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Category: PHY 831

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1. Consider the escape of molecules of a Maxwell-Boltzmann gas through an

opening of area A in the walls of a vessel of volume V. Show that the escaping

molecules have a mean kinetic energy of 2kBT, where T is the quasistatic temperature of the gas (i.e. assuming that the energy lost to the escaping molecules

is small compared to the total energy of the system so the rest of the gas can

be considered to be in equilibrium).

2. In class we discussed the form of the Boltzmann collision term for the MaxwellBoltzmann distribution. For Bosons/Fermions (upper sign/lower sign), the

collision term of the Boltzmann equation takes the form

C[ f ] = Z

d

3 p2

(2πh¯)

3

Z

d

3 p

0

1

(2πh¯)

3

Z

d

3 p

0

2

(2πh¯)

3

r(p1 p2 → p

0

1

p

0

2

)

×

f(~p1)f(~p2)(1 ± f(~p

0

1

))(1 ± f(~p

0

2

)) − f(~p

0

1

)f(~p

0

2

)(1 ± f(~p1))(1 ± f(~p2))

,

where f is the single-particle distribution function and all distribution functions are evaluated at the same point in space. The 1 + f terms account for

stimulated emission for Bosons. For Fermions the 1 − f term accounts for

final state blocking. If there is a high probability that there is a particle already in the final state of the reaction, then the rate at which the process

proceeds is reduced since only one Fermion can be in a single-particle state.

Show this collision integral goes to zero when f takes it’s equilibrium form,

f∓(~p) = (exp[β(p

2/2m − µ)] ∓ 1)

−1

.

3. For an initial one particle distribution function of the form

f1(~x,~p, t = 0) = A exp

−

~x

2

2L

2 −

~p

2

2mT0

, (1)

where A is a constant, find an expression for f1(~x,~p, t > 0) if the distribution

evolves according to the collisionless Boltzmann equation. Show that this has

the form of a Maxwell-Boltzmann distribution with a time dependent temperature T(t), which implies f1 is also a solution of the collisional Boltzmann

equation.

4. Consider a uniform fluid with background density ρ0 and background velocity

u0 = 0. Assume that the disturbance is adiabatic (so that you can neglect the

energy conservation equation of Euler’s equations) and small (so that you only

need to keeps terms to first order in δρ with ρ = ρ0 + δρ).

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