Physics Problems & Solutions: #94

Thermal Physics - Partition Function

A system consists of N weakly interacting subsystems, each with two internal quantum states with energies 0 and $\epsilon$ . The internal energy for this system at absolute temperature T is equal to

A. $N\epsilon$

B. $\frac{3}{2}NkT$

C. $N\epsilon e^{-\epsilon/kT}$

D. $\frac{N\epsilon}{e^{\epsilon/kT} + 1}$

E. $\frac{N\epsilon}{1+ e^{-\epsilon/kT} }$

(GR9677 #94)

Solution:

A. FALSE. It’s not an exponential function

B. FALSE. It’s not an exponential function

C. FALSE. $E\rightarrow 0$ as $T\rightarrow 0$    but   $E\rightarrow N\epsilon$    as    $T\rightarrow \infty$

D. TRUE. $E\rightarrow 0$ as   $T\rightarrow 0$ and   $E\rightarrow N\epsilon/2$ as   $T\rightarrow \infty$
(fits the equipartition theorem for high temperature/classical region/Maxwell-Boltzmann Statistics)

E. FALSE. $E\rightarrow N\epsilon$ as   $T\rightarrow 0$ and   $E\rightarrow \infty$ as   $T\rightarrow \infty$

Alternative Solution #1:

Partition function

$Z=\sum_i e^{-E_i/kT}$

with energies 0 and $\epsilon$ ,

$Z=e^{0/kT}+e^{-\epsilon/kT} = 1+e^{-\epsilon/kT}$

Energy

$E=\frac{\displaystyle \sum_i E_i e^{-E_i/kT}}{Z}$

$E=\frac{0+\epsilon e^{-\epsilon/kT}}{1+e^{-\epsilon/kT}}\times\frac{e^{\epsilon/kT}}{e^{\epsilon/kT}}=\frac{\epsilon }{e^{\epsilon/kT}+1}$

For N subsystems,

$E=\frac{N\epsilon }{e^{\epsilon/kT}+1}$

Answer: D

Alternative Solution #2:

We can also use

$E=-\frac{\partial \ln Z}{\partial \beta}$

with $\beta=1/kT$ , as

$E=\frac{\displaystyle \sum_i E_i e^{-E_i\beta}}{Z}=-\frac{\partial \ln Z}{\partial \beta}$

$\frac{\partial \ln Z}{\partial \beta}=\frac{1}{Z}\frac{\partial Z}{\partial \beta} = \frac{-\epsilon e^{-\epsilon \beta}}{1+e^{-\epsilon \beta}} \times \frac{e^{\epsilon \beta}}{e^{\epsilon \beta}} = \frac{-\epsilon}{e^{\epsilon \beta}+1}$

$E=-\frac{\partial \ln Z}{\partial \beta}= \frac{\epsilon}{e^{\epsilon \beta}+1}$

For N subsystems,

$E=\frac{N\epsilon }{e^{\epsilon/kT}+1}$

Special Relativity - Lorentz transformation

Which of the following is a Lorentz transformation? (Assume a system of units such that the velocity of light is 1.)

A. $\\x'=4x\\ y'=y\\ z'=z\\ t'=.25t\\$

B. $\\x'=x-.75t\\ y'=y\\ z'=z\\ t'=t\\$

C. $\\x'=1.25x-.75t\\ y'=y\\ z'=z\\ t'=1.25t-.75x\\$

D. $\\x'=1.25x-.75t\\ y'=y\\ z'=z\\ t'=.75t-1.25x\\$

E. None of the above

(GR9277 #94)

Solution:

Galilean	Lorentz
x' = x − vt	x' = (x − vt)γ
y' = y	y' = y
z' = z	z' = z
t' = t	t' = (t − vx/c²)γ

A and B are FALSE since t' = t is Galilean.

Check answer C:
x' = (x − vt)γ = 1.25x − .75t
γ = 1.25 and vγ = .75
v = .75/1.25

t' = (t − vx/c²)γ = 1.25t − .75x
γ = 1.25 and (v/c²)γ = .75
v = .75c²/1.25 = .75/1.25, since c = 1

v is consistent → C is TRUE

Answer: C

Electromagnetism - RL Circuit

In the circuit shown above, R₂ = 3R₁ and the battery of emf ξ has negligible internal resistance. The resistance of the diode when it allows current to pass through it is also negligible. At time t = 0, the switch S is closed and the currents and voltage are allowed to reach their asymptotic values. Then at time t₁, the switch is opened. Which of the following curves most nearly represents the potential at point A as a function of time t?

(GR8677 #94)

Solution:

When the switch is closed, at time t = 0, the voltage drops through the circuit. Thus, C, D and E are FALSE.

The time constant in the first process: $\tau_1=\frac{L}{R_1}$

The time constant, τ is the time it takes the voltage across the component to either decrease or increase.

When the switch is opened, at time t₁, the voltage increases with the time constant: $\tau_2=\frac{L}{R_2}=\frac{L}{3R_1}=\frac{1}{3}\tau_1$

Thus, $\tau_2< \tau_1$

It means, it takes less time for the voltage to increase than decrease.

Answer: B