Partition derivatives, energy fluctuations and heat capacity
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| Helmholtz free energy | 亥姆霍兹自由能 | hài mǔ huò zī zì yóu néng |
| energy fluctuation | 能量涨落 | néng liàng zhǎng luò |
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:
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?
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.
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.
Which answer fits this case?
Obtain canonical mean energy and free energy from a fixed energy spectrum
For a fixed two-level nondegenerate spectrum, canonical heat capacity tends to zero as T becomes arbitrarily high.
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.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.