分配関数の導関数、エネルギー揺らぎおよび熱容量
導入済み| English | 意味 |
|---|---|
| Helmholtz free energy/ˈhelmhəʊltz friː ˈenədʒi/ | ヘルムホルツ自由エネルギー |
| energy fluctuation/ˈenədʒi ˌflʌktʃuːˈeɪʃn/ | エネルギーの揺らぎ |
A decision before an answer
- A thermal energy gap can keep heat capacity small at both low and high temperatures, even while the excited-state probability keeps increasing.
- Your goal: Obtain canonical mean energy and free energy from a fixed energy spectrum.
Differentiate a fixed spectrum
- For a system exchanging energy with a bath at T while N and V remain fixed, let β=1/(kBT). Sum Z=Σ_i g_i exp(−βE_i) over energy levels with their degeneracies. Level probability is g_i exp(−βE_i)/Z; mean energy U is the probability-weighted energy sum. Differentiating a temperature-independent spectrum gives U=−∂lnZ/∂β.
- Keep the energy gap and degeneracies fixed during the derivative. Writing Δ=kBT ln3 to describe one evaluation temperature must not be read as making Δ change with T. For N independent distinguishable identical two-level subsystems, Z_total=z^N and U_total=N u. Indistinguishable particles and interactions need their own state counting rather than this product assumption.
Two nondegenerate levels have energies 0 and fixed Δ. At Δ/(kBT)=ln4, excited probability is:
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
p_exc=1/(1+e^x)=1/(1+4)=1/5.
Identify the thermodynamic quantity
- The Helmholtz free energy is F=−kBT lnZ. Canonical entropy follows from S=kB(lnZ+βU), equivalently −kBΣ p_j ln p_j over individual microstates. Constant-volume heat capacity is C_V=(∂U/∂T)_V,N. These quantities carry different units and describe different derivatives.
- Adding a constant energy offset ε to every state multiplies Z by exp(−βε), increases U and F by ε, and leaves probabilities, entropy and heat capacity unchanged when ε is independent of T. A negative chosen mean energy is therefore not evidence of a negative heat capacity. Use the same energy reference in the probabilities and thermodynamic expressions.
A constant temperature-independent energy ε is added to every state. Which quantity stays unchanged?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
All probabilities are unchanged. U and F shift by ε and Z gains e^(−βε), but entropy and heat capacity remain the same.
Relate fluctuations and response
- The second β derivative of lnZ gives variance Var(E)=⟨E²⟩−U². For the same fixed-spectrum canonical model, C_V=Var(E)/(kBT²), so C_V/kB=β²Var(E). Nonnegative variance implies nonnegative C_V within these conditions. The denominator includes kB, not kB², when heat capacity retains its ordinary J/K units.
- This fluctuation formula concerns the canonical energy distribution; it does not say each particle has exactly the mean energy. For independent subsystems variances add, while means add too. Relative energy fluctuations typically shrink like 1/√N when the mean and per-subsystem variance remain finite and nonzero.
For levels 0 and fixed Δ at a temperature where x=ln3, z=1+1/3=4/3 and p_exc=1/4. Thus u=Δ/4, ⟨E²⟩=Δ²/4 and Var(E)=3Δ²/16. C/kB=3(ln3)²/16=0.226303. F=−kBT ln(4/3); S/kB=ln(4/3)+(ln3)/4=0.562335. Raising the reference by ε changes u to ε+Δ/4 but preserves the variance and heat capacity.
For equally probable energies 0 J and 2 J, energy variance is ____ J².
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Mean=1 J and second moment=2 J², so variance=2−1²=1 J².
Test finite-level temperature limits
- For one ground state at E=0 and one excited state at fixed E=Δ>0, put x=Δ/(kBT). Then z=1+e^(−x), p_exc=1/(1+e^x), u=Δp_exc and C/kB=x²e^x/(1+e^x)². At low T, excitation and heat capacity vanish exponentially. At high T, p_exc tends to one half and energy saturates, so heat capacity tends to zero again.
- Entropy rises from zero for the unique ground state toward kB ln2 as the two states become equiprobable. The resulting finite-temperature heat-capacity peak is specific to the finite-level model; it is not the constant classical oscillator value. Ground-state degeneracy or additional levels would change the limiting entropy and temperature response.
Hold Δ fixed when differentiating, include degeneracies in Z, and use energy variance rather than the square of mean energy in the heat-capacity formula.
このケースに当てはまる答えはどれか？
固定されたエネルギースペクトルから正準平均エネルギーおよび自由エネルギーを求める
For a fixed two-level nondegenerate spectrum, canonical heat capacity tends to zero as T becomes arbitrarily high.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Energy approaches Δ/2 and its temperature derivative tends to zero; a finite spectrum saturates.
Keep the distinctions
- energy fluctuation 能量涨落 — Canonical spread of energy about its ensemble mean, quantified by the energy variance.
- Helmholtz free energy 亥姆霍兹自由能 — Thermodynamic potential F=U−TS, equal to −kBT lnZ for the canonical ensemble.
- Obtain canonical mean energy and free energy from a fixed energy spectrum.
- Relate energy variance to constant-volume heat capacity with fixed-spectrum conditions.
- Evaluate entropy and low/high-temperature limits for an original finite-level model.
Match each term with its precise meaning in this lesson.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
難易度 3/5 · 200 コイン + 今日初回完了時に小ボーナス付与
The order follows the stated process; check each stage before the next.