Statistical Physics

Question 1

What is the aim of statistical physics?

Answer 1

To understand macroscopic thermodynamic properties by analyzing the average behavior of a large number of microscopic particles.

Question 2

What is a macrostate?

Answer 2

A macrostate is defined by macroscopic variables such as internal energy (U), entropy (S), temperature (T), pressure (P), and volume (V).

Question 3

What is a microstate?

Answer 3

A microstate is a specific configuration of all the particles in the system.

Question 4

What is the key postulate of statistical physics?

Answer 4

In an isolated system, all accessible microstates are equally probable.

Question 5

How does quantum mechanics justify the postulate of equiprobability?

Answer 5

Because systems transition between microstates at equal rates, the probability of finding the system in any microstate is the same over long periods.

Question 6

In the two-compartment box model, what is a macrostate defined by?

Answer 6

By the number n of atoms on the left-hand side of the box.

Question 7

How many macrostates are possible for N atoms in a two-part box?

Answer 7

There are N + 1 macrostates.

Question 8

How many microstates are there for N atoms?

Answer 8

There are 2^N microstates.

Question 9

What is the probability of all atoms being on the left side?

Answer 9

p = 1 / 2^N.

Question 10

What is the number of microstates with n atoms on the left?

Answer 10

Ω(n) = C(N, n) = N! / (n!(N − n)!).

Question 11

How is the probability of a given macrostate (n atoms on the left) calculated?

Answer 11

p(n) = Ω(n) / 2^N.

Question 12

What is the most probable macrostate in the two-part box model?

Answer 12

n = N / 2, i.e., equal number of atoms on each side.

Question 13

How does the width of the distribution scale with N?

Answer 13

The width scales as √(N), making the distribution sharper for large N.

Question 14

What is the standard deviation of the distribution for p = 1/2?

Answer 14

σ = √(N) / 2.

Question 15

What does the sharpness of the distribution imply for large N?

Answer 15

That deviations from the most probable macrostate become exponentially unlikely.

Question 16

How is entropy defined in terms of the number of microstates?

Answer 16

S = k ln(Ω), where Ω is the number of microstates.

Question 17

What does Boltzmann’s formula S = k ln(Ω) tell us?

Answer 17

Entropy measures the logarithm of the number of microstates compatible with a macrostate.

Question 18

Why is entropy additive for independent systems?

Answer 18

Because the logarithm of a product is the sum of the logarithms: ln(Ω_A * Ω_B) = ln(Ω_A) + ln(Ω_B).

Question 19

What is the physical interpretation of entropy in this model?

Answer 19

Entropy quantifies how many microscopic arrangements correspond to a macroscopic condition.

Question 20

What is Maxwell’s Demon thought experiment?

Answer 20

A hypothetical being that could sort fast and slow molecules to decrease entropy, seemingly violating the second law.

Question 21

Why does Maxwell’s Demon not violate the second law of thermodynamics?

Answer 21

Because the demon must acquire and process information, which ultimately increases entropy.

Question 22

What does Maxwell’s Demon highlight about thermodynamics?

Answer 22

The connection between information theory and entropy.

Question 23

What are the energy levels for a spin-1/2 particle in a magnetic field?

Answer 23

U = ±μB, depending on the spin alignment relative to the magnetic field.

Question 24

How is a macrostate defined for a paramagnet?

Answer 24

By the number n of spins pointing down.

Question 25

What is the total energy of the paramagnet with n spins down?

Answer 25

U = (2n − N)μB.

Question 26

What is the number of microstates for a paramagnet with n spins down?

Answer 26

Ω(n) = C(N, n) = N! / (n!(N − n)!).

Question 27

What is the entropy of a paramagnet with n down spins?

Answer 27

S = k ln(C(N, n)).

Question 28

What is the total entropy of two combined systems A and B?

Answer 28

S_total = S_A + S_B = k ln(Ω_A * Ω_B).

Question 29

What condition must be met at thermal equilibrium between systems A and B?

Answer 29

∂S_A/∂U_A = ∂S_B/∂U_B.

Question 30

How is temperature defined statistically?

Answer 30

1/T = ∂S/∂U.

Question 31

What does thermal equilibrium imply in terms of energy exchange?

Answer 31

That energy will flow until the temperatures are equal.

Question 32

What is the Boltzmann distribution for the probability of a microstate i?

Answer 32

p_i = (1/Z) * exp(−E_i / kT), where Z is the partition function.

Question 33

What is the partition function Z?

Answer 33

Z = ∑ exp(−E_i / kT), summed over all microstates.

Question 34

Why is the partition function important?

Answer 34

It normalizes the Boltzmann probabilities and encodes thermodynamic information.

Question 35

What happens to the Boltzmann distribution as T → 0?

Answer 35

The system tends toward the ground state (lowest energy).

Question 36

What happens to the Boltzmann distribution as T → ∞?

Answer 36

All microstates become equally probable.

Question 37

What is the average energy ⟨E⟩ in the Boltzmann distribution?

Answer 37

⟨E⟩ = ∑ E_i p_i = (1/Z) ∑ E_i exp(−E_i / kT).

Question 38

How can average energy be derived from the partition function?

Answer 38

⟨E⟩ = −∂ ln Z / ∂β, where β = 1 / (kT).

Question 39

How is entropy related to the partition function?

Answer 39

S = −k ∑ p_i ln(p_i) = k ln Z + ⟨E⟩ / T.

Question 40

How is Helmholtz free energy related to Z?

Answer 40

F = −kT ln Z.

Question 41

What is the relation between entropy, internal energy, and free energy?

Answer 41

S = (⟨E⟩ − F) / T.

Question 42

What is the Maxwell-Boltzmann distribution for velocity in 1D?

Answer 42

p(v_x) = √(m / 2πkT) * exp(−mv_x² / 2kT).

Question 43

What is the Maxwell-Boltzmann distribution for speed in 3D?

Answer 43

p(v) = 4πv² * (m / 2πkT)^(3/2) * exp(−mv² / 2kT).

Question 44

What does the most probable speed correspond to in MB distribution?

Answer 44

The speed at which the distribution p(v) is maximized.

Question 45

What is the general expression for entropy in terms of probabilities?

Answer 45

S = −k ∑ p_i ln(p_i).

Question 46

How does the Gibbs entropy formula relate to Boltzmann's formula?

Answer 46

If all states are equally probable, S = k ln(Ω).

Question 47

What is the average energy of a paramagnet in a magnetic field?

Answer 47

⟨E⟩ = −μB tanh(μB / kT).

Question 48

What is the magnetization M of N spin-1/2 particles?

Answer 48

M = Nμ tanh(μB / kT).

Question 49

What is the heat capacity of a paramagnet?

Answer 49

C = Nk (μB / kT)^2 sech²(μB / kT).

Question 50

What is the entropy of a paramagnet as a function of temperature?

Answer 50

S = Nk [ln(2 cosh(μB / kT)) − (μB / kT) tanh(μB / kT)].

Question 51

What happens to the magnetization as T → ∞?

Answer 51

It goes to 0 because spins are randomized.

Question 52

What happens to the magnetization as T → 0?

Answer 52

It approaches Nμ because all spins align with the field.

Question 53

What is adiabatic demagnetization?

Answer 53

A cooling method where entropy is held constant while the magnetic field is reduced, lowering the temperature.

Question 54

How is entropy change related to heat capacity?

Answer 54

ΔS = ∫ (C / T) dT.

Question 55

What is the energy of a quantum harmonic oscillator?

Answer 55

E_n = ℏω(n + 1/2), where n = 0, 1, 2, ...

Question 56

What is the average energy of a quantum harmonic oscillator?

Answer 56

⟨E⟩ = ℏω / (exp(ℏω / kT) − 1) + ℏω/2.

Question 57

What is the high-temperature limit of the quantum harmonic oscillator?

Answer 57

⟨E⟩ ≈ kT, recovering classical behavior.

Question 58

What is the low-temperature behavior of the quantum harmonic oscillator?

Answer 58

⟨E⟩ approaches the ground state energy ℏω/2.

Question 59

What is the heat capacity of a quantum harmonic oscillator?

Answer 59

C = k (ℏω / kT)^2 * exp(ℏω / kT) / (exp(ℏω / kT) − 1)^2.

Question 60

Why does heat capacity go to zero as T → 0?

Answer 60

Because excited states become energetically inaccessible and thermal energy is insufficient to populate them.

Question 61

Why does heat capacity approach a constant at high temperature?

Answer 61

Because all energy levels are populated, recovering the classical Dulong-Petit limit.