Dislocation Nucleation Rate At Finite Temperature In Copper Nanorod

Dislocation Nucleation Rate at Finite Temperature in Copper Nanorod Seunghwa Ryu1 and Wei Cai2 1 Physics, Stanford University 2 Mechanical Engineering, Stanford University 2 Dec 2009 MRS Fall Meeting, Boston

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Sampling Dislocation Nucleation at finite temperature -

Copper Nano Rod under Deformation of nanomaterial is controlled by - Nucleation is a rare event, i.e. Uniaxial Compresson dislocation nucleation.

(waiting time >> reaction time )

- Molecular dynamics => Limited time scale - Transition State Theory with Fb from Nudged Elastic Band method ( Zhu (2008) ) => Zero temperature Need to directly sample calculation T. Zhu, J. Li et al, PRL 100 025502 (2008) dislocation nucleation Embedded Atom Method – Mishin Potential at finite temperature ! 15 x 15 x 20 100 | 010 |100 Cu Nano Rod 2 Dec 2009 MRS Fall Meeting, Boston

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Outline 1. Limitation of Nudged Elastic Band (NEB)

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

3. Results

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Limitation of NEB Dislocation Nucleation Rate from Transition State I TST Number of Nucleation Site

Theory (TST) Fb (σ , T ) = Nν exp(− ) k BT

E( F(λ)

Trial Frequency

State A

Free Energy Barrier

E (σ,T=0) Fb(σ,T) State B λ Size of Partial Slip.

- NEB computes Energy Profile along nucleation path at 0K. Fb (σ , T ) = Eb (σ , T ) − TS (σ , T ) - FromEb (σ , T = 0) , How can we get ~ - Zhu et al (2008) assumes Fb (σ , T ) = (1 − T / Tm ) Eb (σ , T = 0) 2 Dec 2009 MRS Fall Meeting, Boston

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?

E(λ)

Free Energy Barrier

Eb(σ,T=0)

State A

State B λ Size of Partial Slip.

- What is physical reason behind ~ Fb (σ , T ) = (1 − T / Tm ) Eb (σ , T = 0) ?

Free energy decreases linearly with temperature, where Tm is surface melting temperature, approximated to be 1 T bulk ≈ 700 K m

2

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. 2 Dec 2009 MRS Fall Meeting, Boston

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Sample Dislocation Nucleation at finite temperature Define reaction coordinate

λ from state A to state B

λ : number of atoms inside State B slip 0 0   0 0 r = | r r = | r − r | xi = max{| rij − rij |} ij i − ri | ij i j For each atom i, define State A

j

perfect crystal j : over all value A. H. W.neighbor Ngan, Proc.of R.iSoc. A 462, 1662 (2006) 400K 1.7GPa

xi > xc → slip λ : size of the largest xi > xc cluster of atoms with 2 Dec 2009 MRS Fall Meeting, Boston

Red – Surface Yellow – Atoms with

xi>xc Purple – Largest Cluster p. 6/13

Compute Dislocation Nucleation Rate from FFS λ : number of atoms in the largest cluster λ < λA : no slip,

λ > λB : with slip

I FFS = I λ0 P (λ B | λ0 ) Success Probability of small cluster with λ0 evolving into λB

Nucleation rate of 18 16 14

small clusters with λ0

λ0

10

n

λ

max

12

8 6 4 2

0

50

100

150

200

250

300

350

400

450

500

MCS

Time

P (λ B | λ 0 ) =

i = m −1

∏ P (λ

i +1

| λi ) Sample

success

i=0 FFS works for non-Markovian and non-equilibrium configurations process ! Allen et al, JCP (2006).

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

With Backward Switching, F (λ )

2 Dec 2009 MRS Fall Meeting, Boston

can be obtained!

Valeriani, JCP (2007)

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Nucleation Rate from FFS I

FFS

~ Fb (σ , T ) ~ I = Nν exp(− ) k BT

λ0

= I P (λ B | λ 0 )

10

10

700K

5

10

500K

5

10

600K 400K

I

500K

700K 600K

-1

(σ,T) (s )

10

TST

IFFS(σ,T) (s-1)

10

0

400K

0

10

10 1

1.2

1.4 1.6 Stress(GPa)

1.8

2

1

1.2

1.4 1.6 Stress(GPa)

1.8

2

FFS results predict much larger stress dependence (i.e. stiffer slope w.r.t stress change) on the nucleation rate! Where does discrepancy come from? 2 Dec 2009 MRS Fall Meeting, Boston

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Comparison of the Prefactor PrefactorA

FFS

? or Free energy Fbarrier b (σ , T ) FFS

FFS

TST Now, Prefactor of TSTAis = Nν

0.02

AFFS/ATST

0.015

0.01 400K 500K 600K 700K

0.005

0 1

1.2

1.4 1.6 Stress(GPa)

Fb (σ , T ) ) k BT ~ F (σ , T ) ~ because I = Nν exp(− b ) k BT The prefactor is almost independent of T A FFS ≈ 0.01⋅ ATST and σ. Recrossing event would reduce the The prefactor! nucleation rate

I that =A such

A Define Prefactor of FFS

1.8

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FFS ?

FFS

exp(−

discrepancy is Mainly from free energy barrier! 2 p. 9/13

2

Temperature and Stress Dependence of Free Energy Barrier 2

1.5

Ω(σ , T ) 0K NEB

1

0.5

0 1

Fb (eV)

Fb (eV)

1.5

FFS 700K

1.2

1

0K NEB

Ω(σ , T )

0.5 FFS 600K

FFS 500K

1.4 1.6 Stress(GPa)

FFS 400K

500K 600K

1.8

2

Circles : our data, Line : fit to smooth curve What does this imply? ∂Fb (σ , T ) Ω(σ , T ) = −

400K

0 1

700K 1.2

1.4 1.6 Stress(GPa)

1.8

~ Fb (σ , T ) = (1 − T / Tm ) Eb (σ , T = 0)

Eb (σ , T = 0) = 4.8(1 − σ / σ ath ) 4.1

∂σ

Activation Volume Is constant or increasing function of 2T.Dec 2009 MRS Fall Meeting, Boston

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2

Entropy Contribution in Fb 1

1.5

1

1.2 GPa

Fb (eV)

Fb (eV)

0.8

1.2 GPa

1.7 GPa

0.5

0

0.6 0.4

1.7 GPa

0.2

S (σ , T )

S (σ , T ) -0.5 300

400

500 Temperature

Entropy S (σ , T ) = −

600

∂Fb (σ , T ) ∂T

700

0 300

400

500 Temperature

Entropy S (σ , T ) = −

600

700

∂Fb (σ , T ) ∂T

is (slightly) increasing function of is decreasing function of stress. stress. and constant over temperature. 2 Dec 2009 MRS Fall Meeting, Boston p. 11/13 and increases over temperature.

Nucleation Stress(GPa)

Strain Rate Dependence of Nucleation Stress Zhu(2008) ThisStudy

4 3

ε = 108 2

The discrepancy becomes larger at lower strain rate!

ε = 10 −3

1 0 0

200 400 Temperature(K)

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Our results suggest much higher nucleation stress (i.e. yield stress) at high temperature!

600

Experimental Test is needed!

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Conclusion • Free energy barrier and Nucleation Rate for dislocation nucleation at finite temperature is obtained from FFS. • We predict (1) activation volume : constant (or increasing) function of T (2) entropy : increasing function of T constant (or increasing) function of σ • Our results suggest higher nucleation stress at high temperature. 2 Dec 2009 MRS Fall Meeting, Boston

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Free Energy from FFS Stationary Distribution

Backward Flux Sampling Required!

PA IAB = PB IBA

PA : probability being in A ФA : Flux from A boundary Τ+(q;λ0) : AVG time spent on q π+(q;λi) : histogram of q from a trajectory from λi

04/16/09 MRS March Meeting, San Francisco

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Configuration from FFS

700K A lot of fluctuations

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400K Small fluctuations

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Anderson Thermostat κ ⋅ ∆T ⋅ V 1/ 3

Heat Gain Per Unit Time ~

Rate of Energy Gain by collision with frequency ν ~

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3 Nνk B ∆T 2

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