Transcript Slide 1

Heat capacity
• The electronic heat capacity Ce can be found by
taking the derivative of Equation (19.18):
• For temperatures that are small compared with the
Fermi temperature, we can neglect the second term
in the expansion compared with the first and obtain
the electronic specific heat capacity is 2.2 x 10-2 R.
This small value explains why metals have a specific heat
capacity of about 3R, the same as for other solids.
• Thus
• It was originally believed that their free electrons should
contribute an additional (3/2) R associated with their three
translational degrees of freedom. Our last calculation shows
that the contribution is negligible.
•The energy of the electrons changes only slightly with
temperature (dU/dT is small) because only those electrons
near the Fermi level can increase their energies as the
temperature is raised, and there are precious few of them.
• At very low temperatures the picture is different. From the
Debye theory, Cv is proportional to T3 and so the heat capacity
of a metal takes the form Cv = AT + BT3,
where the first term is the electronic contribution
and the second is associated with the crystal lattice.
• At sufficiently low temperatures, the AT term can dominate,
as the sketch of Figure 19.9 indicates.
Figure 19.9 Sketch of
the heat capacity of a
metal as a function of
temperature showing
the electronic and
lattice contributions.
S = 0 at T = 0, as it must be. The Helmholtz function F = U -TS is
The fermion gas pressure is found from
• For silver we found that
N/V = 5.9 x1028 m-3 and TF = 65,000K .
Thus
P = 2/5 *5.9*1028 *(1.38*10-23) (6.5*104)
= 2.1*1010 Pa = 2.1*105 atm.
• Given this tremendous pressure, we can appreciate
the role of the surface potential barrier in keeping
the electrons from evaporating from the metal.
19.5 Applications to White Dwarf Stars
• The temperature inside the core of a typical star is at
the order of 107 K.
• The atoms are completely ionized at such a high T,
which creates a hugh electron gas
• The loss of gravitational energy balances with an
increase in the kinetic energy of the electrons and
ions, which prevent the collapse of star!
Example: The pressure of the electron gas in
Sirius B can be calculated with the formula
Using the following numbers
Mass
M = 2.09 × 1030 kg
Radius
R= 5.57 × 106 m
Volume V= 7.23 × 1020 m3
• Assuming that nuclear fusion has ceased after all the
core hydrogen has been converted to helium!
• The number nucleons =
• Since the ratio of nucleons and electrons is 2:1
there are
electrons
MeV
Therefore, T(=107 K) is much smaller then TF .
i.e.
is a valid assumption !
Thus: P can be calculated as
A white dwarf is stable when its total energy is
minimum
For
Since
can be expressed as
Where
For gravitational energy of a solid
With
In summary
To find the minimum U with respect to R
19.7 a) Calculate Fermi energy for aluminum
assuming three electrons per aluminum atom.
2.69103 Kg
N

V
27 Kg
m 3  6.021026
kilom ole
atom s
 5.991028
kilo  m ole
N
V
 1.8 1029
# Density for electrons 3 
F
6.6310 

 3 1.8 10

31 
2  9.1110 
8
34 2
29



2
3
 5.6  2.1  11.8eV
19.7b) Show that the aluminum at T = 1000 K, μ
differs from εF by less than 0.01%. (The
density of aluminum is 2.69 x 103 kg m-3 and
its atomic weight is 27.)
  2  T 2 
  0  1    
 12  TF  


 2 
1000K

  0  1 
11.8eV
 12 


 8.62105 eVK 1
 0  1  4.38105

less than0.01%







2





19.7c) Calculate the electronic contribution to
the specific heat capacity of aluminum at
room temperature and compare it to 3R.
Using the following equation
T
Ce  Nk 
2
 TF

 2
 kT 
 
Nk  
 2
 F 
19.13. Consider the collapse of the sun into a
white dwarf. For the sun, M= 2 x 1030 kg, R =
7 x 108 m, V= 1.4 x 1027 m3.
(a) Calculate the Fermi energy of the Sun’s
electrons.
2  10
No. of electrons 
30
1.66  10
 27
 1.205 1057
1
of nucleous
2
N 1.205 2  1057

V
1.4  1027
# of electrons 
h2
F 
2me
 3N 


 8V 
2
3
 1.205 7.23 1027
 F  0.33me v  

1
.
26
1.4  1027

 F  0.33me v  6.248 105
 F  20.6eV



2
3
(b) What is the Fermi temperature?
F
20.6eV
5
TF 

 2.39  10 k
5
1
k 8.62  10 eVk
(c) What is the average speed of the electrons in
the fermion gas (see problem 19-4). Compare
your answer with the speed of light.
Chapter 20 Information Theory
20.1 Introduction
• Statistical thermodynamics provides the tool of
calculating entropy.
• Entropy is a measure of the degree of randomness or
disorder of a system.
• Disorder implies a lack of information regarding the
exact state of the system.
• A disordered system is one about which we lack
complete information.
20.2 Uncertainty and Information
• Claude Shannon laid down the foundations of the
information theory.
• Further developed by Leon Brillouin.
• Applied to Statistical thermodynamics by E. T. Jaynes.
• The cornerstone of the Shannon theory is the
observation that information is a combination of the
certain and the uncertain, of the expected and the
unexpected.
• The degree of surprise generated by a certain
event – one that has already occurred – is
zero.
• If a less probable event is reported, the
information conveyed is greater.
• The information should increase as the
probability decreases.
• For a given experiment, consider a set of possible outcomes
whose probabilities are p1, p2, … pn.
• It is possible to find a quantity H(p1 . . . pn) that measures the
amount of uncertainty represented by the given set of
probabilities.
• Only three conditions are needed to specify the function H(p1
. . . pn) to within a constant factor. They are:
1. H is a continuous function of the p.
2. If all the pi’s are equal, pi = 1/n; then H(1/n,…, 1/n) is a
monotonic increasing function of n.
3. If the possible outcomes of a particular experiment
depend on the possible outcomes of n subsidiary
experiments, then H is the sum of the uncertainties of the
subsidiary experiments.
• The above discussion leads to g(R) + g(S) = g(RS), where
one can expect that the function g( ) shall be a logarithm function.
• In a general format, the function can be written as g(x) = A ln(x) + C, where
A and C are constants.
• From the earlier transformation g(x) = x f(1/x), one gets that the
uncertainty quantity, H, shall be (1/p)f(p) = A ln(p) + C, where p = 1/n (n is
the total number of event).
• Therefore, f(p) = A*p*ln(p) + C
• Given that if the probability is 1, the uncertainty H must be zero, the
constant C should be equal to ZERO.
• Thus, f(p) = A*p*ln(p).
• Since p is smaller than 1, ln(p) shall be minus and thus the constant A is
inherently negative
• Following conventional notion, we write f(p) = -K*p*ln(p), where K is a
positive coefficient.
• The uncertainty quantity H(p1, p2, …pn) = Σ f(pi)
• Thus, H(p1, p2, …pn) = Σ –K*pi*ln(pi) = -K Σpi*ln(pi)
• Example:
H(1/2, 1/3, 1/6) = -K*[1/2ln(1/2) + 1/3*ln(1/3) + 1/6*ln(1/6)]
= -K*(-0.346 – 0.377 – 0.299) = 1.01K
from the decomposed procedure,
H(1/2,1/2) + 1/2H(2/3, 1/3) = -K*[1/2ln(1/2) + 1/2ln(1/2)]
-1/2*K*[2/3*ln(2/3) + 1/3ln(1/3)]
= -K(-0.346-0.346) – K/2*(-0.27 – 0.366)
= 1.01K
• For equal probable events, pi = 1/n, H = K*ln(n)
• In a binary case, where two possible outcomes of an experiment with
probabilities, p1 and p2 with p1 + p2 = 1
H = - K*[p1ln(p1) + p2ln(p2)]
• To determine H value when p1 is 0 or 1, one need L’Hopital’s rule
lim[u(x)/v(x)] as x approaches 0 equals lim[u’(x)/v’(x)]
• Therefore, as p1 approaches 0,
lim[p1ln(p1)] = lim[(1/x)/(-1/x2)] = 0
• The uncertainty is therefore 0 when either p1 or p2 is zero!
• Under what value of p1 while H reaches the maximum?
differentiate equation - K*[p1ln(p1) + p2ln(p2)] against p1 and set the
derivative equal 0
dH/dp1 = -K*[ln(p1) + p1/(p1) - ln(1- p1) – (1- p1)/ (1- p1)] = 0
which leads to p1 = 1/2
20.3 Unit of Information
• Choosing 2 as the basis of the logarithm and
take K = 1, one gets H = 1
• We call the unit of information a bit for binary
event.
• Decimal digit, H = log2(10) = 3.32, thus a
decimal digit contains about 3 and 1/3 bits of
information.
Linguistics
•
A more refined analysis works in terms of component syllables. One can test what
is significant in a syllable in speech by swapping syllables and seeing if meaning or
tense is changed or lost. The table gives some examples of the application of this
statistical approach to some works of literature.
Linguistics
• The type of interesting results that arise from such studies
include:
(a) English has the lowest entropy of any major language,
and
(b) Shakespeare’s work has the lowest entropy of any
author studied.
• These ideas are now progressing beyond the scientific level
and are impinging on new ideas of criticism. Here as in
biology, the thermodynamic notions can be helpful though
they must be applied with caution because concepts such as
‘quality’ cannot be measured as they are purely subjective