# Oom Lecture 06

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### 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.
{\displaystyle {\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 ${\displaystyle k_{c}\equiv {3\kappa k^{2}Tn_{e} \over E_{F}}}$. We can estimate some thermal conductivities. We’ll write ${\displaystyle \kappa }$ as ${\displaystyle \lambda _{e}v_{e} \over 3}$, 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:

{\displaystyle {\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 (${\displaystyle n_{e}=n_{ions}}$), (3) is lower by ${\displaystyle {kT \over E_{F}}\sim {1 \over 300}}$, (2) ${\displaystyle v_{e}}$ is a lot bigger than in conductors (${\displaystyle v_{e}\sim {\sqrt {2E_{F} \over m_{e}}}\sim 1000{km \over s}}$). 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 ${\displaystyle a}$ is the spacing between ions, Kittel pg. 302-304 tells us that

${\displaystyle \lambda _{e}\sim {\lambda _{deBroglie}^{2} \over {\bar {d}}^{2}}a\,\!}$

where ${\displaystyle {\bar {d}}^{2}}$ 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 ${\displaystyle a}$ changes ${\displaystyle \lambda _{deBroglie}}$ by ${\displaystyle \pm d}$. Thus, the number of steps for ${\displaystyle \lambda _{deBroglie}}$ to change by order unity is ${\displaystyle \left({\lambda _{deBroglie} \over d}\right)^{2}}$. We can estimate this factor as:

${\displaystyle {\lambda _{deBroglie}^{2} \over {\bar {d}}^{2}}\sim {\left(n_{e}^{-{\frac {1}{3}}}\right)^{2} \over {{\frac {1}{2}}m_{ion}\omega ^{2}\left\langle d^{2}\right\rangle \sim {\frac {3}{2}}kT}}\,\!}$

Now ${\displaystyle \omega ^{2}=(2\pi )^{2}\nu _{max}=(2\pi )^{2}\left({c_{e} \over a}\right)^{2}}$, and ${\displaystyle c_{e}\sim {\sqrt {M_{bulk} \over \rho }}}$, and ${\displaystyle M_{bulk}\sim {E_{c} \over a^{3}}}$, so we end up with

${\displaystyle {\lambda _{deBroglie}^{2} \over {\bar {d}}^{2}}\sim {(2\pi )^{2} \over 3}{E_{c} \over kT}\,\!}$

This means that the mean free path of an electron propagating through a lattice is ${\displaystyle \sim 600a}$ (several hundred lattice spacings). Thus (1) is higher by a factor of 200:

{\displaystyle {\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 ${\displaystyle k_{c}=1}$, and Fe has ${\displaystyle k_{c}=0.2}$. On the other hand, liquid Hg has ${\displaystyle k_{c}=.01}$ because liquids have their long-range order broken, scattering electrons before thermal wiggling does.

• Notice also that though ${\displaystyle \lambda _{e}\propto {1 \over T}}$, (3) includes a factor of ${\displaystyle T}$, 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
${\displaystyle I={VA \over L}\sigma _{e}\,\!}$

where ${\displaystyle \sigma _{e}}$ is the electrical conductivity, ${\displaystyle \sigma _{e}\equiv {1 \over \rho _{e}}}$. Now ${\displaystyle J={I \over A}}$ is the current density, ${\displaystyle {V \over L}=E}$ is the electric field, and ${\displaystyle J=\sigma _{e}E}$.

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

${\displaystyle 1s=9\cdot 10^{9}ohm\cdot m}$, so

${\displaystyle {\rho _{e}\sim 1\cdot 10^{-8}ohm\cdot m}\,\!}$

This estimate is a little low: Al is ${\displaystyle 3\cdot 10^{-8}ohm\cdot m}$, 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:
${\displaystyle {\vec {\nabla }}\times {\vec {B}}={4\pi {\vec {J}} \over c}\,\!}$

and Ohm says:

${\displaystyle {\vec {J}}=\sigma _{e}{\vec {E}}\,\!}$

and this gives us that

${\displaystyle {\vec {\nabla }}\times ({\vec {\nabla }}\times {\vec {B}})={4\pi \sigma _{e} \over c}{\vec {\nabla }}\times {\vec {E}}\,\!}$

Using Faraday’s law and the fact that there are no magnetic monopoles:

${\displaystyle {{\partial {\vec {B}} \over \partial t}={c^{2} \over 4\pi \sigma _{e}}\nabla ^{2}{\vec {B}}}\,\!}$

and ${\displaystyle \kappa _{m}\equiv {c^{2} \over 4\pi \sigma _{e}}}$ is the magnetic diffusivity.

• Thus, the Ohmic decay time is of order:
${\displaystyle t_{decay}\sim {L^{2} \over \kappa _{M}}\,\!}$

where L is the length scale over which B changes.

### Fluid Mechanics

• Navier Stokes
${\displaystyle {d{\vec {v}} \over dt}={\partial {\vec {v}} \over \partial t}+\underbrace {(v\cdot \nabla ){\vec {v}}} _{advective \atop term}=-\nabla \Phi -{\nabla P \over \rho }+\underbrace {\nu \nabla ^{2}{\vec {v}}} _{viscousdrag}\,\!}$

We’ll assume ${\displaystyle \rho }$ is constant ${\displaystyle (\nabla \cdot {\vec {v}}=0)}$, and that we are moving sub-sonically.