Lect1MS133JRG

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Chapter 1

Ficks Laws and MacroscopicDiffusion

1.1 Introduction

In the rst part of this course we will discuss diffusion, which is the fun-damental process for atomic transport in solid state transformations andreactions. We start with a mathematical description of diffusion due to aconcentration gradient. We see that random atomic jumps act to smoothout spatially varying concentrations, with an atomic ux which is propor-

tional to the concentration gradient. Later in the course we show that thisbehavior is a special case, and that, more generally, the atomic ux will beproportional to a gradient in chemical potential.

1.2 Ficks Laws

1.2.1 Ficks First Law

We rst examine a simple one-dimensional model of diffusion. We considerdiffusion of a trace amount of an impurity (or tracer) in a single-phase alloy.If the planer density of impurities at a position z is given by (z) (measuredin atoms / cm2 ), and if the spacing between adjacent planes is z, then thevolume concentration of impurities c(z) is given by:

c(z) =(z) z

1

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2 CHAPTER 1. FICKS LAWS AND MACROSCOPIC DIFFUSION

Further, we assume that we have one-dimensional random nearest neighbor

jumps, and the diffusing atoms are chemically identical to, but distinguish-able from the host atoms. This is the case for radioactive isotope tracers, forexample. We dene J + to be the ux of atoms to the right from the planeat z to the one at z + z. This is given by:

J + =12

(z)

where is the mean jump frequency, and the factor of 1 / 2 accounts for the jumps being able to go in either the plus or minus z direction. We can alsodene J as the ux of atoms to the left from the plane at z + z to the oneat z, and we nd:

J = 12(z + z)

If we assume that is not a function of concentration, then the net ux J isgiven by:

J = J + J =

12

[(z + z) (z)]=

12

z [c(z + z) c(z)]

= 12 ( z)

2 c(z + z)

c(z)

z

12

( z)2cz

= Dcz

(1.1)

where in the last step we have assigned:

D =12

( z)2

The quantity D is known as the diffusivity, and for the three-dimensionalcase we nd:

D =16

( z)2

Equation 1.1 which relates the concentration gradient to the ux is knownas Ficks rst law. As we go into this course, we will nd that Ficks rst

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1.2. FICKS LAWS 3

law does not always hold, but it is in fact a special case of the more general

statement that a ux will be driven by a gradient in chemical potential. Inmany cases, the gradient in chemical potential is proportional to the gradientin concentration.

Lets discuss units. In the cgs system, the various quantities have theunits:

J atomscm2 s

= Dcm2

scz

atomscm4

Actually, the ux is a vector quantity in that you might want to knowthe direction as well as magnitude of the atomic ow. Hence, Ficks rst lawcan be written as a vector equation:

J = D cwhere is the gradient operator.

1.2.2 The Conservation Equation

In practice, atomic uxes are often a difficult quantity to measure. In solids,it is more typical to measure concentration prole changes which result fromthermal treatment. Hence it is helpful to have an expression to describehow a ux will change a concentration prole. Consider the situation in Fig.1.1, where we have a bar with concentration prole c(z), giving rise to uxJ (z) = Dc/z . We consider the volume element A z where A is the crosssection area. If there is curvature in c(z) then the uxes at z and z + z arenot equal. We will have a ux into our volume element given by J (z) and aux going out of J (z + z). Thus the rate of change in concentration in thevolume element due to uxes is:

ct

=J (z) J (z + z)

z

so in the limit of small z we have

c

t=

J

zIn three-dimensions the vectorial relationship is:

ct

= J (1.2)

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where the quantity J is known as the divergence of J , and where now theux is represented by a vector

Jwhich contains information on the directionand magnitude of the atomic ux.

Equation 1.2 is known as the conservation equation, and it holds if thespecies we are considering is neither being created or destroyed. If the speciesis not conserved, then we must add a source term to Eqn. 1.2. For example,we are sometimes interested in the diffusion of vacancies which can be createdor destroyed at a surface or interior defect. If p(r , t ) represents the rate of production of the species of interest as a function of position and time thenthe conservation equation becomes:

c

t

=

J + p(r , t )

It is important to recognize that the conservation equation will hold evenwhen Ficks rst law does not.

1.2.3 Ficks Second Law

If we combine Ficks rst law

J = D cwith the conservation equation

ct

= Jwe nd:

ct

= D c (1.3)Equation 1.3 is known as Ficks second law. We will frequently consider thesituation where the diffusivity D is independent of concentration, and henceposition, so that Ficks second law becomes:

ct

= D 2 c (1.4)

where 2 is known as the Laplacian operator. In a Cartesian coordinatesystem the Laplacian is:

2 = 2

x 2+

2

y2+

2

z 2

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The form of the Laplacian and other vector operators for different coordinate

systems are summarized in a handout from MSE 205.Equation 1.4 is a partial differential equation which is second order inthe spatial coordinate and rst order in time. Hence a unique solution willbe characterized by a statement of the initial state of the system and twoboundary conditions.

1.2.4 Cylindrical Example Using Ficks First Law

For the rst example we consider diffusion through a pipe. This is a commonexperimental situation for determining diffusion rate and its compositionaldependence. We have a pipe of inner radius r 1 and outer radius r 2 (Fig.1.2).

We somehow conspire to maintain a constant concentration c1 on the innerpipe wall at r = r 1 . This can be done, for example, by maintaining a constantchemical potential in the gas inside of the pipe. We also assume that thepipe is long enough so that there are no end effects, and that we have waiteda long time, so the concentration in the pipe is not changing, i.e.

ct

= 0

This later assumption is known as a steady-state condition and with it theconcentration becomes independent of time, i.e. c(r , t ) c(r ).It is natural to use cylindrical coordinates for this problem, and due to thesymmetry of the geometry and boundary condition Ficks rst law becomes:

J = Ddcdr

r

where now we take the total derivative since we are at steady-state, so theconcentration is not a function of time, and the symmetry of the problem (ig-noring end effects) dictates that it is also not a function of spatial coordinatesother than the radius.

A steady-state ux within the cylinder will give rise to a mass ow outsidethe cylinder. In the experimental situation, we measure the rate at whichatoms appear outside the cylinder m (atoms / s). If, inside our pipe, we drawan imaginary cylinder with radius r 1 < r < r 2 we see that the total numberof atoms passing through this cylinder must be the same for any value of rwithin this range. That is, the ux times the area must be m. This yields:

m = ( D c) (2rh )

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1.2. FICKS LAWS 7

h

rr

r2

1

Figure 1.2: Schematic of pipe showing length h, inner radius r 1 , and outerradius r 2 .

= 2rhDcr

If D = D(c) we can integrate to nd:

cc1 dc = rr 1 m2rhD drc c1 =

m2hD

r

r 1

drr

= m

2hDln

rr 1

We see that a plot of c(r ) versus ln r will give a straight line. By measuringthe mass ow m and the concentration at two different r -values, we canobtain D. For example if we measure c1 and c2 , the concentration at theinner and outer radii of the cylinder, we obtain:

D = m2h (c2 c1 )

lnr 2r 1

In the case that the diffusivity depends on concentration ( D = D(c)), we

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cannot perform the integration as above. We have:

m = 2rhD (c)cr

which can be solved for D(c) to nd:

D(c) = m

2rh cr=

m2h c ln r

This gives us a way to determine D(c). We measure the concentration c(r ) asa function of radius r . If D depends on concentration a plot of c versus ln rwill not yield a straight line, but from the slope of the line we can determineD(r ) which can be transformed to D(c). This is shown schematically inFig. 1.3.

1.3 The Thin Film Solution

The next example we consider is the solution of Ficks second law with theinitial condition of an innitely thin lm placed between two semi-innitemedia. This is often a good approximation to real situations, and it forms thebasis for constructing solutions to more complex situations. The boundary

conditions are that the composition must go to zero at z = . The solutionis:c(z, t) =

mB4Dt exp

z2

4Dt(1.5)

where mB is the total amount per area of the diffusing species, D is thecomposition-independent diffusivity, and z is the distance from the originalinterface. It can be shown by direct substitution that Eqn. 1.5 solves Fickssecond law (Eqn. 1.4):

ct

= D 2 cz 2

The solution in Eqn. 1.5 also has the correct normalization, that is, the totalamount of diffusing material is mB :

c(z, t ) dz = mB

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1.3. THE THIN FILM SOLUTION 9

c

ln rln r1 ln r2

c1

c2

D

ln rln r1 ln r2

D1

D2

Figure 1.3: Schematic showing concentration as a function of ln r result-ing from a steady state cylindrical diffusion experiment, and the position-

dependent diffusivity D(r ) which can be extracted.

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Furthermore, we match the boundary conditions, and, at t 0, our solutionhas the values: for |z| > 0, c 0 as t 0for z = 0 , c as t 0which states that initially we have a nite amount of material in an innitelynarrow band. This is the initial condition for the problem we set out to solve.

The concentration has a Gaussian spatial prole, with width w (distancebetween inection points) given by:

w = 2 2DtThis width is a measure of the diffusion distance. The concentration at theorigin ( z = 0) is given by:

c(0, t) =mB4Dt

which decreases linearly with 1 / t.It is instructive to examine the behavior of this solution relative to Ficks

laws. In Fig. 1.4 the composition, ux, and time rate of change of compositionare plotted as a function of distance at some xed time. In region where theslope of the concentration is positive, the ux, given by:

J = Dcz

,

is negative, so that material is moving to the left. The opposite is truewhere the slope of the concentration is negative. The diffusing atoms movedown the concentration gradient. By Ficks second law (Eqn.1.4), the rateof accumulation of material is proportional to the second derivative of theconcentration with respect to position. Thus, where the curvature of theconcentration is positive, diffusing atoms are accumulating. The opposite istrue where the curvature is negative.

1.4 Solutions Based on the Thin-Film Solu-

tionThere are a class of solutions based on the thin lm solution which areexamples of the general problem solving method known as Greens Functions.The general recipe is:

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1.4. SOLUTIONS BASED ON THE THIN-FILM SOLUTION 11

c

-2 -1 0 1 2z

J

-2 -1 0 1 2z

Fluxto Left

Fluxto Right

c /

t

-2 -1 0 1 2z

ccumu a ng

Leaving

Figure 1.4: Composition, ux, and time rate of change of composition forthe thin lm solution.

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Find the response of the system to a delta function input. In the caseof the diffusion equation this amounts to solving Ficks second law forthe innitely thin lm with unit total material ( mB = 1).

Integrate this response over actual input. In the case of the diffusionequation, this amounts to integrating over the initial composition.This technique can be applied to cases when the problem can be framed

with an innite system. In a homework problem, you will show that:

c(z, t) =1

4Dt f (z )exp (z z )2

4Dtdz

solves the diffusion equation, where f (z) is the initial distribution for aninnite system, i. e. :

f (z) = c(z, 0)

and the diffusivity is not a function of composition ( D = D(c)).As an example of the use of this method, we consider the situation of

a pair of semi-innite solids which initially have different composition. Wearbitrarily set the composition in one equal to 0 and the other equal to c0 ,so that the initial condition is:

c(z, 0) = 0 z < 0

c0 z 0The solution is then:

c(z, t) =1

4Dt f (z )exp (z z )2

4Dtdz

=c0

4Dt 0 exp (z z )2

4Dtdz

We can cast this integral into a standard form with the variable substitution:

=z

z

4Dtso that

d = dz4Dt

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1.4. SOLUTIONS BASED ON THE THIN-FILM SOLUTION 13

and the solution becomes:

c(z, t) = c04Dt z/ 4 Dt e2

4Dt d=

c0 0 e2 d +

z/ 4 Dt0

e2

d

This can be represented as a special function known as the error functionerf(y) dened as:

erf(y) 2

y

0e

2

d

This function has the properties:

erf(0) =2

0

0 e2

d = 0

erf() =2

0 e2 d = 1In terms of these denitions, the solution is:

c(z, t) =c02

1 + erf z

4DtIt is interesting to consider what we have done using this solution method.

We have used a collection of innitely thin slabs to represent a semi-innitesolid, and have let the material in each slab diffuse into a Gaussian prolein a manner predicted by the thin lm solution. The composition prole isthen a sum of these Gaussians. This is illustrated schematically in Fig.1.5.

It is also interesting to consider the time evolution of this solution. Thevalue of the composition at the origin for all times is c0 / 2, as is the value atinnite time at all positions. That is:

c(0, t) =c02

c(z, ) =c02

The composition prole spreads as a function of time. For a homeworkproblem you show that the points z1 and z2 dened by:

c(z1 , t ) =c04

c(z2 , t ) =3c04

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1.0

0.5

0.0

c

-10 -5 0 5

zFigure 1.5: Schematic of solution to the step composition prole. Dottedlines are the solutions for thin slabs, and the sum of these solutions, solidline is the actual error function solution.

spread apart with the distance between them being proportional to the squareroot of time. So again, we nd that the diffusion distance is proportional toDt .

We can generalize the error function solution for other cases. The func-tion:

c(z, t) = A + B erf z

4Dtsolves the diffusion equation. If we can choose constants A and B to meetthe initial condition and boundary conditions, we have a solution.

Before leaving this solution method we make some general comments.First of all, we have assumed that the diffusivity is not a function of com-position ( D = D(c)). This is often a bad assumption if there are largecomposition differences in the problem. However this solution is often used

to estimate the behavior of a system assuming some average diffusion con-stant. Furthermore, we have also assumed that the system is innite. Thisassumption can also lead to trouble as explored in one of the exercises. Ingeneral this solution is valuable for the case that the diffusion distances aresmall relative to the dimensions of the system.

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1.5. SEPARATION OF VARIABLES 15

1.5 Separation of Variables

We now turn to a technique which is best used for describing a system atlong times, where it is approaching homogenization. We start by assumingthat the solution can have the form:

c(z, t) = Z (z) T (t) (1.6)where Z (z) is a function only of position z, and T (t) is a function only of time t. We note that this is not always possible, and in fact the solutionswe have considered so far are not separable, since we have found solutions of the form:

c(z, t) = f z/

tNonetheless we proceed. Inserting Eqn. 1.6 into the diffusion equation:

ct

= D 2 cz 2

we obtain:

Z dT dt

= DT d2Z dz2

1

DT

dT dt

=1

Z

d2Z dz

2 (1.7)

Each side of Eqn.1.7 is only a function of one variable, and since both x andt are independent variables which can vary freely, the only way that Eqn. 1.7can hold is if both sides are equal to a constant, for which we choose 2 .Looking at the t dependence we nd:

1

T dT dt

= 2 D

Integrating we nd:

T T 0d

T T = 2

D t

0 dt

ln T T 0

= 2 Dt

T = T 0 e2 Dt (1.8)

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We see from Eqn.1.8 that if is real ( R), then as long as D > 0 thencomposition inhomogeneities will decay with time.The behavior as a function of position is determined by the equation for

Z (z), which is: d2Z dz2

+ 2Z = 0which has solutions:

Z (z) = A cosz + B sin zSo our solution for the composition is:

c(z, t) =

Z (z)

T (t) = ( A cosz + B sin z ) e

2 Dt (1.9)

where we have absorbed the factor T 0 into A and B.In fact, we are free to choose any number of solutions of the form of Eqn. 1.9, and we can write:

c(z, t) =

n = (An cos n z + Bn sin n z) e

2n Dt (1.10)

where A0 is the average composition.At time t = 0 we see that the initial composition prole is:

c(z, 0) = n =

(An cosn z + Bn sin n z)

which is just the Fourier series expansion of the initial composition prole.We can represent any periodic function with this series, and we can representany function on a limited range, say from L to L, by a periodic extension,forming a function with period 2 L. In either case, we nd that n = n/Lso we have:

c(z, 0) =

n = An cos

nzL

+ Bn sinnz

L(1.11)

We can use the standard Fourier analysis tricks to nd An and Bn , namely,multiply both sides of Eqn. 1.11 by sin mz/L and integrate from L to L.Due to the orthagonality of the sin functions we nd:

An =1

2L L

Lc(z, 0) cos

nzL

dz

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Using similar tricks we can nd:

Bn = 12L LL c(z, 0)sin nzL dz.The coefficient A0 is given by

A0 =1

2L L

Lc(z, 0) dz

which is just the average of the initial composition.Lets apply this formalism to a problem. We consider a periodic structure

as shown in Fig. 1.6, where the initial composition prole is a square wave,made up of slabs of thickness L with composition alternating between c0 and0. The coefficient A0 is just the average:

A0 =1

2L L

Lc(z, 0) dz =

c02L

L

0dz =

c02

We integrate to nd the coefficients An and Bn , and for An we nd:

An =1

2L L

Lc(z, 0)cos

nzL

dz

=c02L

L

0cos

nzL

dz

= c02n sin nzL

L

0= 0

and for Bn :

Bn =c02L

L

0sin

nzL

dz

= c02n

cosnz

L

L

0

=c0

2n(1 cosn ) =

c0n n = odd0 n = even

So we can nd the Fourier series representation for the initial compositionprole:

c(z, 0) =c02

+2c0

j =0

12 j + 1

sin2 j + 1

Lz

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c

c

L 2L 3L 4L 5Lz

Figure 1.6: Square composition wave.

The sum of the rst several terms is plotted in Fig.1.7 a. Note that theterms decrease in amplitude as the frequency increases.

It is a simple manner to include the time dependence by using Eqn. 1.10,and we nd:

c(z, t) =c02

+2c0

j =0

12 j + 1

sin(2 j + 1) z

Lexp

(2 j + 1) L

2

Dt

This is plotted in Fig.1.8 for different times. Note that the higher order

terms go away fast. Even at a time as small as 0 .01L2

/D , which correspondsto a diffusion distance of L/ 100, the wiggles associated with high order termsare gone. The prole quickly begins to look like a sine wave, as only the rstorder term is important for longer times. The physical reason behind thisbehavior is clear. Higher order terms correspond to higher spatial frequencies,which have higher concentration gradients and hence faster diffusion.

We can quantify this behavior by examining the ratio of successive terms.If we dene R j,k to be the magnitude of the kth term relative to that of the j th term. We nd:

R j,j +1 =

2 j + 12( j + 1) + 1

exp [2( j + 1) + 1]

L

2

+ (2 j + 1) L

2

Dt

=2 j + 12 j + 3

exp 8( j + 1) 2 Dt

L2

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-0.5

0.0

0.5

F o u r

i e r

C o m p o n e n t s

10z/ L

1.0

0.5

0.0

c ( z )

10z/ L

Figure 1.7: Plot of the rst several terms of the Fourier series expansion of a square wave. a) The individual Fourier components. b) The sum of thecomponents.

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1.0

0.5

0.0

c (

z )

210z/L

t = 0

1.0

0.5

0.0

c (

z )

210z/L

t = 0.01

1.0

0.5

0.0

c (

z )

210z/L

t = 0.05

Figure 1.8: Plot of the rst twelve terms of the Fourier series expansion of asquare wave at different times. Time is in units of L2 /D .

Lect1MS133JRG

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