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E le c tr o p h o re s is o f a P a r tic le o f A r b i t r a r y S h a p eF. A. MORRISON, JR.

University of I l l inois at Urbana-Champaign, Urbana, Ill inois 61801Received December 22, 1969; accepted June 8, 1970

The electrophoresis of an insulating body of arbitrary shape in an unbounded fluidis treated analytically, subject to the restrictions that the local Debye length is muchsmaller than the local radii of curvature and surface conductivity is negligible. Theflow s shown to be irrotational. The Smoluchowskiequation is shown to be valid for aparticle of any shape, as is frequently assumed.INTRODUCTION

The electrophoretic velocity of a particleis known to depend upon several factors.Ordinarily, the velocity is determined in asteady electric field in an unbounded liquid.In particular, it has long been held that theelectrophoretic velocity of an insulatingparticle of arbitrary shape is given by

u = e E o / ~ [ 1 ]provided that the local Debye length ismuch smaller than the local radii of curva-ture and surface conductivity is negligible.In this equation, U is the particle velocity,e the fluid permit tivity, ~ the zeta potential,the fluid viscosity, and E0 the appliedelectric field. This relationship was firstgiven by Smoluchowsld (1). If rationalizedunits are not employed, a factor of 4~r ap-pears in the denominator.The purpose of this paper is to substanti-ate the frequent claim that this result isvalid regardless of the shape of the particle.Smoluchowski first made this assertion in hisdeduction of Eq. [1], but he did not provethis to be true. Henry (2) is usually cited asproving that Eq. [1] holds for a particle ofany shape. He did not, however, demon-strate this generally but made it plausible byshowing that Eq. [1] holds for spheres andcylinders.We shall prove that this result is indeedvalid for a particle of arbitrary shape.

ANALYSISConsider an insulating particle of anyshape immersed in a liquid electrolyte.Under the thin double layer assumption, the

double layer may be treated as a plane layeron the surface of the particle. Analysis of theflow within this plane layer leads to the well-known result for the tangential velocity,relative to the surface, at the outer boundaryof the double layer.u ~ = - E g u , [ 2]

where the subscript t indicates tangential.The tangential velocity and electric field arenot constant, but vary along the suface. Inorder to specify the velocity distributionalong the surface, the electric field distribu-tion must be determined. A unique solutionfor the electric field in the fluid must first befound in order to obtain the boundary con-ditions necessary to permit a unique solutionof the velocity distribution in the fluid.Now, the electric field is related to theelectric potential V. By definition,

E l = - o V / O x ~ . [3 ]The potential distribution is obtained as asolution of Laplace's equation since the fluidis neutral outside the double layer.

02V/Ox~Ox~ = 0. [4]The electric field is rendered unique bythe boundary conditions at the particle sur-face and far from the particle. At the surface

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ELECTROPHORESIS OF A PARTICLE OF ARBITRAI~Y SHAPE 211of an insulating particle, and in the absenceof any surface conductance, the electriccurrent and, consequently, the electric fieldnormal to the surface must be zero.

O VHere n~ is a direction cosine of the outwardnormal from the body.Far from the particle, the potential ap-proaches that of the uniform applied electricfield.

V ~ - E o ~ x i . [6]These two boundary conditions are suffi-

cient to find a unique solution of Laplace'sequation. We know therefore that a uniquepotential distribution and consequently anelectric field distr ibut ion exists. In the de-tailed solution of any particular ease, itremains to find this solution. In the ap-proach used here, it is not necessary to gointo the details of actually finding the solu-tion. For our purposes, it is sufficient tonote that a unique solution exists.To find a solution for the velocity distri-bution, we must find a solution tha t satisfiesthe following conditions:a) The solution must satisfy the governingequations for the fluid flow. These are theNavier-Stokes equation and the continuityequation.b) The solution must satisfy the boundaryconditions for the fluid flow.c) The force exerted by the fluid on theparticle must be zero.d) The moment exerted by the fluid onthe particle must be zero.It will also be shown that the particlemoves without rotating through the fluid.This is not a necessary step of the argumentbut is an interesting result and simplifies theargument.The critical step in the argument is theassertion that the fluid flow is irrotational.

Ou~/Ox j = Ou i /Ox~ . [71This assertion must be justified by showingthat an irrotational flow satisfies the equa-tion of motion and the boundary conditionsand exerts no force or moment on the parti-cle. The assertion of irrotationality implies

that the particle cannot rotate, as can bedemonstrated by StokeS' theorem. This willbe shown first as it justifies the use of a non-rotating coordinate sys tem fixed to theparticle.The angular velocity of the body must bezero in an irrotational flow as a calculationof the circulation about a plane section ofthe body indicates. The circulation is givenby

=- fc u~ dx~ . [8]The curve C bounds the surface of the sec-tion. Equation [2] is valid for the velocityrelative to the surface. Combined with Eq.[3], Eq. [2] shows that this relative velocitymakes no contribution to the circulation.Translation of the particle similarly makesno contribution. A rigid body rotation, how-ever, will produce a circulation of magnitude2~A, wherea is the magnitude of theangular velocity vector normal to the planeand A is the area of the plane section. Thisresult is obtained immediately by applyingStokes' theorem to the plane section.

f A Ouk= nl ~ijk ~ dA. [9]The curve C, however, also bounds manysurfaces lying entirely in the fluid. Applica-tion of Stokes' theorem to one such surfaceyields

r = fs n i e ij k ~ d S = 0 [10]because the flow is irrotational. It followsimmediately tha t the particle cannot rotate.We now wish to show that an irrotationalflow can satisfy the governing equations. Le tus consider the flow relative to a nonrotatingcoordinate system fixed to the body. Use ofthis reference frame simplifies the analysisbut is not essential to the argument.The fluid motion is governed by the con-tinuity equation and the Navier-Stokesequation. For an incompressible fluid, thecontinuity equation is

OU~/OXl = 0. [11]For the steady flow of an incompressible fluid

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2 1 2 MORRISONw i t h c o n s t a n t v i s c o s i t y , t h e N a v i e r - S t o k e se q u a t i o n c a n b e w r i t t e n

2O u l _ O p O u ~ [12]~u j Ox j Ox~ ~ tt Ox j Ox s"

I t h a s l o n g b e e n k n o w n ( 3) t h a t t h eN a v i e r - S t o k e s e q u a t i o n f o r a v i s c o u s i n -compres s ib l e f l u id can be s a t i s f i ed by an J r -ro t a t i on a l mo t ion . 1 Th i s i s ve r i f ied im med i -a t e l y u p o n s u b s t i t u t i o n o f E q . [ 7 ] i n t o E q .[12 ]. The p r e s su re an d v e loc i t y a r e t he nr e l a t e d b y B e r n o u l l i ' s e q u a t i o n ,p + ~ u~ u l = co ns t an t , [13]

t h e c o n s t a n t h a v i n g t h e s a m e v a l u e t h r o u g h -ou t t he f l u id .Fo r j u s t a s l ong , i t ha s been gene ra l l ybe l i eved t ha t on ly t r i v i a l f l ows o f t h i s na -t u r e cou ld ex i s t becau se o f t he no - s l ip cond i -t i o n c o m m o n l y im p o s e d a t a s o l id - li q u idb o u n d a r y i n v i s c o u s f l o w s . T h e o b j e c t i o n i sove rco m e he r e by t h e s l ip cond i t i on , Eq . [ 2] ,o f t he e l ec t ropho re t i c f l ow .T h e a s s u m e d i r r o t a t i o n a l i t y o f t h e f l o wimpl i e s t he ex i s t ence o f a ve loc i t y po t en t i a lu = - -O) / Oz~ , [141

w h e r e , to s a t i s f y t h e c o n t i n u i t y e q u a t i o n f o rt h e v e l o c i t y , t h e p o t e n t i a l m u s t o b e y L a -p l ace ' s equa t i on .O2ep/OxiOx~ = 0. [15]

B y a p p l y i n g t h e b o u n d a r y c o n d i t i o n s f o rt he f l u id f l ow , a so lu t i on w i l l be found t ha ts a t i s f i e s t h i s e q u a t i o n a n d t h e b o u n d a r ycond i t i ons .T h e v e l o c i ty b o u n d a r y c o n d it io n s c a n b es t a t e d i n t e rm s o f t h e v e l o c i t y p o t e n ti a l . A tt h e p a r ti c le , t h e r e c a n b e n o f l o w n o r m a l t othe su r f ace .

0 n ~ = 0. [16]Th e s l i p -f low cond i t i on a t t he su r f ace , Eq .[2] , c an be r e s t a t e d i n t e rm s o f t he ve lo c i t yp o t e n t i a l a n d t h e e l e c t r i c p o t e n t i a l . A t t h e

St okes refe r red t o i r ro t a t i ona l mot i ons by t heexpr ess io n then in use , mot ions where u dx + v dy+ w dz is an exac t differ ential.

J ou r n a l o f Co l loid a* trl ln te r fac e Sc ie nc e , Vol. 34, No. 2, October

sur face ,0 ~ OV

- [171Ox~ # Ox ~"T h e v e l o c i t y m u s t t e n d t o a u n i f o r m f lo wfa r f r om the pa r t i c l e

~ U~ x ~ , [18]w h e r e U is th e m a g n i t u d e o f t h e v e l o c i t y o ft h e f l u i d f a r f r o m t h e b o d y .W e n o t e t h a t b o t h t h e d i f f e r e n t i a l e q u a -t i o n a n d t h e b o u n d a r y c o n d i t i o n s a r e i d e n t i -ca l f o r t he e l ec t r i c po t en t i a l and t he ve loc i t yp o t e n t ia l . T h e r e l a ti o n s h i p b e t w e e n t h epo t en t i a l s i s c l e a r f r om Eq . [ 17 ], t he bou nd-a r y c o n d i t i o n c o u p l i n g t h e v e l o c i t y a n delectr ic f ield.

= - - e ~ V / t L . I191I n t e r m s o f t h e v e l o c i t y

u ~ = - o ~ E ~ / ~ , . [ 2 0 ]Thi s ve loc i t y i s t he f l u id ve loc i t y r e l a t i veto t he pa r t i c le . Th e f l u id ve loc i t y r e l a t i ve t oa s t a t i ona ry f l u id f a r f r om the pa r t i c l e i s ob -

t a i n e d s i m p l y b y c h a n g i n g t h e r e f e r e n c ef r ame . I n pa r t i cu l a r , t he pa r t i c l e ve loc i t yr e l a t i ve t o t he f l u id i sc = ~E o/ , . [ 2 1 ]

T h i s i s t h e S m o l u c h o w s k i e q u a t i o n .T h e p r o o f i s n o t y e t c o m p l e t e . T h e m a g n i -t ude o f t he ve loc i t y U was no t spec i f i ed i nt h e b o u n d a r y c o n d i ti o n , E q . [ 1 5 ], b u t w a st a k e n b y a s s u m p t i o n t o b e t h e v a l u e w h i c ha p o t e n t i a l f l o w s o lu t i o n w o u l d y i e ld . T h es o l u t i o n o b t a i n e d i s a s o l u t i o n b u t i s n o tu n i q u e b e c a u s e U w a s n o t s p e ci fi e d a p r i o r i .A n y v a l u e o f U o th e r t h a n t h a t g i ve n b yEq . [ 21 ] cou ld no t r e su l t i n a po t en t i a l f l ows o lu t io n . T h e v e l o c i t y p o t e n t i a l w o u l d b eo v e r sp e c if i ed b y t h e b o u n d a r y c o n d i t io n s .N o s o l u t i o n t o L a p l a c e ' s e q u a t i o n w o u l de x is t. T h e a s s u m p t i o n o f i r r o t a t i o n a l i tyw o u l d b e i n c o rr e c t.C l e a r l y , a n y v a l u e o f t h e v e l o c i t y U c a nb e o b t a i n e d p h y s i c a l l y , g i v e n t h e b o u n d a r yc o n d i t i o n s a t t h e p a r t i c l e , s i m p l y b y a p p l y -i n g a n e x t e r n a l f o r c e of t h e p r o p e r m a g n i t u d et o t h e p a r ti c le . T h e f lo w w o u l d t h e n b e al i nea r supe rpos i t i on o f a po t en t i a l so lu t i on

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ELECTROPHOI~ESIS OF A PARTICLE OF ARBITRARY SHAPE 213and a Stokes type solution, for sufficientlylow values of the Reynolds number.The velocity U is rendered unique bysatisfying the last two conditions imposed onthe flow. The external force and moment onthe particle must be zero. The particle willthen, without rotating, move steadilythrough, the fluid with a constant velocity.It will be shown that the force on a particleis zero in a viscous potential flow. It willfurther be shown that, in the Stokes ap-proximation, the moment on a particle iszero in a viscous irrotational flow.The general result for the force on a bodyof arbitrary shape alone in an irrotationalflow that is uniform far from the body iseasily obtained by considering the asymp-totic behavior of the velocity far from thebody. It is well known that a potential flowsolution has the asymptotic behavior, as thespherical radial distance r from a point inthe body increases without bound,

u ~ = U ~ + O ( r - ~ ) , [ 2 2 ]O ( r 3 ) denoting a t erm of order r -s.This behavior differs from the asymptoticbehavior ordinarily encountered in the flow,past a body and uniform at infinity, in theStokes or Oseen approximations.

u~ = U i + O( r - 1 ) [23]This relation holds for sufficiently largevalues of r.Corresponding to the asymptotic be-havior of the velocity in an irrotational flow,uniform at infinity, the asymptotic behaviorof the pressure is, from Eq. [13],

p = P + O ( r - ~ ) . [ 2 4 ]The force on the body is now determinedby a generalization of a proof of d'Alem-bert's paradox (4). Consider a sphericalcontrol volume of radius R, centered on thebody. The control volume form of the mo-mentum equation gives, for the forceexerted by the liquid on the body,

where A denotes the surface of the sphericalcontrol volume and ~ij is the stress tensor.~o = --pSo- -t- 2~ei3-, [26]

where 8~i is the Kronecker delta and e~iis the ra te of strain tensore~j = ( o u i / o x j + O u j / O z O / 2 . [ 2 7 ]

For sufficiently large R , the expressionfor the force becomesFi = -PfxnidA

- ~ U i f ~ U j n ~ - d A + 0 ( R - l ) .[ 2 s ]

The integrals are zero and the remainderdecreases as R increases, yieldingFi = 0; [29]d'Alembert's paradox is thus extended toinclude the irrotational flow of a viscousfluid, independent of the Reynolds number.By a similar argument, the moment on anarbitrarily shaped body is shown to be zeroin a low Reynolds number irrotational flow.In the Stokes approximation, the controlvolume form of the moment of momentumequation is

M i = J: ei~k xj c~k~n l d A , [30]where Mi is the moment exerted by the fluidon the body.Because a constant hydrostatic pressureexerts no moment on the body, the ex-pression for the moment becomes, for suffi-ciently large R,

Mi = 0(R-l), [31]and, consequently, the moment must vanishfor an irrotational flow in the Stokes ap-proximation. If inertial terms are retained,the moment is not shown to be zero, but.will depend upon the Reynolds number.The Reynolds nmnber is quite small in atypical electrophoretie flow and the resultantmoment is, in all likelihood, negligible.Each of the conditions imposed on theflow has been satisfied by the irrotationMflow given by Eq. [20]. No assumption of theshape of the particle was made. It followsthat the solution is valid for a particle ofarbitrary shape. The eleetrophoretic veloc-ity is given by the Smoluchowski equation.

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2 1 4 M O R R I S O NCONCLUSIONS R E F E R E N C E S

T h e S m o l u c h o w s l d e q u a ti o n , s u b j e c t t ot h e r e s t r i c t i o r t s g i v e n , h a s b e e n s h o w n t ob e v a l i d f o r a p a r t i c l e o f a n y s h a p e . T h ef lo w a b o u t t h e p a r t i c l e is i r ro t a t i o n a l . Av e l o c i t y p o t e n t i a l e x i s ts t h a t i s s i m p l yr e l a t e d t o t h e e l e c t r ic a l p o t e n t i a l . T h e f o r c ea n d m o m e n t o n a p a r ti c l e ar e s h o w n to b ez e r o i n t h e v i s c o u s i r r o t a t i o n a l f l o w . T h ep a r t i c l e , f u r t h e r , m o v e s w i t h o u t r o t a t i n g .

1. S~OL~JCHOWSK~, M . V ., In L. Grae tz , Ed . ,"Handbuch der Elektrizit /~t und des Mag-net ismus ," Vol . 2 . Bar th, Leipzig, 1914.2. H E N R Y , D . C . , Proc. Roy. Soe. Ser. A 13 3, 106(1931).3. STOKES, G. G., Trans. Cambridge Phil. Soe.

9 , 8 (1851).4. SERRIN, J . , In S . F l u g g e, E d . , " I t an d b u ch d e rPh ysi k," V ol . 8 . Springer-Ver lag, Ber l in,1959.

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