Karlsruher Institut f¨ur Technologie www.tkm.kit.edu/lehre/
Moderne Theoretische Physik III SS 2015
Prof. Dr. A. Mirlin Blatt 01, 100 Punkte
Dr. U. Karahasanovic, Dr. I. Protopopov Abgabe und Besprechung, 24.04.2015
1. Mathematical preliminaries (5 + 5 + 10 + 10 = 30 Punkte, m¨undlich) Stirling theorem
(a) Proof of Stirling theorem.
We begin from
n! = Z ∞
0
xne−x = Z ∞
0
exp (nlnx−x) (1)
It is easy to check that the function f(x) = nlnx−x has a maximum at x = n, (obtained by setting f′(x) = 0, and checking that indeed it is a maximum by computing f′′(n) = −1n < 0). The main contribution to the integral comes from integrating around this maximum. Hence, we can expand the integration limits from (−∞,∞) and Taylor-expand the function f(x) around the maximum atx=n, i.e.
f(x) =f(n) +12f′′(n)(x−n)2+....
This gives
n! = Z ∞
−∞
dxexp
nlnn−n− 1
2n(x−n)2+...
= exp (nlnn−n) Z ∞
−∞
dyexp
−y2 2n
= exp (nlnn−n)√
2πn (2)
After taking the log of (2), and bearing in mind that for n ≫ 1 we have that lnn≪n(so that we can drop it), it follows that
lnn! =nlnn−n. (3)
Differential calculus (b) Jacobi determinants.
The identities (3) and (4) from the exercise sheet follow directly from the definition of the Jacobi determinants
∂(u, v)
∂(x, y) = det
∂u
∂x y
∂u
∂y x
∂v
∂x y
∂v
∂y x
= ∂u
∂x y
∂v
∂y −∂u
∂y x
∂v
∂x y .
To establish Eq. (5) of the exercise sheet we consider first the chain rule for the partial derivatives. We suppose that we are given functionsu(s, t) andv(s, t) and a change of variables s=s(x, y), t=t(x, y). We then have
∂u
∂x
y ≡ ∂u(s(x, y), t(x, y))
∂x
y = ∂u
∂s t
∂s
∂x y+∂u
∂t s
∂t
∂x
y (4)
Considering the derivatives∂u/∂y,∂v/∂xand∂v/∂yin a similar manner we occlude
that
∂u
∂x y
∂u
∂y x
∂v
∂x y
∂v
∂y x
=
∂u
∂s t
∂u
∂t s
∂v
∂s t
∂v
∂t s
∂s
∂x y
∂s
∂y x
∂t
∂x y
∂t
∂y x
.
It is convenient to introduce a special notation J(u, v;x, y) for the Jacobi matrix between variable (u, v) and (x, y). Then
J(u, v;x, y) =J(u, v;s, t)J(s, t;x, y) (5) Taking the determinant of the both sides of Eq. (5) we get
∂(u, v)
∂(x, y) = ∂(u, v)
∂(s, t)
∂(s, t)
∂(x, y). (6)
Interchangings↔x and t↔y in Eq. (6) we get also
∂(u, v)
∂(x, y) =
∂(u, v)
∂(s, t)
∂(x, y)
∂(s, t)
. (7)
A particularly useful form of this relation is (we put,s=u,t=v)
∂(u, v)
∂(x, y) = 1
∂(x, y)
∂(u, v)
. (8)
(c) Let us consider an equationF(x, y, z) = 0 defining a surface in a three dimensional space. We will assume that in the vicinity of the point of interest (x0, y0, z0) (with F(x0, y0, z0) = 0) the function F has non-zero partial derivatives with respect to all the three variables. The equation F(x, y, z) = 0 can then be locally solved with respect to any of the variablesx,y andz. Any pair of the coordinates [(x, y), (x, z) or (y, z)] can be used as coordinates on the surface F = 0 and there exist three functions x(y, z), y(x, z) andz(x, y).
Let us now prove Eq. (7) from the exercise sheet. We consider the change of coor- dinates on the surface from (x, z) to (y, z) and use the properties of Jacobi deter- minants established previously (see Eq. 8)
∂x
∂y
z= ∂(x, z)
∂(y, z) =
∂(y, z)
∂(x, z) −1
= ∂y
∂x z
−1
(9) To prove Eq. (8) from the exercise sheet we consider the Jacobi determinants bet- ween coordinates (x, y), (x, z) and (y, z) and use Eq. (7)
∂y
∂x z
∂z
∂y x
∂x
∂z
y = ∂(y, z)
∂(x, z)
∂(z, x)
∂(y, x)
∂(x, y)
∂(z, y) =−∂(y, z)
∂(x, z)
∂(x, z)
∂(y, x)
∂(x, y)
∂(z, y)
=−∂(y, z)
∂(y, x)
∂(x, y)
∂(z, y) = ∂(y, z)
∂(y, x)
∂(y, x)
∂(z, y) = ∂(y, z)
∂(z, y) =−1. (10) (d) Suppose that now we are given also a functionw(x, y). The two equationF(x, y, z) = 0 andw(x, y) =wcan be (locally) solved with respect tox andy. This means that we can use coordinates (w, z) to describe the surface F = 0.
Let us consider the coordinates (x, z), (y, z) and (w, z). We have
∂x
∂w
z= ∂(x, z)
∂(w, z) = ∂(x, z)
∂(y, z)
∂(y, z)
∂(w, z) = ∂x
∂y z
∂y
∂w
z. (11) We have proven Eq. (9) of the exercise.
Finally, to prove Eq. (10) of the exercise we need to consider the full Jacobi ma- trices of the coordinate transformations between variable (x, z), (y, z) and (y, w).
According to Eq. (5) we have
J(x, z;y, z) =J(x, z;y, w)J(y, w;y, z). (12) Explicitly
∂x
∂y z
∂x
∂z y
∂z
∂y z
∂z
∂z y
=
∂x
∂y w
∂x
∂w y
∂z
∂y w
∂z
∂w y
∂y
∂y z
∂y
∂z y
∂w
∂y z
∂w
∂z y
. (13)
Taking into account relations of the form∂zz|y = 1 and ∂zy|y = 0 we get
∂x
∂y z
∂x
∂z y
0 1
=
∂x
∂y w
∂x
∂w y
∂z
∂y w
∂z
∂w y
1 0
∂w
∂y z
∂w
∂z y
. (14)
Taking the elements in the upper right conner on both sides of this equation we get Eq. (10) of the exercises.
It is a good exercise to check other identities implied by Eq. (14).
2. Ideal Boltzmann gas (10 + 10 + 5 + 10 = 35 Punkte, m¨undlich) (a) To determine the entropy S(T, V) we start from the first law of thermodynamics together with the thermodynamic definition of entropy (we work with fixed number of particles)
dU =T dS−pdV =T∂S
∂T VdT +
T∂S
∂V T −p
dV (15)
On the other hand for an ideal gas dU = 3
2N kBdT (16)
Comparing Eqs. (15) and (16) we get
∂S
∂T
V = 3N kB
2T (17)
∂S
∂V
T = p
T = N kB
V (18)
Integration of Eq. (17) gives
S(T, V) = 3
2N kBlnT+α(V) (19)
with the unknown functionα(V) which can be determined from Eq. (18). The final result reads
S(T, V) =N kBln
V T3/2
+ const (20)
The unknown constant in Eq. (20) actually depends on the number of particles. If number of particles in the system is fixed the constant is irrelevant for thermody- namics and can be set to zero. We will discuss the dpendence of the entropy on the number of particles in the next exercise sheet.
The free energy of the system follows now directly from F(T, V) =U(T, V)−T S.
It is a good exercise to check that
dF =−SdT −pdV (21)
(b) Let us denote the initial state in the Carnot cycle (temperature T1, volumeV1) by 1. We denote by 2 the state reached by the system at the end of isothermal expan- sion (temperatureT1, volume V2). The state at the end of the adiabatic expansion (temperature T1 < T2, volume V2′) is denoted by 3, while the state at the end of isothermal compression (temperature T2, volumeV1′) is denoted by 4.
We analyze the processes 1→2, 2→3, 3→4 and 4→3 one by one.
• The process 1→2.
The work done by the gas in the isothermal expansion 1→2 W12=
Z V2 V1
p(V, T1)dV =N kBT1 Z V2
V1
dV
V =N kBT1lnV2
V1 (22) where we have used the equation of state to express the pressure.
Since the internal energy of the ideal gas does not depend on the volume the internal energy is constant in 1→2 process and the first low of thermodynamics gives for the heat absorbed by the gas
Q12=W12=N kBT1lnV2
V1 (23)
• The process 2→3.
By definition of adiabatic processQ23= 0. The work done by the system equals the absolute value of the change of its internal energy
W23= 3
2N kB(T1−T2). (24)
For the subsequent analysis we will need also the volume V2′ of the system in the state 3. In an adiabatic process the entropy is constant and, according to exercise 2a, we have
V T3/2 = const. (25)
Thus,
V2′ =V2 T1
T2
3/2
. (26)
• The process 3→4.
In full analogy with the process 1→2.
Q34=W34=N kBT2lnV1′
V2′
=N kBT2lnV1 V2
<0. (27) Here we have used the relation (26) and analogous relation betweenV1′ andV1.
• The process 4→1.
In full analogy with the 2→3 process Q41= 0 and W41= 3N kB(T2−T1)/2.
The efficiency of the Carnot cycle is now given by η= W12+W23+W34+W41
Q12
= Q12+Q34 Q12
= T1−T2 T1
. (28)
(c) The integral
∆S = I δQ
T = Q12 T1 +Q34
T2 = 0. (29)
The nullification of the integral is the manifestation of the fact that entropy is the proper function of the thermodynamic state.
(d) The reversed Carnot process is used as a cooling machine. The amount of heat extracted from the cold reservoir in the reversed process can be read of from Eq.
(27), where the volumes V1 andV2 should be exchanged Q43=N kBT2lnV2
V1. (30)
The cooling efficiency
η′= Q43
W12+W23+W34+W41 = T2
T1−T2. (31)
3. Ultrarelativistic Bose gas (15 + 15 + 5 = 35 Punkte, schriftlich) (a) The solution goes along the lines of exercise 2a. The derivatives of entropy are given
by
∂S
∂T
V = 4V σT2, (32)
∂S
∂V
T = 4
3σT3. (33)
Integration of these equations gives S = 4
3V σT3. (34)
The free energy is given by
F =−σV T4
3 . (35)
In the adiabatic process S= const. Accordingly,
V T3= const, (36)
V p3/4 = const. (37)
(b) We use the same notations as in exercise 2b.
• The process 1→2.
The work done by the gas in the isothermal expansion 1→2 W12=
Z V2 V1
p(V, T1)dV = σ
3T14(V2−V1). (38) where we have used the equation of state to express the pressure.
The heat received by the system
Q12= ∆U12+W12= 4
3σT14(V2−V1). (39)
• The process 2→3.
By definition of adiabatic process Q23 = 0. The final volume in the adiabatic expansion is give by
V2′ =V2
T1
T2
3
. (40)
The work done by the system equals the absolute value of the change of its internal energy
W23=σ(V2T14−V2′T24) =σV2T13(T1−T2) (41)
• The process 3→4.
In full analogy with the process 1→2.
W34= σ
3T24(V1′−V2′) = σ
3T2T13(V1−V2)<0 (42) Q34= 4
3σT2T13(V1−V2). (43) Here we have used the relation (40) and analogous relation betweenV1′ andV1.
• The process 4→1.
In full analogy with the 2→3 process Q41= 0 and W41=σV1T13(T2−T1).
The efficiency of the Carnot cycle is now given by η= W12+W23+W34+W41
Q12 = Q12+Q34
Q12 = T1−T2
T1 . (44)
(c) The integral
∆S = I δQ
T = Q12
T1
+Q34
T2
= 0. (45)
The nullification of the integral is the manifestation of the fact that entropy is the proper function of the thermodynamic state.
