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Helium liquefaction converts helium gas into a collectable liquid. It creates a bath for investigating material properties at very low temperatures, including superconductivity. The central question is how the incoming specific enthalpy compares with the specific enthalpies of the outgoing liquid and vapor. The liquefaction enthalpy budget makes that comparison explicit; a temperature decrease by itself does not specify a liquid yield.

In a Joule-Thomson effect cooling stage, a throttling pressure drop approximately preserves enthalpy under the assumptions of the steady-flow energy balance. For an ideal gas, specific enthalpy depends only on temperature: with positive specific heat capacity at constant pressure, equal specific enthalpy means equal temperature. Thus its Joule-Thomson coefficient is zero. An ideal gas also supplies no liquid branch. Cooling by throttling depends on real gas properties; expansion is not an automatic source of refrigeration.

For a local drop $dp<0$, $dT={\mu }_{\mathrm{JT}}dp$ cools only where the Joule-Thomson coefficient is positive. Precooling must bring the real gas into a useful cooling region, with enough cooling to overcome unwanted heat transfer. A finite expansion can cross a Joule-Thomson inversion curve; its final state requires the full enthalpy balance, including phase coexistence if liquid forms. Kamerlingh Onnes's 1908 helium liquefaction experiment combined external precooling with regeneration to reach that regime. The Joule-Thomson effect article develops the throttling identities and their assumptions.

The helium mass fraction is the fraction of a material's mass carried by helium nuclei. Hydrogen burning raises it, whereas core helium burning lowers it while producing carbon and oxygen.

A helium-neon laser is a gas laser using a mixture of helium and neon.

Pyotr Leonidovich Kapitza's 1938 test compared helium flow through optically flat glass disks. Suspension set the tube-to-bath level difference; a cathetometer read levels. The gap was estimated as $0.5\mu \mathrm{m}$. Levels barely moved over minutes above the transition, but equalized in seconds below. A laminar flow viscosity calculation gave only an upper bound.^[ref]

The essential control is to compare the same restriction and level readout at different temperatures. A changed pressure head is not itself evidence for a changed material response: the driving head must be included when interpreting the flow. For an idealized comparison at equal vapor pressure, negligible surface tension corrections and nearly constant mass density $\rho$, hydrostatic equilibrium givesHere $\Delta h$ is the difference in free-surface heights and $g$ is gravitational acceleration. This explains why following a liquid level provides both a driving pressure and a transported volume; it does not turn a finite-resolution observation into proof of zero viscosity.

The Homestake experiment, led by Raymond Davis Jr., used electron neutrino capture on chlorine to produce argon subject to radioactive decay. The 1968 paper reported an upper limit, rather than a resolved positive signal.^[ref]

The underground target contained tetrachloroethylene. Circulated helium swept out argon, which was trapped on cooled charcoal, purified, and put into tiny proportional counters. Electron capture back to chlorine supplies the delayed readout. Added stable argon checked extraction recovery; background counting checked spurious pulses. The capture threshold is $0.814\mathrm{MeV}$ and the product's half-life is about 35 days. The chemical extraction retains neither the original neutrino's arrival time nor its direction.^[ref]

The reaction selects electron neutrinos, and its threshold excludes the lowest-energy part of the solar neutrino spectrum. A low count alone cannot distinguish a low source neutrino flux, a changed neutrino flavor mixture, or an underestimated loss in extraction or counting. This is why chemical recovery tests and an independent water Cherenkov neutrino reconstruction are scientifically valuable.

A hydrogen-burning shell converts hydrogen into helium in a layer surrounding a hydrogen-exhausted stellar core. It moves outward in enclosed mass as it supplies fresh helium to that core; its movement in radius need not have the same sign.

For fully ionized hydrogen and helium with hydrogen mass fraction $x_H$, $n_e=(1+x_H)\rho /(2m_p)$ and ${\sum }_i{Z}_i^2{n}_i=\rho /m_p$. With the same unit Gaunt factor for both species, thermal bremsstrahlung at fixed electron number density is multiplied by $2/(1+x_H)$ relative to pure ionized hydrogen. The factor is one for pure hydrogen and two for pure helium.

Removing helium from a hydrogen-rich envelope lowers He/H and raises its reciprocal H/He. A helium-rain explanation of atmospheric depletion must keep those two ratio conventions distinct. Settling helium-rich material can release gravitational energy while changing atmospheric composition; excess luminosity alone does not uniquely determine the interior process.

On 10 July 1908, Heike Kamerlingh Onnes's Leiden laboratory circulated purified helium through liquid air and evaporatively cooled liquid-hydrogen precooling, a regenerative heat exchanger, and an expansion valve. Returning helium cooled incoming helium; liquid collected in a transparent vacuum vessel. Reflected light revealed its liquid meniscus, crossed by thermocouple wires. Subsequent mass density and spectroscopy checks tested contamination.^[ref]

A constant-volume helium gas thermometer below the valve supplied the temperature indication. An oxygen boiling reference check disagreed with its expected reading; the real gas correction to thermodynamic temperature remained provisional. A temperature plateau suggested immersion before the liquid meniscus was recognized.^[ref]

The measurement principles separate two questions. A visible liquid meniscus establishes an interface; a gas thermometer estimates temperature through an equation of state. A stationary temperature trace can also result from a steady state without phase coexistence, so it cannot replace the interface observation. Conversely, seeing a liquid does not calibrate a thermometer. Thermometer calibration, the gas thermometer dead-space correction, departures from the ideal gas law and incomplete thermal equilibrium all matter when assigning an absolute temperature. A check far from the intended measurement range tests the instrument but does not by itself bound every low-temperature error.

Kapitza resistance is the thermal boundary resistance encountered between a solid and helium in cryogenics. A cold helium heat bath can coexist with a warmer immersed component under continuous heating. The bath's temperature alone therefore does not establish the component's temperature. Kapitza thermal boundary experiment supplied an early direct demonstration of the obstacle at the solid–liquid boundary.

A microscopic route to interfacial heat transfer is transmission of phonons. Transmission depends on the two materials and their interface; scattering and the condition of the surface matter. ^[ref] The fourth-power thermal boundary model isolates one useful low-temperature limiting law without assigning an empirical coefficient to a particular sample.

Pyotr Leonidovich Kapitza's 1941 study suspended cleaned, annealed copper blocks in liquid helium. Sealed, gas-filled cavities held a heater and a bronze resistance thermometer. Electrical readout tracked the heating-induced temperature rise; the earlier readout method assumed a stable bath without a separate bath thermometer. A nearby unheated block warmed much less, localizing the drop near the heated surface. Gold plating and coatings tested surface effects; tight cavities prevented liquid ingress. These were solid-body readings, not probes resolving the microscopic interfacial layer. ^[ref]

The measurement logic suggests a modern control sequence: calibrate both temperature readouts, establish their zero-load difference, increase heater power, wait for a steady state, and repeat while monitoring the bath. Determine lead losses and the bulk temperature gradients before using $r_B=(T_s-T_l)/q$. These are proposed controls derived from the energy balance, rather than additional reported steps in the 1941 apparatus. A nearby unheated solid also has its own interface: its reading need not equal the liquid's local temperature exactly. The experiment supports a drop concentrated near the surface, while a microscopic thickness and a universal interfacial law require further evidence.

The Landau criterion asks whether creating an excitation can lower a moving superfluid's energy. For a weakly coupled, heavy obstacle moving through a uniform liquid initially at zero temperature, let $ϵ(p)>0$ be the excitation energy in the fluid rest frame, with momentum magnitude $p$. Transferring momentum $\mathbf{p}$ from an obstacle moving at velocity $\mathbf{v}$ supplies, to leading order in its negligible recoil, energy $\mathbf{v}\cdot \mathbf{p}$. Emission requiresFor an isotropic dispersion relation, the easiest direction is parallel to $\mathbf{v}$, giving the necessary kinematic conditionAn infimum includes cases in which the limiting slope occurs only as $p\to 0$. A finite-mass obstacle of mass $M$ instead loses $\mathbf{v}\cdot \mathbf{p}-p^2/(2M)$, so the corresponding emission inequality includes the extra positive recoil term $p^2/(2M)$ on the left. This is an energy and momentum balance, not a calculation of an emission rate.

The Landau criterion excludes spontaneous emission into the specified spectrum below its threshold. Above it, kinematic permission alone does not guarantee strong coupling or rapid decay. It is not a universal observed flow threshold: walls, channel geometry and quantum vortex nucleation can enable other decay routes, and finite temperature permits scattering from existing excitations. Using only bulk phonons and rotons leaves those processes outside the model.

Landau's 1941 energy and momentum argument used a roton approximation different from the finite-momentum minimum illustrated below.^[ref]

Landau received his Nobel Prize in Physics in 1962 while hospitalized in Moscow after a car accident.^[ref]

For a deliberately simple Bose-Einstein condensation comparison, stipulate an ideal free-particle excitation law $ϵ=p^2/(2m)$. Then $ϵ/p=p/(2m)$ has infimum zero. Macroscopic one-particle occupation therefore supplies no positive emission threshold by this argument alone. By contrast a linear phonon law $ϵ=cp$ yields $v_L=c$. The dispersion relation and the collective response contain information that the occupation criterion alone does not supply; Bose-Einstein condensation and thermodynamic superfluidity must be tested separately.

To check the picture algebraically, on the stipulated roton branch,Thus tangency occurs at $q_L=\sqrt{4.3}$, slightly above the roton energy minimum. At an energy minimum $e^{\prime }=0$, but a tangent through the origin needs $e^{\prime }=e/q>0$. Minimizing energy and minimizing the energy-to-momentum ratio are different operations. This geometric construction makes the criterion useful even when the full dispersion relation is supplied numerically rather than by a simple formula.

For identical isotropic closed-shell atoms of number density $n$, with $Z$ electrons and mean squared radius $⟨r^2⟩$ per electron, the quadratic magnetic field energy is $Z{e}^2{B}^2⟨r^2⟩/(12m_e)$. It follows from the $e^2{\mathbf{A}}^2/(2m_e)$ term with $\mathbf{A}=\mathbf{B}\times \mathbf{r}/2$. Differentiating the energy with respect to field and using $B\simeq {\mu }_{0}H$ yields the negative SI volume magnetic susceptibility. For helium, $Z=2$; combining it with the independent spin-one-half nuclear response ${\mu }_{0}n{\mu }_a^2/(k_{B}T)$ gives a density-independent sign-change scale $T_0=3m_e{\mu }_a^2/(e^2{k}_B⟨r^2⟩)$.

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.

For a specified age and irradiation, a hydrogen–helium sequence rises at low mass, has radii near Jupiter's across giant planets and brown dwarfs, and declines as electron degeneracy pressure becomes important. Above the hydrogen-burning minimum mass, sustained fusion supports a rising low-mass stellar branch.

Let a projectile have mass $m$, incident speed $v$, and incident kinetic energy $E=m{v}^2/2$. The stationary target has mass $M$. In a head-on elastic collision, the signed outgoing projectile velocity $v^{\prime }$ and target velocity $u$ obey momentum conservation and conservation of energy:Factor the second equation and eliminate $Mu=m(v-v^{\prime })$. The nontrivial collision branch has $u=v+v^{\prime }$, henceThis is the maximum over scattering angles, not the energy of every recoil. To see the bound without assuming a head-on event, set the recoil momentum magnitude to $q$ and its laboratory angle to the incident direction to $\theta$. Conservation of energy with outgoing projectile momentum $\mathbf{p}-\mathbf{q}$ givesThus $T=T_{\max }{\cos }^2\theta$ for a nonzero recoil, and the endpoint occurs at $\theta =0$.

For an original numerical example, stipulate $E=4\mathrm{MeV}$ and $m=m_p$, with $m_p$ the proton mass. Approximate hydrogen, helium and nitrogen target masses by $M=m_p$, $4m_p$ and $14m_p$ respectively. Their maximum fractions and recoil energies areThese stipulated values illustrate the approximation $m_n=m_p$; they are not measurements from 1932. With $m_p{c}^2\simeq 938\mathrm{MeV}$, the incident ratio $E/(m_p{c}^2)\simeq 0.0043$ gives $v/c\simeq 0.092$, consistent with the nonrelativistic limit used here. A million electronvolts is an energy unit, whereas a mass expressed in $\mathrm{MeV}/c^2$ requires the factor $c^2$ when inserted into an energy equation.

The hydrogen endpoint permits essentially complete transfer because the projectile and target masses match. For a very heavy target, $M\gg m$, the fraction approaches $4m/M$, which is small: the target can take substantial momentum with little kinetic energy. The exact inequality follows from $(m-M{)}^2\ge 0$, which implies $(m+M{)}^2\ge 4mM$.

The abundance of chemical elements heavier than helium in astronomical material. A metal mass fraction measures their combined mass; spectroscopic abundance ratios are other measures of metallicity and need not be numerically identical to that fraction.

Heavy elements provide many ionic transitions at temperatures where hydrogen and helium have lost their most effective line emitters. Metal-line cooling can therefore increase the plasma cooling function substantially and broaden its temperature features. Low-energy fine-structure transitions also permit cooling below the hydrogen atomic threshold when the relevant ions and electrons are present.

The fraction of total mass in chemical elements heavier than helium, so the hydrogen mass fraction, helium mass fraction and metal mass fraction satisfy $X+Y+Z=1$. Despite the astronomical name, this includes elements such as carbon and oxygen that are not chemical metals.

Molecular orbital theory describes the electronic arrangement of a molecule using molecular orbitals: one-electron spatial wavefunctions extending over the combined atomic nuclei. Filling bonding molecular orbitals can stabilize a chemical bond; filling antibonding molecular orbitals opposes that stabilization. A molecular orbital is a building block of a many-electron description, rather than the entire many-electron wavefunction. See Robert W. Field, MIT 5.61, lecture 24, pp. 1–4: molecular orbitals and the LCAO approximation.

First make the interference visible with an exactly normalized one-dimensional toy. Replace the three-dimensional $1s$ atomic orbitals of the hydrogen atom by two Gaussian functions centred at $x=\pm a$, with dimensionless $x$ and $a>0$:The normalized sum and difference are ${\psi }_{\pm }=({\varphi }_A\pm {\varphi }_B)/\sqrt{2(1\pm S)}$, following the same linear combination of atomic orbitals algebra. At $a=1$, the orbital overlap is $S=e^{-1}\approx 0.367879$. At the midpoint, ${\varphi }_A(0)={\varphi }_B(0)\approx 0.455581$, givingBoth probability densities have total integral one: interference redistributes probability, rather than creating an extra electron. These are toy wavefunctions, not actual hydrogen atom orbitals or predictions of molecular hydrogen energies.

Now fill the two $1s$-derived molecular orbitals in the minimal model. The lower bonding molecular orbital is labelled ${\sigma }_g$; the upper antibonding molecular orbital is ${\sigma }_u^{\ast }$. The Pauli exclusion principle permits at most two electrons in each spatial molecular orbital, with opposite spin projections along a chosen axis. For the lowest electron configuration in this model, fill ${\sigma }_g$ first. The subscripts $g$ and $u$ denote even and odd behaviour under inversion through the midpoint; the star marks an antibonding molecular orbital. See Flowers et al., Chemistry 2e, 8.4: occupancy and bond order and Field, MIT 5.61, lecture 25, pp. 1–3: bonding, antibonding, and symmetry labels.

With $N_b$ electrons in bonding molecular orbitals and $N_a$ in antibonding molecular orbitals, the model's bond order is $b=(N_b-N_a)/2$. The three elementary counts are:

The zero bond order says this minimal model gives no net covalent bond for ${\mathrm{He}}_2$. It does not rule out every bound pair of helium chemical atoms: a very weak Van der Waals force can bind a neutral helium dimer, which has been observed experimentally. See Zeller et al., Imaging the He2 quantum halo state using a free electron laser, PNAS 113, 14651–14655 (2016), p. 14651: van der Waals binding and Luo et al., The weakest bond: Experimental observation of helium dimer, Journal of Chemical Physics 98, 3564–3567 (1993). The bond order count does not include that weak binding mechanism.

The useful surprise is that sharing alone does not determine a chemical bond: the relative quantum phase and the occupancy matter together. The same two shapes explain why adding an electron to a bonding molecular orbital increases the model's bond order, while adding one to an antibonding molecular orbital decreases it. Molecular orbital theory also accounts for the unpaired electrons and paramagnetism of molecular oxygen, which its all-paired Lewis structure misses. See Chemistry 2e, 8.4: the molecular orbital description of oxygen.