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Inverse temperature is $\beta =1/(k_{B}T)$, where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature. A canonical ensemble weights an energy $E$ by its Boltzmann factor $e^{-\beta E}$. Calling couplings dimensionless means absorbing this factor $\beta$ into their energy coefficients. The alternative convention $k_B=1$ makes inverse temperature $1/T$; physical and dimensionless couplings must not be mixed.

In a thermal Euclidean worldline path integral with $\hslash =1$, $\beta$ is also the circumference of the imaginary-time circle.

The Mayer f-function packages the departure of a pair's Boltzmann factor from one and is the basic bond in the dilute-gas cluster expansion.

For the Gaussian critical exponents, use dimensionless statistical-action conventions, with the Boltzmann factor $e^{-H}$. If $H$ is an energy, apply the following to $H/(k_{B}T)$ and absorb that smooth factor into the couplings. Choose the Fourier transform conventionReality gives $\tilde{\varphi }(-p)=\tilde{\varphi }(p{)}^{\ast }$. The Fourier-space statistical Hamiltonian isThe source term is not multiplied by one half. Completing the Gaussian functional integral gives $⟨\tilde{\varphi }(p)⟩=\tilde{J}(p)/({\kappa }^{-1}{p}^2+m^2)$ and the connected Fourier-space two-point correlation functionAssume $m^2>0$ for the stable massive Gaussian field theory. In the continuum long-distance theory, the real-space correlation function is the Green kernel of $-{\kappa }^{-1}{\nabla }^2+m^2$. Away from its source, a radial ansatz $G(R)\sim R^{-(D-1)/2}{e}^{-R/\xi }$ gives, at large $R$, $m^2-{\kappa }^{-1}{\xi }^{-2}=0$. Equivalently the propagator pole is at $p^2=-\kappa m^2$. Thus the exponential correlation length isAt $m=0$ the correlation length diverges. A sharp momentum regulator itself introduces artificial oscillatory long-distance tails; this length describes the physical continuum pole or a local/smoothly regulated model on distances large compared with $a$.

For a momentum-shell renormalization group step with $b>1$, split the Fourier modes into retained modes $|p|<\Lambda /b$ and eliminated modes $\Lambda /b<|p|<\Lambda$. Define ${\tilde{\varphi }}_B(p)$ as the retained field. Integrating the eliminated modes is exact for a Gaussian field theory; a uniform source acts only on the zero mode, so this integration produces a field-independent free-energy term. Rescale $q=bp$ and set ${\tilde{\varphi }}_B(p)=\tilde{Z}{\tilde{\varphi }}^{\prime }(q)$. The retained quadratic action becomesKeeping the gradient coefficient fixed requires $b^{-D-2}{\tilde{Z}}^2=1$. For $J=h$, the source term is $-h{\tilde{\varphi }}_B(0)$, so the complete Gaussian rescaling isThe corresponding real-space field obeys ${\varphi }_B(x)=b^{(2-D)/2}{\varphi }^{\prime }(x/b)$; confusing that real-space factor with $\tilde{Z}$ would change the source exponent incorrectly.

Let $F$ be free energy per original volume, with $t$ proportional to $m^2$. Rescaled volume is $V^{\prime }=V/b^D$, so the Gaussian free-energy scaling relation isA field-normalisation Jacobian can be included in the field-independent shell term. For any finite step on positive-mass modes, that background is analytic near $t=0$. Subtract regular backgrounds and choose $b=t^{-1/2}$ on the massive side. The usual homogeneous singular scaling then givesThese are Gaussian critical exponents for the stipulated Gaussian approximation. A useful check is its source-dependent contribution $-h^2/(2m^2)$: $D/2-2\Delta =-1$ gives precisely that power of $t$.

There is a qualification to a strictly pure-power singular free energy. The Gaussian free-energy logarithm at effective dimension two in $D=2$ contains $t\log t$ after subtracting analytic terms. Thus the displayed exponent assignments remain the formal power indices, but at $D=2$ the advertised homogeneous expression needs this additive logarithmic term. A purely quadratic theory also cannot stabilise the ordered $t<0$ phase or the zero-mass zero mode at nonzero $h$; the scaling calculation is taken from $t>0$, and an ordered-phase continuation requires stabilising interactions.

At a uniaxial Lifshitz point, the quadratic kernel instead isThe vanishing coefficient of $p_{\parallel }^2$ is an additional tuning that distinguishes this Lifshitz point from an ordinary critical point. Use a factorised cutoff, retain $|p_{\perp }|<{\Lambda }_{\perp }/b$ and $|p_{\parallel }|<{\Lambda }_{\parallel }/c$, and integrate its complement. A rectangular or cylindrical retained region is suitable; its precise boundary is not an exponent. This is a uniaxial Lifshitz Gaussian renormalization step.

Set $q_{\perp }=b{p}_{\perp }$, $q_{\parallel }=c{p}_{\parallel }$, and ${\tilde{\varphi }}_B=\tilde{Z}{\tilde{\varphi }}^{\prime }$. The momentum measure acquires $b^{-(D-1)}{c}^{-1}$. Keeping both kinetic coefficients fixed requiresTherefore the anisotropic blocking and source factors areThe anisotropic effective dimension is $d_{\mathrm{eff}}=D-1+1/2=D-1/2$: one parallel coordinate contributes half the scaling weight of a perpendicular coordinate. The rescaled-volume factor is $b^{-d_{\mathrm{eff}}}$, soChoosing $b=t^{-1/2}$ gives the requested Lifshitz Gaussian exponents:Again $(D-1/2)/2-2\Delta =-1$, checking the uniform-source susceptibility. The associated length powers are ${\xi }_{\perp }\sim t^{-1/2}$ and ${\xi }_{\parallel }\sim t^{-1/4}$. A Gaussian determinant has the analogous logarithmic exception if its effective dimension equals 2. Interactions can alter these exponents; dimensional rescaling here solves the specified Gaussian model.

Take the couplings in the Hamiltonian to include inverse temperature, so the Boltzmann factor is $e^{-H}$. If physical energies are used instead, first replace $J,L,D,K$ by their products with $1/(k_{B}T)$. Split the on-site term equally between its two bonds. For spin states $s,t\in \{1,0,-1\}$ the spin-chain transfer matrix has entriesUsing the specified coupling coordinates, in the order $(1,0,-1)$ this isFor example, $e^{D/2}=x/z$ and $e^{-J+L+D}=y/z$. This is the nearest-neighbour Blume–Emery–Griffiths model with an additive constant and the paper's sign convention for $D$.

Summing the periodic spin chain gives the partition function $Z_N=\mathrm{tr}{W}^N={\sum }_{a=1}^3{\lambda }_a^N$. For finite real couplings all entries of $W$ are strictly positive; the Perron–Frobenius theorem gives a unique positive dominant eigenvalue with ${\lambda }_1>|{\lambda }_a|$ for $a\ne 1$. Since $W$ is a symmetric matrix, all its eigenvalues are real. Thus in the thermodynamic limit,The positivity condition is stronger and more useful than mere ordering by signed value: subdominant eigenvalues can be negative. Also, the printed strict ordering between the other two is not guaranteed for all couplings. For instance, $J=L=D=0$ makes $W$ the positive constant matrix $e^{K}11^T$ with two equal zero eigenvalues. Degeneracy there does not affect the largest-eigenvalue limit; no explicit generic eigenvalues are needed.

For spin magnetization, introduce a dimensionless field $h$ through $-h{\sum }_i{\sigma }_i$, or insert $S=\mathrm{diag}(1,0,-1)$ into the trace. Thenwhere $v_1$ is a normalized eigenvector of the largest eigenvalue and $W_{st}(h)=e^{h(s+t)/2}{W}_{st}(0)$. Spin inversion symmetry gives $v_{1,+}=v_{1,-}$, since the positive eigenvector of the largest eigenvalue is unique. Thereforeat zero field, both at finite $N$ and in the finite-coupling thermodynamic limit. The finite-$N$ result follows directly by pairing each configuration with its spin-reversed partner. Eigenvalues at a single fixed field do not determine a general observable: a field derivative of the largest eigenvalue, or its eigenvector, is required. Positivity and the real analytic dependence on the couplings exclude a finite-temperature spontaneous symmetry breaking transition in this one-dimensional finite-range chain; singular zero-temperature coupling limits require separate treatment.

For even $N$, spin decimation on alternate sites sums the middle spin of each two-bond segment, hence the coarse spin-chain transfer matrix is $W^{\prime }=W^2$. DefineDirect multiplication gives $W^2=e^{2K}{z}^{-2}\left(\begin{array}{ccc}A& B& C\\ B& E& B\\ C& B& A\end{array}\right)$. Matching its entry ratios to the original parameterization yields the spin-1 chain decimation recursionThe remaining overall positive factor is absorbed into $K^{\prime }$. Keeping that factor preserves the free energy as well as normalized spin probabilities. Indeed $\mathrm{tr}(W^2{)}^{N/2}=\mathrm{tr}{W}^N$, exactly; on an odd ring an unmatched boundary segment needs separate handling rather than assuming a uniform two-site block decomposition.

In chemical equilibrium, the reaction $p+e\leftrightarrow H+\gamma$ obeys chemical-potential balance for ionizationbecause a photon gas in equilibrium has photon chemical potential ${\mu }_{\gamma }=0$. Applying the given nonrelativistic Maxwell-Boltzmann distribution to each massive species givesThe Hydrogen binding energy is $I=m_e+m_p-m_H$ in units with $c=1$. Thus Saha's equation isFor ground-state hydrogen, $g_e=g_p=2$ and $g_H=4$, so the degeneracy factor is one. Since $m_p\simeq m_H$, the translational factor is commonly written $(2\pi m_e{k}_{B}T/h^2{)}^{3/2}$.

Charge neutrality gives $n_e=n_p$. Write $n_B=n_e+n_H\simeq \eta n_{\gamma }$. Thenand henceCombining this with the stated photon density produces

In our universe the baryon-to-photon ratio is tiny, so there are roughly $10^9$ photons per baryon. Even when the mean photon energy is far below $I$, the high-energy blackbody radiation tail contains enough ionizing photons to suppress neutral hydrogen. The recombination temperature is reached only after the Boltzmann factor overwhelms this large photon-to-baryon ratio, at $k_{B}T\simeq 0.3\mathrm{eV}$.

For a universe with $\eta =1$, take the degeneracy factor as one, $m_H\simeq m_p$, and define $x=I/(k_{B}T)$. Neutral formation occurs when the boxed ratio is of order one, giving approximatelyWith $I=13.6\mathrm{eV}$ and $m_e=5.11\times 10^5\mathrm{eV}$, this gives $x\simeq 19$ and thereforeto order unity. This is much hotter than recombination in our universe because there is no enormous excess of photons, but it remains below $I$ because the electron's large translational phase space still favours the ionized state.

A subcritical droplet must make a rare thermal fluctuation from the metastable basin over the free-energy barrier $F^{\ast }$. Its probability carries the Boltzmann factor $e^{-F^{\ast }/(k_{B}T)}$, so the nucleation rate has the Arrhenius formFor a volume containing order one candidate subcritical droplet, the waiting time for a supercritical droplet and subsequent macroscopic growth is thereforeup to an algebraic kinetic prefactor. This is the Arrhenius nucleation time.

If a physical tension $f$ acts on a state of extension $l_i$, its mechanical energy is $-f{l}_i$. The canonical ensemble therefore assigns that state the Boltzmann factorComparison with the statistical weight $e^{\tau l_i}$ givesEquivalently, $f$ is the thermodynamic conjugate variable to $L$ in the Gibbs free energy, just as pressure is conjugate to volume.

A spin model assigns a discrete or continuous spin variable to each site of a lattice or vertex of a graph and gives interacting configurations a Boltzmann factor. Spin models describe collective order and phase transitions in magnetic and other many-body systems.

For nearest-neighbour interactions, a two-by-two transfer matrix assigns one Boltzmann factor to every adjacent spin pair. With periodic boundary conditions, summing all spin configurations gives the partition function as the trace of a power of this matrix.