1. What happens to the current if we: 1. add a magnetic field, 2
What happens to the current if we: add a magnetic field, 2. have an oscillating E field (e.g. light), 3. have a thermal gradient H

2. H field apply force to whole wire or just moving carriers?
Add a magnetic field H + - -
Last time :
dp/dt = -p(t)/ + f(t)
H field apply force to whole wire or just moving carriers?

3. Lorentz force = -ev/c x H
Hall Effect
In a current carrying wire when in a perpendicular magnetic field, the current should be drawn to one side of the wire. As a result, the resistance will increase and a transverse voltage develops.
Lorentz force = -ev/c x H H + - -

4. Lorentz force = -ev/c x H
Group
Similar approach:
dp/dt = -p(t)/ + f(t)
If a current flows (of velocity vD) in the positive x direction and a uniform magnetic field is applied in the positive z- direction, use the Lorentz force to determine the magnitude and direction of the resulting Hall field, first in terms of velocity, but then current density.
Ey= vxHy/c = jxHy/nec
Hall coefficient RH= Ey/jxH = 1/nec
R is very small for metals as n is very large.
Useful for calculating carrier density and type
Lorentz force = -ev/c x H
Ez=VxHy/c=jxHy/nec, RH=1/nec Hz - x y

5. The Hall coefficient Ohm’s law contains e2 But for RH the sign of e
is important.
so here we should get one
works fine for the alkalis
does not work at all for Bi (maybe the electron density is somehow wrong?)
but maybe even worse for Be - Al because the sign is wrong
so what if these had positive charge carriers?
A hole is the lack of an electron.
It has the opposite charge so +e.

6. Application:
For a 100-m thick Cu film, in a 1.0 T magnetic field and through which I = 0.5 A is passing, the Hall voltage is V.
For many years the Hall effect was not used in practical applications because the generated voltage in the gold film was extremely low. However, in the second half of the 20th century the mass production of semiconductor chips started. Chips based on the Hall effect became inexpensive and widely available.

7. Not yet prepared to discuss other quantum versions of the Hall effect
weak magnetic fields
Not yet prepared to discuss other quantum versions of the Hall effect
With strong magnetic fields:
The integer quantum Hall effect is observed in 2D electron systems at low temperature, in which the Hall conductance undergoes quantum Hall transitions to take on quantized values
The fractional quantum Hall effect: Hall conductance of 2D electrons shows precisely quantized plateaus at fractional values of e2/h
from D.C. Tsui, RMP (1999) and from H.L. Stormer, RMP (1999)

8. What happens to the electrons if we: have an oscillating E field (e. g
What happens to the electrons if we: have an oscillating E field (e.g. light)

9. plasmon: charge density oscillations
the plasma frequency has a simple interpretation: movement of all the electrons against all the ions’ positive background charge
at low frequencies, the electrons can follow any external field and keep it out of the solid - classical assumption for electrostatics
energy very high. How high is it for optical light? 1.5 eV or so.

10. Longitudinal Plasma Oscillations Equation of Motion: F = ma = -eE
Displacement of the entire electron gas a distance d with respect to the positive ion background. This creates surface charges s = nde & thus an electric field E = 4pnde.
Equation of Motion: F = ma = -eE
Oscillations at the
Plasma Frequency

11. plasmon: charge density oscillations
the plasma frequency has a simple interpretation: movement of all the electrons against all the ions’ positive background charge
at low frequencies, the electrons can follow any external field and keep it out of the solid - classical assumption for electrostatics
energy very high. How high is it for optical light? 1.5 eV or so.
values for the plasma energy

12. Why are metals shiny?
Drude’s theory gives an explanation of why metals do not transmit light and rather reflect it.
Continuum limit:
Where the wavelength is bigger than the spacing between atoms. Otherwise diffraction effects dominate. (Future topic)
Why might you want to think about optical properties? There are at least two reasons. One is that you can make use of known optical materials to design and build devices to manipulate light: mirrors, lenses, filters, polarizers, and a host of others. The second is that you can measure the optical response of a new material and obtain a wealth of information about the low energy excitations in the solid.
With the use of your own eyes, you could see that solids have a wide range of optical properties. Silver is a lustrous metal, with a high reflectance
over the whole visible range. Silicon is a crystalline semiconductor and the basis of modern electronics. With the surface oxide freshly etched off, silicon is also rather reflective, although not as good a mirror as silver. ∗ Salt (sodium chloride) is a transparent ionic insulator, is necessary for life, and makes up about 3.5% (by weight) of seawater. A crystal of salt is transparent over the entire visible spectrum; because the refractive index is about 1.5, the reflectance is about 4%.
If you had ultraviolet eyes, you would see these materials differently. Silver would be a
poor reflector, with at most 20% reflectance and trailing off to zero at the shortest wavelengths.
In contrast, the reflectance of silicon would be better than in the visible, reaching up
to 75%. Sodium chloride would be opaque over much of the spectrum, with a reflectance a
bit higher than in the visible. Those with infrared eyes would also see things differently from
visible or uv-sensitive individuals. Silver would have a reflectance above 99%. Silicon would
appear opaque at the shortest infrared wavelengths but would then become transparent, so
that you could see through even meter-thick crystals.† Sodium chloride remains transparent
over much of the infrared

13. AC Electrical Conductivity of a Metal
Newton’s 2nd Law Equation of
Motion for the momentum of one
electron in a time dependent
electric field. Look for a steady
state solution of the form:
AC conductivity
DC conductivity
Works great for the continuum limit when can treat the force on each electron the same.

14. When we have a current density, we can write Maxwell equations as:
-2E = i/c (4E/c -i  E/c)
To get a more thorough definition for the conductivity that applies for all wavelengths and to help understand the variance of absorption properties of materials with wavelength.
Since dell dot E =0
FYI: N=n+iK= square root of the dielectric tensor. The real part of N, n is the refractive index and the imaginary part κ
-2E = 2/c2 (1 +4 i / )E
( ) =1 + 4 i / 
Usual wave: -2E = 2 ()E/c2
x(xE) = -2E
x(xE) =  x
=  x i  H(,t) /c
= i /c  x H(,t)
x(xE) = -2E = i /c ( )
= -i  H(,t)
j = E

15. Plugging  into : ( ) =1 + 4 0i / (1-i )
From continuum limit
From Maxwell’s equations
( ) =1 + 4 i / 
Plugging  into : ( ) =1 + 4 0i / (1-i )
=1 + 4 0i / ( -i 2)
Do some of this on the board
That’s basically what we have in a metal according to the Drude model
Plugging in 0: ( ) =1 + 4 ne2i / m( -i 2)
For high frequencies can ignore first term in denominator
Ignoring: ( ) =1 - 4 ne2/m2
p is known as the plasma frequency
What is a plasma?

16. Application to Propagation of Electromagnetic Radiation in a Metal
The electromagnetic wave equation
in a nonmagnetic isotropic medium.
Look for a solution with the dispersion
relation for electromagnetic waves
-2E = 2 ()E/c2
(1) e real & > 0 → for w real, K is real & the transverse electromagnetic wave propagates with the phase velocity vph= c/e1/2
E/t= -i  E(,t)
2E/t2= 2 E(,t)
2E = -2 ()E/c2 =2 E

17. E&M waves propagate with no damping when e is positive & real
Transverse optical modes in a plasma
Dispersion relation for
electromagnetic waves
(1) e real & positive, no damping
(2) (p>) e real & < 0 → for w real, K is imaginary & the wave is damped with a characteristic length 1/|K|
(Why metals are shiny)
(3) e complex → for w real, K is complex & the wave is damped in space
(4) e = 0 longitudinally polarized waves are possible
w/wp
(2) (1)
E&M waves propagate with no damping when e is positive & real
E&M waves are totally reflected from the medium when e is negative

18. Ultraviolet Transparency of Metals
Plasma Frequency wp & Free Space Wavelength lp = 2pc/wp
Range Metals Semiconductors Ionosphere
n, cm
wp, Hz 5.7× × ×109
lp, cm 3.3× ×
spectral range UV IF radio
The reflection of
light from a metal
is similar to the
reflection of radio
waves from the
Ionosphere!
The Electron Gas is Transparent when w > wp i.e. l < lp
Plasma Frequency
Ionosphere
Semiconductors
Metals
Metals do a better job of reflecting x-rays than semiconductors
reflects transparent for
metal visible UV
ionosphere radio visible

19. What happens when you heat a metal
What happens when you heat a metal? What do we know from basic thermo (0th law)?
Heat it for a short time and then remove heat
Versus
Keep heating faster than can relax

20. Drude’s Best Success: Explanation of the Wiedemann-Franz law for metals (1853)
Thermal conductivity
Temperature
most convincing piece of evidence in Drude’s time
simple relation works because the electrons contribute most significantly to both el and th conduction. This works only because the phonons are not so important. And they are not because of the metallic bonding. The Debye temperatures are usually low and anharmonicity is important. So we are lucky here.
Electrical conductivity
Wiedemann and Franz observed that the ratio of thermal and electrical conductivity for ALL METALS is constant at a given temperature (for room temperature and above).
Later it was found by L. Lorenz that this constant is proportional to the temperature.
Let’s try to reproduce the linear behavior and to calculate the slope 20

21. Thermal conductivity A material's ability to conduct heat.
(je = I/A)
Electric current density
Heat current density
Fourier's Law for heat conduction. 
Thermal current density
 = Energy per particle
v = velocity
n = N/V 2l

22. About half the particles are moving right, and about half to the left.
Thermal conductivity
Heat current density
Heat Current Density jtot through the plane: jtot = jright - jleft
Heat energy per particle passing through the plane started an average of “l” away.
About half the particles are moving right, and about half to the left. x

23. Thermal conductivity
Heat current density x
Limit as l gets small:

24. Thermal conductivity x v v v

25. Thermal conductivity How does it depend on temperature?
Heat current density x
As you go to three dimensions
How does it depend on temperature?

26. Thermal conductivity 1/3 cvv2 1/3 cvmv2 ne2 ne2/m cvkBT ne2 = =
Drude applied ideal gas law ½ mv2 = 3/2 kBT
cvkBT
The book jumps through claiming a value for cv =
ne2

27. Classical Theory of Heat Capacity
When the solid is heated, the atoms vibrate around their sites like harmonic oscillators.
The average energy for a 1D oscillator is ½ kT.
Therefore, the average energy per atom, regarded as a 3D oscillator, is 3/2 kBT, and consequently the total energy is 3/2 nkBT where n is the conduction electron density and kB is Boltzmann constant.
Differentiation w.r.t temperature gives heat capacity 3/2 n kB
Got to this slide (27), but was running out of time
3/2 n kB2T = =
3kB2T / 2e2
ne2

28. Thermal conductivity optimization
To maximize thermal conductivity, there are several options:
Provide as many conduction electrons as possible
free electrons conduct heat more efficiently than phonons.
Make crystalline instead of amorphous
irregular atomic positions in amorphous materials scatter phonons and diminish thermal conductivity
Remove grain boundaries
gb’s scatter electrons and phonons that carry heat
Remove pores (air is a terrible conductor of heat)

29. Many open questions:
Why does the Drude model work so relatively well when many of its assumptions seem so wrong? In particular, the electrons don’t seem to be scattered by each other. Why?
Why do the electrons not seem to contribute to the heat capacity?
one thing we can understand right now is why Wiedemann Franz works more or less: we get rid of all the funny quantities, lambda, tau and so on.
From Wikipedia: "The simple classical Drude model provides a very good explanation of DC and AC conductivity in metals, the Hall effect, and thermal conductivity (due to electrons) in metals. The model also explains the Wiedemann-Franz law of 1853.
"However, the Drude model greatly overestimates the electronic heat capacities of metals. In reality, metals and insulators have roughly the same heat capacity at room temperature.“ It also does not explain the positive charge carriers from the Hall effect.

30. Failures of the Drude model: electrical conductivity of an alloy
Gitterstruktur und elektrisches Leitvermögen der Mischkristallreihen Au-Cu, Pd-Cu und Pt-Cu
Annalen der Physik
Volume 387, Issue 4, Date: 1927, Pages:
C. H. Johansson, J. O. Linde
the actual data is taken from
Johansson C. H. and Linde J. O., Ann. Phys., 25 (1936) 1.
The resistivity of an alloy should be between those of its components, or at least similar to them.
It can be much higher than that of either component.

31. More detail about Hall resistance
FYI: measurable quantity – Hall resistance
More detail about Hall resistance
for 3D systems
for 2D systems n2D=n
in the presence of magnetic field the
resistivity and conductivity becomes tensors
for 2D: