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\title{Chapter 9: Electronic Transport}
\author{Onsager}
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\maketitle
\tableofcontents
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\Large
As we have seen, transport in insulators (of heat
mostly) is dominated by phonons. The thermal conductivity of
some insulators can be quite large (cf. diamond). However most
insulators have small and uninteresting transport properties.
Metals, on the other hand, with transport dominated by
electrons generally conduct both heat and charge quite well. In
addition the ability to conduct thermal, charge, and entropy currents
leads to interesting phenomena such as thermoelectric effects.
\section{Quasiparticle Propagation}
In order to understand the transport of metals, we must understand how
the metallic state propagates electrons: ie., we must know the electronic
dispersion $\omega (\k )$.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{electron.pdf}}
\caption[]{\em{The dispersion is obtained from band structure
$\E (\k ) = \hbar \omega (\k )$ in which the metal
is approximated as an almost free gas of electrons interacting
weakly with a lattice potential $V(r)$, but not with each other.}}
\end{figure}
The dispersion is obtained from band structure $\E (\k ) = \hbar \omega (\k )$
in which the metal is approximated as an almost free gas of electrons
interacting weakly with a lattice potential $V(r)$, but not with each other.
The Bloch states of this system
\beq
\phi _k(\r ) = U_k(\r )e^{i \k \cdot \r }, \qquad U_k(\r ) = U_k(\r \ +
\r _n)
\eeq
may be approximated as plane waves $U_k(\r ) = U_k$. Then, the state
describing a single quasiparticle may be expanded.
\beq
\psi (x,t) = \frac 1{\sqrt{2\pi }}\int _{-\infty}^{\infty} d\k
U(\k )e^{i(\k \ \cdot \x \ - \omega (\k )t)}
\eeq
If $U(\k ) = c\delta(\k - \k_0)$ then $\psi (x,t) \propto e^{i(\k_0 \cdot \x
- \omega t)}$ and the quasiparticle is delocalized.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{fig2.pdf}}
\caption[]{\em{If $U(\k ) = c\delta(\k - \k_0)$ then $\psi (x,t) \propto e^{i(\k_0 \cdot \x
- \omega t)}$ and the quasiparticle is delocalized.}}
\end{figure}
On the other hand, if $U(\k)=\mbox{constant}$ then $\psi(x,t)\propto\delta (x)$
and the quasiparticle is perfectly localized. This is an expression of the uncertainty
principle
\beq
\Delta \k \ \Delta \x \ \sim \ 1 \quad \mbox{or} \quad \Delta \p \
\Delta \x \ \sim \ \hbar\,,
\eeq
so that we cannot know both the momentum and location of the quasiparticle
to arbitrary precision.
$\omega (\k ) \neq \mbox{constant}$, so the different components propagate
with different phase velocities, so the quasiparticle spreads as it propagates. This is also the reason why the group velocity of the quasiparticle is not
the phase velocity. Consider the propagation of $\psi (x,t)$ which when
$t = 0$.
\beq
\psi (\x,0) = \frac 1{\sqrt{2\pi }}\int _{-\infty }^{\infty } d\k
U(\k )e^{i\k \cdot \x }\,.
\eeq
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{U(k).pdf}}
\caption[]{\em{The distribution of plane wave state that make
up a quasiparticle.}}
\label{fig:U(k)}
\end{figure}
Suppose that $U(\k )$ has a well-defined dominant peak (See Fig.
~\ref{fig:U(k)}) so that
\beq
\omega (\k ) \simeq \ \omega (k_0) +
\left. \grad_\k\omega( \k) \right| _{\k_0}
\cdot(\k - \k_0)t
\eeq
then
\beq
\psi (x,t) \simeq \ \frac 1{\sqrt{2\pi }}\int _{-\infty }^{\infty } d\k
U(\k )e^{i\left( \k \cdot x - \omega _0t -
\left. \grad_\k\omega( \k) \right| _{\k_0}
(\k - \k_0)t\right) }
\eeq
\beqa
\psi (x,t) & \simeq \ & \frac{e^{i\left(\k_0\cdot\left. \grad_\k\omega( \k) \right| _{\k_0} - \omega _0\right)
t}}{\sqrt{2\pi }} \int_{-\infty }^{\infty } d\k
e^{i\k \cdot\left(
\x - \left. \grad_\k\omega( \k) \right| _{\k_0}t\right) }U(\k ) \nonumber \\
& \simeq & \psi \left( \x - \left. \grad_\k\omega( \k) \right| _{\k_0}t,0\right) e^{i\left( k_0 \left. \grad_\k\omega( \k) \right| _{\k_0} - \omega
_0\right) t}
\eeqa
Ie., aside from a phase factor, the quasiparticle travels along with velocity
$\left. \grad_\k\omega( \k) \right| _{\k_0} = v_g$. (If we had
considered higher order terms, we would have seen the quasiparticle distorts as it
propagates. (c.f.\ Jackson p.305). In general,
\beq
\v_g = \grad _k\omega (\k ) = \frac 1{\hbar }\grad _k E (\k )
\eeq
\subsection{Quasiparticle Equation of Motion and Effective Mass}
We are ultimately interested in the transport; i.e.\ the response of
this quasiparticle to an external electric field $\CE$ , from which it
gains energy.
\beq
\delta E = -e\CE \ \cdot \ v\delta t \qquad \mbox{(i.e. force $\times
$ distance)}
\eeq
This energy is reflected by the quasiparticle ascending to higher energy
$\k $ states.
\beq
\delta E = \grad _k E (\k ) \cdot \delta \k \ = \hbar \v \ \cdot \
\delta \k
\eeq
So
\beq
\hbar \delta \k \ = -e\CE \delta t
\eeq
\beq
\hbar \dot{\k } = -e\CE \qquad \mbox{E.O.M}
\eeq
This equation of motion is identical to that for free electrons
(c.f.\ Jackson); however, it may be shown to be applicable to
general Bloch states provided that $\CE$ is smaller than the atomic
fields, and it must not vary in space or time too fast.
We may put this EOM in a more familiar form
\beqa
\vd_i &=& \frac 1{\hbar } \frac{d}{dt} \left( \grad_k E (\k)\right) _i
= \frac 1{\hbar } \sum _j \frac{\partial ^2E }{\partial
k_i\partial k_j} \frac{dk_j}{dt}\nonumber \\
&=& \frac{1}{\hbar } \sum _j \frac{\partial ^2E }{\partial
k_i\partial k_j} \left( -e\CE _j\right)
\eeqa
This will have the form $\F \ = m\a $, if we define the mass tensor
\beq
\left( \frac 1{m^*}\right) _{ij} = \left( \frac 1{m^*}\right) _{ji} =
\frac 1{\hbar ^2}\frac{\partial E }{\partial k_i\partial k_j}
\qquad \mbox{symmetric \& real}
\eeq
which may be diagonalized to define three principle axes.
In the simple cubic case, the matrix will have the same element
along each principle direction and
\beq
m^* = \frac{\hbar ^2}{\frac{d^2E}{dk^2}}\,.
\eeq
In this way the effective mass of electrons on a lattice can vary
strongly, the larger $\frac{d^2E}{dk^2}$ is, the smaller $\frac{m^*}{m}$ is.
Consider the simple 1-d case (See Fig. ~\ref{fig:mass})
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{mass.pdf}}
\caption[]{}
\label{fig:mass}
\end{figure}
\section{ Currents in Bands}
Our previous discussion of the motion of an electron (or a
quasiparticle) in a metal under the influence of an applied field $\CE $, ignored the presence of other electrons and the Pauli principle.
\subsection{Current in an Insulator}
The Pauli principle insures that a {\bf{full band of states}} is
insulating. Consider the electric current due to $d^3\k $ states
\beq
d\J = -e\v(\k ) \left( \frac L{2\pi }\right) ^3 d^3\k
\eeq
The {\it current density} is then
\beq
d{\bf j} = \frac {-e}{\hbar }\grad _k E(\k )\frac 1{(2\pi )^3} d^3\k
\eeq
ie., different occupied states in the Brillouin zone contribute differently to
the current.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{current.pdf}}
\caption[]{\em{Different occupied states make different contributions
to the current density.}}
\end{figure}
The net current density $\j$ is then the integral over all occupied
states, which for our full band is the integral over the first
Brillouin zone
\beq
{\bf j} = -\frac{e}{8\pi ^3\hbar } \int_{\mbox{1st B.Z.}}
\grad_k E(\k ) d^3\k \,.
\eeq
Thus for each $\k $ vector in the integral, there is also $-\k $.
Now consider a lattice with inversion symmetry $\k \rightarrow -\k$
so that $ E(\k ) = E(-\k )$. Alternatively, recall that time-reversal
invariance requires that
\beq
E_{\uparrow }(\k ) = E_{\downarrow }(-\k )\,,
\eeq
but since $E_{\uparrow }(\k ) = E_{\downarrow }(\k )$ due to spin
degeneracy, we must have that $E(\k ) = E(-\k )$. Thus
\beq
\v_{-\k } = \frac 1{\hbar }\grad_{-\k}E(-\k ) = -\grad_\k E(\k ) =
-\v_\k \mbox{!}
\eeq
i.e., for the insulator
\beq
{\bf j} = \frac{-e}{8\pi ^3\hbar }\int_{\mbox{1st B.Z.}}
d^3\k \grad_\k E(\k ) \equiv \ 0
\eeq
Now imagine that the band is not full (See Fig.~\ref{fig:occstates}, left).
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{occstates.eps}}
\caption[]{\em{The Fermi sea of a partially filled band will
shift under the influence of an applied field \CE. This destroys the
inversion symmetry of the Fermi sea, causing a net current.}}
\label{fig:occstates}
\end{figure}
Then, if we apply an external field $\CE$, so that
\beq
\dot{\k } = -\frac{e\CE }{\hbar } \qquad e > 0 \mbox{!}
\eeq
the electrons will redistribute as the Fermi surface shifts
(See Fig.~\ref{fig:occstates} right).
\beqa
{\bf j} & = & \frac{-e}{8\pi ^3} \int _{\k \mbox{occupied}} \v (\k )
d^3\k \nonumber\\
& = & \frac{-e}{8\pi ^3} \int_{\mbox{1st B.Z.}} d^3\k\v (\k ) -
\frac{-e}{8\pi ^3} \int_{\mbox{empty}} d^3\k \v (\k )
\nonumber\\
& = & \frac{e}{8\pi ^3} \int_{\mbox{empty}} d^3\k \v (\k )
\eeqa
{\it Thus the current may be formally described as a current of
positive charge particles (holes) assigned to the unoccupied
states in the band.}
\subsection{Currents in a Metal}
Now imagine that the band is almost full.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{ltfull.pdf}}
\caption[]{\em{Left: A nearly full simple band. States near the Fermi surface
that can be thermally excited have negative mass. Right: Density of states
with holes at the top which have positive charge and mass.}}
\end{figure}
Near the top of the band $\frac{d^2E}{dk ^2} < 0$, so the mass is negative
and the dispersion at the top of the band is always also parabolic, so
\beq
E(\k ) = E_0 - \frac{\hbar ^2k^2}{2\left| m_{\mbox{\^{}}}^*\right| }
\qquad \mbox{k = deviation from top!}
\eeq
or
\beq
\dot{\v } = \frac 1{\hbar } \frac{d}{dt} \grad _k E_\k = -\frac{\hbar
\dot{\k }}{\left| m_{\mbox{\^{}}}^*\right| } = \frac{e\CE
}{\left| m_{\mbox{\^{}}}^*\right| }
\eeq
This is the EOM of a positively charged particle with positive
mass in an electric field $\CE $. {\it I.e., holes at the top of the band
have positive mass.}
We have just shown that a material with full bands is an insulator
(See Fig.~\ref{fig:bands1} left).
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{bands1.pdf}}
\caption[]{\em{An insulator form when the fermi energy falls
in a gap of $D(E)$. As the temperature is raised, electrons
are promoted over the gap, and both the electrons and holes contribute
to the conductivity which increases with temperature.}}
\label{fig:bands1}
\end{figure}
Ie., it carries no current, as least at $T = 0$ and for a small $\CE $.
However we ignored the presence of other bands. If there is a conductiong
band, for $T \neq \ 0$, and a reasonably small $E_g$, there will be
conductivity due to a small number of thermally excited holes and
electrons $n\sim \exp(-E_g/K_BT)$ (See Fig. ~\ref{fig:bands1} right).
Thus perhaps a better definition of an insulator is a material
for which the conductivity increases with $T$.
\section{ Scattering of Electrons in Bands}
According to the EOM for hole at the top of a band
\beq
\dot{\bf v} = \frac e{\left| m_{\mbox{\^{}}}^*\right| }\CE
\eeq
as long as $\CE $ is finite, these holes will continue to accelerate and
$\j$ will increase accordingly. Of course, this does not happen. Rather the
material simply heats up (ie., has a finite $R$). In
addition, if $\CE $ is returned to zero, then $\j$ likewise returns to
zero. Why?
In 1900 Drude assumed that the electrons scatter from the
lattice yielding resistivity.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.3\textwidth,clip=true]{lattice.pdf}}
\caption[]{\em{Drude thought that electrons scatter off the lattice
yielding resistivity. Bloch showed this to be wrong.}}
\end{figure}
Of course, as we have seen the quasiparticle state may be defined from a sum
over Bloch waves (described by $\k $) each of which is a stationary
state and describe the unperturbed propagation of electrons.
Thus a {\it perfect} lattice yields no resistivity. We can get
resistivity in two ways.
\begin{enumerate}
\item Deviations from a perfect lattice
\begin{enumerate}
\item Defects (See Fig. ~\ref{fig:lattice2}a)
\item Lattice vibrations = phonons (See Fig.
~\ref{fig:lattice2}b)
\end{enumerate}
\item Electron - electron interactions (See Fig. ~\ref{fig:interact2})
\end{enumerate}
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{lattice2.pdf}}
\caption[]{\em{Electrons do scatter from defects in the lattice
or lattice vibrations. They contribute to the resistivity, with
the phonon contribution increasing with temperature, and the
defect contribution more-or-less constant.}}
\label{fig:lattice2}
\end{figure}
Due to the strength of the electron-electron interaction and the density of
electrons, (2) should dominate. However, it is easy to show, using the
Pauli principle, that effect of (2) is quite often negligible, so that
we may return to regarding the {\it pure} electronic system as a
(perhaps renormalized) non-interacting Fermi gas.
According to momentum and energy conservation Fig ~\ref{fig:interact2}
\beq
E_1 + E_2 = E_3 + E_4\;\;\;\k _1 + \k _2 = \k _3 + \k _4\,.
\eeq
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{interact2.pdf}}
\caption[]{$E_1 + E_2 = E_3 + E_4$ and $\k _1 + \k _2 = \k _3 + \k _4$.
Electron-electron interactions also contribute to the resistivity (from simple
order of magnitude arguments based on relative strengths of the interactions,
their contribution should dominate--but due to the Pauli principle, it does
not).}
\label{fig:interact2}
\end{figure}
(Of course, momentum conservation is only up to a reciprocal lattice
vector $G, \k _1 + \k _2 = \k _3 + \k _4 = G$; however, as with phonon
conductivity, these processes with finite $G$ involve much higher energies, and
may be neglected near $T = 0$.) Furthermore, since all states up to $E_F$ are
occupied, $E_3; E_4 > E_F$! Suppose $E_1$ is (thermally) excited, so
$E_1 > E_F$ and it collides with an occupied state $E_2 < E_F$.
Then
\beq
(E_1 - E_F) + (E_2 - E_F) = (E_3 - E_F) + (E_4 - E_F) > 0
\eeq
\beq
\epsilon _1 + \epsilon _2 = \epsilon _3 + \epsilon _4 > 0, \qquad
\epsilon _3; \epsilon _4 > 0
\eeq
or $\epsilon _1 + \epsilon _2 > 0$, However, since $\epsilon _2 < 0$, if
$\epsilon _1$ is small, then $\left| \epsilon _2 \right| \ \leq \ \epsilon
_1$, is also small, so only states with
$\frac{\epsilon _2}{E_F} \leq \frac{\epsilon_1}{E_F}$
states may scatter with the state $\k _1$ conserve energy and obey the Pauli
principle, thus restricting $\epsilon _2$ to a narrow shell of width
$\epsilon _1$ around the Fermi surface.
Now consider the restrictions placed on the states 3 and 4
by momentum conservation.
\beq
\k _1 - \k _3 = \k _4 - \k _2
\eeq
I.e. $\k _1 - \k _3$ and $\k _4 - \k _2$ must remain parallel, and since
$\k_1$ is fixed, this restriction on the final states further reduces the
scattering probability by a factor of $\frac{\epsilon _1}{E_F}$.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.65\textwidth,clip=true]{chap6_logo.pdf}}
\caption[]{\em{Momentum and energy conservation severley restrict
the states that can an electron can scatter with and into.}}
\end{figure}
Thus the total scattering cross section $\sigma $ is reduced from the
classical result $\sigma_0 $ by $\left( \frac{\epsilon _1}{E_F}\right) ^2$.
If the initial excitation $\epsilon _1$ is due to thermal effects, then
$\epsilon_1 \sim \ k_BT$ and
\beq
\frac{\sigma }{\sigma _0} \sim \left( \frac{k_BT}{E_F}\right) ^2 \ll 1
!
\eeq
The total scattering due to electron - electron repulsion is {\it very} small.
Therefore, unless $E_F$ can be made small, the dominant contribution to a
material's resistivity is due to defects and phonons.
\section{ The Boltzmann Equation}
The nonequilibrium (but steady-state) situation of an electronic current
in a metal driven by an external field is described by the Boltzmann equation.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{interactions.pdf}}
\caption[]{\em{Electronic transport due to an applied field
$\CE$, is limited by inelastic collisions with lattice defects and phonons.}}
\end{figure}
This differs from the situation of a system in equilibrium
in that a constant deterministic current differs from random
particle number fluctuations due to coupling to a heat and
particle bath. Away from equilibrium ($\CE \ \neq \ 0$) the distribution
function may depend upon $\r $ and $t$ as well as $\k $ (or $E(\k )$).
Nevertheless, when $\CE = 0$ we expect the distribution
function of the particles in $V$ to return to
\beq
f_0(\k ) = f(\r, \k, t)\arrowvert _{\CE =0} = \frac 1{e^{\beta (E(\k) -
E_F)} + 1}
\eeq
As indicated,
To derive a form for $f(\r, \k, t)$, we will consider length scales
larger than atomic distances $\AA $, but smaller than distances in
which the field changes significantly. In this way the system is
considered essentially homogeneous with any inhomogeneity driven
by the external field. Now imagine that there is no scattering
(no defects, phonons), then since electrons are conserved
\begin{figure}[htb]
\centerline{\includegraphics[width=0.65\textwidth,clip=true]{elect.pdf}}
\caption[]{\em{In lieu of scattering, particles flow without decay.}}
\end{figure}
\beq
f(\r, \k, t) = f\left( \r - \v dt, \k \ + \frac{e\CE}{\hbar }dt, t -
dt\right)
\eeq
Now consider defects and phonons (See Fig.~\ref{fig:interactions2}) which can
scatter a quasiparticle in one state at $\r \ - \v \ dt$ and time $t - dt$,
to another at $\r $ and time $t$, so that
$f(\r, \k, t)\neq f\left(\r -\v dt,\k +\frac{e\CE}{\hbar }dt, t dt\right)$.
We will express this scattering by adding a term.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.3\textwidth,clip=true]{interactions2.pdf}}
\caption[]{\em{Scattering leads to quasiparticle decay.}}
\label{fig:interactions2}
\end{figure}
\beq
f(\r, \k, t) = f\left( \r \ - \v \ dt, \k \ + e\CE \frac{dt}{\hbar }, t -
dt\right) \ + \left( \frac{\partial f}{\partial t}\right) _S dt
\eeq
For small $dt$ we may expand
\beq
f(\r, \k, t) = f(\r, \k, t) - \v \ \cdot \ \grad _rf + e\CE \ \cdot \ \grad
_k \frac{f}{\hbar } - \frac{\partial f}{\partial t} + \left(
\frac{\partial f}{\partial t}\right) _S
\eeq
or
\beq
\frac{\partial f}{\partial t} + \v \ \cdot \ \grad _rf - \frac e{\hbar
}\CE \ \cdot \ \grad _kf = \left( \frac{\partial f}{\partial
t}\right) _S \qquad \mbox{Boltzmann Equation}
\eeq
If the phonon and defect perturbations are small, time-independent, and
described by $H$, then the scattering rate from a Bloch state $\k $ to
$\k ^{\prime }$ (occupied to unoccupied) is $w_{k^{\prime
}k} = \frac{2\pi }{\hbar } \left| \left< \k ^{\prime }\left| H\right| \k
\right> \right| ^2$. Then
\beqa
\left( \frac{\partial f(\k )}{\partial t}\right)_S = \left(
\frac{L}{2\pi }\right) ^3 \int d^3\k' && \left\{ \left( 1 -
f(\k )\right) w_{kk^{\prime }}f(\k ^{\prime })\right.\nonumber \\
&-&\left. \left( 1 -
f(\k ^{\prime })\right) w_{k^{\prime }k}f(\k )\right\}
\eeqa
Needless to say it is extremely difficult to solve these last two
coupled equations.
\subsection{Relaxation Time Approximation}
As a result we make a series of approximations and ansatze.
The first of these is the relaxation time approximation that the rate at
which a system returns to equilibrium $f_0$ is proportional to its
deviation from equilibrium
\beq
\left( \frac{\partial f}{\partial t}\right) _S = -\frac{f(\k ) - f_0(\k
)}{\tau (\k )}\,.
\eeq
Here $\tau (\k )$ is called the relaxation time (for a spatially
inhomogeneous system $\tau $ will also depend upon $\r $). Ie., we make the
assumption that scattering merely acts to drive a nonequilibrium system back to
equilibrium.
If $\CE \ \neq \ 0$ for $t<0$ and then at $t = 0$ it is switched off
so that for $t > 0$ $\CE \ = 0$, then for a homogeneous system
\beq
\frac{\partial f}{\partial t} = \left( \frac{\partial f}{\partial
t}\right) _S = -\frac{f - f_0}{\tau }
\eeq
so that
\beq
f - f_0 = \left( f(t=0) - f_0\right) e^{-\frac t{\tau }}
\eeq
ie., $\tau $ is the time constant at which the system returns to
equilibrium.
Now consider the steady-state situation of a metallic system
in a time-independent external field $\CE \ = \CE \hat{\x }$. Then
\beq
\frac{\partial f}{\partial t} = 0
\eeq
Furthermore since the system is homogeneous
\beq
\grad _rf = 0
\eeq
then
\beq
-\frac e{\hbar }\CE \cdot \grad _kf(\k ) = \left( \frac{\partial
f}{\partial t}\right) _S = -\frac{f(\k ) - f_0(\k )}{\tau (\k
)}
\eeq
ie
\beq
f(\k ) = \ f_0(\k ) + \frac e{\hbar }\tau (\k )\CE \ \cdot \ \grad
_k f(\k )
\eeq
which may be solved iteratively, generating a power series in $\CE $ (or $\CE
_x$).
\subsection{Linear Boltzmann Equation}
For small $\CE $ (Ohmic conditions)
\beq
f(\k ) \simeq \ f_0(\k ) + \frac e{\hbar }\tau (\k )\CE \ \cdot \ \grad
_k f_0(\k ) \qquad \mbox{linear Boltzmann Eqn.}
\eeq
I.e. the lowest order Taylor series of $f(\k )$. Or equivalently, if $\CE \ =
\CE _x\hat{\x }$
\beq
f(\k ) \simeq \ f_0\left( \k + \frac{e}{\hbar }\tau (\k )\CE \right)
\eeq
Ie., the effect is to shift the Fermi surface from its
equilibrium position by an amount
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{shift.pdf}}
\caption[]{\em{According to the linear Boltzmann equation, the effect
of a field $\CE_x$ is to shift the Fermi surface by $\delta k_x = -e\tau \frac{\CE _x}{\hbar }$}}
\end{figure}
\beq
\delta k_x = -e\tau \frac{\CE _x}{\hbar }
\eeq
From the discussion in Sec.??, it is clear that a finite current results.
Interesting! Note that elastic scattering $|\k | = |\k ^{\prime }|$ cannot
restore equilibrium. Rather they would only cause the Fermi surface to
expand. Inelastic scattering (i.e.\ from phonons) is needed to explain
relaxation.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{shift2.pdf}}
\caption[]{\em{ Note that elastic scattering $|\k | = |\k ^{\prime }|$ cannot
restore equilibrium. Rather they would only cause the Fermi surface to expand.}}
\end{figure}
\section{ Conductivity of Metals }
\subsection{Drude Approximation}
As mentioned above, Drude calculated the conductivity of metals
assuming that
\begin{itemize}
\item all free electrons participate, and
\item electron-lattice scattering yields a scattering rate $1/\tau$.
\end{itemize}
Under these assumptions, the EOM is
\beq
m\dot{v} + \frac m{\tau }\left( v - v_{\mbox{therm}}\right) \ = -e\CE
\eeq
where $v - v_{\mbox{therm}} = v_D$, the drift velocity, and $\frac m{\tau }v_D$
is friction. Again when $\CE \ = 0$, we again have an exponential decay of $v$
so $\tau $ is again the relaxation time.
In steady-state $\dot{v} = 0$
\beq
v_D = \frac{-e\tau }{m}\CE
\eeq
so that
\beq
j = -env_D = \frac{ne^2\tau }{m}\CE
\eeq
or defining $j = \sigma \CE $, and $\sigma \ = \mu ne$,
\beq
\sigma \ = \frac{ne^2\tau }m \qquad \mu \ = \frac{e\tau }m
\eeq
\subsection{Conductivity Using the Linear Boltzmann Equation}
Of course, this is wrong since all free electrons do {\it not}
participate in $\sigma $ due to the Pauli principle. And a more careful
derivation, using the Boltzmann Equation, is required. Again,
the relationship between ${\bf j}$ and $f(\k )$ is
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{relate.pdf}}
\caption[]{\em{To calculate the conductivity, we apply a
field in the x-direction only and use the linearized Boltzmann Eqn.}}
\end{figure}
\beqa
{\bf j} & = & \frac{-e}{8\pi ^3} \int d^3\k \ \v (\k )f(\k ) \\
& \simeq & \frac{-e}{8\pi ^3} \int d^3\k \ \v (\k )\left\{ f_0(\k ) +
\frac{e\tau (\k )}{\hbar }\CE _x \frac{\partial f_0}{\partial
k_x}\right\}
\eeqa
For an isotopic material $j_z = j_y = 0$, and the equation becomes scalar.
Furthermore, again
\beq
\int \v(\k )f_0(\k )d^3\k \ = 0
\eeq
since $\v _{-k} = -\v_k$. Then as
\beq
\frac{\partial f_0}{\partial k_x} = \frac{\partial f_0}{\partial E}
\frac{\partial E}{\partial k_x} = \frac{\partial f_0}{\partial
E} \hbar v_x
\eeq
\beq
j_x \simeq \ -\frac{e^2}{8\pi ^3}\CE _x \int d^3\k v_x^2\tau (\k
)\frac{\partial f_0}{\partial E}
\eeq
Then as $\frac{\partial f_0}{\partial E} \simeq \ -\delta (E - E_F)$ for $T \ll
E_F$ the integral in $\k $ is confined to the surface of constant $E$, and
\beq
d^3\k \ = dS_E dk_{\perp } = dS_E \frac{dE}{\hbar v(\k )}
\eeq
\begin{figure}[htb]
\centerline{\includegraphics[width=0.6\textwidth,clip=true]{surface.pdf}}
\caption[]{}
\end{figure}
then
\beqa
\sigma & = & \frac{j_x}{\CE _x} = \frac{e^2}{8\pi ^3\hbar } \int dS_E
dE \frac{v_x^2(\k )}{v(\k )}\tau (\k )\delta (E - E_F) \\
& = & \frac{e^2}{8\pi ^3\hbar } \int _{E=E_F} dS_E \frac{v_x^2(\k
)}{v(\k )}\tau (\k )\,.
\eeqa
As expected only the properties of the electrons on the Fermi surface are
relevant.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{shift3.pdf}}
\caption[]{\em{Only the electrons near the fermi surface participate
in the transport. Far below the Fermi surface, pairs of states $\k$ and
$-\k$ are occupied. Their contribution to the conductivity cancels,
leaving contributions to only the occupied states near the fermi surface.}}
\end{figure}
We can now calculate the conductivity of a metal by
averaging $ \frac{v_x^2}v \tau (\k )$ over the Fermi surface. Consider a
simple system with a spherical Fermi surface, then
\beqa
\int dS_E \frac{\tau(\k) v_x^2(\k )}{v(\k )} = \frac{4\pi }3
k_F^2\tau (E_F)v(E_F) \nonumber \\
= \frac{4\pi }3 k_F^2\tau (E_F)\frac{\hbar k_F}{m^*}
\eeqa
or
\beq
\sigma \ = \frac{e^2}{8\pi ^3\hbar } \frac{4\pi }3 k_F^2\tau
(E_F)\frac{\hbar k_F}{m^*}
\eeq
then as $k_BT \ll \ E_F, N = 2 \frac 43\pi \frac{k_F^3}{\left( \frac{2\pi }L
\right) ^3} \Rightarrow \ k_F^3 = 3\pi ^2n$ we find that
\beq
\sigma \ = \frac{e^2\tau (E_F)}{m^*}n \qquad \mu \ = \frac{e\tau
(E_F)}{m^*}
\eeq
For semiconductors where $n$ is $T$ dependent, and for more realistic
material where the Fermi surface $\neq $ sphere, the formula is more complicated.
However, for metals the temperature dependence of $\sigma $ is dominated
by that of $\tau $; ie., by the temperature dependence of phonons.
However, before we can calculate $\sigma (T)$, we must first disentangle
the phonon from the defect scattering. Assuming that the two
mechanisms are independent, they must add
\beq
\frac 1{\tau } = \frac 1{\tau _{\mbox{ph}}} + \frac 1{\tau
_{\mbox{defect}}}
\eeq
i.e.
\beq
\rho \ = \rho _{\mbox{ph}} + \rho _{\mbox{defect}} \qquad
\mbox{Matthiesen's Rule}
\eeq
The defect contribution is proportional to the defect cross section
$\Sigma_{\mbox{defect}}$ and the current, or $ v(E_F)$,
$\frac 1{\tau _{\mbox{defect}}} \propto \ \Sigma_{\mbox{defect}} v(E_F)$.
Is is roughly temperature independent, since the cross section $\Sigma _{\mbox{defect}}$ and $v(E_F)$ are.
The phonon contribution, on the other hand, is highly temperature
dependent since at zero temperature, there are no phonons. The scattering
cross section is roughly proportional to the rms phonon excursion
$\left< S^2(\q )\right> $. However, from the equipartition theorem
\beq
\frac 12M\omega _q^2 \left< S^2(\q )\right> \ = \frac{k_BT}2 \qquad T
\gg \ \theta _D\,.
\eeq
Thus
\beq
\frac 1{\tau _{\mbox{ph}}} \propto \ \left< S^2(\q )\right> \ \propto \
\frac{k_BT}{m\omega _q^2}
\eeq
Ie., at high temperatures, all modes contribute a linear in $T$ scattering
to $\frac 1{\tau_{\mbox{ph}}}$. Therefore, at $T \gg \ \theta_D $
($\theta_D =$ debye temperature)
\beq
\rho = aT + \rho _{\mbox{defect}}
\eeq
\begin{figure}[htb]
\centerline{\includegraphics[width=0.6\textwidth,clip=true]{graphs.pdf}}
\caption[]{\em{The phonon and defect contributions to the
resistivity add (left), and the phonon contribution is linear at
high temperatures $T \gg \ \theta_D $.}}
\end{figure}
\section{ Thermoelectric Effects }
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{thermal.pdf}}
\caption[]{\em{Thermoelctric effects are important in systems
with both electric potential and thermal gradients. We will assume both
are in the x-direction}}
\end{figure}
Until now, we have assumed that the transport system is
thermally homogeneous. Of course this need not be the case since
we can obviously maintain both an electrical and a thermal
current.
Here, each electron can carry a charge current $\sim \ e\v \ \sim \ e^2\CE $
and a thermal current $kT k\grad T$. In fact, a heat current can be used to
induce an electrical potential (Seebeck or thermoelectric effect) and,
conversely, an electric current can be used to move heat (Peltier
effect) which makes the solid state refrigeration possible.
\subsection{Linearized Boltzmann Equation}
To allow for a thermal gradient $\grad T$, our formalism must be
modified. Imagine that $\grad T$ and $\CE $ are fixed in time, then the
Boltzmann equation becomes
\beq
\frac{\partial f}{\partial t} + \v \cdot \ \grad _rf - \frac e{\hbar
}\CE \cdot \ \grad _kf = \left( \frac{\partial f}{\partial
t}\right) _S = -\frac{f(\k ) - f_0(\k )}{\tau (\k )}
\eeq
where in steady state $\frac{\partial f}{\partial t} \rightarrow \ 0$.
After linearizing (replacing $f$ by $f_0$ in the left-hand side), we get
\beq
f(\k ) \simeq \ f_0(\k ) - \tau (\k )\left\{ \v \cdot \ \grad _rf_0 -
\frac e{\hbar }\CE \cdot \ \grad _k f_0\right\}
\eeq
Then, as before
\beq
\frac e{\hbar }\CE \cdot \ \grad _k f_0 = \frac e{\hbar }\CE \cdot \
\frac{\partial f_0}{\partial E}\hbar \v \ = \frac e{\hbar }\CE
_x\frac{\partial f_0}{\partial E}\hbar v_x
\eeq
The spatial inhomogeneity is through $\grad T$, and in a semiconductor
for which $E_F$ depends strongly upon $T$, through $\grad E_F$
\beq
\v \cdot \ \grad _rf_0 = \v \cdot \ \left\{ \frac{\partial
f_0}{\partial T} \grad T + \frac{\partial f_0}{\partial
E_F}\grad E_F\right\} \ = \v \cdot \ \left\{ \frac{\partial
f_0}{\partial T} \grad T - \frac{\partial f_0}{\partial E}\grad
E_F\right\}
\eeq
Apparently $\grad E_F$ only contributes a term which modifies the electric
field dependence
\beq
v_x\frac{\partial f_0}{\partial E}\left\{ e\CE _x + (\grad
E_F)_x\right\} \ \equiv \ v_x\frac{\partial f_0}{\partial
E}e\CE _x^{\prime }
\eeq
Of course, in a metal $\CE ^{\prime } = \CE $.
\subsection{Electric Current}
Thus, we now have
\beq
f(\k ) = f_0(\k ) - \tau \frac{\partial f_0}{\partial
T}v_x\frac{\partial T}{\partial x} + \frac
e{\hbar}\tau \CE _x^{\prime }\hbar v_x\frac{\partial f_0}{\partial
E}
\eeq
Then for
\beqa
& j_x = & -\frac e{8\pi ^3} \int d^3\k v_x(\k )f(\k )\\
& j_x = & -\frac e{8\pi ^3} \int d^3\k v_x(\k )\left\{ f_0(\k ) -
\tau \frac{\partial f_0}{\partial T}v_x\frac{\partial
T}{\partial x} + \frac e{\hbar }\tau \CE
_x^{\prime }\frac{\partial f_0}{\partial E}\hbar
v_x \right\}\nonumber
\eeqa
recall that the last term yielded $\sigma $ last time (and still will)
\beq
j_x = \sigma \CE _x^{\prime } + \frac e{8\pi ^3} \int d^3\k
v_x^2\tau \frac{\partial f_0}{\partial T}\frac{\partial
T}{\partial x}
\eeq
Again we will calculate the second term assuming a spherical
Fermi surface. The term $\frac{\partial f_0}{\partial T}$ confines the
integral
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{confine.pdf}}
\caption[]{\em{The derivative $\frac{\partial f_0}{\partial E}$
is only significant near the fermi surface.}}
\end{figure}
to the Fermi sphere and so again effectively it amounts to a
Fermi-surface average, so $\bar{v_x^2} \rightarrow \frac 13\v ^2 \approx
\frac 2{3m^*} \frac 12 m^*\v ^2 = \frac 2{3m^*}E$
or changing to an integral over the DOS
\beq
j_x = \sigma \CE _x^{\prime } + \frac 23 \frac e{m^*} \int dE \tau
(E)ED(E)\frac{\partial f_0}{\partial T}\frac{\partial
T}{\partial x}
\eeq
Assuming that $\tau (E) \sim \ \tau (E_F)$, we get
\beq
j_x = \sigma \CE _x^{\prime } + \frac 23 \frac e{m^*} \tau
(E_F)\frac{\partial T}{\partial x} \int dE E\frac{\partial
f_0}{\partial T}D(E)
\eeq
\beq
j_x = \sigma \CE _x^{\prime } + \frac 23 \frac e{m^*} \tau
(E_F)\frac{\partial T}{\partial x}c_v(T) \qquad c_v \propto \
m^*
\eeq
In general, this intuitive form is rewritten as
\beq
j_x = \sigma \CE _x^{\prime } + \CL_{xx}^{12}\left( -\frac{\partial
T}{\partial x}\right)
\eeq
and from it we see that both an electric field (or the
generalized field strength $\CE ^{\prime }$), and the thermal gradient
contribute to the electron current, $j_x$.
\subsection{Thermal and Energy Currents}
Of course one can have a thermal current without having an
electric current (same number of electrons moving right and
left, but more of the hot ones moving right). Thermodynamics is
needed to quantify this though since these electrons will also
carry entropy as well as energy and heat.
Imagine that a small subsection of our material is in
thermal equilibrium and then some electrons are introduced/taken
away so that
\beq
dQ = TdS = dU - \mu dN \qquad \mbox{First Law of Thermodynamics}
\eeq
in terms of particle flow
\beq
\j _Q = \j _E - E_F\j _n \qquad -e\j _n = \j
\eeq
where this equation defines $\j _Q$, the thermal current, and
\beq
\j _E = \int \frac{d^3\k }{8\pi ^3}E(\k )\v (\k )f(\k ,\r )\,.
\eeq
Again one could work out the form of $\j _Q$ for the spherical Fermi surface
using the linearized Boltzmann equation. However one must obtain
a form like
\beqa
\j = \CL ^{11}\CE ^{\prime } + \CL ^{12}(-\grad T) \\
\j_Q = \CL ^{21}\CE ^{\prime } + \CL ^{22}(-\grad T)
\eeqa
(The fact that $\CL ^{12} = \CL ^{21}$ is referred as the Onsaser relation.)
These relationships between the $\CL $'s and the transport coefficients
depends upon what experiment is being done. For example in Fig.~\ref{fig:exp1}
there is a potential gradient ($V\neq 0$) but no thermal gradient since
$T_1 = T_2$. The electric field drives both electric and thermal currents.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.6\textwidth,clip=true]{exp1.pdf}}
\caption[]{\em{Here, there is a potential gradient ($V\neq 0$) but
no thermal gradient since $T_1 = T_2$. The electric field drives
both electric and thermal currents. Thus, a heat bath is required to
keep both sides of the sample at the same temperature.}}
\label{fig:exp1}
\end{figure}
\beqa
j &=&\CL^{11} \CE \nonumber \\
j_Q&=&\CL^{12} \CE
\eeqa
Thus, we may identify
\beq
\sigma \ = \frac{ne\tau }{m^*} = \CL ^{11}
\eeq
Note that since there is a thermal current induced by the potential gradient,
a heat bath is required to keep both sides of the sample at the same
temperature.
In Fig.~\ref{fig:exp2} we maintain a thermal gradient, but turn off the electric
current.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{exp2.pdf}}
\caption[]{Here $\j=0=\CL ^{11} \CE ^{\prime } + \CL ^{12}(-\grad T)$
and $\j _Q = \left( - \CL ^{12}\left( \frac{\CL ^{12}}{\CL ^{11}} \right) + \CL
^{22}\right) \left( -\grad T\right) $ where $- \CL ^{12}\left( \frac{\CL
^{12}}{\CL ^{11}} \right) + \CL ^{22} = \kappa T$}
\label{fig:exp2}
\end{figure}
Here,
\beq
\j=0=\CL ^{11} \CE ^{\prime } + \CL ^{12}(-\grad T)
\eeq
and
\beq
\j _Q = \CL ^{21} \CE ^{\prime } + \CL ^{22}(-\grad T) =
\left( - \CL ^{21}\left( \frac{\CL ^{12}}{\CL ^{11}} \right) + \CL
^{22}\right) \left( -\grad T\right)
\eeq
and since (you will show)
\beq
\CL ^{12} = -\frac{2e\tau }{3m^*}c_v = \CL ^{21}
\eeq
\beq
\kappa \ = \CL ^{22} - \frac{\left( \CL
^{12}\right) ^2}{\CL ^{11}}
\eeq
We could also measure the thermal conductivity by driving a heat current
through the sample, maintaining the ends at the same potential
(see Fig.~\ref{fig:exp3} right).
\begin{figure}[htb]
\centerline{\includegraphics[width=0.6\textwidth,clip=true]{exp3.pdf}}
\caption[]{\em{Two methods for measuring $\kappa$.}}
\label{fig:exp3}
\end{figure}
Here, we would find
\beq
\kappa = \CL ^{22}\,.
\eeq
Thus, we can identify
\beq
\kappa \ = \CL ^{22} \mbox{ or } \CL ^{22} - \frac{\left( \CL
^{12}\right) ^2}{\CL ^{11}}
\eeq
depending upon the experiment.
These are the same if the sample is a good metal where $\CL ^{11} =\sigma$ is
large. Many experimentalists measure
$\kappa $ by the method of the left of Fig.~\ref{fig:exp3}.
This yields the more conventional definition of $\kappa $.
\subsection{Seebeck Effect, Thermocouples}
These relations result in some interesting physical effects. Consider a
bimetallic conducting loop with two junctions maintained at temperatures $T_1$
$T_2$. Let metal A be different than B, so that $\CL _A^{ij} \neq \ \CL _B^{ij}$. If no current flows around the loop, then
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{conductlp.pdf}}
\caption[]{\em{A bimetallic conducting loop with junctions maintained
at $T_1$ and $T_2$. If $T_1\neq T_2$, and $\CL_A^{12}/\CL_A^{11}\neq
\CL_B^{12}/\CL_B^{11}$, then the heat current induces a potential
$V\propto T_2-T_1$ }}
\end{figure}
\beq
\j \ = 0 = \CL ^{11}\CE _x + \CL ^{22}(-\grad T) \Rightarrow \ \CE _x =
\left( \frac{\CL ^{12}}{\CL ^{11}}\right) \frac{dT}{dx}
\eeq
where $S = \frac{\CL ^{12}}{\CL ^{11}}$ is called the thermopower and is a
property of a material.
The potential measured around the loop is given by
\beq
V = \int _0^1 \CE _B dx + \int _1^2 \CE _A dx + \int _2^0 \CE _B dx
\eeq
or
\beqa
V & = & S_B\left\{ \int _0^1 \frac{\partial T}{\partial x} dx + \int
_2^0\frac{\partial T}{\partial x} dx\right\} \ + S_A\int _1^2
\frac{\partial T}{\partial x} dx\\
& = & S_B \int _2^1 \frac{\partial T}{\partial x} dx + S_A \int _1^2
\frac{\partial T}{\partial x} dx \\
& = & \left( S_A - S_B\right) \int_{T_1}^{T_2} dT = \left( S_A -
S_B\right) \left( T_2 - T_1\right)
\eeqa
Or if $S_A$ and $S_B$ are not $T$-independent $V = \int_{T_1}^{T_2} dT \left( S_A - S_B\right) $.
So if $T_1 \neq \ T_2$ and $S_A \neq \ S_B$, then the heat current induces an
emf! This is called the Seebeck effect $\Rightarrow $ (solid state thermometer
with ice $H_2O$ as a reference).
\subsection{Peltier Effect}
Now consider the inverse situation
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{conductlp2.pdf}}
\caption[]{\em{An electrical current $\j $ is driven through the loop
which is held at a fixed temperature}}
\end{figure}
where an electrical current $\j $ is driven through the loop which is
held at a fixed temperature $\left( \frac{\partial T}{\partial x} = 0\right)
$. Then
\beq
\j _Q = \CL ^{21}\CE \qquad \j \ = \CL ^{11}\CE
\eeq
\beq
\j _Q = \left( \frac{\CL ^{21}}{\CL ^{11}}\right) \j \ = \pi \j
\eeq
This is known as the Peltier effect whereby heat is carried from
one junction to the other
\begin{figure}[htb]
\centerline{\includegraphics[width=0.6\textwidth,clip=true]{conductlp3.pdf}}
\caption[]{\em{If $dT/dx=0$ and
$\pi_A=\CL_A^{21}/\CL_A^{11}\neq\CL_B^{21}/\CL_B^{11}=\pi_B$
then the electric current also induces a heat from one junction to another.}}
\end{figure}
or an electric current is accompanied by a heat current. One may
use this effect to create an extremely simple (and similarly
inefficient) refrigerator.
\section{ The Wiedemann-Franz Law (for good metals)}
One may independently measure the thermal $\kappa $ and electrical
$\sigma $ conductivities.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.6\textwidth,clip=true]{exp4.pdf}}
\caption[]{\em{The thermal and electrical conductivities may
be measured independently.}}
\end{figure}
However, in general one expects that $\kappa \ \propto \ \sigma T$ since in
electrical conduction each election carries a charge $e$ and is acted on by a
force $-e\CE $. The current per unit electric field proportional to $e^2$. In
thermal conduction each electron carries a thermal energy $k_BT$ and is acted
on by a thermal force $-k_B\grad T$. The heat current per unit thermal
gradient is proportional to $k_B^2T$, thus one expects
\beq
\frac{\kappa }{\sigma } \propto \ \frac{k_B^2}{e^2}T
\eeq
Due to the simplicity of these arguments, our formalism
should reproduce this relationship. As we discussed before
\beqa
\j _Q & = & \j _E - E_F\j _n \\
& = & \frac 1{8\pi ^3}\int d^3\k \ (E- E_F)\v (\k )f(\k )
\eeqa
In the linear approximation to the Boltzmann equation for $\CE _x^{\prime } =
0$, we get
\beq
\j _Q = \frac 1{8\pi ^3}\int d^3\k \ (E- E_F)\frac{\partial
f_0}{\partial T} v_x^2\tau \left( -\frac{\partial T}{\partial
x}\right)
\eeq
where $E- E_F$ and $\frac{\partial f_0}{\partial T}$ are odd in $(E - E_F)$,
for the Fermi liquid
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{plot.pdf}}
\caption[]{\em{The function $(E- E_F)\frac{\partial f_0}{\partial T}$
is sharply peaked at the fermi surface and even. in $E- E_F$.}}
\end{figure}
\beq
\j _Q \simeq \ -\left( \frac{\partial T}{\partial x}\right) \frac
13\v_F^2\tau _F \int \ \frac{d^3\k }{3\pi ^3}(E -
E_F)\frac{\partial f_0}{\partial T}
\eeq
\beq
\j _Q = \left( -\frac{\partial T}{\partial x}\right) \frac 13\v_F^3\tau
_Fc_v
\eeq
\beq
\kappa \ = \frac 13\v _F^3\tau _Fc_v
\eeq
Now recall that for the Fermi liquid $c_v = k_B\frac{\pi^2}{2}
nk_B\frac{T}{k_BT_F}$ so that
\beq
\kappa = \frac 13\frac{m^*v_F^2}{m^*}\tau _F k_B\frac{\pi
^2}{2}nk_B\frac T{E_F} = \frac{\pi ^2}3\tau
_Fn\frac{k_B^2T}{m^*}
\eeq
However, also for the Fermi liquid, we found that $\sigma \ = e^2\tau _F\frac
n{m^*}$, so
\beq
\frac{\kappa }{\sigma } = \frac {\pi ^2}3\left( \frac{k_B}e\right) ^2T
\eeq
Of course this relationship only holds in a good metal. There are two reasons
for this. First we are neglecting terms like $\frac{\left( \CL ^{12}\right)
^2}{\sigma }$ in $\kappa $ which are unimportant for a good metal (or if we
electrically short the sample.) Second, we are assuming that $\kappa $ is
dominated by electronic transport.
\edo