Marder M.P. Condensed Matter Physics

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Plasticity 423

E-,

log[u*i

N/k

Figure 15.4. Storage (Re[G]) and loss (Im[G]) moduli of polystyrene in Θ solvents com-

pared with the results of the Rouse calculation in Eq. (15.50). A Θ solvent is one specially

chosen so that the coefficient B in Eq. (5.74) vanishes. The viscosity η

is the viscosity

of the solvent without any polymer. N is the number of monomers per polymer. The

theoretical curves were obtained by evaluating Eq. (15.50) and multiplying G and ω by

arbitrary constants to obtain a best fit. Better fits are obtained by including hydrodynamic

interactions. [Source: Ferry (1980), p. 197.]

It is possible to use Eq. (15.49) to define a complex viscosity η(ω), but it is more

conventional to define the complex

shear

modulus G*(u) = —ίωη(ω) so that

0{ω)

N-]

i . ■ \

k=\

ωί +

ω^

(15.50)

The real part of G is called the storage modulus, and the imaginary part is called

the loss modulus. Measurements of dilute mixtures of polystyrene compared with

Eq. (15.50) appear in Figure 15.4.

15.4 Plasticity

It is commonplace to think of water flowing, but many metals, however, can also

flow. The boundary between those that can and those that cannot is not easy to

define precisely. For example, even rocks that ordinarily seem brittle can begin to

flow when placed under sufficiently great confining pressure.

The flow of ductile metals resembles in some respects the flow of liquids, but

there are also some important differences. A first difference is that metals do not

begin to flow noticeably until a critical stress level, the yield stress is approached.

424

Chapter 15. Fluid Mechanics

The yield stress corresponds to the point where the force per length / on dislo-

cations reaches the critical value needed to make them mobile. Even at very low

stresses, flow of metals can be observed, but it is extremely slow and it results from

exponentially rare thermal fluctuations that permit rearrangement of small portions

of the metal. As a matter of practice, a definite stress at which a metal begins to

display irreversible deformation is a useful quantity to define.

The possibility of plastic flow after the yield stress is exceeded is explained in

principle by dislocation motion. Knowledge of dislocation mobility has some pre-

dictive power for describing flow. Metals in which dislocations are immobile tend

to be brittle; those in which dislocations are mobile along one or two directions can

be deformed in those directions but still are not fully plastic, while those with five

or more independent slip planes will permit flow in any directions, and if assembled

into a polycrystalline aggregate will deform in a fully plastic fashion. Dislocations

provide only one type of mobile defect permitting flow. Some ceramics can flow

at high temperatures through the motion of vacancies, while amorphous materials

can flow through rearrangement of local groups of atoms.

The theory of plasticity is a phenomenological account of deformation that

does not depend upon the microscopic mechanism that makes it possible. Its con-

struction is an interesting combination of guesswork based upon symmetry princi-

ples,

a desire for simplicity, and constraints imposed by experimental observations.

The most common form of the theory is heavily influenced by the characteristic be-

havior of structural steels, which provided the greatest impetus for its creation, and

the theory might have a different form if aluminum or nickel alloys had an equal

practical importance.

Suppose one has an isotropie solid capable of plastic deformation; a practical

definition would be that the solid is capable of being drawn into wire. Whether the

solid begins to flow at any given location depends upon the state of stress there,

described by the stress tensor σ

β. The tensor σ is symmetric, so there is some

basis in which it is diagonal, with diagonal elements σ\, σ^, and σ^. The solid is

assumed to be isotropie, it cannot matter how these three elements are numbered,

and the onset of yield must be determined by some symmetrical function

£F

of σ\,

CT2, and

στ,.

Demanding that

be symmetrical is equivalent to requiring it to be a

function of symmetrical combinations of σ\ . . . σ^. There are three conventional

combinations of the elements of the stress tensor, the stress

invariants,

similar to

the strain invariants of

Eq.

(12.16), defined by the eigenvalue equation

det|a - el\ = -e

+ e

/, + e/

+ h = 0

(15.51

with

Ιι=Σσ

αα

(15.51b)

Ι2

= Σ {

αβσ

β-σ

αα

σββ} (15.51c)

αβ

=det|cr|. (15.51d)

Plasticity 425

Bridgeman (1949) found that a uniform hydrostatic pressure or tension, so long

as it is much less than Young's modulus, does not much affect the yield properties

of metals. Letting

σ = -^σ

αα

(15.52)

be the hydrostatic component of stress, it follows that yielding depends only upon

the deviatone stress

β = σ

β-σδ

β. (15.53)

Three invariants analogous to those defined by Eq. (15.51) can be defined in terms

The first of these invariants must vanish, while the others are

h =

-z

SaßSaß

(15.54a)

aß

= det|s|. (15.54b)

The requirement that yielding of an isotropie material be independent of hydro-

static stress can therefore be expressed by having flow begin when

3"(72,

J3)

reaches

a critical value. The Mises yielding condition says that flow begins when

— ft. K is some constant that needs to be tabulated (15.55)

for each material. Mises supposed that J3 is

irrelevant.

The characteristic features of metals that particularly distinguish them from flu-

ids are, first, that flow is negligible below this threshhold and, second, that the

threshhold κ itself evolves as the metal deforms, a bit like egg whites stiffening

up as they are beaten, but much more rapid. The tendency for κ to increase is

called work hardening, and implies that stresses on a metal will continually have

to increase if it is to continue to flow.

The theory must now grapple with the question of how precisely a solid de-

forms once the yield stress has been exceeded. The solid always has both reversible

and irreversible components to its deformation, because after removal of stress the

solid will snap back to some extent and may continue relaxing into a final config-

uration over long times. A simplest account of the process keeps only two time

scales, an instantaneous elastic relaxation, and a permanent plastic deformation.

These are distinguished by stating that the strain tensor e

ß is the sum of two

pieces: eig, the elastic strain tensor, and

the plastic strain tensor. The elastic

part of

the

strain is defined by Eq. (12.39); when the stress is given, the elastic strain

is also instantly known. It would be misleading, however, to say that when stress

is relieved the elastic strain instantly vanishes, because after all external stresses

on an object have been released, it may have deformed in such a way that internal

stresses are still present.

It remains to decide how the plastic strain evolves, in a final flurry of guesses.

Guess that flow in any direction is proportional to the deviatoric stress along that

426

Chapter 15. Fluid Mechanics

direction;

„ = \

W-^2 — li\S

ß II V Jl — ^ > U

s some

unknown

constant;

the point

of this

y Q Otherwise. equation is to make an hypothesis about the

tensor

structure of the flow and to make sure

vanishes when yield conditions are not met.

(15.56)

Not only does Eq. (15.56) have the virtue of simplicity, it also predicts Σ

= 0,

which means that volume is preserved by irreversible flow, in accord with experi-

ment.

It is not easy to decide how precisely the increase in flow stress κ should depend

upon the flow history. A common guess makes κ a function of the total plastic

work,

= / dt' y^ è

β. Integral over time. Writing energy as strain (15.57)

^—'

P times stress comes from Eq. (12.22).

aß

Suppose that a plastic body at rest is subject to some small increment in stress

άσ

β. According to Eq. (15.56), the body begins to flow. After a time, it hardens

up,

and flow ceases, resulting in a small increment of plastic strain deL:

aß

Cds

ß. C is a constant of

proportionality.

(15.58)

How should the proportionality constant C be chosen so that flow has stopped when

just the right amount of hardening has occured? The plastic work involved in the

small amount of flow is

= C ^ a

ßds

ß Combine a small change in Eq. (15.57) with (15.59)

^ ap ap

Eq(1558)

aß

C^2

aßds

aß

Use Eq.

(

15.53).

( 15.60)

aß

CdJ

Use Eq. (15.54a). (15.61)

so the change in κ is

= K'C dJ2- ^

prime on κ means to take a derivative (15.62)

with

respect to W.

The only way for

Eqs.

(15.62) and the flow condition (15.55) to be consistent is for

the amount of flow C to be

C=^T= (15.63)

a = ——. Actually, the plastic strain only changes if

the

( 15.64)

n n

~ change in deviatoric stress is such as to '

increase. Otherwise, de

„ vanishes.

aß

For applications of these equations, and for many solutions in special cases,

see Hill (1950) or Lubliner (1990).

Superfluid

427

15.5 Superfluid

Failure of Helium to Solidify. Helium stands out among the elements because

at atmospheric pressure it remains liquid down to a temperature of absolute zero.

A tentative explanation for this perplexing fact, simple but not convincing, follows

from data of Table 11.3. Were helium like the other noble gases, then the van der

Waals attraction would cause it to crystallize in an fee structure whose spacing is

given by Eq. (11.10) as 2.8 Â. The zero-point energy of a helium atom confined

to a region this size is approximately h

/(Irrido), which, for

He, works out to be

2-10

-3

eV. According to Eq. (11.11), the cohesive energy per particle in an fee

structure would be around 7-10

-3

eV. Although from this rough argument crystal-

lizing in fee appears to be favorable, the substance in actuality finds the penalty

of zero-point energy too costly and chooses to remain liquid in the ground state,

rather than solid.

The ground-state liquid has, however, bizarre properties matched by no other

substance on Earth. Poured into a container, it runs up the sides and flows down

over the edges. It flows through tubes as small as 500 Â in diameter without

measurable resistance, while in other respects displaying ordinary liquid viscos-

ity. These facts deserve a more systematic presentation.

Some Experimental Phenomena in

He. Liquid

He undergoes a phase transi-

tion at 2.186 K and atmospheric pressure, called the λ pointby Keesom, its discov-

erer, because of the appearance of the specific heat curve. Below this temperature,

helium begins to display the strange phenomena that lead it to be known as a su-

perfluid, and thought of as the new phase He II. Out of the dozens of ingenious

experiments designed to probe properties of helium II, here are two that are partic-

ularly revealing.

Figure 15.5. Narrowly spaced disks undergo torsional oscillations in

bath of

helium

II to

investigate its viscosity.

The peculiar mechanical properties of superfluid helium are best indicated by

an experiment of Andronikashvili (1948), illustrated in Figure 15.5. A collection

of 100 mica disks of around 4 cm in diameter and spaced by 0.2 mm is lowered

into a bath of superfluid helium, suspended upon a fiber. The experimental quantity

428

Chapter 15. Fluid Mechanics

measured is the rotational frequency of the disks, which is given by

■

/ or Here % is the spring constant of the fiber in

torsion,

/o is the moment of

inertia

the

disks,

4- J

d IF is the effective moment of inertia of

the

fluid dragged along with the

disks.

(15.65)

Calculating the effective moment of inertia of the fluid If is a complicated problem

in hydrodynamics, but to first approximation the result is rather simple; a classical

fluid should act like a solid mass dragged along in the space between the disks, but

not moving outside them. In this approximation, the moment of inertia due to the

fluid is simply proportional to its density and independent of

its

viscosity. When the

experiment was performed, the resonant oscillatory frequency of the disks began

to rise as the temperature decreased below the λ point, rising to

Λ/OC/IQ when the

temperature fell to 1 K. It was as if the density of helium began to drop at the λ

temperature and fell nearly to zero by

Figure 15.6. A thermally isolated container

of helium at temperature

and pressure Pi

is placed inside a second with temperature

and pressure

the only connection being a

micron-scale tube.

A second experiment displays the unusual thermomechanical features of the

superfluid and is illustrated in Figure 15.6. Into a bath of helium at temperature

Ύ2

is placed a thermally isolated chamber that communicates with the outside bath

only because of a small opening, on the order of a micron in diameter. A resistive

coil allows heat to be delivered to the inner chamber, and thermometers measure

temperature both inside and outside of

it.

When heat is supplied to the inner cham-

ber, the pressure Pi rises in direct proportion to the change in temperature 7\. This

fact is not at first surprising, because fluids typically do expand when heated. How-

ever, the phenomenon has nothing to do with ordinary thermal expansion, because

it ceases if

the

microscopic connection between the inner and outer basins is closed.

The inner and outer basins are in thermodynamic equilibrium, although they are at

different temperatures and pressures. A striking example of

this

phenomenon is the

fountain effect. If a thin tube whose base is packed with fine powder is placed into

a basin of superfluid helium, then a slight increase in temperature of the helium

bath causes superfluid to race up the tube and fly out the top, up to a height of tens

of centimeters.

These experimental observations lie behind a phenomenological picture of

called the two-fluid model. In this view, He II consists of two intertwined fluids.

One of them, the normal fluid, has ordinary fluid properties such as viscosity. The

other, the superfluid component, has no viscosity and carries no entropy. There

Superfluid

429

completely accepted microscopic theory

underlie the two-fluid model,

but

clearly helium does

not

actually consist

of two

different fluids.

The

superfluid

component corresponds to

low energy state with quantum-mechanical coherence,

while the normal fluid corresponds

thermal excitations traveling about on top of

the low energy state.

Andronikashvili's experiment

easy

explain from this point

view.

temperature drops below

the λ

point, more

and

of the

fluid converts

to the

superfluid component. Because this component lacks viscosity, the oscillating mica

disks

do not

perturb

it, and

only

the

normal component

of the

fluid

dragged

along with them.

fact,

the

results

of the

experiment

may be

interpreted

as a

measurement,

even

definition,

the density

the superfluid component.

The thermomechanical experiment shows that

two

regions

helium

may be

in contact, particles flowing from one

another,

but

without the two regions shar-

ing

the

same temperature

pressure.

Let G\(T\, P\, N\)be

the Gibbs free energy

of the helium bath, with G

P2,

N2)

the

Gibbs free energy

the internal con-

tainer. Because the two regions

fluid

can

freely exchange particles, equilibrium

demands that

_dGx{N

{

) +

(N-N

)

{

dG\ <9G

andN

(15 66)

«Λ7 «ΛΓ '

μΐ(

1,Ρΐ)=μ2(Τ

Pi)- μι

and

are

the chemical (15.67)

θΝ\

0Ν2

potentials

the two regions.

If entropy could flow between the two fluid regions, then equilibrium would require

the condition

ö£i(Si,Vi) d£

)

i_^_n

\j_ _ £ν_^ ii _^ ^ —T2:

Not true

for

superfluid helium!

dS\

8S2

(15.68)

thus

the

fact that two fluids

can be in

equilibrium

different temperatures while

exchanging particles

is a

demonstration that

the

particles moving between them

somehow contrive

carry

entropy.

According

to Eq.

(15.67), equality

chemical potential

is the

basic require-

ment

for

superfluid equilibrium. When temperature and pressure differ only slightly

in two regions

superfluid, then

can be expanded to first order in the small

dif-

ferences

AT =

Ύ2 — T]

and AP =

—

P\. As

result

T+

p=0

(15

69)

=>· s AT

= —AP

=>-

= OS.

s tne

entropy

per

unit mass,

and p is the

(15.70)

mass density

region 2; obtain these expres-

sions from

μ =

dG/dN.

The entropy

helium can

measured directly

integrating the specific heat

from zero,

so as to

verify (15.70).

430

Chapter 15. Fluid Mechanics

15.5.1 Two-Fluid Hydrodynamics

The superfluid component of helium differs from an ordinary fluid because of the

particular thermodynamic force that makes it move. The thermomechanical effect

demonstrates that superfluid flows so as to equalize the chemical potential. Because

the chemical potential describes the change in energy of

small portion of fluid due

to addition of a single particle, the force per particle acting upon the superfluid is

—V/x, and the velocity v

of a region of superfluid is described by

_j_

(y . \7)v

= . "* is the mass of a helium atom. (15.71)

dt m

As a consequence of the Gibbs-Duhem relation

V S

du=—dP

dT See, for example, Landau and Lifshitz

(

1980) (15 72)

N N '

72.

one may write alternatively

- VP

~ ~Ί~ (fis ' V^)V

\~sVT.

P = mass density, and s = S/Nm is entropy (15.73)

dt p per unit mass.

In accord with Eq. (15.70), superfluid can be in equilibrium in the presence of

simultaneous pressure and temperature gradients. Equation (15.73) predicts that

without pressure gradients, thermal gradients can make a superfluid flow as easily

as pressure differences move an ordinary fluid.

The equation describing motion both of superfluid and normal components of

helium is

(ν

-ν)ν

}+ρ

^ + (ν

-ν)υ

} = -νΡ + ην

. (15.74)

At first glance, Eq. (15.74) appears obvious, but it is not. The relation between

pressure and acceleration is no longer as clear as for a single-component fluid,

because there is no single reference frame in which all the fluid is stationary and

in which one can measure the pressure. An alternative to Eq. (15.74) might be an

equation in which the left-hand side contains only the total density p

—

+ p

and

the mean velocity (p

)/p.

A partial justification for Eq. (15.74) is that it

produces the correct acceleration of superfluid or normal components if either the

normal or superfluid density vanishes, and in addition it is an equation for which the

two fluid components are independent; motion of the superfluid does not produce

convective transport of the normal fluid, and vice versa. Such arguments cannot,

however, rule out the presence of a term such as p

sVT, because p

vanishes in the

normal fluid, while s vanishes in the superfluid. Landau's unambiguous derivation

of Eq. (15.74) is described in the final chapter of Landau and Lifshitz (1987).

Superfluid

431

15.5.2 Second Sound

A remarkable consequence of Eqs. (15.73) and (15.74)

the existence of prop-

agating waves

helium moving

at a

speed

sound, but involving temperature

rather than pressure variations. The origin

these waves can be deduced from

Eq. (15.73). Any gradient in temperature spurs

flow

of superfluid toward the hot

region, trying to cool

down. This motion can occur without noticeable change

in the density

the fluid, because an equal amount of normal fluid flows in the

opposite direction. The region away from which the superfluid flows heats up, but

because the superfluid has already acquired some inertia,

needs some time

reverse course. The consequence is a wavelike oscillation of temperature in which

normal fluid and superfluid are out of phase.

Speeds of First and Second Sound in Liquid

He. Sound is a disturbance trav-

eling at long wavelengths and small amplitudes. Therefore in analyzing the con-

sequences of Eqs. (15.73) and (15.74)

is enough to retain only first order in all

small quantities, and to ignore the dissipative term proportional to V

, because it is

negligible for long-wavelength excitations. In addition to Eqs. (15.73) and (15.74),

two conservation laws are needed to describe sound completely. Conservation of

mass requires

-^ +

V-(p

)=0

(15.75)

while conservation of entropy requires

dps

—

\/■

psv„ Because the viscous term in Eq. (15.74) is ne-

(15.76)

glected, entropy production can be neglected.

Remember that

was defined

be entropy

per unit mass and that it is carried only by the

normal fluid.

Rewriting Eqs. (15.73) and (15.74) to linear order gives

(15.77)

Dropping the contribution from viscosity. (15.78)

Differentiating Eq. (15.75) with respect to time and using Eq. (15.78) to describe

the time evolution of

the

mass current gives the customary relation between density

and pressure

—r-

VP. One is free, forexample, to replace dp„v„/dt

(15.79)

by p„dv„/dt because

is a

small quantity,

and multiplying by dp„/di would produce

second-order quantity, which may be neglected.

A second relation, between entropy and temperature, is obtained by writing

^--Ι

ΐ5

8ο

p at p at

*-d7

-d7

+ sVT

-VP.

432

Chapter 15. Fluid Mechanics

—

1-.

s -

V ·

psv

+ -V ·

V„+

)

Use Eqs. (15.75) and (15.76). (15.81)

—*

= ——V

· (v

—

Pulling p„,

and

outside derivatives be- (15.82)

n cause they multiply

or v

which are

al-

ready small, and combining terms.

Solving Eqs. (15.77) and (15.78) for

8{v

v„)/Bt gives

-{v

-v

)=s^-VT

(15.83)

=>.

—-

£l

Combining Eqs. (15.82) and (15.83), and bring- (15.84)

ing zeroth-order quantities outside the derivative.

To obtain sound speeds from these relations, one needs

observe that

and

are related to

and

by equations

state. The small variations

p and

can be

rewritten

terms

the deviations T

(1)

and P

(1)

temperature and pressure from

equilibrium as

Bn (9

(1)

Cross terms such as 9

p/dP9i

if. I _

1SL I _ =

(1)

9P/9i are second order and

(15 85)

ÖP

BT Bt

negligible. Comes from

Eq. (15.79).

P^ r)ç B

T^ n

|ρ"

\τ ^ + Ί^ \Ρ

~i^=S

T^.

Comes from Eq. (15.84). (15.86)

ÔP dr oT

dr Pn

Suppose now that both

and

(1)

have

the

form

traveling waves

wave

number

and frequency ω

ck. Then

+ % \

=c~

(15.87a)

I PC) , ^1 I

T(i)_

-2„2ft

BT p

Because the determinant

Eq. (15.87) vanishes,

ρΜ + Ρ\

(,)

= c-V^7

(,)

. (15.87b)

\ / \ ds dp

— 1

Ί I _1 _ —

ΘΡ

Bs_

_ _ _

Ì \ BP

J BP

—

(15.88)

Use dp/dT\r/dp/dP\

-dP/8T\

and the standard (15.89)

Cp relation for the difference between constant volume and

constant pressure specific heats Cp and Cv

■

The difference between Cp and Cv is very small for

He. (15.90)

The two values of

solving Eq. (15.88) are therefore

\dP_

and

\\—_\T

(

)

= i / —

Here Cp is the specific heat per unit mass.

(

15.92)

Cp p