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J. [Am. physics. where , H ) h esnre Meissner The 1 c < 1 hspeoeo sdsic rmta of that from distinct is phenomenon This ( T .INTRODUCTION I. 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FIG. nldn h oeho e evidenc Josephson experimental the much including by an supported parameter is realization order mechan- supercurrent complex the quantum The the explained between relation be The metal. only ically. can normal supercurrent the the in whic of phonons resistivity and impurities finite of presence give the p of superconductor spite the in sible in cond current This dissipationless must superconductor. makes the parameter of tion order bulk the the con- in of valued the single phase be The of function pairs. wave Cooper densed macroscopic the describes which ota ntebl ftesprodco h antcflxden flux magnetic the superconductor the superco of the bulk from the expelled in is that field so magnetic applied line, the safnto ftemperature of function a as antcfil ssonas shown met is the the field in penetrates magnetic state a field superconducting applied the to state, temperature normal transition the In zero. is edfo h uko h uecnutraantti a,it law, this against superconductor magnetic the the of expel bulk To the flux. from magnetic field the in change zero. the to works magnetic resist law value Lenz’s the finite electromagnetism, process classical from to this changes According superconductor In the and in expelled. spontaneously, flux is flow field to magnetic superconduct- begins the to supercurrent side the normal side, from ing 1 Fig. in shown boundary e Meissner the explain deeg lrfistene o unu mechanics quantum for need the clarifies energy ld h xsec ftesprurn,hwvr snteog to enough not is however, supercurrent, the of existence The H ff htteMise e Meissner the that tosit oprpisi etudrto sa as understood best is pairs Cooper into ctrons c c seegtclyfvrbedet h energy the to due favorable energetically is ect (0) H oaa oy 5-92 Japan 153-8902, Tokyo Komaba, Superconducting state sciia odsigihn super- distinguishing to critical is d B =0 ff c sepandb classical by explained is ect ff c.We ytmcosstephase the crosses system a When ect. ff T T c ect. . ntesprodcigsaebelow state superconducting the In . caisadpoint and echanics 3 H Normal state c ( T ssonb hc line thick a by shown is ) fthe of T c bec of absence nductor l The al. sity T os- to i- B d h e 2 is required that the superconducting state in which the mag- is the energy of the magnetic field applied externally with µ0 8 netic field is expelled has lower free energy than the state in being permeability of the vacuum, , B = µ0 H is the magnetic which the magnetic field remains in the bulk. In the absence flux density with H being the applied magnetic field and of the magnetic field, the superconducting state has lower free energy than the normal state provided T < T , where T is the 1 2 c c Ek = n(r)mv (r)dV (3) transition temperature to the superconductivity. In the pres- Z 2 ence of the magnetic field, superconducting state that expels is the kinetic energy of the (super) current. In this equation m the magnetic field competes with the normal state in which the is the mass of an electron, n(r) is the number density of elec- magnetic field remains. Superconductivity cannot be realized trons, and v(r) is average velocity of electrons. This energy when the magnetic field penetrates the bulk, so the supercon- can be rewritten as follows. The supercurrent density j(r) is ductor needs to expel the magnetic field. However, expelling expressed as the magnetic field costs energy, so when the applied magnetic field becomes strong enough, the energy of the superconduc- j(r) = n(r)ev(r) . (4) tor becomes higher than that of the normal state, and the su- perconductivity is destroyed. This is the origin of the critical It is also related to rotation of the magnetic flux density owing field. to Ampere’s law that is valid for a steady current: The lowering of the energy in the superconducting state comes from the condensation energy of the electrons into ∇× B(r) = µ0 j(r). (5) Cooper pairs. It is described by the BCS theory,4 which is en- tirely based on the quantum mechanics. The temperature de- Using these equations the kinetic energy is expressed as pendence of the critical field observed experimentally is quan- 1 2 2 titatively explained by the theory. Thus, quantum mechanics Ek = λ (∇× B) dV, (6) is indispensable for the explanation of the Meissner effect. 2µ0 Z In spite of all this, Ess´en and Fiolhais5 claimed that the = 2 Meissner effect can be explained by classical mechanics.6 where λ m/µ0ne is the London penetration depth. Now From our argument above, it is evident that their claim is in- de Gennes assumesp Es to be a constant, and minimizes the + correct. It is however instructive to examine their claim and energy EB Ek with respect to the field distribution B(r), and understand the flaw of their argument. In the next section, obtains the London equation: we introduce their argument. In Sec.III, we see how several + 2∇× ∇× = strange conclusions are derived if we accept their argument. B λ ( B) 0 , (7) In Sec.IV we clarify the flaw of their argument and find that as a condition to minimize the energy. A solution gives a state they treat energy of the magnetic field incorrectly. We need in which supercurrent flows at the surface layer of the conduc- only classical electromagnetism to understand the flaw in their tor, the thickness of the layer being of the order of λ,9 and the argument. No understanding of quantum mechanics or super- magnetic field in the bulk is eliminated. This situation is iden- conductivity is necessary to understand our arguments in this tical to the state realized by the Meissner effect. The essence section. of the argument is that the energy of the magnetic field EB is minimized by making B = 0 in the bulk. In this deriva- II. ARGUMENTBY ESSEN´ AND FIOLHAIS tion quantum mechanics is not used at all. Ess´en and Fiolhais claim based on this argument that the Meissner effect can be explained classically. However, this coincidence is superficial ff Ess´en and Fiolhais try to persuade us that Meissner e ect is as we will see later. a property of a perfect conductor. So let us forget the fact that the supercurrentcan only be explained quantum mechanically, and try to examine whether or not a perfect conductor shows III. STRANGE CONSEQUENCES the Meissner effect. Their argument is based on a description in chapter 1 of a famous textbook by de Gennes,7 where the If this argument by de Gennes is correct we encounter var- London equation is derived using only classical mechanics. ious strange conclusions. We point out some of them in this The derivation is not appropriate as we will see in Sec.IV, but section. we repeat the argument here. (1) In the argument in Sec.II, the condensation energy of For simplicity, the discussion will be limited to T = 0. It the superconductorhas no role. Thus there is no reason for the is assumed that total energy of superconductor in a magnetic superconducting state to be destroyed by an arbitrarily strong field is given by magnetic field. It cannot explain the existence of the critical E = Es + EB + Ek, (1) field Hc above which the superconductivity is destroyed and the magnetic field penetrates the bulk. where Es is the condensation energy of electrons, which ff Furthermore, unphysical latent heat must exist at the phase comes from quantum mechanical e ects. The second term, transition point. Let us consider what happens at the phase 1 2 transition point. We apply a strong enough magnetic field to EB = B dV, (2) 2µ0 Z a sample, and cool it to a temperature near absolute zero. The 3 magnetic field is stronger than the critical field, so the mag- cylinder, supercurrent flows along the surface of the cylinder, netic field penetrates the sample, and the sample is in the nor- and the field in the bulk is canceled. This situation is like a mal state. (See Fig. 1.) Then we reduce the magnetic field. solenoid used to cancel the magnetic field inside. Now be- At H = Hc(T), the sample transitions to a superconductor, cause the current flows only within a thin region at the surface and the magnetic field is expelled. Let us consider a cylin- of the cylinder, nothing will change when a hole is introduced drical sample with radius R, length l, and volume V = πR2l. along the axis of the cylinder. Namely, we will have the same The magnetic field is applied parallel to the axis of the cylin- situation when we replace the bulk cylinder by a hollow tube; der. We investigate what happens to the energy in Sec.II at the Ek stays the same and EB is independent of whether the space transition. Before the transition, there is no current, and the is filled with metal or not. However, experimentally the bulk magnetic field stays in the sample. The energy of the mag- cylinder and the hollow tube behave quite differently. For the (n) (n) netic field EB and the kinetic energy Ek in the normal state hollow tube the magnetic field is not expelled from the hole are easily calculated using eq.(2) and eq.(3), respectively. The in the tube. The field remains as quantized flux with value (n) 2 (n) answers are EB = (1/2µ0)B V, and Ek = 0. quite close to that expected from a normal metal tube. The (s) (s) Next we calculate EB and Ek in the superconducting argument in Sec. II cannot explain this experiment. state. The supercurrent flows at the side wall of the cylinder (5) As we stated in (4) the bulk superconductor can be re- and cancels the magnetic field in the bulk. The current flows placed by a superconducting solenoid whose leads are con- in a thinskin layer ofthe orderof λ. Since λ is typically small, nected together. The supercurrent flows in the solenoid and of the orderof 10−7 m, we can consider that the magnetic field cancels the magnetic field inside. We can easily calculate the is almost completely expelled from the volume V for a sample value of the current flowing. For a solenoid with winding (s) (s) with R ≫ λ. Thus, EB = 0. On the other hand, Ek is not number n per unit length, I0 = B/nµ0. Now what happens zero in this phase. From the London equation eq.(6), we know when the superconductor connecting the leads is replaced by that the magnetic field decays at the surface as B exp[(r−R)/λ] a normal conductor. This replacement can easily be done by with 0 ≤ r ≤ R being the distance from the central axis of raising the temperature locally. We do the operation at time the cylinder. Using this magnetic field, we obtain |∇ × B| ≃ t = 0. In this case, the current I decays in time according to (s) the equation of the classical electrodynamics. (B/λ) exp[(r − R)/λ]. Then Ek is calculated by eq.(6) as 2 dI + = (s) = B L RI 0, (9) Ek exp[2(r − R)/λ]dV dt 2µ0 Z B2 R with the solution ≃ 2πRl exp[2(r − R)/λ]dr 2µ0 Z0 R I(t) = I0 exp − t , (10) B2 B2 λ  L  = πRlλ = V . (8) 2µ0 2µ0 R where L is the self-inductance of the solenoid, and R is the resistance of the normal part. As the current decays, the mag- In the second line we used the fact that the integrand has non netic field penetrates inside the solenoid. At the same time zero value only at r ≃ R. Gathering these results we find that Joule heating occurs in the normal part. The total amount of (n) = (n) + (n) = / µ 2 energy in the normal state E EB Ek (1 2 0)B V is the Joule heat generated is easily calculated. It is same as the (s) = (s) + much larger than that in superconducting state E EB amount of the magnetic field energy that is introduced inside (s) = 2 (n) (s) = 11 Ek (1/2µ0)B V(λ/R), the ratio being E /E R/λ ≫ 1. the solenoid. According to the argument in the previous sec- The difference of these energies and the condensation energy tion, the energy of the magnetic field is increased, since the Es that is neglected in Sec.II must be compensated at the tran- solenoid is filled with the field. At the same time the same sition point in the form of a latent heat of the phase transi- amount of the energy is dissipated as Joule heat. Where does tion. However, such a latent heat is not observed experimen- this energy come from? How can the state realized as the tally. Actually, the latent heat of a phase transition comes from minimum energy state change spontaneously to a higher en- change in the entropy, so it must be zero at T = 0 owing to the ergy state? We cannot answer these questions, if we accept third law of thermodynamics. The argument of the previous the argument in the previous section. section has no experimental support. (2) A perfect conductor has no mechanism to dissipate en- ergy. Thus there is no process to bring the system to the ther- IV. THE ORIGIN OF THE FLAW mal equilibrium state in which the free energy takes its min- imum value. The state in which EB = 0 in the bulk is never As we have seen in the previous section, the argument by realized, if we start from a state in which EB > 0. Ess´en and Fiolhais brings consequences incompatible with ex- (3) In the derivation of the London equation, the fact that periments. Here we explain what is wrong in their argument. the metal is in superconducting state is not used at all. There- We need only knowledge of the electromagnetism to pinpoint fore, if this derivation is correct, any metal must show the their flaw. Meissner effect.10 The essence of their argument is that the Meissner effect (4) Let us consider a long cylindrical bulk superconductor. occurs to lower the magnetic field energy EB in the volume of When the external magnetic field is parallel to the axis of the the conductor. They believe that this energy can be reduced 4 by flowing current at the surface of the conductor. However, plastic material, and we can expand the radius while keep- this is wrong. As we stated in Sec.I, we need energy to cancel ing the field inside to be zero. Since the wall of the solenoid 2 the magnetic field in the conductor. We calculate the energy is pushed from outside by the pressure P = B /2µ0, work is needed to removethe magnetic field in several ways, and show needed for this expansion. The work to realize a solenoid with 2 that it is positive. inner volume V is PV = (B /2µ0)V, the same work as before. We consider a situation in which a magnetic flux density B We have shown that even though a finite amount of mag- 2 is created by an external magnet. The field is assumed to be netic energy is stored in space with density B /2µ0, we need almost uniform in the space in which we do experiments. We the same amount of energy to eliminate this energy. For the bring a solenoid whose length is lA, the cross section is S and Meissner effect, this energy is supplied by the condensation winding number per unit length is n. We use this solenoid energy of the superconducting state. We cannot extract the several times, so we name it as solenoid A. We place this magnetic energy and utilize it without supplying extra energy. solenoid A parallel to the external field B, and try to cancel Ess´en and Fiolhais erroneously argued that the Meissner effect the magnetic flux density inside the solenoid, the volume of occurs to reduce this energy EB without work W. As we have which is V = lAS . It is noted that when the current flowing seen in Sec. III, their argument brings conclusions incompati- in the solenoid is zero, inner space of the solenoid contains ble with experiments. magnetic energy Finally, we explain the energy balance. To eliminate ex- ternal magnetic field B from the volume V, we need work 2 1 2 W = (B /2µ0)V. This work and the energy of the magnetic EB = B V . (11) 2µ0 field EB must be absorbed somewhere. We can show that the energy is absorbed in the electromagnetthat is creating the ex- Let us assume for simplicity that the solenoid A is made of ternal magnetic field B, or as potential energy of the magnet a perfect conductor, so the resistance is zero. The self induc- when the field B is due to a permanent magnet. For simplicity, tance of the solenoid A is let us consider the case in which the field B is created by an- other solenoid (which we label solenoid B) in which current L = µ n2V . (12) 0 IB is flowing. The field at solenoid A should be proportional to IB, so we In order to cancel the magnetic field inside, we need to flow write the relation as B = αIB. Since the magnetic flux that current I0 = B/µ0n. This current does not flow spontaneously. penetrates solenoid A is Φ = nlASB = αnlASIB, the mutual We need to apply electromotive force E(t) to the solenoid. The inductancebetween solenoid B and solenoid A is calculated to equation for the current is be M =Φ/IB = αnlAS = αnV. In the course of our increasing the current IA in solenoid A to cancel the magnetic field B, d 12 E(t) = L I(t) . (13) electromotive force EB is induced in solenoid B: dt dI dI E = M A = αnV A . (15) The work done by the electromotive force is B dt dt t2 t2 d The work done by this electromotive force on the constant W = E(t)I(t)dt = LI(t) I(t)dt Z Z current IB is t1 t1 dt 1 1 t2 = L I(t )2 − I(t )2 = B2V . (14) W = E I dt = αnVI I , (16) 2 2 1 2µ B Z B B B 0   0 t1 = = Here we used the condition that I(t ) = 0, and I(t ) = I = where IA(t1) 0 and IA(t2) I0 is used. Now we use 1 2 0 = = B/nµ0. The amount of energy required is the same as that of the relation αIB B and nI0 B/µ0. Then we obtain = 2 the magnetic field stored in the volume V, but energy must WB (B /µ0)V, which is just sum of the work we used to be supplied externally. Namely, in order to remove magnetic cancel the magnetic field in the solenoid W, Eq. (14), and the field energy EB from volume V, we need to input the same magnetic energy EB, Eq. (11), which existed in the volume. amount of energy to the solenoid. Where does sum of these Our argument is consistent with classical electrodynamics. 2 energies, B V/µ0 = W + EB, go? We will explain the destiny The situation in which the magnetic field inside the solenoid of this energy later. A is canceled by the current I0 is a higher energy state, as we We can calculate the energy another way. A magnetic field have emphasized several times. When we make the solenoid exerts force on an electric current. Thus, the current in a A to have finite resistance R, the work W spent to realize the solenoid is pushed by the magnetic field. This force is known situation is returned from the source of the magnetic field, i.e. as Maxwell stress, and the wall of the solenoid is pushed by from the power source of the solenoid B, as Joule heat gener- 2 pressure B /2µ0, the direction of which is from the high field ated in solenoid A. Actually, it is calculated as follows, side to low field side. Now we consider an infinitesimally ∞ ∞ R thin superconducting solenoid placed parallel to the external Q = RI(t)2dt = RI2 exp −2 t Z Z 0 L magnetic field B. Since the self inductance of the solenoid is 0 0   infinitesimally small, we can flow current to cancel the field = L 2 = 1 2 I0 B V , (17) inside without energy. Suppose that the solenoid is made of 2 2µ0 5

2 where I(t) = I0[−(R/L)t], Eq. (10), L = µ0n V, and I0 = V. CONCLUSIONS B/nµ0 are used. Since the state where the magnetic field is expelled has ff ff As is commonly understdoood, the Meissner e ect requires higher magnetic field energy, the Meissner e ect is possible quantum mechanics for its explanation. We have reviewed only when the condensation energy to the superconducting established evidence for this in the introduction. This fact state is larger than the magnetic field energy. This condi- was challenged by Ess´en and Fiolhais in a recent publication. tion determines the critical field Hc. For ordinary metals the We have explained their argument in Sec.II, and have seen magnetic susceptibility is quite small, so the free energy of in Sec.III that if we accept their argument, unphysical phe- the normal metal, Fn is almost independent of the magnetic nomena are anticipated. We clarified what is wrong in their field applied. If the free energy of the superconducting state argument. in the absence of the magnetic field is Fs, the critical field 13,14 Hc = Bc/µ0 is determined by the condition, 1 F + µ H2V = F . (18) ACKNOWLEDGMENTS s 2 0 c n In this equation V is the volume of the superconductor, and D.Y. acknowledges the hospitality of the Aspen Center for Fn is the free energy when the superconductivity is destroyed Physics, which is supported by the National Science Founda- by the magnetic field. This relation has been experimentally tion Grant No. PHY-1066293. He thanks Steven Girvin and confirmed. Allan MacDonald for critical reading of the manuscript.

1 W. Meissner and R. Ochsenfeld, “Ein neuer Effect bei eintritt der netic field, so that the susceptibility becomes zero in the thermal Superleitf¨ahigkeit,” Naturwisse. 21, 787-788 (1933). equilibrium state. 2 A. A. Abrikosov, Zh. Eksperim. i Teor. Fiz. 32, 1442-1452 (1957) 7 P.G. de Gennes, Superconductivity of Metals and Alloys (Adison- [English transl., “On the Magnetic Properties of Superconductors Wesley, Reading, MA, 1989). of the Second Group,” Soviet Phys. JETP 5, 1174-1182 (1957)]. 8 We use SI instead of cgs-system used in Refs. 5 and 7. 3 B. D. Josephson, “Possible new effects in superconductive tunnel- 9 For typical superconductors, London penetration depth is smaller ing,” Phys. Letters 1, 251-253 (1962). than 10−7 m. 4 J. Bardeen, L. N. Cooper, and J. R. Schrieffer, “Theory of Super- 10 It might be argued that a normal metal has dissipation process conductivity,” Phys. Rev. 408, 1175-1204 (1957). and inclusion of the process invalidates the argument in Sec.II. 5 H. Ess´en and M.C.N. Fiolhais, “Meissner effect, diamagnetism, However, it is easily seen that such an effect does not change the and classical physics – a review”, Am. J. Phys. 80 164-169, situation. If the resistance of the metal stops the current and favors (2012). the state with magnetic field, inclusion of the dissipative process 6 In addition to try to derive the Meissner effect classically, they try must lower the energy of the system compensating the cost of to disprove the Bohr-van Leeuwen theory. This is because once the energy EB. Such an energy must be originated in the region we admit the theorem we conclude that the Meissner effect can- where the current flows. However, since the magnetic energy EB not have a classical explanation. They point out several situations is proportional to the system volume, but the volume current flows where classical systems show magnetic response as evidence for is proportional to the surface area, inclusion of finite resistance the break down of the theorem. However, their argument is based cannot change the energy minimum condition in Sec.II. on incorrect understanding of the magnetic field energy. Further- 11 The calculation of the energy is given in Sec.IV. more, none of their evidence is magnetic response in the thermal 12 The reciprocity theorem of the electromagnetic induction is used equilibrium state. They are confusing dynamical response of the here. system with thermal response of the system. Their arguments do 13 M. Tinkham, Introduction to Superconductivity, 2nd ed. (Dover, not disprove the theorem which states that classical calculation Mineola, N.Y. 1996), pp.22-23. gives a partition function that is independent of the external mag- 14 C Kittel, Introduction to solid state physics, 8th ed. (J.Wiley, Hoboken, N.J., 2005), pp.270-273.