8/14/2019 Dislocation Nucleation Rate at Finite Temperature in Copper_ver3

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Dislocation Nucleation Rateat Finite Temperature in

Copper Nanorod

Seunghwa Ryu1 and Wei Cai2

1

Physics, Stanford University2Mechanical Engineering, Stanford

University

2 Dec 2009 MRS Fall Meeting, Boston p. 1/13

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Sampling DislocationNucleation

at finite temperature

2 Dec 2009 MRS Fall Meeting, Boston p. 2/13

- Deformation of nano-material is controlled bydislocation nucleation.

Need to directly sampledislocation nucleationat finite temperature !

T. Zhu, J. Li et al, PRL 100 025502 (2008)

Copper Nano Rod underUniaxial Compresson - Nucleation is a rare event, i.e.

(waiting time >> reaction time )

- Molecular dynamics=> Limited time scale

- Transition State Theory with Fbfrom

Nudged Elastic Band method( Zhu (2008) )=> Zero temperature

calculation

Embedded Atom Method Mishin Potential15 x 15 x 20 100 | 010 |100 Cu Nano Rod

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Outline

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1. Limitation of Nudged Elastic Band (NEB)

2. Forward Flux Sampling (FFS) forNucleation Rate and Free Energy calculation

3. Results

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Limitation of NEB

2 Dec 2009 MRS Fall Meeting, Boston p. 4/13

Dislocation Nucleation Rate from Transition State

Theory (TST)

Number ofNucleation Site

)),(

exp(Tk

TFNI

B

bTST

=

Trial FrequencyFree Energy Barrier

F()

Fb(,T)

E()

Eb(,T=0)E

b(,T=0)E

b(,T=0)

State A State BSize of Partial Slip.

-NEB computes Energy Profile along nucleation path at 0K.- From , How can we get ?

- Zhu et al (2008) assumes

)0,( =TEb ),(),(),( TTSTETF bb =

)0,()/1(),(~

== TETTTF bmb

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Free Energy Barrier

2 Dec 2009 MRS Fall Meeting, Boston p. 5/13

State A State B

- What is physical reason behind?

)0,()/1(),(~

== TETTTFbmb

Size of Partial Slip.

E()

E

b

(,T=0)E

b

(,T=0)E

b

(,T=0)

Free energy decreases linearly with temperature, whereTm is surface melting temperature, approximated to be

Temperature dependence has the correct limit.

At T=0, it recovers to NEB results, at T=Tm it becomes zero.

This is a reasonable approximation.

KTbulkm

7002

1

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8/14/2019 Dislocation Nucleation Rate at Finite Temperature in Copper_ver3

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Compute DislocationNucleation Rate from FFS

)|( 00

B

FFSPII =

: number of atoms in the largest cluster < A : no slip, > B : with slip

2 Dec 2009 MRS Fall Meeting, Boston p. 7/13

Allen et al, JCP (2006).

FFS works for non-Markovian and non-equilibrium

process !T. Li et al, Nat. Mat. (2009)Sanz et al, J. Phys(2008).Crystallization :

0

0 50 100 150 200 250 300 350 400 450 5002

4

6

8

10

12

14

16

18

MCS

nmax

Time

Nucleation rate of

small clusterswith 0

=

=

+=1

0

10 )|()|(mi

i

iiB PP

Success Probability of smallcluster with

0

evolving into B

Samplesuccessconfigurations

With Backward Switching, can be obtained!Valeriani, JCP (2007))(F

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Nucleation Rate from FFS

2 Dec 2009 MRS Fall Meeting, Boston p. 8/13

FFS results predict much larger stress dependence(i.e. stiffer slope w.r.t stress change) on the nucleationrate!Where does discrepancy come from?

)),(

~

exp(~TkTFNI

B

b

=)|( 00

BFFS PII =

1 1.2 1.4 1.6 1.8 2

100

105

1010

Stress (GPa)

ITST(,T

)(s-1)

400K

700K

600K

500K

1 1.2 1.4 1.6 1.8 2

100

105

1010

Stress (GPa)

IFFS(,T

)(s-1)

400K500K

600K700K

1 1.2 1.4 1.6 1.8 2

100

105

1010

Stress (GPa)

IFFS(,T

)(s-1)

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Comparison of the Prefactor

2 Dec 2009 MRS Fall Meeting, Boston p. 9/13

Prefactor ? or Free energy barrier ?

Define Prefactor of FFS such that

Now, Prefactor of TST is because

),( TFb

FFSA )),(

exp(Tk

TFAI

B

bFFSFFS =

NATST

=)

),(~

exp(~

Tk

TFNI

B

b

=

The prefactoris almost independent of Tand .Recrossing event wouldreduce

the prefactor!

TSTFFS AA 01.0

The nucleation ratediscrepancy isMainly from free energybarrier!

1 1.2 1.4 1.6 1.8 20

0.005

0.01

0.015

0.02

Stress(GPa)

AFFS/ATST

400K

500K

600K

700K

FFSA FFS

d

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Temperature and StressDependence of Free Energy

Barrier

2 Dec 2009 MRS Fall Meeting, Boston p. 10/13

)0,()/1(),(~

== TETTTF bmb

1.4)/1(8.4)0,(

athbTE ==

1 1.2 1.4 1.6 1.8 20

0.5

1

1.5

2

Stress (GPa)

Fb

(eV)

500K

600K

700K

400K

0K NEB

1 1.2 1.4 1.6 1.8 20

0.5

1

1.5

2

Stress (GPa)

Fb

(eV)

0K NEB

FFS600K

FFS500K

FFS700K FFS

400K

1 1.2 1.4 1.6 1.8 20

0.5

1

1.5

2

Stress (GPa)

Fb

(eV)

Circles : our data, Line : fit tosmooth curveWhat does this imply?

Activation Volume

Is constant or increasing functionof T.

=

),(),(

TFT b

),( T

),( T

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Entropy Contribution in Fb

2 Dec 2009 MRS Fall Meeting, Boston p. 11/13

300 400 500 600 7000

0.2

0.4

0.6

0.8

1

Temperature

Fb

(eV)

1.2 GPa

1.7 GPa

Entropy

is decreasing function of stress.

and constant over temperature.

T

TFTS b

=

),(),(

300 400 500 600 700-0.5

0

0.5

1

1.5

Temperature

Fb(eV)

1.2 GPa

1.7 GPa

Entropy

is (slightly) increasing function ofstress.

and increases over temperature.

T

TFTS b

=

),(),(

),( TS ),( TS

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0 200 400 6000

1

2

3

4

Temperature(K)

NucleationS

tress(GPa)

Zhu(2008)

This Study

Strain Rate Dependenceof Nucleation Stress

2 Dec 2009 MRS Fall Meeting, Boston p. 12/13

310=

810=

Our results suggest muchhighernucleation stress (i.e. yieldstress)at high temperature!

The discrepancy becomeslargerat lower strain rate!

Experimental Test is needed!

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Conclusion

Free energy barrier and Nucleation Rate fordislocation nucleation at finite temperature isobtained from FFS.

We predict(1) activation volume :

constant (or increasing) function of T(2) entropy : increasing function of T

constant (or increasing) function of

Our results suggest higher nucleation stress at hightemperature.

2 Dec 2009 MRS Fall Meeting, Boston p. 13/13

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Free Energy from FFS

04/16/09 MRS March Meeting, San Francisco p. 14/11

StationaryDistribution

PA : probability beingin AA : Flux from A boundary

+(q;0) : AVG time spent on q

PA IAB = PB IBA

Backward Flux SamplingRequired!

+(q;i) : histogram of q from

a trajectory from i

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Configuration from FFS

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700KA lot of fluctuations

400KSmall fluctuations

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Anderson Thermostat

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Heat Gain Per Unit Time ~

Rate of Energy Gain by collision withfrequency ~

3/1VT

TkN B2

3