4.6 Transport Coefficients |
255 |
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4.6.4 The Thermoelectric Power (B)
We use the same experimental setup as for thermal conductivity but now we measure the electric field. The absolute thermoelectric power Q is defined as the proportionality constant between electric field and temperature gradient. Thus
E = Q T . |
(4.168) |
Comparing with (4.166) gives |
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Q = − |
b |
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(4.169) |
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a |
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We generally measure the difference of two thermoelectric powers rather than the absolute thermoelectric power. We put two unlike metals together in a loop and make a break somewhere in the loop as shown in Fig. 4.11. If VAB is the voltage across the break in the loop, an elementary calculation shows
Q2 −Q1 |
VAB . |
(4.170) |
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T2 −T1 |
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B |
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Tb |
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T1 |
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2
Fig. 4.11. Circuit for measuring the thermoelectric power. The junctions of the two metals are at temperature T1 and T2
4.6.5 Kelvin’s Theorem (B)
A general theorem originally stated by Lord Kelvin, which can be derived from the thermodynamics of irreversible process, states that [99]
Summarizing, by using (4.162), (4.163), σ = a, (4.165), (4.167), (4.164), and (4.171), we can write
J = σE − |
σΠ |
T , |
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(4.172) |
T |
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2 |
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(4.173) |
H = σΠE − K +σ |
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T . |
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256 4 The Interaction of Electrons and Lattice Vibrations
If, in addition, we assume that the Wiedemann–Franz law holds, then K = CTσ, where C = (π2/3)(k/e)2, and we obtain
J = σE − |
σΠ |
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T , |
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(4.174) |
T |
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Π |
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(4.175) |
H = σΠE −σ CT + |
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T . |
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T |
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We summarize these results in Table 4.5. As noted in the references there are several other transport coefficients including magnetoresistance, Rigli–Leduc, Ettinghausen, Nernst, and Thompson.
Table 4.5. Transport coefficients
Quantity |
Definition |
Comment |
Electrical conductivity |
Electric current density at unit |
See Sect. 4.5.4 and 4.6.1 |
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electric field (no magnetic (B) field, |
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no temperature gradient). |
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Thermal conductivity |
Heat flux per unit temp. gradient |
See Sect. 4.6.3 |
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(no electric current). |
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Peltier coefficient |
Heat exchanged at junction per |
See Sect. 4.6.2 |
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electric current density. |
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Thermoelectric power |
Electric field per temperature gradi- |
See Sect. 4.6.4 |
(related to Seebeck |
ent (no electric current). |
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effect) |
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Kelvin relations |
Relates thermopower, Peltier coef- |
See Sect. 4.6.5 |
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ficient and temperature. |
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References: |
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[4.1, 4.32, 4.39] |
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4.6.6 Transport and Material Properties in Composites (MET, MS)
Introduction (MET, MS)
Sometimes the term composite is used in a very restrictive sense to mean fibrous structures that are used, for example, in the aircraft industry. The term composite is used much more generally here as any material composed of constituents that themselves are well defined. A rock composed of minerals, is thus a composite using this definition. In general, composite materials have become very important not only in the aircraft industry, but in the manufacturing of cars, in many kinds of building materials, and in other areas.
4.6 Transport Coefficients |
257 |
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A typical problem is to find the effective dielectric constant of a composite media. As we will show below, if we can find the potential as a function of position, we can evaluate the effective dielectric constant. First, we want to illustrate that this is also the same problem as the effective thermal conductivity, the effective electrical conductivity, or the effective magnetic permeability of a composite. For in each case, we end up solving the same differential equation as shown in Table 4.6.
Table 4.6. Equivalent problems
Dielectric constant |
Magnetic permeability |
D = εE |
B = µH |
ε is dielectric constant |
µ is magnetic permeability |
E is electric field |
H is magnetic field intensity |
D is electric displacement vector |
B is magnetic flux density |
E = 0 |
B = 0 |
(no changing B) |
(no current, no changing E) |
E = −(φ) |
H = −(Φ) |
· D = 0 |
· B= 0 |
(no free charge) |
(Maxwell equation) |
· (ε(φ)) = 0 |
· (µ(Φ)) = 0 |
B.C. |
analogous B.C. |
φ constant at top and bottom |
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(φ) = 0 on side surfaces |
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Electrical conductivity
J = σE and only driven by E σ is electrical conductivity E is electric field
J is electrical current density
E = 0
(no changing B) E = −(φ)
· J = 0
(cont. equation, steady state)· (s(φ)) = 0
analogous B.C.
Thermal conductivity
J = −K(T) and only driven by T K is the thermal conductivity
T is the temperature J is the heat flux
(T) = 0, an identity
grad dot J = 0
(cont. equation, steady state)· K( (T)) = 0
analogous B.C.
To begin with we must define the desired property for the composite. Consider the case of the dielectric constant. Once the overall potential is known (and it will depend on boundary conditions in general as well as the appropriate differential
258 4 The Interaction of Electrons and Lattice Vibrations
equation), the effective dielectric constant may εc be defined such that it would lead to the same over all energy. In other words
εc E02 = |
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1 |
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∫ε(r)E2 (r)dV , |
(4.176) |
V |
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where |
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= |
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∫ E(r)dV , |
(4.177) |
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where V is the volume of the composite, and the electric field E(r) is known from solving for the potential. The spatial dependence of the dielectric constant, ε(r), is known from the way the materials are placed in the composite.
z
Tt = constant
S
L
Tb = constant
Fig. 4.12. The right-circular cylinder shown is assumed to have sides insulated and it has volume V = LS
One may similarly define the effective thermal conductivity. Let b = − T, where T is the temperature, and h = −K T, where K is the thermal conductivity. The equivalent definition for the thermal conductivity of a composite is
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Kc = |
V ∫ h bdV |
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(4.178) |
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(∫ bdV )2 |
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For the geometry and boundary conditions shown in Fig. 4.12, we show this expression reduces to the usual definition of thermal conductivity.
Note since ·h = 0 in the steady state that − ·(Th) = h·b, and so ∫h·bdV = −(Tt − Tb)∫hzdSz, where the law of Gauss has been used, and the integral is over the top of the cylinder. Also note, by the Gauss law zˆ · ∫bdV = (Tt − Tb)S, where S is the top or bottom area. We assume either parallel slabs, or macroscopically dilute solutions of ellipsoidally shaped particles so that the average temperature gradient will be along the z-axis, then
− KcS(Tt −Tb ) / L = ∫ |
hzdSz , |
(4.179) |
top |
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as required by the usual definition of thermal conductivity.
4.6 Transport Coefficients |
259 |
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It is an elementary exercise to compute the effective material property for the series and parallel cases. For example, consider the thermal conductivity. If one has a two-component system with volume fractions φ1 and φ2, then for the series case one obtains for the effective thermal conductivity Kc of the composite:
1 |
= |
ϕ1 |
+ |
ϕ2 . |
(4.180) |
K |
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K |
c |
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K |
2 |
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1 |
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This is easily shown as follows. Suppose we have a rod of total length L = (l1 + l2) and uniform cross-sectional area composed of a smaller length l1 with thermal conductivity K1 and an upper length l2 with K2. The sides of the rod are assumed to be insulated and we maintain the bottom temperature at T0, the interface at T1,
and the top at T2. Then since T1 = T0 − T1 and T2 = T1 − T2 we have T = T1 + T2 and since the temperature changes linearly along the length of each rod:
K |
T1 |
= K |
T2 |
= K |
T |
, |
(4.181) |
1 |
l |
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2 l |
2 |
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c L |
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1 |
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where Kc is the effective thermal conductivity of the rod. We can thus write:
T |
= |
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K |
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l |
T |
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T |
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= |
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K |
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l |
2 |
T |
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(4.182) |
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K |
L |
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K |
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L |
1 |
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and so |
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K |
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l1 |
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l2 |
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T = |
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(4.183) |
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K |
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and since the volume fractions are given by φ1 = (Al1/AL) = l1/L and φ2 = l2/L, this yields the desired result.
Similarly for the parallel case, one can show: |
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Kc = ϕ1K1 + ϕ2K2 . |
(4.184) |
Consider two equal length slabs of length L and areas A1 and A2. These are placed parallel to each other with the sides insulated and the tops and bottoms maintained at T0 and T2. Then if T = T0 − T2, the effective thermal conductivity can be defined by
K (A |
+ A ) |
T |
= K A |
T |
+ K |
A |
T |
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(4.185) |
1 |
2 |
L |
1 1 |
L |
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2 2 |
L |
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where we have used that the temperature changes linearly along the slabs. Solving for K yields the desired relation, with the volume fractions defined by φ1 = A1/(A1+A2) and φ2 = A2/(A1+A2).
260 4 The Interaction of Electrons and Lattice Vibrations
General Theory (MET, MS)14
Let
u = |
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∫ bdV |
, |
(4.186) |
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∫ bdV |
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and with the boundary conditions and material assumptions we have made, u = zˆ. Define the following averages:
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= |
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1 |
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∫ |
u hdV , |
(4.187) |
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h |
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V V |
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∫ |
u bdV , |
(4.188) |
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V V |
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∫ |
u hdVi , |
(4.189) |
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hi = |
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Vi |
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∫ |
u bdVi , |
(4.190) |
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bi = |
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Vi |
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where V is the overall volume, and Vi is the volume of each constituent so V = ∑Vi. From this we can show (using Gauss-law manipulations similar to that already given) that
will give the same value for the effective thermal conductivity as the original definition. Letting φi = Vi/V be the volume fractions and fi = ¯bi/¯b be the “field ratios” we have
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Ki fi = |
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(4.192) |
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and |
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∑ |
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(4.193) |
hiϕi = h , |
so |
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K = ∑ Ki fiϕi . |
(4.194) |
14This is basically Maxwell–Garnett theory. See Garnett [4.9]. See also Reynolds and Hough [4.36].
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4.6 Transport Coefficients 261 |
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Also |
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∑ fiϕi |
=1 , |
(4.195) |
and |
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∑ϕi |
=1 . |
(4.196) |
The field ratios fi, the volume fractions φi, and the thermal conductivities Ki of the constituents determine the overall thermal conductivity. The fi will depend on the Ki and the geometry. They are only known for the case of parallel slabs or very dilute solutions of ellipsoidally shaped particles. We have already assumed this, and we will only treat these cases. We also only consider the case of two phases, although it is relatively easy to generalize to several phases.
The field ratios can be evaluated from the equivalent electrostatic problem. The b inside an ellipsoid bi are given in terms of the externally applied b(b0) by15
where the i refer to the principle axis of the ellipsoid. With the ellipsoid having thermal conductivity Kj and its surrounding K* the gi are
gi = |
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1 |
, |
(4.198) |
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+ Ni[(K j / K ) −1] |
1 |
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where the Ni are the depolarization factors. As usual,
∑3= Ni =1.
i 1
Redefine (equivalently, e.g. using our conventions, we would apply an external thermal gradient along the z-axis)
u = b0 , b0
and let θi be the angle between the principle axes of the ellipsoid and u. Then
u b = ∑i3=1 gib0 cos2 θi , |
(4.199) |
so |
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f j = ∑i gi cos2 θi , |
(4.200) |
262 4 The Interaction of Electrons and Lattice Vibrations
where the sum over i is over the principle axis directions and j refers to the constituents. Conditions that insure that ¯b = b0 have already been assumed. We have
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f j = ∑i3=1 |
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cos2 θi |
, |
(4.201) |
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1 |
+ Ni[(K j / K ) −1] |
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Kj is the thermal conductivity of the ellipsoid surrounded by K*.
Case 1: Thin slab parallel to b0, with K* = K2. Assuming an ellipsoid of revolution,
N = 0, (depolarization factor along b0 )
f1 =1, f2 =1.
Using
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K = ∑ Ki |
fiϕi , |
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we get |
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K = K1ϕ1 + K2ϕ2 . |
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(4.202) |
We have already seen this is appropriate for the parallel case. |
Case 2: Thin slab with plane normal to b0, K* = K2. |
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N =1 , f = |
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= |
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=1, |
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so we get |
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ϕ2 . |
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(4.203) |
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K |
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Again as before.
Case 3: Spheres with K* = K2 (where by (4.195), the denominator in (4.204) is 1)
N = 1 , |
f = |
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=1 |
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K2ϕ2 + K1ϕ1 |
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+ (K |
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K = |
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1 |
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(4.204) |
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ϕ2 +ϕ1 |
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2 + (K |
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These are called the Maxwell (composite) equations (interchanging 1 and 2 gives the second one).
The parallel and series combinations can be shown to provide absolute upper and lower bounds on the thermal conductivity of the composite.16 The Maxwell
Problems 263
equations provide bounds if the material is microscopically isotropic and homogenous.16 If K2 > K1 then the Maxwell equation written out above is a lower bound.
As we have mentioned, generalizations to more than two components is relatively straightforward.
The empirical equation
is known as Lictenecker’s equation and is commonly used when K1 and K2 are not too drastically different.17
Problems
4.1 According to the equation
K = 13 ∑m CmVmλm ,
the specific heat Cm can play an important role in determining the thermal conductivity K. (The sum over m means a sum over the modes m carrying the energy.) The total specific heat of a metal at low temperature can be represented by the equation
Cv = AT 3 + BT ,
where A and B are constants. Explain where the two terms come from.
4.2Look at Fig. 4.7 and Fig. 4.9 for the thermal conductivity of metals and insulators. Match the temperature dependences with the “explanations.” For (3) and (6) you will have to decide which figure works for an explanation.
(1) |
T |
(a) |
Boundary scattering of phonons |
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K = CV¯λ/3, and V¯, λ approximately constant. |
(2) |
T 2 |
(b) |
Electron–phonon interactions at low temperature |
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changes cold to hot electrons and vice versa. |
(3)constant
(4)T 3
(5)T neβ/T
(6)T 1
(c)Cv T.
(d)T > θD, you know ρ from Bloch (see Problem 4.4), and use the Wiedeman–Franz law.
(e)C and V¯ constant. The mean squared displacement of the ions is proportional to T and is also inversely proportional to the mean free path of phonons. This is high-temperature umklapp.
(f)Umklapp processes at not too high temperatures.
17Also of some interest is the variation in K due to inaccuracies in the input parameters
(such as K1, K2) for different models used for calculating K for a composite. See, e.g., Patterson [4.34].
264 4 The Interaction of Electrons and Lattice Vibrations
4.3Calculate the thermal conductivity of a good metal at high temperature using the Boltzmann equation and the relaxation-time approximation. Combine your result with (4.160) to derive the law of Wiedeman and Franz.
4.4From Bloch’s result (4.146) show that σ is proportional to T−1 at high temperatures and that σ is proportional to T−5 at low temperatures. Many solids show a constant residual resistivity at low temperatures (Matthiessen’s rule). Can you suggest a reason for this?
4.5Feynman [4.7, p. 226], while discussing the polaron, evaluates the integral
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I = ∫ |
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dq |
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q2 f (q) |
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(compare (4.112)) where |
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dq = dqx dqy dqz , |
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and |
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f (q) = |
2 |
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(2k q − q |
2 ) − ωL , |
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2m |
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by using the identity: |
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1 |
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dx |
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K1K2 |
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(1− x)]2 |
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0 [K a + K |
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a.Prove this identity
b.Then show the integral is proportional to
1 |
sin |
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K3k |
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k |
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and evaluate K3. |
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c. Finally, show the desired result: |
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Ek,0 = −αc ωL + |
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2m |
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where |
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m = |
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− |
αc |
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6 |
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and m* is the ordinary effective mass.
