Hauptsatz: dU =δQ−pdV +µdN

Karlsruher Institut f¨ur Technologie Institut f¨ur Theorie der Kondensierten Materie Ubungen zur Theoretischen Physik F¨ SS 12

Prof. Dr. J¨org Schmalian Blatt 3: L¨osungen

Dr. Igor Gornyi Besprechung 4.5.2012

1. Ideales Gas mit f Freiheitsgraden pro Molek¨ul:

(a) 1. Hauptsatz: dU =δQ−pdV +µdN.

Adiabatische Zustands¨anderung: δQ= 0 , konstante Teilchenzahl: dN = 0 ,

also: dU =−pdV . In einem thermisch isolierten Gas in einem geschlossenen Beh¨alter kann sich die innere Energie nur durch mechanische Arbeit ¨andern.

Bekannt ist außerdem: U = f

2N kT →dU = f

2N kdT

und pV =N kT →pdV +V dp=N kdT , zusammen→dU = f

2[pdV +V dp] . Mit dem 1. Hauptsatz von eben gibt das

( 1 + f

2 )

pdV =−f 2V dp,

was jetzt integriert werden kann, von einem thermodynamischen Zustand 0 zu einem anderen 1 ¨uber einen reversiblen Weg,

( 1 + f

2 )∫¹

0

dV

V =−f 2

∫1 0

dp p

Daraus folgt, mit V₁ ≡V , p₁ ≡p, denn der Zustand 1 ist beliebig, (

1 + f 2

) ln V

V₀ = −f 2 ln p

p₀ ⇒ (

1 + 2 f

)

ln(V) + ln(p) = const₁

⇒ p V^(1+2/f) = const₂ Analog ergibt sich aus dU =−pdV und dU = f

2N kdT und pV =N kT ⇒ p= N kT

V unmittelbar N k

V dV =−f 2

N k

T dT , also ln V

V₀ =−f 2 ln T

T₀ ⇒ V T^{f /2} = const₃

(b) Die “Fundamentalbeziehung” lautet:T S =U+pV −µN ⇒S= 1 TU+ p

TV −µ TN. MitS =S(U, V, N) folgt dS = 1

TdU+ p

TdV − µ TdN. Es ist jetzt praktisch, Dichten einzuf¨uhren:

s=S/N , u=U/N , v=V /N ⇒s= 1 Tu+ p

Tv− µ T . MitS =S(U, V, N) ist auch s=s(U, V, N) =s(u, v, N) , s h¨angt offenbar nicht explizit von N ab, also:

s=s(u, v), ds= 1

Tdu+ p Tdv Dass s nicht von N abh¨angt, folgt auch ganz allgemein:

U, V, N, S sind extensiv, das heißt z.B. S(λU, λV, λN) = λS(U, V, N) . Die Dichten sind (trivialerweise) intensiv, alsos(λU, λV, λN) = s(U, V, N) . Dies kann nur erf¨ullt werden, wennU, V, N nur als Quotienten (Dichten) eingehen, alsos=s(U/N, V /N) oders(U/V, U/N) oder s(U/V, V /N) .

ds l¨aßt sich jetzt integrieren, wenn man T und p rauswirft:

U = f

2N kT ⇒u= f

2kT ⇒ 1 T = f

2k1

u, pV =N kT ⇒ p T =k1

v ds= f

2kdu

u +kdv

v ⇒s−s₀ = f 2kln u

u₀ +kln v v₀.

Die Integrationskonstanten sind s0 = S0/N0, u0 = U0/N0, v0 = V0/N0, einsetzen liefert

S = S0

N

N₀ +N k [f

2 (

ln U

U₀ + lnN0

N )

+ (

ln V

V₀ + lnN0

N ) ]

= S₀N

N₀ +N k [f

2ln (U

U₀ )

+ ln (V

V₀ )

− (f

2 + 1 )

ln (N

N₀ ) ]

.

(c) Nehmen wir o.B.d.A. mal an, dass das Gas aus konstantenN =N₀ Teilchen besteht und im konstanten Volumen V = V₀ eingeschlossen ist, dann h¨angt U von der Temperatur ab ¨uberU = ^f₂N₀kT, also, mit U₀ ≡ ^f₂N₀kT₀, T₀ = const.,

S(T) =S₀+ f

2N₀k ln T T₀

F¨ur T →0 divergiert dies, es l¨aßt sich kein S₀ finden, so dass S(T →0) = 0 w¨are.

Das “ideale Gas” ist eine brauchbare N¨aherung f¨ur ideale Gase nur bei hohen Tem- peraturen. BeiT →0 muss die Quantennatur der Teilchen (Fermionen oder Boso- nen) ber¨ucksichtigt werden, die zu einem nichtentarteten Grundzustand des Gases f¨uhrt. Dann folgt aus der informationstheoretischen Fundierung der Entropie sofort S(T →0) = 0 .

2. Velocity and interparticle distance distributions in an ideal gas:

(a) The distribution function f(v) follows immediately from the probability to find a particle with momentump and at position q

ρ(p,q) = ⟨δ(p−p_i)δ(q−q_i)⟩

= 1

Z₁

∫ d³p_id³q_i

h³ ⟨δ(p−p_i)δ(q−q_i)⟩exp (

−βp²_i 2m

)

= λ³ V h³ exp

(

− β 2mp²

) .

Hereλ =h(2πk_BT m)^−1/2 is the thermal de Broglie wavelength following from the

normalization: ∫

d³pd³qρ(p,q) = 1

Integrating over position, we find the momentum distribution function:

f(p) = 1

(2πmkBT)^3/2 exp (

− p² 2mkBT

)

. (1)

Substituting v=p/m, and integrating over the direction of v (which only gives a prefactor 4π:d³v =v²dvsinθdθdϕ) yields the velocity distribution:

f(v) = 4πv²

( m 2πk_BT

)3/2

exp

( mv² 2k_BT

)

. (2)

(b) The most probable velocity, v^∗, follows from ^∂f_∂v

v=v^∗ = 0 yielding v^∗− m

2k_BTv^∗3 = 0 (3)

such that

v^∗ =

√2k_BT

m . (4)

The mean velocity is given as

⟨v⟩ =

∫

dvf(v)v = 4π

( m 2πk_BT

)3/2∫ _∞

0

v³exp

( mv² 2k_BT

) dv

= 2π

( m 2πk_BT

)3/2( 2k_BT

m

)2∫ _∞

0

dze^−zz =

√8k_BT

πm . (5)

Finally, we determine the mean square velocity

⟨v²⟩

= 4π

( m 2πk_BT

)3/2∫ _∞

0

v⁴exp

( mv² 2k_BT

) dv

= 2π

( m 2πk_BT

)3/2( 2k_BT

m

)5/2∫ _∞

0

dze^−zz^3/2

= 3k_BT

m (6)

The inequalities v^∗ < ⟨v⟩ < √

⟨v²⟩ hold even though these three characteristic velocities are of the same order of magnitude.

(c) First we need to find f(r):

f(r) = λ⁶ V²

∫ d³p_id³p_jd³r_id³r_j

h⁶ e^−2mβ (^p2i+p²_j)δ(r− |ri−rj|) (7) where we performed the integration with respect to allp_l and r_l with l̸=i, j. The prefactorλ⁶/V² results from the normalization of the distribution function.

Performing the momentum integration cancels theλ⁶ factor and we obtain f(r) =

∫ d³r_id³r_j

V² δ(r− |r_i−r_j|) (8) Using center of mass and relative coordinates

Q = 1

2(r_i +r_j)

q = r_i−r_j, (9)

we get

f(r) =

∫ d³Qd³q

V² δ(r− |q|) (10)

where the integration limits are determined by |Q±q/2| < R (this is just the condition |ri|,|rj|< R in terms of new variables).

The integral in Eq. (10) can be evaluated in spherical coordinates. One can first fix the length q = |q| of vector q and the direction of vector Q. Clearly, 0 < q < 2R (the maximum is reached when the two points are at the opposite poles of the container), whereas the length Q = |Q| of vector Q satisfies Q < √

R² −q²/4.

Next, we observe that forQ < R−q/2 a small sphere of radius q/2 lies completely within the sphere of radius R and therefore the angle θ between vectors Q and q can be arbitrary. In the rangeR−q/2< Q < √

R²−q²/4 the two spheres intersect and then the condition|Q±q/2|< R translates into the condition on cosθ:

−R²−Q²−q²/4

Qq <cosθ < R²−Q²−q²/4

Qq .

Thus we can re-write Eq. (10) as follows:

f(r) = 9

16π²R⁶

∫ _2π

0

dϕ_Q

∫ _π

0

dθ_Qsinθ_Q

∫ _2π

0

dϕ_q

∫ _2R

0

dqq²δ(r−q)

×





∫ _R−q/2

0

dQQ²

∫ ₁

−1

dcosθ_q+

∫ √

R²−q²/4 R−q/2

dQQ²

∫ ^{R2−Q2−q2/4}

Qq

−^R2−Q_Qq^2−q2/4

dcosθ_q





= 9r² R⁶

{∫ R−r/2 0

dQQ²+

∫ √

R²−r²/4 R−r/2

dQQR²−Q²−r²/4 r

}

(11)

It follows thatf(r >2R) = 0 whereas forr <2R the integration in Eq. (11) yields:

f(r) = 12 R

[( r 2R

)2

− 3 2

( r 2R

)3

+1 2

( r 2R

)5]

. (12)

The calculation of ⟨r⟩=∫_∞

0 drf(r)r is now straightforward and gives

⟨r⟩= 36

35R. (13)

The maximum distance between two particles in the sphere is 2R, the distance of largest probability (determined by ^∂f_∂r = 0) isR and the mean distance is just a bit larger than this value. Thus, particles are on average apart by a distance which is the size of the system.

3. An ideal gas in the field of the Earth:

(a) Since the particles do not interact with each other we only need to calculate the partition function Z(1) of one particle and useZ(N) = Z(1)^N. We find:

Z(1) =

∫ d³p h³ e^−βp

2 2m

∫

d³re^−βmgz

= A

λ³

∫ _∞

0

dze^−βmgz = A

βmgλ³ (14)

whereA = 2πR² is the area of the cylinder. The free energy then reads:

F =−k_BT lnZ =−k_BT Nln A

βmgλ³. (15)

(b) It follows that the density in the cylinder depends only on z:

ρ(z) =

1 λ³

∫_∞

0 dz^′e^−βmgz′δ(z−z^′)

Z(1) = βmg

A e^−βmgz. (16) The density decreases exponentially with height. The length l0 = ^k_mg^BT determines the decay. For the pressure we use

p(z) =ρ(z)k_BT = mg

A e^−βmgz (17)

which assumes that the ideal gas law is valid locally.

Considering a thin horizontal slice of gas between the planes z and z +dz, the difference between the pressures at its bottom and top has to be equal to the weight of the gas within the slice. This gives

mgρ(z)dz =p(z)−p(z+dz) =−dp(z)

dz dz, (18)

so that

dp(z)

dz =−mgρ(z). (19)

This is indeed satisfied by Eqs. (17) and (16):

dp(z)

dz = mg A

de^−βmgz

dz =−mgβmg

A e^−βmgz =−mgρ(z).

4. Logarithmic spectrum:

(a) Using Z_N =Z₁^N we find Z1 =

∑∞ n=1

exp (−β∆ log (n)) =

∑∞ n=1

n^−β∆. (20)

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

Z1 =ζ(β∆) (21)

with the Riemann Zeta function

ζ(s) =

∑∞ n=1

1 n^s, see, e.g.,

http://en.wikipedia.org/wiki/Riemann zeta function

http://mathworld.wolfram.com/RiemannZetaFunction.html Close tos= 1 the function ζ(s) diverges as

ζ(s)≃ 1

s−1. (22)

This behavior can be inferred by considering the integral

∫ _∞

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):

∑∞ n=1

1

(n+ 1)^s <

∫ _∞

1

dx x^s <

∑∞ 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 )

, (23)

the entropy:

S =−∂F

∂T = k_BN∆

∆−k_BT −k_BNlog ( ∆

k_BT −1 )

, (24)

and specific heat:

C =T∂S

∂T = k_BN

(∆−kBT)². (25)

(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 >∆/kB 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 =

∫

dnexp (−βE(n)) (26)

and perform a substitution of variables to

E =E(n) = ∆ log (n) (27)

with

dn

dE = e^E/∆

∆ (28)

follows

Z =

∫ dE

∆ exp (E

∆−βE )

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

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

∆−k_BT (30)

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.