Contains reports on one research project.Joint Services Electronics Programs (U.S. Army, U. S. Navy, and U. S. Air Force) under Contract DAAB07-71-C-030

Herzlinger, G. A.

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GENERAL PHYSICSI. MOLECULAR BEAMSAcademic and Research StaffProf. J. R. Zacharias Dr. G. A. Herzlinger Dr. M. G. R. ThomsonProf. J. G. King Dr. E. H. Jacobsen Dr. J. C. WeaverProf. H. H. Brown F. J. O'BrienGraduate StudentsJ. E. Bergmann H. F. Dylla D. G. PayanS. A. Cohen J. A. Jarrell N. D. PunskyW. B. Davis S. R. Jost B. R. SilverD. G. LysyA. CALCULATION OF THE CROSS SECTION FOR He 3 -ROTONSCATTERING IN SUPERFLUID HELIUM1. IntroductionWe have found from our measurements of diffusion in dilute He 3 - He 4 solutions thatthe cross section for the He 3 -roton interaction is energy-dependent, and increaseswith temperature over the temperature range of our experiment.1 The magnitude ofthe cross section can be approximated from our measurements and we find that it ranges-14 2 -14 2from 1. 6 X 10 cm to 2. 3-2. 4 X 10 cm as the temperature increases from1. 27 ° to 1. 690 K. To understand these results, we have analyzed a possible model forthe interaction.2. Model for the InteractionIn the picture developed by Feynmann and Cohen 2 a solute He3 atom in liquid He IIcan be thought of as a sphere moving in an inviscid background fluid, and the Feynmann-3Cohen wave function for the He quasi particle includes the velocity potential of thedipole flow pattern of a sphere. The energy spectrum3 of a solute He3 quasi particleis consistent with this hydrodynamic picture:2 2E3 - k /2m 3 , (1)where E represents a binding energy to the liquid, and m 3 is an effective mass whichtakes into account the backflow pattern around the sphere. Specific heat measurementsin dilute solutions verify this picture, and measurements of the normal fluid frac-tion4 in solutions indicate that m 3  2. 7 m 3 .The roton excitation involves many atoms of the fluid. In the Feynmann-CohenThis work was supported by the Joint Services Electronics Programs (U.S. Army,U. S. Navy, and U. S. Air Force) under Contract DAAB07-71-C-0300.QPR No. 105(I. MOLECULAR BEAMS)picture, it arises from the backflow pattern of each atom in the liquid in the presenceof the hydrodynamic continuum formed by all of the other identical He4 atoms. At agiven atom, the exact velocity will depend on the pair correlation function, and corre-lation effects involving many atoms. At distances greater than an interatomic distance,however, the dipole form should give a good approximationV(R) = (-A/ i4) v(krwhere kr determines the strength of the dipole and is nearly independent of energy, andthe quantity m 4 /A is approximately 4Twp, where p is the liquid density.Thus, the situation is similar to that of two spheres moving in an inviscid backgroundfluid where the dipole flow pattern of each moving sphere produces an added pressure onthe other. To lowest order in 1/R (neglecting terms of 1/R 6 and higher) this interactionisU = 4rP*l1 .2-3(Pl R)(2 t ) ,where = 1/2 a3 u is the dipole moment, with a the radius, and u the velocity of thesphere. For the case of the He -roton interaction, we use the same form for U, withm -m hk4 3  mk3  3 = hk (3)3 4Tp 3' m 4  r 4TrpAt very small distances, the higher order terms are important, and in addition, thedipole approximation for the roton will probably not be valid, To take account of theseeffects in a crude way, the wave function will be cut off at a small distance a, and Uwill be taken as c for U < a.3. Method of SolutionThe problem is to calculate the cross section of the velocity-dependent interaction (2)for two particles, one having the classical energy spectrum (1), the other having thespectrum Er = A + (h 2/2[)(k r-k )2. Putting the problem into the center-of-momentumsystem does not reduce the Hamiltonian function to a one-body problem as is usually thecase. A partial wave expansion, then, does not seem appropriate. Instead, we usefirst-order perturbation theory, with two-particle wave functions. A perturbation theorycalculation of the He 3 - roton interaction has been previously derived in the well-knownwork of Khalatnikov and Zharkov, who took U(R ) = V 6(R) and found that o was inde-pendent of energy. In the calculation presented here, the dipole-dipole interaction (2)will be used with the cutoff described above. The initial roton velocity will also be takenQPR No. 105(I. \OLECULAR BEAMS)into account. (The roton was assumed to be initially at rest by Khalatnikov and6Zharkov. ) It should be noted that the dipole-dipole interaction (2) has been recently7, 8used in connection with the roton-roton interaction.The He 3 diffusion cross section is the transition rate divided by the input tie 3 cur-rent, which for a single particle is Vrelative /N. The expression for the cross sec-tion isdo- = K 3 U( 1'3 ) 21 6(E3+Er- ' -E'(1 -cos ) d 3 k d3 kx (4)2 2'rel (2?)3 "where f is a factor involving the change in variable to be made in the delta function fromE to k', and y is the scattering angle of the He3 quasi particle.The solution will involve determination of the matrix element and integration ofexpression (4) over the allowed phase space. Since two-particle wave functions areused, momentum conservation is not implicit as with the usual one-body case derivedin the CMI frame. The three equations from momentum conservation, plus the energydelta function, impose 4 constraints on the integration so that the phase space ofallowed states will be two-dimensional. The problem is simplified by changing vari-ables for the final states to d Akd Ak', where Ak k' - k k= k - k and, by3 33  r = - , and, byconservation of momentum, Ak -k' . After adding the appropriate delta functionso3to take this into account, and integrating over d3Ak', we are left with a single integralover d3 k. The relation between iA! and (0, ' ), the angles defining Ar, determinesa surface in phase space over which the integration proceeds.4. Kinetics - Conservation of Energy and Momentum in CollisionsTo first order, kr is constant, equal to ko, and an elementary calculation showsthat Ak = 2kr cos , where is the angle between kr and Ak (see Fig. I-la). To secondorder, let (Ak - 2kr cos ) be a small quantity. With this approximation, conservationof energy leads to2(V cos - V cos 0)cakosO k, (5)V + k /mr r 3A A A Awhere V r  (k -k )/, V3 - k 3/m , and cos 0 - k Ak, cos -kr Ak. With aangle of incidence between k3 and kr' and ¢' the azimuthal angle between Ak andQPR No. 105(I. MOLECULAR BEAMS)the plane formed by k 3 and kr, we havecos c = cos 0 cos a + sin 0 sin a cos 4'.(See Fig. I-lb.)To perform the integration, the argument of the delta function expressing energyTr/2 -0 Kr()/2-(a)Fig. I-i.(a) Lowest order approximation, Ak= 2kr cos P.(b), (c) Angles defining the orientation of thevarious momentum vectors.conservation(c)d(arg)k sincemust be written in terms of Ak with an added factor 1 dAk sincethe energy delta function is of the form 6(f(Ak)). This turns out to bed(arg)d(Ak)=1 2[(kr-ko) cos /L- k3 cos 0/ m3or just Zkr/(numerator in the IAkJ expression) (Eq. 5).QPR No. 105(I. MOLECULAR BEAMS)5. Evaluation of the Matrix ElementThe initial and final wave functions are plane wavesi5 . r3 ik r ik r ik' .r1 33 r r 1 3 3 r ri =-- e e f =-- e eV VTo evaluate (Ui U)Lf) (where U = oc, for r < a, and U is the expression in Eq. 2 forr > a), it is simple to show that what is needed is1 -ia - R (8)V kawith R = r 3 - rr. The integral can be shown to bem [3(k3 Ak)(k , Ak)- k k ]  " (9)rrp 3 r 3 r AkaIf we put A i as a function of (0, 4') in (9), we get a clumsy complicated expres-sion. For Aka < 1, however, 3/ l(Aka)/(Aka) z 1, and the expression is greatly sim-plified. The assumption made in this simplification is that the "hard core" radiusmultiplied by the momentum change in the collision is small. (For Ak of the orderof k 3 , and a of the order of a fraction of an angstrom, the approximation is seen tobe valid.)6. Diffusion Cross-Section Factor 1/2 (1-cosy)The main purpose of this factor is to eliminate the effect of small-angle scatteringand emphasize large-angle scattering in the integral. The problem is being solved byAusing 0, the angle between Ak and k 3 , rather than 4, the angle between k3 and k 3,so an approximation will be used. From Fig. I-1 it is seen thatcos = (k + A k cos 0)/k .  (10)A simple calculation shows that the average thermal energy of a roton is 1/2 kT, whilethat of a He 3 atom is 3/2 kT, so that on the average the energy transferred to a He 3 atomin a collision will not be large, and i 3i will not be too different from l . Then1 Akcos4) 1 + (Ak/k3) cos 0 or - (- cos )= - kcos . (11)3Note that the cos 0 factor eliminates small-angle scattering and increases theamplitude for large-angle scattering, as desired.QPR No. 105(I. MOLECULAR BEAMS)7. Evaluation of the Phase Space IntegralWe have previously derived a relation between Akjl and (0, ' ). The differentialarea of the surface defined by this relation isdA = (Ak)2 sin Odod '/cos y, (12)where y is the angle between the normal to the surface and Ak. If the surface definedby (5) is written as g(Ak, 0, 6 ) = 0, then cosy = Vg - Ak/ vg I and,dA = Ak 2 sin Odd' 1 + + (13)Ak 2  sin Aki 8 dBy using (5) and evaluating (13), we find that2kdA - -- V3 Ak sin @d0d'. (14)(VTr +k /m)Note that the factor of the relative velocity in the input current is canceled by a similarfactor in the differential area of the allowed phase space.The cross section can now be evaluated by substituting the various factors in (4)from (7), (8), (9), (11), and (14), using (5) and (6), and the approximation of our eval-uation of the matrix element. The result isScr (2/T 3 ) 2 k 5 k / +Vr I,and I is a somewhat involved angular integral. After evaluating the integral, the crosssection is determined for a given k 3 and k r. By integrating the result over all anglesof incidence a, and substituting the average values of k 3 and kr at a given tempera-ture, we determine - (T). \We now use8 1/2 -1k3 = 0. 71 x 108 T cm(k -k )/k = 0. 0 60 Tn 3 = 2. 7 m 3S= 0. 1 6 ni4 (curvature parameter in the roton spectrum)QPR -No. 105(I. MOLECULAR BEAMS)k = 1. 92 x 108 crnp = 0. 141 gm cm 3 (He 4 density).The result is125- 14 (1+.06 T1 ) 2cr = 2.7 X 10 T crnm12 3(1+.82 T1/2To compare this result with the values for o- derived from our diffusion measure-ments, we evaluate 0- at the end points of our experimental temperature range:-14 2T 1. 27 o = 0. 67 X 10 cm-14 2T = 1. 690 a = 0. 76 X 10 cm .-14 'Thus cr increases with energy and is of order 10- cm . The experimental andcalculated values for a- agree within a factor of two and a half to three in magnitudeand the calculated values increase less rapidly with temperature. Note that the short-range part of the interaction has been dealt with very crudely, and only first-order per-turbation theory has been used. Nevertheless, a plausible model has been presentedwhich leads to an expression for the cross section which increases with energy, andis of the correct order of magnitude, without containing adjustable parameters.G. A. HerzlingerIe ferences1. G. A. Herzlinger and J. G. King, to be published; G. A. Ilerzlinger, Ph. D. Thesis,Department of Physics, -I. I. T. , August 1971 (unpublished).2. R. P. Feynmann and AI. Cohen, Phys. Rtev. 102, 1189 (1956).3. I. Pomeranchuck, Zh. Eksp. Teor. Fiz. [Sov. Phys. - JETP] 19, 42 (1949).4. t. De Bruyn Ouboter, K. Taconis, C. LePair, and J. J. Beenakker, Physica 26,853 (1960).5. See discussion by G. V. Chester, in G. Careri (Ed.), Liquid Helium (AcademicPress, Inc., New York, 1963).6. I. I. Khalatnikov and V. N. Zharkov, Sov. Phys. - JETP 5, 905 (1957).7. J. Yau and -XI. Stephen, Phys. Rtev. Letters 27, 482 (1971).8. F. Toigo, Nuovo Cimento 62B, 103 (1969).QPR No. 105 7

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