VN (2πmE)3N/2 Γ (3N which is the result of the integration Ω (E

Karlsruher Institut f¨ur Technologie www.tkm.kit.edu/lehre/

Ubungen zur Theoretischen Physik F¨ SS 14

Prof. Dr. J¨org Schmalian Blatt 4

Dr. Una Karahasanovic, Dr. Peter Orth Besprechung 16.05.2014

1. Entropie aus der Anzahl der Zust¨ande:

English:

(a) The volume of the region in phase space with energy less than E is Ω (E) = V^N (2πmE)^3N/2

Γ (3N/2 + 1) (1)

which is the result of the integration Ω (E) = V^N

Z ^N Y

i=1

d³p_iθ 2mE −

N

X

i=1

p²_i

!

. (2)

The integral is most easily done my noticing

∂Ω (E)

∂E = 2mV^N Z ^N

Y

i=1

d³piδ 2mE −

N

X

i=1

p²_i

!

(3) which is easier to do. The number of states is determined via

N(E) = 1 N!

Ω (E)

h^3N (4)

The term h^3N is just the elementary volume in phase space and the prefactor _N!¹ takes care of the indistinguishability of the particles. It follows:

S =k_Blog V^N (2πmE)^3N/2 N!Γ (3N/2 + 1)h^3N

!

. (5)

(b) Using logN!'NlogN −N and Γ (3N/2 + 1)'(3N/2)! '(3N/2e)^3N/2 gives S = 3N k_B

2 +N k_Blog V

4πmE 3h²N

3/2!

+N k_B−N k_BlogN

= 5N k_B

2 +N k_Blog V N

4πmE 3h²N

3/2!

(6)

(c) We can determine the pressure as

p=− ∂E

∂V V,N

(7)

which leads top= ²₃^E_V. Next we find the temperature from T = ∂E

∂S V,N

= 3 2

N k_B

E (8)

which gives the familiarE = ³₂N k_BT and leads to p= ^{N k}_V^BT. Deutsch:

(a) Bei einem klassichen nichtrelativistischen idealen Gas mit N Teilchen gilt f¨ur die Gesamtenergie:

H =

N

X

i=1

p_i² 2m

Dabei ist das Spektrum f¨ur pi kontinuierlich, man kann also bereits bei zwei Teil- chen die Zust¨ande mit einer bestimmten Gesamtenergie E nicht mehr abz¨ahlen. Die einzige M¨oglichkeit, N(E) zu berechnen, geht also ¨uber die Zustandsdichte (siehe auch Aufgabe 3).

Wenn ν(E) die Zustandsdichte ist, so gibt ν(E)dE an, wie viele Zust¨ande pro Vo- lumen sich im Energieintervall [E, E+dE] befinden. Damit folgt dann:

ν(E) = 1 V

X

p

δ(E − H(p)) ⇒ N(E) = X

p

δ(E − H(p))

Da unser Spektrum kontinuierlich ist, benutzen wir wieder, wie in der Vorlesung X

p

→ V^N h^3N

N

Y

i=1

Z d³p_i

wobei unser Impulsraum nat¨urlich 3N-dimensional ist.

Damit gilt dann:

N(E) = V^N N!h^3N

Z ^N Y

i=1

d³p_iδ E −

N

X

i=1

p_i² 2m

!

= V^N N!h^3N

Z

dΩ_(3N)p²_(3N)dp_(3N)δ E − p²_(3N) 2m

!

∝ E^(3N−2)/2

Vom 1. zum 2. Schritt wurden sph¨arische Koordinaten p²_3N =

N

X

i=1

p_i² =

i=N,j=3

X

i=1,j=1

p²_i,j

benutzt und die Integration ¨uber ein 3N-dimensionales Kugelvolumen durchgef¨uhrt.

Dabei m¨ussen wieder die Rechenregeln f¨ur δ[g(x)], die in der L¨osung zu Aufgabe 3 n¨aher erl¨autert sind, ber¨ucksichtigt werden.

Das Wesentlich hieran ist nun, dass die Zustandsdichte f¨ur große N sehr stark mit der Energie E ansteigt. Dann liefern bei großen N also die Zust¨ande mit der Energie nahe bei E den Hauptbeitrag, kleinere Energien tragen nur unwesentlich bei. Deswegen

kann man in guter N¨aherung N(E) mit der Anzahl aller Zust¨ande mit Energie kleiner als E betrachten, was die Rechnung etwas vereinfacht:

N(E) ≈ Ω(E)

N!h^3N = V^N N!h^3N

Z ^N Y

i=1

d³p_iθ E −

N

X

i=1

p_i² 2m

!

= V^N N!h^3N

π^3N/2(2mE)^3N/2 Γ ^3N₂ + 1

Dabei wurde wieder wie bei der Zustandsdichte verfahren und benutzt, dass das Volumen einer d-dimensionalen Kugel mit Radius R gerade

V_d = π^d/2R^d Γ ^d₂ + 1

betr¨agt. (Alternativ h¨atte man nat¨urlich auch N(E) ¨uber E integrieren k¨onnen, doch die direkte Berechnung von Ω(E) ist leichter nachvollziehbar.) Nun kann man im thermodynamischen Limes großer N wieder die Stirling-Formel N! ≈

√2πN ^N_eN

≈ ^N_eN

und mit der Definition der Gammafunktion ¨uber die Fa- kult¨at die N¨aherung Γ ^3N₂ + 1

≈ ^3N₂

! ≈ ^3N_2e^3N₂

verwenden und erh¨alt:

N(E) = 8e^5/2V (πmE)^3/2 3^3/2h³N^5/2

!N

Daraus ergibt sich dann die Entropie zu:

S = kBlnN(E) ≈ N kBln 8V (πmE)^3/2 3^3/2h³N^5/2

! + 5

2N kB

(b) Wenn man nun annimmt, dass E = U = ³₂N k_BT f¨ur ein ideales Gas gilt, dann erh¨alt man:

S ≈ N k_Bln

V(2mπkBT)^3/2 N h³

+ 5

2N k_B = N k_Bln V

N λ³

+ 5 2N k_B wobei die thermische de Broglie-Wellenl¨ange λ = ^√_2πmk^h

BT eingesetzt wurde. Ver- gleichen wir dies mit der Entropie im kanoischen Ensemble, wie sie in der Vorlesung hergeleitet wurde:

S_K = N k_Bln V

N λ³

+ 5 2N k_B

so sieht man, dass im thermodynamischen Limes die Entropie in der urspr¨unglichen statistischen Definition und die in der kanonischen Gesamtheit gleich sind.

(c) Wir gehen aus von der Entropie, die wir in a) berechnet haben:

S ≈ N k_Bln 8V (πmE)^3/2 3^3/2h³N^5/2

! + 5

2N k_B

Also gilt f¨ur die Funktion S offensichtlich S = S(U, V, N). Dann benutzen wir die

”thermodynamische Fundamentalbeziehung“:

T S = U + pV − µN ⇒ S = U + pV − µN T

Wir berechnen das Differential dS = ∂S

∂U V,N

dU + ∂S

∂V U,N

dV − ∂S

∂N V,U

dN = 1

TdU + p

TdV − µ TdN damit erhalten wir dann:

1

T = ∂S

∂U _V,N

= 3 2

N k_B

U ⇒ U = 3

2N kBT

und: p

T = ∂S

∂V U,N

= N k_B

V ⇒ pV = N k_BT wobei wir wiederE =U gesetzt haben.

Wir haben also jeweils f¨ur die Temperatur und den Druck die ideale Gasgleichung reproduziert, wobei wir von der statistischen Definition der Entropie ausgegangen sind.

2. Logarithmic spectrum:

(a) Using Z_N =Z₁^N we find Z₁ =

∞

X

n=1

exp (−β∆ log (n)) =

∞

X

n=1

n^−β∆. (9)

This series is convergent only for β∆>1, i.e. k_BT <∆. Then it follows that

Z₁ =ζ(β∆) (10)

with the Riemann Zeta function

ζ(s) =

∞

X

n=1

1 n^s.

Close tos= 1 the function ζ(s) diverges as ζ(s)' 1

s−1. (11)

This behavior can be inferred by considering the integral Z ∞

1

dx

x^s = 1 s−1.

Indeed, one can estimate this integral from above and from below by replacing it by the discrete sums (dx→∆x= 1):

∞

X

n=1

1

(n+ 1)^s <

Z ∞

1

dx x^s <

∞

X

n=1

1 n^s, ζ(s)−1 < 1

s−1 < ζ(s).

This gives for k_BT close to ∆ the following behavior of the free energy:

F 'N k_BT log ∆

k_BT −1

, (12)

the entropy:

S =−∂F

∂T = k_BN∆

∆−k_BT −k_BNlog ∆

k_BT −1

, (13)

and specific heat:

C =T∂S

∂T = k_BN∆²

(∆−kBT)². (14)

(b) The specific heat diverges at temperatures larger than ∆/k_B which makes it im- possible to heat this system up to temperatures larger than this value. The regime T >∆/k_B is not reachable.

Physically this is caused by the huge number of states which occur per energy interval. To see this we approximate the sum over n by an integral

Z = Z

dnexp (−βE(n)) (15)

and perform a substitution of variables to

E =E(n) = ∆ log (n) (16)

with

dn

dE = e^E/∆

∆ (17)

follows

Z = Z dE

∆ exp E

∆−βE

(18) thus the exponentially diverging density of states causes the integral to diverge if

∆⁻¹ > β, which is just the above criterion. Indeed it follows forkBT < ∆ Z = k_BT

∆−k_BT (19)

which agrees with the exact behavior for k_BT ' ∆. The approximation of a sum by an integral is simply very good if the states are exponentially dense.

3. Maxwell’s relations

(a) Relation between c_p, c_V: We have that

c_V = T ∂S

∂T

V,N

c_p = T ∂S

∂T

p,N

(20)

We can then write

c_V = T ∂S

∂T

V

=T∂(S, V)

∂(T, V) =T

∂(S,V)

∂(T ,p)

∂(T,V)

∂(T ,p)

= T

∂(S,V)

∂(T ,p)

∂V

∂p

T

= T

∂V

∂p

T

∂S

∂T _p

∂V

∂p _T

− ∂S

∂p _T

∂V

∂T _p

!

= cp− T

∂V

∂p

T

∂S

∂p T

∂V

∂T p

(21)

where the functional determinants have been defined in Q2 in Blatt 1, and we used the identities derived in that question (quantities p, V, T are linked to each other by equation of state).

Next, we derive Maxwell’s relations. We begin by writing the expression for Gibbs function (at constant N)

δG=−SδT +V δp = ∂G

∂T _p

δT + ∂G

∂p _T

δp (22)

Then if we use the commutativity of second derivatives, we get that

− ∂S

∂p T

= ∂V

∂T p

(23) Putting (23) into (21) we get that

c_V =c_p+T ∂V

∂T

p

2

∂V

∂p

T

(24)

(b) From (24) and the fact that κ =−_V^{1 ∂V}_∂p

T > 0, it follows that always cp > cv, i.e the heat capacity at constant pressure is always bigger than the heat capacity at constant volume.

(c) We are given the values of

α = 1 V

∂V

∂T _p

= 1 V

∂(V, p)

∂(T, p) κ = −1

V

∂V

∂p T

=−1 V

∂(V, T)

∂(p, T) (25)

We can then write

α κ =−

∂(V,p)

∂(T,p)

∂(T ,V)

∂(T,p)

=−∂(V, p)

∂(T, V) = ∂p

∂T V

(26) where the first identity that we used above has been proven in Blatt 01, Q2. After putting in the numbers, we get the answer of 34 bar/K, which is a huge number.

4. Zweiatomiges Gas

(a) Partition function. the partition function of N molecules can be written as

Z = (Z_tZ_r)^N N!

Zt = 1 h³

Z

d³pd³qe^−βp2/(2m)=V

2πm h²β

³₂

= V λ³ λ =

r2πm h²β Z_r =

∞

X

l=0

(2l+ 1)e^{−l(l+1)2Tθrot} (27)

HereZt is the translational part of the partition function of a single molecule (see lecture notes for more details),Z_r is the rotational part of the partition function of a single molecule. State with the angular momentuml, is (2l+ 1)-degenerate (since ml =−l,−l+ 1, ...l – there 2l+ 1 such states). In priciple the summation over l is difficult to calculate and we will perform this summation only in the limiting cases of very low or very high temperatures.

(b) Low temperature limit. For T θrot, we can only keep the first two terms in the l summation of the partition function (27), to obtain

Zr ≈1 + 3e^−θrotT (28)

For the internal energy and heat capacity we find that find that

U = −∂lnZ

∂β = 3

2N k_BT

| {z }

transl. contribution

+3N k_Bθ_rote^−θrotT

c_v =

∂U

∂T

V

= 3

2N k_B+ 3N k_B θ_rot

T 2

e^−θrotT (29) where we used (27) and (28). We see that i) every translational degree of freedom contributesN k_B/2, and ii) the rotational spectrum is gapped, hence the contributi- on is proportional toe^−∆T, where ∆ =θ_rot is the value of the gap (energy diffrerence between the ground state and the first excited state).

(c) High temperature limit, T θ_rot. In this case we can replace the sum overl by an integral

Z_r =

∞

X

l=0

(2l+ 1)e^{−l(l+1)2Tθrot}

= 1

∆l Z ∞

l=0

(2l+ 1)e^{−l(l+1)2Tθrot}

= 2T θ_rot

Z ∞

0

dxe^−x= 2T θ_rot x = l(l+ 1)θ_rot

2T (30)

where ∆l= 1 is the spacing between adjacent l. The replacement of the sum by an integral is justified since ∆l = 1χ, where χ'T /θ_rot is the characteristic length of the exponential function in the integral.

We obtain

U = −∂lnZ

∂β = 3

2N k_BT

| {z }

transl. contribution

+N k_BT

c_v =

∂U

∂T

V

= 3

2N k_B+N k_B (31)

This is in agreement with the equipartition theorem – 3 translation degrees of freedom, each contributingN kBT /2 to internal energy (and hence byN kB/2 to heat capacity), 2 rotational degrees of freedom, each contributing N k_BT /2 to internal energy.

(d) Spin singlet case: S = 0. In this case the spin part of the wave-function is anti- symmetric, which means that the spatial part of the wave-function Yml must be symmetric (in order to get overall antisymmetric wave-function). This is the case only for even l = 0,2,4,6.... ThenZ_r =P∞

l=0,2,4(2l+ 1)e^{−l(l+1)2Tθrot}.

Spin tripletS = 1 case. In this case the spin part of the wave-function is symmetric, which means that the spatial part of the wave-function Yml must be antisymmetric (in order to get overall antisymmetric wave-function). This is the case only for odd l= 1,3,5,7.... Then Z_r =P∞

l=1,3,5...(2l+ 1)e^{−l(l+1)2Tθrot}.

We have seen that the spin singlet S = 0 allows l = 0,2,4.., while spin triplet S = 1, allows only l = 1,3, .... The l = 0 state is the state with lowest rotation energyE_r ∝l(l+ 1); and this state is only allowed ifS = 0. Hence spin singlet state is the ground-state.