Electric field and electric forces in a spontaneously polarized nonpolar isotropic dielectric
aa r X i v : . [ c ond - m a t . o t h e r] A ug Electric field in spontaneously polarized isotropic dielectric
Maksim TomchenkoBogolyubov Institute for Theoretical Physics,14b, Metrolohichna Str., Kyiv 03143, UkraineE-mail: [email protected]
Abstract
As a rule, the electric field in a spontaneously polarized dielectric is determined from the Maxwellequations in a medium under the condition that the spontaneous polarization P s is known and coincideswith the total polarization P . However, such approach is not quite accurate, since the polarization P s arising in an ordered collection of dipoles creates some electric field E . The latter generates the inducedpolarization P i , which in turn affects the value of E . The field E is created by the spontaneous dipolesof the whole dielectric and, therefore, is nonlocal. As a result, P i and P s are generally not codirectedin a dielectric of finite sizes. Therefore, the account for P i changes the polarization P = P s + P i qualitatively. We construct a method of determination of the field E that takes the polarization P i into consideration self-consistently. For an isotropic dielectric, the account for P i leads to the change P s → P s /ε ( i ) in the formula E = E ( P s ) (where ε ( i ) is the dielectric permittivity of a dielectricin an external electric field). In this case, D = E + 4 π ( P i + P s ), P i = ε ( i ) − π E , D j = P l ε jl E l , ε jl = δ jl ε ( i ) + ε ( s ) jl , where ε ( s ) jl ( r ) is the effective dielectric permittivity associated with the spontaneouspolarization. Moreover, we find the exact solution E ( r ) for several typical examples and argue that, forthe ferroelectric, one should set P = P s + P i ( P s ) while minimizing of the thermodynamic potential. Keywords: spontaneous polarization, dielectric, ferroelectric.
As known, the polarization of dielectrics can be of two types: the polarization inducedby an external electric field and the spontaneous one [1, 2, 3, 4, 5, 6]. The spontaneouspolarization is characteristic of ferroelectrics, pyroelectrics, and piezoelectrics. It can alsoarise in “customary” dielectrics not related to the materials mentioned above. In particular,electric signals were observed in He II under torsion oscillations [7] and in standing half-waves of the second [8, 9, 10] and first [11] sounds. In those experiments, the electricfield can be associated only with the spontaneous polarization, since He atoms possess nointrinsic dipole or multipole moment.The main point is as follows. The spontaneous polarization P s ( r ) is usually consideredto be given. Taking it into account, the field E ( r ) is determined. However, such approachis not quite accurate. Indeed, the polarization P s ( r ) as a collection of dipoles creates theelectric field E ( r ) in the medium. This field, in turn, induces the additional polarization P i ( r ) (which is usually not codirected with P s ( r )). Such polarization affects the field E ( r ).The physical nature of the polarizations P s and P i is different. To find the field E ( r )correctly, we need to take P i ( r ) into account self-consistently. In what follows, we proposea method of determination of the field E ( r ) for such problems. For example, the gravity,acceleration, or sound wave induce the density gradient in a nonpolar isotropic dielectric.Due to the mutual polarization of two nonpolar atoms [12, 13, 14], this gradient createsthe spontaneous polarization P s in the medium. If P s ( r ) is unknown (in particular, for theferroelectrics), then E ( r ) can be determined with the help of additional reasonings (e.g.,the minimization of the thermodynamic potential). We emphasize that the field P i should e considered for any mechanism of spontaneous polarization. Apparently, this point wasnot clearly considered in the literature previously.Below, we consider isotropic dielectrics (fluids or gases) in more details and discussferroelectrics in brief. In the electrodynamics of continua (see [4], Chapters II, IV, IX) the electromagnetic field ofisotropic dielectrics is described by the local (i.e., valid at each point of a medium) relations D = E + 4 π P = ε E , (1) B = H + 4 π M = µ H (2)and by the Maxwell equations in a medium: div D = 4 πρ f , (3) rot E = − c ∂ B ∂t , (4) div B = 0 , (5) rot H = 1 c ∂ D ∂t + 4 π j f c . (6)Here, ρ f is the density of foreign charges, and j f is the foreign current.The spontaneous polarization of a dielectric arises due to the internal “nonelectric”mechanism causing the appearance of a set of dipoles (or multipoles) in the medium;we will call them “spontaneous” dipoles. We will see below that, in order to describea spontaneously polarized isotropic dielectric, one needs to make some changes in Eqs.(1)–(6). Let us deduce the necessary equations from the microscopic Maxwell equations.First, we find the formula for the polarization P ( r ), since the formulae available inthe literature are not quite accurate sometimes. The density of polarization charges ¯ ρ ( r ) , averaged over the physically infinitesimal volume, satisfies the relation Z V + ¯ ρ ( r ) dV = 0 . (7)Therefore, ¯ ρ ( r ) can be presented in the form ¯ ρ ( r ) = − div P ( r ), where P = 0 outsidethe dielectric. Here, V + is the volume confined inside the surface that covers the wholedielectric and oversteps its limits by an infinitesimal distance. In [4] (Chapt. 2, § P ( r )is determined with the use of the following formulae: Z V + r ¯ ρ ( r ) dV = − Z V + r ( ∇ P ( r )) dV = − I S r ( P ( r ) d S ) + Z V + P ( r ) dV = Z V + P ( r ) dV. (8)However, these formulae do not allow us to find P ( r ): relation (8) does not yield P ( r ) = r ¯ ρ ( r ). Indeed, let the polarization be uniform: P ( r ) = P . Then ¯ ρ ( r ) = − div P = 0 insidethe body. On the surface P ( r ) decreases by jump to zero, and ¯ ρ ( r ) is singular. For suchdielectric, the equality P ( r ) = r ¯ ρ ( r ) is violated at all points of the body. In this case, theintegral R V + r ¯ ρ ( r ) dV is defined by the surface part of ¯ ρ ( r ). These properties are natural,since a uniformly polarized body can be considered as two bodies that possess uniformlydistributed (+) and ( − ) charges and are shifted relative to each other by an infinitesimal istance. The jump of ¯ ρ ( r ) on the surface is usually large for the nonuniform polarizationas well. In this case, the equality P ( r ) = r ¯ ρ ( r ) should also be significantly violated.In order to obtain the formula for P ( r ) we note that, for a polarized dielectric withoutforeign charges, ρ ( r ) reads ρ ( r ) = N X j =1 [ q ( j ) δ ( r − r j ) − q ( j ) δ ( r − r j − r ( j )0 )] , (9)where N is the number of atoms in the dielectric. We set q ( j ) < j . Then Z V + r ρ ( r ) dV = − N X j =1 r ( j )0 q ( j ) = N X j =1 d j = Z V + n ( r ) d ( r ) dV, (10)where n ( r ) is the microscopic concentration of dipoles, and d ( r ) is the dipole moment ofa volume, which contains one atom and has the coordinate r . Since R V + ρ ( r ) dV = 0, wemay write ρ ( r ) = − div P mic ( r ), where P mic is a microscopic quantity. Similarly to (8), weobtain Z V + r ρ ( r ) dV = Z V + P mic ( r ) dV. (11)The relations ¯ ρ ( r ) = − div P mic ( r ) = − div ¯ P mic ( r ) yield P ( r ) = ¯ P mic ( r ). Since equalities(10) and (11) hold for a dielectric of any shape, they yield finally P mic ( r ) = n ( r ) d ( r ) , P ( r ) = n ( r ) d ( r ) . (12)The possible jump of ρ ( r ) on the surface does not violate equalities (12), since in the right-hand side of (10) this jump is smeared over the whole volume. Therefore, the function n ( r ) d ( r ) is smooth. The formula ¯ ρ ( r ) = − div P ( r ) sets P ( r ) to within rot g ( r ), where g ( r )is any function. Therefore, we can define P ( r ) in the infinite number of ways, but thechosen way must be consistent with the equation ¯ ρ ( r ) = − div P ( r ). For formulae (12) thisholds. According to (12), P ( r ) is the dipole moment of a unit volume of the dielectric.Relation (11) yields also P ( r ) = r ρ ( r ), though P mic ( r ) = r ρ ( r ) (since ρ ( r ) is a sum of δ -functions, but P mic ( r ) = n ( r ) d ( r ) is smoothly varied in space).We now pass to the description of the spontaneous polarization. Consider an isotropicdielectric, which is characterized in an external electric field by the dielectric permittivity ε .Let it contain a macroscopic number of spontaneous dipoles d ( j ) s , and let it be surroundedby a dielectric with dielectric permittivity ε (without the intrinsic electromagnetic field).We will write the equations only for the first dielectric and take the second one into accountin boundary conditions. The spontaneous dipoles are associated with some mean chargedensity ¯ ρ s ( r ) and the polarization P s . These dipoles create the electric field, which polarizesthe surrounding atoms (including the atoms, being the carriers of spontaneous dipoles).This leads to the appearance of induced charges with mean density ¯ ρ i ( r ) and the inducedpolarization P i ( r ). The polarizations P i and P s have different physical nature . Therefore,they should be considered separately. It is clear that, for the densities ¯ ρ s ( r ) and ¯ ρ i ( r ) , thetotal charge is zero: Z V + ¯ ρ s ( r ) dV = 0 , Z V + ¯ ρ i ( r ) dV = 0 . (13)Similarly to the above analysis, relations (13) yield the formulae¯ ρ s ( r ) = − div P s ( r ) , ¯ ρ i ( r ) = − div P i ( r ) , (14) s ( r ) = n s ( r ) d s ( r ) , P i ( r ) = n i ( r ) d i ( r ) , (15)where n s ( r ) and n i ( r ) are the microscopic concentrations of spontaneous and induceddipoles.Further, the averaging of the microscopic Maxwell equation div E mic ( r ) = 4 πρ ( r ) over aphysically infinitesimal volume gives the equation div E ( r ) = 4 π ¯ ρ ( r ) = 4 π [ ¯ ρ s ( r ) + ¯ ρ i ( r )] . (16)It is the basic equation, which should be solved. Let us try to solve the problem byconsidering P i and P s together. In view of (14), we can write (16) in the form div D ( r ) = 0 , (17) D = E + 4 π ( P i + P s ) = ε E . (18)Without a magnetic field, we have rot E = 0. Therefore, E = −∇ ϕ . Let us substitute D = − ε ∇ ϕ in (17). For ε ( r ) = const, we obtain the Laplace equation △ ϕ = 0 (19)with the boundary conditions D n = D n , E t = E t , (20)which are equivalent to the conditions ε ∂ϕ∂n = ε ∂ϕ ∂n , ϕ = ϕ provided that Eq. (18) holds.However, such approach is incorrect by two reasons. (A) The field E ( r ) is nonlocal, sinceit is the sum of fields created at a point r by all spontaneous dipoles. Therefore, E ( r ) isgenerally not codirected with P s ( r ). Then ε is a tensor ε jl ( r ), which is unknown and cannotbe found from the boundary conditions (20). (B) If ε jl ( r ) is a scalar, the second difficultyarises. The solution of Eq. (19) for the potential ϕ ( r ) inside a dielectric can be found intwo cases. (i) If the value of ϕ or ε ∂ϕ∂n on the boundary is known (in these cases, the Laplaceequation has the unique solution [2, 3]). (ii) If these values are unknown, but the externalfield E (0)2 is given. Then, for a number of problems, the solutions for E and E can beindicated to within constants, which can be determined from (20). However, in our case,the source of the field is the spontaneous dipoles inside the system, and there are no fieldsources outside the system. The spontaneous dipoles create some field E inside the system.Therefore, a nonzero field E should exist outside the system. But E and E are unknown.In other words, all quantities in the boundary conditions (20) are unknown. Therefore, thevalues of ϕ, ϕ , ∂ϕ∂n , and ∂ϕ ∂n on the boundary are also unknown. Hence, the method doesnot allow us to solve the problem. The consideration of a possible nonuniformity of ε ( r )does not change the situation.We note that the source of the electric field in the initial equation (16) is “spontaneouscharges” with density ¯ ρ s ( r ) (we consider it known), whereas E ( r ) and ¯ ρ i ( r ) should bedetermined. It is natural to combine the unknown quantities into one quantity and to keep¯ ρ s ( r ) in the right-hand side as a source: div D i ( r ) = 4 π ¯ ρ s ( r ) , (21) D i = E + 4 π P i = ε ( i ) E . (22)Below we show that such approach allows us to solve the problem. Here, ε ( i ) is the “in-duced” dielectric permittivity. That ε is a characteristic of the given dielectric withoutthe spontaneous polarization, but in the presence of an external electric field. This follows rom the physical meaning of the induced polarization P i as a local response to the electricfield E created in the dielectric. It is obvious that the response is independent of whichforce has created the field E at the given point of the dielectric. E can be created by freecharges outside the system or by spontaneous charges inside the system. In both cases, thevalue of P i must be the same. The resulting field is E = E + E i , where E i is the fieldcreated by induced dipoles. Since P i is identical in both cases, E should be also the same.Therefore, the relation 4 π P i = ( ε ( i ) − E valid for the induced polarization should also besatisfied, if E is created by spontaneous dipoles.Thus, one should solve Eqs. (21), (22) with regard for the boundary conditions and therelation E = −∇ ϕ − c ∂ A ∂t , (23)which follows from (4). In a stationary problem, the magnetic field is absent: A = 0 , B = rot A = 0. If spontaneous charges are moving, then a current j s and a magnetic field arise.The charge moving with a velocity v creates the potential ϕ and the vector potential A = ϕ v /c [15] (in the immovable reference system). The last relation indicates that the magneticfield is weak for the processes, which are slow as compared with the electromagnetic wave.Therefore, for the stationary and slow processes, we may set in (23) ∂ A /∂t = 0, and Eq.(21) takes the form of the Poisson equation △ ϕ = − π ¯ ρ s /ε ( i ) . (24)We neglect the nonuniformity of ε ( i ) , which is justified for uniform and weakly nonuniformfields. Let the spontaneous dipoles d ( j ) s = | q ( j )0 | r ( j )0 ( q ( j )0 <
0) be distributed in dielectric.The average density of effective spontaneous charges reads¯ ρ s = ≺ N s X j =1 [ q ( j )0 δ ( r − r j ) − q ( j )0 δ ( r − r j − r ( j )0 )] ≻ , (25)where N s is the number of spontaneous dipoles, r j and r j + r ( j )0 are the coordinates of theeffective charges q ( j )0 , − q ( j )0 of the j -th dipole, and ≺ f ≻≡ ¯ f . The solution of Eqs. (24),(25) is known: ϕ ( r ) = ≺ N s X j =1 q ( j )0 ε ( i ) | r − r j | − q ( j )0 ε ( i ) | r − r j − r ( j )0 | ≻ . (26)For the points r far from the spontaneous dipoles ( | r − r j | ≫ r ( j )0 ), we can make expansionin r ( j )0 . As a result, we have ϕ ( r ) = ≺ N s X j =1 d ( j ) s · ( r − r j ) ε ( i ) | r − r j | ≻ = ≺ Z V d ´ r n s (´ r ) d s (´ r ) · ( r − ´ r ) ε ( i ) | r − ´ r | ≻ == Z V d ´ r n s (´ r ) d s (´ r )( r − ´ r ) ε ( i ) | r − ´ r | = Z V d ´ r P s (´ r ) · ( r − ´ r ) ε ( i ) | r − ´ r | , (27)where V is the volume of the dielectric. This formula is applicable to the points inside asystem of dipoles, if the distance between adjacent dipoles is much larger than the size ofa dipole.Formula (27) without ε ( i ) in the denominator is well known [5]. The quantity ε ( i ) in(27) takes the polarization P i into account. Thus, taking P i into consideration leads tothe replacement d s → d s /ε ( i ) (or P s → P s /ε ( i ) ) in formula(27). This change has a simple hysical meaning: it shows that the medium weakens the field of a dipole by ε ( i ) times. Inview of this reasoning, formula (27) was intuitively guessed in [16, 17].For the uniform spontaneous polarization, we have P s = ( N s /V ) d s = ¯ n s d s , ϕ ( r ) = Z V d ´ r P s · ( r − ´ r ) ε ( i ) | r − ´ r | = − P s ∇ r Z V d ´ r ε ( i ) | r − ´ r | . (28)With regard for the formula [18] ∇ ( FG ) = ( F ∇ ) G + ( G ∇ ) F + F × ( ∇ × G ) + G × ( ∇ × F ) , (29)we find from (28) that E ( r ) = −∇ r ϕ ( r ) = ( P s ∇ r ) ∇ r f ( r ) , f ( r ) = Z V d ´ r ε ( i ) | r − ´ r | . (30)Formulae (27) and (30) give the exact solution (at any spatial point r ) for the field E ( r )created by a spontaneously polarized isotropic dielectric of volume V , which is surroundedby vacuum. It is seen from (30) that E ( r ) = const · P s only in particular cases, for example,if f ( r ) = c + c r or if P s = P s i z and f ( r ) = c + c z + c z . The symmetry-based reasoningimplies that, for a finite V, the first case is possible only for the dielectrics with shape ofa ball, but the second case is impossible for a three-dimensional system. That is, E ( r ) isnot codirected with P s in the general case. Moreover, E can be nonuniform, when P s isuniform. We have verified these properties for a cylindrical dielectric with P s = P s i z . Theseproperties suggest that, for an isotropic spontaneously polarized dielectric, the relationshipbetween the electric induction D = E + 4 π ( P i + P s ) = ε ( i ) E + 4 π P s (31)and the strength E takes the tensor form in the general case: D j = X l ε jl E l , ε jl = δ jl ε ( i ) + ε ( s ) jl , (32)where ε ( s ) jl ( r ) is the effective dielectric permittivity associated with the spontaneous polar-ization. It is of importance that ε ( s ) jl is a nonlocal quantity, since its value is determinednot by local properties of the medium, but by the distribution of the spontaneous polariza-tion in the whole volume of the dielectric and by the boundary conditions. Therefore, thevalues of ε ( s ) jl can be arbitrary, generally speaking: negative or positive. For an anisotropicdielectric, polarized by an external electric field, the principal values of the tensor ε jl mustbe > ε ( s ) jl is a consequence of the fact that the field E ( r ) (30)created by spontaneous dipoles is not codirected with P s ( r ) in the general case. If E ( r )and P s ( r ) are codirected, then ε ( s ) ( r ) is a scalar: ε ( s ) jl ( r ) = δ jl ε ( s ) ( r ).In the solution of specific problems, it is better to start from the representation D = ε ( i ) E + 4 π P s and to find ε ( s ) jl from the solution for E and from the equation P l ε ( s ) jl E l =4 πP s,j . In particular, relations (30) and (32) yield ε ( s ) jl = 4 πα − jl , where α jl = ∇ l ∇ j f ( r ).Thus, in order to find the field E ( r ) in a spontaneously polarized dielectric, we need tosolve Eqs. (21), (22) with the boundary conditions (20), in which D is given by formula(31). If the dielectric is surrounded by vacuum, then formula (27) gives the exact solutionfor the potential (in this case, the boundary conditions (20) are satisfied automatically). et us turn to the Maxwell equations (3)–(6). We noted above that in the presenceof the spontaneous polarization one should solve Eq. (21) (with the replacement ¯ ρ s ( r ) → ¯ ρ s ( r ) + ρ f ( r ), if foreign charges are present) instead of Eq. (3). In this case, Eq. (21) issimply another form of Eq. (3). Equations (4) and (5) remain the same. Equation (6)can be verified similarly to the analysis in [4] (Chapt. IX, § D is given by formula (31). However, while solving Eq. (6), weshould separate P s from D by means of the replacements D → D i and j f → j f + ¯ j s , where¯ j s = ρ s v s is the current related to spontaneous dipoles. Thus, the Maxwell equations(3)–(6) conserve formally their validity. However, while solving them, it is necessary toseparate P s from D , since the spontaneous dipoles are the source of the field. In addition, D is given by formulae (31), (22), (32) instead of (1).In the experiments [7, 8, 9, 10, 11], the potential ϕ ( r ) was measured during a macroscopictime. For such time interval, a configuration of spontaneous dipoles change a huge numberof times. Therefore, we need to average ϕ ( r ) over the measurement time. According toJ. Gibbs [19], the average over the time can be replaced by the average over the ensemble.Quantum statistics gives the following formula for the average over the ensemble [20]: h ˆ ϕ i = Z d Ω Z − X n e − E n /k B T Ψ ∗ n ˆ ϕ Ψ n , Z = X n e − E n /k B T , (33)where ˆ ϕ is given by (26) or (27), E n is the energy of the system in the n -th state, and { Ψ n } is the complete collection of wave functions of the system. The operator ˆ ϕ and thefunctions Ψ n should be written in terms of the coordinates of the nucleus and electrons ofeach atom. Since the integral of a sum is the sum of integrals, formula (33) shows thatthe polarization P s (´ r ) in (27) must be additionally averaged statistically. In this case, P s (´ r ) = h n s (´ r ) d s (´ r ) i = h n s (´ r ) d s (´ r ) i . The ferroelectrics possess peculiar and interesting properties [4, 6, 21, 22]. The distributionof the polarization P in a ferroelectric is usually found by means of the expansion of thethermodynamic potential Φ in degrees of the components P and by the calculation of theminimum of Φ. In our opinion, such method does not consider the local balance of forces.Indeed, let we have found some domain structure by means of the minimization of Φ,and let the dielectric be in vacuum without external fields. It is clear that the collection of“elementary dipoles” forming the field P ( r ) creates in the space some field E ( r ). The latteracts, in turn, on each elementary dipole. Such dipole knows nothing about the principle ofminimum Φ, but it feels a force and turns slightly in the direction of E ( r ). For an isotropicdielectric surrounded by vacuum we have P i = ε ( i ) − π E = − ε ( i ) − π ∇ ϕ. (34)Here, ϕ ( r ) is determined by formula (27), where the volume V includes all domains. For aferroelectric, the analysis of the previous section should be generalized, since the dielectricpermittivity of ferroelectric is a tensor ˘ ε ( i ) ≡ ε ( i ) jl . For each domain, the matrix ε ( i ) jl shouldbe the same as that of a ferroelectric, in which all domains have a codirected polarization(this corresponds to the regime of saturation). The solutions for domains can be obtained,apparently, with the help of the minimization of the thermodynamic potential [4, 6, 21].In this case, it is necessary to take components of P as order parameters and to take intoaccount that P = P s + P i , where P i ( r ) = ˘ ε ( i ) − π E ( r ), and the field E ( r ) is caused by the olarization P . That is, one needs to seek a minimum of Φ not on all possible configurationsof P , but on a narrower class of configurations, for which P = P s + P i ( P s ). We are not surethat there exists such distribution of P s ( r ), for which the resulting field P ( r ) is uniform ineach domain. Most probably, P ( r ) is not quite uniform in each domain at any P s ( r ). Theneed to consider P i in the description of ferroelectrics was apparently not noticed in theliterature. We have constructed a method of description of an isotropic spontaneously polarized di-electric that takes into account the induced polarization P i ( r ), which always accompaniesthe bare spontaneous polarization P s ( r ). In Appendix we have also obtained (as examples)several solutions for the electric field in a spontaneously polarized dielectric under differentboundary conditions.The author is grateful to A. S. Rybalko for the detailed discussion of the experiments[8, 9]. I also thank A. Morozovska for the discussion of the problem. The below-presented examples can be useful for the comprehension of properties of the field E inside of a resonator in experiments similar to [7, 8, 9, 10, 11]. In those experiments, theboundary conditions (BCs) are complicated. We will consider more simple BCs and willsee how the shape of a resonator and the presence of an external metallic shell affect thefield E (in experiments [8, 9], a resonator was placed in metallic shell for the protectionagainst external electric signals). Consider a dielectric plate, which occupies the space region x, y ∈ [ − L/ , L/ , z ∈ [ − H/ , H/ P = P i z (for sim-plicity, we consider the polarizations P i and P s together, as P ). The plate is surroundedby vacuum.The knowledge of the solution for the field E in this system is essential for the experi-ments like [7, 8, 9, 10, 11] and for the ferroelectrics. In many works (including well-knownones [6, 21]), the field in such system was described by the formulae E ( int ) = − π P , E ( ext ) = 0 , (35)where E ( int ) and E ( ext ) are the fields inside and outside of the plate, respectively. However,in reality the solution of this problem is different.One may solve the equation div D = 0 separately outside and inside of a dielectric withregard for BCs (20). The general solution takes the form D = f x ( x, y, z ) i x + f y ( x, y, z ) i y + f z ( x, y, z ) i z + C , where the functions f x , f y , f z and the constant C are different for vacuumand the dielectric. However, it was noted in Section 2 that such approach does not allowone to solve the problem. Since for a dielectric of finite sizes the connection between D and E has generally a tensor form D j = P l ε jl E l , but the tensor ε jl is unknown.The problem can be solved accurately, if we start from Eq. (16). The total polarizationis constant and known. Therefore, we can write relation (16) as div E ( r ) = 4 π ¯ ρ ( r ) , (36) ,0 0,2 0,4 0,6 0,8 1,0 1,2 1,4 1,6 1,8 2,0024681012141618 E x / P x/L Fig. 1: [Color online] E x ( x ) for a dielectric plate with L = 2 H at y = 0 and 1) z = H/ z = H/ z = H (triangles). It is clear from the symmetry that the same curves correspondto E y ( y ) for x = 0 and z = H/ H/ H . The lateral surface of the plate corresponds to x/L = 0 . where ¯ ρ ( r ) = 0 in the space outside of a dielectric. Equation (36) is true in the whole space .Therefore, BCs (20) are satisfied automatically (we remark that the equation div D = 0can be violated on the boundary, since D t can be changed by jump there). The solutionof Eq. (36) is given by formulae (27), (30), where V is the volume of a dielectric, and weshould set ε ( i ) = 1, P s = P . This solution has the properties ϕ ( x, y, − z ) = − ϕ ( x, y, z ), E x ( x, y, − z ) = − E x ( x, y, z ), E y ( x, y, − z ) = − E y ( x, y, z ), and E z ( x, y, − z ) = E z ( x, y, z ).We have found the solution of Eq. (27) numerically, see Figs. 1, 2 (the field near the plateedge is found approximately; the field E y ( r ) for | y | ≪ L is very weak and is not shown inthe figures). It is clear from the symmetry of the problem that E x ( r ) coincides with E y ( r ′ ),where r ′ is obtained by the rotation of r around the axis z by ± ◦ . As is seen from theplots, the field E ( r ) is nonuniform and has a complicated structure. Inside of the dielectric, E ≈ − π P = − πP i z (roughly). More accurately, E x , E y = 0; therefore, the dependence D ( E ) has a tensor form D j = P l ε jl E l . Outside of the dielectric, the field E ( r ) is ratherstrong near the dielectric and decreases with distance. The direct checking indicates thatBCs (20) hold on all surfaces of the plate.As L/H → ∞ , the solution passes into (35). That is, solution (35) is valid only in thecase of infinite plate.In the examples given below, we consider that the bare polarization P s is known andtake P i into account separately. Let the ball of radius R with the induced dielectric permittivity ε ( i ) is uniformly spon-taneously polarized: P s = P s i z . The spontaneous dipoles create the electric field, whichproduces the induced polarization P i = ( ε ( i ) − E ( int ) π . (37)Consider the system as a set of spontaneous and induced dipoles in vacuum. The field E is determined by Eq. (16). The solution for a domain inside the ball is [5] E ( int ) = − π P − π ( P i + P s )3 (38) ,0 0,2 0,4 0,6 0,8 1,0 1,2 1,4 1,6 1,8 2,0-10-8-6-4-20246 E z / P x/L Fig. 2: [Color online] E z ( x ) for a dielectric plate with L = 2 H at y = 0 and 1) z = H/ z = 0 . H (stars), 3) z = 0 . H (rhombs), 4) z = H (triangles). The symmetry implies that the function E z ( y ) at x = 0 and z = H/
4; 0 . H ; 0 . H ; H is described by the same curves. The upper surface of theplate corresponds to z = 0 . H . That is, the stars and rhombs show the field at the very boundary (underand above the surface, respectively). The jump of E z on the upper surface is equal to 4 πP in agreementwith BCs (20). (see also Eq. (50) below, which is the solution of Eq. (21)). Relations (37) and (38) yield P i = − ε ( i ) − ε ( i ) + 2 P s , E ( int ) ( r ) = − πε ( i ) + 2 P s , (39) ϕ ( int ) ( r ) = 4 πε ( i ) + 2 P s r . (40)Setting 4 π P s = ε ( s ) E ( int ) , we have ε ( s ) = − ε ( i ) − ε = ε ( i ) + ε ( s ) = − ϕ ( ext ) ( r ) outside the ball creates spontaneous and induced dipoles located insidethe ball. This field can be determined in the following way. The uniformly polarized ballcentered at the point r = 0 can be considered as two uniformly charged balls: the ball withcharge − Q < r − = ( x = y = 0 , z = − z /
2) and the ball with charge Q centered at the point r + = ( x = y = 0 , z = z / z is an infinitesimal value,and Qz = P V = 4 πP R /
3. The ball with charge − Q creates the potential ϕ ( r ) = − Q | r − r − | around itself, and the ball with charge Q generates the potential ϕ ( r ) = Q | r − r + | arounditself. The total potential outside the polarized ball at z → ϕ ( ext ) ( r ) = − Q | r − r − | + Q | r − r + | = 4 πR Pr r = 4 πR ε ( i ) + 2 P s r r , (41)where P = P s + P i = P s ε ( i ) +2 . The BCs (20) are satisfied.We can solve the problem in a different way. If we set ϕ ( int ) ( r ) = A P s r , ϕ ( ext ) ( r ) = B P s r /r and find the constants A and B from the BCs (20), we obtain the same result. Consider a grounded metallic sphere of radius R m . Let it contain an isotropic dielectricwith uniformly distributed over the volume spontaneous dipoles d s = | q | r , corresponding o the mean spontaneous polarization P s = n s d s = P s i z ( n s = const ). The BC in thespherical coordinates ρ, θ, φ reads [3, 4] ϕ ( ρ = R m ) = 0 . (42)If the polarization P s is perfectly uniform , then, for any shape of a resonator, the problemhas the known solution ϕ ( r ) = 0 , E ( r ) = 0 , (43)since the equation ε ( i ) △ ϕ = 4 πdiv P s for P s = const and BCs (42) has the unique solution ϕ ( r ) = 0. However, here there is a difficulty: On the boundary, P s decreases to zero byjump. Therefore, div P s = ∞ . The solution ϕ ( r ) = 0 neglects this property. Below, we willfind the solution within the method, which allows us to avoid this difficulty.It was assumed in some works that the relation D ( r ) = 0 holds for a dielectric surroundedby a metallic shell. It was not substantiated or was substantiated by that the equality D ( r ) = 0 holds in a metal. In our opinion, this reasoning is incorrect. It is well knownthat a metal should be considered as a dielectric with ε = ∞ (see [2] Chapt. IV, §
6; [4]Chapt. II, § § D ( r ) = 0 inside a metal. In addition,the condition D ( r ) = 0 for dielectrics leads to the mathematical contradiction. Indeed,the equality D ( r ) = 0 yields D n = 0 on the surface. If ϕ is given on a closed surface ofthe dielectric, then we can find a solution of the equation − div [ ε ( r ) ∇ ϕ ( r )] = 4 π ¯ ρ ( r ) (with¯ ρ ( r ) to be known) inside the dielectric uniquely [2, 3]. This was proved for ε = const ([3],Chapt. 1, §
9) and for ε = ε ( r ) ([2], Chapt. III, § D n = 0, then the problem becomes overspecified and has no solutions for ϕ ( r ).Thus, we need to solve Eq. (24) with the BC (42). The solution for the potential is givenby formula (28). In order to satisfy (42), we need to consider “images” that are reflectionsof dipoles in the metal. This is the most complicated part in a problems of this kind.In our case, the simplest way to consider the images is, apparently, the following. A realuniformly polarized ball can be represented as two balls with radius R shifted relative eachother by the size r of the dipole. The first ball is uniformly charged negatively (so thatits total charge is Q − = N s q = Q , where N s is the total number of spontaneous dipolesin helium). Let its center have the coordinates x = y = 0, z = − r /
2. The second ball ischarged positively and has the total charge Q + = − Q − . The coordinates of its center are x = y = 0, z = r /
2. The center of the segment joining both balls is the coordinate origin: x = y = z = 0. We consider that these two balls are placed in a metallic sphere so that the( − ) and (+) balls touch the internal surface of the sphere. In this case, R m = R + r / < ∼ r /
2) ofvacuum. The potential created by a uniformly charged dielectric ball with radius R at apoint located at the distance R from the ball center is ϕ ( R ) = Z V gdxdydzε ( i ) q ( x − X ) + ( y − Y ) + ( z − Z ) = (44)= R Z r dr π Z sin θdθ π Z dφ · gε ( i ) q r + R − rR cos θ = 2 πgε ( i ) R R Z rdr ( r + R − | r − R | ) . This formula yields ϕ ( R ) = πgε ( i ) (cid:16) R − R (cid:17) R ≤ R, Qε ( i ) R R ≥ R. (45) ere, g = Q/V = 3 Q/ (4 πR ) is the charge density, ε ( i ) is the induced dielectric permittivityof the ball. The dielectric weakens the field by ε ( i ) times, in accordance with Eq. (24).Solution (45) is well known [5].All points at a distance of r ≤ R m − r from the coordinate origin belong to both balls:( − )-ball and (+)-ball. Let us consider this domain. According to (45), these two ballscreate at the point r the potential ϕ ( r ) = 2 πg − ε ( i ) R − r − ! + 2 πg + ε ( i ) R − r ! . (46)Here, g − = Q /V , g + = − g − , V = (4 π/ R , and r − , r + are the distances from the pointof observation r to the centers of the ( − ) and (+) balls, respectively: r − = r + ( r / − rr cos ( π − θ ) , (47) r = r + ( r / − rr cos θ. (48)In this case, the vector r is directed from the coordinate origin to the point of observation,and θ = ( d i z , r ) = ( d r , r ). Formulae (46)–(48) yield the exact solution ϕ ( r ) = − Q rr cos θε ( i ) R = − Q rr ε ( i ) R = 4 π P s r ε ( i ) , (49)which is also well known [5] (in [5], formula (49) was deduced without ε ( i ) , since the responseof the medium to an external field was calculated). It is essential that relation (49) followsdirectly from formula (28). It is easy to see, making use of the relations (28), (44), and(45). Relation (49) implies that, in the domain r ≤ R m − r , the electric field strength isconstant: E ( r ) = −∇ r ϕ ( r ) = Q r ε ( i ) R = − π P s ε ( i ) . (50)According to (45), the ( − ) and (+) balls located inside the metallic sphere create onthe internal surface of the sphere the potential ϕ ( r ) = Q − ε ( i ) r − + Q + ε ( i ) r + . (51)It coincides with the potential, which is obtained, if the ( − )-ball and the (+)-ball arereplaced by the point charges Q − and Q + located at the centers of the ( − ) and (+) balls,respectively. In this case, the BC (42) is easily satisfied: potential (51) can be exactlycompensated on the whole surface of the cavity, if we introduce two additional charges thatare the images of the point charges Q − and Q + .Let the point charge Q − be placed inside a conducting sphere with radius R m at adistance of r / l − = 2 R m /r from the sphere center, and the charge ofthis image is q − = − Q − R m /r . Moreover, the sphere center, charge, and image are locatedon the same line, and the charge is placed between the sphere center and the image. Thepotential created by the charge Q − and its image is equal to zero on the whole surface ofthe sphere, which can be directly verified.The ( − ) and (+) balls induce polarization charges on the internal surface of the metallicsphere. The field created by these charges inside the sphere coincides with the field ofimages. With regard for this, the total potential at a point r inside the sphere is the sum of otentials created at this point by the ( − )-ball, (+)-ball, and images of the point charges Q − and Q + . The solution for r ≤ R m − r is as follows: ϕ ( r ) = − Q rr cos θε ( i ) R + q − ε ( i ) r q − + q + ε ( i ) r q + = − Q rr cos θε ( i ) R − R m Q ε ( i ) r r q − + 2 R m Q ε ( i ) r r q + . (52)Here, r q − and r q + are the distances from the point of observation r to the images of thecharges Q − and Q + , respectively: r q − = r + ( l − ) − rl − cos ( π − θ ) , (53) r q + = r + ( l + ) − rl + cos θ, (54)where l − = 2 R m /r = L and l + = 2 R m /r = L are the distances from the image of thecharge Q − and the image of the charge Q + to the center of the spherical cavity. Sincethe charges in (51) are decreased by ε ( i ) times, the charges of the images in (52) are alsodecreased by ε ( i ) times.For r /R ≪ r ≤ R m , we have r/L ≪
1. Let us use r/L as a small parameter.Then relations (53) and (54) yield1 r q − = 1 L " − r cos θL + r L (cid:18)
32 cos θ − (cid:19) + r L (cid:18)
32 cos θ −
52 cos θ (cid:19) + O r L ! , (55)1 r q + = 1 L " r cos θL + r L (cid:18)
32 cos θ − (cid:19) + r L (cid:18)
52 cos θ −
32 cos θ (cid:19) + O r L ! . (56)Substituting expansions (55), (56) in (52) and taking the relation L = 2 R m /r into account,we get ϕ ( r ) = Q rr cos θε ( i ) R m − R ! + Q r r R m ε ( i ) (5 cos θ − θ ) + O r L ! . (57)In view of the formulae rr cos θ = rr , R m = R + r / r /R , relation(57) yields finally: ϕ ( r ) = − Q rr ε ( i ) R r R + O r R ! , (58) E ( r ) = −∇ r ϕ ( r ) ≈ Q r ε ( i ) R r R = − π P s ε ( i ) r R . (59)Two last formulae imply that the spherical conductor decreases potential (49) and thefield strength (50) approximately by
R/r times. For example, the known mechanisms ofpolarization of He II give the value of r comparable with (or much less of) the interatomicdistance. Taking the realistic values R ∼ r ∼
10 ˚A , we get R/r ∼ . In otherwords, the images almost completely suppress the electric field E inside a spontaneouslypolarized dielectric ball.We remark that at r = 0 the dielectric possesses a perfectly uniform polarization. Inthis case, relations (58) and (59) yield solution (43). However, for real bodies r is small,but nonzero (since r is the size of an elementary dipole, e.g., a molecule). Therefore, nearthe boundary P is nonuniform in a layer with finite thickness ∼ r . The result given by(58), (59) is apparently new. ormula (59) and the relation 4 π P s = ε ( s ) E yield ε ( s ) = − ε ( i ) R/r . The “spontaneous”dielectric permittivity ε ( s ) turns out very large in modulus, which leads to the smallness of E . In this case, D = ( ε ( i ) + ε ( s ) ) E = ε ( i ) E (1 − R/r ).We now determine the field inside a spherical conductor in a thin layer of width r nearthe sphere surface. This layer is divided into two regions: the region lying outside the( − )-ball and inside the (+)-ball (or vice versa) and the region lying outside the ( − ) and(+) balls.For the region outside the ( − )-ball and inside the (+)-ball, the above formulae give ϕ ( r ) = 2 πg − ε ( i ) R − r − ! + Q + ε ( i ) r + + q − ε ( i ) r q − + q + ε ( i ) r q + ≈ Q ε ( i ) R + Q rr ε ( i ) R (cid:18) − r R (cid:19) −− Q rr ε ( i ) r − Q (4 r + r )8 ε ( i ) R − Q ε ( i ) r − Q r (3 cos θ − ε ( i ) r + O r R ! , (60) E ( r ) ≈ − Q ( rr ) ir ε ( i ) r + Q ir ( r − R ) ε ( i ) R r , ir = r r . (61)Since r ≈ R for this region, it is seen that the strength E is comparable by magnitude withstrength (50) of the problem without a resonator.For the region between the dielectric and the conductor (outside the ( − ) and (+) balls),we have ϕ ( r ) = Q − ε ( i ) r − + Q + ε ( i ) r + + q − ε ( i ) r q − + q + ε ( i ) r q + ≈ Q rr ε ( i ) R m − Q rr ε ( i ) r , (62) E ( r ) ≈ − Q r ε ( i ) R m + Q r ε ( i ) r − Q ( rr ) ir ε ( i ) r . (63)Here, the field strength is also comparable with (50). In this case, strength (61) is directedradially, (63) has the radial and z components, and strength (50) is directed along the z -axis.In the region r ≤ R m , the potential is continuous, whereas the field strength undergoesa jump on the surface of the dielectric and the internal surface of the conductor. It is seenfrom (62) that the BC (42) is satisfied.Thus, the electric field is strong (comparable with the field in the absence of a conductor)only in the narrow space of thickness ≤ r near the internal surface of the conductor. Inthe remaining volume inside the conductor (i.e., in almost whole volume of the dielectric),the field E is almost completely suppressed. This effect is caused by the presence of theconductor around the dielectric. It has a simple visual explanation. Without a conductor,the field E inside the dielectric is uniform (see (50)). The field created by the conductorin the region r ≤ R m coincides with the field of images. But the images are remote fromthe dielectric by the distance L = 2 R m /r , which is much larger than the size 2 R of thedielectric. Therefore, the field created by the images inside the dielectric is almost uniform,but is directed against the intrinsic field (50) of the dielectric and compensates it. If thecompensation would be absent, then the condition ϕ = 0 would not be satisfied: thecondition ϕ = const requires E t = 0 on the internal surface of the conductor. Since thefield E is uniform inside the dielectric, the condition E t = 0 on the surface requires that E ≈ e performed in order to transport two metallic hemispheres from infinity and to enclose apolarized dielectric by these hemispheres.As a rule, a dielectric is in the gravity field. Therefore, such dielectric is slightly spon-taneously polarized. Let us consider He II. If the walls of a vessel are vertical, and thebottom is horizontal, the polarization is uniform. In this case, the electric field is weak, butis measurable [16]. However, if the shape of a vessel is a sphere, then the polarization shouldhave the z - and r -components and, therefore, should be nonuniform. In other words, it isapparently impossible to get the uniform spontaneous polarization in a spherical vessel.According to the above-presented solutions, the boundaries affect strongly such bulkproperty of a macroscopic system as the electric field. This is related to the long-rangecharacter of the Coulomb interaction.We mentioned above that in experiments [8, 9] the resonator was placed in a metallicshell. The polarization of helium in those experiments is due to the second sound and,therefore, should be nonuniform . In this case, we may expect that the shell weakens thefield E only by several times or by one order of magnitude.[1] J.C. Maxwell, A Treatise on Electricity and Magnetism , Clarendon Press, Oxford(1892).[2] W.R. Smithe,
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