Methylation of lysine 9 of histone H3. Different methylation states can recruit specific reader proteins and contribute to heterochromatin regulation.

A bound-state configuration with one electron predominantly in $1s$ and the other in a different orbital $n\ell$. The two assignments of identical electrons to these orbitals must be combined with the appropriate exchange symmetry of two identical spin-one-half fermions. Symmetric spatial combinations accompany the singlet state; antisymmetric combinations accompany the triplet state.

A helium atom electronic state with total spin zero. Its antisymmetric two-spin function requires a symmetric spatial wavefunction. Its approximate spectroscopic family is distinct from the triplet family under spin-conserving electric dipole transitions.

A helium atom electronic state with total spin one. Its symmetric spin function requires an antisymmetric spatial wavefunction, excluding two electrons in the same spatial orbital. The three spin projections are magnetic substates of the same spin multiplet, not three separately distinguishable particles.

For fixed orthonormal orbitals $a,b$, symmetrizing or antisymmetrizing $a(1)b(2)$ gives electron-repulsion expectations $J_{ab}+K_{ab}$ and $J_{ab}-K_{ab}$. Antisymmetry of the complete electronic state therefore places the singlet above the triplet by $2K_{ab}$, with $K_{ab}$ the positive Coulomb exchange integral. Orbital relaxation and further correlation refine this comparison. Exchange splitting here is a consequence of Fermi statistics and the electrostatic interaction, not a separate classical spin-spin force.

The low $1s2s$ levels of a helium atom lack an allowed direct electric dipole decay to the ground $1s^2{}^1{S}_0$ level. The triplet is excluded by the spin selection rule, and the singlet-to-singlet $S$ transition is excluded by parity and orbital selection. Weaker radiative channels and collisions can still de-excite these states; metastable does not mean infinitely lived.

A fluid surface has no equilibrium static in-plane shear modulus, because material can rearrange after shear. This does not set its membrane tension or membrane bending modulus to zero. Time-dependent deformation can also produce viscous in-plane shear stress; absence of a static elastic coefficient is not absence of dissipation.

In a bound-cap model, an adhered sphere of radius $r$ covers area $A=2\pi r^2\xi$ with $\xi =1-\cos \alpha \in [0,2]$. Its excess area over its circular footprint is $\pi r^2{\xi }^2$. The Helfrich energy of that spherical cap is $4\pi \kappa \xi$, adhesion gives $-wA$, and membrane tension gives $\sigma \pi r^2{\xi }^2$. Setting $\bar{w}=w{r}^2/\kappa$ and $\bar{\sigma }=\sigma r^2/\kappa$ yields the displayed expression. Free-surface deformation, neck structure, Gaussian curvature and topology changes are omitted. Full wrapping alone does not prove scission.

The 1975 measurements of CeAl3 below $0.2\mathrm{K}$ found an electronic heat capacity coefficient of $1.62\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-2}$ and a quadratic electrical-resistivity coefficient of $35\mu \mathrm{\Omega }\mathrm{cm}{\mathrm{K}}^{-2}$. These are complementary thermal and transport signatures of an unusually dense low-energy electronic spectrum.^[ref] A temperature sweep and a $C/T$ versus $T^2$ fit separate the electronic intercept from a Debye lattice term; the transport comparison tests a Fermi liquid temperature law.

For quasiparticles with density of states $g(E_F)$ including both spins, the leading electronic heat capacity is $C=\gamma T$ with $\gamma ={\pi }^2{k}_B^{2}g(E_F)/3$. A narrow heavy band makes this coefficient large. The density of states and heat capacity must use consistent extensive, molar or volumetric normalization.

For a sinusoidal ground-temperature model whose coldest surface day is $t=0$, source temperature on that day is $T_0-T_A{e}^{-x}\cos x$, where $x=z/\delta$. The first and largest thermal maximum occurs at $x=3\pi /4$, not at infinite depth. This maximizes the winter coefficient of performance for a reversible pump with a fixed indoor temperature. With a heating load even about that day, the same optimum minimizes annually integrated reversible heating work. Excavation cost, finite heat extraction and real heat-pump behaviour can change the optimum.

An ideal cyclic heat engine producing no entropy overall. Between two equilibrium reservoirs its efficiency reaches the Carnot bound; finite reservoir temperatures can be treated as varying through successive reversible cycles.

The heat input to a cycle sums positive transfers of heat into its working substance. It is not the net heat over the entire cycle, which equals net work. A single geometric branch can contain both heat absorption and rejection, so grouping the branch into one net transfer can overestimate a heat engine's thermal efficiency.

Consider an ideal-gas cycle with constant heat-capacity ratio $\gamma >1$: expand isobarically from $V_1$ to $r{V}_1$, cool isochorically, then compress reversibly and adiabatically to the initial state. With $P_{\ast }=p_1{V}_1=nR{T}_1$, closure gives $T_2=r{T}_1$ and $T_3=r^{1-\gamma }{T}_1$. The heat input is $Q_h=\gamma P_{\ast }(r-1)/(\gamma -1)$ and the rejected heat magnitude is $Q_c=P_{\ast }(r-r^{1-\gamma })/(\gamma -1)$. Thus $\eta =1-(r-r^{1-\gamma })/[\gamma (r-1)]$. Reservoir temperatures determine feasibility and entropy generation but need not coincide with the gas temperatures at heat-exchange endpoints.

A heat generation rate specifies the power deposited into microscopic thermal degrees of freedom, usually through conversion of electrical, chemical or mechanical energy. Its unit is the watt. The term is used as a source in a thermal energy balance; it does not imply creation of total energy.

A resistor supplied with current $I$ has electrical power $I^{2}R$ converted by Joule heating. If that power is deposited approximately uniformly in a body of volume $V$, the volumetric heat generation rate is $q^{\prime \prime \prime }=I^{2}R/V$. A steady body can export the same total power through heat transfer and retain constant temperature. If it is approximately isolated for a short interval, the deposited energy instead raises its internal energy, giving $C\dot{T}=P$ when its heat capacity is approximately constant.