An exact sampling scheme for Brownian motion in the presence of a magnetic field

5 Pages·0.11 MB·English

by  Mini P. Balakrishnan

3
0 An exact sampling scheme for Brownian motion in the presence of a magnetic field
0
2
Mini P. Balakrishnan1, M. C. Valsakumar2 and P. Rameshan1

n 1 Department of Physics, University of Calicut, Calicut University PO, Pin 673635, India
a 2 Materials Science Division, IGCAR, Kalpakkam, Pin 603102, India
J
(Dated: February 2, 2008)

0
1 Langevin equation pertinent to diffusion limited aggregation of charged particles in thepresence
of an external magnetic field is solved exactly. The solution involves correlated random variables.

] A new scheme for exactly sampling the componentsof theposition and velocity is proposed.
h
c
e
I. INTRODUCTION II. SOLUTION OF LANGEVIN EQUATION IN
m
THE PRESENCE OF MAGNETIC FIELD
-
t The diffusion limited aggregation(DLA)model[1] and
a someofitsvariantshavebeenusedtomodelpatternfor- ConsidertheLangevinequationdescribingtheBrown-
t
s mation in diversecontexts such as electrochemicaldepo- ian motionof a chargedparticle of unit mass and charge
t. sition, dissolutionand erosionof porous media, gelation, q,in presenceof a magneticfield B~ alongthe z-direction
a
fracture, dielectric breakdown etc [2, 3, 4, 5, 6]. These
m
patterns, in general, have a fractal structure and the d
~r(t) = ~v(t) (1)
- studyofDLAclustershaveresultedintheunderstanding dt
d
n of various aspects of fractals such as fractal dimension, d~v(t)+γ~v(t) q~v(t) B~ = ~η(t) (2)
o multifractalityandthescalinglaws. Theinsightobtained dt − ×
c fromthe analysisofDLA clusters has ledto the applica- In the above equation, ~η(t) = (η1(t),η2(t),η3(t)) is a
[ Gaussian white noise of zero mean and its components
tion ofthis model to the study ofthe scaling behaviorof
satisfy the normalization condition
1 two-dimensional quantum gravity. Yet a complete theo-
v retical understanding of DLA has not been achieved to
6 date. Recently, a few groups [7, 8] have carried out ex-
3 perimental investigation of the influence of an external hηi(t)ηj(t′)i=2Aδijδ(t−t′) (3)
1
magnetic field on DLA of charged particles. They ob-
1 Sincethe magneticfieldisorientedalongthe z-direction,
served a bending of the branches of the DLA cluster,
0 itaffectsthemotioninthexandydirectionsonly. Defin-
3 which could be attributed to the Lorentz force, as well ing
0 as a conspicuous change in the morphology of the clus-
/ ters with increase in the strength of the magnetic field.
t X =x+iy, V =v1+iv2, Γ=γ+iω, η =η1+iη2(4)
a As a prelude to a detailed study of DLA in presence of
m
an external magnetic field, we solved the corresponding we get
- Langevin equation exactly. It is found that the solution
d
involves correlated Gaussian random variables. In order d
n X = V (5)
o tocarryoutnumericalsimulationofDLA,itisnecessary dt
c to sample these random variables,at discrete instants of d
V +ΓV = η(t) (6)
v: time, consistent with the correlation relations. In this dt
paper, we describe an exact sampling procedure that is
i
X valid for arbitrary values of all the parameters that ap- The formal solution is given by
r pear in the problem. The results of the DLA simulation
a t
using this prescription are in qualitative agreement with V(t)=e−ΓtV(0)+ e−Γ(t−t1)η(t ) dt (7)
1 1
the experimental results and will be published elsewhere
Z0
[9].
1 t t
X(t) X(0)= (1 e−Γt)V(0)+ η(t ) dt e−Γ(t−t1)η(t ) dt (8)
1 1 1 1
− Γ − −
(cid:18) (cid:20)Z0 Z0 (cid:21)(cid:19)
Usingtheaboveformalsolution,wecanimmediatelyob- v (t) are given by
2
tain x(t),y(t) by taking the real and imaginary parts of
Xfas(ht)io.nv.1T(th)eatnwdovt2i(mt)eccaonrraellasotiobnefoubntcatiinoends oinf va1(sti)mailnadr hv1(t)v1(t′)i = Aγ e−γ|t−t′|cos(ω(t−t′)) (9)
(cid:18) (cid:19)
2
v (t)v (t′) = v (t)v (t′) (10) j−1
h 2 2 i h 1 1 i ψ(t) = ψ (15)
hv1(t)v2(t′)i = Aγ e−γ|t−t′|sin(ω(t−t′)) (11) j Xl=0 l
(cid:18) (cid:19) j−1
φ(t) = e−(j−1−l)Γτφ (16)
The other correlation functions can also be calculated j l
easily. Xl=0
The interesting point is that unlike in the case of the We can then write
motion without the magnetic field, now the components
v (t) and v (t′), of the velocity, are correlated. But, the 1
1 2 X = (1 e−jΓτ)V(0)+ψ(t) φ(t) (17)
second moment < vi(t)2 >, of each of the components j Γ − j − j
of the velocity, has the same value A/γ as in the case of h i
zero magnetic field. This can be easily understood once
We can perform the numerical simulations if we can re-
we recognize that < v (t)2 > is twice the kinetic energy
i liably sample ψ andφ . since ψ andφ depend onη in
j j j j
and that does not change by application of a magnetic
the interval [jτ,(j +1)]τ] alone, and since the η are
field. However, < x(t)2 > and < y(t)2 > asymptotically delta-correlated,wecanimmediatelyseethat ψ{ψ′i}and
(i.e.,ast )tendto 2A t= 2A tsothatthedif- j j
fusion coe−ffi→cie∞nt D = (γ2|+AΓ|ω22) de(cγr2e+aωs2e)s with increasing reφljaφte′jd.vaTnhieshrefwohreenitjis6=imj′p.oHrtoawnetvetor,eψvjolavneda(cid:10)φpjroarcee(cid:11)dcuorre-
(cid:10) (cid:11)
magnetic field. to sample these correlated random variables.
Inthe numericalsimulationswewillbe interestedinget-
It is convenient to work with the following real random
ting realizations of x(t), y(t), etc. at discrete instants of
variates:
time t = j τ. We have developed such a scheme that
j
∗
is valid for arbitrary values of A, ω, γ and τ.
Define
τ
ψ = (ψ ) = du η (jτ +u) (18)
xj =x(tj), yj = y(tj), Xj =xj +iyj (12) jR ℜ j 1
Z0
τ τ
ψj = du η(jτ +u) (13) ψjI = (ψj) = du η2(jτ +u) (19)
Z0 ℑ Z0
τ
φ = du e−Γ(τ−u)η(jτ +u) (14)
j
Z0
τ
φ = (φ ) = du e−γ(τ−u)[cos(ω(τ u))η (jτ +u)+sin(ω(τ u))η (jτ +u)] (20)
jR j 1 2
ℜ − −
Z0
τ
φ = (φ ) = du e−γ(τ−u)[ sin(ω(τ u))η (jτ +u)+cos(ω(τ u))η (jτ +u)] (21)
jI j 1 2
ℑ − − −
Z0
[Intheaboveequations, (z)and (z)aretherealandimaginarypartsofthecomplexnumberz]. Sinceψ andφ
j j
ℜ ℑ
are Gaussian randomvariates of zero mean, the knowledge of their covarianceis adequate for describing them. They
are given below:
ψjRψj′R = 2Aτδj,j′, ψjIψj′I =2Aτδj,j′, ψjRψj′I =0 (22)
h i h i h i
A A
φjRφj′R = (1 e−2γτ)δj,j′, φjIφj′I = (1 e−2γτ)δj,j′, φjRφj′I =0 (23)
h i γ − h i γ − h i
2A
hφjRψj′Ri = (γ2+ω2) γ(1−e−γτcos(ωτ))+ωe−γτsin(ωτ) δj,j′ (24)
2A (cid:2) (cid:3)
hφjRψj′Ii = (γ2+ω2) ω(1−e−γτcos(ωτ))−γe−γτsin(ωτ) δj,j′ (25)
φjIψj′R = φjRψj′I(cid:2) , φjIψj′I = φjRψj′R (cid:3) (26)
h i −h i h i h i
Our aim is to provide a scheme for sampling ψ , ψ , We propose to achieve this by expressing ψ , ψ , φ
jR jI jR jI jR
φ and φ that is consistent with the above equations. and φ as sums of n identically distributed Gaussian
jR jI jI
3
random variables (pj1,pj2,...,pjn) with zero mean and A(1 e−2γτ) n
unit variance in the following manner: φjI = − νlpjl (30)
s γ
l=1
X
n
(31)
ψ = √2Aτ α p (27)
jR l jl
l=1
X
n
ψ = √2Aτ β p (28)
jI l jl
l=1 Thesubstitutionoftheaboveresultsinequations22-26
X
A(1 e−2γτ) n gives 10 constraintsthat the parametersαl, βl, ǫl and νl
φ = − ǫ p (29) should satisfy.
jR l jl
s γ
l=1
X
n n n
α2 = 1, β2 =1, α β =0 (32)
l l l l
l=1 l=1 l=1
X X X
n n n
ǫ2 = 1, ν2 =1, ǫ ν =0 (33)
l l l l
l=1 l=1 l=1
X X X
n 2γ [γ(1 e−γτcos(ωτ))+ωe−γτsin(ωτ)]
ǫ α = − C (34)
l l sτ(1 e−2γτ) (γ2+ω2) ≡ 1
l=1 −
X
n 2γ [ω(1 e−γτcos(ωτ)) γe−γτsin(ωτ)]
ǫ β = − − C (35)
l l sτ(1 e−2γτ) (γ2+ω2) ≡ 2
l=1 −
X
n n n n
ν α = ǫ β = C , ν β = ǫ α =C (36)
l l l l 2 l l l l 1
− −
l=1 l=1 l=1 l=1
X X X X
α , β , ǫ and ν maybeconsideredasunimodu- where R= (1 C2 C2) and θ is a free parameter.
{ i} { i} { i} { i} − 1 − 2
larvectorsα~,β~,~ǫand~ν inann-dimensionalspace. Such
p
a geometrical interpretation of the above relations im-
mediately leads to the conclusionthat 4 is the minimum
value ofn, for whichthe aboverelationscanbe satisfied, B. Solution 2
seetheAppendix. Weconsiderthecasen=4,henceforth.
Since there are 16 parameters and only 10 constraints, Another choice for α~ and β~ is
there will be a 6-parameter family of solutions. We can
choose α~ and β~ consistent only with the constraints 1
α =α =α = α = (41)
1 2 3 4
represented by eq. 32, and then solve for ~ǫ and ~ν. 2
In what follows, we show two solutions explicitly and 1
β = β = β = β = (42)
alsodiscusstheprescriptiontoobtainanyothersolution. 1 − 2 − 3 4 2
We then obtained~ǫ and~ν to be
ǫ =D +D cos(θ), ǫ =D +D sin(θ), (43)
A. Solution 1 1 1 3 2 2 3
ǫ =D D sin(θ), ǫ =D D cos(θ) (44)
3 2 3 4 1 3
− −
A simple choice for α~ and β~ would be ν1 =D2+D3sin(θ), ν2 = D1 D3cos(θ),(45)
− −
ν = D +D cos(θ), ν =D +D sin(θ), (46)
3 1 3 4 2 3
α =1, α =0, α =0, α =0, (37) −
1 2 3 4
β1 =0, β2 =1, β3 =0, β4 =0 (38) where D1 = (C1 + C2)/2, D2 = (C1 − C2)/2,
D = 1 (C2+C2)/2 and θ is a free parameter
This choice leads to wh3ich was se−t to1be π/24 in our simulations.
p
ǫ1 =C1,ǫ2 =C2,ǫ3 =Rcos(θ),ǫ4 =Rsin(θ), (39) Any set of vectors α~′, β~′,~ǫ′ and ~ν′ obtained by simul-
ν = C ,ν =C ,ν = Rsin(θ),ν =Rcos(θ), (40) taneouslyrotating(SO(4)rotation)agivensetofvectors
1 2 2 1 3 4
− −
4
α~,β~,~ǫand~ν willalsobeavalidsolutionoftheequations is given which is valid for arbitrary values of the friction
32 - 36. It remains to be seen if a particular choice of coefficient of the medium, the strength of the magnetic
these vectors is better than others for numerical simula- field and the time step of evolution. The components of
tions with finite number of realizations or not. the position and velocity are represented as linear com-
binationsoffour independentandidentically distributed
Gaussian random variates. A six parameter family of
III. CONCLUSIONS solutions is obtained for the expansion coefficients. Fur-
ther studies are required to see if a particular choice of
The Langevin equation for diffusion of charged parti- theexpansioncoefficientsisbetterthanalltheotherpos-
cles in presence of an external magnetic field is solved sibilities.
exactly. Ascheme forsamplingthe positionandvelocity
[1] T.A.WittenandL.M.Sander,Phys. Rev. Lett.47(1981) The expansion coefficients a1, a2, b1 and b2 can be ob-
1400 tained by substituting eq. 52 in eqs. 49 and 50. The
[2] A.Bunde,H.J.Hermann,A.MargolinaandH.E.Stanley, result is
Phys. Rev. Lett. 55 (1985) 653
[3] R.F.Xi,J.I.D.AlexanderandF.Rosenberger,Phys. Rev. ~ǫ=C α~ +C β~, ~ν = C α~ +C β~ (53)
1 2 2 1
A38 (1988) 2447 −
[4] M. Ausloos and J.M. Kowalski, Phys. Rev. B45, (1992) Eq. 53 implies
12830
[5] P. Meakin, Phys. Rep. 235 (1993) 189 ~ǫ.~ǫ=~ν.~ν =C2+C2 (54)
1 2
[6] L. Nietmeyer, L.Pietronero and H.W. Wiesmann, Phys.
Rev. Lett. 52 (1984) 1033 which is not equal to 1 in general. Thus the ~ǫ and ~ν
[7] Matsushita et. al,Phys. Rev. Lett. 53 (1984) 286 fail to satisfy the unimodularity condition (see eq. 48).
[8] Sawada et. al., Phys. Rev. Lett. 56 (1984) 1260 Hence we see that it is impossible to obtain the four
[9] Mini P. Balakrishnan, M. C. Valsakumar and P. Rame- vectors α~, β~, ~ǫ and ~ν satisfying the constraints eqs.
shan, (unpublished)
47-50 in two dimension.
IV. APPENDIX
B. 3-dimensional case
Consider four unimodular vectors α~, β~,~ǫ, ~ν satisfying
the conditions The most general vector α~ can be written down as
α~.α~ =1, β~.β~ =1, α~.β~ =0 (47) α~ =(sinθ1cosφ1,sinθ1sinφ1,cosθ1) (55)
~ǫ.~ǫ=1, ~ν.~ν =1, ~ǫ.~ν =0 (48) where θ and φ are arbitrary parameters. The most
1 1
~ǫ.α~ =C , ~ǫ.β~ =C (49) general vector β~ perpendicular to α~ is given by
1 2
~ν.α~ = C2, ~ν.β~ =C1 (50) β~ =cos(ζ)β~+sin(ζ)(α~ ~η) (56)
−
×
with
A. 2-dimensional case
~η =(cosθ cosφ ,cosθ sinφ , sinθ ). (57)
1 1 1 1 1
−
Let us first consider the 2-dimensional case. Here, the and ζ is a free parameter. It is obvious that the vectors
most general pair of vectors α~ and β~ satisfying eq. 47 α~,β~,α~ β~ formabasisinthreedimensionalspace. Then
×
can be easily constructed as follows ~ǫ can be defined as
α~ =(cosθ1,sinθ1), β~ =(−sinθ1,cosθ1). (51) ~ǫ=a1α~ +a2β~+a3α~ ×β~. (58)
Using eqs. 49 and then 48, we get
where θ is an arbitrary parameter. Since α~ and β~ are
1
orthonormal,theyformabasisin2-dimensionsandhence ~ǫ=C α~ +C β~+a α~ β~. (59)
1 2 3
anyothervectorcanbeexpressedaslinearcombinations ×
of these vectors. Therefore~ǫ and ~ν can be written as with C2+C2+a2 =1. Similar analysis gives
1 2 3
~ǫ=a α~ +a β~, ~ν =b α~ +b β~ (52) ~ν = C α~ +C β~+b α~ β~ (60)
1 2 1 2 2 1 3
− ×
5
with C2+C2+b2 =1. Substituting for~ǫ and~ν, we get straints eqs. 47-50 in three dimensions as well. It is,
1 2 3
however,possible to obtain solutions in four dimensions,
~ǫ.~ν = C1C2+C1C2+a3b3 =a3b3 (61) as explicitly demonstrated in the text.
−
which is not zero in general. Hence it is impossible to
obtain the four vectors α~, β~,~ǫ and ~ν satisfying the con-