Thermodynamics and Statistical Mechanics 2 Lecture notes Dr. O. Krichevsky August 14, 2007

Thermodynamics and Statistical Mechanics 2
Lecture notes
Dr. O. Krichevsky∗
August 14, 2007
∗ Ben
Gurion university physics department; [email protected]
1
List of Contributors:
Lior Yosub, Liana Diesendruck, Gitit Feingold, Yaacov Kleeorin, Alexander
Elikashvili, Mattan Frish, Ilan Shemesh, Ofir Gana, Roei Cohen, Zahi Levi,
Maya Rot, Yuri Khodorkovsky, Michael Guzman, Avraham Aloni, Meny Gabbay, Noa Benjamini, Tal Peer, Miri Gelbaor, Yair Bar, Yoav Pollack, Oleg
Shneider, Dimitri Bukchin, Itamar Gurman, Roi Tzaig, Nessi Benishti, Guy
Yafe, Yotam Soreq.
2
Contents
I
Phase Transitions
5
1 Review of thermodynamics ensembles (Thermodynamics and
Statistical Mechanics I)
5
1.1 Micro-Canonical Ensemble . . . . . . . . . . . . . . . . . . . . .
5
1.2 Canonical Ensemble . . . . . . . . . . . . . . . . . . . . . . . . .
6
1.3 Grand Canonical Ensemble . . . . . . . . . . . . . . . . . . . . .
7
2 Intensive thermodynamics potentials
7
3 Introduction to phase transitions
3.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 The physical theory of phase transitions using the free energy
potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
9
11
4 Gas Model - id, hc and vdw interactions
14
4.1 Helmholts free energy for the gas model . . . . . . . . . . . . . . 15
4.2 Example: the water critical temperature and phase transition . . 19
5 Fluctuations
20
6 Clausius Clapyeron equation
23
7 Phase transitions at different temperatue scales
24
7.1 Low temperature approximation . . . . . . . . . . . . . . . . . . 24
7.2 Finding binodal points for τ << τc . . . . . . . . . . . . . . . . . 26
7.3 Finding binodal points and other physical entities for τ � τc . . . 26
8 Spin lattice
30
9 Surface energy
34
9.1 Nucleation and grows process . . . . . . . . . . . . . . . . . . . . 35
II
Kinetic Theory
35
10 Boltzmann and Maxwell velocity distribution
11 The
11.1
11.2
11.3
11.4
11.5
kinetic theory of gases
Introduction . . . . . . . . . . . . . . . . . . . . . .
Random walks . . . . . . . . . . . . . . . . . . . .
Mean free path . . . . . . . . . . . . . . . . . . . .
Probability evolvement in time . . . . . . . . . . .
Definition of pressure from dynamical point of view
3
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40
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12 Diffusion
46
12.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
12.2 Derivation of flux caused by potential gradient . . . . . . . . . . 49
12.3 Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
4
Part I
Phase Transitions
1
Review of thermodynamics ensembles (Thermodynamics and Statistical Mechanics I)
Let’s begin with a small review of Thermodynamics and Statistical Mechanics
I. We defined 3 types of ensembles:
1.1
Micro-Canonical Ensemble
In this ensemble we set three macroscopic parameters - Energy (U), Volume
(V) and Number of Particles (N). The probability of a inaccessible state is 0
and the probabilities of the accessible states are defined as all equal to g1 , where
g(U, V, N ) is the total number of accessible states, called multiplicity function
or degeneracy
�
1
P (accessible) = , where g =
1,
g
accessible
P (inaccessible) = 0
We define the entropy of micro-canonical ensemble as σ(U, V, N ) = ln g. The
Second Law of Thermodynamics states that in a closed (or isolated) system the
entropy may only rise while the system approaches thermodynamic equilibrium
or stay constant (in equilibrium). Thus equilibrium entropy is a function of
system parameters: σ(U, V, N ). The energy of an open system may change as
a result of heat δQ transferred to the system or work δW performed on the
system: dU = δQ + δW . The are numerous ways to carry out work. A general
type of work is mechanical work associated the change in the system volume
δW = −pdV . The transferred heat is related to entropy change as: δQ = τ dσ.
Then the law of conservation of energy reads:
dU = τ dσ − pdV
(1)
In addition (as discussed below in the Grand Canonical ensemble), the energy
of an open system can change as a result of exchange of particles with the
surrounding:
dU = τ dσ − pdV + µdN,
(2)
where µ is the chemical potential of the particles.
Looking at the same formula from a different perspective, we see that it
defines the change in equilibrium entropy resulting from variation in system
parameters U, V, N :
dU
p
µ
dσ =
+ dV − dN.
(3)
τ
τ
τ
5
From the above different Maxwell relations follow, e.g.:
�
�
�
�
�
�
1
∂σ
p
∂σ
µ
∂σ
=
=
=−
τ
∂U V,N
τ
∂V U,N
τ
∂N U,V
(4)
One can show that two otherwise isolated systems put into thermal contact,
reach equal temperature in equilibrium.
1.2
Canonical Ensemble
In this ensemble we put our system in thermal contact with a heat reservoir.
We assume that the heat reservoir is much larger than the system and its temperature is not affected by the heat transferred from (or to) our system. Thus,
in equilibrium the temperature of the system is fixed by the temperature of the
reservoir. In addition to setting the temperature τ we also set constant the
volume (V ) and the Number of Particles (N ).
The probability of finding the system in particular state {si } depends on the
energy �({si }) of that state:
�
�
1
�({si })
P ({si }) = exp −
,
(5)
Z
τ
�
�
�
�({si })
where Z(τ, V, N ) =
exp −
is the partition function of the system.
τ
{si }
From Z, we can determine free energy F of the system:
F (τ, V, N ) = −τ ln Z
(6)
Free energy can be expressed through other thermodynamic function as:
F = U − στ
(7)
The free energy can only decrease (or remain at the same level) in the system
which is part of Canonical ensemble, and reaches its minimum in equilibrium.
The minimization of the free energy can be seen as representing two opposing
tendencies in the statistical system: 1) the mechanical tendency to energy minimum, reaching which in usually accompanied by decrease in entropy, and 2)
the law of increasing entropy, which drives the system away from any special
configuration.
The differential of F is:
dF = dU − d(τ σ)
(8)
Previously we found dU = τ dσ−pdV +µdN , therefore: dF = −σdτ −pdV +µdN .
This gives relations for derivatives of F with respect to the parameters defining
the canonical ensemble τ, V, N :
�
�
�
�
�
�
∂F
∂F
∂F
σ=−
p=−
µ=
(9)
∂τ V,N
∂V τ,N
∂N τ,V
6
One can show that two systems systems put into thermal contact with heat
reservoir and diffusional contact between themselves, reach equal chemical potentials in equilibrium.
1.3
Grand Canonical Ensemble
In this ensemble the system to into thermal and diffusional contact with a
heat-particles reservoir, i.e. the boundary between the system and the reservoir
conducts heat and is porous to particles. In other words, the boundary can be
purely imaginary. As usual, the reservoir is assumed to be much larger than
the system at least in two respects: it has much more internal energy and many
more particles than the system.
� dF �
The chemical potential is defined as µ = dN
. In previous ensembles we
V,τ
fixed number of particles N in the system. Now we let it change, so now we
should rectify the expression of the energy and all its derivatives.
dU = τ dσ − pdV + µdN
dF = −σdτ − pdV + µdN
dσ =
dU
τ
+ τp dV − µτ dN
We start by defining the probability of a certain state as:
P (N, {si }) =
where
ζ=
��
N
1 µN −�s
e τ
ζ
e
µN −�s
τ
(10)
(11)
{s}
Directly from ζ, the Grand Partition Function, we shall derive the definition
of the Grand Thermodynamical Potential, Ω.
Ω = −τ lnζ
(12)
Another form of Ω is: Ω = F − µN . The differential form of Ω is:
Ω = −σdτ − pdV − N dµ
2
Intensive thermodynamics potentials
We should now elaborate a specific characteristic of physical parameters of a
thermodynamic system. An extensive parameter is a parameter whose value
is proportional to the size of the system (number of particles), an intensive
parameter is a parameter which is independent of the system size.
The thermodynamic potentials incountered so far are:
• U - Energy
7
• F - Free Energy
• H- Enthalpy
• Ω - Grand Potential
We have expressed all of these potentials in a differential form, where the number
of elements depends on the number of ways to change the system’s energy. Each
term involves one extensive parameter and one intensive parameter. The pairs
we have encountered so far were (τ, σ), (p, V ), (µ, N ), (B, M ), where the first
element of each pair is the intensive parameter and the second is the extensive
one. The parameters under differential sign have a special meaning: when they
are fixed, this thermo potential tends to minimum/maximum. For example:
dσ =
dU
τ
+ τp dV − µτ dN ⇒ σ ↑ max, when U, V, N = const.
or
dF = −σdτ − pdV + µdN ⇒ F ↓ min when τ, V, N = const.
In order to get from one potential to another one we simply need to add the
proper pair of parameters to the first potential. For example, let’s construct an
ensemble where we fix the temperature, the number of particles and the pressure
of the system. Like in the previous ensembles we’ll look for a matching potential
for our new ensemble and we shall see that our matching potential tends to a
minimum.
G(τ, p, N ) = U (V, σ, N ) + pV − τ σ = H(p, σ, N ) − τ σ = F (V, τ, N ) + pV
dG = −σdτ + V dp + µdN
We attach to the system a heat-particle reservoir and will show that Gibbs
potential is minimum at τ , p and N - constant. The total entropy σR+S increase
since R + S is a closed system.
We use Taylor expansion:
σR (U0 − U, V0 − V ) + σS (U, V ) ↑ max
σR (U0 , V0 ) −
∂σR
∂U |U0
Using Maxwell relations we get:
� ∂σR �
∂U
N,V
=
1
τ
U−
;
∂σR
∂V |V0
V + σS ↑ max
� ∂σR �
∂U
N,V
=
p
τ
σR (U0 , V0 ) − Uτ − p Vτ + σS ↑ max
� �� �
constant
Now we multiply by −τ , so the expression should tend to minimum:
G = U + pV − τ σ ↓ min
8
We call this potential Gibbs free energy. All of our potentials so far are extensive
quantities.
Sometimes we want to characterize the system by an intensive quantity with
a meaning of thermodynamic potential. The easiest way define it is to divide the
potential by an extensive parameter. If we divide the potential by, for example,
N , we’ll get potential per particle, which is an intensive size. We shall see now
that in the special case of the potential G we’ll get:
�
�
∂G
G
µ=
=
∂N p,τ
N
i.e. µ is Gibbs free energy per particle
We start with the Energy (U ):
U = (σ, V, N ) = N U (
σ V
, , 1)
N N
The above equation simply states that the energy of a system is equal to the
number of particles times the energy of one particle.
The same can be done with the Free Energy (F ):
F (τ, V, N ) = N F (τ,
V
V
, 1) = N f1 (τ, )
N
N
(The temperature is an intensive quantity so it’s the same for every particle.)
Now we do the same to Gibbs Free Energy:
G(τ, p, N ) = N G(τ, p, 1) = N g1 (τ, p)
From the differential form of G we calculate the Chemical Potential:
�
�
∂G
G
µ=
= g1 (τ, p) ⇒ µ =
∂N p,τ
N
From here we receive:
G = F + pV = U + pV − τ σ = µN
Ω = F − µN = F − G = −pV
3
3.1
Introduction to phase transitions
Definitions
We’ll begin the discussion of phase transition with an example of real gas compression. From this example the definitions of phase and phase transitions will
be made.
Consider vessel which contain water vapor with volume V and at low pressure
so that the gas is an ideal gas. We start decrease the volume so the pressure
9
increases (p ∝ V1 ). At a certain point a small layer of water will appear and
from this point, the following decreasing of volume wil not change the pressure,
it will be constant. So what happens in the system? If we continue to reduce the
volume, the water level will rise, but the pressure will remain constant. When
all the vapor turn into water, reducing the volume will increase the pressure
drastically and the proportion between the pressure and volume will change. In
range of constant pressure we observe coexistence of two phases: liquid and gas.
The constant pressure with the change of volume can be explained by the
phase changes of the matter. When reducing the volume, molecules shift from
gas phase to liquid phase (from less dense interior to more dense one). The shift
”absorbs”’ the pressure. After a certain volume for which all the molecules have
shifted - any further reduction in volume will cause pressure to rise again.
While performing such an experiment, it is possible to distingiush between
two different forms of matter - each of one differs from the other by its density.
The density of each of these two froms will stay constant with any change of
volume. Each of the two forms is called a phase. In this example these would be
the vapour phase and the water phase. A phase transition is the shift from one
phase to the other, which might or mogith not have a stage with mixed phases.
The determination of this depends on the temprature of the system, as we shall
see below. The system is shown in Figure 1
Figure 1: ’snapshoots’ of the real gas compression process. The phase seperation
is seen on stages B and C.
The process we described above is done at a constant temperature. If the
experiment is done in a higher temperature, the volume range of phase separation will shrink(the volume range in which there is a constant pressure). Raising
the temperature high enough, will yield only a point of constant pressure. This
temperature is called critical temperature, and for a phase transformation process at this temperature we don’t have phase separation, we have everytime
coexistence of gas and liquid, this named - fluid. The appropriate chart of p − V
diagram for such a process under constant temperatures is Figure 2. That kind
of transition is called: first order phase transition.
10
Figure 2: pressure as function of volume for different temperatures.
3.2
The physical theory of phase transitions using the free
energy potential
We can describe the thermodynamics of phase separation by considering the
free energy F. We already know that the free energy tends to minimum at
equilibrium with τ, V = const. So it is needed to show that a preferable state of
a system in a certain conditions is when it has phase separation. We will mark
the density of molecules in gas state VN� and the density of molecules in liquid
stat VN�� .
Given a system (τ ,V,N) n = N
V , and we want to look both on the given
system and the system after it has been separated; liquid state (τ ,V1,N1), gas
state (τ ,V2,N2).
F (τ, N1 , V1 ) + F (τ, N2 , V2 ) = F � (τ, N, V )
(13)
F � - for system in phase separation region.
Reminder: F is an extensive parameter.
For phase separation we require:
V · fv (n, τ ) > V1 · fv (n1 , τ ) + V2 · fv (n2 , τ )
(14)
Where fv is the free energy per unit volume for homogeneous system - intensive parameter.
Now; we divide the equation by V.
fv (n, τ ) >
V1
V2
· fv (n1 , τ ) +
· fv (n2 , τ ) ≡ f �
V
V
(15)
With: N = N1 + N2 and V = V1 + V2 . Rewriting the constrains on the system,
we get: 1 = VV1 + VV2 ⇒ VV2 = 1 − VV1
N = N1 + N2 ⇒ nV = n1 V1 + n2 V2 ⇒ n = n1 VV1 + n2 VV2
⇒ n = n1 VV1 + n2 (1 −
11
V1
V )
⇒
V1
V
=
n−n2
n1 −n2
placing: f � =
V1
V fv (n1 )
+ (1 −
V1
V )fv (n2 )
f � = fv (n2 ) +
f � = fv (n2 ) +
V1
V (fv (n1 ) − fv (n2 )) ⇒
2
( nn−n
)(fv (n1 ) − fv (n2 ))
1 −n2
Looking at the last expression as a function of n we receive a straight line
between the marked dots on figure 3.
n = n1 ; f � = fv (n1 ) ,
n = n2 ; f � = fv (n2 )
(16)
The equation says that the free energy density f � (n) for a system, which has
an average density n and which separates into two phases with densities n1 and
n2 each, is a point on this line corresponding to density n. This energy density
should be compared to the energy f (n) of the homogeneous system, which is a
point on the original curve corresponding to density n.
Figure 3: free energy per unit volume fv as a function of the density n. The
graph is convex.
2
∂ f
�
We see in figure 3 that for ∂n
2 > 0, f (n) > f (n), i.e. free energy of the system
increases when system undergoes phase separation, and therefore in that case
there will be no phase transition.
Now, we will use the concave graph:
12
Figure 4: free energy per unit volume fv as a function of the density n. The
graph is concave.
Figure 4 represents the free energy per unit volume fv as a function of the
density n but this time the graph is concave.
2
∂ f
�
Respectively, in figure 4 we see that for ∂n
2 < 0, f (n) < f (n), i.e. free
energy of the system decreases when system undergoes phase separation. In
that case we do have phase separation. We� can �conclude that the condition
∂2f
to see phase separation is the following: for ∂n
< 0 we can observe phase
2
τ
� 2 �
∂ f
separation, and for ∂n2
> 0 there is no phase separation- this is local stability
τ
� 2 �
∂ f
requirement. Now let’s try to find the conditions under whose ∂n
< 0.
2
τ
We have reached the point where we can understand the phase transition
through the graph of free energy per volume unit as a function of density (see
fig. 5). We notice 4 points that represent the graph:
• 2 spinodal points (s1 , s2 ) which are calculated by
∂2f
∂n2
= 0.
• 2 binodal points (b1 , b2 ) which are found by the common tangential line.
We shall follow the graph (see fig. 5), starting with the gas phase at low density.
Once we reach the first binodal point we may start to see the phase separation, but not necessarily. if the process of densing the system is done carefully
we can get past the binodal point and still not have phase separation, but by
the time we reach the spinodal point, the phase separation will have started.
Then the gas phase will have the density of n(b1 ) and the liquid phase will have
the density of n(b2 ).
For finding the common tangential line we must demand the gradient of the
tangent line at both points b1 and b2 be equal :
∂f
∂f
f (n2 ) − f (n1 )
|n=n2 =
|n=n1 =
∂n
∂n
n2 − n1
The phase separation will occur in the density range n(b1 ) to n(s1 ).
13
The reason is that when we get past the binodal point and come closer to
the spinodal point the fluctuation of the density will grow and the system will
notice there is a lower energy state (global minimum) and will procede with the
phase transition in order to reach the lowest energy possible.The system will
never be able to reach the spinodal point because the density fluctuations will
grows to infinity.
� 2�
τ
∆n = � 2 �
∂ f
V ∂n
2
This range between the binodal point and the spinodal point is called metastable.
4
Gas Model - id, hc and vdw interactions
From previous semester we know how to deal with a system of ideal gas, now
we want to implement phase separation to it: The partition function for one
molecule in one dimension Z1 = nq V , where; nq is the quantum density nq =
mτ 32
( 2Π�
and V is the volume. And the partition function for N molecules is:
2)
ZN
ZN = N1! . We also know that Helmholts free energy in canonical ensemble can
be calculated form the equation: F = −τ ln ZN = N τ (ln VNnq − 1). Then:
fvid (n, τ ) = nτ (ln
n
− 1)n < nq :
nq
(17)
Figure 5: A scheme of fv as function of n, with concave and convex parts
14
Figure 6: fvid - Helmholtz free energy per unit volume for the ideal gas as
function of density n
We see in figure 6 that the second derivative of that function is always
∂2f
positive ∂n
2 > 0, no matter how condensed the gas is, there will never be
phase separation. We already know that both free energy F and energy U seeks
its minimum value at equilibrium, while entropy σ seeks its maximum value,
F = U − στ . Discussing a system of ideal gas, the energy comes from kinetic
energy, in that case the entropy is dominant, system has no reason undergo phase
separation and thus we need interactions. Let’s introduce two possibilities for
interaction between molecules:
• Short range interaction - repulsion, results from Pauli’s exclusion principle
in quantum discription, or from the finite radius of the molecules in the
mechanical approximation.
• Long range interaction- attraction. Their origin will be explain below.
Repulsion: We look at the system made of small ”‘billiards balls”’ - Hard core
interaction, in such that they cannot come closer to each other then twice their
radii.
�
�
0 ,r > d
U hc =
∞ ,r ≤ d
Attraction: It was suggested by Van der Waals to introduce an attracting
force of the magnitude represented by the following potential: U vdw (r) ∝ − r16 .
The interaction as function of distance between molecules is shown in figure
7.
4.1
Helmholts free energy for the gas model
The free energy per unit volume can be described using the expression :
fv (n, τ ) = τ g hc (n) − an2
15
Figure 7: The total potential as a function of distance between the molecules
The first part of the expression is due to entrophy and hard core interaction,
the second part is due to Van Der Waals attraction. Where a is:
a = −2π
�∞
UV DW (r) r2 dr
d
d - molecule diameter.
Free energy per unit volume of ideal gas as we have showed it earlier in the
course:
�
�
N
fvid (n, τ ) = nτ ln
−1
nq · V
When we account for the volume each molecule captures : V → V − N · b
and the attraction force, we will recieve the free energy per unit volume of Van
Der Waals gas:
�
�
n
fv (n, τ ) = nτ ln
− 1 − an2
nq (1 − n · b)
Again : The first part of the expression is due to entrophy and hard core interaction, the second part is due to Van Der Waals attraction.
In this gas the repulsion interaction is the hard core interaction due to the
volume each molecule captures. n is the density, number of molecules per unit
16
volume can be calculated using the expression:
n=
N
V −N ·b
b represents the ”molecule volume” however its value changes with the density. In fact b is the volume a single molecule captures to itself and does not
allow to other molecules to penetrate into that volume. This volume is bigger
than the actual physical volume of the molecule. A molecule of radius R will
not allow to another identical molecule to get closer than 2R distance from their
center. If we consider the molecule to be spherical than its volume is:
ω=
4 3
πR
3
Thus the volume each molecule captures in space is:
b=
1 4
3
· π · (2R) = 4ω
2 3
The factor of 1/2 is due to the fact that each such volume is being devided
among 2 molecules. The expression for b is good for use only when the volume
is splitted between 2 molecules, meaning the density is small enough.
The maximal density that can be achieved in this model is:
n=
1
1
=
b
4ω
For high density not all the interactions are dual, b is smaller and approaches
the actual volume of the molecule, w. The molecule wavelength can be then
calculated (at room temperature):
λ=
�
�
∼
<< R
p
m·v
Thus we can neglect quantum effects. We notice that this is an inaccurate
model, however its advantage is in its simplicity, and the derived results are
good enough. In figure 8 fv is plotted for diffrent temperatures. It can be seen
that (only) for τ < τc , fv has a part with concave shape, which allows phase
seperation as explained above.
To find the spinodal line we will differentiate the free energy twice and equal
it to 0.
� 2 �
∂ fv
τ
=
2 − 2a = 0
∂n2
n (1 − n · b)
2
τs = 2an (1 − n · b)
In the area captured between the x-axis and the spinodal line there will be
no homogeneous phase, each phase will be devided into 2. In order to find the
critical point (nc , τc ) which is on the spinodal line we will demand:
� 3 �
∂ fv
=0
∂n3
17
Figure 8: fvid+hc+vdw for different temperatures
Figure 9: The spinodal line in a tempreature-density plot
18
Thus:
1
3b
8a
τc =
27b
nc =
4.2
Example: the water critical temperature and phase
transition
g
The water density is known to us : 1 cm
3
g
The water’s molar mass is : 18 mol
1
Avogadro constant : NA = 6 · 1023 mol
To calculate the volume of water molecule we will calculate the number of
molecules per unit volume and inverse it:
ω=
4 3
πR =
3
�
g
1 cm
3
23 1
g · 6 · 10
18 mol
mol
�−1
=
�
3.33 · 1022
1
cm3
�−1
= 3 · 10−23 cm3
R ≈ 1.5 · 10−10 m
Thus b is :
16 3
πR
3
Let us calculate the the Van Der Waals interaction, assuming that the molecules
are spherical:
−αd2e
UV DW =
2
(4πε0 ) r6
b = 4ω =
α = 4πε0 R3
de = eR
a = −2π
�∞
2R
−αd2e
2
(4πε0 ) r6
· r2 dr =
e2 R2
48ε0
And finelly we will get :
τc =
8a
e2
≈
27b
392πε0 (2R)
After we assign the number and substitute the units we will have Tc ≈ 600K
which is a very good approximation for the water’s critical temperature. For
most of the materials the critical temperature is between 100k to 1000k. When
q
heating water the vapor pressure acts accordingly to P ∼ e�− τ . Untill
certain
�
temperature which depends on the air pressure is reached 1000 C , the vapor
is created only on the water surface. At certain temperature when the vapor
pressure reaches equlibrium with the atmospheric pressure then bubbles will
appear inside the water itself.
19
5
Fluctuations
Let us place a Gas into a cylinder, now, let us compress it carefully, if we look
at the graph P − V we will see the pressure increase until it arrive to the first
Spinodal point, any disturbance will cause it drop down to the parallel curve
and reduce the volume V rapidly to the left Binodal point. We can see all the
above at the f − n graph, here it can seem the density n increase until it arrive
to the first Spinodal point. Physical, the gas change its’ Phase to liquid, but
why?
First, let us quantify the fluctuations at some volume V . As been taught at
Thermodynamics and statistical mechanics 1 we know that
∂<N >
)τ,V
∂µ
< (δN )2 >= τ (
This equation is at constant volume thus it can be divided with V 2 , then, one
has
τ ∂<n>
τ ∂<n>
τ
< (δn)2 >= 2 (
)τ,V = (
)τ =
∂µ
V
∂µ
V
∂µ
V ( ∂<n> )τ
µ=(
∂f
∂µ
∂2f
)τ ⇒
= ( 2 )τ
∂n
∂n
∂n
combining the two:
< (δn)2 >=
τ
V
∂2f
( ∂n
2 )τ
= kτ
Where k is the compressibility of the system.
It easy to see now that the fluctuations are infinite when the density reach to
∂2f
the binodal point where ∂n
2 → 0, Thus the system will ’probe’ its stable points,
until it will find the second binodal point which it is more stable then the current
one. That what we call a golbal stability point. Even if the density never arrive
to the spinodal point, as a result of external disturbance, a fluctuations will
begin at the binodal point, thus the system will find the global stability point.
Let us analyze the phase diagram, with temperature as function of density.
Let’s discuss what can we obtain from figure 10:
• The critical temperature τc and the critical density nc are at the peak of
the parabola.
• The area between the spinodal and the density axis represent a definite
phase separation.
• The Spinodal points are always inside the Binodal envelope, and they only
cross each other at the critical point (nc , τc ).
• For τ = τ1 the phases that will appear in the system are gas and liquid.
• Above the critical temperature the system is only in one phase - fluid
phase. However, that could not have been concluded from the graph.
20
• If we pass along the fluid path, as shown in the graph below, then we will
not be able to see a phase transition. The changes are continuous.
� 2 �
For a Spinodal line: ∂∂nf2v
=0
τ�
�
�
�
∂fv
v (n1 )
v
For a Binodal line : ∂f
|
=
|
= fv (nn22)−f
n
n
1
2
∂n
∂n
−n1
τ
τ
� 2 �
In order to calculate nc and τc two conditions must be satisfied: ∂∂nf2v
=0
τ
� 3 �
∂ fv
and ∂n3
= 0.
τ
• Underneath and above the line in figure 12 there is a change in the Chem-
Figure 10: availible phases as function of temperature and density
Figure 11: Shows the a possible trace (’fluid path’) in which there is no phase
seperation
21
Figure 12: pressure as function of temperature
ical Potential-µ so that:
�
∂µg
∂τ
�
p
�=
�
∂µl
∂τ
�
p
• On the line plotted the system is at equilibrium between the 2 phases (gas
and liquid) so that: µg (τ, p) = µl (τ, p)
For small changes along the equilibrium curve we can derive:
�
�
�
�
� �
� �
∂µg
∂µ
l
l
dτ + ∂pg dp = ∂µ
dτ + ∂µ
dp
∂τ
∂τ
∂p
p
dp
dτ
=
τ
�
p
� ∂µ �
− ∂τg
� ∂µ �p �
�p
∂µ
g
− ∂pl
∂p
∂µl
∂τ
τ
Remembering that:
µ=
G
N
τ
�
τ
� ∂G �
; dG = d(F + pV ) = −σdτ + V dp;
∂τ p,N = −σ
� �
�
�
∂µ
σ
⇒ N1 ∂G
= −s
∂τ p,N = − N = ∂τ
p
where s - entropy per one molecule.
Considering that the space of one molecule
� � is different
� �for liquid compared to
∂µ
1
V
gas we can derive that for one molecule: ∂p
= N ∂G
= −N
= −Vsmall
∂p
τ
substituting the formulas we get:
τ,N
dp
∆s
τ ∆s
L
dp
L
=
=
=
⇒
=
dτ
∆Vsmall
τ ∆V
τ ∆V
dτ
τ ∆V
22
where L - latent heat.
Assuming that L does not depend on temperature and that the gas behaves
like an ideal gas (Vg >> Vl ) we can derive:
L
p = p0 · e− τ
These conditions are usually satisfied at τ << τc .
6
Clausius Clapyeron equation
Next we shall explore the graph of pressure as a function of temperature (see
fig. 13). When we look at the coexistence line on the p − τ graph, there
is equilibrium between the two phases. the rest of the grapf has a uniform
phase. When both phases are in equilibrium, it means that they both have
equal chemical potentials:
µg (p, τ ) = µl (p, τ )
from this we can reach:
dp
sg − sl
τ ∆p
L
=
=
=
dτ
vg − vl
τ ∆s
τ ∆v
σ
V
where s = N
is the entropy per one molecule, and v = N
is the volume per
one molecule. L = τ ∆s is the latent heat (for particle) of the phase transition.
it means that this is the heat needed for one gas molecule to transfer into the
liquid phase. For the latent heat be positive we will define it as the heat the
system needs to move one molecule from the more ordered state (liquid) to the
less order state (gas) ∆s = sg − sl > 0.
Figure 13: pressure as function of temperature
23
7
Phase transitions at different temperatue scales
7.1
Low temperature approximation
Looking at temperatures below critical τ � τc we will prove the following
approximations:
1. We will assume that the latent heat is independent of the temperature ⇒
L �= L(τ ). We will explain this assumption later.
2. The differences between liquid and gas become much more evident. The
liquid will be much denser than the gas ng � nl and from the relation
v = n1 we can easily see that the gas will take much more volume than the
liquid vg � vl .
3. ⇒ We can neglect vl and make the approximation of ∆v ≈ vg .
4. We shall assume that the gas is diluted enough so we can treat it as ideal
gas.
5. p = ng τ =⇒ ∆v = vg =
=
1
ng
τ
P
Now we will solve the Clausius Clapeyron equation :
dp
dτ
Lp
L
τ ∆v = τ 2
dp
Ldτ
p = τ2
=
We get differential equation, which can be solved by integration of both
sides. For solving the integral we must define a range. We shall assume we
know a specific point (p0 ,τ0 ) on the p − τ graph .
�p
� τ dτ �
L
� τ0 τ �2�
1
L τ0 − τ1
�
�
�
− Lτ = p∗ exp − Lτ
dp�
p0 p�
ln pp0 =
�
p = p0 exp
L
τ0
=
L
Where we defined p∗ to be p∗ = p0 e(− τ ) .
We got that the pressure of saturated gas depends exponentially on the
temperature (see fig. 14), whereas the presure of ideal gas depends linearly on
the temperature. The reason that saturated gas behaves differently than ideal
gas is that in ideal gas the number of the molecules is constant as opposed to
saturated gas where the number of molecules grows exponentialy.
We will now see that the latent heat L is actually the difference in enthalpy
(∆h) between the gas phase and the liquid phase. The enthalpy is defined as:
H = U + pV
24
we will go back a few steps and remember that:
dU = τ dσ − pdV ⇒ τ dσ = dU + pdV
if we perform the experiment at a constant presure we have:
p = const ⇒ τ dσ = d(U + pV ) = dH
dividing by the number of molecules we finelly get: dh = d (u − pv) = τ ∆S = L,
where h is the enthalpy for one molecule.
∆h = hg − hl = (ug − ul ) + p (vg − vl )
Let’s try to understand what ∆h is composed of and prove that for τ << τc ,
∆h does not depends on temperature.
For the case of τ � τc and vg � vl we will make the ideal gas and low
temperature approximation:
p = ng τ and vg = n1g ⇒ pvg ∼
= τ.
The total energy of a liquid molecule, including the potential energy which
3
a molecule has from its neighbors is: ul = −�z
2 + 2τ
From equipartition theorem, the energy of a gas molecule is : ug = 32 τ
Finally we get the difference in enthalpy:
∼ �z + τ
L = ∆h = (ug − ul ) + p (vg − vl ) =
2
� �� � � �� �
�z
2
τ
but since we are making a low temperature approximation � � τ , so we can
neglect τ and remane with ∆h ∼
= �z
2 .
So we see that for first order approximation the latent heat is not dependent
of the temperature,it is constant. This is logical because of the big diffrences
between the two phases at low temperatures.
Figure 14: p-T for saturated gas
25
7.2
Finding binodal points for τ << τc
Lets go back to the free energy density for Van Der Waals gas:
�
�
n
f = nτ ln
− 1 − an2
nq (1 − nb)
(18)
another way of finding the binodal points is by using :
f (n2 ) − f (n1 )
∂f
∂f
|n =
|n =
∂n 1
∂n 2
n2 − n1
(19)
Usually it will be difficult to solve it, but when τ << τc the densities are low for gas,
and high for liquid and we can use some approximations.
First assumption is that the gas has such a density that he acts as an ideal gas
which means
∂f
n
|n = µ = τ ln
(20)
∂n
nq
and that the liquid density high enough to approximate it to nL ≈ 1b .
When τ << τc we have n1 << n2 , |f (n1 )| << |f (n2 )| so we can write equation
19 as
f (n2 )
∂f
|n =
(21)
∂n 2
n2
the solution of this gives us a correction to the liquid density:
nL ≈
1
τ
−
b
a
(22)
(simple algebra and using nL ≈ 1b )
When we put these back with the chemical potential of ideal gas:
f (n2 )
∂f
n1
|n = τ ln
=
∂n 1
nq
n2
(23)
we get the density and pressure
� a�
a
exp
−
(24)
τ b2
τb
� a�
a
p = nτ = 2 exp −
(25)
b
τb
We got the same expression using latent heat and Clausius-Clapeyron equation.
This is explained well by the meaning of latent heat far from the critical temperature: Its the bond energy between a molecule and its neighbors.
When a molecule is changing phase from liquid to gas all the bonds with its
neighbors must be broken. The heat (energy) invested in breaking these bonds is
the latent heat.
n1 =
7.3
Finding binodal points and other physical entities for
τ � τc
Now lets look at a temperature close the the critical temperature. we define:
∆τ = τ − τc
∆n = n − nc
26
When τ ≈ τc :
|∆τ | << τc
|∆n| << nc
We want to expand this expression near the critical temperature until the forth
order ∆n4 this is a lot of work since we have two variables to expand by - both ∆τ
and ∆n. Instead of that we will expand until second order the second derivation of f :
τ
∂2f
=
− 2a
∂n2
n(1 − nb)2
(26)
and then itegrate twice in order to get to the forth order expantion.
First we’ll wirte f as a function of ∆n and ∆τ :
∂2f
τc + ∆τ
=
− 2a
∂n2
(nc + ∆n)(1 − (nc + ∆n)b)2
(27)
1
We take nc = 3b
⇒ b = 3n1 c
τc
8 a
and nc = 27
· 3b = 89 a
b
�
�
2a 1 + ∆τ
τc
∂2f
��
� − 2a
= �
∆n 2
∂∆n2
1 + ∆n
1 − 2n
)
nc
c
� � � �
� ∆τ � � ∆n �
we expand when the small parameters are � τc � , � nc � << 1
using (1 + x)α = 1 + αx we get:
= 2a
∆τ
3
+ a
τc
2
�
∆n
nc
�2
(28)
(29)
Now we integrate twice and get
f (∆n, ∆τ ) = a
∆τ
a
∆n2 +
∆n4
τc
8nc 2
(30)
What can we learn from this expression?
Lets look at it in different temperatures: τ1 > τ2 > τ3
Lets look at τ1 - we can see a parabolic behavior for small ∆n (around nc ) and a
bigger influence of ∆n4 for bigger ∆n.
As τ grows the parabula opens. When τ = τc we have only ∆n4 and we get a
flater graph.
Lets look at the density fluctuations:
< ∆n2 >≈
τ
∂2f
∂n2
(31)
When the temprature gets closer to τc , the second derivation decrease hence the
fluctuations increase. As the potential becomes flater, the fluctuations grow, there will
27
Figure 15:
∂2f
∂n2
as a function of n for different temperatures
28
be more separations to different phases, and we will see the phenomenon called critical
opalescence Where light is scattered from the substance.
Why does the light scatter?
As we get closer to τc the average density stays constant, but there are big fluctuations in the refractive index - because the refractive index is dependant of the density.
As the light moves through the different refractive indexes few times it scatters.
We can see that also in water: In low temperature we will have two phases and
in critical temperature the substance will be homogeneous and will scatter more and
more light.
Why is the milk white?
milk is a mix of fat (oily substance) and water. since the two substances don’t
mix, and have different refractive indexes we get the same phenomenon of light going
through two different refractive indexes few times and get scattered - htis means they
cannot pass through the liquid in strait lines and it will not be transparent bkut white
and the white light that is refracted.
We are continuing the discussion of a Wan der Valaas gas described by the
equation :
n
f = nτ (ln( nq (1−nb)
) − 1) − an2
We are interested in studying the behavior of the system near the critical
point (τc , nc ) In the last lesson we arrived at the expression :
f (∆τ, ∆n) = τac ∆τ ∆n2 + 8na 2 ∆n4
c
We observe that as we approach τc from above the curvature decreases and
the fluctuations increase.
< ∆n2 >∼
1
∂2 f
∂n2
When we approach the critical point (τc , nc ) from below meaning: τ < τc
we get:
f (∆τ ∆n) = − τac |∆τ |∆n2 + 8na 2 ∆n4
c
∂f
In order to find the binodal line we can demand: ∂∆n
= 0 = − 2a
τc |∆τ |∆n +
a
2
2n2c ∆n
This because of the symetry of the graph of the free energy near the critical
point.
�
∆n = ±2nc ∆τ
nc
�
n = nc + ∆n = nc (1 ± 2 τcτ−τ
)
c
where a plus sign will refer to a liquid phase and a minus to a gas phase. One
can see that at the critical point the densities are equal and when we diverge
from that point they separate.
how does the entropy change?
In order to calculate the entropy near the critical point τ ∼ τc
notation: s ≡ σpermolecule
We expand and recieve:
σ
2
4
N ≈ ln2 − 2s − s
|∆τ |
|
s = ln2 − 3/2 τc − 1/10 |∆τ
τc
29
Figure 16: second degree phase transition
Thus we see that until the critical temprature the entropy rises lineary(due
to first aproximation) then past τc in is constant ln2
Heat capacity :
dσ
C = δQ
δτ = τc dτ
We see that it is constant(in first aproximation) untill τc and then drops to
zero.
first and second order transitions
If we examine the system with nc and above τc keeping the concetration
constant, we will go by second degree phase transition. There will be no discontinuity
.
On a different senario if we are not at critical density and lower the temperture we have a discontinuity . This is called first order phase transition.
8
Spin lattice
It can be deduced that the free energy of a spin in a latice model in which:
• Each spin can be up or down
• There are N spins
30
Figure 17: first order phase transitions
• Each magnetic dipole has z relatives which can affect it
• All the resorts are full
• When two relative magnetic dipoles which has the same sign don’t contribute energy, but two in opposite signs contribute � > 0 energy
is:
F
= τ (ϕ ln ϕ + (1 − ϕ) ln (1 − ϕ)) − aϕ (1 − ϕ) ,
N
a = zε,
ϕ=
N↑
N
(32)
We can easily see that the free energy is symmetrical with respcet to ϕ = 0.5,
therefore we shall build a magnetic spin as:
1
2
1
ϕ=s+
2
s=ϕ+
−1 < s < 1 Puting that into 32 we get:
��
� �
� �
� �
��
�
�
F
1
1
1
1
1
=τ
+ s ln
+s +
− s ln
−s
+a
− s2
N
2
2
2
2
4
(33)
(34)
(35)
There is a difference between the free energy of Van Der Wals gas to the one of
the magnetic depoles latice because when we change the density in one of the
31
phases in the gas model it has to change respectively in the other phase because
the number of particles is constant, but in the magnetic depoles every density
of each phase isn’t related to the other phase.
We shall derive the free energy to find its minimas.
�
�
� 1 + 2s � 2a
1 ∂F
�=
= 0 ⇒ ln ��
s
(36)
N ∂s
1 − 2s �
τ
It is easy to see that the at s=0 the free energy is maximal because the entropy
in ϕ = 0 is maximal.
Examing how the left side of expreesion 36 is acting near s=0.
lim (ln |1 + 2s| − ln |1 − 2s|) = {ln (1 + x) � x} � 4s
s→0
(37)
Close to s=0 the the function is acting like a linear function and is aspiring to
infinity and negative infinity as s is getting close to 1 and −1 accordingly.
Therefore for:
a
≥τ
(38)
2
has only one acceptable phase as a minima for equation 36
For:
a
<τ
2
There are two acceptable phases as minimas for the same equation.
thus we shall call:
a
= τc
2
(39)
(40)
the critical tempature.
At the tempature lower from the critical tempature but close to it, s << 1.
The closer we get to τc the left side of eq. 36 becomes flatter and more
fluctuations happen.
��
� �
� �
� �
��
σ
1
1
1
1
1
=−
+ s ln
+s +
− s ln
−s
≈ − ln − 2s2 − s4 = ln 2 − 2s2 − s4 (41)
N
2
2
2
2
2
We can find the spinodal points near τc from eq. 36:
�
�
�
�
� 1 + 2s �
�
� ≈ 4s + 16 s3 = τc = a = 4 τc s
ln �
�
1 − 2s
3
2
τ
⇓
�
�
3 τc − τ
3 τc − τ
s=±
≈
4 τ
4 τc
(42)
(43)
(44)
s(τ ):
�
s = ± 34 τcτ−τ
,
c
s = 0, τc < τ
32
τc > τ
(45)
Figure 18: illustration of eq.36 for different temperatures
Figure 19:
33
We only estimated how the spin is going to react at tempatures closed to τc
therefore it passes s=0.5, which physicly it can’t. Without this estimation the
function supposed to drop faster.
The phase transition is a phase transition of the 2nd order, which mean the
function is continous but its derivative isn’t. Placing s(τ ) from equation (15) in
the entropy for τ < τc :
�
�2
σ
3 ∆τ
9 ∆τ
3 ∆τ
≈ ln 2 −
−
≈ ln 2 −
(46)
N
2 τc
16 τc
2 τc
for τ > τc s=0 therefore:
σ
= ln 2
N
The entropy is continous at phase transition.
The heat reception is:
� 3
δQ
τc ∂σ
τ < τc
2τc ,
Cv =
=−
≈
0, τ > τc
dτ
∂∆τ
(47)
(48)
The heat reception isn’t continous at phase transition.
9
Surface energy
When talking about a glass of water, for example, so far we referd to the the
water molecules whithin the glass and not on the surface, the molecule on the
surface has less neighbors. We will assume the surface-molecule has half the
neighbors and therefor it’s energy will be u = −�z
4 .
The energy diffrece between an inner-molecule and a surface-molecule will
then be:
∆u ≈
−�z
−�z
�z
L
−
=
≈
4
2
4
2
We shall define the surface area energy
γ=
∆Fs
L
≈
∆A
2πR2
where Ris molecular radius.
We can find ∆Fs by deviding the surface area energy γ of the molecule by
the surface aree of the molecule.
In some cases we can neglect γ because the ratio between the number of the
inner molecules and the number of the surface molecules : Nv ≈ D3 , Ns ≈ D2
Ns
1
=⇒ N
=D
.
v
In other cases we must not neglect γ this occurs when treating small objects,
or if we are talking about a surface phenomenon. For small objects with relatively large volume, surface effects are important because the volume is fixed
and all of the changed in free energy are due to changes of the surface.
34
9.1
Nucleation and grows process
We come back to the question of why the phase separation does not start in a
small volume right away after passing the binodal point. We will look at the
creation of a new phase (a drop of liquid in phase).
The constants in the system are p and τ , G = F + pV is the Gibbs thermodynamic potential and all the changes in the system should work in direction of
reducing it G(τ, P ) �
4
∆G = |∆µ| πR3 nl + γ4πR2
3
The first term is the volume energy of the drop and the second term is the
surface energy of the drop.
The system is always driving the reduce the Gibbs energy, (see fig. 7).
- If the radius of the drop is smaller than the critical radius Rc the system
will shrink it until it disappears.
- If the radius of the drop is bigger than Rc , then in order to reduce the gibs
potential the system will enlarge the drop of the liquid phase.
2γ
from d∆G
dR = 0 we get the critical radius to be : Rc = nl ∆µ .
Part II
Kinetic Theory
10
Boltzmann and Maxwell velocity distribution
The expression for the kinetic energy of a particle moving in the x direction:
m < vx2 >
τ
=
2
2
< vx2 >=
(49)
τ
m
(50)
The expression for a particle moving in a general direction in a three dimensional space:
< v 2 >= 3 < vx2 >=
3τ
m
(51)
The speed of Nitrogen molecules which make up most of earth’s atmosphere:
N2 : Mm = 14.2[
gr
28 · 10−3
]m=
kg
mol
NA
35
NA - Avugadro number
√
< v 2 > ∼ 500
m
sec
Velocities Diverge according to the Boltzman Divergence.
Let’s look at only one Dimension (Particles move only along the x axis).
Figure 20: illustration of a particle free to move in 1 degree of freedom
We’ll also assume that only the velocity determins the energy of a particle
�=
mVx2
2
�
P (vX ) ∼ e− τ
It is meaningless to ask what is the probability of each velocity since the
span of possible velocities is not only infinite but also innumerable.
Figure 21: veocity distribution
We’ll find instead the value of the probability density.
dp = P (vx )dvx
P (vx ) =
dp
dvx
p(V1 < Vx < V2 ) =
�v2
v1
36
(52)
(53)
P (Vx )dvx
(54)
Figure 22: distribution of the velocity in 1 dimension
We’ll use the Boltzman probability distrbution:
p(vx ) = c · e
2
−mvx
2τ
(55)
We’ll find the constant c by normalizing:
�
+∞
+∞
�
�
2
−mvx
2πτ
p(−∞ < vx < +∞) =
P (vx )dvx = 1 ⇒ c
e 2τ dvx = 1 ⇒ c
= 1 (56)
m
−∞
−∞
and Finally we get:
c=
�
m
⇒ p(vx ) =
2πτ
�
m −mvx2
e 2τ
2πτ
(57)
We can see that the average velocity is obviously zero
We’ll now consider a three dimensional world and define a volume in the
space of velocities.
�v = (vx , vy , vz )
37
Figure 23: a Cartezian integration
p(�v ∈ Ω) =
�
P (�v )d�v :=
�
P (�v )dvx dvy dvz
(58)
The energy, corresponding to the velocity in each axis, appears independently
in the expression for the probability distribution.
Therefore the probability for a velocity defined using all three axes is equal
to the product of the three separate probabilities (for each axis).
p(�v ) = c3 e
−mvx 2
2τ
e
−mvy 2
2τ
e
−mvz 2
2τ
= c3 e
−mv 2
2τ
+∞ �
+∞ �
+∞
�
P (�v )dvx dvy dvz = 1 ⇒ c3 = c3
(59)
(60)
−∞ −∞ −∞
3
−mv 2
m 2
p(vx ) = ( 2πτ
) e 2τ
Now that we have seen how the velocity probability is distributed, we’ll want
to see how the speed (scalar) probability is distributed. The speed probability
is called Maxwell probability, PM .
To do that we’ll integrate over shells in the three dimensional velocity space.
38
Figure 24: a thin sphere integration
The volume of each shell is defined as the region: [v,v+dv]
p(�v ∈ Ω) =
�
P (�v )d�v
(61)
Ω
dpM = PM (v)dv
PM (v) =
�
Ω
3
m 2
P (�v )d�v = ( 2πτ
) e
−mv 2
2τ
·
�
Ω
(62)
3
m 2
d�v = 4πv 2 ( 2πτ
) e
−mv 2
2τ
Figure 25: velocity distribution shows a double gaussian
We can see that the most probable value for speed is not zero even though
the average velocity we’ve encountered earlier was zero.
39
Figure 26: Illustration of molecules and a barrier
11
11.1
The kinetic theory of gases
Introduction
Kinteic Theory examins How a system changes from one state to another. The
kinetic theory is more general than statistical theory. Until now we have been
studying about states of equilibrium. Now we’ll talk about the process of getting
from one equilibrium state to another. This is a more complicated theory but
it will give us more. From the statistical point of view we derived for ideal gas
� �
∂f
N
p=−
= nτ ; n =
∂τ τ,N
V
Let’s try to understand this relation from the point of view of kinetic theory.
Kinetically, the pressure is formed from gas molecules hitting the wall . Let’s
assume the wall is a YZ plane. The molecules hitting it have velocity component
vx .
If the collision is elastic and the wall stays stationary, the x component of
the momentum will change from mvx before the collision to −mvx after it. The
momentum that one molecule passes to the wall equals 2mvx .
For time interval ∆t and wall area S, the molecules near the wall will pass
a distance vx ∆t in the x direction. Therefore: The momentum that all the
molecules pass to the wall equals
� ∞
� ∞
∆P =
nvx ∆t S (2mvx ) P (vx ) dvx = 2S∆t nm
vx2 P (vx ) dvx
vx =0
0
Now according to Newton’s 2nd law:
� ∞
� ∞
F
∆P
2
p=
=
= 2nm
vx P (vx ) dvx = nm
vx2 P (vx ) dvx
S
S∆t
0
−∞
The last equation holds because the integrand is symetrical.
� �
p = nm vx2
� �
According to the equipartition theorem m vx2 = τ and we get the desired
p = nτ .
40
11.2
Random walks
As we �
have seen previously a molecule’s velocity variance in room tempreture is
m
about �v 2 � ≈ 500 sec
. However, the average molecule does not cross a distance
of 500m in a second, and that is because it will hit a few other molecules before
it will move a much shorter distance. In order for us to understand this random
process of collision between molecules we start by introducing random walks.
The easiest way to start explaining this term is by looking at a drunk person
walking along a one dimensional line, with every step being the same size (a)
and the probabilty for him to go left is equal to the one to go right. We shall
also assume that the steps are uncorrelated, hence �ai aj � = 0 for every i �= j.

probabilty 12
 a
ai =

−a probabilty 12
1
1
a + (−a) = 0
2
2
The total distance the drunk person passed
�ai � =
x=
N
�
ai
i=1
⇓
�x� =
�
�
x2 =
�N N
��
ai aj
i=1 j=1
�
N
�
i=1
�ai � = 0
N
�
� 2� �
� �
=
ai +
�ai aj � = N a2i = N a2
i=1
⇒
�
i�=j
√
�x2 � = a N
As we
see, as the amount of steps N grows bigger so does the relative
√ can
�x2 �
variace N = √aN decreases, hence the drunk hardly leaves the origin point.
We can use this mathematical model in order to describe the lattice model
for gas, where a molecule arriving at a new cell might collide and change its
velocity’s direction, or keep on moving in the same direction.
The rate a molecule moves from one cell to another will be defined :
Γ=
number of steps
∆N
=
time
∆t
� 2�
2
⇒ x = Γa t
Defining the diffusion coefficient:
D≡
Γa2
m2
; [D] =
2
sec
41
Figure 27: path cylinder of one moelcule
� �
⇒ x2 = 2Dt
Going back to our drunk person model, we’ll now calculate the probability
to find him at distance of x from the origin after N steps - P (x, N ). We can
define x as :
x = a [(steps to the lef t) − (step to the right)] = a(N← −N→ ) ; N = N← +N→
Therefore P (x, N ) = P (N← , N ).
We can see that we got the same expression as he had for a system of N
spins, so the probability density function is:
�
2 −x2 /(2a2 N )
P (x, N ) =
e
πN
The values that x can have are discrete, so in order for us to move back to the
continuous world we shall define
N = Γt
⇒ P (x, t) =
11.3
2
P (, x, N )
1
|D= Γa2 = √
e−x /(4Dt)
2
2a
4πDt
Mean free path
Let’s calculate the mean square free path (< l2 >), and learn about the distribution of free paths.
First we need to find the probability of 1 molecule ,passing the distance dl,
having a collision (dpc ).
If the center of another molecule is in the cylinder, then the molecules will
collide. Therefor we will find the probability for the molecules presents in a 2d
42
Figure 28: long path cylinder of one moleclue
Figure 29: Illustration of 1D lattice with length l
radius cilinder (where d stands for the diameter of the molecule).
dpc = πd2 ndl
1
λ=
nπd2
dl
⇒ dpc =
λ
(63)
(64)
(65)
now lets draw a longer ”Cilinder”,
and look at the path the molecule travels (N collisions, L path length):
N = πd2 Ln
L
1
λ=
=
N
πd2 n
(66)
(67)
now we ask what is the probability that the molecule won’t collide until the
distance l(Pnc (l)).
k distances of the length ∆l.
for each ∆l, the probability there won’t be a collision:
∆l =
lim (1 −
k→∞
l
k
l
∆l k
l k
) = lim (1 −
) = e− λ
k→∞
λ
λk
43
(68)
(69)
Figure 30: substance-barrier contact
the probability is decreasing exponentialy.
The probability there is a free path from l to l + dl (move the distance l
without collisions and collide within dl):
l
e− λ dl
P (l)dl =
λ
� ∞
l
e− λ
⇒ P (l) =
(
P (l)dl = 1)
λ
� ∞
� 0∞
l −l
< l >=
lP (l)dl =
e λ dl = λ
λ
0
0
(70)
(71)
(72)
as we would anticipated
< l2 >=
�
∞
l2 P (l)dl =
0
�
∞ 2
0
l −l
e λ dl = 2λ2
λ
(73)
Diffusion index:
D=
1 v < l2 >
1
= vλ
6
λ
3
(74)
Diffusion definition:
∆N
∆S∆t
= −D∇n
J=
JD
(75)
(76)
where J is the diffusion flux.
11.4
Probability evolvement in time
If we know the distribution of the probabilities in time t, we can predict how
the distribution of the probability will change in ∆t.
The lattice model will assist us:
44
-the difference between the coordinates is a.
∆P (x, t) = P (x, t+∆t)−P (x, t) = P (x − a, t)Γ→ ∆t + P (x + a, t)Γ← ∆t − (Γ→ + Γ← )P (x, t)∆t
�
��
� �
��
� �
��
�
(1)
(2)
1. The probability to jump from x-a to x coordinate.
2. The probability to jump from x+a to x.
3. The probability to escape from x (to the left or to the right).
The distribution of the probabilities in x coordinate might change as a result of
a jump from x-a coordinate.
P (x+a,t)−P (x,t)
(x−a,t)
− P (x,t)−P
∆P (x, t)
Γa2
a
a
=
·
← Its a second order
∆t
2
a
derivative.
1 ∂2P 2
It is also can be done by Taylor series P (x = a) = P (x, a) ± ∂P
∂x a + 2 ∂x2 a +
...
∂p
∂2p
∂P
∂2P
← Fick’s law
∂t = D ∂x2 ⇒ ∂t = D ∂x2
We can get a distribution in any time.
There are� few ways to do that, one of them is Fourier.
∞
p(x, t) = −∞ P (q, t)eiqx dq
∂p(q,t)
∂t
= −Dq 2 P (q, t)
p(q, t) = p(q, 0)e−qDt
p(�r) = p(x, y, z)
2
∂P
2
2
y 2 > + < z 2 > = 6Dt
∂t = D∇ P < r >= <
� x
�� >� + <
� �� � � �� �
2Dt
2Dt
2Dt
P (�r) = P (x)P (y)P (z)
In average a molecule passes a distance λ between collisions.
D ∝ Γa2 ∝ λv λ2 ∝ λv , with v being the average velocity. Let’s assume that
after every collision the molecule goes to a random direction (that is not true for
same mass molecules) it is true for a small molecule that collides with a heavy
molecule. A molecule that moves in time t and had been in N collisions.
�
�N
�
< r2 >= ij < li lj >= i=1 li2 + i�=j < li , lj >= N < l2 >=
<l2 >vt
λ
= 6Dt
2
D = 16 v<lλ > , where D is coefficient of diffusion.
11.5
Definition of pressure from dynamical point of view
Now we shall look for an expression for pressure from the dynamic point of view,
instead of the statistical one as we used before.
Let us examine a molecule in a certain volume, assuming that when it hits
one of the walls the collision in elastic so that the system’s energy in preserved.
Therefore if the molecule hits the wall with a velocity of vz it shall continue
moving with a velocity of −vz . The molecule’s momentum changes by ∆Pz =
45
(3)
−2mvz (m being the molecule’s mass), and as a result of the law of conservation
of momentum the wall also receives a momentum change of ∆Pz = 2mvz .
Clearly, when looking at a volume with many molecules that collide with
the wall we understand that the momentum the wall recieves is not negligible.
According the Newton’s second law :
dP�
F� =
dt
Pressure is defined as the force applied on a surface:
p=
|F |
∆P
=
S
S∆t
We shall calculate how much momentum is passed on to the wall during a
period of ∆t. Looking at all the molecules with a certain velocity vz , we know
that all the ones in a distance of ∆z ≤ vz ∆t from the wall will hit it and pass
on a momentum of ∆Pz = 2mvz . The amount of molecules who will do so is :
N (∆t · vz ) = ∆tvz SnP (vz )dvz
when :
• ∆tvz S - the volume which contains the molecules that will hit the wall
• n- density of molecules
• P (vz )dvz - the probabilty of a molecule having a velocity vz .
Therefore the change of momentum will be:
� ∞
∆P =
2mvz ∆tvz SnP (vz )dvz
0
And the pressure on the wall :
� ∞
� ∞
� �
∆P
τ
= 2mn
vz2 P (vz )dvz = mn
vz2 P (vz )dvz = mn vz2 = mn = nτ
p=
S∆t
m
0
−∞
As we can see, we have received the same expression for pressure as we got
from the statistical point of view.
12
12.1
Diffusion
Definitions
In the following lecture we will talk about diffusion. First we will define the
flux:
∆N
jN =
∆S∆t
46
The direction of the flux is opposite to the direction of the density gradient, so
we can define Fick law in the following manner:
j�N = −D∇ n
In this lecture we will see that diffusion is a liner transport phenomena. Transport phenomena cause a un-equilibrium system to get to a equilibrium state.
This phenomena is liner because the system is near equilibrium and therefore
the flux depend on the gradient in a liner way.
We can observe another liner transport phenomena when we inspect a temperature gradient in a system. In that kind of system we can define a heat flux
:
∆Q
jQ =
∆S∆t
Again the direction of the flux is opposite to the temperature gradient, and
therefore we can define the Fourier law in the following manner:
j�Q = −k∇ τ
Where k is the Thermal Conductivity.
Another liner transport phenomena related to the viscosity of the fluid is
described in the following figure:
We can define Shear Stress :
F
∂Vx
=η
S
∂y
(77)
Where η is the viscosity.
We remember that we can define force in the following manner :
∆P�
F� =
(78)
∆t
Using the flux definition with the aid of eq.77 and eq.78 we can derive the
momentum flux equation :
∆P
∂Vx
j�P =
= −η
∆S∆t
∂y
47
It is important to say that the collective diffusion coefficient is not always equal
to the diffusion coefficient that was defined in the random walk lecture. In gases
it is happening while in liquids it is most of the time not.
When we are looking at a system with temperature gradient:
The kinetic energy of the molecules from the left side of the arbitrary border
is higher from that from the left side. The concentration gradient is zero and
therefore the same amounts of molecules are cross the border from each side.
Therefore there is a kinetic energy flux from the left side to the right. This
energy is as we know - thermal energy.
In the same manner, when we look at a fluid with viscosity, the same amounts
of molecules are moving across the layers in the different directions. In the upper
layers there is more momentum to the particles and therefore in total we observe
a momentum flux trough the bottom.
Now we are ready to develop the theory behind that phenomena. The theory
will be more accurate if we will look at the gradient of a light molecules in a
heavy molecules environment.
Thermalization : The molecule should collide with its neighbors several
times in order to adopt the layer temperature.
48
We would like to know where the last collision has happened. It is more
comfortable to check molecules inside small volumes and to see whether they
cross the surface(S) without to collide in their way with other molecules and
after that to calculate the integral over the entire domain.
12.2
Derivation of flux caused by potential gradient
In this subsection we’ll derive the equations of motion of linear transport phenomena, starting with diffusion.
The assumption which will lead to a linear description of the phenomena
is based on the simplest model that describes this phenomena - Small distortion from equilibrium, meaning that the gradient of the density (or any other
parameter) might not be zero, but constant.
First we need to define the geometry of the problem: Consider a small volume
element dV located in space, at distance x along the x axis, and angle α above
it. Also a small aperture with size s is located at x = 0.
Figure 31: Finite element in space
The element is located at distance r = cosx α from the aperture, and seemed
with spatial angle dΩ. Since the element is not perpendicular to �r, but with
2
r 2 dΩdx
angle α, its actual surface is rcosdΩ
α , hence the volume of this element is cos α .
Note that we are assuming that each part of dV is seemed with angle α from
each part of s and vice versa, meaning that r � s and r � (The diameter of
dV ).
This assumption is legitimate, because both s and dV are infinitesimal.
Now we need to understand the kinetics of the problem, by counting the
molecules that will collide inside this element, thus receive its temperature, and
pass through the aperture without colliding with any other molecule. The
reason is that we want each molecule from dV to pass through the aperture,
with the same temperature it got in the element dV .
49
A single molecule that moves with average velocity v, and random free walk
λ will collide with other molecules N times in time ∆t. The number of free
walks λ in this time is therefore the number of collisions. It is also the distance
that the molecule will pass and is equal to the average velocity times ∆t:
Nλ
N
is:
= v∆t
v∆t
=
λ
Now, if the density is n, the number of collisions per unit volume in time ∆t
v∆t
λ
The directions of the molecules’ movements in dV are distributed uniformly
α
over 4π steradians. The spatial angle which dV “sees” the aperture s is s cos
r2 ,
s cos α
therefore the part of the molecules that are in the direction of s is 4πr2 .
Finally, the probability for a single molecule to pass the distance r without
r
any collision through its path is e− λ .
Now we can derive the number of molecules that will collide inside dV at
time ∆t and pass through s without colliding with any other molecule outside
dv through their path:
N =n
dN (x, Ω) =
nv∆t r2 dxdΩ s cos α − x
e λ cos α
λ
cos α 4πr2
Next we will define the flux from dV to be the number of molecules that are
passing x = 0 from right to left per unit surface per unit time:
dj←
dj←
dN
s∆t
nv − x
= −
e λ cos α dΩdx
4πλ
= −
In order to find the total flux through a small aperture at (0, 0, 0) we need
to sum all of the contributions from each volume element in space.
Remember that we are using first order approximation so v is constant and
∇n is also constant all over the space.
Each volume element at x > 0 contributes flux from right to left, and each
element at x < 0 contributes flux from left to right. Therefore one can integrate
over the right half-space at x > 0 while adding the negative contribution from
−x:
�
J�n =
(dj (x) − dj (−x))
x+
�
� ∞
|x|
v
J�n = −
dΩ
[n(x) − n(−x)]e− λ cos α dx
4πλ α< π2
0
50
According to the assumption of constant density gradient:
n(x) − n(−x)
= ∇n
2x
n(x) − n(−x) = 2x∇n|(x=0)
Thus we get:
v∇n
= −
2πλ
J�n
�
dΩ
α< π
2
�
∞
|x|
xe− λ cos α dx
0
Dividing and multiplying by λ2 cos2 α:
�
� ∞
� x �
|x|
v∇n
x
J�n = −
λ2 cos2 αdΩ
e− λ cos α d
2πλ α< π2
λ cos α
λ cos α
�0
��
�
J�n
J�n
= −vλ∇n
�
�
1
α< π
2
1
= − vλ∇n
3
cos2 α
dΩ
2π
��
�
1
3
One can easily see that the same diffusion coefficient as in Random walks: D =
1
3 vλ has been derived.
Now taking into account temperature gradient:
Since the temperature is derived from the energy of the molecules, the calculation will be similar to the above,replacing the density n with the energy �,
and the flux would be flux of heat.
�
� ∞
|x|
v
�
JQ = −
dΩ
[�(x) − �(−x)] e− λ cos α dx
π
�
��
�
4πλ α< 2
0
�(τ (x))
Here:
�(x) − �(−x) = 2x
∂�
∂τ
=
2x
∂�
∇τ |(x=0)
∂τ
∂�
Let us remember that ∂τ
is C1 - the heat capacity of a single molecule.
Repeating the same calculations as above we get:
1
J�Q = − vλC1 n∇τ
3
The expression nC1 equals to CV - the heat capacity per unit volume and thus
we get again fourier law:
1
J�Q = − vλCV ∇τ
3
�
JQ = −Dc CV ∇τ = −κ∇τ
51
Where κ is the thermal conductivity.
Moving to viscosity - we are looking for the momentum on x direction flux
along y axis. A small change of the molecules’ velocity at y0 causes momentum
gradient down through the y axis direction:
Figure 32: Momentum flux
Down through the system, the momentum flux must be constant (matter
conservation), therefore the gradient which is proportional to the flux, is also
constant.
�
� ∞
|y|
v
J�p = −
dΩ
[mvx |y − mvx |−y ]e− λ cos α dy
4πλ α< π2
0
Wehre m is the mass of the molecule.
Although the system is not spread all over the space, it is legitimate to
integrate all over the space, because the significant values of mvx remains in
distance of order λ. Therefore we get:
= 2y
vx |y − vx |−y
∂vx
∂y
Notice that although v is constant, its x component vx might not be.
Again repeating the calculations, and taking into account the viscosity coefficient η, that had defined in previous lessons, we get:
1
∂vx
= − λm
3
∂y
= Dc nm = Dc ρ
J�p
η
So far we got:
η
=
κ
=
1
vλnm
3
1
vλnCV
3
52
Noticing that λ ∝ n1 we get that with first order approximation and as long
as the free move is much smaller than the dimensions of the system, both the
viscosity and the thermal conductivity do not depend on the density.
12.3
Solids
When we talk about solids we have a periodic potential:
Figure 33: potential of a solid
Each molecule is vibrating in frequency of ν. The energy distribution is
Boltzman’s (∝ e−�/τ ), so the rate that the molecule will move between the sites
is:
Γright = νe−�/τ
Γlef t = νe−�/τ
⇒ D = Γa2 = νa2 e−�/τ
When we add the external force the barriers change so the well is not symmetric
and therefore the rate in which the molecule moves changes:
53
Figure 34: potential of a solid in an external force field
Γright = ν exp(−
v¯D = Γright −Γlef t
� − 0.5|∇ϕ|
)
τ
� + 0.5|∇ϕ|
Γlef t = ν exp(−
)
τ
�
�
|∇ϕ|
|∇ϕ|
|∇ϕ|
D
≈ aνe−�/τ 1 +
−1+
= νa2 e−�/τ
= ∇ϕ
2τ
2τ
τ
τ
And we find Einstein’s relation:
D
=b
τ
54
Lecture notes on Ideal Quantum gas
When we dealt with ideal classical gas we assumed that the concentration number of particles
� M τ � 32
”n” is small compared with what we defined as the quantum concentration n � nq = 2π�
.
At room conditions that is usually true, for example, Nitrogen is the most abundant gas in
−3
kg
Earth’s atmosphere with a molecular mass of 28g/mol or 28·10
at room temperature of
6·1023
�
� 32
J
−23
5·10−26[kg]·1.38·10
[ K ]·300
300[K] translates to nq =
which is about 2 · 1032 [ m13 ] however the
2π·10−68
concentration it has is only 2.5 · 1025 [ m13 ] (found from the law of ideal gases n = τt ). 7 orders
of magnitude is a big difference, even after compressing air to the maximum, considering air
molecules as billiard balls with a diameter of d = 3Å ,
n∗ =
1
d3
≈ 3 · 1028 [ m13 ].
There are ways to reduce the quantum concentration, one is by reducing the mass of the
particles, another is by lowering temperature. For instance, should we want to reduce nq by
a factor of 4 by means of temperature change it should be
τ∗
τ
2
= (104 ) 3 = 450, for Helium
atoms that would be about 1[K]. Should we like to reduce the mass, by addressing electron
gas in metal (ignoring internal interactions) me = 10−30 [kg]nq = 1025 [ m13 ] which at room
conditions n � nq .
One way of categorizing elementary particles is by spin ”s” which is a fundamental characteristic property of elementary particles analogues to a particles angular momentum. Quantum
gas is divided to two categories:
Fermions with half-integer spin number ( 12 , 32 ...) who’s characteristic property is that two
fermions cannot occupy the same state, the most known fermions are the Electron, Electron
neutrino, Muon and others.
Bosons with integer spin number (0, 1, 2...) who’s basic property is that any number of bosons
can occupy the same state, the most known bosons are: Photons, He 4, W boson, Z boson
two other famous yet unconfirmed bosons are the Higgs boson and Graviton.
According to quantum mechanics, one can describe a particle as a wave function, so we shall
define
p2
�=
2m
,
p = ��k
,
�
πn
k = kx2 + ky2 + kz2 =
L
,
n = 1, 2, 3...
where L is the dimensions of the system taking particle-in-a-box boundary conditions.
1
(1)
1
Quantum ideal gas
Density of states and concentration of energy
We consider the arrangement of electrons around an atom, the electrical force strives to get
each electron as close to the nuclei , but because electrons are fermions only two can have
the same state, one for ”spin up” denoted as s =
1
2
and one for ”spin down” s = − 12 . Our
first aim is to find the distribution of energy, and total energy.
For that we will first define the total number of states in one octant of a sphere of radius n as
√
p2
π 2 �2 2
L
Γ = (2s + 1) · 18 · 4πn2 . By using the relations: � = 2m
= 2mL
2m�.
2 n which gives us n = π�
A derivative of this relation gives dn =
L √m
d�.
π� 2m�
Assigning n and dn we get:
√ � �3
3
(2s + 1) �
L
m2
√
·
· 2 d� ≡ D� d�
�
π
2
(2)
We can now define the total number and total energy:
� �f
� �f
N=
fT D� d� , U =
�fT D� d�.
0
(3)
0
Here fT is the Fermi-Dirac function describing the average population of states as a function
of temperature. At the absolute zero 0[K] f(T ) is a step function with 1 up to the Fermi
energy �f and 0 after (meaning that all the particles are at their lowest possible energy) at
slightly higher temperatures there will be a smother transition from 1 to 0 around �f . It is
noteworthy remind that �f = Kb · Tf .
Fermi and Bose statistics.
Once dealing with low yet nonzero temperatures we must treat the function describing the
average population of states as a function of temperature - fT . For Fermions fT is:
fT =
1
eβ(�−µ)
(4)
+1
For bosons f is:
fT =
1
eβ(�−µ)
(5)
−1
For both cases at high temperatures (Maxwell-Boltzmann distribution)
fT = e−β(�−µ)
(6)
2
The energy difference between two close energy levels can now be calculated:
�=
π 2 �2 2
n
2mL2
(7)
∆� =
π 2 �2
= 0.5 · 10−33 [J]
2mL2
(8)
∆τ =
∆�
= 10−10 [K].
Kb
(9)
The Fermi energy: �f
The Fermi energy is defined as the highest energy populated By Fermions at T = 0. We shall
find �f by using the result from integration:
N=
3
V
2
·
(2m
·
�
)
f
π 2 �3
(10)
2
3
(3π 2 �3 N
)3
2 �
V
�f =
= (3π 2 ne ) 3
2m
2m
At this energy the temperature will be Tf =
to
(11)
�f
Kb
c
.
100
= 104 − 105 [K], which translates for electrons
The total energy: U at T=0
The equipartiton law allows us to estimate that U =
�∞
done by U = 0 �ft D� d�, become At T = 0
� �f
U=
�D� d�
�f
N,
2
a more rigorous computation is
(12)
0
U=
√
3
v
2 2 3 m2
π �
�
�f
3
� 2 d� =
(13)
0
√
3 5
2 2 v
3
U=
m 2 � 2 = N �f .
2
3
5 π �
5
(14)
3
Pressure at T=0
From the energy dependence of the volume we will find the pressure in the system. The
thermodynamic relation F = U − στ is transformed at T = 0[k] to F = U and so:
�
�
�
�
∂F
∂U
P =−
=−
∂v τ,N
∂V τ,N
3 2 �f
2
P = − N N = �f n
5 3 V
5
(15)
(16)
Whereas, for classical ideal gas P = nτ = 0 at those conditions. An electron gas pressure in
a lattice is of the order of 1010 [P a] which is why we were able to use the Hard core approximation.
Low nonzero temperatures
At temperatures slightly higher than T = 0 but much lower than the Fermi temperature
only the electron which occupy the Fermi level will be able to use the thermal energy to
rise to higher levels. For electrons of lower levels that will not be possible because the full
occupancy above them. The energy difference between two close levels can be computed as
∆U =
τ
N
N τ = τ 2.
�f
�f
The heat capacity is defined as CV =
Cv =
(17)
δU
δτ
∂∆U
N
= τ
∂τ
�f
(18)
From Cv we see that: The heat capacity is temperature dependant. The heat capacity is
very small compared to that of ideal gas.
An exact treatment for heat capacity
One can use the grand-canonical cluster in which there is a reservoir of heat and particles,
the temperature and chemical potential remain constant and while the number of electron
4
number on an orbital change.
E−
P �i =
µN −�
τ
ζ
< N >=
ζ=
∞
�
e
µN
τ
N PN = τ
e
−�1
τ
(19)
�i
N =0
�
�
�
∂ ln(ζ)
∂µ
�
f� =< N >�
(20)
τ,V
Each state is defined by an orbital: nx , ny , nz , Sz the most notable property of Fermions is
that do not share states, so one can easily write
ζ =1+e
f =τ
µ−�
τ
1 µ−�
e τ
τ
1+e
µ−�
τ
(21)
=
1
1+e
(22)
µ−�
τ
It can be seen that at the absolute zero �f = µ, while at room temperature one must use the
Fermi statistics. It is possible to represent f as a tanh. Defining Z =
z
µ−�
τ
and calculating
−z
1
2 − eZ − 1
1 − eZ
e2 − e 2
Z
f(Z) − 0.5 = Z
− 0.5 =
=
=
0.5
(23)
−z = 0.5 tanh
z
Z
Z
e −1
2(e − 1)
2(e − 1)
2
e2 + e 2
A major difference is that tanh(θ) is defend for −∞ < θ < ∞, where as f is only positive.
As previously mentioned at 0[K] �f = µ but this is true for higher temperatures as well.
N=
�
∞
0
f(�,µ,τ ) D� d� =
�
∞
f(�,µ,τ =0) D� d�
(24)
0
And by rearranging the RHS we can get:
� ∞
�
�
f(�,µ,τ ) − f(�,µ,τ =0) D� d� = 0
(25)
considered as a constant and as such be removed from the integrand.
� ∞
�
�
D�
f(�,µ,τ ) − f(�,µ,τ =0) d� = 0
(26)
0
At the area ∆t where f(t) is changed from 1 to 0 D� is changed much slower, so D� can be
0
5
D� (µ − �f ) = 0
(27)
Using the definitions of the energy and heat capacity:
� ∞
U=
�f� D� d� = �N
(28)
0
CV =
Using
∂N
∂τ
�
�
∞
�
0
�
∂f�
∂τ
�
D� d�
(29)
N,V
=0
∞
0
∂f�
D� d�
∂τ
(30)
Hence once can express the heat capacity in using
� ∞
∂N
∂f�
�f
=
�f
D� d� = 0
∂τ
∂τ
0
CV =
�
∞
0
∂f�
�
D� d� −
∂τ
�
∞
0
∂f�
�f
D� d� =
∂τ
�
∞
0
(31)
(� − �f )
∂f�
D� d�
∂τ
(32)
Once again we assume that the change in f is much faster then the change in D� which case
can be taken as a constant:
D�
�
∞
0
(� − �f )
1
Assigning f� =
e
CV = D�f
∂f�
d�
∂τ
�−�f
τ
�
∞
0
(33)
and multiplying by
+1
�
� − �f
τ
�2
�
e
e
�−�f
τ
�−�f
τ
+1
τ
τ
�2 d�
We can now change the differential to be d(� − �f ) and set
Riemann Zeta function:
� ∞
Z 2 ez
π2
π2 N
D� f τ
= D� τ =
τ
z
3
2 �f
−∞ (e + 1)
(34)
�−�f
τ
= Z We are thus left with a
(35)
6
Magnetic properties of quantum Fermions gas.
At the absolute zero temperature the energy of a spin possessing particle is described as
�=
p2
�
− �s · Bµ
2m
(36)
� is an outer magnetic field. When the spin’s direction is
Where s is the spin of the particle, B
that of the magnetic filed the energy is lower than that where this spin opposes the external
filed. In order to find the magnetization we must find how many are with and how many
oppose the field.
�
+
N =V ·
N+ =
�f −µB
−
N =V ·
ft D� d�
0
�
�f +µB
ft D� d�
(37)
0
�
3
V �
2
2m(�
−
µB)
f
3π 2 �3
(38)
The magnetization is defined as:
M = µ(N + + N − ) =
� �
��
3
3
V µ ��
2
2
2m(�
−
µB)
−
2m(�
+
µB)
f
f
3π 2 �3
3 1
1 ∂M
µ2
χ=
= 2 3 (2m) 2 �f2
V ∂B
π �
(39)
(40)
At higher temperatures still much lower than Tf (for metals Tf is of the order of 104 [K])
f is no longer a step function, it is now described by the statistics described earlier. The
magnetization will now be:
+
−
M = µ(N + N ) = µ
And we approximate
�
�
�f
0
�
�
D� f(�f +µB) − f(�f −µB) d�
� ∂f
f(�f +µB) − f(�f −µB) =
· 2µB
∂�
(41)
(42)
The first order of the magnetization isM = 2µ2 BD� which is call Pauli’s paramegnetisem,
its first order does not depend on temperature.
7
The energy exerted on an ion in a magnetic filed
We describe the energy exerted on an ion in a magnetic filed as
� ·S
�
E = µB
(43)
�=
For a single electron S
Z1 =
�
eβ E = e
1
2
µBE
2
We can compute the distribution function Z from the energy:
+e
−µBE
2
�
F1 = −τ ln(Z1 ) = −τ ln e
µBE
2
(44)
+e
−µBE
2
�
1
= −τ ln(2 cosh( µBβ)
2
1
FN = −N τ ln(2 cosh( µBβ))
2
M =−
(45)
(46)
∂f
1
1
= N µtgh( µBβ)
∂B
2
2
(47)
When τ >> µB we can approximate tgh(x) = x, and so we find Curie’s law:
M =−
∂f
1
1
= N µ( µBβ)
∂B
2
2
(48)
At high temperatures the arrangement of the spin will be random.
2
Bosons
Bosons are with spin number of 0, 1, 2, 3 who’s basic property is that any number of bosons
can occupy the same state, the most known bosons are: proton, Photon, W boson, Z boson
two other famous yet unconfirmed bosons are the Higgs boson and Graviton, molecules are
also considered Bosons. The model system for bosons is He4 with a molecular mass of 4
gr/mol, at T=4[K] it becomes a liquid with a density of n = 2 · 1028 m13 . At T=2.17[K] (the
λ point) the liquid’s viscosity nullifies, at that temperature n > nq this is when we get the
Bose-Einstein condensation.
8
At low temperature most molecules are at the lowest energy level, one can compute the
� π �2
�2
energy difference for Bosons just the same as for Fermions. ∆� ≈ 2m
which can be
L
transformed to a temperature difference ∆T =
�
KB
≈ 1014 [K] (for a system of the size of
L = 1[cm]. For this reason we will not expect to have a difference between the energy levels
at 2[k], howeverwe find experementaly that almost all the particles in the system are at the
ground level.
Two state energy system, expectation from classical system
If we assume a two state system with two different energies �1 and �2 the distribution function
�1
�2
is Z1 = e− τ + e− τ and for two particles Z1 =
�
�
� +�
1 −2 �1 1 −2 �2
− 1τ 2
τ
τ
e
+ e
+e
2
2
Z12
2!
which is
(49)
However the 2! factor originates from the demand to separete each state from one enother,
where as Bosons can occupy the same state at the same time. So the distributtion function
we now expect
� �1
�
�2
�1 +�2
e−2 τ + e−2 τ + e− τ
(50)
This is our motivation for creating the new statistics called the Bose - Einstein condensate.
We can now make detailed computations using the distribution function, addressing two
energy levels �0 = 0 and � so that:
�
Z1 = 1 + e − τ
ZN =
(51)
(Z1 )N
N!
(52)
�
f(�)
e− τ
=N
�
1 + e− τ
(53)
When τ >> � then f(�) =
function is SN = 1 + e
− τ�
N
For
2
− 2�
τ
+e
a system of two energy levels of N Bosons the Distribution
+ + e−
N�
τ
much the same as a harmonic oscillator, by treating
9
it as series we find that:
N +1
1 − e− τ
Z=
�
1 − e− τ
�
(54)
The orders of magnitude of the system are:
�
τ
that
ZN =
≈ 1014 while N ≈ 1022 so that
N�
τ
>> 1 so
1
�
1 − e− τ
(55)
From which we can find that:
< U >= −
f=
∂ ln Z
∂β
(56)
<U >
1
1
�
= −�
≈
≈ ≈ 1014
�
�
1+ τ −1
τ
e τ −1
(57)
which is why most atoms are in the lower energy level (1014 is much smaller then 1022 .
We may now estimate the number of exited molecules:
�
�3 �
√
� ∞
1
V
2m 2 ∞
�
Ne =
D� = 2
d�
�
�
2
4π
�
eτ − 1
eτ − 1
0
0
� �3
We now multiply and divide by ττ 2 and asign Z we get:
� ∞√
2
ZdZ
Ne = n q V √
Z
π 0 e −1
(58)
(59)
Where the integral is the Ryman zeta function with a value of 0.5. the number of molecules
at the ground state is N0 = N − Ne .
N0
Ne
nq
=1−
= 1 − ζ( 3 ) ≈ 1 −
2
N
N
n
� � 32
τ
τc
So we can estimate the critical temperature to be
�
� 23
2π�2
n
τc =
m
ζ( 3 )
(60)
(61)
2
In such instance the diffrentiation between classical to quantum gas is either because τ > τc
or n << nq the transition here is a second order phase transition.
10