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Line tension is energy per unit length of a boundary, such as the edge between two phases of a lipid bilayer. Positive line tension tends to minimize the perimeter at fixed domain area. Its dimensions and geometry differ from surface tension, which is energy per unit area.

Lipids are a chemically diverse family of chemical substances, conventionally associated with biological origin and solubility in nonpolar solvents. They include triglycerides, phospholipids, and cholesterol. The name does not specify one molecular formula, one repeating monomer, or one kind of polymer. Some lipids have both hydrophilic and hydrophobic regions rather than being uniformly nonpolar. See IUPAC Gold Book: lipids.

Structure explains the different jobs. A triglyceride has three fatty acid-derived acyl groups attached to glycerol and serves as an energy store. A typical glycerophospholipid instead combines two such chains with a phosphate-containing head, making it an amphiphile suited to a lipid bilayer. Cholesterol has a different structure and also contributes to cell membrane properties. Thus a lipid is a chemical building material, while a lipid bilayer is an organization of many molecules. See Clark, Douglas and Choi, Biology 2e, 3.3: lipid structures and functions.

Lipid metabolism includes synthesis, modification, transport, storage and breakdown of lipids. Stored triacylglycerols can release fatty acids through lipolysis, and subsequent oxidation can supply acetyl-CoA and reducing equivalents. Other lipids are assembled into membranes or signalling molecules. These distinct destinations explain why more lipid synthesis does not necessarily imply more energy storage. ^[ref]

Lipolysis hydrolyses stored triacylglycerols to release fatty acids and glycerol through successive cleavage steps. It mobilizes stored carbon but is distinct from fatty-acid oxidation, which subsequently extracts energy. Released fatty acids can be transported, oxidized or re-esterified depending on tissue state. Thus an increased rate of lipolysis need not equal the rate of net loss of body lipid stores. ^[ref]

Consider an original model of a cold section enclosing a throttle and collecting region, with the precooling heat exchanger outside its boundary. The inlet has mass flow rate $\dot{m}>0$ and specific enthalpy $h_{\mathrm{in}}$. The outlets are equilibrium liquid and vapor at one fixed pressure, with specific enthalpies $h_l$ and $h_g$. Let $y$ be the outlet liquid mass fraction, and let $q=\dot{Q}/\dot{m}\ge 0$ be incoming unwanted heat per unit incoming mass. Assume steady flow, no shaft work, and negligible bulk kinetic energy and gravitational potential energy changes. The steady-flow energy balance givesHere $L$ is the latent heat per unit mass at the chosen outlet pressure. The mass flow rates of the outlets are $y\dot{m}$ and $(1-y)\dot{m}$. The formula describes a flowing cold section; a filling vessel with appreciable energy or mass accumulation requires storage terms.

For a deliberately dimensionless example, choosePer unit incoming mass, one quarter leaves as liquid and three quarters as vapor. With zero incoming heat, the same inlet would give $y=0.30$. Increasing $q/L$ to $0.20$ reduces $y$ to $0.10$, and $q/L=0.30$ reaches the no-liquid boundary. These are chosen model values, not helium property data or a historical yield. In normalized enthalpy units with $h_l=0$ and $h_g=1$, conservation checks directly: $0.70+0.05=0.25\times 0+0.75\times 1$.

The admissible interval is $h_l\le h_{\mathrm{in}}+q\le h_g$, or $0\le y\le 1$. Outside it, the assumed equilibrium two-phase outlet fails: an enthalpy above $h_g$ requires a hotter single-vapor outlet under these assumptions, rather than a negative amount of liquid. Below $h_l$, a colder liquid or a different thermodynamic phase must be considered. The balance does not predict its own outlet pressure, temperature or helium property functions.

At fixed outlet pressure, hence fixed $L$, the derivatives are $\partial y/\partial h_{\mathrm{in}}=-1/L$ and $\partial y/\partial q=-1/L$. Lowering inlet specific enthalpy by regeneration and reducing incoming heat therefore buy the same increase in liquid mass fraction per unit energy per unit mass in this model. Adding any constant to all three specific enthalpies leaves $y$ unchanged. This independence of the arbitrary enthalpy reference makes the balance useful even when property tables choose different zero points.

A liquid crystal is a phase with orientational or partial positional order while retaining some fluid-like motion. Its structure lies between a fully disordered isotropic fluid and a crystal with positional order in all spatial directions.

In a nematic liquid crystal, molecular axes prefer a common unoriented nematic director without forming a three-dimensional lattice of fixed positions. The nematic order parameter measures that alignment and can vary across the sample. Orientational distortions cost elastic energy and can form topological defects. Applied fields and boundary anchoring can change the director pattern, thereby changing optical propagation. The phase should be specified, since different liquid-crystal phases break different symmetries and need different order parameters.

A liquid that produces particle scintillation when deposited energy excites its components. Photomultipliers can convert the emitted light into detector pulses.

The nuclear reactionconserves electric charge and total nucleon count: $3+1=2+2$ and $7+1=4+4$. Its nuclear reaction Q-value is the initial minus final rest energy. It is additional to the incoming beam kinetic energy; the accelerator triggers a rearrangement whose products can carry much more kinetic energy than the incident proton.

Use one consistent set of neutral-atom isotope masses, in unified atomic mass units, from the National Institute of Standards and Technology:These are isotope masses, not element-averaged atomic weights.^[ref] With $\mathrm{u}{c}^2=931.49410372\mathrm{MeV}$,^[ref]The four electron rest masses cancel between sides. Differences of total electronic binding energies do not cancel exactly: this atomic-mass result approximates the bare-nucleus nuclear reaction Q-value at the displayed $0.001\mathrm{MeV}$ scale, not at arbitrary precision. No mixture of rounded nuclear binding energies is needed. At chosen incident kinetic energy $T=0.500\mathrm{MeV}$ on a stationary target, the two outgoing alpha particles together carry approximately $Q+T=17.846\mathrm{MeV}$ before losses in the target and detector window.

Equal product masses imply equal kinetic energies and opposite momenta in the center of mass frame. The laboratory frame has nonzero initial momentum because the proton is moving. Conservation of momentum therefore forbids the two laboratory momenta from always being equal and opposite. The products are equal in laboratory kinetic energy only for particular emission directions, not for every event.

An exact two-body calculation makes this distinction quantitative. Write $a=m_p{c}^2$, $b=m_{\mathrm{Li}}{c}^2$, $h=m_{\alpha }{c}^2$, and $T$ for incident kinetic energy. Express every quantity below in the same energy units; $\mathcal{P}=pc$ is momentum times the speed of light. For a stationary target,Here $W$ is the center-of-mass energy, and $\beta$ is the center-of-mass speed divided by the speed of light. Each outgoing alpha particle has center-of-mass total energy $E_{\ast }=W/2$ and momentum-times-$c$ magnitude ${\mathcal{P}}_{\ast }=\sqrt{E_{\ast }^2-h^2}$. A Lorentz transformation gives the paired laboratory kinetic energies, with ${\theta }_{\ast }$ measured from the incident beam in the center of mass frame:Their sum is independent of ${\theta }_{\ast }$ and equals $Q+T$. Their difference is $2\gamma \beta {\mathcal{P}}_{\ast }\cos {\theta }_{\ast }$; the symmetry belongs to the product interchange and the center of mass frame.

For the chosen $T=0.500\mathrm{MeV}$ model, subtract $Z$ electron rest masses from each neutral-atom mass, neglect electronic binding energies, and use $m_e{c}^2=0.51099895069\mathrm{MeV}$.^[ref] At ${\theta }_{\ast }=0$, the paired laboratory kinetic energies are approximately $9.979$ and $7.867\mathrm{MeV}$; at ${\theta }_{\ast }=90^{\circ }$ both are approximately $8.923\mathrm{MeV}$. In the latter case both products have a forward laboratory momentum component, so even equal laboratory kinetic energies do not imply back-to-back laboratory trajectories. A real thick target and finite detector acceptance would broaden or select these ideal kinematic possibilities.

Self-gravitating energy is nonlocal. Shell assembly assigns $-GM(<r)dM/r$ and counts pairs once; the density $\rho \Phi /2$ is another total-energy allocation. Removing a test parcel from a fixed potential instead requires $-\Phi dM$. These prescriptions need not give the same radial feedback threshold, even though their units agree.

A density-functional approximation using the energy per particle of a uniform electron gas at the local density. Its exchange-energy-per-electron tail decays exponentially with an exponentially decaying density.

The local density of states resolves the density of states by position as well as energy. With a consistent counting of spin states, its integral over the specimen's volume gives the total density of states, $g(E)=\int \rho (\mathbf{r},E)d^3\mathbf{r}$. It describes the spatial distribution of available states, rather than their occupation.

In occupied bulk regions where density-gradient energy is negligible, the stationary relation $g\rho =\mu -V$ gives speed of sound $c=\sqrt{g\rho /m}$. Empty regions do not have an imaginary physical speed of sound; the bulk condensate formula is simply inapplicable there.

For disturbances constant along the horizontal magnetic field, the minimized energy density is $\rho g[-(\ln \rho {)}^{\prime }-\rho g/(\gamma p+C)]|{\xi }_z{|}^2$. If its coefficient is nonnegative everywhere, the full energy is nonnegative. If it is strictly negative at a point of a smooth equilibrium, a compactly supported vertical displacement in a neighbouring interval gives negative energy. These interchange motions compress gas and magnetic pressure together.

The electric field does work on matter at local power per volumeHere $\mathbf{J}$ is current density and the dot product includes the relative direction. Positive $\mathbf{J}\cdot \mathbf{E}$ means energy passes from the field to matter; negative means matter supplies energy to the field. In general this is a work rate, not automatically an irreversible heating rate.^[ref]

For a stationary isotropic Ohmic conductor with real scalar electrical conductivity $\sigma >0$, local Ohm's law gives $\mathbf{J}=\sigma \mathbf{E}$. Thus the local Joule heating rate isThe electrical resistivity ${\rho }_{\mathrm{el}}$ has units $\Omega \mathrm{m}$, and electrical conductivity has units $(\Omega \mathrm{m}{)}^{-1}$. These formulas use the material's rest frame and dissipative conduction current density. Moving media, complex conductivity, and additional storage or transport processes require their appropriate energy balance rather than this scalar substitution.

For uniform current in a straight homogeneous wire of length $L$, wire cross-sectional area $A$ and constant electrical conductivity, $E=V/L$, $J=I/A$ and $R=L/(\sigma A)$. The volume integral recovers terminal power:As a numerical check, the wire already used in electrical resistance has $L=1\mathrm{m}$, $A=10^{-6}{\mathrm{m}}^2$, $\sigma =5\times 10^5(\Omega \mathrm{m}{)}^{-1}$ and $V=5\mathrm{V}$. It gives $J=2.5\times 10^{6}A/{\mathrm{m}}^2$, $p_{\mathrm{vol}}=1.25\times 10^{7}W/{\mathrm{m}}^3$ and $P=12.5\mathrm{W}$, agreeing with $I=2.5\mathrm{A}$ and $R=2\Omega$. Concentrating the same electric current in a smaller wire cross-sectional area therefore raises the local heating rate if electrical resistivity stays fixed.

Under CPT, a Hermitian local tensor monomial picks up a minus sign for each Lorentz vector index, including derivative indices. A Lorentz scalar contracts all such indices, so the total sign is positive; charged terms are exchanged with their Hermitian conjugates. Therefore a local Hermitian Lagrangian density obeys $\Theta \mathcal{L}(x){\Theta }^{-1}=\mathcal{L}(-x)$, leaving its action invariant. This field-level argument illustrates the CPT theorem for ordinary scalar, spinor and vector theories; the axiomatic theorem also uses positive energy and microcausality.

For a constant-mass Schrödinger equation with a finite step in potential energy, the reduced radial wavefunction $u(r)$ and its first derivative are continuous at the interface $r=R$. When $u(R)\ne 0$, the matching can be written as equality of logarithmic derivatives:The ratio removes the arbitrary amplitude of each regional solution. For quantum scattering, the exterior solution contains the scattering phase shift, so this equation determines that phase after matching to a regular interior solution.

At a wavefunction node, division by $u(R)$ is undefined. A form that also handles such nodal matches isThis is a vanishing Wronskian of the two interface value/derivative vectors. For nontrivial regional solutions it says those vectors are proportional, allowing amplitudes to be chosen so that both continuity conditions hold.

For a bound state, the exterior solution is chosen to decay, and the same matching quantizes the allowed energy. Singular delta-function interactions instead impose a derivative jump and require that modified matching condition.

Fill the lower energy band of the dimerized tight-binding chain with both spin states. Take zero on-site energy and measure ground-state energy per doubled primitive unit cell. State counting in a Bloch band and $q=ka$ giveEquivalently, for $|\delta |\le t$, $\mathcal{E}(\delta )=-(8t/\pi )E(\sqrt{1-{\delta }^2/t^2})$, where $E(k)$ is the complete elliptic integral of the second kind in the modulus convention.

For fixed $t>0$ and $|\delta |/t\to 0$, the change isThe logarithm comes from states near the edge of the Brillouin zone. Put $u=\pi /2-q$. To leading order the square root there is $\sqrt{t^2{u}^2+{\delta }^2}$. For a fixed small cutoff $\eta >0$, direct integration givesThe rest of the band, and the higher-order corrections near its edge, contribute only $O({\delta }^2/t)$; multiplying by $-8/\pi$ gives the displayed asymptotic expansion. Here the remainder uses Big O notation. The negative logarithmic term exceeds any fixed quadratic elastic cost $K{\delta }^2$, $K>0$, for sufficiently small nonzero $\delta$. This explains the zero-temperature instability to a Peierls distortion in the ideal chain model.

With constant amplitude outside a core, the planar nematic divergence-elasticity identity reduces the elastic density to $K{\lambda }_0^2|\nabla \chi {|}^2/2$. A topological charge of a two-dimensional nematic disclination $q$ has $\chi =q\theta +{\chi }_0$ and $|\nabla \chi {|}^2=q^2/r^2$. Integrating over an annulus gives the displayed energy, plus finite core and boundary terms. The outer cutoff is the sample size, compensating-defect separation, or another screening length. An isolated defect in an infinite ordered plane has divergent energy. A neutral pair $q,-q$ separated by $R$ instead has $E_{\mathrm{pair}}=2\pi K{\lambda }_0^2{q}^2\log (R/r_0)$ plus core terms, since its far-field gradients cancel.

A single two-dimensional quadratic saddle with $A={\hslash }^2/(2m^{\ast })$ contributes this leading joint density of states when both contour branches are included and $g_s$ counts independent spin channels. The contour integral is proportional to ${\int }_{-\Lambda }^{\Lambda }dy/\sqrt{y^2+\delta /A}=2\mathrm{arsinh}(\Lambda \sqrt{A/\delta })$ above the saddle. Its asymptote gives the logarithm with $E_{\Lambda }$ of order $A{\Lambda }^2$. The cutoff-dependent constant is not universal. This is a density per area per energy; changing to a density per frequency requires a Jacobian. ^[ref]