Intuition The big pictureThe partition function Z Z Z master quantity of statistical mechanics: it counts, with proper Boltzmann weighting, how many ways a system can arrange its energy. Once you have Z Z Z Helmholtz free energy F F F logarithm of Z Z Z .
WHY a logarithm? Because energies add when you join systems, but partition functions multiply (independent systems → product of Z Z Z
Definition Canonical partition functionFor a system in contact with a heat bath at temperature T T T N N N V V V canonical partition function is
Z = ∑ i e − β E i , β ≡ 1 k B T Z = \sum_i e^{-\beta E_i}, \qquad \beta \equiv \frac{1}{k_B T} Z = ∑ i e − β E i , β ≡ k B T 1 microstates i i i E i E_i E i
Definition Helmholtz free energyF = U − T S F = U - TS F = U − T S U U U S S S thermodynamic potential whose natural variables are = = T , V , N = = ==T, V, N== == T , V , N == T , V , N T,V,N T , V , N
The goal: show the deceptively simple result
F = − k B T ln Z \boxed{F = -k_B T \ln Z} F = − k B T ln Z derive it from scratch , not just quote it.
Intuition Strategy (WHAT we'll do)We compute U U U S S S Z Z Z F = U − T S F = U - TS F = U − T S − k B T ln Z -k_BT\ln Z − k B T ln Z
The Boltzmann probability of microstate i i i p i = e − β E i Z . p_i = \frac{e^{-\beta E_i}}{Z}. p i = Z e − β E i . Why this step? The bath fixes T T T Z Z Z ∑ i p i = 1 \sum_i p_i = 1 ∑ i p i = 1
U = ⟨ E ⟩ = ∑ i p i E i = 1 Z ∑ i E i e − β E i . U = \langle E \rangle = \sum_i p_i E_i = \frac{1}{Z}\sum_i E_i\, e^{-\beta E_i}. U = ⟨ E ⟩ = ∑ i p i E i = Z 1 ∑ i E i e − β E i . ∂ ∂ β e − β E i = − E i e − β E i \dfrac{\partial}{\partial\beta}e^{-\beta E_i} = -E_i e^{-\beta E_i} ∂ β ∂ e − β E i = − E i e − β E i ∑ i E i e − β E i = − ∂ Z ∂ β . \sum_i E_i e^{-\beta E_i} = -\frac{\partial Z}{\partial\beta}. ∑ i E i e − β E i = − ∂ β ∂ Z . U = − 1 Z ∂ Z ∂ β = − ∂ ln Z ∂ β . U = -\frac{1}{Z}\frac{\partial Z}{\partial\beta} = -\frac{\partial \ln Z}{\partial \beta}. U = − Z 1 ∂ β ∂ Z = − ∂ β ∂ l n Z . Why this step? We turned a weighted sum into a derivative of Z Z Z differentiate ln Z \ln Z ln Z
Start from the Gibbs/Shannon entropy
S = − k B ∑ i p i ln p i . S = -k_B \sum_i p_i \ln p_i. S = − k B ∑ i p i ln p i . Why this step? This is the statistical definition of entropy; it reduces to k B ln Ω k_B\ln\Omega k B ln Ω ln p i = − β E i − ln Z \ln p_i = -\beta E_i - \ln Z ln p i = − β E i − ln Z S = − k B ∑ i p i ( − β E i − ln Z ) = k B β ∑ i p i E i ⏟ U + k B ln Z ∑ i p i ⏟ = 1 . S = -k_B\sum_i p_i(-\beta E_i - \ln Z) = k_B\beta\underbrace{\sum_i p_i E_i}_{U} + k_B\ln Z\underbrace{\sum_i p_i}_{=1}. S = − k B ∑ i p i ( − β E i − ln Z ) = k B β U i ∑ p i E i + k B ln Z = 1 i ∑ p i . S = k B β U + k B ln Z = U T + k B ln Z . S = k_B\beta\, U + k_B\ln Z = \frac{U}{T} + k_B\ln Z. S = k B β U + k B ln Z = T U + k B ln Z .
F = U − T S = U − T ( U T + k B ln Z ) = U − U − k B T ln Z . F = U - TS = U - T\left(\frac{U}{T} + k_B\ln Z\right) = U - U - k_B T\ln Z. F = U − T S = U − T ( T U + k B ln Z ) = U − U − k B T ln Z . F = − k B T ln Z \boxed{F = -k_B T\ln Z} F = − k B T ln Z Why this is beautiful: U U U pure ln Z \ln Z ln Z F F F Z Z Z
Because d F = − S d T − p d V + μ d N dF = -S\,dT - p\,dV + \mu\,dN d F = − S d T − p d V + μ d N
Self-consistency check: using U = F + T S U = F+TS U = F + T S F = − k B T ln Z F=-k_BT\ln Z F = − k B T ln Z S S S U = − ∂ β ln Z U=-\partial_\beta\ln Z U = − ∂ β ln Z
Worked example (1) Single two-level systemEnergies 0 0 0 ε \varepsilon ε Step: Z = e 0 + e − β ε = 1 + e − β ε Z = e^{0} + e^{-\beta\varepsilon} = 1 + e^{-\beta\varepsilon} Z = e 0 + e − β ε = 1 + e − β ε Why? Just sum Boltzmann weights over the two states.
Step: F = − k B T ln ( 1 + e − β ε ) F = -k_BT\ln(1 + e^{-\beta\varepsilon}) F = − k B T ln ( 1 + e − β ε ) Step: U = − ∂ β ln Z = ε e − β ε 1 + e − β ε = ε e β ε + 1 U = -\partial_\beta\ln Z = \dfrac{\varepsilon e^{-\beta\varepsilon}}{1+e^{-\beta\varepsilon}} = \dfrac{\varepsilon}{e^{\beta\varepsilon}+1} U = − ∂ β ln Z = 1 + e − β ε ε e − β ε = e β ε + 1 ε Why? Differentiate ln Z \ln Z ln Z β \beta β Sanity: as T → ∞ T\to\infty T → ∞ β → 0 \beta\to0 β → 0 U → ε / 2 U\to\varepsilon/2 U → ε /2 T → 0 T\to0 T → 0 U → 0 U\to0 U → 0
N N N Why: independent ⇒ Z tot = Z 1 N Z_{\text{tot}} = Z_1^{N} Z tot = Z 1 N F = − k B T ln Z 1 N = − N k B T ln ( 1 + e − β ε ) . F = -k_BT\ln Z_1^{N} = -N k_BT\ln(1+e^{-\beta\varepsilon}). F = − k B T ln Z 1 N = − N k B T ln ( 1 + e − β ε ) . Key lesson: F F F extensive (∝ N \propto N ∝ N ln \ln ln Z Z Z
Worked example (3) Classical harmonic oscillator (1D)H = p 2 2 m + 1 2 m ω 2 x 2 H = \frac{p^2}{2m} + \frac12 m\omega^2 x^2 H = 2 m p 2 + 2 1 m ω 2 x 2 d x d p / h dx\,dp/h d x d p / h Z = 1 h ∫ e − β p 2 / 2 m d p ∫ e − β m ω 2 x 2 / 2 d x . Z = \frac{1}{h}\int e^{-\beta p^2/2m}\,dp\int e^{-\beta m\omega^2 x^2/2}\,dx. Z = h 1 ∫ e − β p 2 /2 m d p ∫ e − β m ω 2 x 2 /2 d x . Step: Gaussian integrals: ∫ e − a x 2 d x = π / a \int e^{-ax^2}dx=\sqrt{\pi/a} ∫ e − a x 2 d x = π / a Z = 1 h 2 π m β 2 π β m ω 2 = 2 π h β ω = k B T ℏ ω . Z = \frac{1}{h}\sqrt{\frac{2\pi m}{\beta}}\sqrt{\frac{2\pi}{\beta m\omega^2}} = \frac{2\pi}{h\beta\omega} = \frac{k_BT}{\hbar\omega}. Z = h 1 β 2 π m β m ω 2 2 π = h β ω 2 π = ℏ ω k B T . Why this step? Each Gaussian gives a β − 1 / 2 \beta^{-1/2} β − 1/2 β − 1 \beta^{-1} β − 1 Step: U = − ∂ β ln Z = − ∂ β ( − ln β + const ) = 1 β = k B T U = -\partial_\beta\ln Z = -\partial_\beta(-\ln\beta + \text{const}) = \frac{1}{\beta} = k_BT U = − ∂ β ln Z = − ∂ β ( − ln β + const ) = β 1 = k B T Equipartition! (1 2 k B T \frac12 k_BT 2 1 k B T F = − k B T ln ( k B T / ℏ ω ) F = -k_BT\ln\!\big(k_BT/\hbar\omega\big) F = − k B T ln ( k B T /ℏ ω )
F = − k B T Z F = -k_BT\, Z F = − k B T Z Why it feels right: F F F Z Z Z Fix: Without the log, F F F extensive (joining systems multiplies Z Z Z add F F F ln Z \ln Z ln Z F = − k B T ln Z F=-k_BT\ln Z F = − k B T ln Z
U = − ∂ β Z U=-\partial_\beta Z U = − ∂ β Z − ∂ β ln Z -\partial_\beta\ln Z − ∂ β ln Z Why it feels right: ∂ β Z = − ∑ E i e − β E i \partial_\beta Z = -\sum E_i e^{-\beta E_i} ∂ β Z = − ∑ E i e − β E i Fix: but U U U average energy, so you must divide by Z Z Z 1 Z ∂ β Z = ∂ β ln Z \frac{1}{Z}\partial_\beta Z = \partial_\beta\ln Z Z 1 ∂ β Z = ∂ β ln Z
S S S S = + ∂ F / ∂ T S=+\partial F/\partial T S = + ∂ F / ∂ T Why it feels right: differentiation looks symmetric. Fix: d F = − S d T − ⋯ dF=-S\,dT-\cdots d F = − S d T − ⋯ S = − ∂ F / ∂ T S=-\partial F/\partial T S = − ∂ F / ∂ T F F F decreases with T T T S > 0 S>0 S > 0
Common mistake "For independent particles,Z = Z 1 + Z 2 Z = Z_1 + Z_2 Z = Z 1 + Z 2 Why it feels right: energies add, so maybe Z Z Z Fix: Z = ∑ e − β ( E a + E b ) = ( ∑ e − β E a ) ( ∑ e − β E b ) = Z 1 Z 2 Z=\sum e^{-\beta(E_a+E_b)} = (\sum e^{-\beta E_a})(\sum e^{-\beta E_b}) = Z_1 Z_2 Z = ∑ e − β ( E a + E b ) = ( ∑ e − β E a ) ( ∑ e − β E b ) = Z 1 Z 2 energies in the exponent ⇒ multiplying exponentials.
Recall Feynman: explain to a 12-year-oldImagine a big bag of dice that each can land on a number, but cold dice "prefer" small numbers and hot dice are wild. The partition function Z Z Z F F F Z Z Z Z Z Z add their free energies instead of multiplying. From that one stretched number you can predict the bag's temperature behavior, its energy, and its messiness (entropy) — without ever counting every die one by one.
Mnemonic Remember the formula"Free Logs Are Negative, Times Temperature" → F = − k B T ln Z F = -k_BT\ln Z F = − k B T ln Z "U is the beta-slope of log-Z, downhill" → U = − ∂ β ln Z U=-\partial_\beta\ln Z U = − ∂ β ln Z
#flashcards/physics
Define the canonical partition function. ::: Z = ∑ i e − β E i Z=\sum_i e^{-\beta E_i} Z = ∑ i e − β E i β = 1 / k B T \beta=1/k_BT β = 1/ k B T
Helmholtz free energy in terms of Z Z Z F = − k B T ln Z F=-k_BT\ln Z F = − k B T ln Z
Why a logarithm in F F F Z Z Z Z Z Z ln \ln ln F F F
Internal energy from Z Z Z U = − ∂ ( ln Z ) / ∂ β = 1 Z ( − ∂ β Z ) U=-\partial(\ln Z)/\partial\beta = \frac{1}{Z}(-\partial_\beta Z) U = − ∂ ( ln Z ) / ∂ β = Z 1 ( − ∂ β Z )
Entropy from Z Z Z U U U S = U / T + k B ln Z = − ∂ F / ∂ T S=U/T + k_B\ln Z = -\partial F/\partial T S = U / T + k B ln Z = − ∂ F / ∂ T
Pressure from Z Z Z p = k B T ∂ ( ln Z ) / ∂ V = − ∂ F / ∂ V p=k_BT\,\partial(\ln Z)/\partial V = -\partial F/\partial V p = k B T ∂ ( ln Z ) / ∂ V = − ∂ F / ∂ V
In deriving F = U − T S F=U-TS F = U − T S U U U − k B T ln Z -k_BT\ln Z − k B T ln Z
For N N N F = ? F=? F = ? F = − N k B T ln ( 1 + e − β ε ) F=-Nk_BT\ln(1+e^{-\beta\varepsilon}) F = − N k B T ln ( 1 + e − β ε )
Classical 1D oscillator Z Z Z Z = k B T / ℏ ω Z=k_BT/\hbar\omega Z = k B T /ℏ ω U = k B T U=k_BT U = k B T
Common sign error in entropy formula? ::: S = − ∂ F / ∂ T S=-\partial F/\partial T S = − ∂ F / ∂ T d F = − S d T − p d V + μ d N dF=-SdT-pdV+\mu dN d F = − S d T − p d V + μ d N
U equals minus d lnZ / d beta
S equals minus kB sum p ln p
log gives extensive potential
Boltzmann probability p_i
Intuition Hinglish mein samjhoDekho, statistical mechanics ka asli "boss" quantity hai partition function Z = ∑ i e − β E i Z=\sum_i e^{-\beta E_i} Z = ∑ i e − β E i Z Z Z
Sabse seedha connection hota hai Helmholtz free energy F = U − T S F=U-TS F = U − T S F = − k B T ln Z F=-k_BT\ln Z F = − k B T ln Z Z Z Z Z = Z 1 Z 2 Z=Z_1Z_2 Z = Z 1 Z 2 add honi chahiye (extensive). Log multiplication ko addition me badal deta hai — isiliye F F F ln Z \ln Z ln Z Z Z Z
Derivation simple hai: U = − ∂ β ln Z U=-\partial_\beta\ln Z U = − ∂ β ln Z S = U / T + k B ln Z S=U/T + k_B\ln Z S = U / T + k B ln Z F = U − T S F=U-TS F = U − T S U U U − k B T ln Z -k_BT\ln Z − k B T ln Z S = − ∂ F / ∂ T S=-\partial F/\partial T S = − ∂ F / ∂ T p = − ∂ F / ∂ V p=-\partial F/\partial V p = − ∂ F / ∂ V
Do common galtiyaan: (1) log bhool jaana — phir F F F U = − ∂ β Z U=-\partial_\beta Z U = − ∂ β Z Z Z Z 1 Z ∂ β Z = ∂ β ln Z \frac{1}{Z}\partial_\beta Z=\partial_\beta\ln Z Z 1 ∂ β Z = ∂ β ln Z
