Marder M.P. Condensed Matter Physics

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Other Constitutive Laws 333

terms available that are invariant when the sample as a whole is rotated are

(n-V)n

(12.51a)

V-n (12.51b)

n-Vxn. (12.51c)

The first of these, Eq. (12.51a) can be rewritten as V(h-h)/2, which vanishes

because h is

unit vector. The second cannot have

nonzero coefficient because

it does not respect the symmetry h

—>

—h.

The third changes sign under reflection

about the x, y, or z planes, and it must be discarded on this account. Therefore, one

is driven to terms that are quadratic in gradients of h.

There is a discouraging number of terms to consider. A general term involving

two gradients is of the form

P^P,

(12.52)

and as each index can adopt three values, there is a total of

different contribu-

tions at the outset. So the free energy is

dr iF(r)

= | / dr

~S~] C

5 —— ——. 3(7) is the free energy per volume.

αβ

9r&

(12.53)

One can always interchange αη

<->

βδ, because Eq. (12.53) does not change when

this switch is performed, but this still leaves 45 independent values of

further reduce this number by application of symmetry, pick

particular point in

space, r, and choose the

axis to coincide with the direction of the director n(r).

Most of the calculation will be carried out with this specific choice of coordinate

system, until the very end, when the results will be interpreted in a fashion that is

independent of it.

A first simplification results immediately from the fact that « is

unit vector.

All gradients of n

vanish because

7Γ"

=Έ-(η·«) H

)

dr,

,—

- ..

—..

have to be calculated

Although

points along

right

at r, it

will

not generally point precisely along

z a

small

: 2n

——n

2——n

distance away, and therefore gradients of n

(12.55)

There are 21 independent coefficients

remaining.

The remaining simplifications are deduced with greater difficulty. Whatever

terms enter the free energy must

invariant

one rotates the whole system

slightly around the

axis. Problem

shows that following rotation about the

axis through a small angle

Θ,

one obtains

8η

0η

^ θη

drs

drs ^ dr

μ ιμ

334

Chapter 12. Elasticity

where

(°

°\

R= 1 0 0 . (12.57)

\0 0 0/

Placing Eq. (12.56) into the free energy (12.53) and demanding that all terms pro-

portional to

vanish gives

ο=Σ

\dn

dn~

8η

3η

dy dr

^αχ-γδ

αηδ

•s—^

δη

dx drs

ayjS

drs

■yßr/δ

δη

d~r~

~dr~s

Cxß

The coefficient of every term, such as

dz dy '

(12.58)

(12.59)

has to vanish independently. For example, demanding that the coefficient of (12.59)

vanish gives

The tedious task of constructing all 21 such equations and using them to eliminate

as many coefficients as possible is best turned over to any computer algebra system,

which within a few minutes will show that only five independent coefficients now

survive. These are the coefficients of

(12.61a)

(12.61b)

(12.61c)

(12.61d)

(12.61e)

~dn

[dz\

dy i

~~d~l

' dUy

[dz\

driy dn

dy dx

dx dy dy dx

The final task is to identify each of these quantities in terms of operators that

no longer depend upon the particular choice of z as the local axis of the nematic.

The quantities appearing in Eq. (12.61) may be rewritten as

(V·«)

\n x (V xn)|

(«•(VXAÌ))

ή· (V x n)V-h

V·

(«· V)n

—

/ì(V ·η)

This quantity is difficult to deduce from

Eq. (12.6le), although easy to verify

once one has made the right guess.

(12.62a)

(12.62b)

(12.62c)

(12.62d)

(12.62e)

Other Constitutive Laws

335

The coefficient of (12.62d) must vanish because this term is odd in h. The final

term, (12.62e) need not be included in the free energy, because as a divergence it

contributes to the surface energy but not to the bulk energy. The free energy density

has finally only three terms,

3-=y(V-n)

+y(MVx«))

+f(rcx(Vx«))

{ηω)

splay twist bend

Problem 10 provides an example of how this free energy can be employed to find

the mechanical response of liquid crystals. Many other calculations are found in

de Gennes and Prost (1993) Chandrasekhar (1992), and Chaikin and Lubensky

(1995).

12.4.2 Granular Materials

Granular materials, such as sand, are peculiar hybrids of solids and liquids. Placed

in a pile, they retain their shape like a solid. Poured from glass, they flow like

liquids. Their mechanical properties are challenging to describe. An introduction

to some of the issues involved is presented by Bergman and Stroud (1992) and

Jaeger et al. (1996a,b).

One of the classic ways to examine mechanical properties of a granular solid

is in a triaxial text. A tube of sand sustains compressive stress σ\ from the top,

and compressive stress 02 from the sides, with \σ\\ > |cr

|. Using the idea of static

friction, one can calculate when grains will begin to slide and the tube of sand fall

apart.

The idea of the calculation is to imagine slicing through the tube of sand with

imaginary planes pointing in every possible direction. For every plane, calculate

the normal force

and the shear force r. If for any plane r > μ

Ν\ where μ

the coefficient of static friction, then sand should begin to slip along that plane. So

the stability of a granular column comes down to finding normal and shear forces

in all possible directions.

Specialize to two dimensions. The stress tensor has the simple form

What is the stress when viewed along some arbitrary direction

ΘΊ

In a frame rotated

through angle

Θ,

the stress tensor becomes

cos

sin

\ / σ\ 0 W cos

Θ —

sin

—

sin

cos

) V 0 σ2 J \ sin

cos

θ σ\ + sin

θ 02 (σ\

—

02) sin

cos

(σι

—

σ2) sin

cos

Θ 02

+ sin

θ σ\ '

336

Chapter 12. Elasticity

Ασ, subcriticai

Figure 12.5. Diagram describing the Mohr-Coulomb circle.

The components of the first row give normal N and shear stress r across the plane

at angle

Θ.

They can be rewritten

yV =

— (<J

+ Δσ COS 20) Introduce the minus sign because the convention with

stresses is that they are positive in tension, while in this

problem it is natural to take the normal force positive in

compression.

T = -Δσ sin 2Θ

where the average stress σ and differential stress Δσ are

σι + σι σ\— σ

—τ—;

Δσ = —-—.

(12.66)

(12.67)

Thus a plot of r versus

describes a circle centered on the point (|σ|, 0),

as shown in Figure 12.5. The figure also shows a line of slope τ/Ν = μ, which

describes the threshhold beyond which the granular tube will be unstable. When

stresses are just at the point where instability is about to occur, the circle parametrized

by Eq. (12.67) is just tangent to the line. This critical condition occurs when

Δσ

μ.

(12.68)

Problems

Symmetries of

Complete the comment attached to Eq. (12.21). Write the additional 6 terms

needed to symmetrize C, and explain why it is legitimate to replace C by this

symmetrized quantity.

Rotations in linear elasticity: Because the nonlinear term has been discarded

from Eq. (12.19), linear elasticity does not treat rotations very well. Suppose

that a solid would begin to crumble inside when subjected to a uniform exter-

nal pressure of (λ + μ)/5. Subject this solid to rigid rotations, compute the

Problems 337

(spurious) pressure predicted by linear elasticity, and find the rotation angle at

which the solid would begin to come apart.

Isotropie solids:

(a) Derive Eqs. (12.31), (12.32), and (12.33).

(b) Derive Eq. (12.38).

Equation of

motion:

Derive Eq. (12.35).

Rotation of nenia

tics:

Derive Eqs. (12.56) and (12.57).

6. Elastic constants:

(a) Express the bulk modulus B of an isotropie material in terms of the Lamé

constants.

(b) Find Young's modulus Y, as defined in Figure 12.4, along [100] for a crystal

of cubic symmetry.

Waves in cubic crystals:

(a) Find the equation of motion Eq. (12.35) in a form appropriate for a cubic

crystal. One way to proceed is to begin with Eq. (12.24), evaluate δ^/δϋ

and then generalize the result to the other two components of

Ü.

(b) Assuming that u(r, t) =w e\p[ik -r

—

ϊωή, find the matrix equation for w.

evaluate them for undoped silicon.

(d) Now assume that k = ko[l 1 1]. Again find the two sound speeds, and

evaluate them for undoped silicon.

8. Surface waves: Consider a two-dimensional isotropie elastic solid stretching

from — oo to oo in the x direction, but from — oo to 0 in the y direction. Surface

waves, also called Rayleigh waves, can run along the free surface, vanishing

exponentially as they reach down into the bulk.

To find the dispersion relation for these surface waves, begin with the obser-

vation that any displacement field u can be decomposed into longitudinal and

transverse parts,

_ _,i _,

The decomposition can be defined by taking the Fourier

U — U + U . transform of «; the longitudinal part of u is the part par- ( 12.69)

allei to k, and the transverse part is whatever is left over.

In two dimensions, u

and u' derive from potentials φ

and φ' such that

S' = V^and«'=(-^,^Y (12.70)

V ov ox I

338

Chapter 12. Elasticity

(a) Assume that φ

= A e\p[ikx + g/y - ϊωή and φ' = B exp[ikx + g,y

—

ίωή.

Determine gi and g

. Use the transverse and longitudinal wave speeds Q and

c, to carry information about the elastic constants.

(b) The boundary conditions on the free surface at y = 0 are that σ

νν

= σ^ =

0. Find two homogeneous equations involving A and B by imposing these

boundary conditions.

CRIC,

and determine the Rayleigh wave speed

in terms of the longitudinal and transverse wave speeds c/ and c

. Verify

that when c, = \/3c,, c

= 0.9194c,.

9. Rubber: Consider a solid cylinder of rubber, of original length L and radius r.

What restoring force does it exert as L is stretched beyond its original length?

10.

Friedrichs transition:

Figure 12.6. A nematic crystal is held between two plates. In (A) the liquid crystal is in

equilibrium with no external

field,

while in (B) it is in equilibrium in the presence of

field

along t

Consider a nematic liquid crystal caught between two plates, as shown in

Figure 12.6. The director h points along y at x = 0 and x = L. Between the

plates is applied a uniform magnetic field of strength H, along z. When it

reaches a critical value H

, the magnetic field causes the director n to begin

twisting in the y-z plane, as shown in Figure 12.6(B).

(a) Assume that the magnetic field enters the free energy of the liquid crystals

-XH

J dr(H-fi)

. (12.71)

Show that the complete free energy takes the form

?KAn^-n,^-\

df(H-iìY.

(12.72)

(b) Write h = (0, cos

Ö,

sin

Θ).

Find the Euler-Lagrange equation corresponding

to Eq. (12.72).

for d9/dx.

References

339

(d) Assume that for the critical magnetic field H

, the angle

is always very

close to zero. Making use of this approximation, solve explicitly for

(up to

an overall constant multiplier), and calculate the critical field H

11.

Mohr-Coulomb Failure Criterion Suppose one has a vacuum-packed bag

of coffee. Calculate the heaviest person that can stand on it without it giving

way. Neglect the strength of the packaging and treat the coffee is a granular

medium under atmospheric pressure. Assume that the top of the coffee bag

has an area of 200 cm

and that the coefficient of friction of coffee beans is

0.2.

References

J. Bergman and D. Stroud (1992), Physical properties of macroscopically inhomogeneous media,

Solid State Physics: Advances in Research and Applications, 46, 147-269.

M. Born (1914), On the space lattice theory of diamond, Annalen der

Physik,

44, 605-642. In Ger-

man.

G. S. Brady and H. R. Clauser (1991), Materials Handbook, McGraw-Hill, New York.

M. Chaikin and T. C. Lubensky (1995), Principles of Condensed Matter Physics, Cambridge

University Press, Cambridge.

S. Chandrasekhar (1992), Liquid Crystals, 2nd ed., Cambridge University Press, Cambridge.

P.-G. de Gennes and J. Prost (1993), The Physics of Liquid Crystals, 2nd ed., Clarendon Press,

Oxford.

I. S. Grigonev and E. Z. Meilkhov, eds. (1997), Handbook of Physical Quantities, CRC Press, Boca

Raton, FL.

H. M. Jaeger, S. R. Nagel, and R. P. Behringer (1996a), Granular solids, liquids and gases, Reviews

of Modern Physics, 68, 1259-1273.

H. M. Jaeger, S. R. Nagel, and R. P. Behringer (1996b), The physics of granular materials, Physics

Today,

49(4), 32-38.

L. D. Landau and E. M. Lifshitz (1970), Theory of Elasticity, 2nd ed., Pergamon Press, Oxford.

H. Landolt and R. Börnstein (New Series), Numerical Data and Functional Relationships in Science

and

Technology,

New Series, Group III, Springer-Verlag, Berlin.

M. Mooney (1940), A theory of large elastic deformation, Journal of Applied Physics, 11, 582-92

I. S. Sokolnikoff (1956), Mathematical Theory of Elasticity, 2nd ed., McGraw-Hill, New York.

S. Timoshenko and J. N. Goodier (1970), Theory of Elasticity, 3rd ed., McGraw-Hill, New York.

L. R. G. Treloar (1975), The Physics of Rubber Elasticity, 3rd ed., Oxford University Press, London.

X. Xing, P. M. Goldbart, and L. Radzihovsky (2007), Thermal fluctuations and rubber elasticity,

Physical Review Letters, 98, 075 502.

13.

Phonons

13.1 Introduction

Continuum elasticity is the only theory one needs to find stresses in skyscrap-

ers.

However, this description of the motion of solids fails in a qualitative manner

as soon as one considers deformations whose wavelength is comparable to inter-

atomic distances. When deformations are small, they must also be fast: taking

the speed of sound to be 10

m/s, one has vibrations of 10~

m at frequencies

of 10

Hz. No ordinary motor can generate vibrations this fast. However, they

provide the microscopic underpinning of phenomena ranging from specific heat to

electrical resistance.

The quantitative theory of microscopic deformations is by far most advanced in

crystalline lattices. The study of phonons is the study of how to catalog and name

the vibrating modes, how to calculate their frequencies, and of how they interact

with mechanical, electromagnetic, and other forces. Phonons are traveling waves

in crystals, and many of the techniques used to study electron waves in Chapter 7

carry over immediately to phonons as well.

The first work on lattice dynamics, by Born and van Karman preceded the

experimental proof by Laue, Friedrich, and Knipping that solids were crystalline

lattices. Born says:

The first paper by Karman and myself was published before Laue's dis-

covery. We regarded the existence of lattices as evident not only because

we knew the group theory of lattices as given by Schoenflies and Fedorov

which explained the geometrical features of crystals, but also because a

short time before Erwin Madelung in Göttingen had derived the first dy-

namical inference from lattice theory, a relation between the infra-red

vibration frequency of a crystal and its elastic properties.... Von Laue's

paper on X-ray diffraction which gave direct evidence of

the

lattice struc-

ture appeared between our first and second paper. Now it is remarkable

that in our second paper there is also no reference to von Laue. I can

explain this only by assuming that the concept of the lattice seemed to

us so well established that we regarded von Laue's work as a welcome

confirmation but not as a new and exciting discovery which it really was.

—Born (1965), p. 2

341

Condensed Matter

Physics,

Second Edition

by Michael P. Marder

342

Chapter 13. Phonons

Figure 13.1. A one-dimensional chain of ions connected by springs is the setting for the

simplest discussion of phonons.

13.2 Vibrations of a Classical Lattice

13.2.1 Classical Vibrations in One Dimension

Phonons are easiest to understand in one dimension. Consider a one dimensional

chain of ions of mass M sitting at equilibrium distance a connected by springs

of constant X, as shown in Figure 13.1. The calculations simplify as much as

possible when one adopts periodic boundary conditions, since no ions then need to

be treated differently from any others. There are N ions, and the total length of the

chain is L = Na.

Denote by u

the deviation of ion /

[0. .

N—

from its equilibrium location,

with periodic boundary conditions implying that

= u°. (13.1)

Then

Mü

= X(u

l+x

-u

)+X(u

^ -u

). (13.2)

This system of equations is solved by plane waves. Let

It is purely a matter of convention which constants one

.,/ __ ikla—iuit takes to multiply / in the exponent. One could choose Π 3 3Ϊ

instead Inikl/N or ikl. Different choices would amount ^ ' '

to a redefinition of k.

The periodic boundary condition (13.1) requires

kNa = 2Kn^k = 2wn/{Na),n<E [0 . . .N-l]. (13.4)

Substituting Eq. (13.3) into Eq. (13.2) gives

-Muj

ikla

X{e

ika

-\)+X{e-

ika

-\)

ikia-iu;

t (135)

-Miü

= X{2co<i{ika)-2)^uj = 2\l —

sin(*a/2)|; (13.6)