Lectures on Theoretical Physics Continuum Physics · 2011-03-15 · examples. This postgraduate level course assumes a basic acquaintance with classical mechanics, electrodynamics,

Lectures on Theoretical Physics

Continuum Physics

Peter Hertel

Universitat Osnabruck

Ordinary matter consists of material points. On the one hand, each materialpoint is so small that it has no further structure. On the other hand, eachmaterial point is made up of so many particles that the laws of thermody-namics for infinitely large systems apply. It makes sense to speak of content,flow, and production of certain quantities like mass, charge, momentum, en-ergy and entropy. The corresponding balance equations are valid irrespectiveof the special material under consideration. They include precise versionsof the first and second law of thermodynamics.Additional constitutive equations define entire branches of physics like fluiddynamics, elastomechanics, heat transport, diffusion, reaction kinetics andso forth. The Navier-Stokes equation for Newtonian fluids, Hooke’s law forelastic media, Fourier’s law of heat and Ohm’s law of charge conduction areexamples.This postgraduate level course assumes a basic acquaintance with classicalmechanics, electrodynamics, thermodynamics and vector analysis, but notquantum mechanics. Emphasis is laid on deriving equations, not on solvingthem. However, a few examples are worked out to familiarize the readerwith applications.

January 1, 2008

CONTENTS ii

Contents

1 Balance Equations 11.1 Ordinary matter . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Content, flow, and production . . . . . . . . . . . . . . . . . . 11.3 Local formulation . . . . . . . . . . . . . . . . . . . . . . . . . 2

2 Mass and Charge 42.1 Number of particles . . . . . . . . . . . . . . . . . . . . . . . 42.2 Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.3 Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

3 Convection and Conduction 73.1 Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2 Transformation properties of currents . . . . . . . . . . . . . 83.3 Substantial time derivative . . . . . . . . . . . . . . . . . . . 9

4 Momentum 114.1 Stress and external forces . . . . . . . . . . . . . . . . . . . . 114.2 Angular momentum . . . . . . . . . . . . . . . . . . . . . . . 12

5 Energy and the First Law 135.1 Kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . . 135.2 Potential energy . . . . . . . . . . . . . . . . . . . . . . . . . 145.3 Internal energy . . . . . . . . . . . . . . . . . . . . . . . . . . 14

6 Entropy and the Second Law 166.1 Local equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . 166.2 The second law . . . . . . . . . . . . . . . . . . . . . . . . . . 17

7 Fluid Dynamics 197.1 Hydrodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . 197.2 Aerodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . 207.3 Example: Hagen-Poiseuille equation . . . . . . . . . . . . . . 217.4 Reynolds’ number . . . . . . . . . . . . . . . . . . . . . . . . 23

8 Elastomechanics 248.1 Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248.2 Hooke’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258.3 Example: Torsion . . . . . . . . . . . . . . . . . . . . . . . . . 278.4 Elastic waves . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

CONTENTS iii

9 Heat Transport 309.1 Heat equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 309.2 Example: Soil temperature . . . . . . . . . . . . . . . . . . . 31

10 Diffusion 3210.1 Diffusion equation . . . . . . . . . . . . . . . . . . . . . . . . 3210.2 Brownian movement . . . . . . . . . . . . . . . . . . . . . . . 34

11 Charge Transport 3611.1 Maxwell’s equations . . . . . . . . . . . . . . . . . . . . . . . 3611.2 Ohm’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3711.3 Electro-chemical potential . . . . . . . . . . . . . . . . . . . . 38

12 Onsager’s Relation 3912.1 Forces and fluxes . . . . . . . . . . . . . . . . . . . . . . . . . 3912.2 Onsager’s theorem . . . . . . . . . . . . . . . . . . . . . . . . 3912.3 An example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

13 Thermoelectric Effects 4213.1 Charge and heat transport . . . . . . . . . . . . . . . . . . . . 4213.2 Thomson effect . . . . . . . . . . . . . . . . . . . . . . . . . . 4313.3 Seebeck and Peltier effect . . . . . . . . . . . . . . . . . . . . 44

1 BALANCE EQUATIONS 1

1 Balance Equations

In this series of lectures we are concerned with ordinary matter. In the firstpart we study properties like mass, charge, momentum, energy and entropywhich add if two subsystems are joined. Such properties can also flow, andthey may appear or vanish. We shall call them quantities in contrast toqualities like temperature or pressure.

1.1 Ordinary matter

Although we know that matter is made up of atoms or molecules, there areso many of them that averaging is allowed. The notion of a ’material point’is of central importance. A material point is a region of space so small thatit appears to be a point to the engineer. The same region is so large, as seenby a physicist, that the number of particles in it approaches infinity suchthat the laws of thermodynamics apply.A cubic millimeter of water is a point to someone who tries to optimize aship’s body. The same cubic millimeter contains 3 × 1019 H2O molecules,a truly huge number. Even after scaling down to a cubic micrometer thefiction of a material point makes sense.Fluctuations, e.g. of the number of particles in a given region, will playno role. As a consequence, the Boltzmann constant1 kB will never appearagain.Likewise, the quantization of energy and angular momentum—a pronouncedeffect for single particles—is irrelevant for material points. Therefore, Planck’sconstant2 ~ is absent in the following text.We also assume that relative velocities are small as compared with the ve-locity of light c which does never show up again. After all, we deal withordinary matter.

1.2 Content, flow, and production

With these preliminaries in mind we will concentrate on the notion of aphysical quantity Y . The number N of molecules in a system may serve asan example.The quantity of Y at time t in a region R of space may be written as

Q(Y ; t,R) =∫x∈R

dV ρ(Y ; t,x) . (1)

A material point at x with volume dV contains a quantity dY which is1Ludwig Boltzmann, 1844-1906, Austrian physicist2Max Planck, 1857-1974, German physicist


proportional to dV . The proportionality factor ρ(Y ) is the Y -density, ascalar field ρ(Y ) = ρ(Y ; t,x), which depends on time t and location x.Let us now speak about an areaA. It is made up of area elements dA = n dAwhere n is a unit vector normal to the area. It points from the back to thefront side ofA. dA = |dA| is the size of the area element under consideration.The amount of Y flowing, per unit time, from the back to the front side ofA may be written as

I(Y ; t,A) =∫x∈A

dA · j(Y ; t,x) . (2)

Here j(Y ) = j(Y ; t,x), a vector field, denotes the current density of Y attime t at location x. The scalar product dA · j indicates that only thenormal component of the current density contributes to the net transitionof Y through the area element.The quantity Y may also be generated or annihilated. We denote by

Π (Y ; t,R) =∫x∈R

dV π(Y ; t,x) (3)

the amount of Y produced per unit time within the region R. It is the sum(integral) over contributions dΠ from material points with volume dV . Theproduction efficiency π(Y ) = π(Y ; t,x), a scalar field, describes how muchY is produced per unit time, at time t, per unit volume, at location x.The amount Q(Y ; t,R) of Y within a region R will change if there is anoutflow I(Y ; t, ∂R) through the surface ∂R of the region R. It may alsochange because Y is produced. Therefore the balance equation for Y maybe stated as

d

dtQ(Y ; t,R) = −I(Y ; t, ∂R) + Π (Y ; t,R) . (4)

I(Y ) is the outflow, −I(Y ) the inflow of Y because the surface ∂R of a regionR is oriented such that the normal vector points from inside to outside.

1.3 Local formulation

The global balance equation (4) may be rewritten into a local form. Weinvoke the Gauss3 theorem∫

x∈RdV ∂if(x) =

∫x∈∂R

dAif(x) , (5)

3Carl Friedrich Gauß, 1777-1855, German mathematician and physicist


where ∂i denotes partial differentiation with respect to the i-th argument.∂R is the surface of the region R. The result is

∂tρ(Y ) + ∂iji(Y ) = π(Y ) . (6)

∂t denotes the partial derivative with respect to time. Note that we useEinstein’s4 summation convention: if, within a term, the same index appearstwice, the sum over it is silently assumed. Thus, ∂iji is the divergence ofthe current density field ji = ji(t,x).Whereas (4) holds for any regionR, its local equivalent is true at any locationx. Both are true at all times.In the following sections we shall discuss balance equations for the followingquantities Y :

• number Na of particles of species a,

• mass M ,

• electric charge Q e,

• the three components Pk of linear momentum,

• the three components Lk of angular momentum,

• kinetic energy E k, potential energy E p, and internal energy U ,

• entropy S.

4Albert Einstein, 1879-1955, German physicist

2 MASS AND CHARGE 4

2 Mass and Charge

Although we want to investigate continuum properties, we should not forgetthat matter is an aggregation of particles. Protons and neutrons form nucleiwhich are dressed by electrons to become atoms or ions, which in turn formmolecules or solids. It is import to bear in mind that identical particles areprincipally indistinguishable. Therefore, we may distinguish only betweenspecies of particles.

2.1 Number of particles

Let us label by a = 1, 2, . . . the species of particles in our system. Na is thenumber of particles of species a. We denote by

na = ρ(Na) resp. jai = ji(Na) (1)

the density and current density of a-particles, respectively. In the followingexample we will refer to H2O, H+, OH−, H2, and O2 which are enumeratedby a from 1 to 5.The number of particles of a certain kind is not conserved. Particles mayvanish and appear in chemical reactions.Let us label by r = 1, 2, . . . the possible chemical reactions. If one reactionof type r takes place, then νra particles of species a will be created orannihilated, depending on the sign.

νra particle a H2O H+ OH− H2 O2

reaction r (1) (2) (3) (4) (5)

H2O→H++OH− (1) -1 1 1 0 0

2H2O→2H2+O2 (2) -2 0 0 2 1

ν21 = −2 says that two H2O molecules will disappear in one chemical reac-tion 2H2O→2H2+O2. The stoichiometric coefficients νra, for a given reac-tion r, are integers without a common divisor.We denote by Γ r the number of chemical reactions of type r per unit volumeper unit time. With this we may write the following expression for theproduction term:

π(Na) =∑r

Γ rνra . (2)

(2) is the number of a-particles produced per unit volume per unit time. Thebalance equation for the number of particles of a given species a therefore

2 MASS AND CHARGE 5

reads

∂tna + ∂ij

ai =

∑r

Γ rνra . (3)

2.2 Mass

If particles move, they carry their properties with them, in particular massand electric charge. With ma the mass of a particle of species a we shouldwrite the following expression for the density of mass M :

ρ = ρ(M) =∑a

mana . (4)

Where there is no mass, there is no mass current. Consequently, the masscurrent density may be represented as the product of mass density and avelocity,

ji(M) = ρvi . (5)

The vector field v = v(t,x) describes the velocity of the local center of mass.Since∑

a

νrama = 0 (6)

holds for any chemical reaction, we conclude

π(M) =∑r

∑a

Γ rνrama = 0 . (7)

Mass cannot be produced or annihilated, it is conserved. This fact is ex-pressed by

∂tρ+ ∂iρvi = 0 . (8)

2.3 Charge

Chemical reactions also respect the law of charge conservation,∑a

νraqa = 0 , (9)

where qa is the charge of an a-particle. With

ρ e = ρ(Q e) =∑a

qana (10)

2 MASS AND CHARGE 6

as density of electric charge Q e,

j ei =

∑a

qajai (11)

as electric current density and

π(Q e) =∑r

∑a

Γ rνraqa = 0 (12)

we obtain

∂tρe + ∂ij

ei = 0 , (13)

the well-known continuity equation of electrodynamics.It is interesting to note that (13) is a consequence of Maxwell’s equations5

for the electromagnetic field E,B which read

ε0∇ ·E = ρ e , (14)

−ε0∂tE +1µ0

∇×B = j e , (15)

∇ ·B = 0 , (16)

∂tB + ∇×E = 0 . (17)

Adding the time derivative of (14) to the divergence of (15) results in (13)because the divergence of a curl vanishes.

5James Clerk Maxwell, 1831-1879, British physicist

3 CONVECTION AND CONDUCTION 7

3 Convection and Conduction

While matter flows it takes its properties along with it. Therefore there isalways a contribution ρ(Y ) vi to the current density ji(Y ). This contributionis termed ’convection’ because the quantity in question is simply conveyed.By definition, mass is transported by convection only, ji(M) = ρvi.

3.1 Conduction

However, there may be an additional contribution to the current density,

ji(Y ) = ρ(Y )vi + Ji(Y ) . (1)

This additional current density Ji(Y ) describes the conduction of a quan-tity Y .As said above, there is no conduction of mass.The conductive contributions to the particle currents are

Jai = Ji(Na) = jai − navi . (2)

There is no particle transport if there are no particles. We may thereforesplit off the particle density:

jai = navai . (3)

Obviously, vai is the average velocity of particles of species a. With it wemay write

Jai = na(vai − vi) . (4)

Particle conduction, or diffusion, takes place if the average velocity of acertain kind of particles va differs from the center of mass velocity v. Ifparticles of a certain species move relative to the local center of mass, wespeak of diffusion.The situation is slightly more complicated for the electric charge Q e. Thecharge density ρ e may vanish although there are charged particles. In ametal, the mobile electrons and the positive ions of the lattice compensateeach other, and the charge density ρ e vanishes. Nevertheless, there may bean electric current. The conductive contribution J e

i to the electric currentdensity may be expressed in terms of diffusion current densities:

J ei = j e

i − ρ evi =∑a

qaJai . (5)


3.2 Transformation properties of currents

As we have just explained, transport is more than convection. There areadditional mechanisms like diffusion of particles, electric conduction, me-chanical stress, or heat flow.The decomposition (1) of a current density into a convective and a conduc-tive contribution is useful as well from a formal point of view. We will shownow that Ji(Y ) transforms properly as a vector field, but not ji(Y ).Events (what happened where and when) are parameterized by a time coor-dinate t and three space coordinates x1, x2, x3 referring to an inertial systemΣ . The same event has to be parameterized by (t ′, x ′1, x

′2, x

′3) if another

inertial system Σ ′ is used. The relation is described by a Galilei6 transfor-mation:

t = τ + t ′ and xi = ai +Rijx′j + uit

′ with RijRkj = δik . (6)

A scalar field S, such as ρ(Y ), is transformed into S ′ according to

S(t,x) = S ′(t ′,x ′) . (7)

A vector field Vi transforms as

Vi(t,x) = RijV′j (t ′,x ′) . (8)

We conclude that the time derivative of a scalar field is not a scalar fieldbecause of

∂ ′tS′ = ∂tS + ui∂iS . (9)

Here and in the following text we drop the arguments; primed fields dependon primed coordinates, unprimed fields are to be evaluated at the corre-sponding unprimed coordinates.However, the divergence of a vector field is a scalar field:

∂ ′i V′i = Rji∂

′i Vj = RjiRki∂kVj = ∂iVi . (10)

Because of

dxidt

= Rijdx ′jdt ′

+ ui (11)

we conclude, for an arbitrary scalar field S,

Svi = Rij(Svj) ′ + Sui . (12)

6Galileo Galilei, 1564-1642, Italian physicist


The divergence of this expression transforms as

∂ ′i (Svi)′ = ∂i(Svi)− ui∂iS . (13)

(9) and (13) say that the sum

∂tρ(Y ) + ∂iρ(Y )vi (14)

transforms as a scalar field.We conclude:

• the density ρ(Y ) is a scalar field,

• the production term π(Y ) is a scalar field,

• the conductive current density Ji(Y ) is a vector field.

These transformation properties guarantee that balance equations

∂tρ(Y ) + ∂iρ(Y ) vi = −∂iJi(Y ) + π(Y ) (15)

hold true in all inertial systems.

3.3 Substantial time derivative

The partial time derivative ∂t measures the rate of change as observed bya spectator at rest. The substantial time derivative Dt measures the rateof change as observed by someone who flows together with matter. It isdefined by

Dtf(t,x) =f(t+ dt,x+ dtv)− f(t,x)

dt= ∂tf(t,x) + vi∂if(t,x) (16)

or

Dt = ∂t + vi∂i . (17)

With (17) we may rewrite the mass balance equation as

Dtρ+ ρ∂ivi = 0 (18)

or

ρDtρ−1 = ∂ivi . (19)


The left hand side is the relative rate of change of the specific7 volume asobserved by a co-moving spectator, the right hand side is the divergence ofthe velocity field.It is sometimes advisable to introduce the specific Y -quantity σ(Y ) whichis defined by

ρ(Y ) = ρσ(Y ) . (20)

With it we may write the general balance equation as

ρDtσ(Y ) = −∂iJi(Y ) + π(Y ) . (21)

This form is very suggestive from a physical point of view as will becomeevident in the following sections.

7per unit mass

4 MOMENTUM 11

4 Momentum

The momentum of a system is the sum of momenta of its constituents. Thethree components Pk of momentum are therefore quantities Y in the senseof the preceding sections.

4.1 Stress and external forces

The density of momentum Pk is

ρ(Pk) = ρvk (1)

whereby we have argued ’mass times velocity per unit volume’. Note thatthe specific momentum σ(Pk) is the same as the velocity vk.The momentum current density is

ji(Pk) = ρvkvi − Tki . (2)

The conductive contribution Ji(Pk) = −Tki transforms as an object withtwo indices, it is a tensor. Tki describes the stress acting on the material.dFk = TkidAi is the force exerted by the medium on the tiny area elementdA from its front side. Recall that the unit vector n in dA = n dA pointsfrom the back to the front size. The negative sign in front of the stresstensor inverts this convention.The amount of momentum produced per unit time is a force. The amountof momentum π(Pk) produced per unit time per unit volume therefore isthe external force acting on the unit volume which we denote by fk. Disre-garding magnetic effects, there are just two such forces: gravitational andelectrostatic.Denote by φ g the gravitational and by φ e the electric potential. The gravi-tational field strength −∂iφ g couples to mass while the electric field strengthEi = −∂iφ e couples to charge.The momentum balance equation therefore reads

∂tρvk + ∂i(ρvkvi − Tki) = fk (3)

with

fk = −ρ∂kφ g − ρ e∂kφe . (4)

An alternative form of (3) is

ρDtvk = ∂iTki + fk . (5)

The left hand side should be interpreted as ’mass times acceleration per unitvolume’. There are two causes of acceleration: long range external forces(fk) and momentum conduction (∂iTki) because of short range interactions.

4 MOMENTUM 12

4.2 Angular momentum

Let us now discuss angular momentum Lk = εkrsXrPs where Xr denotes theposition of a material point. εkrs is the totally antisymmetric Levi-Civita8

symbol. L = X × P is a possibly more familiar notation.The specific angular momentum is given by

σ(Lk) = εkrsxrvs . (6)

We shall likewise write

Ji(Lk) = −εkrsxrTsi (7)

and

π(Lk) = εkrsxrfs . (8)

The balance equation for angular momentum

ρDtσ(Lk) = −∂iJi(Lk) + π(Lk) (9)

holds provided

εkrs(ρvsDtxr − Tsi∂ixr) = 0 (10)

is met. We have made use of the rule Dtab = bDta+ aDtb which is true fora derivative operator. With ∂ixr = δir and Dtxr = vr we arrive at

εkrs(ρvsvr − Tsr) = 0 (11)

or

Tsr = Trs . (12)

To summarize: the balance equation for angular momentum demands thestress tensor to be symmetric.A real symmetric tensor can be diagonalized by an orthogonal coordinatetransformation. At each location we may introduce a local Cartesian coor-dinate system such that the stress tensor has the form

T =

T1 0 0

0 T2 0

0 0 T3

. (13)

Positive eigenvalues Tj mean tension while negative values indicate pressure.8Levi-Civita, 1873-1941, Italian mathematician

5 ENERGY AND THE FIRST LAW 13

5 Energy and the First Law

In this section we will derive balance equations for kinetic and for potentialenergy. In fact, they are consequences of the momentum balance equation.In order to guarantee energy conservation another form of energy has to beintroduced: internal energy. Its balance equation is a very precise formula-tion of the first law of thermodynamics.

5.1 Kinetic energy

Let us multiply the momentum balance equation

ρDtvk = ∂iTki + fk (1)

by vk. The result is

ρDt12vkvk = vk∂iTki + vkfk . (2)

Because of

vk∂iTki = ∂ivkTki − Tki∂ivk (3)

we have derived the balance equation for kinetic energy E k if we identify

ρ(E k) =12ρvkvk , (4)

Ji(E k) = −vkTki (5)

and

π(E k) = −Tki∂ivk + vkfk . (6)

Since the stress tensor is symmetric, the velocity gradient ∂ivk can be re-placed by the symmetrized expression

Gik =∂ivk + ∂kvi

2, (7)

and we may write

π(E k) = −TkiGik + vifi (8)

instead of (6).


5.2 Potential energy

Besides kinetic energy E k there is potential energy E p. Its density is givenby

ρ(E p) = ρφ g + ρ eφ e . (9)

Recall that φ g and φ e are the gravitational and the electrostatic potential,respectively. They have been introduced before when detailing the externalforce per unit volume:

fk = −ρ∂kφ g − ρ e∂kφe . (10)

For simplicity we shall assume that both potentials do not depend on time.The external forces do not vary with time, and the energy of our systemshould be conserved.We conclude

∂tρ(E p) = φ g∂tρ+ φ e∂tρe = −φ g∂iji(M)− φ e∂iji(Q e) . (11)

By rewriting

∂tρ(E p) = −∂i(φ gji(M) + φ eji(Q e)) + ji(M)∂iφ g + ji(Q e)∂iφ e (12)

we can read off the current density of potential energy:

ji(E p) = φ gji(M) + φ eji(Q e) , (13)

a plausible expression.The remaining contributions to (12) describe the production of potentialenergy. With ji(M) = ρvi and j e

i = ρ evi + J ei we arrive at9

π(E p) = −vifi − J ei Ei . (14)

5.3 Internal energy

In general, π(E k)+π(E p) does not vanish. The sum of kinetic and potentialenergy is not conserved. There is another energy contribution, internalenergy U , which is needed to guarantee energy conservation.We denote the specific internal energy by σ(U) = u. The internal energycurrent density is written as

ji(U) = ρuvi + J ui . (15)

9Recall that −∂iφe is the electric field strength Ei.


The conductive contribution J ui to the internal energy current density is

called the heat current density.The production of internal energy must be described by

π(U) = TkiGik + J ei Ei (16)

in order to warrant energy conservation.Any current density Ji(Y ) may be split into a reversible, or elastic part J ′i (Y )and an irreversible, or inelastic part J ′′i (Y ). They differ by their behaviourwith respect to time reversal. The reversible part transforms similar to theconvection term ρ(Y )vi while J ′′i (Y ) acquires an additional minus sign.Charge transport may serve as an example. The convection term ρ evichanges sign upon t → −t. So does J e

i′ = DtPi where Pi denotes the

polarization. Ohm’s law states J ei′′ = σEi. This contribution to the electric

current does not change sign upon t→ −t, it describes an irreversible effect.The following equation

ρDtu = T ′kiGik + J ei′Ei − ∂iJ u

i + T ′′kiGik + J ei′′Ei (17)

is a precise formulation of the first law of thermodynamics.There are five reasons why the internal energy changes, as observed by aco-moving spectator:

• deformation, T ′kiGik,

• polarization, J ei′Ei,

• inflow of heat, −∂iJ ui ,

• friction, T ′′kiGik,

• Joule’s law10, J ei′′Ei

Traditionally, the first two terms are classified as work while the remainingthree terms are considered to be heat. We prefer the distinction betweenreversible and irreversible: only the last three expressions are linked withentropy production. This will be the subject of the following section.

10James Prescott Joule, 1818-1889, British physicist

6 ENTROPY AND THE SECOND LAW 16

6 Entropy and the Second Law

We now want to elaborate on the notion of a material point. These smallpieces of matter are so tiny that they appear to be points (from a macro-scopic viewpoint). On the other hand, they contain such a huge number ofparticles that the laws of thermodynamics for infinitely large systems apply.It is well known that smaller systems reach their thermodynamic equilibriumfaster. We assume that material points are so small that they are alwaysinfinitely close to their equilibrium state. Thus, a continuum is locally inan equilibrium state, but the parameters describing the equilibrium maydepend on location and time.

6.1 Local equilibrium

Let us single out a particular material point of fixed mass M = ρV . In thecourse of time, its volume V may change because the mass density ρ mightchange.We will compare the equilibrium states of our material point at time t andat t+ dt. The change of internal energy is given by

dU = TdS + dW +∑a

µadNa . (1)

T is the temperature of the equilibrium state. The µa denote the chemicalpotentials of a-particles. dU , dS and dNa are changes of internal energy,entropy, and particle numbers, respectively. dW stands for the amount ofwork spent on the material point.Translating into the language of balance equations we may write

dU = MdtDtu = V dtρDtu . (2)

With the specific entropy s we may likewise write

dS = MdtDts = V dtρDts . (3)

The changes of number of particles are

dNa = MdtDtna

ρ= V dt(−∂iJai +

∑r

Γ rνra) . (4)

Recall that we denote by Jai the diffusion currents, Γ r stands for the numberof reactions of type r per unit time per unit volume, and the νra are stoi-chiometric coefficients (how many a-particles are produced in one reactionof type r).


As we have discussed in section 5, the amount of work is given by

dW = V dt(T ′kiGik + J ei′Ei) . (5)

T ′ki and J ei′ are the reversible contributions to the stress tensor and the

charge current density, respectively. Gik stands for the symmetric velocitygradient while Ei denotes the electric field strength.Let us now combine (1) with (2), (3), (4), and (5):

TρDts = −∂iJ ui + T ′′kiGik + J e

i′′Ei +

∑a

µa∂iJai +

∑r

Γ rAr . (6)

We have introduced the concept of chemical affinity

Ar = −∑a

νraµa . (7)

If Ar is positive, the system may lower its free energy by reactions of type r.Likewise, if the affinity is negative, backward reactions will prevail. There isequilibrium with respect to a chemical reaction of type r if the correspondingaffinity Ar vanishes.We divide (7) by the temperature field T and replace a∂ib by ∂iab − b∂ia.In this way a balance equation for entropy may be derived:

∂tρs+ ∂i(ρsvi + J si ) = π(S) . (8)

The entropy conduction current density is

J si =

1TJ ui −

∑a

µa

TJai . (9)

6.2 The second law

Entropy production is described by

π(S) = + J ui ∂i

1T

(10)

−∑a

Jai ∂iµa

T(11)

+1TT ′′kiGik (12)

+1TJ ei′′Ei (13)

+1T

∑r

Γ rAr . (14)


The entropy production cannot be negative,

π(S) ≥ 0 . (15)

This finding, the second law of thermodynamics, is difficult to prove fromfirst principles. In the framework of continuum physics the message is clearand simple:

• heat flows from higher to lower temperature, (10)

• at constant temperature, particles migrate into regions with lowerchemical potential, (11)

• by friction, momentum flows from higher to lower velocities, (12)

• an electric field causes a conduction current in field direction, (13)

• a positive chemical affinity drives forward reactions, (14).

There is much ado about the second law of thermodynamics. It is not truethat entropy increases incessantly. The correct statement is

Q(S;R) + I(S; ∂R) ≥ 0 (16)

where Q(S;R) is the entropy content of a region R and I(S; ∂R) the outflowof entropy through its surface.Many machines work periodically. The entropy Q(S) of the machine remainsconstant although irreversible processes take place. If heat enters at a highand leaves at a lower temperature level, then I(S) will be positive, cf. (9).This net entropy export must compensate the production of entropy.

7 FLUID DYNAMICS 19

7 Fluid Dynamics

Matter in its gaseous or liquid state is a fluid. One cannot draw at a fluidand it is not possible to exert shear forces. The force on an area element isalways anti-parallel to the surface normal. This is expressed by

T ′ki = −pδki . (1)

p = p(t,x) is the pressure field.We obtain

T ′kiGik = −p ∂ivi (2)

for the work done per unit time per unit volume by reversible momentumtransfer. Recall that

∂ivi = ρDtρ−1 (3)

is the relative change of volume per unit time as observed by a co-movingspectator, see (3.19).Consider a material point with mass M = ρV . During the time dt its volumewill change by dV = V dtρDtρ

−1. The work done to it is

dW = −V dtpρDtρ−1 = −pdV , (4)

a familiar expression. By the way, (4) justifies why we have associated theterm T ′kiGik in (5.17) with deformation. For a fluid, deformation withoutchange of volume is for nothing. In fact, that is the proper characterizationof a fluid. In contrast, it costs work to deform a solid even if the volumeremains unchanged.So far there was no need to distinguish liquids from gases, both are fluidsas defined by (1). Let us now concentrate on the two most widely studiedmedia, water and air.

7.1 Hydrodynamics

Water11 can hardly be compressed, just as most other liquids, in particularoils. We idealize this behaviour by demanding Dtρ = 0 which is equivalentto

∂ivi = 0 . (5)

The velocity field of an incompressible fluid is divergence-free.11greek: υδωρ

7 FLUID DYNAMICS 20

The irreversible contribution to the stress tensor may be written as

T ′′ki = 2ηGki = η(∂kvi + ∂ivk) (6)

which states that the force of friction is proportional to the symmetric ve-locity gradient. η is the viscosity of the fluid.Note that a contribution ∝ δkiGjj is formally correct as well, but has tovanishes for an incompressible liquid. A liquid obeying (5) and (6) is calledNewtonian12. Water and hydraulic oils are Newtonian fluids to a very goodapproximation.The momentum balance equation for a Newtonian liquid reads

ρDtvk = ρ {∂tvk + vi∂ivk} = −∂kp+ η∆vk + fk . (7)

Together with (5), the famous Navier13-Stokes14 equation (7) describes thedynamics of a Newtonian liquid.A material point is accelerated (ρDtvk) by a negative pressure gradient(−∂kp), if the surrounding medium is faster (η∆vk), or if an external forcefk is present.

7.2 Aerodynamics

A gas, like air, is very well described by the ideal gas law:

p = nkBT with n =∑a

na . (8)

n is the number of particles per unit volume, kB the Boltzmann constant,and T denotes absolute temperature (in Kelvin).We denote by za = na/n the molar fraction of the constituent a. Becauseof ρ =

∑mana = n

∑zama we may rewrite (8) into

p = R ρ T . (9)

The constant R of proportionality is given by

R =kB∑a z

ama. (10)

Its value for air15 is 288 m2s−2K−1 .12Isaac Newton, 1642-1727, English physicist13Claude-Louis-Marie-Henri Navier, 1785-1836, French engineer14George Gabriel Stokes, 1819-1903, British physicist15approximately 21% oxygen, the rest mostly nitrogen

7 FLUID DYNAMICS 21

The Navier-Stokes equation applies to gases as well, with one modification.Friction should now be described by two terms,

T ′′ki = η(∂kvi + ∂ivk) + ηδki∂jvj . (11)

The η contribution vanishes for incompressible media, as explained above. Itis not clear whether this term is really required for gases. The phenomenonof turbulence overshadows the effect of internal friction in many cases.

7.3 Example: Hagen-Poiseuille equation

Let us study the laminar flow of an incompressible fluid in a circular tubeof radius R. r =

√x2

1 + x22 is the distance from the pipe’s axis. For reasons

of symmetry we postulate v1 = v2 = 0 and v3 = v3(r). The divergence ofthis field vanishes automatically. Physical intuition tells that there must bea pressure gradient along the pipe axis, and we try p = p(x3) = p0 − p ′x3.The Navier-Stokes equation (without external forces) reduces to

0 = p ′ + η

(v ′′3 +

1rv ′3

). (12)

The corresponding two boundary conditions are v3(R) = 0 and v3 beingregular at r = 0. The solution is

v3(r) =p ′

4η(R2 − r2) . (13)

We denote by V the volume of fluid passing the pipe per unit of time. With(14) we obtain

V =∫dAv3 = 2π

∫ R

0drr v3 =

πp ′

8ηR4 . (14)

This equation has been derived independently by Hagen16 and Poiseuille17.Hagen was responsible for the municipal water supply of the city of Berlin.Poiseuille, a physician, investigated the human blood vessel system.Which power P is required to pump this amount of water through a pipe oflength L?One way to answer this question is as follows. A small volume dV of sub-stance is pressed into the system (the pipe’s interior) the volume of whichdiminishes by dV . With p1 as pressure at entry we have to expend the workp1dV . At exit the same volume of substance leaves the system at a pressure

16Gotthilf Hagen, 1797-1884, German engineer17Jean-Louis-Marie Poiseuille, 1799-1869, French physician

7 FLUID DYNAMICS 22

p2, so the work p2dV is regained. Therefore, per unit of time we have toexpend the work

P = V ∆p . (15)

Another, more professional way of reasoning is to work out the energy losscaused by internal friction which must be compensated by the pump andresults in a radial heat current.The symmetrized velocity gradient belonging to (13) is

Gki = − p′

4η

0 0 x1

0 0 x2

x1 x2 0

. (16)

It is associated with the stress tensor field (6) so that we obtain

π(U) = T ′′kiGik =p ′2

4ηr2 . (17)

Integration over the pipe’s interior yields∫dV π(U) = L 2π

p ′2

4ηR4

4. (18)

Inserting the Hagen-Poiseuille equation results in (15).A third line of argument is to calculate the power as force times velocity.v πR2 = V defines the average velocity v = V /πR2.Forces are described by the stress tensor Tki associated with (13),

Tki = −(p0 − p ′x3)

1 0 0

0 1 0

0 0 1

− p ′

2

0 0 x1

0 0 x2

x1 x2 0

. (19)

The effect of pumping the liquid through a resting pipe is equivalent todrawing the pipe with velocity v over a resting pile of liquid. The powerto achieve this is P = F3v. F3 is the shear force exerted by the liquid onthe pipe, to be calculated by integrating the stress tensor over the interiorsurface of the pipe, F3 =

∫dAiT3i.

The integration results in F3 = πR∆p. If this force on the pipe is multipliedby the average velocity v we again arrive at the expression (15).It is reassuring that these three different lines of reasoning lead to the sameresult.

7 FLUID DYNAMICS 23

7.4 Reynolds’ number

The partial differential Navier-Stokes equation (7) is difficult to solve ingeneral. Besides linear terms, there is the contribution ρvi∂ivk which isquadratic in the velocity field. Proven mathematical methods such as thetheory of Green’s functions or Fourier analysis are not applicable. In fact,the Navier-Stokes equations or simplified versions thereof serve as examplesfor chaotic behaviour, or turbulence.The term η∆vk damps velocity differences while turbulence is stirred up byρvi∂ivk. Their ratio can be estimated as follows. Denote by ` a typicallength, by v a typical velocity and by ρ a typical mass density. The frictionterm is of order ηv`−2, the competing turbulence contribution will be v2ρ`−1.The ratio

Re =ρ ` v

η(20)

is called Reynolds’ number18.A small Reynolds number signifies that friction prevails. Neigbouring lay-ers of matter remain neighbors, and one speaks of laminar flow. The flowbecomes turbulent if Reynolds’ number exceeds values of 1000. . . 10000. De-tails depend on subtleties such as the the roughness of surfaces or the pres-ence of obstacles like welds. However, there is no doubt that the Navier-Stokes equation correctly describe even turbulent flow.Here is an example.A circular water pipe of d = 2.0 cm diameter transports one litre per second.The average velocity will be v = 3 m s−1. With `= 0.01 m, ρ= 1000 kg m−3

and η= 1.00 × 10−3 kg m−1 s−1 we obtain Re = 30000. This flow is ratherturbulent, and the Hagen-Poiseulle law will not apply.

18Osborne Reynolds, 1842-1912, British engineer and physicist

8 ELASTOMECHANICS 24

8 Elastomechanics

In this section we discuss the balance of forces within an elastic solid body.The dependence on time is of secondary importance only and we will dropthe time argument whenever appropriate.

8.1 Strain

A solid is characterized by fixed neighborhood relations. Particles willslightly change their relative positions upon stress, but they return to theiroriginal position when the stress is removed.Let us consider two neighboring material points, one at x, the other one atx+ dx. These coordinates shall refer to the medium without any stress. Ifthe medium suffers stress, the first material point is moved to x ′ = x+u(x),the second to x ′ + dx ′ = x + dx + u(x + dx). u is the displacement fieldwhich characterizes the dislocation of material points because of stress.The distance between the to material points was ds = |dx| before disloca-tion. It is given by

ds ′ = |dx ′| = |dx+ u(x+ dx)− u(x)| (1)

after stress has been applied. We calculate

ds ′2 = {δjk + ∂juk + ∂kuj + (∂jui)(∂juk)} dxjdxk . (2)

With

Sjk =∂juk + ∂kuj

2(3)

we may write

ds ′ − dsds

=Sjkdxjdxk

ds2. (4)

Sjk is the strain tensor. It describes how much the distances between adja-cent material points changes if the medium suffers stretching or quenching.A rigid translation or a rigid rotation does not change distances, hence Sjkvanishes. Likewise, if the strain tensor vanishes everywhere, all distanceswithin the medium remain unchanged which is possible only for a rigidtranslation or a rigid rotation. Therefore, the strain tensor truly describesthe deformation of a solid medium. The additional contribution (∂jui)(∂juk)has been dropped because we assume the displacement gradients to be small.


The process of deforming a solid body is described by a coordinate transfor-mation x → x ′ = x + u where u denotes the displacement field. Therebythe volume element changes according to

dV ′ =∂(x ′1, x

′2, x

′3)

∂(x1, x2, x3)dV . (5)

The functional determinant

∂(x ′1, x′2, x

′3)

∂(x1, x2, x3)= det{δjk + ∂kuj} (6)

is 1 + ∂juj if the displacement field gradients are assumed to be small.Therefore, the relative change of volume is given by the divergence of thedisplacement field, or the trace of the strain tensor,

dV ′ − dVdV

= ∂juj = Sjj . (7)

8.2 Hooke’s law

There is no strain if there is no stress, by definition. Hence, the strain willdepend on stress as

Sjk = ΛjkrsTrs + . . . . (8)

The linear relationship between strain Srs and stress Tjk is known as Hooke’slaw19. For most solids it breaks down when the stress surpasses a limitingvalue beyond which the material is irreversibly deformed. Rubber or rubberlike substances (elastomers) however require non-linear modifications.Imagine a column of diameter w and height h. It is firmly grounded at itslower end and pressed upon at its upper end. It will get shorter and broader.The relative change of height will be proportional to the pressure p,

δh

h= − p

E. (9)

E is the elastic (or Young’s) modulus20 characterizing the material underinvestigation. The relative diameter change is written as

δw

w= ν

p

E. (10)

19Robert Hooke, 1635-1703, English physicist20Thomas Young, 1773-1829, British physician, physicist and egyptologist (deciphering

the stone of Rosetta)


ν, Poisson’s ratio21, is a dimensionless number varying between 0 and 1/2.We may describe the deformation of our column by the following displace-ment field:

u1 =δw

wx1 , u2 =

δw

wx2 and u3 =

δh

hx3 . (11)

The strain tensor belonging to this displacement is

Sjk =

δw/w 0 0

0 δw/w 0

0 0 δh/h

(12)

while the stress tensor is obviously

Tjk =

0 0 0

0 0 0

0 0 −p

. (13)

(9), (10), (12), and (13) are compatible if

Sjk =1 + ν

ETjk −

ν

EδjkTii (14)

holds, Hooke’s law for an isotropic medium. It may be solved for the stresstensor,

Tjk =E

1 + ν

{Sjk +

ν

1− 2νδjkSii

}. (15)

In most applications of elastomechanics one is interested in static solutionsonly. The momentum balance, with all time derivatives vanishing, reducesto

∂iTki + fk = 0 . (16)

As usual, fk is the external force per unit volume, in most cases gravitation(f1 = f2 = 0, f3 = −gρ).Not all solutions of (16) are admissible. A valid stress tensor field must alsobe related, by Hooke’s law, to a strain tensor field which in turn must bederived from a displacement vector field according to (3).

21Simeon-Denis Poisson, 1781-1858, French mathematician


Note the similarity with electrostatics. It is not sufficient to derive a solutionof ∂iDi = ρ e. The electrostatic displacement Dj must be related to anelectric field Ei by Dj = εjkEk which in turn has to be derived from a scalarfield φ (the potential) by Ei = −∂iφ.In most cases the deformation of the solid as described by the strain tensoris of minor interest. The normal approach in structural mechanics is to an-alyze the stress tensor locally. It is automatically symmetric because it isderived from a symmetric strain field by Hooke’s law. Being symmetric, itcan be diagonalized. Its eigenvalues must be within certain limits which arecharacteristic of the material. Concrete, for example, can stand high pres-sure, but cracks easily under tension. Therefore, concrete will be reinforcedby embedded steel rods which may resist strong tension.The three equations (3), (15) and (16) are the foundation of structural me-chanics.

8.3 Example: Torsion

We consider a cylinder of radius R and length L. Its base, at x3 = 0 is fixed.Its upper end, at x3 = L is twisted by an angle α.We guess the following displacement field (with α ′ = α/L):

u =

+x1 cosα ′x3 − x2 sinα ′x3 − x1

+x1 sinα ′x3 + x2 cosα ′x3 − x2

0

≈ α ′x3

−x2

+x1

0

. (17)

From it we derive the strain tensor

Sjk =α ′

2

0 0 −x2

0 0 +x1

−x2 +x1 0

. (18)

Its trace Sjj vanishes, as expected, since twisting does not change volume.Hooke’s law (15) provides the stress tensor:

Tjk =α ′ER

2(1 + ν)1R

0 0 −x2

0 0 +x1

−x2 +x1 0

. (19)

It has been derived from a displacement field and its divergence ∂iTki van-ishes, hence (17) describes an admissible solution.


It remains to relate α ′ with the torque N exercised on the upper end. Wecalculate

N =∫

x21+x2

2≤R2

dA3(x1T23 − x2T13) =α ′E

2(1 + ν)πR4

2, (20)

or

σ =α ′ER

2(1 + ν)=

2NπR3

. (21)

It is obvious that the stress increases linearly with the distance r from theaxis. It is maximal at the surface. Without loss of generality we can inves-tigate it at x = (R, 0, 0). The matrix

Tjk = σ

0 0 0

0 0 1

0 1 0

(22)

has the following eigenvectors and eigenvalues:

• (1, 0, 0) with eigenvalue 0. There is neither stress nor pressure in radialdirection.

• (0, 1, 1)/√

2 with eigenvalue σ. There is stress in the tangential plainat 45◦.

• (0, 1,−1)/√

2 with eigenvalue −σ. There is pressure in the tangentialplain at -45◦.

If the torque exceeds a critical value, the material will crack at the surface.

8.4 Elastic waves

Note that we did not distinguish between the stress tensor Tki and its re-versible (or elastic) contribution T ′ki so far since the velocity field vi wasassumed to vanish. We have discussed elastostatics only. There is no irre-versible (or inelastic) contribution to stress without velocity differences.The velocity field is vi = Dtui. Since displacements (and therefore velocities)remain small, this is the same as vi = ∂tui. Disregarding external forces andneglecting friction we have to solve

ρ uk =E

2(1 + ν)

{∆uk +

11− 2ν

∂k∂iui

}. (23)


This equation describes elastic waves. We have indicated the derivative withrespect to time by a dot since there is no need to distinguish between restingand co-moving observers. ∆ = ∂i∂i is the Laplacian.There are two types of waves, longitudinally and transversely polarized.A longitudinally polarized elastic wave is described by a displacement field

u(t,x) = A

0

0

cos(ωt− kx3)

. (24)

Inserting this into (23) yields ω = c‖k where

c2‖ =1− ν

(1 + ν)(1− 2ν)E

ρ. (25)

The following displacement fields

u(t,x) = A

cos(ωt− kx3)

0

0

or A

0

cos(ωt− kx3)

0

(26)

describe transversely polarized elastic waves. Again (23) will be satisfiedprovided we set ω = c⊥k where

c2⊥ =1

2(1 + ν)E

ρ. (27)

By measuring the speed c‖ and c⊥ of longitudinally and transversely po-larized elastic waves one can determine Young’s module E and Poisson’sratio ν. However, beware that both quantities are adiabatic, not isothermalvalues.By adding to (23) an inelastic term η∆uk or something more complicatedwe may take friction into account which describes damping of elastic waves.

9 HEAT TRANSPORT 30

9 Heat Transport

Recall the first law of thermodynamics,

ρDtu = T ′kiGik + J ei′Ei − ∂iJ u

i + T ′′kiGik + J ei′′Ei . (1)

The first two terms describe work (deformation and polarization), the re-maining three contributions characterize irreversible effects, namely heatconduction, friction, and Joule’s heat.

9.1 Heat equation

Here we will concentrate on a solid so that Dtu = ∂tu = u holds as well asGik = 0. We may therefore write

ρu+ ∂iJui = πu (2)

where ρ is the mass density, u the specific internal energy, J ui the heat cur-

rent density, and πu describes the production of internal energy by variousmechanisms, such as Joule’s heat, radioactive decay, etc.In a solid, the specific internal energy depends on time via temperature T .We express this as

ρu = ρCpT (3)

where Cp is the specific heat capacity at constant pressure.Temperature differences cause heat transport, and in a solid this is the only22

mechanism. We therefore write

J ui = −λ∂iT , (4)

Fourier’s law23 of heat transport. λ is the heat conductivity24 of the material.With these specializations we arrive at the following heat equation:

ρCpT − ∂iλ∂iT = πu . (5)

This equation simplifies further if the temperature differences are smallenough. It is then allowed to neglect ∂iλ = (∂iλ/∂T )∂iT , and we arriveat

T − κ∆T = T ∗ . (6)22not quite, as we shall see in the section on thermoelectric effects23Jean-Baptiste-Joseph Fourier, 1768-1830, French egyptologist, mathematician and

physicist24possibly a tensor, J u

i = −λik∂kT

9 HEAT TRANSPORT 31

The coefficient κ of temperature conductivity is given by

κ =λ

ρCp(7)

while T ∗ is shorthand for

T ∗ =1ρCp

πu , (8)

the rate of spontaneous temperature increase.

9.2 Example: Soil temperature

By the way, the Fourier transformation technique was developed by the sameauthor who investigated the problem of heat transfer. Among other things,Fourier studied the variation of temperature in soil. With z as depth belowsurface he wanted to solve the following equation (in today’s notation):

∂tT = κ∂2zT . (9)

The temperature T0(t) = T (t, 0) at the surface was assumed to be a periodicfunction, the period τ being one day or one year.Fourier proved that any (well behaved) periodic function may be expandedinto trigonometric functions,

T (t, z) =∑n

e2iπnt/τ

ϑn(z) . (10)

The z-dependent coefficient ϑn must obey an ordinary differential equation,namely

2iπnτ

ϑn(z) = κϑ ′′n (z) , (11)

its initial value is

ϑn(0) =1τ

∫ τ

0dt e−2iπnt/τ

T (t, 0) . (12)

Define

qn =

√π|n|τκ

(13)

so that the general solution may be expressed as

T (t, z) =∑n

ϑn(0) e i(2πnt/τ − qnz) e−qnz . (14)

Note the exponential decrease with depth of temperature departures fromthe average.

10 DIFFUSION 32

10 Diffusion

We have introduced the term ’diffusion’ when the particle current densitiesjai were split into a convective contribution navi and the remainder, thediffusion currents Jai . Recall that a labels the particle species and vi denotesthe local center of mass velocity.

10.1 Diffusion equation

In the absence of other non-equilibrium conditions, diffusion currents aredriven by gradients of the chemical potentials µa. After all, particle exchangebetween two systems stops if the chemical potentials have become equal. Wemay therefore write

Jai = −∑b

dab∂iµb . (1)

Recall the contribution

π(S) = −∑a

Jai ∂iµa

T+ . . . (2)

to the production of entropy per unit time per unit volume. We concludethat the matrix dab in (1) must be positive.The chemical potentials µa depend on temperature T , pressure p, and par-ticle densities n1, n2, . . .. As said above, we assume the temperature and thepressure be constant everywhere. We then may write

∂iµb =

∑c

∂µb

∂nc∂in

c (3)

which leads to

∂tna + ∂in

avi = ∂i∑c

Dac∂inc . (4)

The matrix of diffusion constants is given by

Dac =∑b

dab∂µb

∂nc. (5)

I know of no general argument why this matrix should be positive.Our description of diffusion is, however, unnecessarily complicated. In real-istic situation one has a solid or a solvent and one or more species of diffusing

10 DIFFUSION 33

particles the number of which is small. Think about titanium ions diffusingin a lithium niobate substrate or sugar molecules diffusing in water.The interaction between different species of diffusing particles may usuallybe neglected such that

Jai = −da∂iµa (6)

holds where the coefficients da are positive. Likewise,

∂iµa =

∂µa

∂na∂in

a . (7)

Since the particle number increases with the chemical potential we concludethat the diffusion constants

Da =∂µa

∂nada (8)

are positive. In the absence of mass flow, diffusion is described by

∂tna = ∂iDa∂in

a , (9)

or

∂tna = Da ∆na , (10)

if the diffusion constants Da are the same everywhere.Usually, there is just one species of diffusing particles. We denote its numberdensity by n and its diffusion constant by D, and the diffusion process isdescribed by

∂tn = D∆n . (11)

Note the formal similarity with the heat equation

∂tT = κ∆T (12)

which we have derived and discussed earlier. No wonder, heat transfer insolids means phonon diffusion.

10 DIFFUSION 34

10.2 Brownian movement

As we have made clear, the diffusion equation (12) applies to particles whichmove in a substrate and do not interact with each other or with other dif-fusing particles. In fact, diffusion is Brownian movement25 of many inde-pendent particles.Consider a particular Brownian particle which collides with the moleculesof its host liquid. The time intervals τk between consecutive collisions areindependent, equally distributed random variables, and we assume the samefor the shifts sk. We denote by τ = 〈τk〉 the average time between twocollisions. The average shift vanishes, 〈sk〉 = 0, its variance is denoted by〈s2k〉 = a2.

The law of large numbers says

limn→∞

1n

n∑k=1

τk = τ . (13)

The central limit theorem of probability theory states that

1√n

n∑k=1

ska

(14)

has a distribution which converges, with n → ∞, towards a normal distri-bution.Assume that the Brownian particle was located at x = 0 at time t = 0. Theprobability density to find the particle after time t = nτ at x is proportionalto

e−x2/2na2

= e−x2τ/2a2t

. (15)

Introducing the abbreviation

D =a2

τ(16)

allows to write

dW = dVe−x2/2Dt

(2πDt)3/2. (17)

This is the probability that the particle will be found in a volume dV at xafter a random walk of duration t.

25Robert Brown, 1773-1858, Scottish botanist

10 DIFFUSION 35

Denote by n the density of Brownian particles. If they walk independently,we may write

n(t,x) =∫d3y n(0,y)

e−(x− y)2/2Dt

(2πDt)3/2. (18)

It is a trivial exercise to show that n obeys the diffusion equation with (16)as diffusion constant, so we have indeed shown that diffusion is masswiseBrownian motion.As a by-product, we have found the Green’s function26 for solving the initialproblem of the diffusion (and any similar) equation. Note that (18) assumesthat the diffusing particles may spread unhindered everywhere. Differentboundary conditions lead to different Green’s functions.

26George Green, 1793-1841, English mathematician

11 CHARGE TRANSPORT 36

11 Charge Transport

In this section we will attempt to clarify the role of various contributions tothe density and the current density of electric charge.

11.1 Maxwell’s equations

Let us recall Maxwell’s equations for the electromagnetic field E,B:

ε0∇ ·E = ρ e , (1)

∇ ·B = 0 , (2)

1µ0

∇×B = ε0E + j e , (3)

∇×E = −B . (4)

Assume that a continuum is exposed to an electromagnetic field. One effectwill be that matter gets polarized: there is a certain density of electric dipolemoment which we denote by P . The negative divergence of this polarizationis a contribution to the charge density,

−∇ · P = ρ p . (5)

We call it the polarization charge. Likewise, there is a current if the polar-ization changes with time,

P = j p . (6)

Another effect of the electromagnetic field will be magnetization. Denoteby M the density of magnetic moments. Its curl is another contribution tothe current density,

∇×M = jm . (7)

In general, the total charge density consists of two terms,

ρ e = −∇ · P + ρ f , (8)

the current density of three,

j e = P + ∇×M + j f . (9)

The superscript ’f’ stands for ’free’ or ’to be freely manipulated’, in contrastto ’p’ (forced by polarization) or ’m’ (forced by magnetization).


This is a vicious circle: it is the electric field which causes polarization, thelatter then contributes to the electric charge density which, in turn, must beknown for working out the electric field. The same applies to the magneticinduction field.The well known way out is to define auxiliary fields, the dielectric displace-ment

D = ε0E + P (10)

and the magnetic field strength

H =1µ0B −M . (11)

With it (1) may be rewritten as

∇ ·D = ρ f (12)

while (3) changes into

∇×H = j f + D . (13)

(2) and (4) remain unchanged.The prize to be paid is that we have to cope with two new vector fields. Werequire additional functional relations between the displacement and themagnetic fields on the one hand and the electric and induction fields on theother.We will not discuss magnetization further since magnetism was ignored inthe preceding sections.

11.2 Ohm’s law

We know that electric charge is conserved,

ρ e + ∇ · j e = 0 . (14)

This is a consequence of Maxwell’s equations (1) and (3). However, it is alsotrue that the free charge is conserved:

ρ f + ∇ · j f = 0 . (15)

This follows from (12) and (13).We should now compare the decomposition (9) with our previous decompo-sition in terms of convection as well as reversible and irreversible conduction.

j e = P + j f = (−∇ · P + ρ f)v + J e ′ + J e ′′ (16)

says


• ρ fv accounts for the convection of free charge,

• reversible (or elastic) conduction is described by J e ′ = DtP , thesubstantial time derivative of the polarization,

• the remainder j f characterizes the irreversible (or inelastic) conductioncontribution J e ′′ to the electric current density. It is well describedby Ohm’s law.

As mentioned before, magnetic effects have been left out of the discussion.Ohm’s law27 states

j f = σE (17)

for isotropic media. σ is the electric conductivity. For anisotropic media wehave to introduce a corresponding tensor.

11.3 Electro-chemical potential

The above formulation of of Ohm’s law must be modified since the electricfield strength may not be the only driving force. If the chemical potential µ∗

of free, or mobile electrons has a gradient, then electrons will also migrateby diffusion. In fact, the electric potential φ e must be supplemented by theelectron chemical potential divided by the electron charge. The sum

ψ = φ e − µ∗

e(18)

is known as the electro-chemical potential. Therefore, (17) should be re-placed by

j f = −σ∇ψ . (19)

Differences of the electro-chemical potential drive electric currents, whetherthey are maintained by a dynamo or a battery plays no role.

27Georg Simon Ohm, 1789-1854, German physicist

12 ONSAGER’S RELATION 39

12 Onsager’s Relation

Irreversible effects are characterized by entropy production. Each terms isa product of a generalized force and a generalized flux. No forces, no fluxes,small forces, small fluxes. This idea is expressed by linear relations betweenfluxes and forces.

12.1 Forces and fluxes

Let us again look at the expressions (6.10) to (6.14) for entropy produc-tion. Each term is the product of a generalized force Xα and a generalizedflux Φα. The forces will vanish in thermodynamic equilibrium: constanttemperature, constant chemical potentials, no velocity gradients, vanishingchemical affinities. The fluxes describe changes of state. Thus we may write

π(S) =1T

∑α

ΦαXα . (1)

If there are no forces, then the fluxes will vanish. If the forces are sufficientlysmall, the fluxes will also be small. Therefore, a linear relationship betweenforces and fluxes is plausible, and we shall assume

Φα =∑β

LαβXβ . (2)

12.2 Onsager’s theorem

Onsager28 has shown that, in the absence of external magnetic fields, thematrix L of generalized conductivities is symmetric,

Lβα = Lαβ . (3)

We shell not give a proof of this statement here. Onsager’s original reasoningappears questionable today, and a proper treatment within the frameworkof linear response theory is beyond the scope of this text.However, (2) remains true if the entropy production term is reformulated.Assume a new definition of generalized forces by

Xα =∑β

SαβX′β . (4)

The new generalized fluxes are

Φ ′α =∑β

S†αβΦβ (5)

28Lars Onsager, 1903-1976, Norwegian-USA chemist


which choice assures

π(S) =1T

∑α

ΦαXα =1T

∑α

Φ ′αX′α . (6)

The new matrix of generalized conductivities is L ′ = S†LS which is sym-metric if L is. It is therefore allowed to replace the generalized forces bylinear combinations.

12.3 An example

Let us study an example. With

∂i1T

= − 1T 2∂iT (7)

and

∂iµa

T=

1T∂iµ

a − µa

T 2∂iT (8)

the entropy production can be rewritten as

Tπ(S) =∑

a µaJai − J u

i

T∂iT (9)

−∑a

Jai ∂iµa (10)

+ T ′′kiGik (11)− J e

i′′∂iφ

e (12)

+∑r

Γ rAr . (13)

In this representation of entropy production the generalized forces are

• the temperature gradient ∂iT with corresponding flux (∑

a µaJai −

J ui )/T

• the gradients ∂iµa of chemical potential µa with the negative diffusioncurrent −Jai as flux

• the velocity gradients Gik = (∂ivk + ∂kvi)/2 which is associated withthe irreversible momentum current density T ′′ki

• the gradient ∂iφ e of the electric potential to which belongs the negativeconduction contribution −J e

i′′ of free charges


• the affinities Ar for reactions of type r with the reaction rates per unitvolume Γ r as associated fluxes

If there are no mass flows, if temperature and all chemical potentials arethe same everywhere and if no electric field is present, we have a situationwhich is characteristic for chemical reactions. Close to chemical equilibriumwe may write

Γ r =∑s

RrsAs . (14)

The second law of thermodynamics demands that R be a positive29 matrix.Onsager’s theorem says that R be also symmetric. Ar drives Γ s and As willdrive Γ r, both with the same proportionality constant Rsr = Rrs.

29∑rs Rrsxrxs ≥ 0 for xr ∈ R

13 THERMOELECTRIC EFFECTS 42

13 Thermoelectric Effects

In this section we will discuss cross effects: there is more than one generalizedforce driving more than one generalized flux. To be specific, we want toinvestigate solid electrical conductors. There is no mass flow, and there areno chemical reactions.Electrons are the only species of mobile charged particles, we index them byan asterisk. Recall that electric charge transport (j e) is by convection (ρ ev)or conduction (J e), the latter being reversible (J e ′) or irreversible (J e ′′).The reversible part coincides with the time derivative of the polarizationwhich is of no interest here. The irreversible part corresponds to the motion,by drift or diffusion, of free electrons, J f . In this context we may write theelectric current density as

J e = −eJ∗ . (1)

There is a current (J∗) of electrons each carrying its charge −e. Electronswill transport not only charge, but also internal energy.Phonons, the quanta of lattice vibrations, are also mobile quasi-particles,they transport internal energy. Phonon diffusion, or heat conduction in thiscontext, is described by the associated heat current J u.There are two species of mobile particles (phonons and electrons), two cor-responding currents (J u and J e), and two driving forces, namely ∇T and∇ψ, the gradients of temperature T and electro-chemical potential

ψ = φ e − µ∗

e. (2)

13.1 Charge and heat transport

Let us recall the expression

π(S) = J u ·∇ 1T

+ J∗ ·∇ µ∗

T+

1TJ e ·E (3)

for the amount of entropy produced per unit volume per unit time. Theentropy conduction current is given by

J s =1T

(J u +

µ∗

eJ e

). (4)

It is a simple exercise to rewrite (3) into

π(S) = − 1T

(J s ·∇T + J e ·∇ψ) . (5)


This sum of ’flux times force’ terms suggests the following linear relationship J s

J e

= −

A B

B C

∇T

∇ψ

. (6)

We have already taken into account Onsager’s relation by postulating a sym-metric generalized conductivity matrix. This matrix must also be positivewhich amounts to A ≥ 0, C ≥ 0, and AC ≥ B2.By comparing

J e = −σ(∇ψ + α∇T ) (7)

with (5) we recognize σ = C, therefore the electric conductivity is alwayspositive. The sign of α = B/C cannot be predicted from first principles.If the temperature is the same everywhere, (7) coincides with Ohm’s law(11.19). Moreover, the chemical potential, which depends on compositionand temperature, cannot have a gradient within a homogeneous material,and we may write J e = −σE, as in (17). (7) is a generalization of Ohm’slaw, it allows for a temperature gradient which will also drive a chargecurrent.The second relation contained in (6) may be expressed as

J u = −λ∇T + γJ e (8)

where λ = T (A − B2/C) is the heat conductivity. It is guaranteed to bepositive. The coefficients γ in (8) and α in (7) are related by

γ = Tα− µ∗

e. (9)

Without electric current we obtain Fourier’s law J u = −λ∇T . An addi-tional electric current causes additional heat transport. However, the signof γ (just as the sign of α) cannot be predicted from first principles.

13.2 Thomson effect

Let us work out the heating of a particular material point. We have tocalculate

∂tρu = −∇ · J u + J e ·E . (10)

This expression takes into account that there is no difference between or-dinary and substantial time derivative and that only Joule’s heat J e · Econtributes to the production of internal energy.


By solving (7) for the electrical field strength we arrive at

E =J e

σ+ α∇T −∇ µ∗

e. (11)

This is to be multiplied by J e. On the other hand, because of ∇ · J e = 0,we obtain

−∇ · J u = ∇ · λ∇T − J e∇γ . (12)

We refer to (9) and conclude

∂tρu = ∇ · λ∇T +J e2

σ− J eT∇α . (13)

The first term describes the net inflow of internal energy as caused by atemperature gradient. The second contribution is the rate of internal en-ergy production per unit volume if an electric current flows in a conductingmedium, it is always positive30. The third term describes the Thomson ef-fect31. In a homogeneous material it takes the form ρJ e ·∇T where theThomson coefficient is given by

ρ = −T ∂α∂T

. (14)

This contribution to the heating of a material point is proportional to theelectric current and to the temperature gradient. It changes sign if thedirection of the electric current is reversed.

13.3 Seebeck and Peltier effect

So far we have always assumed that the medium under investigation is spa-tially homogeneous. We now mention two related effects which appear ifdifferent conducting or semi-conducting materials are involved.Imagine a simple circuit made up of two homogeneous wires (I and II) ofdifferent composition soldered at points P1 and P2. These soldering contactsare kept at temperatures T1 and T2, respectively. If the circuit is open, sothat no electric current can flow, (7) reads −∇ψ = α∇T . The line integralaround the circuit is the electromotive force U ,

U = −∮ds∇ψ =

∮dsα∇T =

∫ T2

T1

dT (αI − αII) . (15)

30it vanishes in a super-conductor where 1/σ = 031William Thomson (Lord Kelvin), 1824-1907, British phycisist


The circuit is assumed to run from point P1 via wire I to point P2 and backvia wire II to point P1. This thermoelectric electromotive force requiresdifferent materials and different temperatures at the soldering contacts. Ifthe circuit is electrically closed by an external load it will drive a current

I =U

Ri +Re, (16)

where Ri and Re denote the internal and the external resistance, respectively.The existence of an electromotive force (15) caused by soldering differentconductors is known as the Seebeck32 effect. Ironically, Seebeck himselfdetected the magnetic field associated with (16). He never spotted it downto a current caused by an electromotive force. The Seebeck effect may servefor the direct heat to power conversion. It is, however, rather inefficient.With today’s materials the degree of efficiency is much below the Carnotlimit (T2 − T1)/T2 where T2 is the higher temperature. Best results areobtained with p-n-semiconductor couples.The Peltier effect33 is related to the Seebeck effect. If an electric currentflows through a thermocouple, heat is delivered to or subtracted from thesoldering contact, depending on the direction of the electric current.Consider the interface between two different conducting materials I andII. Assume the temperature to be constant on both sides. The normalcomponent of the electric current must be continuous as well as the electricand the electro-chemical potential. We conclude that the coefficient γ of (8)jumps by

Π = T (αII − αI) (17)

at the interface between material I and II. Therefore, the heat current jumpsby

J uII − J u

I = ΠJ e . (18)

This heat current difference is supplied by the heat reservoir to which the in-terface is coupled. Π is the Peltier coefficient which depends on temperatureand on the composition of the two materials soldered together. The Peltiereffect allows to heat or cool a soldering contact as desired by controlling anelectric current. However, Peltier refrigerators are rather inefficient.

32Thomas Johann Seebeck, 1770-1831, Baltic-German physicist33Jan-Charles-Athanase Peltier, 1785-1845, French physicist

Lectures on Theoretical Physics Continuum Physics · 2011-03-15 · examples. This postgraduate level course assumes a basic acquaintance with classical mechanics, electrodynamics,

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