Lubrication Forces. Ranga University of Edinburgh, UK (Dated: June 21, 2017) I. COMPUTING FORCES AND TORQUES

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1 Lubrication Forces Ranga University of Edinburgh UK Dated: June I. COMPUTING FORCES AND TORQUES The lubrication force between a pair of particles say p = 1 q = are given as a product between the Gr Resistance Matrix which only depends on the positions of the two particles a vector containing particle background fluid velocities in Refs. [1 3]. Our task below is to simplify the relation between the lubrication forces torques given in Refs. [1 3] for spheres of different sizes into a form similar to the one given by Ball&Melrose for spheres of the same size [4]. 1 a FIG. 1. Sketch of the two spheres with the normal vector indicated by the arrow. We have sphere p = 1 of radius a at position x 1 separated by a distance h from sphere q = of radius b at position x as shown in Fig. 1. The superscripts denote the particle number later the subscript will denote the component index. The total center to center distance is given by R = x 1 x = a + b + h. The separation between the two spheres is non-dimensionalised as ξ = R a b a + b the normal vector connecting the two spheres is given by = n = x x 1 x x 1 = x x 1 R h a + b/ 1. The velocity vectors of the spheres are given by U 1 U the angular velocities by ω 1 ω. The background fluid velocity is assumed to be linear with uniform velocity given by v angular velocity given by Ω rate of strain tensor given by E. According to Ref. [] the lubrication force acting on sphere 1 is given by h b F 1 /µ =A v x 1 U 1 + A 1 v x U + B Ω ω 1 + B 1 Ω ω + G E + G 1 E 3 the torque is given by T 1 /µ =B v x 1 U 1 + B 1 v x U + C Ω ω 1 + C 1 Ω ω + H E + H 1 E 4 where the bold symbols denote resistance functions which operate on a vector say χ = χ j as specified below A pq ij = Xpq A n in j + Y pq A δ ij n i n j B pq ij = Y pq B ɛ ijkn k C pq ij = Xpq C n in j + Y pq C δ ij n i n j G pq ijk = Xpq G n in j δ ij /3n k + Y pq G n iδ jk + n j δ ik n i n j n k H pq ijk = Y pq H ɛ ikln l n j + ɛ jkl n l n i. 5

2 Further Bpq ij = B qp ji Gpq ijk = Gqp formulation. Therefore we also have jki Hpq ijk = Hqp jki are required due to symmetry reasons of the resistance matrix B ij = Bji = YB ɛ ijk n k B ij 1 = Bji 1 = YB 1 ɛ ijk n k G ijk = G jki = XG n j n k δ jk /3n i + YG n j δ ki + n k δ ji n j n k n i G 1 ijk = G 1 jki = XG 1 n j n k δ jk /3n i + YG 1 n j δ ki + n k δ ji n j n k n i H ijk = Hjki = YH ɛ ijl n l n k + ɛ ikl n l n j H ijk 1 = Hjki 1 = YH 1 ɛ ijl n l n k + ɛ ikl n l n j. 6 The different components of the resistance functions specified by symbols such as X pq A Y pq A... are given in Sec. VI the relations between them are specified in Eq. 54. It follows that the lubrication force given in Eq. 3 can be written in component form as F 1 i /µ = X A n i n j + A δ ij n i n j U j U 1 j ɛ ijk Y B Ω j ωj 1 + YB 1 Ω j ωj n k + XA n i n j + YA δ ij n i n j vj x 1 vj x + XG + XG 1 n j n k δ jk /3n i Ejk + YG + YG 1 n j δ ki + n k δ ji n j n k n i Ejk 7 the torque given by Eq. 4 in component form is T 1 i /µ =ɛ ijk B U j U 1 j n k ɛ ijk B v j x v j x 1 n k + δ ij n i n j C Ω j ω 1 j + Y 1 C Ω j ω j H + Y 1 H ɛ ijl n l n k + ɛ ikl n l n j E jk. 8 Our objective is to make use of the relations between the different terms in Eq. 54 the simplifications in Sec. V to convert the forces torques above into simpler forms.

3 3 II. COMPUTING THE FORCE ACTING ON THE SPHERES A. Grouping terms in Eq. 7 Making use of the relations in Eq. 54 Sec. V we can get X A n i n j v j x 1 v j x + X G + X 1 G n j n k δ jk /3n i E jk =X A E jkn i n j n k R + a + b = X A E jkn i n j n k a + bξ/ 0 9 A δ ij n i n j vj x 1 vj x + YG + YG 1 n j δ ki + n k δ ji n j n k n i Ejk = YA δij n i n j Ejkn k + ɛ ikl Ω k n l R + Y A δ ij n i n j n k Ejka + b = YA ɛ ikl Ω k n l R YA δ ij n i n j Ejkn k a + bξ/ YA ɛ ikl Ω k n l R 10 A ɛ ijk Ω j n k R B + Y 1 B ɛ ijk Ω j n k = A ɛ ijk Ω j n j R a + b = A ɛ ijk Ω j n j a + bξ/ 0 Using the results from Sec. II A Eq. 7 simplifies to B. Simplified Lubrication force acting on sphere 1 F 1 i /µ = X A n i n j + A δ ij n i n j U j U 1 j + ɛ ijk Y B ω 1 j + Y 1 B ω j nk. 1 Eq.1 can be written as F 1 i /µ = X A n i n j + A δ ij n i n j Uj Uj 1 1 a + b ɛ ijk A ω 1 j + ω j n k + ba + 4b a4a + b B ɛ ijk ω 1 j ω j n k /. 13 C. Force acting on sphere According to the form given by Ref. [] F /µ =A 1 v x 1 U 1 + A v x U + B 1 Ω ω 1 + B Ω ω 14 + G 1 E + G E And similar to Eq. 7 we can write Eq. 14 after using the relations given in Eq. 54 as F i /µ = X A n i n j + A δ ij n i n j U j U 1 j + ɛ ijk Y B Ω j ωj 1 + YB 1 Ω j ωj n k XA n i n j + YA δ ij n i n j vj x 1 vj x XG + XG 1 n j n k δ jk /3n i Ejk YG + YG 1 n j δ ki + n k δ ji n j n k n i Ejk. 15 Comparing Eq. 7 Eq. 15 we get F i = F 1 i.

4 4 III. LUBRICATION TORQUES A. Grouping terms in Eq. 8 Making use of the relations in Eq. 54 Sec. V we can get which implies that H + Y 1 H ɛ ijl n l n k + ɛ ikl n l n j E jk =a + b B ɛ ijl n l n k E jk YC + YC 1 δ ij n i n j Ω j = a + byb δ ij n i n j Ω j B ɛ ijl v j x v j x 1 n l = a + b1 + ξ/ B ɛijl n k n l E jk δ ij n i n j Ω j 16 YC + YC 1 δ ij n i n j Ω j ɛ ijk YB vj x vj x 1 n k YH + YH 1 ɛ ijl n l n k + ɛ ikl n l n j Ejk = ξa + byb ɛijl n k n l Ejk + δ ij n i n j Ω 17 j / = Oξ 0. Using the above equations we get B. Torque acting on sphere 1 T 1 i /µ = B ɛ ijk U j U 1 j n k δ ij n i n j C ω 1 j + Y 1 C ω j. 18 We can try further simplification by noting Therefore Eq. 18 becomes simply T 1 i /µ = B ɛ ijk U j U 1 j n k + a + b B δ ij n i n j ω 1 j + ω j / + 1 4a/bY 1 C δ ij n i n j ω 1 j ω j / 19 C. Torque on sphere For the torque acting on sphere we have which turns out to be T /µ =B 1 v x 1 U 1 + B v x U + C 1 Ω ω 1 + C Ω ω + H 1 E + H E T i /µ =ɛ ijk Y 1 B U j U 1 j n k ɛ ijk Y 1 B v j x v j x 1 n k + δ ij n i n j Y 1 C Ω j ω 1 j + Y C Ω j ω j Y H + Y 1 H ɛ ijl n l n k + ɛ ikl n l n j E jk. Analogous to the previous calculations this makes Eq. 1 to be Therefore Eq. becomes simply T i /µ = Y 1 B ɛ ijk U j U 1 j n k δ ij n i n j Y 1 C ω 1 j + Y C ω j. 0 1 T i /µ =Y 1 B ɛ ijk U j U 1 j n k + a + by 1 B δ ij n i n j ω 1 j + ω j / 1 4b/aδ ij n i n j Y 1 C ω 1 j ω j / 3 Equally Ti ba + 4b /µ = a4a + b Y B ɛ ijk Uj Uj 1 n k + 1 4b/aY 1 C δ ij n i n j ω 1 j ω j / ba + ba + 4b YB δ ij n i n j ωj 1 + ωj ab + 4a 4

5 5 IV. TESTING THE SIMPLIFIED EXPRESSIONS A. Comparison of the forces torques to Ref. [4] For equal spheres b = a Eq. 13 for the force becomes F 1 i /µ = X A n i n j + A δ ij n i n j U j U 1 j ɛ ijk A a ω 1 j + ω j nk = X A n i n j + A δ ij n i n j U 1 j + a ω 1 n j U j a ω n j. 5 is equal to the form given in Ref. [4]. For the torques we have T 1 i /µ = a A ɛ ijk U j U 1 j n k a A δ ij n i n j ω 1 j + ω j 3Y 1 C δ ij n i n j ω 1 j ω j / T i /µ = a A ɛ ijk U j U 1 j n k a A δ ij n i n j ω 1 j + ω j + 3Y 1 C δ ij n i n j ω 1 j ω j /. The first two terms are the same as Ref. [4]. However the third term is which is 4 times the value given in Ref. [4]. 3Y 1 C / = 3 5 πa3 ln1/ξ B. Shearing motion of two rigid surfaces This is an example problem in Chap. 9 of Ref. []. Sphere 1 is moving past a fixed sphere along the x-axis at a constant velocity U. We have U 1 = U ˆx U = 0 ω 1 = 0 ω = 0 v = 0. The force on particle 1 given by Eq. 1 after substituting for the different values of velocities becomes after substituting for A F 1 x = µ A U 6 from Sec. VI we get Fx 1 6πµaU = 4β + β + β β 3 ln ξ 1. 7 The torque on particle 1 given by Eq. 18 after substituting for the different values of velocities becomes T 1 y = µ B U 8 1 z a b U x FIG.. Sketch of the two spheres with the normal vector indicated by the arrow. after substituting for B from Sec. VI we get T 1 y 8πµa U These results are the same as that Eq. 9.4 Eq. 9.5 in Ref. []. = β4 + β β. 9

6 6 C. Shearing motion of two rigid surfaces due to rotation This is an example problem in Chap. 9 of Ref. []. Sphere 1 is rotating near a fixed sphere about the y-axis at a constant angular velocity ω. We have U 1 = 0 U = 0 ω 1 = ωŷ ω = 0 v = 0. The force on particle 1 given by Eq. 1 after substituting for the different values of velocities becomes after substituting for B F 1 x = µ B ω 30 from Sec. VI we get Fx 1 β4 + β 8πµa = ω β. 31 The torque on particle 1 given by Eq. 18 after substituting for the different values of velocities becomes after substituting for C T 1 y = µ C ω 3 from Sec. VI we get These results are the same as that Eq. 9.6 Eq. 9.7 in Ref. []. T 1 y 1 z a b x FIG. 3. Sketch of the two spheres with the normal vector indicated by the arrow. 8πµa 3 ω = β β. 33 D. Comparison of the expressions with lubricate/poly pair style in LAMMPS In the LAMMPS implementation of lubricate/poly we have where the velocities V 1 j Fi 1 /µ = XA n i n j + YA δ ij n i n j Vj Vj 1 34 are defined as The previous equations imply that Vj 1 = Uj 1 vj x 1 + aɛ jlm ωl 1 Ω l n m aejmn m 35 Vj = Uj vj x bɛ jlm ωl Ω l n m + bejmn m. 36 V j V 1 j = U j U 1 j ɛ jlm aω 1 l + bω l n m v j x v j x 1 + a + be jm + ɛ jlm Ω l n m U j U 1 j ɛ jlm aω 1 l + bω l n m = U j bɛ jlm ω l n m U 1 j + aɛ jlm ω 1 l n m 37 by using the definition for the infinite velocities given in Sec. V noting that R a + b. Using the above Eq. 34 becomes Fi 1 /µ = XA n i n j + YA δ ij n i n j Uj Uj 1 ɛ ilm YA aωl 1 + bωl n m 38 which is equal to Eq. 5 when both particles are of the same size a=b if X A = X A Y A = Y A. 1. Difference between the terms in lubricate/poly Eq. 1 If we set XA = X A Y A = Y A in Eq. 38 we compare it with Eq. 1 we obtain that the pre-factors to the tangential forces due to rotation of the spheres are different between them while the force contribution due to linear velocity components are the same. The difference in the rotational contributions can be found out to be B + a A = 1πa b b a 5a + b 3 log 1/ξ Y 1 B + b A = 1πa b a b 5a + b 3 log 1/ξ 39 for ω 1 ω contributions to the rotational forces respectively. For a typical case of particles with the aspect ratio β = b/a = 1.4 we can calculate that lubricate/poly will under-estimate force due to rotation on particle 1.% over-estimate the force due to rotation on particle by 8.% even if other terms in the force class were corrected.

7 7 V. SIMPLIFICATIONS By definition of the rate of strain tensor E angular velocity vector ω we know that v j x = E jk x k + ɛ jkl Ω k x l. It implies Taking the dot product of Eq. 40 by n i n j we have v j x v j x 1 = E jkx k x 1 k + ɛ jlk Ω l x k x 1 k = E jkn k + ɛ jlk Ω l n k R. 40 n i n j vj x vj x 1 = Ejkn k n i n j + ɛ jlk Ω l n k n i n j R = Ejkn i n j n k R by dot product with the unit tensor δ ij vj x vj x 1 = Ejkn k δ ij + ɛ jlk Ω l n k δ ij R = Eik n k + ɛ ilk Ω l n k R. From the above two equations we have 41 4 Similarly we can also get δ ij n i n j v j x v j x 1 = E jkδ ij n i n j n k + ɛ ilk Ω l n k R. 43 ɛ ijk vj x vj x 1 n k = Rɛ ijk Ejmn m + ɛ jlm Ω l n m n k = R ɛ ijk Ejmn m n k δ kl δ im δ km δ il Ω m n l n k = R ɛ ijk Ejmn m n k δ im n m n i Ω m = 1 + ξ/a + b ɛ ijk E jmn m n k δ ij n i n j Ω j. 44 VI. NEAR FIELD FORMS OF THE RESISTANCE FUNCTIONS For the sake of completeness the resistance functions as specified in Chap. of Ref. [] up to Oln ξ 1 are given in this section. Ref. [3] only includes the leading terms of the following expressions its normal vector n points in the opposite direction to ours. Note that β = b/a for all the functions below the symmetry relations for the resistance matrix Chap. 7 Ref. [] are satisfied by the expressions below. The components of the A resistance functions are β XA = 6πa 1 + β 3 ξ 1 + β1 + 7β + β 51 + β 3 ln ξ 1 β XA = 6πb 1 + β 1 3 ξ 1 + β β 1 + β 51 + β 1 3 ln ξ 1 4β + β + β YA 45 = 6πa β 3 ln ξ 1 4β YA 1 + β 1 + β = 6πb β 1 3 ln ξ 1 The components of the B resistance functions are XA = XA = XA 1 = XA 1 YA = YA = YA 1 = YA 1. β4 + β B = 4πa 51 + β Y β B = 4πb β β

8 8 YB = YB 1 YB 1 = YB. 48 The components of the C resistance functions are β YC = 8πa 3 YC 1 = 8πa 3 β 51 + β β β YC = 8πb β 1 YC 1 = 8πb 3 β β 1 49 X pq C = O1. 50 The components of the G resistance functions are 3β XG = 4πa 1 + β 3 ξ 1 + 3β1 + 1β 4β β 3 ln ξ 1 3β XG = 4πb 1 + β 1 3 ξ 1 + 3β β 1 4β β 1 3 ln ξ 1 β4 β + 7β YG = 4πa β 3 ln ξ 1 Y G = 4πb β 1 4 β 1 + 7β β 1 3 ln ξ 1 51 XG = XG 1 XG = XG 1 YG = YG 1 YG = YG 1. 5 The components of the H resistance functions are Y β β H = 8πa β β H = 8πb 3 1 β β 1 Y 1 β H = 8πa β 01 + β Y β H = 8πb β β 1. 53

9 9 A. Relations between different components of the resistance functions To make the analysis simple we can note that the following relations based on the formulae given in Ref. [] for the leading terms Oξ 1 Olnξ 1 of the lubrication force C = 4 a b Y 1 C Y C = 4 b a Y 1 C C = a b Y C Y 1 C = Y 1 C X G + X 1 G = a + bx A G + Y 1 G = a + by A B + Y 1 B = a + b A YC + YC 1 = a + byb YC 1 + YC = a + byb 1 YC + YC 1 = YH + YH 1 YC 1 + YC = YH 1 + YH. Other simplifications can be done based on the above relations 54 YB ωj 1 + YB 1 ωj = YB + YB 1 ωj 1 + ωj / + YB YB 1 ωj 1 ωj / 55 = a + bya ωj 1 + ωj ba + 4b / + 1 YB ωj 1 ωj / a4a + b 56 C ω 1 j + Y 1 C ω j = C + Y 1 C ω 1 j + ω j / + C Y 1 C ω 1 j ω j / = a + b B / + 4a/b 1Y 1 C / Y 1 C ω 1 j + Y C ω j = Y 1 C + Y C ω 1 j + ω j / + Y 1 C Y C ω 1 j ω j / = a + by 1 B / + 1 4b/aY 1 C / VII. STRESSLETS AND STRESS According to Ref. [] the stresslet acting on particle 1 is given by S 1 /µ =G v x 1 U 1 + G 1 v x U + H Ω ω 1 + H 1 Ω ω + M E + M 1 E 59 where the resistance function M is given by M pq ijkl = Xpq M d0 ijkl + Y pq M d1 ijkl + Zpq M d ijkl 60 d 0 ijkl = 3 n in j δ ij /3n k n l δ kl /3 d 1 ijkl = 1 n in k δ jl + n j n k δ il + n i n l δ jk + n j n l δ ik 4n i n j n k n l 61

10 10 we can neglect Z pq M = O1 contributions. We can also find the following relationships to simplify our calculations including terms only upto Olog 1/ξ XM + XM 1 = a + bxg /3 YM + YM 1 = a + byg YH + YH 1 = a + byg. 6 Eq. 59 can be written in index notation as S 1 ij/µ = X G n i n j δ ij /3n k + G n i δ jk + n j δ ik n i n j n k v k x 1 U 1 k + XG 1 n i n j δ ij /3n k + YG 1 n i δ jk + n j δ ik n i n j n k vk x Uk + YH ɛ ikl n l n j + ɛ jkl n l n i Ω k ωk 1 + YH 1 ɛ ikl n l n j + ɛ jkl n l n i Ω k ωk + + XM d 0 M d 1 ijkl ijkl + XM 1 d 0 ijkl + Y M 1 d 1 ijkl Ekl Ekl. 63 The above can be simplified to be S 1 ij/µ = X G n i n j δ ij /3n k + G n i δ jk + n j δ ik n i n j n k U k U 1 k ɛ ikl n l n j + ɛ jkl n l n i YH ωk 1 + YH 1 ωk XG n i n j δ ij /3n k + YG n i δ jk + n j δ ik n i n j n k Ekl n l + ɛ kql Ω q n l R + a + byg ɛ ikl n l n j + ɛ jkl n l n i Ω k + a + bxg d 0 ijkl /3 + a + byg d 1 ijkl Ekl. 64 Grouping G terms in the above equation we have 1 = n i δ jk + n j δ ik n i n j n k E kl n l + ɛ kql Ω q n l R + d 1 ijkl E kl + ɛ ikl n l n j + ɛ jkl n l n i Ω k which we can simplify using the following identitities a + b 65 d 1 ijkl E kl = 1 n in k δ jl + n j n k δ il + n i n l δ jk + n j n l δ ik 4n i n j n k n l E kl = n i n k δ jl n j n l E kl n i δ jk + n j δ ik n i n j n k E kl n l = n i n k δ jl n j n l E kl ɛ ikl n l n j + ɛ jkl n l n i Ω k = n i ɛ jkl Ω k n l n i δ jk + n j δ ik n i n j n k ɛ kql Ω q n l = n i ɛ jkl Ω k n l 66 to Grouping some of the X G terms in Eq. 64 we have 1 = Oξ. 67 = n i n j n k n k δ ij /3 E kl n l + ɛ kql Ω q n l R + /3d 0 ijkl E kl a + b 68 which can again be simplified based on the following relation /3d 0 ijkl E kl = n i n j δ ij /3n k n l δ kl /3d 0 ijkl E kl = n i n j n k n l E kl 69 to be = Oξ. 70

11 From 1 Eq. 64 becomes S 1 ik/µ = X G n i n k δ ik /3n j + n k G δ ji n j n i U j U 1 j n i ɛ kjl n l H ω 1 j + Y 1 H ω j. 71 Similarly we can get S ik/µ = X 1 G n i n k δ ik /3n j + n k Y 1 G δ ji n j n i U j U 1 j n i ɛ kjl n l Y 1 H ω 1 j + Y H ω j. 7 Note that the stresslets on particles 1 are not equal in magnitude i.e. S 1 ik S ik. The stress acting between the two particles can be obtained as a sum of the individual stresslets since the total stress is a sum of all the stresslets acting on all the particles. So we have S 1 ik + S ik/µ = X G + X 1 G n i n k δ ik /3n j + n k G + Y 1 G δ ji n j n i U j U 1 j n i ɛ kjl n l H + Y 1 H ω 1 j + Y 1 H + Y H ω j =a + b X A n i n j + A δ ij n i n j U j U 1 j n k + a + bɛ kjl B ω 1 j + Y 1 B ω j n l n i a + bx A U j U 1 j n j δ ik /3 73 The anti-symmetric part of the stress acting between two particles is given by A 1 ij/µ = X G + X 1 G n i n k δ ik /3n j + n k G + Y 1 G δ ji n j n i U j U 1 j n i ɛ kjl n l H + Y 1 H ω 1 j + Y 1 H + Y H ω j =a + b X A n i n j + A δ ij n i n j U j U 1 j n k + a + bɛ kjl B ω 1 j + Y 1 B ω j n l n i a + bx A U j U 1 j n j δ ik /3 74 Let us now compare the stress tensor obtained by dyadic of force of particle 1 to the distance vector which is F 1 i n k R/µ = X A n i n j + A δ ij n i n j U j U 1 j n k R + Rɛ kjl Y B ω 1 j + Y 1 B ω j nl n i. 75 The difference between the two equations apart from the Oξ differences is only in the isotropic part δ ik which should vanish if the stress satisfies continuity equation σ = 0. [1] D. J. Jeffrey Physics of Fluids A: Fluid Dynamics [] S. Kim S. J. Karrila Microhydrodynamics Butterworth-Heinemann [3] R. Mari R. Seto J. F. Morris M. M. Denn Journal of Rheology [4] R. Ball J. Melrose Physica A: Statistical Mechanics its Applications