Karlsruher Institut f¨ur Technologie www.tkm.kit.edu/lehre/
Moderne Theoretische Physik III SS 2014
Prof. Dr. J. Schmalian Blatt 01, 100 Punkte
Dr. U. Karahasanovic, Dr. P. P. Orth Abgabe und Besprechung, 25.04.2014
1. Stirling theorem (25 Punkte)
(a) Proof of Stirling theorem.
We begin from
lnn! = 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 f0(x) = 0, and checking that indeed it is a maximum by computing f00(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) +12f00(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 lnnn(so that we can drop it), it follows that
lnn! =nlnn−n. (3)
(b) Die Anzahl aller m¨oglichen F¨alle ist f¨ur n Personen m = 365n, wobei alle F¨alle gleich wahrscheinlich sind. Von diesen m¨oglichen F¨allen beinhalten
u= 365·364· · ·(365−(n−1))
nur unterschiedliche Geburtstage. F¨ur die erste Person kann der Geburtstag frei gew¨ahlt werden, f¨ur die zweite gibt es dann 364 Tage, an denen die erste nicht Geburtstag hat, etc. Damit ergibt sich die Wahrscheinlichkeit von
u
m = 365·364· · ·(365−(n−1))
365n ,
dass allenPersonen an unterschiedlichen Tagen Geburtstag haben. Die Wahrschein- lichkeit f¨ur mindestens einen doppelten Geburtstag im Verlauf eines Jahres ist somit
P = 1− u
m = 1−365·364· · ·(365−(n−1))
365n .
Der Ausdruck f¨urP kann weiter umgeformt werden:
P = 1− u
m = 1− 365!
(365−n)!·365n. Mit der Stirling-Formel,N!≈√
2πN(N/e)N (Aufgabe 2), l¨asst sich dies gut n¨ahern zu
P ≈ 1−e−n
365 365−n
365+1/2−n
= 1−exp
−
365 +1 2 −n
ln
1− n
365
−n
≈ 1−exp
−
365 +1
2 −n − n 365−1
2 n
365 2
−n
≈ 1−exp
−n(n−1) 2·365
. (4)
P = 1
2 =⇒ 1
2 = exp
−n(n−1) 2·365
n(n−1) = 730·ln 2 =⇒ n= 1 2
1 +√
1 + 2920 ln 2
'22.9. (5) F¨urn= 23 ergibt sich:
1−365·364· · ·343
36523 ≈0.5073> 1 2.
(c) The number of ways of getting n1 heads aftern tosses (and hence n−n1 tails), is given by
w(n1, n−n1, n) = n
n1
= n!
(n−n1)!n1!. (6)
After taking the log we have that
lnw(n1, n−n1, n) = ln (n!)−ln (n−n1)!−ln (n1!)
= nlnn−(n−n1) ln (n−n1)−n1lnn1 (7) where we have applied Stirling’s theorem in the last line.w(n1, n−n1, n) will have a maximum when lnw(n1, n−n1, n) has a maximum. The maximum is obtained by differentiating the expression above lnw(n1, n−n1, n) with respect to n1 and setting
dlnw(n1, n−n1, n)
dn1 = 0
→n1 = n
2. (8)
(d) The number of waysw(n/2−m, n/2 +m, n) in which one can get the outcome with n/2−m heads is
w(n/2−m, n/2 +m, n) =
n n/2−m
= n!
(n/2 +m)!(n/2−m)! (9) Taking the log and applying Stirling’s theorem we have
lnw(n/2−m, n/2 +m, n) = ln (n!)−ln (n/2−m)!−ln (n/2 +m)!
= nlnn−(n/2−m) ln (n/2−m)−(n/2 +m) ln (n/2 +m) (10) We expand ln (n/2±m) for m n using the identity ln (1 +x) ≈ x− x22, which gives
lnw(n/2−m, n/2 +m, n) = nln 2−2m2
n (11)
After re-exponentiating (11), we obtain
w(n/2−m, n/2 +m, n) = 2n
|{z}
w(n/2,n/2,n)
exp
−2m2 n
˜
w(p,1−p, n) = 2nexp −2n(p−1/2)2
(12) From the gaussian function above we immediately see that the most likely outcome is p= 1/2, and that its width scales like 1/√
n, i.e. that the maximum is sharp for largen.
2. Funktionaldeterminantenkalk¨ul:
∂(u, v)
∂(x, y) = ∂u
∂x y
∂v
∂y x
− ∂u
∂y x
∂v
∂x y
Def. der Determinante (a)
1. v(x, y) =y ⇒ ∂v∂y
x = 1, ∂v∂x
y = 0 ⇒ ∂(u,v)∂(x,y) = ∂u∂x y
2. u(x, y) =x ⇒analog ∂(u,v)∂(x,y) = ∂u∂y x
3. folgt trivial aus Determinanteneigenschaft (Vertauschen von u ↔ v oder x ↔ y vertauscht Vorzeichen)
4.
∂(u, v)
∂(x, y) =∂(u(s(x, y), t(x, y))), v(s(x, y), t(x, y))
∂(x, y)
= ∂u
∂s t
∂s
∂x y
+ ∂u
∂t s
∂t
∂x y
!
∂v
∂s t
∂s
∂y x
+ ∂v
∂t s
∂t
∂y x
− ∂u
∂s t
∂s
∂y x
+ ∂u
∂t s
∂t
∂y x
∂v
∂s t
∂s
∂x y
+ ∂v
∂t s
∂t
∂x y
!
= ∂u
∂s t
∂v
∂s t
∂s
∂x y
∂s
∂y x
− ∂s
∂y x
∂s
∂x y
! +∂u
∂s t
∂v
∂t s
∂s
∂x y
∂t
∂y x
− ∂s
∂y x
∂t
∂x y
!
+ ∂u
∂t s
∂v
∂s t
∂t
∂x y
∂s
∂y x
− ∂t
∂y x
∂s
∂x y
! +∂u
∂t s
∂v
∂t s
∂t
∂x y
∂t
∂y x
− ∂t
∂y x
∂t
∂x y
!
1. und 4. Term sind Null. 2. und 3. Term ergeben ∂u
∂s t
∂v
∂t s
− ∂u
∂t s
∂v
∂s t
∂s
∂x y
∂t
∂y x
− ∂s
∂y x
∂t
∂x y
!
= ∂(u, v)
∂(s, t)
∂(s, t)
∂(x, y)
Durch Umbenennen von x, y→s, tund s, t→x, y folgt
∂(u, v)
∂(s, t) = ∂(u, v)
∂(x, y)
∂(x, y)
∂(s, t)
aufl¨osen
=⇒ ∂(u, v)
∂(x, y) =
∂(u,v)
∂(s,t)
∂(x,y)
∂(s,t)
(b)
Die erste Relation spiegelt einfach den Satz der lokalen Umkehrbarkeit aus Analysis wieder. Z.B. so: Seiφ(x, y) = const., also x(y). Dann ist das Differentialdx(y) = ∂x∂ydy.
Andererseits ist aberdy(x) = ∂x∂ydx. Einsetzen liefert danndx= ∂x∂y∂y∂xdx, also 1 = ∂x∂y∂y∂x. Und daraus folgt schließlich
∂x
∂y = ∂y
∂x −1
.
Da wir ja von Anfang an φ = const. angenommen haben, d¨urfen wir auch noch bei beiden partiellen Ableitungen dieφ’s dazuschreiben und erhalten das gesuchte Resultat.
∂y
∂x φ
= ∂(y, φ)
∂(x, φ) = ∂(y, φ)
∂(y, x)
∂(y, x)
∂(x, φ) =− ∂φ
∂x y
1
∂(x,φ)
∂(x,y)
=−
∂Φ
∂x
y
∂φ
∂y
x
{Hier wird φ=φ(x, y) =const. als unabh¨angige Variable aufgefasst, d. h.y=y(x, φ)}.
(c)
Wir betrachtenxundzals unabh¨angige Variable, so dassy=y(x, z). Das erfolgt durch Aufl¨osen vonF(x, y, z) = 0 nach y. Das Differenzial f¨ury ist dann
dy= ∂y
∂x
zdx+∂y
∂z xdz Weiterhin ist w=w(x, y), also
dw= ∂w
∂x
ydx+ ∂w
∂y xdy.
Wir substituieren y(x, z) f¨ury inw(x, y), und erhalten als Differenzial f¨urw(x, z):
dw = ∂w
∂y x
∂y
∂z xdz+
∂w
∂x
y+ ∂w
∂y x
∂y
∂x z
dx.
Daraus folgt
∂w
∂x
z = ∂w
∂x y+∂w
∂y x
∂y
∂x
z. (13) Analog erhalten wir aus dem Differenzial f¨urw(x(y, z), y):
∂w
∂y
z = ∂w
∂y
x+∂w
∂x y
∂x
∂y
z. (14) Wir eliminieren ∂w/∂y|x (d.h. Gl. (14) in Gl. (13) einsetzen), und erhalten
∂w
∂x
z= ∂w
∂y z
∂y
∂x z.
Wir l¨osen nun w = w(x, y) nach x auf, betrachten also x als Funktion von y und w, d.h. x=x(y, w). Wir erhalten
dx= ∂x
∂y
wdy+ ∂x
∂w ydw
Dir Gleichung F(x, y, z) = 0 l¨osen wir nachxauf, betrachten alsoxals Funktion vony und z, und erhalten daraus w=w(x(y, z), y). Das Differential f¨urw(y, z) ist
dw= ∂w
∂y
zdy+∂w
∂z ydz.
Wir substituieren dies in die Gleichung f¨urdxund erhalten schließlich
∂x
∂y z= ∂x
∂y
w+ ∂x
∂w y
∂w
∂y z. 5. Rubber band
(a) Equation of state.
If we integrate the first equation with respect toT, at fixed l, we obtain
J(T, l) =Tal l0
1− l0
l 3!
+h1(l) (15)
whereh1(l) is unknown function.
Similarly, if we integrate the second equation with respect tol, at fixedT, we obtain
J(T, l) =Tal l0
1− l0
l 3!
+h2(T) (16)
whereh2(T) is an unknown function.
Comparing (15) and (16), we conclude thath1(l) =h2(T) =c, wherecis a constant.
We see that forl=l0 the tensionJ vanishes iff c= 0.
Therefore the equation of state is
J(T, l) =Tal l0 1−
l0 l
3!
. (17)
(b) Adiabatic stretching.
The Helmholtz free energy of the rubber band is give by F =U−T S, where
dU =T dS+J dl (18)
Then dF =−SdT +J dl, leading to Maxwell relations ∂S
∂l
T
=− ∂J
∂T
l
(19) From (18) we have that
∂U
∂l
T
= T
∂S
∂l
T
+J
= (19)−T ∂J
∂T
l
+J
= (17) = 0 (20)
where in the last line we used the equation of state (17) to deduce that ∂T∂J
l=J/T. The identity ∂U∂l
T = 0 tells us that the internal energy is a function of temperature only (like for an ideal gas). ThendU =cldT. Combining this with (18) leads to
dS=CldT T − J
T
|{z}
g(l)
dl. (21)
whereg(l) = all
0
1−
l0
l
3
. Next, we can integrate (21) to find the entropy diffe- rence ∆S between states (T1, l1) and (T2, l2):
∆S = cl Z T2
T1
dT T −
Z L2
L1
g(l)dl
= clln T2
T1
−al0
u l2
l0
−u l1
l0
(22) where we have defined the function u(x) = x2/2 + 1/x. For adiabatic stretching
∆S= 0, and hence we obtain that
Tf =Tiexp al0
cl
u lf
l0
−3 2
, (23)
– the temperature of the band will increase.
(c) Rubber contraction
During the contraction no work is done and no heat is exchanged with sourroun- dings, which results in unchanged internal energy, and hence temperature (we have shown that the internal energy is a function of temperature only). Using the results in b) we have that
∆S =al0
u
lf
l0
− 3 2
>0. (24)
4. Carnot cycle for an ultrarelativistic gas of bosons It follows fromU =σV T4 andp=U/3V that the pressure
p= σT4
3 (25)
is independent of V. Therefore, an isothermal expansion does not change the pressure (i.e., here the isotherms appear to bep=const.lines in the p−V diagram).
We start the analysis of the Carnot cycle with the isotherm. The work of the system is given by
∆W = Z V2
V1
pdV = σT4
3 (V2−V1). (26)
The change in internal energy is
∆U = Z V2
V1
dU dV T
dV =σT4(V2−V1). (27) Thus, the heat change during the isothermal expansion is
∆Q= ∆U+ ∆W = 4
3σT4(V2−V1) (28)
At T =Th one gets
∆Qh= 4
3σTh4(V2−V1). (29)
Next we consider the adiabatic expansion. Of course, ∆Q= 0 holds. We only need to determine the value of the lower temperature Tl. From
p=− ∂U
∂V S
= 1 3
U V it follows forS =const that
dU U =−1
3 dV
V . (30)
This yields U ∼V−1/3. Since U is extensive, and the only other extensive quantity is entropy, it must behave as U = αS4/3V−1/3 with constant α. This is equal toσV T4, which gives
σ α
3/4
V T3 =S (31)
Thus,V T3 =const along the adiabat. This gives for the expansion a volume V3 =V2Th3
Tl3 (32)
and leads to an isothermal compression from V3 to V4 = V1TTh33 l
(such that the final adiabat reaches the initial state). Thus the heat change on the last isotherm is
∆Ql = 4
3σTl4(V4−V3) =−4 3σTl4Th3
Tl3 (V2−V2)
= −4
3σTlTh3(V2−V1). (33)
There is a finite total heat change ∆Qh+ ∆Ql >0 (and thus work done). At the same time,
∆S = ∆Qh
Th +∆Ql
Tl = 0, (34)
which means that the entropy is indeed a state variable.
