UCSD NANO106 - 07 - Material properties and tensors

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Properties of Materials and Tensors Shyue Ping Ong Department of NanoEngineering University of California, San Diego

Transcript of UCSD NANO106 - 07 - Material properties and tensors

Page 1: UCSD NANO106 - 07 - Material properties and tensors

Properties of Materials and TensorsShyue Ping OngDepartment of NanoEngineeringUniversity of California, San Diego

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Readings¡Chapter 1 of Properties of Materials

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Einstein notation¡ Summation implied over repeated

indices

¡ Examples

¡ Refer to handout 8 for more details.

NANO 106 - Crystallography of Materials by Shyue Ping Ong - Lecture 7

xiyi = xiyii=1

3

∑ = x1y1 + x2y2 + x3y3

Dot product: x ⋅ y = xiyiMatrix product: y = Axyi = aij x j

“I have made a great discovery in mathematics;

I have suppressed the summation sign every

time that the summation must be made over an

index which occurs twice...”

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Material properties¡A physical property is a relationship between two

measured quantities.

NANO 106 - Crystallography of Materials by Shyue Ping Ong - Lecture 7

Equilibriumproperties

Transportproperties

Hystereticproperties

Irreversiblebreakdownproperties

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Heckman Diagram

NANO 106 - Crystallography of Materials by Shyue Ping Ong - Lecture 7

Electrical properties

Thermal properties

Mechanical properties

Tensors are the mathematics of choice to represent properties.

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Tensors¡Tensors are geometric objects that describe linear

relations between vectors, scalars, and other tensors.

¡Can be thought of as multi-dimensional arrays. The rank of a tensor is the number of independent indices:¡ Rank 0: Scalar¡ Rank 1: Vector¡ Rank >= 2: Tensors

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Examples of Tensors

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Q =CΔTRank Property Equation

0 Heat capacity

1 Pyroelectricity

2 Electrical conductivity

3 Piezoelectricity

4 Elasticity and compliance

Q =CΔT

ΔPi = piΔT

ji =σ ikEk

Pi = dijkσ jk

σ ij = cijklεklHow many components are there for each rank?

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Transformation of Orthogonal Axes¡ In the first part of the course, we went through the

transformation of basis vectors / reference frames. This is a similar, but simpler concept.

¡ Material properties are typically measured in Cartesian axes. For example, you typically measure the current and electric field in a set of orthogonal axes and the conductivity tensor relates the current in a direction to the components of the electric field in each direction.

¡ To understand the effect of symmetry on properties, we need to know how tensors transform under various operations, e.g., rotation, reflection, inversion, etc.

NANO 106 - Crystallography of Materials by Shyue Ping Ong - Lecture 7

ji =σ ikEkTypically, indices refer to Cartesian directions

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Notation¡We will use Z1, Z2, and Z3 to refer to the x, y and z

Cartesian axes. Transformed axes are denoted with as Zi’.

¡As per standard notation, the cartesian basis vectors are denoted as ex, ey, ez or e1, e2, e3, the latter is used with the more compact Einstein summation notation.

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Direction cosine¡ The direction cosines of a

vector are the cosines of the angles between the vector and the three coordinate axes.

¡ Component contributions of the basis to the unit vector.

NANO 106 - Crystallography of Materials by Shyue Ping Ong - Lecture 7

vv= cosa ⋅ ex + cosb ⋅ ey + cosc ⋅ ez

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Direction cosine tensor¡ Consider the transformation of a set of

orthogonal axes Z1, Z2, and Z3 to another set of orthogonal axes Z1’, Z2’, and Z3’.

¡ The relationship between the new basis vectors and old basis vectors are given by:

¡ where aij is the direction cosine between ei’and ej.

NANO 106 - Crystallography of Materials by Shyue Ping Ong - Lecture 7

e1!

e2!

e3!

"

#

$$$$$

%

&

'''''

=

a11 a12 a13

a21 a22 a23

a31 a32 a33

"

#

$$$$

%

&

''''

e1

e2

e3

"

#

$$$$

%

&

'''' or simply ei! = aijej

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Properties of the direction cosine tensor¡ Orthogonal¡ Norm of each row and column must = 1, e.g.,

¡ Rows and columns are mutually perpendicular, e.g.,

¡ More compactly, in Einstein notation,

¡ Inverse is simply the transpose. Hence,

NANO 106 - Crystallography of Materials by Shyue Ping Ong - Lecture 7

a112 + a12

2 + a132 =1 and a11

2 + a212 + a31

2 =1

a11a21 + a12a22 + a13a23 = 0 and a11a12 + a21a22 + a31a32 = 0

aikajk = δik = akiakj

e1

e2

e3

!

"

####

$

%

&&&& =

a11 a21 a31

a12 a22 a32

a13 a23 a33

!

"

####

$

%

&&&&

e1'

e2'

e3'

!

"

#####

$

%

&&&&&

or simply ei ='= ajiej'

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Quick Review of Point Symmetry Operations¡ Rotation

¡ Reflection/Mirror

¡ Inversion

NANO 106 - Crystallography of Materials by Shyue Ping Ong - Lecture 7

e1!

e2!

e3!

"

#

$$$$$

%

&

'''''

=cosθ sinθ 0−sinθ cosθ 00 0 1

"

#

$$$

%

&

'''

e1e2e3

"

#

$$$$

%

&

''''

e1!

e2!

e3!

"

#

$$$$$

%

&

'''''

=1 0 00 1 00 0 −1

"

#

$$$

%

&

'''

e1e2e3

"

#

$$$$

%

&

''''

e1!

e2!

e3!

"

#

$$$$$

%

&

'''''

=−1 0 00 −1 00 0 −1

"

#

$$$

%

&

'''

e1e2e3

"

#

$$$$

%

&

''''

Anti-clockwise rotation about z-axis

Reflection about xy-plane

e1!

e2!

e3!

"

#

$$$$$

%

&

'''''

=cosθ −sinθ 0sinθ cosθ 00 0 1

"

#

$$$

%

&

'''

e1e2e3

"

#

$$$$

%

&

''''

Clockwise rotation about z-axis

Note that this is opposite from the definition for rotation matrices for points / vectors!

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Standard settings¡ Crystallographic axes and cartesian property measurement

axes do not always coincide.

¡ IEEE standard settings¡ Triclinic: Z3 is chosen parallel to c, and Z2 is normal to the ac plane

(010). Z1 lies in the (010) plane and is perpendicular to Z2 and Z3.¡ Monoclinic: Z2 is chosen parallel to the b [010] axis with twofold

symmetry and perpendicular to mirror plane. Z3 is chosen along c, and Z1 is perpendicular to Z2 and Z3.

¡ Orthorhombic: Select unit cell in which c < a < b. Z1, Z2, and Z3 are chosen as a, b, and c, respectively.

¡ Trigonal and Hexagonal: Z3 is along the c-axis [001] with three-fold / six-fold symmetry. Z1 axis is parallel to a-axis [100], a crystallographic direction, with Z2 completing the right-handed orthogonal set.

¡ Tetragonal: Z3 is parallel to c-axis with four-fold symmetry. Z1 and Z2are parallel to a and b, respectively.

¡ Cubic: Z1, Z2, and Z3 parallel to a, b, and c, respectively.

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Curie Symmetry Principle¡Thus far, we have dealt only with symmetry of the

crystal.

¡When dealing with properties, we need to consider both the symmetry of the crystal, as well as the symmetry of the “cause”.

¡Examples:¡ Most materials are polycrystalline, and the processing of

these materials create anisotropy in properties.¡ The externally applied “force”, e.g., an electric field, has a

different symmetry from the underlying symmetry of the crystal.

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Curie Symmetry Principle¡ The Curie symmetry principle is the causality relation between the

symmetry of the cause and that of the effect. The principle is composed of three parts:¡ If certain causes yield the known effects, the symmetry elements of the

causes should be contained in the generated effects.¡ If the known effects manifest certain dissymmetry (absence of symmetry

elements), this latter should be contained in the causes which have generated those effects.

¡ The converse to these two previous propositions is not true, at least in practical: i.e., the effects may have higher symmetry than the causes which generate these effects.

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!K = K∩G

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Curie symmetries

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Example of application of the Curie Symmetry Principle¡ Consider crystal with space group . It has a fourfold inversion axis (which

contains in itself also a two-fold rotation axis), and two two-fold rotation axes. The symmetry of the electric field is , which has an infinite-fold rotation axis containing every possible rotation axes of lower symmetry, and an infinite number of intersecting mirror planes.

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42m

∞m

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Curie symmetries of processed materials

NANO 106 - Crystallography of Materials by Shyue Ping Ong - Lecture 7

Material Curie Symmetry

Polycrystalline metals / ceramics with randomly oriented grainsOrganic left- or right-handed crystallitesUniaxially hot-pressed polycrystallinematerialsUniaxially hot-pressed left-or right-handedpolycrystallinematerials

∞∞m

∞∞

∞mm

∞2

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Transformation of Tensors

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Materials Properties as Tensors¡A material property is an expression of the response

of a material to the application of a generalized “force” by a generalized “displacement”.

¡Generalized “Forces”:¡ Stress¡ Temperature gradients¡ Electric fields

¡Generalized “Displacements”¡ Strain¡ Heat flow¡ Current density

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Example: Rank-2 tensor¡Consider the response displacement q to a force

p.

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p

qIn general, p ≠ q (not parallel)but we can make the assumption thatq ∝ pfor small forces and displacements (think of Taylor's series). If we denote rij as the response of component i of p to component j of q, then

q =r11 r12 r13

r21 r22 r23

r31 r32 r33

#

$

%%%%

&

'

((((p

or qi = rij pj

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Recap: Transformation of Orthogonal Axes¡ Recall that in the previous lesson, we showed

that for a transformation of orthogonal axes, we have

¡ where aij is the direction cosine between ei’ and ej. Note the use of the orthogonality property of the direction cosine tensor, which allows us to use the transpose aji as the inverse transformation.

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ei! = aijej

ei = ajiej!

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Transformation of a 2nd rank tensor

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Consider qi = rij pjWhen we switch to a new set of orthogonal axes (and hence basis vectors), the vectors q, p and the tensor r must transform to new values. Let's denote this as

qi! = rij!pj!

How can rij! be evaluated in terms of the original rij using the direction cosines?Let's see what happens to the components qi and pj first. Using the direction cosinematrix, we can write

qi! = aimqm = aimrmn pn = aimrmnajn pj!

Note the use of new dummy indices m and n to distinguish it from the i and jwe have used so far. It is absolutely important to keep proper track of the indices!!

Hence, rij! = aimajnrmn

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Matrix form¡Now that we have gone through the exercise

using Einstein notation, can we do the same using matrices?

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rij! = aimajnrmn is equivalent to !R = ARAT

Dummy indices

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How do higher rank tensors transform?

¡For a second rank tensor,

¡Using the above as a template, we can deduce that an n-rank tensor with transform as

¡(capital letters are dummy indices)

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rij! = aimajnrmn

rijklm… ' = aiIajJakKalLamM…rIJKLM…

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Example: 4th rank tensor¡Consider the compliance tensor

¡ It transforms as

¡Each element has 5 x 81 terms. You have 81 elements. So a total of 5 x 81 x 81 = 32805 terms! (clearly the Einstein notation is a lot more compact…)

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εij = sijklσ kl

sijkl ' = aiIajJakKalLsIJKL

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Method of Direct Inspection¡Useful algorithm to deduce value of a transformed

tensor element.

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X1X2X3

Consider a point with coordinates X1X2X3

Let's consider how a product of this point XiX j will transform under an orthogonal axis transformation.

Xi!Xj! = aimXmajnXn = aimajnXmXn

Recall that the law of transformation of a 2nd rank tensor is

rij! = aimajnrmni.e., the transformation of a 2nd rank tensor is analogous tothe transformation of a product of coordinates! Note that retaining the order is *important*.

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Application of Method of Direction Inspection

¡Consider a 4-fold rotation about the Z3 axis.

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X1

X2

X3

X1’

X2’X3’

Blackboard

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Application of Method of Direction Inspection, contd.¡ Algorithm can be generalized to higher ranked tensors.

¡ Consider the same coordinate transformation, but let’s say we want to know the value of a 4th rank compliance tensor s1231 after the transformation

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Blackboard

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Neumann’s Principle¡Neumann's principle, or principle of symmetry,

states that, if a crystal is invariant with respect to certain symmetry elements, any of its physical properties must also be invariant with respect to the same symmetry elements, or otherwise stated, the symmetry elements of any physical property of a crystal must include the symmetry elements of the point group of the crystal.

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Demonstration of the power of applying Neumann’s principle

¡ Consider the transformation of any tensor under inversion symmetry. We have

¡ A general tensor will transform as

¡ We see that for odd-rank tensors,

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X1! = −X1X2! = −X2X3! = −X3

By the method of direct inspection, we see that

Xi!Xj!Xk!…= (−1)n XiX jXk…

rijk… ' = (−1)n rijk…

rijk… ' = −rijk…

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Demonstration of the power of applying Neumann’s principle¡ If a crystal has inversion symmetry, that means that the tensor

property measured under that symmetry must be invariant under inversion. Hence,

¡ For even-ranked tensors, we have

¡ For odd-ranked tensors, we have , which is true only if .

¡ Hence, all odd-ranked tensor properties disappear in all crystals with an inversioncenter, i.e., centrosymmetric point groups!

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rijk… ' = (−1)n rijk… = rijk…

rijk… ' = rijk… = rijk…

rijk… ' = −rijk… = rijk…rijk… = 0

We have just derived a very general and broad rule about material properties just by considering a single

symmetry operation!

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Effect of symmetry on rank 2 tensors¡ To make things more concrete, let’s consider the conductivity tensor

(though the results we derive are going to be applicable to all rank-2 tensors).

¡ Recall the definition of the conductivity tensor as

¡ From the previous analysis, we have seen that for even-ranked tensors, the inversion symmetry does not impose any restriction on the tensor elements.

¡ Let us now analyze what forms the conductivity tensor can take for the various crystal systems. Note that we are only going to analyze the restrictions imposed by symmetry arguments only. Thermodynamic arguments can impose further restrictions (e.g., many rank-2 tensors are “symmetric” in that aij = aji) that we shall not consider here.

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ji =σ ikEk

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Rank-2 tensors in triclinic crystals¡ The triclinic crystal system essentially has no further

symmetry (except for perhaps inversion).

¡ This means that the conductivity tensor has the general form:

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σ ij =

σ11 σ12 σ13σ 21 σ 22 σ 23

σ 31 σ 32 σ 33

!

"

####

$

%

&&&&

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Rank-2 tensors in monoclinic crystals¡ For monoclinic crystals, we have a two-fold

[010] rotation axis.

¡ Assume that the two-fold axis corresponds to the X2 direction (as per standard settings), under the two-fold rotation operation, we have

¡ Let’s consider how each of the nine elements in the conductivity tensor will transform under the two-fold rotation.

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Blackboard

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Rank-2 tensors in monoclinic crystals, contd.

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Blackboard

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Rank-2 tensors in orthorhombic crystals

¡ In orthorhombic systems, we have 2-fold rotations as well as mirror planes. To save some effort, let us just use the tensor that we have already derived for monoclinic crystals (the same symmetry restrictions apply) and impose additional symmetry restrictions from the mirror plane. Let’s consider the mirror plane in the a-b plane:

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Blackboard

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Rank-2 tensors in tetragonal, hexagonal and trigonal crystals ¡ In these crystals, we have a >2-fold

rotation rotation about the Z3 axis. Let’s analyze these in a general manner.

¡ Recall that the rotation direction cosine tensor is of the form:

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aij =cosθ sinθ 0−sinθ cosθ 00 0 1

"

#

$$$

%

&

'''

Hence,

X1( = X1 cosθ + X2 sinθ

X2( = −X1 sinθ + X2 cosθ

X3( = X3

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Rank-2 tensors in tetragonal, hexagonal and trigonal crystals

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Blackboard

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Rank-2 tensors in tetragonal, hexagonal and trigonal crystals

¡Final form of rank 2 tensor in these crystal systems

NANO 106 - Crystallography of Materials by Shyue Ping Ong - Lecture 7

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σ ij =

σ11 0 00 σ11 00 0 σ 33

!

"

####

$

%

&&&&

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Rank 2 tensors in cubic systems¡ Start from the tetragonal crystal system, and consider what would

happen if we impose additional four-fold rotation symmetry in the other two directions. Now, the four-fold rotation in Z3 enforces that

¡ Logic dictates that four-fold rotation axes in Z1 and Z2 will enforce respectively. Hence,

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σ ij =

σ11 0 00 σ11 00 0 σ11

!

"

####

$

%

&&&&

σ11 =σ 22, σ 33 non zero and all other elements 0.

σ 22 =σ 33, and σ11 =σ 33

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Summary of rank 2 tensors for all crystal systems

¡ Note that rank 2 tensors take these forms only when the measurement axes are based on the crystal axes as per the analysis. In a general direction, a rank 2 tensor has 9 non-zero components (though considerably fewer independent components, as we shall see later)

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σ ij =

σ11 σ12 σ13σ 21 σ 22 σ 23

σ 31 σ 32 σ 33

!

"

####

$

%

&&&&

Triclinic

σ ij =

σ11 0 σ130 σ 22 0σ 31 0 σ 33

!

"

####

$

%

&&&&

Monoclinic

σ ij =

σ11 0 00 σ 22 00 0 σ 33

!

"

####

$

%

&&&&

Orthorhombic

σ ij =

σ11 0 00 σ11 00 0 σ 33

!

"

####

$

%

&&&&

Tetragonal, Trigonal, Hexagonal Cubic

σ ij =

σ11 0 00 σ11 00 0 σ11

!

"

####

$

%

&&&&

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Lecture Summary¡ In this lecture, we started a foray into the transformation of

tensors.

¡ We looked at Neumann’s Principle, and demonstrated how just based on this very simple concept and symmetry considerations, we can know a lot about the form of a material property tensor.

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