Download UW Physics PhD. Qualifying Exam Spring 2008, problem 11

UW Physics PhD. Qualifying Exam
Spring 2008, problem 11


Consider a parallel LCR circuit maintained at a temperature T  h kb LC . Use
Maxwell-Boltzmann statistics to calculate the mean-square flux threading the inductor
and the mean-square charge on the capacitor.

Solution 1: by means of partition function
Recall that the energy of a capacitor is E 
1 Q2
1
1 2
and of an inductor is E  L I 2 
.
2 C
2
2 L
1 Q 2 1 2
The energy stored in the circuit is E 
.

2
C
2
L


Assert that the relative probability that a circuit element has energy E is well-described
by the Boltzmann factor: P(E) e E kT . (Note that this assertion about the system energy
 of what statistics are obeyed by the particles in the system).
is completely independent
1 Q 2 1  2 
i
i
  
2 C  2 L 



The partition function is then Z   e
, where  = 1/kT. How are we to decide
i

what the charge and flux of the ith state are??
To proceed we must convert the sum to an integral over phase space. What variables
define phase space?
Phase space is defined by generalized coordinates p and momenta q. Here is one recipe
for assigning the coordinates and momenta.
1. Write down the Lagrangian.
In the circuit above, note that VC=VL, that is
Q
dI Ý
 L  
, since   LI .
C
dt
Ý2 1 2
C
.

2
2 L
Observe that the energy, written in this fashion, is analogous to the energy of an

harmonic oscillator, with  taking the role of position. Therefore identify the inductor
flux  as the generalized coordinate. View the first term as a kinetic energy term, the

Ý2 1 2
C
second as a potential energy term, and write the Lagrangian as L 
.

2
2 L
Rewrite the energy E 

2. Find the momentum p that is conjugate to the coordinate.
The conjugate momentum is defined to be p 
L L
Ý Q.
 Ý  C
qÝ 
Phase space is thus defined for this problem by the variables , Q. The phase space

volume element is therefore ddQ.
Recall the recipe for converting sums over quantum states to integrals (this was discussed
in connection with the Bose-Einstein Condensation problem):
t
dpdq 
1  
 ,
 h 
where t is the number of sets of (p,q) in the problem (in this case t =1). Thus, the correct
form for the partition function is
1 Q 2 1  2 
i
i
  
2 C  2 L 



Z  e

1 Q2 1 2 
  
2 C  2 L 



 Z   e
i
1 Q2 
1 2 
  
2 L 

ddQ 1  2 C 
 e
dQ  e  d
h
h
Now we can easily calculate the expectation values that we want:
1 Q2 
1 2 
1 Q2 
  
  
2 C 

2 L 

1



 Q 2 e  dQ  e  d  Q 2 e  2 C dQ
Q2  h


1 Q2 
1 2 
1 Q2 
  
  
2 L 

2 C 

1  2 C 
dQ  e  d
e
 e  dQ
h



C

Qe
1 Q2 
  
2 C 



e

C

e
1 Q2 
  
2 C 



dQ
1 Q2 
  
2 C 



dQ
 CkT,
2  LkT.

Solution 2: by means of equipartition

Follow the solution above, until arriving at this expression for the energy:
Ý2 1 2
C
.
E

2
2 L
Observe this has the same form as the energy of a 1-D harmonic oscillator.
Deduce that this system must have two degrees of freedom, like the harmonic oscillator.

Use the classical theorem of equipartition of energy to conclude that there must be ½ kT
associated with each degree of freedom, that is ½ kT for kinetic energy (first term) and ½
kT for potential energy (second term).
Ý2
C
1 Q2
kT
1 2
kT


,

, and 2  LkT, Q2  CkT , as before.
Thus
2
2 C
2
2 L
2


Solution 3: Johnson Noise (following Nyquist, Phys. Rev. 32 (1928), 110-113).
The method of Nyquist involves imagining the resistor is connected across an ideal
transmission line, which is terminated at the far end by an identical resistor:
The normal modes of the system are then the normal modes of the transmission line, with
wavelength n=2l/n, where l is the length of the transmission line. The energy
contained in each mode is calculated by using Bose-Einstein statistics in the limit
kT>>En, where En is the energy of the nth mode En  h n  h c  n . The result is that the
energy contained in each mode is kT, which is identical to result expected from classical
equipartition, on the assumption that there are two degrees of freedom per normal mode.
One degree of freedom represents a travelling wave carrying energy down the
 2, and the other degree of freedom represents
transmission line from resistor 1 to resistor
a travelling wave carrying energy from resistor 2 to resistor 1. Since all energy is
contained in travelling waves the power delivered to each resistor per unit time may be
equated with the V2/R power dissipated in the resistor. This gives the Nyquist result:
V 2  4RkT f
See Nyquist’s (very readable) paper for details.

Only circuit elements with non-zero
real impedances (ie resistors) serve as voltage
sources. Thus a resistor at a finite temperature may be modeled as an ideal resistor in
series with a voltage source:
The fact that finite-temperature resistors act as voltages sources was noticed
experimentally by Johnson (before being explained almost immediately by Nyquist, the
theorist in the next office), so the phenomenon is known as Johnson noise.
The factor f indicates that all frequency bands contribute equally to <V2>. Thus
f may be interpreted equally well as in infinitesimal (in which case the left-hand side
ought to be d<V2>), or as a bandwidth, or as a maximum frequency on the assumption
that the minimum frequency is zero.
If we now model the LRC circuit problem 11 like this:
,
the voltage across the inductor and capacitor in a particular frequency range df is related
to the Johnson noise voltage < V2> by:
1
1
d VL2  d VC2  d V 2
 4 RkT df
1 2
1 2
1 R 2 (C 
)
1 R 2 (C 
)
L
L
We find the mean-square flux linking the inductor by using
dI
V  L  V 2  L2 2 I 2   22 ,
dt

which implies that
1
d 2  2 d VL2 ,


so
1
  4kTR
2

2

 d
1

1 
1 R C 


L 
2
2
.
2
0

This integral may be solved by the substitution x=C-1/L,   x  x 2  4 C L
 2C to
give L/2R, so that 2  LkT , as before.

The total fluctuating voltage is
2
L
V

1
 V  4RkT
2
2
C


0

d

1 
1 R 2 C 


L 
2

kT
C
The charge on the capacitor is then Q2  C 2 V 2  CkT .

One conclusion that may be drawn from this is that while the mean square charge or flux
on an isolated capacitor or an inductor is non-zero, its output impedance (ability to drive

current in other circuit elements)
is zero; only resistors have non-zero output impedances.
Addendum: if the inequality kT  h LC does not apply (high frequencies and/or low
temperatures) the Nyquist argument may easily be extended, with the result
4hf Rf
.
V 2  hf /kT
e
1

Solution 3b: a slightly different circuit model

It turns out that if one replaces the resistor R with a complex impedance Z=R+iX, the
Nyquist result still holds. In this case <V2>=4R kTf.
Thus our parallel RLC circuit could be modeled like this
Then the resistive part of the impedance is:
R =
1
1 2
)
L
It is now no longer true that all frequency bands contribute equally to <V2>. In
particular, high frequencies don’t contribute at all, because the capacitor becomes a short
circuit at high frequencies, and the resistive impedance goes to zero.

1+ R 2 (C 
In this case it makes sense to calculate the total <V2> as
 V 2  4kT


R df 
2kTR



0
d

1 
1 R C 


L 
2
2
This is the same integral as before, and yields the same answers. The mean-square
charge and flux do not depend on the details of the circuit.
Method 4: diffusion
Consider the difference between electron movement in a perfect conductor and electron
movement in a resisitor:
--in a perfect conductor, the electrons never collide with anything, and their number
density is determined by the local value of the potential
--in a resistor, the electrons collide with other particles (“the lattice”), and their number
density is determined not only by the local value of the potential but also by diffusion.
In the absence of an external electric field, electron movement through a resistor is
entirely due to diffusion. The net current is zero, but this may be thought of as the sum of
a leftward-going current and a rightward-going current that are each non-zero.
Consider a resistor with no external voltage:
Red
Blue
Suppose that electrons that pass through the resistor going to the right reach the “blue”
end, turn blue, and head back through the resistor going to the left. Blue electrons that
reach the “red” end of the resistor turn red, and head back through the resistor going to
the right.
A diffusive process is may be thought of as a process in which each particle makes a
random walk through the system. It may be characterized by a time  and a length .
Every  the particle takes a “step” of length  in a random direction. In a onedimensional diffusive process, the direction can be either left or right.
Let us assume that <<L, where L is the length of the resistor. Then it is not true that all
particles going to the right are necessarily red. We expect each electron to change
direction multiple times before passing through the resistor.
n
n
nred
nblue
0
0
0
x
L
0
x
L
The total density of electrons at all points is n. Diffusive processes are described by the
equation:
  D n ,
where  is the particle flux, n is the density, and D is the diffusion constant (dimensions
of length2/time), which can be approximated as 2/.

Consider the red and blue electrons separately. In the 1-d case n has units of particles per
length and  has units of particles per time. Specifically:
Dn
qDn

, I  q; I 
L
L
Einstein showed that (with certain assumptions) D for charged particles in the absence of
an electric field was related to the mobility  those same charged particles would have in
the presence of an electric field by

 kT
D
,
q
where the mobility is defined by
v   E,
v being the terminal velocity the particles attain in an electric field E. Thus the current

becomes
 kTn
.
I

L
In this model the completion of the passage of an electron through the resistor may be
treated as an event subject to Poisson statistics. Choose a time interval dt such that the
number N of electrons that complete the passage through the resistor during dt is N0
 distribution simplifies to the Gaussian distribution, and
>1000 or so. Then the Poisson
we conclude the probability P(N) that exactly N electrons will complete the passage is
2
P(N) e(NN0 ) /2N0 ,
q N qN0
q N0 q Idt / q
from which we write N  N0 ;  N  N0 ; I 
.

; I 

dt
dt
dt
dt
Using the previous result for I the variance of the current becomes

qI q kTn
.
 I2  
dt
L dt

The quantity qn may be recognized as the conductivity  (defined by J=E), which has
as its inverse the resistivity , (defined by E=J. In our 1-d model the resistance
R=L, so

kT
.
 I2 
Rdt
It is customary to write the averages of fluctuating quantities in terms not of dt but of df,
an infinitesimal frequency. The formal way to relate dt to df involves taking a Fourier
transform of the current from the time interval 0 to dt, and then to calculate the spectral

intensity from the Fourier components.
Informally, we note that the number of electrons which arrive in time dt will be affected
by fluctuations with time scales much greater than dt, but will not be affected by
fluctuations with time scales much less than dt. Thus dt and df are related somehow.
The results of the formal calculation (involving spectral intensity) are that 1/dt should be
replaced by 2 df. Thus we have:
2 kT df
(true for “red” or “blue” electrons)
 I2 
R

This type of noise is called “shot noise”. In this case “shot noise” is entirely due to
electron diffusion.
The total current (sum of “red” and “blue” currents) is zero, but the current fluctuations
add in quadrature (this is an example of the Central Limit Theorem), so:
4kTdf
2
2
I total  0; I 2   I2   red
  blue

R
V 2  R2 I 2  4kT Rdf
Therefore

This is the same result for Johnson Noise (solution 3, above), so the calculations for <Q2>
and 2 will give the same result as solution 3.


Comments on solution 4. Solution 4 depends on the Einstein relation D=kT/q. Under
what conditions does this hold?
Einstein’s relation certainly holds if the electrons follow a Maxwell-Boltzmann
distribution.
However, the electrons in a metal certainly do not follow a Maxwell-Boltzmann
distribution. Electrons are Fermi-Dirac particles, and the chemical potential for a typical
metal is a few eV (compare with room temperature ~ 1/40 eV).
(Note that the chemical potential  (not the Fermi energy Ef, which is the highest
occupied state at T=0) appears in the Fermi-Dirac distribution function,
1
ni  ( Ei ) /kT ;
e
1
note that the chemical potential adjusts itself as a function of temperature to keep the total
number of particles N 
n
i
a constant; and finally note that at T=0, =Ef while at

higher temperatures  increasingly deviates from Ef, becoming smaller than Ef.)
i
The only way to recover Maxwell-Boltzmann statistics from Fermi-Dirac statistics is to
limit one’sinterest to electrons with energy considerably larger than the chemical
potential, ie (Ei-) >> kT.
A more qualitative model for why Fermi-Dirac particles might obey the Einstein relation
may be found in Guidoni and Aldao, Eur. J. Phys. 23 (2002) 395-402, “On Diffusion,
Drift, and the Einstein Relation”. Also see the discussion in section 5.2 in Robinson,
Noise and Fluctuations, Oxford (1974); TK7867.5 R62.