what happens to the current if we: 1. add a magnetic field, 2. have an oscillating e field (e.g....

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

2

Last time :

--+

Add a magnetic field H

dp/dt = -p(t)/ + f(t)

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.

Lorentz force = -ev/c x H

--+

H

++++++++++++++++

- - - - - - - - - - - - - - - - - -

GroupIf 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

Hz

-++++++++++++++++

xy

Similar approach:dp/dt = -p(t)/ + f(t)

- - - - - - - - - - - - - - - - - -

The Hall coefficient

Ohm’s law contains e2

But for RH the sign of eis 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.

from D.C. Tsui, RMP (1999) and from H.L. Stormer, RMP (1999)

weak magnetic fieldsnec

HH

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

What happens to the electrons if we:

have an oscillating E field (e.g. light)

9

plasmon: charge density oscillations

Longitudinal Plasma Oscillations

0

)4(

2

2

2

2

2

dt

d

ndeNeNeEt

dNm

p

m

nep

22 4

Oscillations at thePlasma Frequency

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 = nde & thus an electric field E = 4nde.

11

plasmon: charge density oscillations

values for the plasma energy

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

i

mne

m

ne

i

e

et

et

tet

dt

td

ti

ti

/1

)()/()()(

/1

)()(

)(),(

)(),(

),(),(),(

2 Epj

Ep

pp

EE

EppNewton’s 2nd Law Equation of

Motion for the momentum of one electron in a time dependent electric field. Look for a steadystate solution of the form:

m

ne

i

2

0

0

1)(

)()()(

Ej

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:

j = E

x(xE) = x

= -i H(,t)

= x i H(,t) /c = i /c x H(,t)

x(xE) = -2E

x(xE) = -2E = i /c ( )

-2E = i/c (4E/c -i E/c)

-2E = 2/c2 (1 +4 i / )E

( ) =1 + 4 i / Usual wave: -2E = 2 ()E/c2

15

( ) =1 + 4 i / m

ne

i

2

00 ,

1)(

From continuum limit From Maxwell’s equations

Plugging into : ( ) =1 + 4 0i / (1-i )

=1 + 4 0i / ( -i 2)

Plugging in 0: ( ) =1 + 4 ne2i / m( -i 2)

For high frequencies can ignore first term in denominator1Ignoring: ( ) =1 - 4 ne2/m2

p is known as the plasma frequency

What is a plasma?

The electromagnetic wave equation in a nonmagnetic isotropic medium.Look for a solution with the dispersion relation for electromagnetic waves

222 /),(

)exp(

Kc

iti

K

rKE

Application to Propagation of Electromagnetic Radiation in a Metal

-2E = 2 ()E/c2

E/t= -i E(,t)

2E/t2= 2 E(,t)

2E = -2 ()E/c2 =2 E

real & > 0 → for real, K is real & the transverse electromagnetic wave propagates with the phase velocity vph= c/

(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)

(3) complex → for real, K is complex & the wave is damped in space

(4) = 0 longitudinally polarized waves are possible

Transverse optical modes in a plasma222),( Kc K

Dispersion relation forelectromagnetic waves

2222 Kcp

rKeE

(2) (1)

/p

E&M waves are totally reflected from the medium when is

negative

E&M waves propagate with

no damping when is

positive & real

( )

Ultraviolet Transparency of Metals

m

nep

22 4

Plasma Frequency p & Free Space Wavelength p = 2c/p

Range Metals Semiconductors Ionospheren, cm-3 1022 1018 1010

p, Hz 5.7×1015 5.7×1013 5.7×109

p, cm 3.3×10-5 3.3×10-3 33 spectral range UV IF radio

The Electron Gas is Transparent when > p i.e. < p

Plasma FrequencyIonosphereSemiconductorsMetals

The reflection of light from a metal is similar to the

reflection of radio waves from the

Ionosphere!

reflects transparent formetal visible UVionosphere radio visible

What happens when you heat a metal?

What do we know from basic thermo (0th law)?

19

20

Drude’s Best Success:Explanation of the Wiedemann-Franz law for

metals (1853)

•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 slope20

Thermal conductivity

Electrical conductivity

Temperature

Thermal conductivity A material's ability to conduct heat.

area

Ejt sec

jt vn

Thermal current density

= Energy per particle

v = velocity

n = N/V

Electric current density

Heat current density

Fourier's Law for heat conduction.

(je = I/A)

2l



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

jt vn



x

Limit as l gets small:


x

v v

v



x

Tx

T

22222 3 xzyx vvvvv

Tcvj vt

2

3

1vcv 2

3

1

How does it depend on temperature?

1/3 cvv2= =

1/3 cvmv2

ne2ne2/m

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

=

ne2

3/2 n kB2T = 3kB

2T / 2e2

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)

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.

http://en.wikipedia.org/wiki/Drude_model#Accuracy_of_the_model

30

Failures of the Drude model: electrical conductivity of an alloy

•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 HRHH

ecn

H

I

V

Dx

yH

2

zD nLn 2for 3D systems for 2D systems n2D=n

jE Ej in the presence of magnetic field the resistivity and conductivity becomes tensors

yxcy

xycx

jjE

jjE

0

0

yyyx

xyxx

0 0

0 0

1

1

cx x y

cy x y

E j j

E j j

for 2D:

00

00

1

1

c

c

nec

H

ne

m

cxy

xx

0

20

1

22

22

xyxx

xyxy

xyxx

xxxx

20

20

)(1

)(1

c

cyxxy

cyyxx

1

yyyx

xyxx

yyyx

xyxx

x xx xy x

y yx yy y

E j

E j

0 0

0 0

1

1x xc

y yc

E j

E j

x xx xy x

y yx yy y

j E

j E

More detail about Hall resistance

what happens to the current if we: 1. add a magnetic field, 2. have an oscillating e field (e.g....

Documents