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13 Dec 2018: b/ø (x,h) = exp{ik [xS h (x) cos 0 ] }. •ei•$[ 1 ikh(x)cos O- ( k2/2 )h2(x )cos 2 0]. (9). ... k2 •b• ø (x,zt) = - 1 ikh (x) cos 0- • cos 2 0 X [h2(x) (Zl-h) 2] )e ikxs -(za-h) •xx ikcos0.

25 Jul 2018: 708 J. B. KIRKEGAARD AND R. E. GOLDSTEIN. Fig. 5. Drift efficiency 〈cos θ 〉 with finite tumbling time. ... 1 I1(k)I0(k). cos μ. )(〈cos θ 〉 〈q cos θ 〉) , (B.2).

22 Jan 2018: b "arctan ðcos2 W þ q sin2 WÞR. 2ð1 " qÞsin W cos Wh! : ... Q2 1 þ 2hð1 " qÞsin W cos WRðcos2 W þ q sin2 WÞ!

2 Nov 2018: 12 e2iωγerf. (iω(γ 1). (iω)12. )(1 γ). ]. 27. Let. Ca,b(α) :=. ba. cos(αt)eiωt2dt =. π12 exp. ( iα24ω. )4(iω). 12. [erf. (a(iω). ... 2(iω)12. )],. C1,a,b(α) :=. bat cos(αt)eiωt. 2dt =. i. 2ω. [cos(aα) exp(iωa2) cos(bα) exp(iωb2). ]

19 Oct 2018: µs(θ)=. l=0. µslPl(cos θ), µp(θ)=. l=0. µplPl(cos θ). (2.8a,b). In the case of our spherical particle, the slip velocity due to either one ... s0 =sb(rc)r sR3. 2r2(sz cos θ s. y sin θ sin φ).

1 Nov 2018: CA =. 0. B@cos sin 0sin cos 0. 0 0 1. ... X cos Y sin. Again, U () must act on X this way if it is indeed to correspond to arotation.

3 Apr 2018: 1/C0 sin(. C0κ1) cos κ3φ2. /C0 sin(. C0κ1) sin κ3φ1. ]J2 cos(. ... Aκ1 = φ2J1 φ1J2. Aκ2 =1C0. sin(C0κ1)φ1J1. 1C0. sin(C0κ1)φ2J2+. cos(C0κ1)J3. A direct calculation shows that this gauge potential is indeed GC0-equivariant.