Comment on "Eshelby twist and correlation effects in diffraction from nanocrystals" [J. Appl. Phys. 117, 164304 (2015)]

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by  Jean-Marc Roussel

Comment on “Eshelby twist and correlation effects in diffraction from
nanocrystals “ [J. Appl. Phys. 117, 164304 (2015)]
Jean-Marc Roussel1,a) and Marc Gailhanou1,b)
Aix Marseille Universit´e, CNRS, IM2NP UMR 7334, 13397, Marseille, France
(Dated: 24 January 2017)
The aim of this comment is mainly to show that anisotropic effects and image fields should not be omitted
as they are in the publication of A. Leonardi, S. Ryu, N. M. Pugno, and P. Scardi (LRPS) [J. Appl. Phys.
117, 164304 (2015)] on Palladium (cid:104)011(cid:105) cylindrical nanowires containing an axial screw dislocation . Indeed,
according to our previous study [Phys. Rev. B 88, 224101 (2013)], the axial displacement field u along
z
the nanowire exhibits both a radial and an azimuthal dependence with a twofold symmetry due the (cid:104)011(cid:105)
7
1 orientation. uz is made of the superposition of three anisotropic fields : the screw dislocation field in an
0 infinite medium, the warping displacement field caused by the so-called Eshelby twist and an additional
2 image field induced by the free surfaces. As a consequence by ignoring both anisotropy and image fields,
the deviatoric strain term used by LRPS is not suitable to analyze the anisotropic strain fields that should
n
a be observed in their Molecular Dynamics simulations. In this comment, we first illustrate the importance
J of anisotropy in (cid:104)011(cid:105) Pd nanowire by calculating the azimuthal dependence of the deviatoric strain term.

3 Then the expression of the anisotropic elastic field is recalled in term of strain tensor components to show
2 that image fields should be also considered.
The other aspect of this comment concerns the supposedly loss of correlation along the nanorod caused
]
i by the twist. It is claimed for instance by LRPS that : “As an effect of the dislocation strain and twist, if
c
the cylinder is long enough, upper/lower regions tend to lose correlation, as if the rod were made of different
s
- sub-domains.”. This assertion that is repeatedly restated along the manuscript appears to us misleading
rl since for any twist the position of all the atoms in the nanorod is perfectly defined and therefore prevents
t any loss of correlation. To clarify this point, it should be specified that this apparent loss of correlation
m
can not be ascribed to the twisted state of the nanowire but is rather due to a limitation of the X-ray
.
t powder diffraction combined with the Whole Powder Pattern Modeling (WPPM). Considering for instance
a
coherent X-ray diffraction, we show an example of high twist where the simulated diffractogram presents a
m
clear signature of the perfect correlation.
-
d
n
o I. ANISOTROPIC STRAIN FIELD INDUCED BY AN imuth θ. Clearly the azimuthal exploration shows that
c AXIAL SCREW DISLOCATION IN A (cid:104)011(cid:105) FCC METAL
[ NANOWIRE
Anisotropictwiststrainonly(cid:15)tdwevist
Anisotropicscrewstrainonly(cid:15)screw
1 dev
θ=0
v The displacement field uz induced by an axial screw 100 θ=π/2
0 dislocation in a (cid:104)011(cid:105) fcc metal nanowire has been stud- Isotropic(cid:15)screw and(cid:15)twist
dev,iso dev,iso
6 iedindetailrecently.1 Foracircularcrosssection,theuz ‰) Isotropic(cid:15)dscerve,iws+otwist
4 field presents the two-fold symmetry of the (cid:104)011(cid:105) orien- n(
06 tationwithanazimuthalθ dependencethatiscontrolled Strai
. bytheanisotropyoftheshearmodulus. Thislatterissig- oric 10
1 nificant for Palladium since like in the case of Copper1 at
0 thevaluesoftheelasticmoduliaresimilarwithC44 ≈28 Devi
7
GPa and C ≈ 82 GPa in the {[100],[011],[011]} coor-
1 55
dinate system.
:
v To illustrate the importance of these anisotropic ef-
Xi fects, let us calculate the azimuthal dependence of two 1
0 1 2 3 4 5 6 7 8
particular quantities discussed in the article of LRPS2
r RadialDistancer(nm)
a (and reported in their Figure 6d), namely the deviatoric
strain term due to the screw deformation only and the
FIG.1. Isotropicdeviatoricstraintermsreproducedfromthe
one due to the twist only. These latter, denoted here Figure 6d of LRPS2 (black dotted lines) and compared to
(cid:15)sdcervew and (cid:15)tdwevist respectively, are plotted in Figure 1 as the same terms (cid:15)sdcervew and (cid:15)tdwevist calculated from anisotropic
a function of the radial distance r but also for all az- elasticity for all azimuth θ from Ref.[1]. Clearly for Palla-
dium, values of (cid:15)screw and (cid:15)twist spread over large domains
dev dev
bounded by extrema (for θ = 0 and θ = π/2) that differ by
a factor C /C . The isotropic (cid:15)screw+twist is also reported
55 44 de√v,iso
a)Electronicmail: [email protected] (red dotted line), it vanishes for R/ 2.
b)Electronicmail: [email protected]
2
both (cid:15)screw and (cid:15)twist belong to large domains bounded term of strain becomes :
dev dev
by extremum values (for θ = 0 and θ = π/2) that differ
−br c (θ) br c (θ)
by a ratio equal to C55/C44 ≈2.93. (cid:15)twist = 55 (cid:15)twist = 45 (3)
Incidentally, we wish to comment the analysis made θz πR2C44+C55 rz πR2C44+C55
of the deviatoric strain terms in the Figure 6d, that
Finally, in the present case of an anisotropic (cid:104)011(cid:105)
leads the authors to conclude at the end of section III:
nanowire of circular cross section containing a coaxial
“..., so that the combined effect (screw and twist) gets
screw dislocation, an image stress field σimg is neces-
closer to the MD simulation.”. This assertion is dou-
sary to fulfill the condition of a vanishing traction at
bly misleading. First of all because the MD simula-
the lateral surface. Formally, this condition reduces to
tion curve must contain the above mentioned anisotropy (cid:12)
which is not shown on this graph (some clarification on σrimzg(cid:12)r=R + σrszcrew|r=R = 0 because σrtwzist is null for a
circular cross section.
the method used to get the MD curve would be help-
Thus,lookingforanimagefieldthatobeysbothtothe
ful). Andsecondlybecausetheisotropicdeviatoricstrain
boundaryconditions,theequilibriumandthecompatibil-
term designated as “Screw and Twist deformation field”
ityequations,wecouldobtainanumericalsolutionofthe
in Figure 6d does not match a calculation of the com-
stress field based on a Fourier series analysis. Approxi-
bined effect of both the dislocation and the torsion. The
plot of this term denoted as (cid:15)screw+twist in Figure 1 of mateexpressionsofσθimzg andσrimzg werealsoproposedin
dev,iso Ref.[1]. Using Eqs.(1), these latter can be converted in
the present work reveals a very different behavior since
√
(cid:15)screw+twist should vanish for r = R/ 2, R being the term of strain and written as :
dev,iso
nanowire radius. This result can be directly understood br (cid:20) (cid:18)c (θ)(cid:19) c2 (θ)(cid:21)
by examining the (cid:15) and (cid:15) strain components in this (cid:15)img =− √ c (θ)ln 44 − 45
θz rz θz 4π C C R2 55 C c (θ)
isotropic case: the (cid:15) are null for both the dislocation 44 55 0 44
rz (cid:20) (cid:18) (cid:19)(cid:21)
and the torsion but the (cid:15)θz components have opposite (cid:15)img =− √ br c (θ) 1−ln c44(θ) (4)
signs with (cid:15)screw = b and (cid:15)twist = −1 br where b rz 4π C C R2 45 C
θz,iso 4πr θz,iso 2πR2 44 55 0
is the magnitude of the Burgers vector.3,4 Consequently,
since (cid:15)dev,iso = √46(cid:112)(cid:15)2θz+(cid:15)2rz, one gets (cid:15)dscerve,iws+otwist = with C0 is equal to C55/2.
(cid:113) √
2√b ( 1 − r )2. Thus, for r approaching R/ 2 the
π 6 2r R2
combined effect(screwand twist)gets faraway fromthe
MD simulations shown by the authors.
To conclude this section, we provide the expressions
of the strain components (cid:15) and (cid:15) leading to the
θz rz
anisotropicbehaviorreportedinFigure1. Wealsoderive
fromourpreviouswork1 theadditionalimagestrainfield
that results from the interaction of the screw dislocation
with the lateral surfaces of the anisotropic cylinder.
Having determined the equilibrium stress components
σ and σ in Ref.[1], the derivation of the strain field
θz rz
becomesstraightforwardbyusingthefollowingrelations:
(cid:20) (cid:21)
1
(cid:15) = σ c (θ)−σ c (θ)
θz 2C C θz 55 rz 45
44 55
(cid:20) (cid:21)
1
(cid:15)rz =2C C −σθzc45(θ)+σrzc44(θ) (1) FIG. 2. (cid:15)θz (×) and (cid:15)rz (+) strain components calculated
44 55 fromourMolecularStatics(MS)simulationsinRef.[1]atdif-
ferentrvaluesinthecaseofanuntwisted[110]circularcopper
wheretheelasticmodulicanbewrittenasc44(θ)=C⊕+ nanowire of radius R = 30nm containing an axial screw dis-
C(cid:9)cos2θ,c55(θ)=C⊕−C(cid:9)cos2θandc45(θ)=C(cid:9)sin2θ location. These results are compared to the expressions of
with C =(C +C )/2 and C =(C −C )/2. (cid:15)screw+(cid:15)img and (cid:15)screw+(cid:15)img (solid lines) proposed in Eqs.
⊕ 44 55 (cid:9) 44 55 θz θz rz rz
Thus, the strain field induced by a perfect Volterra (2)and(4). Theboundaryproblemisalsosolvednumerically
screw dislocation, with Burgers vector b = 1/2 a(cid:104)110(cid:105) is through the Fourier series analysis described in Ref.[1].
inversely proportional to r with a marked θ dependence:
InRef.[1],theimagefieldderivedintermofstresscom-
√
b C C ponentswascomparedtotheonecalculatedfromMolec-
(cid:15)screw = 44 55 (cid:15)screw =0 (2)
θz 4πrc (θ) rz ular Statics simulations (MS). Similarly in the present
44
comment,theMSsimulationscanserveasareferencefor
The twist of the nanowire that is necessary to cancel the testing the validity of the approximate expression given
torque due to the dislocation produces a σtwist stress in Eqs.(4) of the image strain. In practice, the analytic-
θz
component (σtwist is null for a circular cylinder) that in ity of the Tight Binding potential used in our atomistic
rz
3
simulationsallowsastraightforwarddeterminationofthe cluding their relative orientation. This method was used
strain components per atom. This is illustrated in Fig- recently to determine the structure of inversion domains
ure2wheretheradialandtheazimuthaldependenciesof in a Gallium Nitride nanowire6, a system which presents
thestrainfieldcomponents(cid:15) and(cid:15) resultingfromour similarities with the one discussed here.
θz rz
MS simulations are plotted in the case of an untwisted
Cu nanowire of radius 30 nm containing a screw disloca-
tion at its center (the torsion can be treated separately
sinceitdoesnotaffecttheimagefieldforacircularcross
section). As for the stress analysis, the same conclusions
can be drawn. The dislocation field in Eq. (2) combined
with the image field in Eq. (4) capture well the radial
dependence and the azimuthal anisotropy of the strain
field found in our simulations. This anisotropy is par-
ticularly pronounced for Copper (as for Palladium). It
controls for instance the shape of the Eshelby potential
well that traps the screw dislocation at the center of the
twisted nanowire.5
II. DIFFRACTION FROM A TWISTED CYLINDER
At the end of their article, LRPS arrive at the con-
clusion that “the twist weakens the correlation between
more distant regions of the cylindrical domain, up to the
pointthatneedle-likenanocrystalsappearasmadeofsub-
domains (...) which scatter incoherently.”.
Fundamentally, torsion does not introduce any ran-
domness of the atomic positions and therefore can not
be the cause of a loss of correlation.
We believe rather that this apparent loss of correla-
tion should be presented as a limitation of the technique
employed (i.e., the WPPM analysis combined with X-
raypowderdiffraction)thatdoesnotpermittodiscernif
theabovesub-domainsscattercoherentlyorincoherently
FIG. 3. (a) Simulated coherent X-ray diffraction from two
in such twisted samples. Besides, it is worth complet-
Coppercylinders(height16nm,diameter16nm)withonero-
ing that there are other techniques like X-ray coherent tated by 3 degrees around their common [011] axis. The re-
diffractionthatarecapabletoshowtheinterferencephe- ciprocal space maps are in the (011)* plane around the 222
nomena that occur from the different sub-domains. reciprocal space point. (b) Same as (a) but the intensities
Toillustratethispoint,letusforinstanceconsiderthe diffractedbythetwocylindersareaddedasifthetwoobjects
model system envisaged by LRPS made of two identical wereseparatedbyadistancemuchlargerthantheX-raybeam
Pd cylinders with the upper one rotated by different an- coherence length (and therefore scatter incoherently).
gles around the common [hh0] axis. According to these
authors “A WPPM analysis of the corresponding pow-
der patterns shows that for tilt angles > 1.5 ◦ coherence 1M. Gailhanou and J.-M. Roussel, “Displacement field of a screw
between the two half-cylinders is completely lost, so that dislocation in a (cid:104)011(cid:105) Cu nanowire: An atomistic study,” Phys.
Rev.B88,224101(2013).
powder diffraction “sees” completely separate (incoher-
2A.Leonardi,S.Ryu,N.M.Pugno, andP.Scardi,“Eshelbytwist
ently scattering) domains”. Considering now the same and correlation effects in diffraction from nanocrystals,” Journal
samplestudiedwithX-raycoherentdiffraction, thissup- ofAppliedPhysics117,164304(2015).
posed“lossofcoherence”isnotobserved. Figure3shows 3J.D.Eshelby,“Screwdislocationsinthinrods,”J.Appl.Phys.24,
176(1953).
an example of large tilt angle (3 degrees) where clearly
4J.HirthandJ.Lothe,Theoryofdislocations,2nded.(JohnWiley
one can make the difference between the real diffraction
andsons,1982).
patternfromthetwocylinders[Fig.3(a)]andtheonethat 5J.-M.RousselandM.Gailhanou,“Stabilityofascrewdislocation
would correspond to incoherent diffraction [Fig.3(b)]. ina(cid:104)011(cid:105)coppernanowire,”Phys.Rev.Lett.115,075503(2015).
Finally, let us mention that Fig.3(a) is only a slice 6S. Labat, M.-I. Richard, M. Dupraz, M. Gailhanou, G. Beutier,
M. Verdier, F. Mastropietro, T. W. Cornelius, T. U. Schu¨lli,
of a three dimensional reciprocal space structure. From
J. Eymery, and O. Thomas, “Inversion domain boundaries in
themeasurementofthislatter, associatedwithmeasure-
GaN wires revealed by coherent bragg imaging,” ACS Nano 9,
ments around other reciprocal space points, an inversion 9210–9216(2015).
method should provide the two cylinders structure in-