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An isolated Newtonian system in virial equilibrium has total energy $E=K+W=-K$. Along an equilibrium sequence at fixed particle number, define the kinetic temperature by $K=3N{k}_B\Theta /2$. Its heat capacity is therefore negative, as displayed: losing total energy increases the kinetic temperature. A collisionless stellar system need not have a thermodynamic temperature; this is a kinetic-temperature statement unless thermal equilibrium is additionally specified. Boundary pressure and external confinement can change the energy relation.

A negative temperature describes an internally equilibrated distribution that favors higher energy levels. For Boltzmann entropy $S=k_B\ln \Omega$, with $\Omega$ counting states in an energy window, defineA decreasing branch of $S(E)$ therefore has $T<0$. In a finite collection of independent two-level systems, the number of configurations is largest near equal populations and decreases as the upper level becomes more fully occupied. For a level separation $\Delta >0$, the Boltzmann factors give $p_{\mathrm{upper}}/p_{\mathrm{lower}}=e^{-\beta \Delta }>1$ when $\beta <0$.

A bounded upper energy makes these inverted canonical weights normalizable in the usual finite-spin models. An ordinary free-particle kinetic energy is unbounded above: negative $\beta$ would make its momentum integral diverge. The spin subsystem must also equilibrate internally before uncontrolled exchange with such other degrees of freedom destroys the inversion. Heat flow from negative to positive temperature shows why this state is hotter than every positive-temperature reservoir.

Negative-temperature motional states have also been prepared for ultracold bosons in a tailored Bose-Hubbard model. The original experiment reports a quasi-momentum distribution peaked at the upper band edge. ^[ref]

Nematic bend curves the integral lines of the nematic director. For a unit field, $(\mathbf{n}\cdot \nabla )\mathbf{n}=-\mathbf{n}\times (\nabla \times \mathbf{n})$, so the bend energy measures curvature along those lines.

Here $b^2=b_0^2+a^2-a_0^2$. Substitute the neo-Hookean solid energy into the twist and inflation of an incompressible tube work identity. Differentiating with respect to $a$ gives the integrand $\mu \rho [{\rho }^{-2}-{\rho }^2/({\rho }^2+c{)}^2+{\alpha }^2]$, with $c=a^2-a_0^2$. Integrating between the two reference radii proves the displayed pressure when the outer surface is traction free.

High-temperature photodisintegration of neon-20 releases an alpha particle and oxygen-16. Capture of that alpha particle by another neon-20 nucleus produces magnesium-24. The breakup consumes energy, but the combined reactions can release net energy. Other captures modify the ashes in a real network.

For a small, uniform surface at temperature $T$, enclosed by much larger black walls all at temperature $T_s$, the outward net heat transfer by thermal radiation isAssume an opaque grey body that is a diffuse thermal emitter, with the same emissivity $ϵ$ and radiometric absorptance at the surface's current temperature. Its emitting area $A$ sees only the enclosure: the view factor to the walls is one, with no self-viewing surface, opening or intervening obstruction. The intervening medium must neither absorb nor emit appreciable thermal radiation; a vacuum supplies a useful idealization. The enclosure is maintained at uniform $T_s$. The small-body assumption makes that reservoir description physically reasonable; genuinely black isothermal walls supply the incoming irradiance even without an infinite area ratio. Positive ${\dot{Q}}_{\mathrm{rad}}$ means net loss by the surface; negative means net gain. This expression counts thermal radiation alone, not all heat transfer.^[ref]

The reasoning is a two-way energy balance. Black walls supply irradiance $E=\sigma T_s^4$, so the surface absorbs $P_{\mathrm{abs}}=ϵAE$ and emits $P_{\mathrm{emit}}=ϵA\sigma T^4$. Subtract these to obtain ${\dot{Q}}_{\mathrm{rad}}$. For an opaque surface the remaining fraction $1-ϵ$ of incoming radiant flux is reflected; counting emitted plus reflected outgoing radiant flux minus total incoming radiant flux gives the same result. Kirchhoff's law of thermal radiation justifies equal spectral absorption and emission factors; the grey body assumption lets one factor be taken outside both spectral integrals.

Choose $ϵ=0.8$, $A=0.01{\mathrm{m}}^2$, $T=500\mathrm{K}$ and $T_s=300\mathrm{K}$. These are stipulated model inputs, not measured material properties. Using $\sigma \approx 5.670374419\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$ givesThe incoming irradiance is $459.300\mathrm{W}{\mathrm{m}}^{-2}$, corresponding to $4.5930\mathrm{W}$ incident on the surface; only $80\%$ is absorbed. At $T=T_s=300\mathrm{K}$, emitted and absorbed powers are both $3.6744\mathrm{W}$, so radiative equilibrium has zero net exchange despite continuing emission. Changing only emissivity to $0.2$ reduces both one-way powers and the net loss by a factor of four, giving $6.1694\mathrm{W}$ net. A low-emissivity coating therefore suppresses both radiative cooling to colder walls and radiative heating from hotter walls.

Near equal temperatures, write $T=T_s+\Delta T$. Taylor's theorem givesFor constant emissivity, the radiative heat transfer coefficient $h_{\mathrm{rad}}$ is approximately $4.8992\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$ at $T_s=300\mathrm{K}$ and $ϵ=0.8$. This local linear approximation is useful for small temperature differences; the $200\mathrm{K}$ difference in the worked example needs the full fourth-power expression.

A cooling experiment requires a wider first law of thermodynamics balance. In a uniform-temperature model without a heater,where $C(T)$ is the sample's heat capacity and the last two terms are signed outward convection and thermal conduction losses. A vacuum suppresses convection but does not eliminate thermal conduction through supports or sensor leads. A cooling slope therefore cannot establish emissivity unless these paths and heat capacity are independently constrained. Multiple wall temperatures, incomplete view factors, wavelength-selective materials or appreciable intervening absorption require a more complete radiative transfer calculation.

The weak neutral current couples the Z boson to a fermion without changing its electric charge. Its vector and axial-vector coefficients depend on weak isospin, electric charge, and the weak mixing angle.

The neutral current transfers no electric charge at a fermion vertex; “neutral” does not mean that every participating particle is uncharged. For example, ${\nu }_{\mu }{e}^{-}\to {\nu }_{\mu }{e}^{-}$ transfers energy and momentum to an electron through Z boson exchange. By contrast, a charged current can convert a muon neutrino into a muon, as in ${\nu }_{\mu }n\to {\mu }^{-}p$. Both processes conserve total electric charge. A photon also carries no electric charge, but an active neutrino has no electromagnetic charge coupling: its neutral current is a distinct weak interaction.

The experimentally useful contrast on a nucleon target iswhere $N$ is a nucleon and $X$ is an allowed final state of hadrons. The charged current makes a charged lepton available for identification; the outgoing neutrino in a neutral current leaves no bubble-chamber track. A sample of hadronic events without an identified muon therefore poses an experimental question: how many could instead be neutron interactions, or charged current events whose muon was missed? The answer requires a detector-response and background analysis, not merely an inspection for an absent line.

For canonical chiral superfields, nonzero $\lambda$, a gauged $U(1)$ with charges $(0,+1,-1)$, and no Fayet–Iliopoulos term, F-flatness requires the charged product to vanish and D-flatness requires equal charged magnitudes. Together they force both charged scalar values to zero, leaving the neutral scalar arbitrary. The vacuum moduli space of a supersymmetric gauge theory has zero energy, unbroken gauge symmetry and unbroken supersymmetry. At $\lambda =0$, equal nonzero charged magnitudes instead permit a gauge-Higgsed supersymmetric branch. A neutral modulus alone does not imply internal gauge breaking.

How much does a small directional uncertainty help? Consider an original idealized model: a source direction fixed in advance, uniform background directions over the full sphere, and circular acceptance radius $\psi$. The accepted solid angle and single-event background probability areThe derivation integrates the area element on a unit sphere. For small $\psi$ in radians, $q\simeq {\psi }^2/4$: halving the radius quarters the accidental acceptance, to leading order. At $psi=0$ acceptance vanishes; at $\psi =\pi$ it covers the full sphere. The solid angle has units of steradians; $q$ is dimensionless.

For $0<\psi \le \pi$, take a deliberately crude point spread function: stipulate that every signal direction lies uniformly inside that cap. The signal and background probability density functions per steradian are $S=1/\Omega$ inside the cap and zero outside, and $B=1/(4\pi )$ everywhere. Both integrate to one. For an event inside the cap their likelihood ratio isThis ratio compares two specified directional models; it is not the probability that the event came from the source. Such a conditional probability would additionally require prior source odds and the other selection information. For fixed source fraction $f$ among selected events, a normalized mixture distribution has density $(1-f)B+fS$. For independent events the likelihood is the product of those densities; its ratio to background alone isOutside-cap events contribute factors $1-f$ rather than being silently discarded. At $f=0$, $\Lambda =1$; for a full-sky cap, $S=B$ and every $f$ gives $\Lambda =1$. A directional model that cannot distinguish source from background carries no directional evidence.

As an original numerical example, radii $1^{\circ }$ and $10^{\circ }$ give $q\simeq 7.615\times 10^{-5}$ and $7.596\times 10^{-3}$, respectively: nearly a hundredfold difference. For a fixed number $N$ of independent background events, the binomial distribution givesWith chosen $N=100$, those radii give approximately $0.00759$ and $0.5335$. A small single-event acceptance can become a common coincidence after many opportunities. This probability concerns a fixed cap, rather than a scan for the most favorable cap anywhere. For $m$ independent searches with common local tail probability $p$, the corresponding global tail probability is $1-(1-p{)}^m$; overlapping sky searches generally violate that independence. Actual source tests must include directional response, nonuniform exposure, energy information and the look-elsewhere effect in their null calibration.

A neutrino-dominated accretion flow is a hot, dense accretion disk in which weak-interaction emission removes much of the dissipated energy. At extreme accretion rates, escaping photons can become inefficient at cooling the inner gas, while reactions involving nucleons and leptons emit neutrinos. Hyperaccreting black holes have therefore been investigated as engines for gamma-ray bursts. Whether cooling wins over inward energy transport depends on the flow parameters; the name describes a regime, not every rapidly accreting disk.^[ref]

An elastic weak interaction of a neutrino with an electron. For a stationary electron of mass $m$ and an effectively massless incident neutrino of energy $E$, the maximum recoil kinetic energy in natural units is $T_{\max }=2E^2/(m+2E)$. This kinematic limit does not specify the interaction probability.

A lepton-number-violating nuclear transformation emitting two electrons without neutrinos. A signal appears near the nuclear Q value in summed electron energy, after recoil and detector response are accounted for. Its interpretation requires the decay mechanism and nuclear matrix elements.

Neutron activation is the production of radioactivity by irradiating a material with neutrons. One route is neutron capture producing an unstable isotope; other nuclear reactions can also produce unstable atomic nuclei. A specimen's later radioactive decay provides a way to detect that a transformation occurred during irradiation. It does not by itself identify the product or reconstruct the incoming neutron energy distribution.

This makes neutron activation a useful bridge between an invisible radiation field and a delayed measurement. The neutron flux and collision cross-section determine production; radioactive decay and detection efficiency determine what remains observable later. Neutron activation buildup and decay keeps these stages separate.

Let $\overline{N}$ be the expected value of the number of radioactive product atomic nuclei, initially zero, produced at constant rate $R$. Let their nuclear decay constant be $\lambda >0$. Assume one product species, negligible target depletion and negligible capture by the product. During irradiation, production adds $Rdt$ while radioactive decay removes $\lambda \overline{N}dt$ in expectation. Multiplying the resulting ordinary differential equation by $e^{\lambda t}$ givesThe expected radioactive activity $\overline{A}=\lambda \overline{N}$ tends to the production rate $R$. This saturation is a balance of continuing production and radioactive decay, not exhaustion of target atomic nuclei.

If irradiation ends at $t_i$, let $s$ denote elapsed time afterward. Setting the production term to zero givesFor detection efficiency $\epsilon$, constant background count rate $b$, delay $d$ and counting-window duration $\Delta$, integrating the expected count rate givesThis assumes at most one recorded pulse per product transformation and negligible counter losses. Detector response, branching fractions and background must be calibrated independently. Subtracting an estimated background does not remove its counting uncertainty. Several products give a sum of these terms with distinct nuclear decay constants; a single exponential decay fit can then mislead.

For a wholly stipulated example, choose $R=120{\min }^{-1}$, half-life $2\min$ and $t_i=4\min$. Then $\lambda =\log 2/(2\min )$, $\overline{A}(t_i)=90{\min }^{-1}$ and $\overline{N}(t_i)\simeq 259.69$. Choose $\epsilon =0.25$, $b=2{\min }^{-1}$ and $d=\Delta =2\min$. The result isThe background contributes four expected counts. Multiplying the initial net count rate in that window by its duration would give $22.5$ net counts, overestimating the integrated $16.23$ because the product decays appreciably during counting. All these values and the following curves are original examples, not measured historical data or parameters assigned to a particular silver isotope.

After neutron capture on $(A,Z)$, the compound nucleus is $(A+1,Z)$. With negligible incoming kinetic energy, conservation of energy gives $E^{\ast }=M(A,Z)+m_n-M(A+1,Z)$ in natural units. This is the neutron separation energy of the compound nucleus. At finite incident energy, add the kinetic energy in the initial centre-of-momentum frame; in a nonrelativistic heavy-target approximation it is $A/(A+1)$ times the laboratory neutron kinetic energy. Nuclear pairing can significantly change capture excitation between neighboring even- and odd-mass targets.

A neutron capture resonance is an energy interval where a matching excited state of the combined atomic nucleus enhances capture. Its collision cross-section can have a peak rather than a monotone dependence on energy. The resonance pole description connects this to quantum scattering. Slowing a neutron through such a region can first increase and then decrease a particular reaction response. The capture probability also competes with other reaction channels.

In neutron diffraction, coherent elastic scattering adds scattering amplitudes from regularly positioned scatterers. Since $E_f=E_i$, the neutron scattering kinematics gives $k_f=k_i=2\pi /\lambda$ andFor lattice planes of interplanar spacing $d_{\mathrm{p}}$, constructive addition requires $Q=2\pi n/d_{\mathrm{p}}$ along the plane normal. Equating these expressions proves Bragg's law:For example, stipulate $d_{\mathrm{p}}=2$ ångströms and $\lambda =2$ ångströms. The first diffraction order has $\theta =30^{\circ }$ and detector deflection $60^{\circ }$. A peak position constrains a period; its intensity also depends on the crystal basis structure factor.

Clifford Glenwood Shull describes powdered sodium chloride, a crystal monochromator, and a shielded counter recording angular counts. Inserting cadmium measured background; substantial diffuse scattering complicated peak interpretation.^[ref]

Powdering supplies many crystal orientations. Comparing the angular peaks with Bragg's law constrains crystal structure; a background estimate helps distinguish specimen peaks from unwanted counts. A two-axis diffraction instrument has no separate outgoing energy analysis, so an angular pattern alone does not separate every inelastic contribution. The richer magnetic example, including a warm/cold comparison and its limits, is the manganese monoxide neutron diffraction experiment. Its extra peaks reveal magnetic order beyond the chemical periodicity; the associated magnetic neutron diffraction model is developed there.

Let $n(E)dE$ be the number of neutrons per unit volume in an energy interval. The scalar neutron flux per unit energy is $\varphi (E)=v(E)n(E)$, with $v(E)$ the speed. Its integral has units of inverse area per unit time. It counts path length traversed per volume per time, including all directions; it is not the signed current through an oriented surface.

For a small uniform specimen containing $N_T$ target atomic nuclei, a reaction-specific collision cross-section ${\sigma }_a(E)$ gives the expected production rateAssume the incident field is little changed within the specimen, target depletion is negligible and each counted reaction creates the product of interest. If that product has a production branching fraction below one, include it in ${\sigma }_a$ or as an additional factor. A single neutron activation measurement supplies one weighted integral of an energy distribution, not a unique reconstruction of that distribution.

For an original two-group example, stipulate response collision cross-sections ${\sigma }_s=10{\sigma }_f$. With the same total scalar neutron flux $\Phi$, changing from all fast neutrons to half slow and half fast changes $R/(N_T{\sigma }_f\Phi )$ from $1$ to $5.5$. No additional neutrons need be emitted by the source. The chosen factor ten is not a historical silver measurement.

For some capture reactions over a limited low-energy interval, ${\sigma }_a\propto 1/v$ is a useful approximation. At fixed neutron flux, it gives increasing reaction probability per path length as speed decreases. At fixed neutron number density, ${\sigma }_{a}v$ is constant in this approximation, so the rate per target is unchanged. A real neutron moderator changes the local density, energy distribution, escape and absorption together. Neither all collision cross-sections nor all neutron activation rates vary monotonically as $1/v$.

Neutron moderation reduces the kinetic energy of fast neutrons through repeated collisions in a neutron moderator. In a simple elastic collision model, hydrogen is especially effective because a proton and a neutron have nearly equal masses: the maximum elastic recoil energy fraction can approach one. The mean logarithmic energy decrement describes typical accumulated slowing, whereas the maximum transfer describes a single endpoint.

Slowing, absorption and escape compete. A material can remove neutrons by neutron capture before they become thermal neutrons, and some neutrons escape its boundaries. Describing a neutron moderator therefore requires more than collision kinematics. Near thermal equilibrium, target thermal motion can also give energy back to a neutron; the initially stationary target approximation must then be replaced.

Neutron physics studies the production, interaction, transport and detection of neutrons. A neutron has no net electric charge, so it does not experience the long-range Coulomb repulsion that impedes charged particles approaching a positively charged atomic nucleus. Its nuclear interactions nevertheless depend strongly on energy and target isotope.

Elastic scattering can slow neutrons in a neutron moderator, capture can change a nucleus, and suitable reactions can initiate nuclear fission. Neutron magnetic moments also couple to magnetic structure, enabling magnetic neutron diffraction. A measured attenuation therefore combines different reaction channels and must not be interpreted as one universal absorption probability.