/2− χ). (13)Taking the divergence of Eq. (13) gives corresponding δ functions on the surfaces and leads immediately to∆Eg =6Eca∫dηdχ [ξζ(ζ0 − a/2, η, χ)− ξζ(ζ0 + a/2, η, χ)]− 2Eca∫dζdχ [ξη(ζ,−L/2, χ)− ξη(ζ, L/2, χ)]− (η ↔ χ).(14)This transformation entirely accounts algebraically forthe difference between the forces in isotropic andSchwarzchild coordinates, but it does not yet explainphysically why there are two different answers, nor tellus which, if either, is correct.There seem to be two possible ways to proceed. First,it is clear that the energy-momentum tensor of the com-plete physical situation must be conserved, and there-fore the expression (3) would be gauge-invariant if weincluded a physical mechanism holding the plates apartagainst the Casimir attraction. That road probably leadsto complicated, model-dependent calculations. The al-ternative is to find a physical basis for believing that onecoordinate system is more realistic than another. Fortu-nately, that problem apparently has a natural solution.The crux of the difficulty is that the relations between co-ordinate increments and physical distances depend uponthe distance from the gravitating center in the most com-mon coordinate systems.Of course, a perfect coordinate system is not possiblein a curved space, but the kind that comes closest to rep-resenting distances accurately all along a timelike world-line is a Fermi coordinate system, the general-relativisticextrapolation of an inertial coordinate frame. Such asystem has been given by Marzlin [18] for a resting ob-server in the field of any static mass distribution. Here wegive a simple rederivation for the case at hand. Startingfrom the isotropic metric (10), first eliminate the con-stant term by rescaling the coordinates, t→ (1 + GMR ) t,r→ (1− GMR ) r, and expand to first order:ds2 = −(1 + 2gz)dt2 + (1− 2gz)dr2. (15)But we need r to measure physical displacements evenwhen z 6= 0, so we writex = x′ + gx′z′, y = y′ + gy′z′,z = z′ + g2(z′2 − x′2 − y′2). (16)Then to first order in coordinates we obtain the Fermimetric (4). The corresponding gravitational force istherefore given by Eq. (6), after all!Now we can use the method described above to trans-form the energy in isotropic coordinates to that in Fermicoordinates. We use Eq. (14) to compute the additionalgravitational energy, in terms of the gauge field ξµ thatcarries us from isotropic coordinates to Fermi coordi-nates,hFµν = hIµν + ∂µξν + ∂νξµ . (17)Here hI00 = −gz, hIij = −gzδij , hF00 = −gz, hFij = 0,hI,F0i = 0. The gauge field turns out to beξζ =12g(12ζ2 cosα+ ζη sinα)+ f(η, χ),ξη =12g(ζη cosα+12η2 sinα)+ g(ζ, χ),ξχ =12g (ζ cosα+ η sinα)χ+ h(ζ, η), (18)4where the functions f , g, and h are irrelevant. Now fromEq. (14) we obtain a ∆Eg(z0) that yields an additionalforce, ∆F/A = gEc . When this is added to the isotropicforce (12), we obtain the result given in Eq. (6)F I +∆FA= −2gEc + gEc = −gEc = FFA. (19)The conceptual reason why other coordinates give dif-ferent answers is that under the virtual displacement in-volved in defining ∂∂z0 one is stretching the apparatus aswell as moving it. Correcting for the spurious changes inL and a restores the Fermi result in all cases. The im-portance of distinguishing a from the physical gap wasnoted by Sorge [11] in studying a related problem.As noted above, Calloni et al. [9] find a result 4 timesours, which is the only result from Ref. [9] cited in thelater paper [12] with which it shares authors. How-ever, Ref. [9] states that that force formula has twoparts, in the ratio of 3/1, and that only the smallerpiece is “Newtonian,” or “to be tested against observa-tion.” Our understanding of what that means is thefollowing. Start from Eq. (2) and consider a generalcoordinate transformation, x′µ = xµ + δxµ, so thatg′µν(x′) = gµν(x) + δgµν(x), whereδgµν = δxλ∂λgµν + gαν∂δxα∂xµ+ gµβ∂δxβ∂xν. (20)For a rigid translation, δxλ is a constant, so only the firstterm in Eq. (20) is present, which gives the result (6).However, if we do not make this restriction, we obtainfrom Eq. (2) (after integration by parts) a surface-termcorrection to the force:∫Ω(dx)√−gfλ = δWmδxλ−∫∂ΩdΣν√−gT νλ , (21)where the force vector density is [9] −∇νT νλ or√−gfλ = −∂ν(√−gT νλ)+ 12√−gT µν∂λgµν . (22)Note that the surface term identically cancels the firstterm in fλ . Now if Ω refers to all space, the surfaceterm vanishes (as we have shown explicitly). But if Ωis just the space volume between the plates, and we in-clude this correction for the Fermi metric (4) for which√−g = 1+gz, we obtain an additional term −3gEc cosα.Adding this to the previous result (6), we obtain the re-sult of Ref. [12] if α = 0. However, in general the resultdepends on the angle between the apparatus and the ver-tical. Is this consistent with the equivalence principle(the scalar nature of mass)? A similar angle dependencewill now occur with the isotropic Schwarzchild metric.Why should one trust the formula (22) over the morefundamental variational principle when boundaries arepresent? Omission of the surface term resolves the dis-crepancy, giving the equivalence-principle result (6).After circulation of the present paper asarXiv:hep-th/0702091, Bimonte et al. [19] respondedthat (i) our result is implicit in their earlier analysis;(ii) our Eq. (2) implicitly assumes the equivalenceprinciple, which they identify with the conservation law∇µT µν = 0. We reply: (i) Ref. [12] states without quali-fication that the force is 4 times the correct value. (ii)By “equivalence principle” we merely mean that gravitytends to enforce geodesic motion. This is not obviouslyequivalent to conservation, as shown by the existinguncertainty in the literature. Contrary to statementsin Ref. [19], it is generally accepted in quantum fieldtheory in curved space-time that the renormalized totalT µν must be conserved.This work is supported in part by Collaborative Re-search Grants from the US National Science Foundationand in part by the US Department of Energy. It wascompleted while S.A.F. enjoyed partial support at the I.Newton Institute for Mathematical Sciences, Cambridge.We thank Gabriel Barton, Michael Bordag, Robert Jaffe,and Ron Kantowski for helpful conversations and emails.∗ Electronic address: fulling@math.tamu.edu;URL: www.math.tamu.edu/~fulling/† Electronic address: milton@nhn.ou.edu;URL: www.nhn.ou.edu/~milton/[1] H. B. G. Casimir, Proc. K. Ned. Akad. Wet. 51, 792(1948).[2] D. Deutsch and P. Candelas, Phys. Rev. D 20, 3063(1979).[3] P. Candelas, Ann. Phys. (N.Y.) 143, 241 (1982); Ann.Phys. (N.Y.) 167, 257 (1986).[4] N. Graham, R. L. Jaffe, V. Khemani, M. Quandt,O. Schroeder and H. Weigel, Nucl. Phys. B 677, 379(2004) [arXiv:hep-th/0309130], and references therein.[5] N. Straumann, in 18th IAP Colloquium: Observa-tional and Theoretical Results on the Accelerating Uni-verse arXiv:gr-qc/0208027; in 1st Se´minaire Poincare´arXiv:astro-ph/0203330.[6] S. Weinberg, Rev. Mod. Phys. 61, 1 (1989).[7] K. A. Milton, R. Kantowski, C. Kao and Y. Wang, Mod.Phys. Lett. A 16, 2281 (2001).[8] M. Karim, A. H. Bokhari, and B. J. Ahmedov, Class.Quantum Grav. 17, 2459-2462 (2000).[9] E. Calloni, L. Di Fiori, G. Esposito, L. Milano, and L.Rosa, Phys. Lett. A 297, 328 (2002).[10] R. R. Caldwell, arXiv:astro-ph/0209321.[11] F. Sorge, Class. Quant. Grav. 22, 5109 (2005).[12] G. Bimonte, E. Calloni, G. Esposito, and L. Rosa, Phys.Rev. D 74, 085011 (2006).[13] L. S. Brown and G. J. Maclay, Phys. Rev. 184, 1272(1969).[14] K. A. Milton, The Casimir Effect: Physical Manifesta-tions of Zero-Point Energy (World Scientific, Singapore,2001), Sec. 11.1.[15] J. Schwinger, Particles, Sources, and Fields (Addison-Wesley, Reading, MA, 1970).5[16] M. T. Jaekel and S. Reynaud, Journal de Physique I 3,1093-1104 (1993) [arXiv:quant-ph/0101082] and relatedreferences.[17] S. Weinberg, Gravitation and Cosmology: Principles andApplications of the General Theory of Relativity (Wiley,New York, 1972).[18] K.-P. Marzlin, Phys. Rev. D 50, 888 (1994).[19] G. Bimonte, E. Calloni, G. Esposito and L. Rosa,arXiv:hep-th/0703062.

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How Does Casimir Energy Fall?
Stephen A. Fulling,1, ∗ Kimball A. Milton,2, † Prachi Parashar,2 August Romeo,3 K.V. Shajesh,2 and Jef Wagner2
1Departments of Mathematics and Physics, Texas A&M University, College Station, TX 77843-3368 USA
2Oklahoma Center for High-Energy Physics and Homer L. Dodge Department of Physics and Astronomy,
University of Oklahoma, Norman, OK 73019-2061 USA
3Societat Catalana de F´ısica, Laboratori de F´ısica Matema`tica (SCF-IEC), 08028 Barcelona, Catalonia, Spain
(Dated: July 9, 2018)
Doubt continues to linger over the reality of quantum vacuum energy. There is some question
whether fluctuating fields gravitate at all, or do so anomalously. Here we show that for the simple
case of parallel conducting plates, the associated Casimir energy gravitates just as required by the
equivalence principle, and that therefore the inertial and gravitational masses of a system possessing
Casimir energy Ec are both Ec/c
2. This simple result disproves recent claims in the literature. We
clarify some pitfalls in the calculation that can lead to spurious dependences on coordinate system.
PACS numbers: 03.70.+k, 04.20.Cv, 04.25.Nx, 03.65.Sq
The subject of quantum vacuum energy (the Casimir
effect) dates from the same year as the discovery of renor-
malized quantum electrodynamics, 1948 [1]. It puts the
lie to the naive presumption that zero-point energy is
not observable. On the other hand, it continues to be
surrounded by controversy, in large part because sharp
boundaries give rise to divergences in the local energy
density near the surface (see Refs. [2, 3, 4]). The most
troubling aspect of these divergences is in the coupling
to gravity. Gravity has its source in the local energy-
momentum tensor, and such surface divergences promise
serious difficulties. The gravitational implications of
zero-point energy are an outstanding problem in view of
our inability to understand the origin of the cosmological
constant or dark energy [5, 6, 7].
As a prolegomenon to studying such issues, we here ad-
dress a simpler question: How does the completely finite
Casimir energy of a pair of parallel conducting plates cou-
ple to gravity? The question turns out to be surprisingly
less straightforward than one might suspect! Previous
authors [8, 9, 10, 11, 12] have given disparate answers,
including gravitational forces, or gravitationally modi-
fied Casimir forces, that depend on the orientation of the
Casimir apparatus with respect to the gravitational field
of the earth. We will here resolve some of this confusion
with a convincingly calculated result consistent with the
equivalence principle. That is, the renormalized Casimir
energy couples to gravity just like any other energy. In
our opinion, this fact is evidence that vacuum energy
must be taken seriously in gravitational theory and that
the problem of boundary divergences must be resolved by
a better understanding of the modeling and renormaliza-
tion processes.
We start by recalling the electromagnetic Casimir
stress tensor between a pair of parallel perfectly con-
ducting plates separated by a distance a, with transverse
dimensions L≫ a, as given by Brown and Maclay [13]:
〈T µν〉 = Ec
a
diag(1,−1,−1, 3), (1)
where the third spatial direction is the direction normal
to the plates. This is given in terms of the Casimir en-
ergy per unit area, Ec = −π2~c/(720a3). Outside the
plates, 〈T µν〉 = 0. Omitted here is a constant divergent
term that is present both between and outside the plates,
and also in the absence of plates, which cannot have
any physical significance. Because the electromagnetic
field respects conformal symmetry, there is no surface di-
vergent term such as is present for a minimally coupled
scalar field subject to Dirichlet conditions on the plates,
or more generally for curved surfaces [14]. (Henceforth
we will set ~ = c = 1.)
Now we turn to the question of the gravitational in-
teraction of this Casimir apparatus. It seems to us that
this question can be most simply addressed through use
of the gravitational definition of the energy-momentum
tensor, as a variation of the matter part of the action,
Wg ≡ δWm ≡ 1
2
∫
(dx)
√−g δgµνT µν . (2)
Following Schwinger (note the factor of 2 in the defini-
tion), for a weak field we define gµν = ηµν+2hµν. To first
order we can ignore
√−g. The gravitational energy, for
a static situation, is therefore given by (δW ≡ − ∫ dt δE)
Eg ≡ δEm = −
∫
(dx)hµνT
µν . (3)
We then replace T µν here by the one-loop expectation
value of the electromagnetic stress tensor (1).
Calloni et al. [9] and Bimonte et al. [12] use the Fermi
metric
g00 = − (1 + 2gz) , gij = δij , (4)
in terms of the gravitational acceleration g. This is evi-
dently appropriate for a constant gravitational field. We
will discuss its relation to the field due to the earth below.
Let the apparatus be oriented at an angle α with respect
to the direction of the gravitational field. The Cartesian
2y
z
ζ η
α
FIG. 1: Relation between two Cartesian coordinate frames:
One attached to the earth (x, y, z), where −z is the direction
of gravity, and one attached to the parallel-plate Casimir ap-
paratus (ζ, η, χ), where ζ is in the direction normal to the
plates. The parallel plates are indicated by the heavy lines
parallel to the η axis. The x = χ axis is perpendicular to the
page.
coordinate system attached to the earth is denoted by
(x, y, z), where, as noted above, z is the direction of −g.
The Cartesian coordinates associated with the Casimir
apparatus are (ζ, η, χ), where ζ is normal to the plates,
and η and χ are parallel to the plates. (See Fig. 1.) The
center of the apparatus is located at (ζ0 , η = 0, χ = 0).
Now we calculate the gravitational energy from Eq. (3):
Eg =
∫
(dx) gzT 00 =
gEc
a
L2a ζ0 cosα, (5)
up to an additive constant, independent of ζ0 . (Any
constant added to the gravitational potential gz simply
shifts the value of this constant.) Thus, the gravitational
force per area A = L2 on the apparatus is
F
A
= − 1
A
∂Eg
∂z0
= −gEc ≡ − ǫ
2a
Ec , (6)
because z0 = ζ0 cosα. Note that on the earth’s surface,
the dimensionless number ǫ ≡ 2ga/c2 is very small. For
a plate separation of 1µm, ǫ = 2 × 10−22, so the effects
that we are considering are very small compared to the
gravitational forces on the plates. (However, Calloni et
al. [9] discuss the possibility of experimentally observing
this force.)
It is a bit simpler to use the energy formula (3) to
calculate the force by considering the variation in the
gravitational energy directly, that is,
δEg = −
∫
(dx) δhµνT
µν. (7)
It is easy to verify that this gives the correct force on a
mass point, F = mg. If we use this formula to calculate
the gravitational force on the rigid Casimir apparatus, by
considering a virtual displacement upward by an amount
δz0, we find the same α-independent result (6).
Alternatively, we can start from the definition of the
gravitational field [15],
δWg =
∫
(dx) δT µνhµν , (8)
which can again be checked to yield the correct force on
a mass point. For the constant field (4) the force on a
Casimir apparatus is obtained from the change in the
energy density T 00; that is, recalling that z0 = ζ0 cosα,
δT 00 =
Ec δz0
a
1
cosα
[δ(ζ−ζ0−a/2)−δ(ζ−ζ0+a/2)], (9)
where the δ functions arise from the step functions at the
boundaries. This yields from Eq. (8) the same result (6).
Our answer is consistent with the principle of equiva-
lence, and with the second analysis of Jaekel and Rey-
naud [16], who state that the inertia of Casimir energy
(at least in two dimensions) is Ec/c
2. However, it is only
1
4
that found by Bimonte et al. [12], which is also the
first force formula [Eqs. (7) and (8)] provided by Cal-
loni et al. [9]. Our Eq. (6) does, however, reproduce the
second formula [Eq. (9)] given in Ref. [9], which those
authors describe as the one that should be observable.
We discuss this situation further below.
We now digress to consider whether the constant-
field approximation (4) is adequate for an apparatus
suspended above the earth or some other pointlike
mass. Should we instead use the perturbation of the
Schwarzschild metric? One might expect that the result-
ing curvature corrections are of order LR ≪ 1, at worst,
relative to the main term, where R is the earth’s radius.
The point, however, is that naive attempts to do the cal-
culation in curved space change the answer by factors like
2, and also differ among themselves, and the resolutions
of the fallacies are sufficiently instructive to justify our
belaboring the point.
The Schwarzschild metric in isotropic coordinates [17]
and for weak fields (GM/r ≪ 1) is
ds2 = −
(
1− 2GM
r
)
dt2 +
(
1 +
2GM
r
)
dr2. (10)
If we expand a short distance z above the earth’s sur-
face, we find the nonzero components of the gravitational
field to be h00 = h11 = h22 = h33 = GM/R − gz, in
terms of the acceleration of gravity, g = GM/R2. (It
is important to recognize that the constant GM/R is ir-
relevant in the following, and that correspondingly the
results do not depend on the origin of z.) The virtue
of isotropic coordinates is that the spatial line element
(apart from an overall factor) has the usual Cartesian
form dr2 = dx2 + dy2 + dz2. Now when we compute the
gravitational energy from Eq. (7) each component of the
Casimir stress tensor contributes with equal weight:
δEg = gA δz0
∫ ζ0+a/2
ζ0−a/2
dζ (T 00+ T 11+ T 22+ T 33), (11)
which gives the force
− 1
A
δEg
δz0
=
F
A
= −2gaT 00 = −2gEc , (12)
3since T = T λλ = 0. This is twice the previous result (6).
Note that again the result is independent of α. The same
result is obtained if we start from Eq. (8).
We should be able to obtain the same result using
the original Schwarzchild coordinates, where h00 = −gz,
hρρ = −gz, and all other components of hµν are zero.
However, now if we use the first method (7), the result
is F/A = −4gEc cos2 α, so now the force depends on the
orientation of the apparatus. Even if α = 0, the magni-
tude differs from Eq. (12) by an additional factor of 2.
(It thereby fortuitously agrees with the result in Ref. [12]
for that angle.)
What is going on here? The reason we get differ-
ent answers in different coordinate systems is that our
starting point (3) is not gauge-invariant. Under a co-
ordinate redefinition, which for weak fields is a gauge
transformation of hµν [15], hµν → hµν + ∂µξν + ∂νξµ ,
where ξµ is a vector field, Eq. (3) is invariant only if
the stress tensor is conserved, ∂µT
µν = 0 (in the weak-
field context). Otherwise, there is a change in the action,
∆Wg = −2
∫
(dx)ξν∂µT
µν . Now in our case (where we
make the finite size of the plates explicit, but ignore edge
effects on T µν because L≫ a)
T µν =
Ec
a
diag(1,−1,−1, 3)θ(ζ − ζ0 + a/2)θ(a/2− ζ + ζ0)θ(η + L/2)θ(L/2− η)θ(χ + L/2)θ(L/2− χ). (13)
Taking the divergence of Eq. (13) gives corresponding δ functions on the surfaces and leads immediately to
∆Eg =
6Ec
a
∫
dηdχ [ξζ(ζ0 − a/2, η, χ)− ξζ(ζ0 + a/2, η, χ)]− 2Ec
a
∫
dζdχ [ξη(ζ,−L/2, χ)− ξη(ζ, L/2, χ)]− (η ↔ χ).
(14)
This transformation entirely accounts algebraically for
the difference between the forces in isotropic and
Schwarzchild coordinates, but it does not yet explain
physically why there are two different answers, nor tell
us which, if either, is correct.
There seem to be two possible ways to proceed. First,
it is clear that the energy-momentum tensor of the com-
plete physical situation must be conserved, and there-
fore the expression (3) would be gauge-invariant if we
included a physical mechanism holding the plates apart
against the Casimir attraction. That road probably leads
to complicated, model-dependent calculations. The al-
ternative is to find a physical basis for believing that one
coordinate system is more realistic than another. Fortu-
nately, that problem apparently has a natural solution.
The crux of the difficulty is that the relations between co-
ordinate increments and physical distances depend upon
the distance from the gravitating center in the most com-
mon coordinate systems.
Of course, a perfect coordinate system is not possible
in a curved space, but the kind that comes closest to rep-
resenting distances accurately all along a timelike world-
line is a Fermi coordinate system, the general-relativistic
extrapolation of an inertial coordinate frame. Such a
system has been given by Marzlin [18] for a resting ob-
server in the field of any static mass distribution. Here we
give a simple rederivation for the case at hand. Starting
from the isotropic metric (10), first eliminate the con-
stant term by rescaling the coordinates, t→ (1 + GMR ) t,
r→ (1− GMR ) r, and expand to first order:
ds2 = −(1 + 2gz)dt2 + (1− 2gz)dr2. (15)
But we need r to measure physical displacements even
when z 6= 0, so we write
x = x′ + gx′z′, y = y′ + gy′z′,
z = z′ + g
2
(z′2 − x′2 − y′2). (16)
Then to first order in coordinates we obtain the Fermi
metric (4). The corresponding gravitational force is
therefore given by Eq. (6), after all!
Now we can use the method described above to trans-
form the energy in isotropic coordinates to that in Fermi
coordinates. We use Eq. (14) to compute the additional
gravitational energy, in terms of the gauge field ξµ that
carries us from isotropic coordinates to Fermi coordi-
nates,
hFµν = h
I
µν + ∂µξν + ∂νξµ . (17)
Here hI00 = −gz, hIij = −gzδij , hF00 = −gz, hFij = 0,
hI,F0i = 0. The gauge field turns out to be
ξζ =
1
2
g
(
1
2
ζ2 cosα+ ζη sinα
)
+ f(η, χ),
ξη =
1
2
g
(
ζη cosα+
1
2
η2 sinα
)
+ g(ζ, χ),
ξχ =
1
2
g (ζ cosα+ η sinα)χ+ h(ζ, η), (18)
4where the functions f , g, and h are irrelevant. Now from
Eq. (14) we obtain a ∆Eg(z0) that yields an additional
force, ∆F/A = gEc . When this is added to the isotropic
force (12), we obtain the result given in Eq. (6)
F I +∆F
A
= −2gEc + gEc = −gEc = F
F
A
. (19)
The conceptual reason why other coordinates give dif-
ferent answers is that under the virtual displacement in-
volved in defining ∂∂z0 one is stretching the apparatus as
well as moving it. Correcting for the spurious changes in
L and a restores the Fermi result in all cases. The im-
portance of distinguishing a from the physical gap was
noted by Sorge [11] in studying a related problem.
As noted above, Calloni et al. [9] find a result 4 times
ours, which is the only result from Ref. [9] cited in the
later paper [12] with which it shares authors. How-
ever, Ref. [9] states that that force formula has two
parts, in the ratio of 3/1, and that only the smaller
piece is “Newtonian,” or “to be tested against observa-
tion.” Our understanding of what that means is the
following. Start from Eq. (2) and consider a general
coordinate transformation, x′µ = xµ + δxµ, so that
g′µν(x
′) = gµν(x) + δgµν(x), where
δgµν = δx
λ∂λgµν + gαν
∂δxα
∂xµ
+ gµβ
∂δxβ
∂xν
. (20)
For a rigid translation, δxλ is a constant, so only the first
term in Eq. (20) is present, which gives the result (6).
However, if we do not make this restriction, we obtain
from Eq. (2) (after integration by parts) a surface-term
correction to the force:∫
Ω
(dx)
√−gfλ = δWm
δxλ
−
∫
∂Ω
dΣν
√−gT νλ , (21)
where the force vector density is [9] −∇νT νλ or
√−gfλ = −∂ν
(√−gT νλ)+ 1
2
√−gT µν∂λgµν . (22)
Note that the surface term identically cancels the first
term in fλ . Now if Ω refers to all space, the surface
term vanishes (as we have shown explicitly). But if Ω
is just the space volume between the plates, and we in-
clude this correction for the Fermi metric (4) for which√−g = 1+gz, we obtain an additional term −3gEc cosα.
Adding this to the previous result (6), we obtain the re-
sult of Ref. [12] if α = 0. However, in general the result
depends on the angle between the apparatus and the ver-
tical. Is this consistent with the equivalence principle
(the scalar nature of mass)? A similar angle dependence
will now occur with the isotropic Schwarzchild metric.
Why should one trust the formula (22) over the more
fundamental variational principle when boundaries are
present? Omission of the surface term resolves the dis-
crepancy, giving the equivalence-principle result (6).
After circulation of the present paper as
arXiv:hep-th/0702091, Bimonte et al. [19] responded
that (i) our result is implicit in their earlier analysis;
(ii) our Eq. (2) implicitly assumes the equivalence
principle, which they identify with the conservation law
∇µT µν = 0. We reply: (i) Ref. [12] states without quali-
fication that the force is 4 times the correct value. (ii)
By “equivalence principle” we merely mean that gravity
tends to enforce geodesic motion. This is not obviously
equivalent to conservation, as shown by the existing
uncertainty in the literature. Contrary to statements
in Ref. [19], it is generally accepted in quantum field
theory in curved space-time that the renormalized total
T µν must be conserved.
This work is supported in part by Collaborative Re-
search Grants from the US National Science Foundation
and in part by the US Department of Energy. It was
completed while S.A.F. enjoyed partial support at the I.
Newton Institute for Mathematical Sciences, Cambridge.
We thank Gabriel Barton, Michael Bordag, Robert Jaffe,
and Ron Kantowski for helpful conversations and emails.
∗ Electronic address: fulling@math.tamu.edu;
URL: www.math.tamu.edu/~fulling/
† Electronic address: milton@nhn.ou.edu;
URL: www.nhn.ou.edu/~milton/
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How Does Casimir Energy Fall?

How Does Casimir Energy Fall?

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How Does Casimir Energy Fall?

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