# Oom Lecture 06

### Thermal Conductivity

• The thermal flux in metals is carried by mobile free electrons, and this makes them much better heat conductors than the insulators we have discussed previously.
{\begin{aligned}F&=-\kappa \nabla U_{thermal,e^{-}}\\&=-\kappa \nabla n_{e}\left({\frac {3}{2}}k{kT \over E_{F}}\right)T\\&=-{\kappa \cdot 3k^{2}Tn_{e}\nabla T \over E_{F}}\\&=-k_{c}\nabla T\\\end{aligned}}\,\! Thus the thermal conductivity is $k_{c}\equiv {3\kappa k^{2}Tn_{e} \over E_{F}}$ . We can estimate some thermal conductivities. We’ll write $\kappa$ as $\lambda _{e}v_{e} \over 3$ , with the factor of 3 reducing the diffusivity for 3 dimensions (it takes longer to randomly get somewhere than in 1D). Thus, the important factors are:

{\begin{aligned}(1)&\lambda _{e}\\(2)&v_{e}\\(3)&k{kT \over E_{F}}\\(4)&n_{e}\\\end{aligned}}\,\! In comparison to insulators (4) is the same ($n_{e}=n_{ions}$ ), (3) is lower by ${kT \over E_{F}}\sim {1 \over 300}$ , (2) $v_{e}$ is a lot bigger than in conductors ($v_{e}\sim {\sqrt {2E_{F} \over m_{e}}}\sim 1000{km \over s}$ ). This makes (2) higher by about a factor of 300. Finally, (1), the mean-free-path of an electron in a solid, must be much larger than the mfp of a phonon (as we know by the fact that metals are excellent thermal conductors). If $a$ is the spacing between ions, Kittel pg. 302-304 tells us that

$\lambda _{e}\sim {\lambda _{deBroglie}^{2} \over {\bar {d}}^{2}}a\,\!$ where ${\bar {d}}^{2}$ is the rms thermal displacement of an ion (their jiggling around). If all the ions were on a perfect lattice, a plane-wave of electrons would never scatter. An ad hoc rational for the above equation is as follows: each step in $a$ changes $\lambda _{deBroglie}$ by $\pm d$ . Thus, the number of steps for $\lambda _{deBroglie}$ to change by order unity is $\left({\lambda _{deBroglie} \over d}\right)^{2}$ . We can estimate this factor as:

${\lambda _{deBroglie}^{2} \over {\bar {d}}^{2}}\sim {\left(n_{e}^{-{\frac {1}{3}}}\right)^{2} \over {{\frac {1}{2}}m_{ion}\omega ^{2}\left\langle d^{2}\right\rangle \sim {\frac {3}{2}}kT}}\,\!$ Now $\omega ^{2}=(2\pi )^{2}\nu _{max}=(2\pi )^{2}\left({c_{e} \over a}\right)^{2}$ , and $c_{e}\sim {\sqrt {M_{bulk} \over \rho }}$ , and $M_{bulk}\sim {E_{c} \over a^{3}}$ , so we end up with

${\lambda _{deBroglie}^{2} \over {\bar {d}}^{2}}\sim {(2\pi )^{2} \over 3}{E_{c} \over kT}\,\!$ This means that the mean free path of an electron propagating through a lattice is $\sim 600a$ (several hundred lattice spacings). Thus (1) is higher by a factor of 200:

{\begin{aligned}k_{c,insulator}&\sim 10^{-2}{cal \over cm\cdot s\cdot K}\\k_{c,metal}&\sim 2\\\end{aligned}}\,\! And as it turns out, Cu and Al have $k_{c}=1$ , and Fe has $k_{c}=0.2$ . On the other hand, liquid Hg has $k_{c}=.01$ because liquids have their long-range order broken, scattering electrons before thermal wiggling does.

• Notice also that though $\lambda _{e}\propto {1 \over T}$ , (3) includes a factor of $T$ , so thermal conductivity should be temperature-independent, which it is until at low energies there aren’t any phonons to scatter electrons, and then instead scatter off of the same impurities that scatter phonons in insulators.

### Electrical Conductivity (Resistivity)

• The current through a wire of cross-section A, length L, and voltage V is
$I={VA \over L}\sigma _{e}\,\!$ where $\sigma _{e}$ is the electrical conductivity, $\sigma _{e}\equiv {1 \over \rho _{e}}$ . Now $J={I \over A}$ is the current density, ${V \over L}=E$ is the electric field, and $J=\sigma _{e}E$ .

• Electrons encounter resistance from ions: $I\equiv A\cdot J=A\cdot n_{e}\cdot e\cdot v_{drift}$ . The drift velocity is much smaller than the average forward velocity. $v_{drift}\sim at_{free}\sim {eE \over m_{e}}{\lambda _{e} \over v_{forward}}$ . Thus:
$\sigma _{e}\sim {e^{2}n_{e}\lambda _{e} \over m_{e}v_{F}}\sim 8\cdot 10^{17}\left({T_{room} \over T}\right)s^{-1}\,\!$ $1s=9\cdot 10^{9}ohm\cdot m$ , so

${\rho _{e}\sim 1\cdot 10^{-8}ohm\cdot m}\,\!$ This estimate is a little low: Al is $3\cdot 10^{-8}ohm\cdot m$ , Cu is 2, Fe is 10, and liquid Hg is 100.

### Magnetic Diffusivity

• Magnetic fields decay because currents decay. This decay is governed by Ohmic diffusion. Ampere’s law says:
${\vec {\nabla }}\times {\vec {B}}={4\pi {\vec {J}} \over c}\,\!$ and Ohm says:

${\vec {J}}=\sigma _{e}{\vec {E}}\,\!$ and this gives us that

${\vec {\nabla }}\times ({\vec {\nabla }}\times {\vec {B}})={4\pi \sigma _{e} \over c}{\vec {\nabla }}\times {\vec {E}}\,\!$ Using Faraday’s law and the fact that there are no magnetic monopoles:

${{\partial {\vec {B}} \over \partial t}={c^{2} \over 4\pi \sigma _{e}}\nabla ^{2}{\vec {B}}}\,\!$ and $\kappa _{m}\equiv {c^{2} \over 4\pi \sigma _{e}}$ is the magnetic diffusivity.

• Thus, the Ohmic decay time is of order:
$t_{decay}\sim {L^{2} \over \kappa _{M}}\,\!$ where L is the length scale over which B changes.

### Fluid Mechanics

• Navier Stokes
${d{\vec {v}} \over dt}={\partial {\vec {v}} \over \partial t}+\underbrace {(v\cdot \nabla ){\vec {v}}} _{advective \atop term}=-\nabla \Phi -{\nabla P \over \rho }+\underbrace {\nu \nabla ^{2}{\vec {v}}} _{viscousdrag}\,\!$ We’ll assume $\rho$ is constant $(\nabla \cdot {\vec {v}}=0)$ , and that we are moving sub-sonically.