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The gravity of magnetic stresses and energy Giuseppe Bimonte, Enrico Calloni and Luigi Rosa Dipartimento di Scienze Fisiche, Università di Napoli Federico II, Complesso Universitario MSA, Via Cintia I-80126 Napoli, Italy; INFN, Sezione di Napoli, Napoli, ITALY (Dated: February 1, 2008) In the framework of designing laboratory tests of relativistic gravity, we investigate the gravita- tionalfieldproducedbythemagneticfieldofasolenoid. Observingthisfieldmightprovideameanof testingwhetherstressesgravitateaspredictedbyEinstein’stheory. Apreviousstudyofthisproblem byBraginsky, Cavesand Thornepredictedthat thecontribution tothegravitational field resulting 8 from thestresses of themagneticfield andofthesolenoid walls would cancelthegravitational field 0 producedbythemass-energy ofthemagneticfield,resultingin anullmagnetically-generated grav- 0 itational force outside the solenoid. They claim that this null result, once proved experimentally, 2 would demonstrate the stress contribution to gravity. We show that this result is incorrect, as it n arises from an incomplete analysis of the stresses, which neglects the axial stresses in the walls. a Oncethestressesareproperlyevaluated,wefindthatthegravitationalfieldoutsidealongsolenoid J is in fact independent of Maxwell and material stresses, and it coincides with the newtonian field 6 produced by the linear mass distribution equivalent to the density of magnetic energy stored in a 1 unit length of the solenoid. We argue that the gravity of Maxwell stress can be directly measured in thevacuum region inside the solenoid, where thenewtonian noise is absent in principle, and the ] gravity generated by Maxwell stresses is not screened by the negative gravity of magnetic-induced c stresses in thesolenoid walls. q - r PACSnumbers: 04.20.-q,04.80.Cc,04.25.Nx g Keywords: GeneralRelativity,laboratory,magneticstresses [ 2 I. INTRODUCTION esting inthis respectbecause, accordingto Maxwellthe- v ory, the energy density of a magnetic field has the same 7 7 magnitude as the trace of the Maxwell stress tensor and According to General Relativity, material and field 0 therefore this type of experiment may provide an excel- stresses are sources of gravity, because the active gravi- 1 lent tool to probe the gravity of stresses. With this pur- tationalmassdensity,intherelativisticanalogueofPois- . pose, the authors of Ref. [3] considered a simple setup, 7 son’s equation, is proportional to ρ+T, where ρ is the 0 in which the gravityproduced by the magnetic field of a density of energy and T is the trace of the stresses [1]. 7 long solenoid would be measured by means of a torsion Stress-generated gravity is very important in a number 0 balance, having one of its test masses near the solenoid. : of problems. For example, in astrophysics it affects the Of course, the difficulty of the experiment is due to the v maximum mass of neutron stars, but if one intends it, i factthatmagnetically-generatedgravityisveryweak,for X in a broad sense, as the gravity produced by the spatial experimentallyattainablemagneticfields. Togetanesti- components of the momentum-energy tensor, it displays r mateoftherequiredmagneticfieldsandbalancesensitiv- a its full powerin cosmology,where itmay wellbe respon- ity,one maytemporarilyneglectallstressesandassume, sible of the recently discovered accelerated expansion of onthebasisoftheequivalencebetweenmassandenergy, the Universe [2]. that the magnetically-generated gravitational field near As of today, there exists no direct experimental proof a long solenoid is the same as that of a cylindrical road, that stresses indeed gravitate, and it is clearly of great with a linear mass density equal to the magnetic energy interest to investigate the possibility of a laboratory ex- (divided by the square of the speed of light c) stored periment to test this prediction of General Relativity. in a unit length of the solenoid. Even for very strong Unfortunately, this is very difficult because in ordinary magnetic fields, the effect is very small, if one considers material bodies, of a size that can be handled in a labo- that the mass density equivalent to the energy density ratory, the trace of stresses is many orders of magnitude of a magnetic field of 105 G is as small as 4.4×10−13 smallerthantheenergydensityassociatedwiththemass gcm−3. However, it was argued in [3] that the demands density of the body, and therefore its effects are negligi- of the experiment could have soonbe met, imagining re- ble. However, it was realized thirty years ago [3] that a alistic improvements of the technology available in the possible way to circumventthis difficulty is by observing seventies, in cryogenic low-noise torque-balancesand su- the gravityofmagnetic fields,whichone expects to exist perconducting solenoids. because in General Relativity all forms of energy (and stresses)aresourcesofgravity. Magneticfields areinter- When considering the effect of stresses, one notices 2 that two types of stresses may contribute to the grav- alizedthatthenullresultfoundin[3]wasdeterminedby itational field of the solenoid: Maxwell stresses of the amistakenevaluationofthe stressesthatbuildupinside magnetic field and material stresses that build up in the the solenoid walls when the magnetic field is present. In walls of the solenoidin response to the applied magnetic particular,the authors overlookedthe axialstresses that field. In Ref [3] it was correctly stated that the walls of ariseinresponsetotheaxialelectrodynamiccompression thesolenoidcanbeconsideredtobeininstantaneousme- of the solenoid. Besides leading to an incorrect result chanicalequilibrium,becauseintheconsideredsetupthe forthemagnetically-generatedgravitationalfieldoutside modulation frequency of the magnetic field is extremely the solenoid, this error led the authors to overlook the low (around 10−3 Hertz, which represents the typical largenewtoniannoiseoriginatingfrommagnetic-induced resonance frequency of a torque balance). The conclu- changes in the length of the solenoid. siondrawnin[3]wasthatthe inclusionofstresseswould lead to a null magnetically-generated gravitational force (apart from the newtonian noise caused by the stress- After stresses are properly accounted for, our analy- induced modulation of the mass-density of the solenoid sis shows, in a general way, that the total magnetically- walls),because ofa purportedcancellationoccurringbe- generated gravitational mass, measured far from the tweenthe gravityof stressesand the gravityof magnetic solenoid, is independent of the stresses and is just equal energy. to the total magnetic energy (divided by the square of the speed of light c2), in accordance with one’s intu- This result appears suspicious, from the point of view ition and in agreement with earlier studies on the Tol- of a well-known paradox, that was pointed out long ago man paradox. We then consider the field near a long by Tolman[4]in his investigationsonthe roleof stresses solenoid, and we show that the magnetically-generated as source of gravity. Tolman found the paradox while gravitationalfield is different fromzero,and as expected considering the gravitational field of a static spherical itis equivalenttothe newtonianfieldgeneratedby alin- impermeable box filled with a fluid, which undergoes a earmass-densitythat is equalto the instantaneousmag- spherically symmetric transformationthat conserves the neticenergyperunitlength(dividedbyc2)storedinthe total energy, but causes a change of pressure, like mat- solenoid. Since the near field outside the solenoid, like ter and antimatter annihilating into radiation. One may the far field, is independent of the stresses, we conclude think that, since the total energy of the system is pre- that observation of the external field cannot be used to served, the change in pressure determines a change in test the gravity of stresses. Moreover, measuring this the active gravitational mass of the box, and a conse- magnetically-generated field will be very hard, because quent change in the gravitational field outside the box. weestimatethatmagnetic-inducedchangesinthelength However,thisinferenceisincontradictionwithBirkhoff’s of the solenoid produce a newtonian noise that is many theorem,whichstatesthattheexternalgravitationalfield order of magnitudes larger than the magnetically gener- ofasphericallysymmetric body is staticandthereforeit ated gravity. This by no means implies, however, that isinsensitivetowhateversphericallysymmetrictransfor- the gravityofstressesisnotobservableinthis setup, be- mations may occur inside the box. The Tolman paradox causeinthevacuumregioninsidethesolenoidthegravity wasinvestigatedin[5],wherethecrucialroleofthewalls producedbyMaxwellstressesisnotscreenedbytheneg- that keep the fluid confined was realized. It was shown ative gravity of the material stresses in the walls, and there that the stresses that build up in the walls in re- therefore it contributes to the field as much as the den- sponse to the transformation, give a negative contribu- sityofmagneticenergy. Moreover,itisexpectedthatthe tion to the active gravitationalmass of the system, that newtonian noise will be much less of a problem, because justcompensatesthepressurecontributionfromthefluid in the ideal case of a infinitely long and perfectly axially inside, resulting in an overall unchanged total gravita- symmetric solenoid, newtonian noise inside the solenoid tional mass across the transformation, as expected from is strictly zero. Birkhoff’s theorem. The same problem has been investi- gated again in a recent paper [6], leading to analogous conclusions (The key role of the stresses in the walls bounding a relativistic gravitatingsystems has been dis- The paper is organized as follows: in Sec. 2 we de- cussed by us very recently, in connection with the prob- rive, within Linearized Theory for General Relativity, lemofdeterminingtheweightofaCasimirapparatusina the magnetically-generatedgravitationalpull exerted on weakgravitationalfield [7]). The generallessonthat one a test particle by a solenoid carryinga quasi-static mag- learns from these studies is that the gravitational field netic field. In Sec. 3 we analyze in detail the contribu- outside a spherical body is independent of the stresses tions from Maxwell and material stresses and we prove in its interior, and it is determined solely by the mass- that outside the solenoid they cancel each other. Sec. 4 energy content of the body. Since there is no reason deals with the problem of newtonian noise, while Sec. 5 to imagine that this is true only for the spherical case, contains a discussion of the results and our conclusions. one is led to suspect that the results of [3] may not be Finally,intheAppendix weprovideexplicitformulaefor correct. This motivated us to reconsider in detail the the materialstressesthatbuildup withinthe wallsofan analysis of [3], and we present here our findings. We re- idealized solenoid. 3 II. THE GRAVITY OF A QUASI-STATIC Tij =Tij +Tij . (2.8) walls mag MAGNETIC FIELD In the above equations, δρ represents the change in walls the(classical)mass-densityofthesolenoidwallsresulting In this Section we estimate the pull F exerted on a i frompossibledeformationsofthesolenoiddeterminedby test particle atrest, by the magnetically-generatedgrav- the magnetic field [12], while Tij denote the extra me- itational field of a solenoid S, producing a quasi-static walls chanical stresses that build up within the solenoid walls magnetic field B. Since the gravitational fields involved when the field is turned on. Note that Tij does not are extremely small, non-linear effects are negligible and walls includethemechanicalstressesresultingfromtheweight we can safely study the problem using the simple Lin- of the solenoid and from the external forces exerted on earized Theory for Einstein’s General Relativity [1]. In the solenoid walls by the mounts that hold it. Finally, this approximation, the gravitationalfield g is written µν E =B2/(8π)denotes the density ofmagneticenergy, as [11]: mag while Tij is the Maxwell tensor: mag g =η +h , (2.1) µν µν µν 1 1 whereηµν =diag{−c2,1,1,1}isthe flatMinkowskimet- Tmijag = 4π 2B2δij −BiBj . (2.9) ric, and h represents a weak gravitational field. We (cid:18) (cid:19) µν further split hµν as Upon solving Eqs. (2.4) it is easy to obtain for the pull F the following expression: h =h | +γ , (2.2) i µν µν B=0 µν F =−m∂ (δΦ +ψ), (2.10) whereh | isthefieldthatexistswhenthesolenoidis i i walls µν B=0 turned off, while γ is the magnetically-generated field µν where that is present when the magnetic field B is turned on. The field hµν|B=0 includes the background gravitational δΦ =−G d3yδρwalls , (2.11) existing in the laboratory, together with the small field walls |x−y| Z generated by the walls of the solenoid when no currents flow in it. and To linear order, the pull F on a test particle of mass m arising from the magneticailly-generated gravitational ψ =−G d3y Emag+Twiialls+Tmiiag . (2.12) field is: |x−y| c2 ! Z 1 F =−m(Γi −Γi | )= m∂ γ , (2.3) Of the two terms appearing on the r.h.s. of Eq. (2.10), i 00 00 B=0 2 i 00 that involving δΦ just represents a purely classical walls where Γi are Christoffel symbols. For a quasi-static ”newtonian noise”, and we postpone to Sec.5 a discus- 00 magnetic field, Linearized Theory gives the following sion of its consequences. The interesting term for us is Equations for γ : the contribution proportional to ψ, that represents the µν magnetically-generated gravitational field. We see that 16πG ψ coincides with the classical gravitational field gener- △γ¯ =− T . (2.4) µν c4 µν ated by and effective mass distribution ρ equal to: eff In these Equations, △ denotes the flat space-time lapla- 1 cian, and γ¯µν is the field: ρeff = c2 Emag+Twiialls+Tmiiag . (2.13) 1 (cid:0) (cid:1) γ¯ =γ − η γ, (2.5) This is a rather complicated formula, for it involves the µν µν µν 2 traceofthestressesTij inthesolenoidwalls. Itiscon- walls where γ =ηµνγ . The above equations have to be sup- venient to define the total ”effective gravitationalmass” µν plemented by the Lorenz gauge conditions, which for a Meff, as the integral over all space of ρeff: static field imply: M = d3xρ . (2.14) ∂ γ¯i =0. (2.6) eff eff i µ ZAllspace It is important to bear in mind that, according to the We split M as: eff definition of γ , the energy-momentum tensor T ap- µν µν pearing on the r.h.s. of Eqs. (2.4) represents the sole M =M +M , (2.15) eff magen str contribution to the total energy-momentum tensor that ariseswhenthemagneticfieldisturnedon. Thesolenoid whereMmagen isthemassassociatedwiththetotalmag- being at rest, and the magnetic field being quasi-static, netic energy Emag the non-vanishing components of Tµν read: 1 E M = d3xE ≡ mag , (2.16) T00 =δρ +E /c2 , (2.7) magen c2 mag c2 walls mag ZAllspace 4 while M is associated with Maxwell and material of the total divergence by Gauss theorem, we obtain for str stresses: M the expression: str 1 Mstr ≡ c2 d3x(Twiialls+Tmiiag). (2.17) Mstr = d2σxjni[(Twijalls+Tmijag)|ins−Tmijag|out]+ ZAllspace Z∂S Note that both the integrals for M and M exist, magen str because atlargedistancesR fromthe solenoid,the mag- + lim d2σTij xjni , (3.5) neticfieldfallsofflikeR−3 andthenEmag andTmijag both R→∞ZSR mag decay as R−6. The existence of a contribution to M , eff where S denotes a two-sphere of radius R centered at such as M , arisingfrom the magnetic energy is not R magen any point inside the solenoid. Now, the first integral on surprisinginviewofthe equivalencebetweenenergyand the r.h.s. is zero because of the boundary condition Eq. mass,establishedintheTheoryofSpecialRelativity. On (3.3), and the second vanishes because Tij nixj falls off thecontrary,thecontributionM fromthestressesrep- mag str as R−5. Therefore, as promised, we obtain resents a true General Relativistic effect. In the next Sectionit will be proventhat M is alwayszero,at me- str M =0. (3.6) chanical equilibrium. str Theconclusionisthat,independentlyoftheshapeofthe solenoid and of the detailed distribution of the stresses III. THE CONTRIBUTION FROM STRESSES insideitswalls,thegeneralconditionsofmechanicalequi- librium as encoded in Eqs. (3.1) and Eqs. (3.3) imply Tobe definite,weimaginethatthe solenoidS ishang- that the combinedcontribution of Maxwelland material ing by a suitable set of threads, and that apartfrom the stresses to the total gravitational mass of the solenoid suspension points its surface is free. Now, upon taking vanishes. Therefore,thetotaleffectivegravitationalmass thespatialdivergenceofbothsidesofEq. (2.4),andthen associated with the magnetic field is equal to M : magen using the gauge condition Eq. (2.6), we obtain M =M . (3.7) eff magen ∂ (Tij + Tij )=0. (3.1) i walls mag The gravitational field that is observed far from the Outside the solenoid walls, where Tij = 0, the above solenoid when the magnetic field is turned on is then walls equaltothatofapointchargewithmassM ,placed equations are satisfied as a consequence of the static magen at the position of the solenoid. Maxwell Equations in vacuum: Obviously, Eq. (3.6) does not imply that the magnet- ∇·B=0, ∇×B=0. (3.2) ically generated stresses produce no gravity at all, because it only states that Maxwell and material stresses Inside the solenoid walls, instead, Eqs. (3.1) express the cancel each other on average, namely after integrating localbalancebetweenelectrodynamicforcesandmaterial over all space. While this is sufficient to conclude that stresses, at mechanical equilibrium. At points on the stressesdo not contribute to the far-field,it still remains boundary ∂S of the solenoid walls, Eqs. (3.1) must be the possibility that stresses produce significant gravita- supplemented by the following boundary condition tional effects in the vicinity of the solenoid, for the near field probes also the detailed spatial distribution of the n (Tij +Tij )| =n Tij | , (3.3) stresses. Thestudyofthenearfieldisclearlymuchmore i walls mag ins i mag out complicatedingeneral,becauseitrequiresadetailedde- where ni is the normal to the surface of the solenoid terminationofthe mechanicalstressesinside the wallsof walls,orientedoutwardsthesolenoid,andthesuffixesins the solenoid. The study of the stresses that arise in a (out) denote the values of the fields immediately inside solenoid generating a strong magnetic field has received (outside) the solenoid walls. Eq. (3.3) expresses the fact much attention in the literature over the years, in view that the total electrodynamic self-force on the solenoid of its greatpractical importance (see for example Ref.[8] is zero, and therefore the threads that support it do not and References therein), and in general it is a difficult apply any extra force when the magnetic field is turned problem, that involves making a definite model for the on. Using Eq. (3.1) and the boundary condition Eq. constitutive equations characterizing the material, and (3.3), we can now show that M is always zero. For it usually requires numerical tools. We shall not discuss str this purpose, we note that at all points not lying on the thisdifficultproblemhere,andwecontentourselveswith boundary ∂S of S, Eq. (3.1) implies the identity: a few simple considerations that can be drawn on the basis of general mechanical equations, without any con- (Tii +Tii )=∂ [(Tij +Tij )xi]. (3.4) sideration of specific constitutive equations. To simplify walls mag j walls mag the problem, we consider below a very long cylindrical Upon substituting this expression for Tii +Tii into solenoidandwe discussseparatelythe gravitationalfield walls mag ther.h.s. ofEq. (2.17),andthenperformingtheintegral outside and inside the solenoid. 5 A. The external near field AddingupEq. (3.10)andEq. (3.11),weobtainforEext mag the expression: We consider, as in Ref.[3], a very long cylindrical πR B2 solenoid, constituted by a (non magnetic) pipe with in- Eext ≃ 2A+ 1 in R3 (3.12) ner and outer radii R and R respectively, and length mag L 8π 1 1 2 (cid:18) (cid:19) L ≫ R . We suppose for simplicity that the electric 2 current producing the magnetic field flows along the in- On the other hand, the internal magnetic energy Eminatg ner surface of the pipe, in the positive azimuthal direc- can be estimated to be tion, and that it has a uniform surface density j. We B2 let {x,y,z} a cartesian coordinate system whose z axis Eint = in ×πR2L, (3.13) mag 8π 1 coincides with the solenoidaxis, and whose originlies at the center of the solenoid, and we let r = x2+y2 the and therefore we obtain for the ratio of Eext /Eint the distance from the solenoid axis. mag mag p estimate: Axial symmetry obviously implies that the effective massdensityρeff inEq. (2.13)isafunctiononlyofr and Emexatg 2A R1 R1 2 z. As a first step, we show that ρ is significantly dif- = + , (3.14) eff Eint π L L ferent from zero only inside the solenoid, i.e. for r ≤R mag (cid:18) (cid:19) 2 and|z|≤L/2. Thisisobviousforthecontributiontoρ eff which shows that Eext becomes negligible with respect arising from the material stresses, because Tij vanish mag walls to Eint for R /L≪1. outside the solenoid walls. Then, upon noting that mag 1 ConsidernowapointP inthe vicinity ofthe solenoid, Tii =E , (3.8) but far from its ends. The above estimation of the mag mag external magnetic stresses and energy shows that the ascanbeseenbytakingthetraceoftheMaxwellstresses magnetically-generated gravitational field at P is deter- in Eq. (2.9), we see that the contribution to ρeff arising mined by the stresses and the magnetic energy that are from the magnetic field is equal to twice Emag/c2. We present inside the solenoid and within its material walls. can estimate the integral of Emag outside the solenoid as Since far fromthe solenoid’s ends the magnetic field and follows: the external magnetic field coincides with the the material stresses are approximately independent of field of a cylindrical magnet having length L and radius the z coordinate,we see fromEq. (2.12) that the field ψ R1, carryingauniformmagnetizationm=j/calongthe atP coincideswith the classicalfieldofaninfinite cylin- positivez-direction. Thefieldofsuchamagnetcoincides drical rod, with a uniform linear mass density σ equal eff with the sum of the fields B1 and B2 produced by the to: oppositesurfacedistributionsofmagneticchargesonthe oppositecapsofthemagnet(atz =±L/2),withuniform 1 R2 σ = dr2πr E +Tii +Tii . (3.15) surface densities σm = ±j/c. The total energy Emexatg of eff c2 Z0 mag walls mag the external field can then be estimated to be (cid:0) (cid:1) Now, we can split σ analogously to what we did with 1 1 eff Eext = d3x(B2+B2)+ d3xB ·B . M in Eq. (2.15): mag 8π 1 2 4π 1 2 eff Zoutside Zoutside (3.9) σ =σ +σ , (3.16) eff magen str Thefirstintegralonther.h.s. oftheaboveEquationrep- resents the sum of the magnetic energies of two isolated where pole distributions at z = ±L/2. Therefore, it is independent of the solenoid length L, and on dimensional 1 R2 E˜ mag grounds one expects it to be of the form: σmagen = c2 dr2πrEmag ≡ c2 , (3.17) Z0 1 B2 d3x(B2+B2)= in 2AR3 , (3.10) with E˜ the magnetic energy per unit length of the 8π 1 2 8π 1 mag Zoutside solenoid, and where B = 4πj/c is the magnetic field inside the in solenoid, and A is some numerical constant. As for the 1 R2 σ = dr2πr Tii +Tii . (3.18) secondintegralonther.h.sofEq. (3.9),itrepresentsthe str c2 walls mag Z0 interaction energy among the two poles of the magnet, (cid:0) (cid:1) and it can be approximated as the interaction energy of We can easily see that σstr vanishes. Indeed, neglect- two opposite point-like magnetic charges of magnitude ingthe contributionstoMstr fromthe externalmagnetic q =πR2j/c at distance L: field, which we have seen to be small, as well as the con- m 1 tribution from the small region near the solenoid’s ends, 1 d3xB ·B ≃ qm2 = 2π2j2R14 = Bi2nR14 . we can then express Mstr as 4π 1 2 L c2L 8L Zoutside (3.11) M =Lσ . (3.19) str str 6 Since, according to Eq. (3.6) M is zero, it follows at the regionofspaceoccupiedbythe solenoidwallsis neg- str once ligible, we have σstr =0. (3.20) E˜minatg(R1)≃E˜mag , (3.24) Weconcludethatalsonearthesolenoidthemagnetically- and therefore Eq. (3.23) implies that a test mass placed generated gravitational field ψ is independent of the immediately inside the solenoid would feel an oscillating stresses, and it simply coincides with the field generated pull towardsthe solenoid’saxis that is twice as strongas byacylindricaldistributionofmass,havingalinearden- the pull observed just outside the solenoid: sity that is equal to the instantaneous magnetic energy stored in the solenoid (divided by c2) per unit length: F(R )=2F(R ). (3.25) 1 2 E˜ This result arises because, for r < R , the gravity origi- σ = mag . (3.21) 1 eff c2 natingfromMaxwellstresses(the secondterminsidethe brackets in Eq. (3.22)) is not screened by the negative These results are in sharp contrast with the findings gravity of the magnetic-induced stresses in the walls of of Ref.[3], where it was concluded that the contribution the solenoid. fromMaxwellandmaterialstressesisdifferentfrom zero, and of such a magnitude as to cancel the gravitational field produced by the mass-energy of the magnetic field, IV. THE NEWTONIAN NOISE resulting in a null magnetically-generated gravitational field ψ outside the solenoid. A detailed analysis of the Producing strong magnetic fields and designing sensi- sketchycomputationsin[3]showsthatthisincorrectcon- tive torque balances may not be enough to ensure that clusion arose from an incomplete evaluation of the ma- one would be able to actually observe the magnetically terial stresses that build up inside the solenoid walls, as generated gravitational field ψ. For that to be possible, the authors only considered the effect of the radial elec- one has to make sure that the newtonian noise δΦ trodynamic forces pushing to increase the radius of the walls is not exceedingly large compared to ψ. The order-of- solenoid, but they overlooked the existence of an axial magnitudeestimatepresentedbelowshowsthatthereare force tending to compress the solenoid [9]. When the little prospects of measuring ψ outside the solenoid, for contribution fromthe axialstresses is accountedfor, our we estimate that outside the solenoid δΦ is about result Eq. (3.21) is recovered. As a further check of the walls nine order of magnitudes larger that ψ. At the end, we fundamental Eq. (3.20), in the Appendix we providethe shall briefly comment on the chances of measuring ψ in- explicit formulae for the material stresses that build up side the solenoid, where the newtonian noise is expected within the walls of an idealized solenoid. to be much smaller. As we pointed out in the previous Sections, δΦ walls comes about because electrodynamic forces deform the B. Internal field solenoid walls, resulting in a change of shape and density of the walls. An accurate determination of δΦ walls We consider now the gravitationalfield in the vacuum requiresa detailed model for the solenoid,and is beyond region in the interior of the solenoid, i.e. for r < R . 1 thescopeofthepresentpaper. Weshallcontentourselves Since Tij is zero for r <R , the field ψ coincides with walls 1 with simple considerations based on order-of-magnitude the classicalpotentialgeneratedbyalinearmassdensity estimates. σint(r): eff We consider separately radial electrodynamic forces, 1 r that tend to increase the radius of the solenoid, and ax- σint(r) = dr′2πr′ E +Tii . (3.22) ialelectrodynamicforces,thattendtomakethesolenoid eff c2 mag mag Z0 shorter. Radial forces were the only source of newto- (cid:0) (cid:1) Differentlyfromtheexternalregion,intheinteriorofthe nian noise that was considered in [3], because, as ob- solenoid Maxwell stresses are not screened by material served earlier, the authors did not take account of the stresses,andthereforethey do contribute to the internal axialcompressionofthesolenoid. Inprinciple,radialde- gravitationalfield. Upon recalling that Tii =E , see formations are innocuous because, for a perfectly cylin- mag mag Eq.(3.8), we see that Maxwell stresses contribute to the dricalsolenoid, a symmetric radialdeformationdoes not internal field as much as magnetic energy, and then we alter the axial mass-density of the solenoid, and there- can rewrite Eq. (3.22) as fore it produces no newtonian noise. Real solenoids of course are not perfectly symmetrical, and therefore one 1 r 2E˜int (r) expects that slightly asymmetrical radial deformations σeinfft(r) = c2 dr′2πr′2Emag ≡ mca2g . (3.23) will actually produce some noise. A possible remedy for Z0 this problem was pointed out in Ref. [3], and consists in It is interesting to consider a solenoid with thin walls. averaging over azimuthal inhomogeneities in the radial Since in such a case the magnetic energy contained in deformation, by setting the solenoid in rotation around 7 its axis, with an angular frequency much larger than the and we see that the newtonian noise is over nine order modulation frequency of the magnetic field. magnitudeslargerthanthe magnetically-generatedfield. As we shall now see, the real trouble comes from the This elementary analysis shows that it will be ex- axial compression of the solenoid. To estimate the new- tremely difficult to observe the oscillating field ψ out- toniannoiseintroducedbythis compression,weconsider side the solenoid. However, the newtonian noise should a cylindrical long solenoid of length L, whose walls have be much less of a problem inside the solenoid, which we a cross-sectional area A . We assume for simplicity showed to be the interesting region for the purpose of walls that the axial compression Tzz is uniform throughout testing the gravity of stresses. This is so because, in the walls the section of the walls, and that it does not exceed the idealcase of an infinitely long and perfectly axially sym- elastic limit of the material. If we let F the total axial metric solenoid, the newtonian noise inside the solenoid ax compression is strictly zero. R2 F = dr2πrTzz , (4.1) ax walls ZR1 V. CONCLUSIONS from Hook’s law we estimate that the length of solenoid will suffer a fractional change of magnitude: According to General Relativity, stresses act as a source of gravity on the same footing as energy. While δL 1 F =− × ax , (4.2) stress-generated gravity is normally negligible, it is L E Awalls though to play an important role in astrophysics, where where E is the Young modulus for the material of walls. itcontributesto determining the maximummassofneu- In the Appendix we show that, sufficiently far from the tron stars, and it is perhaps determinant in cosmology, end-points, the axial compression F has magnitude: where ”negative” pressure-generatedgravity may be the ax cause of the recently discoveredacceleratedexpansionof Fax =E˜mag. (4.3) the Universe. The importance of these problems makes it highly desirable to design a laboratory test, still lack- Using this formula in the r.h.s. of Eq. (4.2), we obtain ing as we write, to verify if stresses actually gravitate anestimate of the relativechange inthe solenoidlength: as predicted by GeneralRelativity, or not. A test of this δL 1 E˜ sortwasproposedlongagoin[3],anditinvolvedmeasur- mag =− × . (4.4) ing the gravitational pull on a test mass placed outside L E A walls a long solenoid, carrying a slowly alternating current. Considernowthe totalmassσwalls per unit lengthofthe The conclusion was that in General Relativity the oscil- solenoid. Obviously, under a change δL in the solenoid lating magnetic field inside the solenoid produces a null length, σwalls changes by the amount gravitational force on the test mass, because the attractive gravity generated by the energy and stresses of the δL δσ =− σ . (4.5) magnetic field was found to cancel against the negative walls walls L gravity generated by the material stresses that build up Then, from Eq. (4.4) we obtain: inside the solenoid walls. In this paper we demonstrated that this result is incorrect, as it hinges on a mistaken E˜mag σwalls E˜mag analysis of the material stresses, in which the electrody- δσ = × = ρ , (4.6) walls walls E A E namic axial compression of the solenoid was overlooked. walls After amending this mistake, we found that the contri- whereρ =σ /A isthemassdensityofthema- walls walls walls bution to the external gravitational field from Maxwell terial for the walls. Having estimated the change δσ walls stressesandmaterialstresseswithinthewallscanceleach in the linear mass-density of the solenoid, we can eas- other,andthereforetheresultinggravitationalfieldisde- ily obtain an estimate for the ratio ψ/δΦ among the walls termined solely by the linear density of magnetic energy magnetically-generated field ψ and the newtonian noise. stored inside the solenoid. Thus observation of the ex- Since the former is proportionalto σ and the latter to eff ternalfieldcannotbe usedtotest the gravityofstresses. δσ , we find walls Moreover, observing this field is extremely unlikely be- ψ σ E˜ E E cause of the enormous newtonian noise that results from eff mag = = × = , smallchangesinthe lengthofthe solenoidcausedbythe δΦwalls δσwalls c2 E˜magρwalls c2ρwalls axial electrodynamic compression. (4.7) Theinterestingregionfortestingthegravityofstresses whereinthesecondpassageweusedEq. (3.21). Itshould is the one inside the solenoid, because there the gravity be noted that the result is independent of the strength of Maxwellstresses is not screenedby the gravityof ma- ofthe magneticfield. Inthe caseofstainlesssteel,which terialstressesexistinginthesolenoidwalls,andtherefore has E =2×1011 N/m2 and ρ=8g/cm3, we obtain: theycontributeasmuchasthemagneticenergyingener- ψ ating gravity. In the internal region the newtonian noise =2.8×10−10 , (4.8) δΦ should also be much less of a problem, because in the walls 8 ideal case of a long solenoid, with perfect axial symme- To obtain it, we observe that far from the end points, try, newtonian noise is zero. The major experimental stresses are independent of z, and therefore Eqs. (3.1) difficultythatweforesee,apartfromcontroloftheresid- reduce to: ual noise resulting from asymmetries of the solenoid, is to find means of accurately measuring the gravitational ∂k(Twjkalls+ Tmjkag)=0. (6.4) field in the presence of strong magnetic fields. k=x,y X Therefore, we have the identity: Acknowledgments (Tjj +Tjj )= ∂ [(Tjk+Tjk )xj]. (6.5) sol mag k sol mag j=x,y j,k=x,y X X The authors would like to thank Prof. K.S. Thorne forvaluablecommentsthathelpedgreatlyimprovingthe Upon integrating both sides of the above equation on a manuscript. TheworkofL.Rosahasbeenpartiallysup- cross section Σ of the solenoid, we obtain: ported by PRIN FISICA ASTROPARTICELLARE. R2 dr2πr (Tjj +Tjj )= walls mag Z0 j=x,y X VI. APPENDIX 2π In this Appendix we provide the explicit formulae for =R dθ (Tjk +Tjk )xjnk| . (6.6) 2 walls mag r=R2 thestressesthatbuildupwithinthewallsoftheidealized Z0 j,k=x,y X solenoidconsideredinSec. 3.A,consistingofacylindrical pipe carrying a uniform azimuthal current concentrated Theintegralonther.h.s. vanishes,becauseTjk nk| walls r=R2 on its inner face. The expressions presented below pro- is zero in view of Eq. (3.3), while Tjk nk| vanishes mag r=R2 vide an explicit verification of the important formulae, because the magnetic field is negligible outside a long Eq. (3.20) and Eq. (4.3). solenoid(seethediscussionoftheexternalfieldfollowing Weconsiderfirsttheeffectoftheradialmagneticpres- Eq. (3.7)). Therefore, the l.h.s. of Eq. (6.6) is zero and sure P, on the inner face of the pipe. Far from the this proves Eq. (6.3). Indeed, it is easy to verify that solenoid’s ends, P is uniform and its magnitude is equal Eq.(6.3) is satisfied by the explicit expressions for the totheradialcomponentoftheMaxwellstresstensorTmrrag transverse material stresses given in Eqs. (6.2), together inside the solenoid: with the Maxwell stresses Eq. (2.9). By using Eq. (6.3), we can now easily obtain F . To B2 ax P = in . (6.1) do this, we recall the identity: 8π R2 This radial pressure determines transverse stresses dr2πr(Tii +Tii )=0, (6.7) walls mag Trr (r) and Tφφ (r) in the pipe’s walls, whose expres- Z0 walls walls sions are well known [10] and read: which is a direct consequence of Eq. (3.20). Upon sub- tracting Eq. (6.3) from Eq. (6.7), we then obtain: R2 R2 Trr =P 1 2 −1 , walls R2−R2 r2 R2 2 1 (cid:18) (cid:19) dr2πr(Tzz +Tzz )=0. walls mag Z0 R2 R2 It follows from the above Equation that Tφφ =−P 1 2 +1 . (6.2) walls R2−R2 r2 2 1 (cid:18) (cid:19) R2 R2 F = dr2πrTzz =− dr2πrTzz . (6.8) Besides these transverse stresses, the magnetic field de- ax walls mag ZR1 Z0 termines alsoaxialstressesTzz (r) inside the walls. We walls derive below the average value F of Tzz , as given in Upon using into the r.h.s of the above formula the ex- Eq. (4.3). Inviewofthekeyrolepalxayedbwyaltlsheaxialcom- pression of Tmzzag inside the solenoid: pression F , and in order to explain its physical origin, ax we provide two different derivations of Eq. (4.3). The Tzz =−Bi2n =−E , (6.9) first derivation is based on the general equilibrium con- mag 8π mag ditions Eq. (3.1). Indeed, using Eq. (3.1), one canprove we immediately obtain Eq. (4.3). thefollowingidentityholdingfarfromthesolenoid’sends Inorderto clarifythe physicaloriginofthe axialforce F ,itisusefultoprovideamoredirectderivationofEq. R2 ax dr2πr (Tjj +Tjj )=0. (6.3) (4.3). For this purpose, we consider the cylindricalsheet walls mag Z0 j=x,y ΣofradiusR1 andheightLthatcontainsallthecurrent X 9 flowinginthesolenoid,andweimaginesplittingitintwo integral can be rewritten as: parts Σ and Σ , consisting respectively of the points of 1 2 Σ that lie above and below a plane of equation z =z¯. If 2πj2R2 2π zcosθ dF(Amp) =− 1 dz dz dθ , we imagine Σ1 and Σ2 as consisting of a largenumber of z c2 1 2Z0 z2+2R12(1−cosθ) closed circular current loops, it is clear by Ampere’s law (6.11) that an attractive axial force F(Amp)(z¯) arises between p with z =z −z . Since the integrand is positive, we see 1 2 Σ and Σ , and we show below that for z¯ far from the 1 2 thatthe tworingsattracteachother,as expected. Upon end points F(Amp)(z¯) has a constant magnitude equal to integrating over z and z we then obtain for F(Amp)(z¯) E˜ . 1 2 z mag the expression Indeed,axialsymmetryimpliesthatF(Amp)(z¯)isalong the z axis, and we let F(el)(z¯) its z-component. Now, 2π2R2j2 z F(Amp)(z¯)=− 1 I(z¯), (6.12) Ampere’s law gives the following expression for the ele- z c2 mentary force dF(Amp)(z ,z ) between two infinitesimal z 1 2 where I(z¯) is the integral circular current loops within Σ and Σ : 1 2 dj dj ~x −~x 1 L/2 z¯ 2π zcosθ dF(Amp)(z ,z )=− 1 2 (d~l ·d~l ) 1 2 , I(z¯)= dz dz dθ . z 1 2 c2 I I 1 2 |~x1−~x2|3 π Zz¯ 1Z−L/2 2Z0 z2+2R12(1−cosθ) (6.10) (6.13) where dj = jdz , and d~l are line elements tangential The integrals over z and z in Ip(z¯) can be done by the i i i 1 2 to the surface elements, and parallel to the surface cur- change of variables (z ,z ) → (w,z), where w = (z + 1 2 1 rent density~j. Using cylindrical coordinates, the above z )/2. The result is: 2 1 2π 2z¯ 2z¯ 2 R2 2z¯ 2z¯ 2 R2 I(z¯)= dθ log 1− + 1− +8 1(1−cosθ) +log 1+ + 1+ +8 1(1−cosθ) + π   L s L L2   L s L L2  Z0  (cid:18) (cid:19) (cid:18) (cid:19)      R2 1 −log 1+ 1+2 1(1−cosθ) − log(1−cosθ) cosθ (6.14) L2 2 " r # ) where we omitted a few terms that are zero upon inte- This formula can be conveniently expressed in terms of grating over θ. The positive quantity I(z¯) reaches its the magnetic energy density by noticing that inside a maximumvalueatthe centerofsolenoid(forz¯=0),and long solenoid, the strength of the magnetic field B is in monotonicallydecreasestowardszerowhenz¯approaches related to the surface current density as the end-points at ±L/2. For a long solenoid, R /L≪1, 1 4πj and far from the end points, (L/2−|z|)/R1 ≫ 1, I(z¯) Bin = . (6.17) becomesindependentofz¯anditslimitingvalueforanin- c finitely long solenoid can be obtained by observing that Using this formula, we can rewrite Eq. (6.16) as: for R /L → 0 the first three terms between the curly 1 brackets of the above integral become independent of θ B2 lim F(Amp)(z¯)=πR2 × in =E˜ . (6.18) andtherefore,aftermultiplicationbycosθ,theyintegrate R1/L→0 1 8π mag to zero, leaving us with Obviously, since the current sheet Σ is anchored to 2πdθ the inner face of the solenoid, the electrodynamic force lim I(z¯)=− cosθlog(1−cosθ) =1. (6.15) R1/L→0 Z0 2π Fz(Amp)(z¯) compresses the pipe and, at mechanical equilibrium, it is balanced by the axial stresses Tzz inside Upon inserting this value into Eq. (6.12), we see, as walls the pipe walls: expected, that in the limit of a long solenoid, and for z¯ far from the end-points, the current sheets Σ1 and Σ2 Fax(z¯)=F(Amp)(z¯). (6.19) attract each other with a force of magnitude It should be noted that according to this formula the 2π2R2j2 compressionF (z¯) vanishes atthe ends ofthe pipe, and lim F(Amp)(z¯)= 1 . 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