Acoustic modes in He I and He II in the presence of an alternating electric field
aa r X i v : . [ c ond - m a t . o t h e r] J a n Acoustic modes in He I and He IIin the presence of an alternating electric field
Maksim D. TomchenkoBogolyubov Institute for Theoretical Physics of the NAS of Ukraine14-b, Metrolohichna Str., Kyiv 03143, Ukraine
Abstract
By solving the equations of ordinary and two-liquid hydrodynamics, we study the oscillatory modesin isotropic nonpolar dielectrics He I and He II in the presence of an alternating electric field E = E i z sin ( k z − ω t ). The electric field and oscillations of the density become “coupled,” since the den-sity gradient causes a spontaneous polarization P s , and the electric force contains the term ( P s ∇ ) E . Theanalysis shows that the field E changes the velocities of first and second sounds, propagating along E , bythe formula u j ≈ c j + χ j E (where j = 1 , c j is the velocity of the j -th sound for E = 0, and χ j is a con-stant). We have found that the field E jointly with a wave of the first (second) sound ( ω, k ) should createin He II hybrid acousto-electric (thermo-electric) density waves ( ω + lω , k + lk ), where l = ± , ± , . . . .The amplitudes of acousto-electric waves and the quantity | u − c | are negligibly small, but they shouldincrease in the resonance way at definite ω and ω . Apparently, the first resonance corresponds to thedecay of a photon into two phonons with the transfer of a momentum to the whole liquid. Therefore, thespectrum of an electromagnetic signal should contain a narrow absorption line like that in the M¨ossbauereffect. Keywords: first sound, second sound, spontaneous polarization, dielectric.
It is well known that the external electric field E ext polarizes an isotropic dielectric [1, 2].The measure of such polarization is the dielectric permittivity ε . In addition, the isotropicnonpolar dielectric can polarize itself spontaneously. The spontaneous polarization related tothe acceleration and the density gradient was theoretically studied, respectively, in [3, 4, 5, 6,7, 8] and [3, 5, 9, 10, 11, 12]. The density gradient causes the spontaneous polarization, becausetwo nonpolar atoms polarize each other [13, 14, 15]. It was shown [16] how the spontaneouspolarization of an isotropic nonpolar dielectric should be taken in Maxwell equations for amedium into account.The electric properties of such isotropic nonpolar dielectrics as He I and He II were ex-perimentally studied in a number of works. In the experiment by A.S. Rybalko it was foundthat the standing half-wave of the second sound in He II is accompanied by an electric signal not observed at T > T λ . It is natural for the second sound (that simply does not exist at T > T λ ),but it is strange for the first one. We note also that a supernarrow absorption line at theroton frequency was found in experiments with an electromagnetic resonator imbedded in HeII [27, 28]. Several models were proposed to explain this line [29, 30, 31].In work [17], Rybalko mentioned the observation of a second-sound wave induced by analternating electric field. However, this effect was not revealed in the recent experiment [32].In what follows, we will study theoretically the influence of an external alternating electricfield E ext on the oscillatory modes of He I and He II and will show that the field leads toseveral interesting effects. In Sections 2 and 3 we present the results of calculations. Thephysical consequences and experiments will be considered in Section 4. Consider an isotropic nonpolar liquid dielectric in an alternating electric field. The motion ofan ideal liquid is described by the equations [33, 34] ρ∂ v /∂t + ρ ( v ∇ ) v = −∇ p + F , (1) ∂ρ/∂t + div ( ρ v ) = 0 , (2)where ρ is the density, p is the pressure, v is the velocity, and F is a nonmechanical force perunit volume. In our case, the force F is induced by the electric field E : F = ∇ " E π ρ ∂ε∂ρ | T − E π ∇ ε + (3)+ ( P s ∇ ) E + ( a − ∇ ( P s E ) + 12 rot ( P s × E ) . Here, two first terms were obtained by H.L. Helmholtz (see [1]), and the rest ones are relatedto the spontaneous polarization and are found in [16]. a is the parameter from the formula P s ( r ) = const · ρ a ∇ ρ [3, 5, 9, 10, 11, 12]. We will consider only nonpolar liquids and gases.Then the dielectric permittivity ε satisfies the Clausius–Mossotti formula [2] ε − ε +2 = π nβ (here, β is the polarizability of a molecule, and n = ρ/m ). For the gases and some liquids,including He I and He II, ε is close to 1 [35]. Then ρ∂ε/∂ρ ≈ ε −
1, and formula (3) takes theform F = ε − π ∇ E + ( P s ∇ ) E + ( a − ∇ ( P s E ) + 12 rot ( P s × E ) . (4) s was mentioned above, the spontaneous polarization P s of an isotropic nonpolar dielectriccan be related to the acceleration and the density gradient. We note that the availablecalculations of the polarization caused by the acceleration are rather crude [3, 4, 5, 6, 7, 8].The motion of an element of the liquid dielectric volume in an acoustic wave is accompaniedby the acceleration and the density gradient. If we subtract the contribution related to thedensity gradient from the total polarization, we get the part of the polarization which is causedonly by the acceleration. It is worth noting that the available works contain no proof that thispart is nonzero. According to the estimations made in [5], this part should be much less thanthe polarization caused by the density gradient. This is related to the fact that the electronshell of a nonpolar atom is hard to be stretched. In view of this fact we will neglect a possiblepolarization caused by the acceleration. The polarization caused by the density gradient was studied theoretically in works [3, 5,9, 10, 11, 12]. It can be accurately evaluated [5, 9, 11, 12] on the basis of the formula for amutual polarization of two nonpolar atoms [13, 14, 15]. Additionally, we need to average overdifferent configurations of atoms, which gives the following formulas [5]: P s ( r ) = ξ ∇ n ( r ) , ξ ≈ ς (7 / d ¯ r ( n ( r ) /n ) a , (5)where a = 1, ¯ r = n − / is the mean interatomic distance, d = − D | e | a B ¯ r , a B = ¯ h me is Bohr’sradius, D is the atomic constant [13, 14, 15], and ς ≈ . ς should be almost the same. Indeed, ς depends on the pair correlation function g ( r ) [5],and the latter is almost independent of the temperature, if T = 1–4 . K [36]. We note thatthe value of ξ was determined in [5] with wrong sign. In formula (5), we took the relation S ≈ n/n ) / into account [see Eqs. (28) and (29) in [5]] and assumed that the deviationsof the particle number density n ( r ) from the mean value n are small.The spontaneous polarization (5) is caused by the interaction of atoms and, therefore,exists at any temperature: in He II, He I, and gaseous helium at high temperatures.We now determine the influence of an external field E ext = E i z sin ( k z − ω t ) on theoscillatory modes of a liquid. The total field E in (4) is the sum E ext + E own , where E own is the field created by the dipoles of a dielectric. We can roughly consider that E own ∼ P , where P = P i + P s , and P i = ( ε − E / (4 π ) is the induced polarization [16]. Since( ε − / (4 π ) < ∼ . E ext . Below, we will see that the field created by the spontaneousdipoles is also low. Therefore, we set E ≈ E ext .Next. The force (4) contains three terms, in which the field E “couples” with P s . Since P s ∼ ∇ n ( r ), the electric field must create some density wave in the medium. We considerit to be weak and the perturbations of parameters of the system to be small. Therefore,we seek the deviations from the unperturbed values in the linear approximation. Let theunperturbed system be characterized by the parameters ρ , p = const , and v , E = 0. Forthe perturbed system, we take the velocity v to be nonzero and small. We set ρ = ρ + ρ ′ , = p + p ′ , E = E i z sin ( k z − ω t ) and take into account that the ideal liquid movesadiabatically ( s = const ). Therefore, p ′ = ∂p ′ ∂ρ | s ρ ′ = c ρ ′ , where c is the sound velocity[33, 34]. First, we set P s = 0 in (4). Equations (1) and (4) imply that the velocity must bedirected along the field E : v = v i z . In (1) and (2), we remain only the terms linear in ρ ′ , p ′ ,v and get the following equations for small perturbations: ρ ∂v∂t + c ∂ρ ′ ∂z = ε − π ∂E ∂z , (6) ∂ρ ′ ∂t + ρ ∂v∂z = 0 . (7)We set v = ˜ v cos ( kz − ωt + α ), ρ ′ = ˜ ρ cos ( kz − ωt + α ). For E = 0 Eqs. (6) and (7) yield − kc ˜ ρ + ωρ ˜ v = 0 , (8) ω ˜ ρ − kρ ˜ v = 0 . (9)From whence, we obtain the sound dispersion law: ω = c k . For E = E i z sin ( k z − ω t )Eqs. (6), (7) contain the driving force. The corresponding solution takes the form v =˜ v , cos (2 k z − ω t ), ρ ′ = ˜ ρ , cos (2 k z − ω t ). Then − k c ˜ ρ , + 2 ω ρ ˜ v , = ε − π E k , (10)2 ω ˜ ρ , − k ρ ˜ v , = 0 , (11)and we get the amplitudes˜ ρ , = ρ k ω ˜ v , , ˜ v , = ε − πρ E k ω ω − c k . (12)Now, we consider the terms with P s in (4). In this case, (2) leads, as before, to (7), andthe linearized equation (1) with force (4) takes the form ρ ∂v∂t + c ∂ρ ′ ∂z = ε − π ∂E ∂z + aξm ∂ρ ′ ∂z ∂E∂z + ( a − ξm E ∂ ρ ′ ∂z . (13)Any real liquid has a nonzero temperature. Therefore, it contains an ensemble of thermalphonons, including those moving along the field E . The latter create the density wavesof the form ρ ′ = ˜ ρ cos ( kz ± ωt + α ), where ω >
0. We will consider only the wave ρ ′ =˜ ρ cos ( kz − ωt + α ), by considering that ω can be positive or negative. For such wave, theright-hand side of (13) reduces to the form ε − π E k sin ( k , z − ω , t ) − ξE ˜ ρ m ( akk + ( a − k ) sin ( k , z − ω , t + α ) −− ξE ˜ ρ m ( akk − ( a − k ) sin ( k , − z − ω , − t + α ) . (14)Here and below, we use the notations k l,i = lk + ik , ω l,i = lω + iω . It shows that ρ ′ shouldbe sought in the form ρ ′ = ˜ ρ , cos ( k , z − ω , t + α ) + ˜ ρ , cos ( k , z − ω , t ) ++ ˜ ρ , cos ( k , z − ω , t + α ) + ˜ ρ , − cos ( k , − z − ω , − t + α ) . f we substitute this expansion to the right-hand side of (13), the latter will generate severalnew harmonics. They should be taken into account in solutions for ρ ′ , v . Then we againsubstitute ρ ′ , v in (7), (13), and so on. Finally, we have found that a solution of Eqs. (7) and(13) should be sought in the form ρ ′ = X i =1 , ,... ˜ ρ ,i cos ( k ,i z − ω ,i t ) + X i =0 , ± , ± ,... ˜ ρ ,i cos ( k ,i z − ω ,i t + α ) , (15) v = X i =1 , ,... ˜ v ,i cos ( k ,i z − ω ,i t ) + X i =0 , ± , ± ,... ˜ v ,i cos ( k ,i z − ω ,i t + α ) , (16)where the phase α is any real number. We substitute expansions (15) and (16) in Eqs. (7),(13) and take (14) into account. Then (7) and (13) take the form X l,i A l,i sin ( k l,i z − ω l,i t + q l α ) = 0 , X l,i B l,i sin ( k l,i z − ω l,i t + q l α ) = 0 , (17)where l, i run the values l = 0 , i = 0 , ± , ± , . . . (except for l = i = 0). In this case, q = 0 , q = 1. Equations (17) are satisfied, if A l,i = 0 and B l,i = 0 for all l, i . As a result, (7)yields the equations A l,i ≡ ω l,i ˜ ρ l,i − k l,i ρ ˜ v l,i = 0 . (18)Moreover, (13) leads to B , ≡ ρ ω ˜ v , − c k ˜ ρ , + ξE m ˜ ρ , (2 ak − ( a − k ) = 0 , (19) B , ≡ ρ ω ˜ v , − c k ˜ ρ , − ε − π E k ++ ξE m ˜ ρ , ( ak + ( a − k ) + ξE m ˜ ρ , ( a k − ( a − k ) = 0 , (20) B ,j ≡ ρ ω ,j ˜ v ,j − c k ,j ˜ ρ ,j + ξE m ˜ ρ ,j − ( ak k ,j − + ( a − k ,j − ) ++ ξE m ˜ ρ ,j +1 ( ak k ,j +1 − ( a − k ,j +1 ) = 0 , (21) B , ≡ ρ ω ˜ v , − c k ˜ ρ , + ξE m ˜ ρ , − [ ak ( k − k ) + ( a − k − k ) ] ++ ξE m ˜ ρ , [ ak ( k + k ) − ( a − k + k ) ] = 0 , (22) B ,i ≡ ρ ω ,i ˜ v ,i − c k ,i ˜ ρ ,i + ξE m ˜ ρ ,i − [ ak k ,i − + ( a − k ,i − ] ++ ξE m ˜ ρ ,i +1 [ ak k ,i +1 − ( a − k ,i +1 ] = 0 , (23)where j = 3 , , . . . , i = ± , ± , . . . . For Eqs. (18)–(23) and similar equations of the followingsection, the small parameter is ϑ = ξE k mc . (24) ven in strong fields E , we have ϑ ≪ k . The smallness of ϑ ensures theconvergence of series (15). It is convenient to introduce the phase velocity u l,i = ω l,i /k l,i foreach harmonic. Then (18) takes the form˜ v l,i = u l,i ˜ ρ l,i /ρ . (25)The system of equations (19)–(25) is separated into two independent systems: for theharmonics (0 , i ) and for the harmonics (1 , i ). In Eqs. (19)–(21) we present ˜ v ,i in terms of ˜ ρ ,i with the help of (25). The solutions for the harmonics (0 , , (0 , ,
3) are as follows:˜ ρ , ≈ ε − π E c − c , ˜ v , = c ˜ ρ , ρ , (26)˜ ρ , ≈ ξE k ( a − m ( c − c ) ˜ ρ , , ˜ v , = c ˜ ρ , ρ , (27)˜ ρ , ≈ ξE k (2 / − a ) m ( c − c ) ˜ ρ , , ˜ v , = c ˜ ρ , ρ , (28)where c = ω /k is the velocity of light in a dielectric. The mode (0 ,
2) is dominant, andthe remaining modes are weak: ˜ ρ , ∼ ˜ ρ , ∼ ϑ ˜ ρ , , ˜ ρ , ∼ ϑ ˜ ρ , , ˜ ρ , ∼ ϑ ˜ ρ , , and so on.Moreover, the amplitudes ˜ ρ ,i ∼ ϑ i − with i ≥ i ≥ ρ ,i ˜ v ,i and ˜ ρ ,i ˜ v ,i ˜ v ,i [from Eqs. (1) and (2)], whichwere neglected. Solution (26) coincides with (12).At the zero temperature the liquid contains no acoustic waves, and the electric field gen-erates in a liquid only oscillations of the density of the type (0 , i ) [see (26)–(28)] that havethe phase velocity equal to the velocity of light. If an acoustic wave is generated at T = 0artificially, hybrid modes obtained below should additionally appear in the system.Let us consider the chain of equations (22), (23) for the harmonics (1 , i ). Let us set˜ v ,i = u ,i ˜ ρ ,i /ρ and ω ,i = u ,i k ,i . Then (22) and (23) take the form˜ ρ , k ( u − c ) = − ξE m ˜ ρ , − [ ak ( k − k ) + ( a − k − k ) ] −− ξE m ˜ ρ , [ ak ( k + k ) − ( a − k + k ) ] , (29)˜ ρ ,i k ,i ( u ,i − c ) = − ξE m ˜ ρ ,i − [ ak k ,i − + ( a − k ,i − ] −− ξE m ˜ ρ ,i +1 [ ak k ,i +1 − ( a − k ,i +1 ] , (30)where i = ± , ± , . . . . With regard for the smallness of ϑ , relation (30) gives the recurrencerelations ˜ ρ ,i ≈ − ξE ˜ ρ ,i − m ak k ,i − + ( a − k ,i − ( u ,i − c ) k ,i , (31)˜ ρ , − i ≈ − ξE ˜ ρ , − i +1 m ak k , − i +1 − ( a − k , − i +1 ( u , − i − c ) k , − i (32) i = 1 , , , . . . ). Using them, we can express all ˜ ρ , ± i in terms of ˜ ρ , . We consider the quantity˜ ρ , to be known. It represents small fluctuations of the density related to thermal phonons( ω, k ).Substituting ˜ ρ , ± [31), (32] in (29), we get the formula for the sound velocity u , ≡ u : u = c + χ ξE m ! , (33) χ ≈ [ k + ( a − k ][ ak − ( a − k ] u , − − c + [ k − ( a − k ][ ak + ( a − k ] u , − c . (34)For a = 1 we get χ ≈ k u , − − c + k u , − c . (35)We consider the quantities k and ω to be positive (this can always be attained in theformula E = E i z sin ( k z − ω t ) by the choice of a direction of the axis z ). We also consider k of phonons in ˜ ρ , cos ( kz − ωt + α ) to be positive. In this case, the angular frequency ω = uk can be positive or negative, since the phase velocity u can have different signs. From (33), weget two solutions: u ≈ ± c + χ c ξE m ! . (36)Thus, we have found the solutions for small oscillations of the density for a nonsuperfluidliquid dielectric placed in an alternating electric field E = E i z sin ( k z − ω t ).As was mentioned above, the electric field E s induced by spontaneous dipoles can beneglected. This is seen from the formula E s ∼ P s = ( ξ/m ) ∇ ρ ′ and from the fact that themain contribution to ρ ′ is given by ˜ ρ , and ˜ ρ , . The latter leads to E s ∼ ( ξ/m ) ∇ ˜ ρ , ∼ ϑ ( ε − c πc E ext ≪ E ext , and the quantity ˜ ρ , gives E s ∼ ( ξ/m ) k ˜ ρ , . Since the density ˜ ρ , ofthermal phonons with momentum ( k x , k y , k z ) = (0 , , k ) is very small, we have E s ≪ E fornot too small E .The above solutions have interesting properties. The modes (0 , , (0 , , (0 ,
3) (26)–(28)correspond to weak oscillations of the density with parameters ( ω , k ), (2 ω , k ), and(3 ω , k ). These waves have a phase velocity equal to the velocity of light in the medium c = c v / √ εµ (where c v is the velocity of light in vacuum), which is larger by 6 orders thanthe velocity of sound. The modes (1 , ±
1) are hybrid acousto-electric modes. They exist,if the phonon mode ˜ ρ , cos ( kz − ωt + α ) and the electric field E = E i z sin ( k z − ω t ) arepresent. Phonons exist always at T >
0. According to the solutions, the modes (1 , ±
1) shouldbe stronger than the modes (1 , ± , ±
1) haverather unusual properties. The mode (1 ,
1) is a wave with frequency ω + ω and with wavevector k + k . If ω ∼ ω , then k ≪ k and k + k ≈ k . Therefore, if the wave vector k + k is close to k , the frequency ω + ω can be any one, in fact, from the interval ] ω, ω [. Inparticular, for 10 ω < ∼ ω < ∼ ω, we have k + k ≈ k and ω + ω ≈ ω . Such acouelon has he wave length close to that of a sound wave (phonon) and the frequency close to that of anelectromagnetic wave (photon). The modes (1 , −
1) have similar properties as well.In addition, the solutions ˜ ρ ,i and ˜ ρ , − i (31), (32) are characterized by a parametric reso-nance , respectively, at | u ,i | = c (1 + δ ) , δ → | u , − i | = c (1 + δ ) , δ → . (38)At i = 1 , if any of these conditions is satisfied, we get a resonant growth of χ (34). As is seen,the resonance arises, if the phase velocity u , ± i of a hybrid wave coincides with the soundvelocity c for the medium without a field E .Of course, solutions (31) and (32) do not work near the resonance point (i.e., as δ → ρ , ± i following from (1), (2). In suchapproach, the solutions ˜ ρ , ± i and χ should be finite at the resonance point. We will restrictourselves by solutions (31)–(35) which are true for values of k not too close to the resonancepoint. Therefore, we consider | δ | to be small ( | δ | ≪ | δ | ≪ , relations (35), (37), and (38) yield χ ≈ k δ · c . (39)We consider | δ | to be not too small, if | δ | ≫ ϑ /
16. In this case, | χ | c (cid:16) ξE m (cid:17) ≪ c and | u | ≈ c ,according to (33) and (39).For the modes (1 ,
1) and (1 , − , we consider a neighborhood of the resonance correspondingto not too small | δ | . Condition (38) is equivalent to two conditions: u , − = − c (1 + δ ) or u , − = c (1 + δ ). In the first and second cases, the phase velocity u , − is, respectively,negative and positive. Let u > . The first condition yields the relations k , − ≈ k ≈ k c c , u ≈ c , ω , − ≈ − ω . (40)From the second condition we get k , − ≈ k ≈ ζ k cδ I c , δ I = χ c ξE m ! − δ, u ≈ c , ω , − ≈ ζ ω δ I , (41)where ζ = 1. Since | δ I | ≪
1, the value of k , − (41) is much larger than k , − (40). For thereal electric waves, the values of k , − (41) are very large and should go beyond the phononregion of the spectrum. Therefore, we do not consider solution (41).Condition (38) with i = 2 , , . . . leads to solution (41) with ζ = 1 and with the changes(1 , − → (1 , − i ) and k → ik , as well as to solution (40) with the changes (1 , − → (1 , − i )and k → ik : k , − i ≈ k ≈ ik c c , u ≈ c , ω , − i ≈ − iω . (42) ormulas (42) describe a near-resonance solution for the modes (1 , − i ) with i = 1 , , . . . .For u <
0, condition (38) gives solutions with k < k > k are too large and go beyond the phonon region).Let us turn to condition (37). It can be written in the form u ,i = c (1 + δ ) or u ,i = − c (1 + δ ). In the first case for u > i = 1 , we get solution (41) with ζ = − , − → (1 , i = 2 , , . . . . For u < k ,i ≈ k ≈ ik c c , u ≈ − c , ω ,i ≈ iω . (43)It differs from (42) by signs of the phase velocities u and u ,i . In this case, the value of χ isset by formula (39), like for solution (42). The second case, u ,i = − c (1 + δ ), is possible for u <
0. But here, the solutions are characterized by k outside the phonon region.Thus, we have found two near-resonance solutions: (42) and (43). For clarity, let us considertheir behavior, as the phonon wave vector k increases. For k ranging from the smallest value k = π/L ( L is the resonator length) to k = 10 k (suppose that 10 k > π/L ), we have | u , ± i | ≫ c at any i . Therefore, the values of ˜ ρ , ± i and χ are small, and the sound velocity | u | ≈ c . However, at k ≈ k c c ∼ k , the relation | u , ± | ≈ c holds, and the quantities ˜ ρ , ± and χ increase in the resonance way. In this case for solutions (42) and (43), we have δ, χ > k < k c c and δ, χ < k > k c c . Therefore, by (36), the phonon energy | ω ( k ) | must besomewhat higher than c k at k < k c c and somewhat lower than c k at k > k c c . Near the point k ≈ k c c , these deviations can be large . And at the very point k ≈ k c c , the phonon dispersioncurve | ω ( k ) | should be discontinuous, and the amplitude | ˜ ρ , − | (or | ˜ ρ , − | , depending on thesign of u ) should sharply increase. In this case, the velocity u , − (or u , ) becomes equal tothe sound velocity c . In other words, at the resonance point the hybrid mode becomes similarto a phonon, and vice versa . At k > k c c , we leave the resonance region, as k increases. Nearthe points k ≈ ik c c ( i = 2 , , . . . ), the amplitudes ˜ ρ , ± i have resonances. We now consider the analogous problem for superfluid He II. The equations of hydrodynamicsfor He II describe the motion of the normal and superfluid components [34, 37]: ∂J i /∂t + X j =1 , , ∂∂r j ( pδ ij + ρ n v n,i v n,j + ρ s v s,i v s,j ) = F i , (44) ∂ρ/∂t + div J = 0 , (45) ∂ ( ρs ) /∂t + div ( ρs v n ) = 0 , (46) ∂ v s /∂t + ( v s ∇ ) v s = −∇ ( µ + Ω) , (47) here ρ = ρ s + ρ n , J = ρ n v n + ρ s v s , and F /ρ = −∇ Ω is a nonmechanical force per unitmass. Such force acting on the superfluid component must be the gradient of some function(according to (47), this ensures the potentiality of the motion of the superfluid component, rot v s = 0). The microscopic substantiation of Eqs. (44)–(47) was proposed in [38].Let the equilibrium system be characterized by the parameters ρ , p , s , T = const, v s = v n = 0, and Ω = 0. We now find the oscillatory modes of the system in the presence of aforce F = − ρ ∇ Ω. As usual, the sound and thermal waves are considered as small deviationsfrom the equilibrium. Therefore, we consider v s and v n to be small and ρ, p, s, and T to beclose to the equilibrium values. Then, from (44)–(47) we can pass to the linearized system ∂ J /∂t + ∇ p = − ρ ∇ Ω , (48) ∂ρ/∂t + div J = 0 , (49) ∂ ( ρs ) /∂t + ρs · div v n = 0 , (50) ∂ v s /∂t = −∇ ( µ + Ω) . (51)Equations (49)–(51) and the thermodynamic relation [34] dp = ρdµ + ρsdT + ( ρ n / d ( v n − v s ) (52)(we neglect the last term) yield the equation [37] ∂ s∂t = s ρ s ρ n △ T. (53)In addition, Eqs. (48) and (49) lead to the equation [37] ∂ ρ∂t = △ p + ρ △ Ω . (54)We set ρ = ρ + ρ ′ , p = p + p ′ , s = s + s ′ , T = T + T ′ , where ρ ′ , p ′ , s ′ , and T ′ are small. Thenit is convenient to write Eqs. (53), (54) in the form [37] ∂s∂p | T ∂ p ′ ∂t + ∂s∂T | p ∂ T ′ ∂t − s ρ s ρ n △ T ′ = 0 , (55) ∂ρ∂p | T ∂ p ′ ∂t + ∂ρ∂T | p ∂ T ′ ∂t − △ p ′ = ρ △ Ω . (56)These are the basic equations which will be analyzed in what follows. They differ from theLandau’s equations [37] by the additional term ρ △ Ω chracterizing the influence of the electricfield on the oscillatory modes.We write the perturbations p ′ and T ′ in (55), (56) as p ′ = ˜ p cos ( kz − ωt + α ) and T ′ =˜ T cos ( kz − ωt + α ) and use the relations [34] ∂ρ∂p | T = C p C V c , ∂s∂T | p = C p T . (57) hen, instead of (55) and (56) we get " ˜ p − u ∂s∂p | T ! + ˜ T s ρ s ρ n − u C p T ! k cos ( kz − ωt + α ) = 0 , (58) " ˜ p − u C p c C V ! + ˜ T − u ∂ρ∂T | p ! k cos ( kz − ωt + α ) = ρ △ Ω , (59)where u = ω/k . For Ω = 0 , Eqs. (58) and (59) and formula [34] ∂ρ∂T | p ∂s∂p | T = C p T c (cid:18) C p C V − (cid:19) (60)yield the well-known equation for the velocities of first and second sounds [34]: u c − ! u c − ! + C V C p − , (61)where c = ∂p∂ρ | s , c = ρ s s Tρ n C V . (62)We note that the electric field E = E i z sin ( k z − ω t ) depends only on the coordinate z and the time. It is clear from the symmetry of the problem that, for the infinite system, ρ ′ , p ′ , s ′ , and T ′ should depend only on z and t as well. Let us set ρ ′ = ˜ ρ cos ( kz − ωt + α ).Then the force (4) can be represented in the form F = − ρ ∇ Ω, whereΩ = − ε − πρ E (1 − cos (2 k z − ω t )) − ξE k ˜ ρ mρ ·· ( k k , + a − ! cos ( k , z − ω , t + α )+ (63)+ k k , − − a + 1 ! cos ( k , − z − ω , − t + α ) ) and k , − = 0. For k , − = 0 the term k k , − cos ( k , − z − ω , − t + α ) should be replaced by k z sin( ω , − t − α ).We seek the solutions for ρ ′ , p ′ , s ′ , and T ′ in the form of expansions analogous to (15), (16).In this case, Ω acquires a rather awkward form, but it can be easily found with the help of(63). We substitute the formula for Ω and the expansions for p ′ and T ′ in (55), (56). Withregard for formula (57), Eqs. (55) and (56) take, respectively, the forms X l,i A l,i cos ( k l,i z − ω l,i t + q l α ) = 0 , X l,i B l,i cos ( k l,i z − ω l,i t + q l α ) = 0 . (64)Here, analogously to the previous section, l = 0 , i = 0 , ± , ± , . . . (the case l = i = 0 isexcluded), and q = 0 , q = 1. Equations (64) are valid for A l,i = 0 , B l,i = 0 for all l, i . In sucha way, relation (55) yields ˜ T l,i = ˜ p l,i u l,i ∂s∂p | T s ρ s ρ n − u l,i C p T , (65) here u l,i = ω l,i /k l,i . Moreover, Eq. (56) yields the following chain of equations:˜ p , − u , C p c C V ! − ˜ T , c ∂ρ∂T | p ! = ξE m ˜ ρ , k (cid:18) − a + 1 (cid:19) , (66)˜ p , − u , C p c C V ! − ˜ T , c ∂ρ∂T | p ! = − ε − π E ++ ξE m (cid:20) ˜ ρ , k (cid:18)
12 + a − (cid:19) + ˜ ρ , k (cid:18) − a + 1 (cid:19)(cid:21) , (67)˜ p ,j − u ,j C p c C V ! − ˜ T ,j c ∂ρ∂T | p ! = ξE m " ˜ ρ ,j − k ,j − j + a − ! ++ ˜ ρ ,j +1 k ,j +1 j − a + 1 ! , (68)˜ p , − u , C p c C V ! − ˜ T , c ∂ρ∂T | p ! = ξE m " ˜ ρ , − ( k − k ) k k + a − ! ++ ˜ ρ , ( k + k ) k k − a + 1 ! , (69)˜ p ,i − u ,i C p c C V ! − ˜ T ,i c ∂ρ∂T | p ! = ξE m " ˜ ρ ,i − k ,i − k k ,i + a − ! ++ ˜ ρ ,i +1 k ,i +1 k k ,i − a + 1 ! , (70)where j = 3 , , . . . , i = ± , ± , . . . , and u , = u , = . . . = u ,j = c . We solve Eqs. (66)–(70)similarly to Sec. 2. We substitute˜ ρ l,i = ∂ρ∂p | T · ˜ p l,i + ∂ρ∂T | p · ˜ T l,i = C p c C V ˜ p l,i + ∂ρ∂T | p · ˜ T l,i (71)into the right-hand sides of those equations and then present ˜ T l,i in terms of ˜ p l,i with the helpof formula (65).At small ϑ Eqs. (66)–(68) yield ˜ p , ≈ ε − π E C V c C p c , (72)˜ p , ≈ ξE k ( a − mc C V C p ˜ p , , (73)˜ p , ≈ ξE k (2 / − a ) mc C V C p ˜ p , . (74)The remaining ˜ p ,j are very small: ˜ p , ∼ ϑ ˜ p , , ˜ p , ∼ ϑ ˜ p , , etc. ith the help of formulas (65) and (71), we write (69), (70) as the equations for ˜ p ,i :˜ p , G , C V C p − u , c = ξE m " ˜ p , − ( k − k ) k k + a − ! C p ( c − u , − ) c ( c C V − u , − C p ) ++ ˜ p , ( k + k ) k k − a + 1 ! C p ( c − u , ) c ( c C V − u , C p ) , (75)˜ p ,i G ,iC V C p − u ,i c = ξE m " ˜ p ,i − k ,i − k k ,i + a − ! C p ( c − u ,i − ) c ( c C V − u ,i − C p ) ++ ˜ p ,i +1 k ,i +1 k k ,i − a + 1 ! C p ( c − u ,i +1 ) c ( c C V − u ,i +1 C p ) , (76)where i = ± , ± , . . . , and we denoted G l,i = u l,i c − ! u l,i c − ! + C V C p − . (77)At small ϑ , formula (76) leads to the recurrence relations˜ p ,i ≈ ξE m ˜ p ,i − k ,i − k k ,i + a − ! c C V − u ,i C p c C V − u ,i − C p c − u ,i − c c G ,i , (78)˜ p , − i ≈ ξE m ˜ p , − i +1 k , − i +1 k k , − i − a + 1 ! c C V − u , − i C p c C V − u , − i +1 C p c − u , − i +1 c c G , − i (79)( i = 1 , , . . . ). These relations allow us to present ˜ p , ± i in terms of ˜ p , . Like in Section 2,we consider the quantity ˜ p , to be known. We now substitute ˜ p , (78) and ˜ p , − (79) in Eq.(75), reduce both sides of the equation by ˜ p , , and determine the dispersion relation G , ≡ u c − ! u c − ! + C V C p − χc ξE m ! , (80)where u ≡ u , and χ ≈ c − u c ( [ k + ( a − k ][ ak − ( a − k )] c − u , − c G , − ++ [ k − ( a − k ][ ak + ( a − k ] c − u , c G , ) . (81)For a = 1 we get χ ≈ k c ( c − u ) ( c − u , − c G , − + c − u , c G , ) . (82)For liquid He the quantity C p /C V − < C p /C V − < ∼ . < ∼ . u c − ! u c − ! = 1 − C V C p + χc ξE m ! ≡ δ u , (83) here δ u is small (0 < δ u ≪
1) at sufficiently small ϑ . From (83) we get the solutions for thevelocities of first and second sounds: | u | ≈ c δ u c /c − ! , (84) | u | ≈ c δ u c /c − ! . (85)We have found the solutions for small oscillations of the pressure in a superfluid dielectricplaced in the electric field E = E i z sin ( k z − ω t ).We now verify whether the solutions for He II pass into solutions for He I as ρ s → c → C p = C V and c → , (80) yields (33). Turning c → u → c (81), itis easy to see that formula (81) passes in (34). In reality, we have C p = C V . Therefore, thesolutions for He II do not pass exactly into solutions for He I, which is related to the fact thatthe first sound in He II is not quite identical to the ordinary sound in He I.Formulas (78), (79), (81) imply that the quantities ˜ p , ± and χ should resonantly increaseas G , ± →
0. To find solutions in a neighborhood of the resonance, we set G , − = 2 δ (or G , = 2 δ ). Like in the previous section, we consider | δ | to be small, but not too small(1 − C V /C p ≪ | δ | ≪ G , − = 2 δ is equivalent to four possible solutionsfor u , − : u , − ≈ ± c δ , − c /c − ! , (86) u , − ≈ ± c δ , − c /c − ! , (87)where 2 δ , − = 2 δ + 1 − C V /C p . The situation is analogous for the condition G , = 2 δ . Byanalyzing these solutions, we should take into account that the phase velocity u in (84) and(85) can be positive or negative. In such a way, we find the following near-resonance solutionswith positive and not too large (phonon) values of k for the modes ˜ p , − : k , − ≈ k ≈ ck c + c , ω , − ≈ − c ω c + c , u , − ≈ − c , u ≈ c , (88) k , − ≈ k ≈ ck c − c , ω , − ≈ − c ω c − c , u , − ≈ − c , u ≈ − c , (89) k , − ≈ k ≈ ck c + c , ω , − ≈ − c ω c + c , u , − ≈ − c , u ≈ c , (90) k , − ≈ k ≈ ck c , ω , − ≈ − ω , u , − ≈ − c , u ≈ c , (91) k , − ≈ k ≈ ck c − c , ω , − ≈ c ω c − c , u , − ≈ c , u ≈ c , (92) k , − ≈ k ≈ ck c , ω , − ≈ − ω , u , − ≈ − c , u ≈ c . (93) esonance (93) is characterized by the resonance-like increase in the value of χ ≈ k δ ( c − c ) c c as δ →
0. In this case, according to (83), (84), (93), the velocity of the first sound resonantlyvaries as δ →
0. For solutions (88) and (89), we find χ ≈ − k c (1 − C V /C p )[2 δ + ϑ c / (4 c )] − .Here, the resonances are possible for χ and the velocity of the second sound. For solutions(90)–(92), the value of χ is close to a constant as δ → χ ≈ − k /c for (90) and (92), and χ ≈ k c ( c − c ) (1 − C V /C p )[1 − ϑ c c − c ) ] − for (91)); that is, there is no resonance for χ .For the mode ˜ p , we get the following near-resonance solutions: k , ≈ k ≈ ck c + c , ω , ≈ c ω c + c , u , ≈ c , u ≈ − c , (94) k , ≈ k ≈ ck c + c , ω , ≈ c ω c + c , u , ≈ c , u ≈ − c , (95) k , ≈ k ≈ ck c , ω , ≈ ω , u , ≈ c , u ≈ − c , (96) k , ≈ k ≈ ck c − c , ω , ≈ − c ω c − c , u , ≈ − c , u ≈ − c , (97) k , ≈ k ≈ ck c , ω , ≈ ω , u , ≈ c , u ≈ − c , (98) k , ≈ k ≈ ck c − c , ω , ≈ c ω c − c , u , ≈ c , u ≈ c . (99)The function χ ( δ ) is not constant as δ → χ ≈ k δ ( c − c ) c c ) and (99), (94)( χ ≈ − k c (1 − C V /C p )[2 δ + ϑ c / (4 c )] − ). In the last case due to the smallness of 1 − C V /C p it will be apparently difficult to observe χ near the resonance ( δ → k → ik and ω → iω in formulas (88)–(93) and (94)–(99), we get thenear-resonance solutions for ˜ p , − i and ˜ p ,i , respectively ( i ≥ p ,i and ˜ p , − i ( i ≥
1) should sharplyincrease also as c C V − u ,i − C p → c C V − u , − i +1 C p → , respectively (in this case,there is no resonance for χ and other ˜ p , ± i ). For i = 2 , this leads to solutions (90)–(92) and(95)–(97) (if we replace k → ( i − k and ω → ( i − ω in them, we get solutions for i > We will try to understand the physical nature of solutions and discuss which of the above-foundpeculiarities can be observed.Let us estimate the intensity of the above-obtained modes for He I (for He II, the resultsare analogous). At small ϑ , the mode (2 ω , k ) is the most intense from the modes ( iω , ik ) both for He I and He II). It is a density wave whose phase velocity is equal to the velocity oflight. For this mode, the frequency and the wave vector are two times larger than for the field E . Let us consider critical the field E = E c for which ˜ ρ , = 0 . ρ (for E > E c the densityperturbation ˜ ρ , becomes sufficiently high so that our approximation of small perturbationsfails). With the use of the parameters of He II ¯ r = 3 .
58 ˚A, d ≈ − . · − | e | ˚A [5] andformula (26), we find E c ≃ . · q g / cm / sec ≃ V / m. This is a very strong field. Forcomparison, in experiments [27, 28] the field E near a resonator was 11 orders of magnitudeweaker. For E = E c , a = 1 , and L = 1 cm , formulas (27), (28) yield ˜ ρ , ∼ ˜ ρ , ∼ − ˜ ρ , (in this case, we used the resonance relations k ≈ c k/c and k = π/L , see below).From the hybrid modes ( ω ± iω , k ± ik ) , the modes ( ω ± ω , k ± k ) are the most intense.From formula (31) for E = E c , a = 1, k ≈ c k/c , k = π/L , L = 1 cm, and | u , | = c (1 + δ )(resonance for the (1 , ρ , ≃ ( − ϑ/ δ ) ˜ ρ , ≃ (6 · − /δ ) ˜ ρ , . Here we also seethe smallness of | ϑ | : ϑ ≃ − . · − . For the resonance of the (1 , − ρ , . Theseestimates show that, in real fields E ≪ E c , all waves-satellites of the (0 , i )- and (1 , i )-modesare extremely weak. In this case, the (1 , i )-modes can in principle become observable, if thefrequency ω is very close to the resonance one or if the amplitude ˜ ρ , of a bare acoustic (ora thermal one, for He II) wave ( ω, k ) is artificially made very high.For the phase velocity u of an acoustic wave in He I, we have found that u = c + ( δu ) ,where δu = √ χ ξE m . The estimate with the use of the parameters above gives δu ∼ − / √ δ .Such value can be observable (i.e., ∼ c ) only in a small vicinity of the resonance. Forthe second sound, χ ∼ − C V /C p , which suppresses δu . As is known from the theory ofoscillations [40], the amplitude of oscillations at the resonance point should increase with thetime until the growth is terminated by the nonlinear viscosity and nonlinear corrections whichwere neglected in the solution of (1), (2). In addition, the linear viscosity leads to that theresonance exists only in a field E higher than some threshold one.Near the resonances, the hybrid modes ( ω ± ω , k ± k ) sharply increase. We note that thehybrid modes at resonances are characterized by the phase velocity equal to the velocity ofthe first or second sound. It is natural that the energy of an electric wave easily transits intothe energy of a hybrid mode, if this mode is similar to an eigenmode of the system. Thus, theresonance point corresponds to the intersection of dispersion curves of the sound and hybridmodes (see Fig. 1). In this case, apparently, the reconnection (hybridization) of two curvesshould occur. To clarify this point, one needs to accurately find a solution near the resonance.The absorption of the energy of an electromagnetic wave that occurs at the resonanceamplification of a hybrid mode has the quantum origin. We can try to establish the characterof the process from the conditions for a resonance. For example, for resonances (42) and (93),we have u , − i ≈ − c and u ≈ c . Therefore, ω , − i ≡ ω − iω = uk − ick = u , − i k , − i ≈ u , − i k ≈ − c k (here, we took into account that k ≪ k ), which yields ic ¯ hk ≈ c ¯ hk . The ω k Fig. 1: [Color online] The dispersion laws ω , ± ( k , ± ) of the hybrid modes (1 , −
1) and (1 ,
1) for He Iat different ω . 1) The mode (1 , −
1) for ω = 2 πc /L (open circles), ω = 6 πc /L (open triangles), and ω = 20 πc /L (open diamonds); 2) the mode (1 ,
1) for ω = 2 πc /L (filled circles), ω = 6 πc /L (filledtriangles), and ω = 20 πc /L (filled diamonds). Resonance points (stars) correspond to the intersection of thecurves of the modes (1 , ±
1) with the acoustic dispersion law ω = ± c k (solid lines), which is characteristicof the medium without a field E ; ω , ± and k , ± are given in dimensionless units, for which c = 2 π/L = 1;values of ω correspond to formula (104) with i = 1 and j = 1 , ,
10. In the formula ω , ± = uk ± ck , we set u = c for the mode (1 , −
1) and u = − c for the mode (1 , | u | at the resonance points should significantly differ from the values of c . same equation can be obtained for resonances (43) and (98). For i = 1 , we get c ¯ hk ≈ c ¯ hk .This indicates that such resonance corresponds to the decay of a photon with energy c ¯ hk intotwo phonons, each possessing the energy c ¯ hk . Since k ∼ − k , the momentum conservationlaw can hold for low k < ∼ π/L only if the phonons have the momenta ¯ h k and − ¯ h k , and thephoton momentum is transferred to the whole liquid. Therefore, we assume that resonances(42) and (93) for i = 1 and low k correspond to the following exact equations: c ¯ hk = u ¯ hk + u ¯ hk + P liq / (2 M ) , (100)¯ h k = ¯ h k − ¯ h k + P liq = P liq , (101)where P liq / (2 M ) and P liq are, respectively, the energy and momentum of the liquid as a whole, M is the liquid mass, and u is the velocity of the first sound in the medium with field E . Suchprocess is similar to the M¨ossbauer effect [41].The M¨ossbauer effect is observed in crystals. In this effect, the momentum is transferredto the whole crystal due to its stiffness. In our case, the momentum should be transferred tothe liquid which has no stiffness. Apparently, this is possible due to that the process has thequantum origin and involves the whole system (because the wavelength λ of a phonon is ofthe order of system size L , and λ of a photon is much larger than L ).Resonances (42), (43), (93), (98) with any i correspond to the conditions ic ¯ hk = u ¯ hk + u ¯ hk + ( P liq ) / (2 M ) , (102) ¯ h k = ¯ h k − ¯ h k + P liq . (103)Here, i photons with the same momentum are transformed into two phonons with oppositemomenta, and the recoil momentum i ¯ hk is transferred to a liquid as a whole. It is clear thatthe process with i ≥ i should be unlikely. We note thatsince the field E has the form of a running wave, the momenta of all photons must be directedto the same side.The conditions for resonances (88) and (94) for any i lead to the relation ic ¯ hk ≈ c ¯ hk + c ¯ hk . This can be interpreted as the coalescence of i photons with the formation of a phononand a “quantum” of the thermal wave with the momenta ¯ h k and − ¯ h k . In a similar way,resonances (89) and (99) give the relation ic ¯ hk + c ¯ hk ≈ c ¯ hk which can be interpreted as thecoalescence of i photons and a “quantum” of the thermal wave with the creation of a phonon.Such interpretations are questionable, because a thermal wave is a classical structure, namely,a wave in the gas of quasiparticles. Nevertheless, it is worth to verify in experiments whetherthe spectrum of electromagnetic waves has the absorption lines at the corresponding ω .The other resonances correspond to the second sound: u , ± i ≈ ± c . Consider the resonancesfor the modes ( ω ± ω , k ± k ) for He II. These are solutions (90)–(92) and (95)–(97). Theycan be joined in pairs: (90), (95); (91), (96); and (92), (97). In each pair, the waves have thesame wave vector, and the phase velocities of waves differ from one another only by a sign.The sum of such waves forms a standing wave (we assume that the constants α for both wavesare close; this is possible, if the system contains many phonons or second-sound waves). Sucha standing wave is stable, if its wavelength is λ = 2 L/j , where j = 1 , , , . . . , and L is theresonator length. Therefore, the following equalities must be valid: k , ± = 2 π/λ = πj/L , | ω , ± | = c πj/L . It follows from (90)–(92) that ω = 2 πjc /L or ω = πj ( c ± c ) /L . Theaccount for the resonances for the modes ( ω ± iω , k ± ik ) with i > ω = 2 πjc / ( iL ) or ω = πj ( c ± c ) / ( iL ), i = 1 , , . . . . However, all theseresonances are suppressed by the factors c C V − u ,i C p in (78) and c C V − u , − i C p in (79) whichare close to zero at u , ± i ≈ c (the condition which should be satistied for the second sound).In the experiment [32], the frequency band ω = 2 πjc / L , j = 1 ÷
8, including our theoreticalfrequencies 2 πjc / ( iL ) with j = 1 , i = 1 , , ,
4, was measured. In this case, no second soundwas registered. This result agrees with our analysis in view of the above-indicated suppressionand the fact that the second-sound wave can exist only at resonances ω (note that the searchfor such narrow bands at ω was not performed in [32]).Such suppression is absent for the hybrid modes with the velocity of the first sound ( u , ± ≈± c ). Therefore, such modes should be intense near a resonance. From formulas (88), (89),(93), (94), (98), and (99), we get that the standing wave of the first sound with λ = 2 L/j ,corresponding to the hybrid mode, is possible for k , ± = πj/L ≈ k , | ω , ± | = c πj/L , ω =2 πjc /L (or ω = πj ( c ± c ) /L ). The account for resonances for the modes ( ω ± iω , k ± ik ) ith i > ω = 2 πjc / ( iL ) ≈ ω/i, (104)or ω = πj ( c ± c ) / ( iL ) ≈ ( ω ± ω ) /i, (105)where i, j = 1 , , . . . . For He I, we have only relation (104). Relations (104) and (105) forfrequencies are characteristic of the parametric resonance [40]. Thus, if the frequency of theelectric field is close to (104) or (105), then an acoustic gage should register the weak firstsound with the frequency | ω , ± i | = c πj/L equal to iω / iω c / ( c ± c ). As i increases, thewidth △ ω of the resonance strongly decreases, as usual [40] (for our solutions, △ ω i,j ∼ | ϑ i | ω resi,j ;though the account for small nonlinear corrections, including the friction, strongly affects △ ω i,j and can increase it by many orders of magnitude). Therefore, only the resonances with i = 1can be apparently observed. For j such limitation is absent. We believe that one needs toseek in experiments firstly the modes with small j ( < ∼ j are more stable. For example, the frequency ω = 2 πc /L corresponds to formula (104)with i, j = 1 , or i, j = 2, and so on. In this case, a first-sound wave with ω = c π/L and veryweak waves with ω = jc π/L ( j = 2 , , . . . ) should arise. The hybrid mode corresponding tothe very resonance point should get the highest amplification. We did not find the amplitudes˜ ρ , ± i at the resonance point. If ˜ ρ , ± ≫ ˜ ρ , at the resonance point, then the hybrid wave with ω = c π/L is intense enough and can be observed.Note that since the hybrid modes accompany always the acoustic (or thermal) one (1 , E = E i z sin ( k z − ω t ). We have considered only the acoustic modes runningalong the field E . It is clear that the hybrid waves-satellites should arise also for the modesrunning not in the line of E . Most likely, the resonances also exist for them. The neutronpassing through a liquid should create namely a “dressed” phonon. Therefore, based on thespectrum of scattered neutrons, we will find the dispersion law of dressed phonons. In thiscase, k of a phonon should be quantized as usual, because the law of quantization is definedby boundary conditions. But the energy of a dressed phonon should differ from the energy ofa “bare” one according to the formula E phon ( k ) = E phonn ( k ) + α ( k ) E + α ( k ) E + . . . , where E phonn = c k is the energy of a bare phonon, and the constants α j can be determined from amicroscopic calculation. Such constants should be negligibly small (except for the cases where k and k are resonance quantities) and have no influence on the heat capacity of the system.Formulas (104) and (105) are obtained on the basis of the classical approach in Sections 2and 3. The quantum formulas (102) and (103) yield the condition ω = 2 πjuiL " u ¯ hkM c , (106)where k = πj/L . The distinction of formulas (104) and (106) is mainly related to the difference f the values of u and c (velocities of dressed and bare phonons, respectively), since thecorrection u ¯ hk/ ( M c ) is negligible for a macroscopic body. According to our analysis, thevalue of u near resonances should significantly differ from c . Nevertheless, we think thatformulas (104) and (106) describe the same resonance. The difference of these formulas canbe due to the fact that the classical approach in Sections 2 and 3 somewhat distorts the exactquantum solutions.These resonances can be observed by means of the measurement of an electric signal aswell, since the mode ( ω ± iω , k ± ik ) must generate the electric field satisfying the equations[1] div D = 0 , D = ε E + 4 π P s (107)(the last equation was obtained in [16]). For an infinite system whose properties depend onlyon the coordinate z , Eqs. (107) have the solution D = D i z , εE i z = D i z − πP s i z , where D = const . In view of (5), it is clear that the density wave must induce a wave of the field E with the same ω, k . Therefore, the resonance for a density wave ( ω ± iω , k ± ik ) mustbe accompanied by the electric field E with the same frequency ω , ± i and the same wavevector k , ± i [see formulas (42), (43) for He I and (88), (89), (93), (94), (98), (99) for He II; inorder to get solutions with i ≥ k → ik and ω → iω in theformulas; in this case, ω is given by formula (104) or (105)]. One can try to register mode(0 ,
2) in the same way (according to the estimates in Section 2, this mode creates the field E s ∼ ϑ ( ε − c πc E i z with frequency 2 ω ). These properties also show that a phonon can possessa very weak coat without an external field E as well, because a phonon creates oscillations ofthe density, which induces the electric field and the spontaneous polarization of a medium.It is of importance that the wave vector of a phonon at the zero boundary conditions isquantized by the law k = πj/L [42, 43] which coincides with the above condition λ = 2 L/j for a standing wave of the first sound. Therefore, this condition should be satisfied.We wrote no solutions for the resonances with large values of | k ,i | [except for (41)]. How-ever, it is possible that they can be experimentally realized for very large λ of an electricwave.Above we have found the solutions for oscillatory modes of an infinite system, i.e.,without consideration of the boundaries. We note that the one-dimensional field E = E i z sin ( k z − ω t ) can be created only in a resonator with sizes L z ≪ L x , L y . Otherwise, thefield E should depend on three coordinates [16, 44], and the solutions should differ from theabove-presented ones. Since the boundaries change the frequency of the second sound onlyby 2–10 % [20, 32], we expect that the above-presented solutions for frequencies will not bestrongly changed, if the boundaries are taken into account. If L z ≪ L x , L y is satisfied, thenour solutions should be true with good accuracy.If the solutions will be experimentally confirmed, it will be interesting to elucidate whetherthe relation a = 1 is satisfied. Conclusion
According to our study, the external field E = E i z sin ( k z − ω t ) in the presence of phonons(or temperature waves, for He II) should create a set of hybrid acousto-electric or thermo-electric waves (acouelons/“thermoelons”) in a nonpolar liquid dielectric. Such a set accom-panies each wave of the first (second) sound. Such waves-satellites are very weak and un-observable. However, at certain frequencies of a phonon ( ω ) and a field E ( ω ), one of thewaves-satellites should be amplified in the resonance way and can become observable. Accord-ing to our solutions, the hybrid wave should be intense in the case where ω is very close tothe resonance frequency ω resi,j : | ω − ω resi,j | ≡ △ ω i,j ∼ | ϑ i | ω resi,j . In this case, the field E should besufficiently strong in order that many quanta of the field with the frequencies in the interval[ ω resi,j − △ ω i,j , ω resi,j + △ ω i,j ] exist.The resonances are of the parametric nature. Apparently, the absorption lines should beobserved in the electromagnetic spectrum at the energies ¯ hω resi,j equal to the double energies oflowest levels of a system (see (104)). The registration of such lines would mean the observationof a discrete structure of the energy spectrum of a liquid. This is of particular interest, sinceno such structure was directly observed, though its existence raises no doubts. In this case, theresonances should correspond to the absorption of one or several quanta of an electromagneticfield with the momentum recoil to the whole liquid, like the M¨ossbauer effect. The M¨ossbauereffect was earlier observed only in crystals, to our knowledge.We have found the resonance frequencies and the values of amplitudes far from resonances.It is important to find the amplitudes at the resonance points with regard for small nonlinearcorrections in the equations, including the friction. Such solutions would correspond to realsystems. The main questions are the following: Will the above-obtained singularities bepreserved, and will they be observable?[1] L.D. Landau, E.M. Lifshitz, Electrodynamics of Continuous Media , Pergamon, NewYork (1984).[2] I.E. Tamm,
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