aa r X i v : . [ c ond - m a t . s up r- c on ] J un On Sound Reflection in Superfluid
L.A. Melnikovsky ∗ Abstract
We consider reflection of the first and the second sound waves by a rigidflat wall in superfluid. Nontrivial dependence of the reflection coefficientson the angle of incidence is obtained. Sound conversion is predicted atslanted incidence.
Bulk superfluid has twice as many variables as a normal fluid[1]. Direct conse-quence is the existence of two independent sound modes in helium II referredto as the first sound and the second sound. Due to anomalously small thermalexpansion of helium, these modes can be viewed as purely pressure and temper-ature waves respectively. The “purity” is essential, e.g. , for the effectiveness ofsound emission by various sources.Restricted geometry effectively eliminates some of the hydrodynamic vari-ables. Particularly, a steady cell wall eliminates the normal component of themass flux j and the normal velocity v n : the boundary condition at the wall is j ⊥ = 0 , v n = 0 . (1)The fourth sound [2] is one result of such elimination. This is the only soundmode in a narrow channel, both temperature and pressure oscillate coherentlyin this wave. In general, the sound modes, independent in bulk liquid, begin tointeract at the boundary. ∗ This necessitates complete two-fluid considerationof the sound reflection in superfluid which is the subject of the present paper.To avoid complications associated with the wall deformation consider a per-fectly rigid flat wall. We therefore ignore numerous peculiarities of the soundtransmission into solids and limit ourselves to linear hydrodynamic equationsonly. In Sec.2 we find all three nontrivial harmonic ( i.e. , proportional to e i kr − i ωt )solutions of the equations. It is important to mention that unbounded solutions(those with complex wave vector k ) should not be overlooked in the restrictedgeometry. ∗ E-mail: [email protected] ∗ Sound reflection at the free helium surface and at the solid helium boundary is extensivelyexplored [3],[4], [5], and [6]. ω . Specific solutions correspond to the first soundreflection (Sec.3) and the second sound reflection (Sec.4). Interestingly enough,some part of the incident wave energy is transformed between the first and thesecond sound. Consider the linearized equations of superfluid hydrodynamics[8]:˙ ρ + ∂j i ∂x i = 0 , (2)˙ j i + ∂p∂x i = η ∂∂x k (cid:18) ∂v i n ∂x k + ∂v k n ∂x i − δ ik ∂v l n ∂x l (cid:19) + ∂∂x i (cid:18) ζ ∂v l n ∂x l − ζ ∂ρ s w l ∂x l (cid:19) , (3)˙ v k s + ∂µ∂x k = ∂∂x k (cid:18) ζ ∂v l n ∂x l − ζ ∂ρ s w l ∂x l (cid:19) , (4) T (cid:18) ˙ σρ + σ ˙ ρ + σρ ∂v l n ∂x l (cid:19) = κ ∂ T∂x l ∂x l , (5)where η , ζ = ζ , ζ , ζ , κ are dissipative coefficients, σ is entropy per unitmass, p , µ , T are pressure, chemical potential, and temperature, v s , v n , and w = v n − v s are superfluid, normal, and relative velocities, ρ and j are mass andmomentum densities. The velocities and the momentum density are coupled bythe equation j = ρ v s + ρ n w = ρ v n − ρ s w . (6)Further simplification is facilitated by ignoring thermal expansion (we there-fore disregard the difference between specific heats c = T ∂σ/∂T at constantpressure and at constant volume). Namely put p ′ = s ρ ′ , (7) ρµ ′ = − σρT ′ + p ′ = − σρT ′ + s ρ ′ , and (8) T σ ′ = cT ′ , (9)where s = ( ∂p/∂ρ ) / is the first sound velocity. The prime denotes smalldeviation of the variables from their equilibrium values.In a harmonic perturbation, space and time dependence of all deviationshas a form of exp(i kr − i ωt ). To find all possible harmonic excitations in bulksuperfluid we substitute this exponential term in Eqs.2-5 and keep linear termsonly.The mass conservation (2) gives ωρ ′ = k i j i . (10)The momentum conservation law (3) can be transformed as follows: − i ωj i + i p ′ k i = − ηk k (cid:18) v i n k k + k i v k n − δ ik k l v l n (cid:19) − k i k k (cid:0) − ρ s ζ w k + ζ v k n (cid:1) , ω + i ηk /ρ ) j i + (cid:0) i A − s /ω (cid:1) k i k k j k + i ηk ρ s w i /ρ + i Bk i k k w k = 0 , (11)where the constants A = ( η/ ζ ) /ρ and B = ( A − ζ ) ρ s .From the energy conservation law (5) for harmonic deviation we get T ωσ ′ ρ + T ωσρ ′ − T σρv l n k l + i k κT ′ = 0 . Using (6), (9), and (10) this can be reduced to
T σρ s k i w i = ( cωρ + i κk ) T ′ . (12)Finally, substituting exponential term and (6) in (4) we obtain − i ω ( j i − ρ n w i ) /ρ + i k i µ ′ = − k i (cid:0) − ρ s ζ w k k k + ζ (cid:0) j k + ρ s w k (cid:1) k k /ρ (cid:1) Combining this with (8), (10), and (12) ωj i + (cid:0) i ζ − s /ω (cid:1) k i k k j k − ωρ n w i + Gk i k k w k = 0 , (13)where G ≈ T ρ s σ cω − i ρ s (cid:18) T κk σ c ω ρ − ( ζ − ρζ ) (cid:19) Eqs. (11) and (13) can be written together asˆ L (cid:18) jw (cid:19) = 0 , where ˆ L is a square 6 × L = 0. Due to the system isotropy, thedeterminant can not depend on individual components of k i . Instead it dependson k = k i k i only. We therefore can put k y = k z = 0 and treat ˆ L as a 4 × L = ω + (cid:0) − s /ω + i A + i η/ρ (cid:1) k ηρ s /ρ + B ) k ω + i ηk /ρ ηk ρ s /ρω + (cid:0) − s /ω + i ζ (cid:1) k − ωρ n + Gk ω − ωρ n After factorization det ˆ L simplifies to (cid:12)(cid:12)(cid:12)(cid:12) ω + (cid:0) i A + i η/ρ − s /ω (cid:1) k i( ηρ s /ρ + B ) k ω + (cid:0) i ζ − s /ω (cid:1) k − ωρ n + Gk (cid:12)(cid:12)(cid:12)(cid:12) · (cid:12)(cid:12)(cid:12)(cid:12) ω + i ηk /ρ i ηk ρ s /ρω − ωρ n (cid:12)(cid:12)(cid:12)(cid:12) . All nontrivial solutions immediately follow: ω ≈ k (cid:18) s − i ωA − i ωηρ (cid:19) ≈ k (cid:18) s − i ωρ (cid:18) η ζ (cid:19)(cid:19) ≈ k s , (14) ω ≈ k ω ( G − i ηρ s ρ − i B ) /ρ n ≈ k T ρ s σ cρ n ≡ k s , (15) ω = − i ηk /ρ n , (16)3Sfrag replacements k I xy k R k R φ φ φ φ Figure 1: First sound reflectionwhere s is the second sound velocity.Roots k (14) and k (15) correspond to “longitudinal” solutions where j i ∝ w i ∝ k i , , while the root k (16) corresponds to a “transverse” one j i k i = w i k i = 0. The approximation in (14) and (15) is based on an assumption of lowbulk damping, i.e. , | k | ≫ | k | > | k | . This implies complete splitting betweenthe first and second sound, namely w = 0 for (14) and j = 0 for (15). In thethird solution (16), the superfluid velocity vanishes v s = 0, i.e. , the mass fluxand the relative velocity are coupled by the relation j = ρ n w . Consider the first sound wave incident towards the impervious rigid plane wallat an angle φ (see Fig.1). The subscripts I R
1, and R x axisruns along the wall and the y axis is directed into the liquid.The heat transfer through the interface at low temperature can be neglecteddue to Kapitza resistance. Appropriate boundary conditions are j y = 0, w y = 0, j x + ρ s w x = 0. They can be written in the matrix form j I sin φ − j I cos φ + j R sin φ j R cos φ + w R sin φ w R cos φ + ρ n w x ρ n w y w x w y = − ρ s w w , (17)where cos φ = 1 − ( s /s ) sin φ , to satisfy the condition k xI = k xR . The lastterm in the left-hand side of (17) represents the transverse surface wave witha wave vector k . The wave must decay away from the boundary, thereforeIm k y >
0. This requirement selects the sign in (3), which is the transversality4elation w ⊥ k : (cid:18) w x w y (cid:19) ∝ (cid:18) − k y k x (cid:19) ≈ (cid:18) − k k sin φ (cid:19) = (cid:18) − κ e i π/ k sin φ (cid:19) = (cid:18) − κ e i π/ k sin φ (cid:19) , where κ = p ωρ n /η . Substituting this in (17) we get j I sin φ − j I cos φ + j R sin φ j R cos φ + ρ s w R sin φ w R cos φ + ρw x ρ n w y w y = 0and 2 j I + w R (cid:18) ρ s sin φ sin φ + ρ n cos φ cos φ + e i π/ κ ρ cos φ k sin φ (cid:19) = 0Second sound is slower than the first one s > s , consequently φ < π/ φ = 0. One can therefore neglect the first term in parenthesis w R = − j I sin φ cos φ k ρ n k sin φ tan φ + e i π/ κ ρ . (18)Similarly, the amplitude of the reflected first sound is obtained from the equation (cid:18) j I ρ n k sin φ − j I ρ κ e i π/ cos φ (cid:19) + (cid:18) j R ρ n k sin φ j R ρ κ e i π/ cos φ (cid:19) + (cid:18) ρρ n k sin φ w x ρρ n κ e i π/ w y (cid:19) = 0 . From this we have j R j I = ρ κ cos φ − ρ n k sin φ e − i π/ ρ κ cos φ + ρ n k sin φ e − i π/ . (19)Reflection and conversion efficiency must be characterized by appropriatecoefficients R = F R /F I and R = F R /F I respectively. Here F and F are the energy fluxes in the first and second sound waves. They are given bythe expressions F = s ρ | j | , F = s ρ s ρ n ρ | w | . Using (19) and (18) we get R = (cid:12)(cid:12)(cid:12)(cid:12) ρ κ cos φ − ρ n k sin φ e − i π/ ρ κ cos φ + ρ n k sin φ e − i π/ (cid:12)(cid:12)(cid:12)(cid:12) , (20) R = s s sin φ cos φ ρ s ρ n k (cid:12)(cid:12) ρ n k sin φ tan φ + ρ κ e i π/ (cid:12)(cid:12) . (21)Sample graph of these functions is illustrated on Fig.2. The reflection coefficient R has a minimum of min R = 3 − √ PSfrag replacements R R φ Figure 2: Reflection and conversion coefficients R and R vs. the angle ofincidence φ . The same approach can be used to investigate the second sound wave incidentat an angle φ (see Fig.3). The boundary conditions in this case are w I sin φ − w I cos φ + w R sin φ w R cos φ + j R sin φ j R cos φ + ρ n w x ρ n w y w x w y = − ρ s w w . After simplification this gives2 ρ s ρ n w I = j R (cid:18) − ρ κ e i π/ cos φ k sin φ − ρ n sin φ sin φ − ρ s cos φ cos φ (cid:19) The last term in parenthesis is always negligible (last equation is meaningfulonly if sin φ < s /s ). This gives j R = − ρ s w I ρ n k sin φ ρ κ e i π/ cos φ + ρ n k sin φ sin φ . (23)The conversion coefficient R = F R /F I is therefore given by R = 4 ss ρ s ρ n k sin φ (cid:12)(cid:12) ρ κ e i π/ cos φ + ρ n k sin φ sin φ (cid:12)(cid:12) . (24)Its maximum max R = 4 ρ s s ρ n s (25)6Sfrag replacements xy k R k R k I φ φ φ φ Figure 3: Second sound reflectionis reached at the critical angle sin φ = s /s .Amplitude of the reflected second sound wave is found from the relation w R w I (cid:18) κ ρ e i π/ k sin φ + ρ n tan φ + ρ s tan φ (cid:19) = (cid:18) κ ρ e i π/ k sin φ + ρ n tan φ − ρ s tan φ (cid:19) , (26)where tan φ = s sin φ (cid:30)q s − s sin φ and Im tan φ ≤ k y ≥ R = (cid:12)(cid:12)(cid:12)(cid:12) ρ κ e i π/ + ρ n k sin φ tan φ − ρ s k sin φ tan φ ρ κ e i π/ + ρ n k sin φ tan φ + ρ s k sin φ tan φ (cid:12)(cid:12)(cid:12)(cid:12) . (27)These functions for sample parameters are plotted on Fig.4. It is shown that sound reflection at slanted incidence by a plane impervious wallis suppressed for both first and second sound. This phenomenon is similar toKonstantinov effect in ordinary gases.Coincidentally with the reflection suppression a sound conversion takes place.The effect is predicted to have strong angle dependence and should allow exper-imental verification. Moreover, there is a number of the heat pulse propagationmeasurements ( e.g. , [9] and [10]) where the pulse transit time was often muchshorter than that for the second sound. This phenomenon is usually explainedby anomalously long phonon free path at low temperatures or by sound con-version in bulk (due to nonlinear effects) or at liquid-vapour interface. It seemsprobable that fast propagation is in fact the manifestation of the sound conver-sion described in this paper, so that the heat pulse is transformed at some wallinto the pressure pulse and is later transformed back near the receiver. The7
PSfrag replacements R R φ Figure 4: Reflection and conversion coefficients R and R vs. the angle ofincidence φ .signal therefore travels (some part of) the path with the velocity of the firstsound. Acknowledgements
I thank A.F. Andreev and V.I. Marchenko for fruitful discussions. The work wassupported in parts by RFBR grants 06-02-17369, 06-02-17281 and RF presidentprogram 7018.2006.2.
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