Download Hall Effect, AC Conductivity and Thermal Conductivity

What happens to the current if we:
1. add a magnetic field,
2. have an oscillating E field (e.g. light),
3. have a thermal gradient
H
Add a magnetic
field H
-
-
+
Last time :
dp/dt = -p(t)/ + f(t)
2
H field apply force to whole
wire or just moving carriers?
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.
H
Lorentz force =
-ev/c x H
------------------
++++++++++++++++
-
+
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 zdirection, use the Lorentz force to determine the
magnitude and direction of the resulting Hall field, first
in terms of velocity, but then current density.
Group
•
•
•
•
Ey= vxHy/c = jxHy/nec
Hall coefficient RH= Ey/jxH = 1/nec
Lorentz force =
-ev/c x H
R is very small for metals as n is very large.
Useful for calculating carrier density and type
Hz
-----------------++++++++++++++++
-
x
y
The Hall coefficient
Ohm’s law contains e2
But for RH the sign of e
is important.
A hole is the lack of an electron.
It has the opposite charge so +e.
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 0.737 V.
H  
H
nec
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
from D.C. Tsui, RMP (1999) and from H.L. Stormer, RMP (1999)
The fractional quantum Hall effect:
Hall conductance of 2D electrons
shows precisely quantized plateaus at
fractional values of e2/h
What happens to the electrons if we:
have an oscillating E field (e.g. light)
plasmon: charge density oscillations
9
Longitudinal
Plasma Oscillations
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 = 4nde.
Equation of Motion: F = ma = -eE
d
Nm 2   NeE   Ne(4nde)
t
2
Oscillations at the
d
2
Plasma
Frequency


d

0
p
2
t
2
4ne
2
p 
2
m
plasmon: charge density oscillations
values for the plasma energy
11
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)
AC Electrical Conductivity of a Metal
Newton’s 2nd Law Equation of
dp( , t )
p( , t )

 eE( , t )
Motion for the momentum of one
dt

electron in a time dependent
E( , t )  E( )e it
electric field. Look for a steady
p( , t )  p( )e it
state solution of the form:
eE( )
p( ) 
j( )  s ( )E( )
AC conductivity
DC conductivity
s0
s ( ) 
1  i
ne 2
s0 
1 /   i
nep( ) ( ne 2 / m )E( )
j( )  

m
1 /   i
m
Works great for the continuum limit when can treat
the force on each electron the same.
When we have a current density, we can write Maxwell
equations as:
-2E = i/c (4sE/c -i  E/c)
-2E = 2/c2 (1 +4 si / )E
( ) =1 + 4 si / 
Usual wave: -2E = 2 ()E/c2
x(xE) = -2E
x(xE) =  x
= -i  H(,t)
14
=  x i  H(,t) /c = i /c  x H(,t)
x(xE) = -2E = i /c (
j = sE
)
From continuum limit
s0
ne 
s ( ) 
,s 0 
1  i
m
From Maxwell’s equations
2
( ) =1 + 4 si / 
Plugging s into : ( ) =1 + 4 s0i / (1-i )
=1 + 4 s0i / ( -i 2)
Plugging in s0: ( ) =1 + 4 ne2i / m( -i 2)
For high frequencies 
 1can ignore first term in denominator
Ignoring: ( ) =1 - 4 ne2/m2
p is known as the plasma frequency
15
What is a plasma?
Application to Propagation of Electromagnetic Radiation in a Metal
E  exp( it  iK  r )
The electromagnetic wave equation
in a nonmagnetic isotropic medium.
2
2
2

(

,
K
)

/
c

K
Look for a solution with the dispersion
2E = 2 ()E/c2
-
relation for electromagnetic waves
(1)  real & > 0 → for  real, K is real & the
transverse electromagnetic wave propagates with the
phase velocity vph= c/1/2
E/t= -i  E(,t)
2E/t2= 2 E(,t)
2E = -2 ()E/c2 =2 E
Transverse optical modes in a plasma
Dispersion relation for
 (, K ) 2  c2 K 2
electromagnetic waves
 2  p 2  c2 K 2
(1)  real & positive, no damping
(2) (p>)  real & < 0 → for  real, K
 ( )
is imaginary & the wave is damped with
a characteristic length 1/|K|
(Why metals are shiny)
E e
Kr
(3)  complex → for  real, K is
complex & the wave is damped in space
(4)  = 0 longitudinally polarized waves
are possible
 / p
(2)
E&M waves are totally
reflected from the
medium when  is
negative
(1)
E&M waves
propagate with
no damping
when  is
positive & real
Ultraviolet Transparency of Metals
Plasma Frequency p & Free Space Wavelength lp = 2c/p
Range
Metals
n, cm-3
1022
p, Hz
5.7×1015
lp, cm
3.3×10-5
spectral range UV
Semiconductors
1018
5.7×1013
3.3×10-3
IF
Ionosphere
1010
5.7×109
33
radio
The Electron Gas is Transparent when  > p i.e. l < lp
p
2
4ne

m
2
The reflection of
light from a metal
is similar to the
reflection of radio
waves from the
Ionosphere!
Plasma Frequency
Ionosphere
Semiconductors
Metals
reflects transparent for
metal
visible
UV
ionosphere radio
visible
What happens when you heat a metal?
What do we know from basic thermo
19
th
(0
law)?
Drude’s Best Success:
Explanation of the Wiedemann-Franz law for
metals (1853)
Thermal conductivity
Temperature
Electrical conductivity
•
•
•
20
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
Thermal conductivity
A material's ability to conduct heat.
Electric current density
(je = I/A)
Heat current density
Fourier's Law for heat conduction.
 
E
 Thermal current density
jt  

sec

area


jt   vn
 = Energy per particle
v = velocity
n = N/V
2l
Thermal conductivity
Heat current density
jt   vn
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
Thermal conductivity
Heat current density
x
Limit as l gets small:
Thermal conductivity
x
v
v
v
Thermal conductivity
Heat current density
x

T
 T
x
v
2
 vx

1 2
jt   v cv T
jt  T
3
2
 vy
2
 vz
2
 3 vx
2
1 2
  v cv
3
How does it depend on temperature?
Thermal conductivity
1/3 cv
=
2
v
2
ne /m
1/3 cvmv2
=
2
ne
Drude applied ideal gas law ½ mv2 = 3/2 kBT
=
cvkBT
ne2
The book jumps
through claiming a
value for cv
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
=
3/2 n kB2T
ne2
=
3kB2T / 2e2
Thermal conductivity optimization
To maximize thermal conductivity, there are several options:
•
Provide as many conduction electrons as possible
•
•
Make crystalline instead of amorphous
•
•
irregular atomic positions in amorphous materials
scatter phonons and diminish thermal conductivity
Remove grain boundaries
•
•
free electrons conduct heat more efficiently than
phonons.
gb’s scatter electrons and phonons that carry heat
Remove pores (air is a terrible conductor of heat)
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?
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.
Failures of the Drude model: electrical
conductivity of an alloy
•
•
30
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.
FYI: measurable quantity – Hall resistance  H  RH H
Vy
H
More detail about
H   
Ix
n2 D ec
Hall resistance
E  j
j  sE
in the presence of magnetic field the
resistivity and conductivity becomes tensors
  xx  xy 
 Ex    xx  xy  jx 

for 2D:   



 j 


E


yy 
yy  y 
 yx
 y   yx
s 0 Ex  cj y  jx
s 0 E y  cjx  j y
c
Ex 
jx 
jy
s0
s0
c
1
Ey  
jx 
jy
s0
s0
1
c s 0 
 1 s0

  
  c s 0 1 s 0 
c s 0   jx 
 Ex   1 s 0
E   
 j 



s
1
s
0
0  y 
 y  c
1
m
 xx 
 2
s 0 ne 
c
H
 xy 

s 0 nec
for 3D systems n2 D  nLz
for 2D systems n2D=n
 jx   s xx s xy  Ex 
 j   s
 E 
s
yy  y 
 y   yx
 s xx s xy    xx
  
s  
 s yx s yy    yx
 xy 

 yy 
s0
s xx  s yy 
1  (c ) 2
 s 0c
s xy  s yx 
1  (c ) 2
 xx
s xx  2
 xx   xy 2
 xy
s xy   2
 xx   xy 2
1