The internal energy of the gas equals U

1 SPA Homework solutions – week IV

Marta Wac lawczyk & Maciej Lisicki

1. Classical gas Consider a classical gas in volume V . Gas consists of N particles, each of mass m. The internal energy of the gas equals U . Calculate the probability that the energy of a selected particle in this gas is contained within E and E + dE.

Solution Let Γ(U, V, N, ∆U ) be the number of microstates corresponding to the system’s total energy between U and U + ∆U The number of microstates such that the selected particle has the energy between E and E + dE is

Γ(U − E, V, N − 1, ∆U ) − Γ(U − E − dE, V, N − 1, ∆U ) ≈ ∂ω(U − E, V, N − 1)

∂(U − E) ∆U dE,

where ω is the density of microstates. To calulate the probability it is necessary to divide this quantity by Γ(U, V, N, ∆U ), which is the number of microstates of the system of energy between U and U + ∆U . We have

Γ(U, V, N, ∆U ) = V^N N ! Γ(1 + 3N/2)

 mU 2πh²

^3N/23N

2U∆U = ω∆U

ω(U − E, V, N − 1) = V^{N −1}

(N − 1)! Γ(1 + 3(N − 1)/2)

 m(U − E) 2πh²

3(N −1)/2

3(N − 1) 2(U − E) dividing the derivative of ω with respect to (U − E) by Γ(U, V, N, ∆U ) we obtain the probability

P (E)dE = C(V, N )(U − E)^3N/2−2/U^3N/2−1 where C(V, N ) contains all terms that do not depend either on U or U − E.

2. Gas fluctuations in a cubic box A cubic box with isolating (adiabatic) walls of length L contains N particles of an ideal gas. Find the dispersion of the centre of mass of the system in equilibrium. How does the dispersion behave with the increasing number of particles in the box?

Solution The system is isolated, so each state of a fixed energy E is equally probable (the distribution of states is microcanonical). Thus

p(r1, . . . , rN, p1, . . . , pN) = 1 ω(E, V, N, δE) where

ω(E, V, N, δE) = Z

E<H<E+δE

dΓ.

Integrating over the momenta, we get the distribution for positions p(r1, . . . , rN) = 1

L^3N

We introduce a coordinate system with the origin in the middle of the cube. The centre of mass is defined as R =

P

im_ir_i P

im_i , so by symmetry

hRi = P

imihrii P

im_i = 0.

Now

R² =

*1 N

N

X

i=1

ri

!²+

=

* _N X

i=1

1 Nri+

N

X

i6=j

1 N²rirj

+

2 because the variables are independent and hrii = 0, we have

*_N X

i6=j

1 N²rirj

+

= 0

and thus

R² = 1

N r²₁ = 3

N x²₁ = L² 4N so we find

σ = L 2√

N.