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For equal exterior potential energies and constant mass, the exact transmission probability isBelow the barrier, the hyperbolic sine is positive for $a>0$, so $0<T<1$. At $E=V_0$, the interior wavefunction is linear in $x$; either neighbouring formula has the displayed finite limit, since $\sinh z/z$ and $\sin z/z$ tend to one. Above the barrier, the interior wavefunction oscillates: the quantum reflection probability generally remains nonzero, but $qa=n\pi$ for a positive integer $n$ gives perfect transmission probability $T=1$ through constructive wave interference. These are scattering resonances, not below-barrier quantum tunnelling.

At half height, $E=V_0/2$, we have $k=\kappa$ and $T={\mathrm{sech}}^2(\kappa a)$, consistent with finite square barrier transmission at half barrier height after replacing this width $a$ by that example's width $2a$. In the illustration, the width is four times $\ell =\hslash /\sqrt{2m{V}_0}$ and $E/V_0=1/2$, giving $T\simeq 0.01388$. The plotted wavefunction comes from solving the four interface equations; the transmission probability curve uses the closed formula.

The repulsive inverse-square force $\mathbf{F}=k\mathbf{r}/r^3$ has potential energy $V(r)=k/r$ and scatters positive-energy trajectories along hyperbolic conics.

This potential energy generates a repulsive central force $F_r=-V^{\prime }(r)=2c/r^3$. It is an inverse-square potential but an inverse-cube force; the powers must not be confused with the Coulomb or gravitational potential proportional to $1/r$.

Write the RF electric field as ${\mathbf{E}}_{\mathrm{amp}}(\mathbf{r})\cos \Omega t$, where ${\mathbf{E}}_{\mathrm{amp}}$ is a real peak-amplitude vector with a common drive phase. Decompose the trajectory into a slow position $\mathbf{R}$ and fast displacement $\mathbf{\xi }$. Over one short RF cycle, hold $\mathbf{R}$ fixed. The leading fast equation and solution areThe homogeneous slow part is absorbed into $\mathbf{R}$. Expand the electric field to first order in $\mathbf{\xi }$ and average, using $⟨{\cos }^2\Omega t⟩=1/2$:The second equality uses zero curl in the quasistatic RF electric field. This derives an effective potential energy, not an extra static electric potential. Its $Q^2$ dependence confines either sign of electric charge towards a field-amplitude minimum. Using an RMS field instead would give a coefficient $1/(2m{\Omega }^2)$ multiplying $Q^2|{\mathbf{E}}_{\mathrm{rms}}{|}^2$.

The approximation needs fast displacement small compared with the field's variation length and slow angular frequencies small compared with $\Omega$. In the stated electric quadrupole model, small $|q_i|$ and $a_i$ of order $q_i^2$ provide a controlled regime, away from stability boundaries. Add $Q{\Phi }_{\mathrm{dc}}$ when $U\ne 0$. For $U=0$ the result isFor the pure RF case, ${\omega }_z\simeq 2{\omega }_x=2{\omega }_y$. The square-root condition describes leading local restoring curvature; it is not a substitute for full Mathieu equation stability. Collisions, departures from the ideal electric quadrupole, and a displaced RF null require further analysis.

A rigid body undergoes translation when every point has the same displacement at each time, so its orientation stays fixed. For a freely moving system of particles coupled only by internal springs, the common displacement is a zero-frequency normal mode: internal spring lengths and potential energy are unchanged. Its mode coordinate is $A+Bt$, representing constant centre-of-mass velocity, rather than an oscillation at nonzero frequency.

A scalar-field vacuum is a constant field at a minimum of its potential energy, with the minimum shifted to zero when discussing finite-energy classical field-theory solitons. Distinct isolated vacua allow a scalar-field kink to approach different values at the two ends of space.

The Schrödinger equation governs the evolution of a quantum state between instances of measurement in quantum mechanics. For a particle of mass $m>0$ moving in a prescribed real potential energy $V(\mathbf{r},t)$, its position wavefunction satisfiesHere $h$ is the Planck constant, ${\nabla }^2$ is the Laplace operator, and the Hamiltonian operator combines kinetic energy with potential energy. Specify an initial wavefunction, an operator domain and any boundary conditions: the differential expression alone does not completely specify a physical model. The Born rule then turns the evolving probability amplitude into position probabilities through $\rho =|\psi {|}^2$ and integration over a region.

This scalar Schrödinger operator describes nonrelativistic quantum mechanics, with relevant momenta much smaller than $mc$, where $c$ is the speed of light. It omits the constant rest energy $m{c}^2$, assumes fixed particle number, and suppresses active spin dynamics and coupling to a magnetic vector potential. A free electron with a fixed, uncoupled spin state can still use this scalar model; the Pauli equation supplies the additional spin components and magnetic coupling when needed. A prescribed real $V$ describes a closed evolution model or an approximation in which the source of that potential energy is treated as external. Dissipation, particle creation and a general measurement in quantum mechanics need further modelling.

For a time-independent self-adjoint Hamiltonian operator, the spectral theorem defines unitary time evolution byInitial states in the operator domain satisfy the differential equation in the Hilbert space sense; the unitary operator also evolves all other states in that Hilbert space. Time-dependent Hamiltonian operators need suitable regularity and domain hypotheses; their evolution is not generally the exponential of $-itH(t)/\hslash$.

The local reason for conservation is especially tangible. Multiply the Schrödinger equation by the complex conjugate ${\psi }^{\ast }$, subtract its conjugate multiplied by $\psi$, and use the fact that the real potential energy terms cancel. For smooth wavefunctions this gives the probability continuity equationThe divergence theorem identifies the change in probability inside a region with minus the outward probability current. Decay at infinity, or appropriate probability-conserving boundary conditions, preserves wavefunction normalization. Conservation of total probability does not require a stationary probability density.

Its usefulness comes from being a linear partial differential equation: quantum superpositions of solutions are solutions, while different quantum phases can produce interference. For a fixed Hamiltonian operator, the time-independent Schrödinger equation finds the energy eigenstates; each then acquires just its own time-dependent quantum phase. One eigenvalue problem therefore organizes both allowed energies and the evolution of arbitrary quantum superpositions, with a continuous spectrum included when required. This is the attractive bridge between a spatial wave problem and a clock: energy differences determine the rate at which relative quantum phases change. It makes atomic spectral lines, quantum tunnelling and quantum scattering calculable from a specified model.

For an explicit example without confinement, take a free particle on the whole line, $V=0$, initially in a Gaussian wave packet with position standard deviation ${\sigma }_0$ and mean momentum $\hslash k_0$:Unlike a single plane wave, this wavefunction satisfies wavefunction normalization. A Fourier transform writes it as a quantum superposition of plane waves. Their dispersion relation is $\omega (k)=\hslash k^2/(2m)$, so each Fourier component gains the factor $e^{-i\hslash k^{2}t/(2m)}$. Completing the square in the resulting Gaussian integral giveswith the square root chosen continuously from $t=0$. Its position probability density and standard deviation areThe centre moves with the group velocity ${\omega }^{\prime }(k_0)=v_0$ and the wave packet spreads because different wavenumbers have different group velocities. This is unitary wave dispersion, without a random force or loss of probability. Independently, the initial momentum standard deviation is $\Delta p=\hslash /(2{\sigma }_0)$ and the initial position-momentum covariance vanishes. Free evolution gives $x(t)=x(0)+p(0)t/m$ in the Heisenberg picture, henceThis independently checks the spreading law. The Gaussian integral of $\rho$ is one at every time; the decreasing peak is exactly offset by increasing width.

As hypothetical preparation parameters, choose a free electron with ${\sigma }_0=1\mathrm{nm}$ and $k_0=2{\mathrm{nm}}^{-1}$. Using $\hslash \simeq 1.054571817\times 10^{-34}\mathrm{J}\mathrm{s}$ and $m\simeq 9.1093837\times 10^{-31}\mathrm{kg}$ givesThe rounded electron mass uses the 2022 adjustment.^[ref] At $t=t_s$, the centre is $4\mathrm{nm}$ and ${\sigma }_t=\sqrt{2}\mathrm{nm}$; at $t=2t_s$, they are $8\mathrm{nm}$ and $\sqrt{5}\mathrm{nm}$. The mean kinetic energy staysIt includes both mean motion and the momentum spread. The mean speed is about $7.72\times 10^{-4}c$ and the momentum spread corresponds to $1.93\times 10^{-4}c$, making the nonrelativistic model accurate for the appreciably occupied part of this ideal Gaussian distribution. Its mathematically unbounded tails are an idealization.

The complex width also explains where the probability goes. Write $\psi =\sqrt{\rho }{e}^{i\theta }$, where $\theta$ is the dimensionless quantum phase. Differentiating the exact solution givesRelative to the moving centre, the probability current $j_{\mathrm{rel}}=j-v_0\rho$ points outward on both sides for $t>0$. This is a flux statement, not an additional assumption about individual particle trajectories. At $t=t_s$ and $x-v_{0}t=\pm {\sigma }_0$, the relative flow speeds $j_{\mathrm{rel}}/\rho$ are $\pm {\sigma }_0/(2t_s)$, about $\pm 2.8942\times 10^4\mathrm{m}{\mathrm{s}}^{-1}$. The probability within a moving interval of half-width ${\sigma }_0$ falls from about $0.6827$ initially to $0.5205$ at $t_s$, exactly because of this outward probability current.

Historically, the 1926 Quantisierung als Eigenwertproblem (Erste Mitteilung) by Erwin Schrödinger began with the stationary hydrogen atom problem. It distinguishes discrete negative energies from a positive continuous spectrum; it is not yet the general evolution equation displayed above.^[ref] The opening of quantum mechanics discusses its original vibrating-string analogy.

In the 12 December 1933 lecture The fundamental idea of wave mechanics by Erwin Schrödinger, Erwin Schrödinger explains the move from independent rays to waves through diffraction, then applies that analogy to atomic physics. His closing pages explicitly acknowledge the difficulty of reconciling wave language with observed particle tracks. This retrospective physical argument motivates the model; it does not derive the modern Born rule from classical mechanics.^[ref]

The dates separate proposal from measurement: the de Broglie wavelength belongs to the 1924 thesis, the Schrödinger equation to 1926, and the Davisson–Germer experiment's electron diffraction beams to 1927.^[ref] That linked experiment describes the prepared nickel crystal, evacuated electron gun, angular collector scans, retarding-potential selection and heating controls. Its measured collector electric current tests wave-dependent scattering probabilities; it neither measures a wavefunction directly nor establishes this equation uniquely.

Sensible heat is transferred heat associated with a temperature change within a single thermodynamic phase. By contrast, latent heat can change the phase fraction while the temperature stays fixed during equilibrium phase coexistence. Neither name denotes an additional stored substance: both describe how transferred heat participates in a process.

For fixed mass $m$ and composition, use the heat capacity appropriate to the heating constraint. Under the quasistatic process assumptions in heat capacity, with only pressure-volume work and unchanged bulk kinetic energy and potential energy, fixed-pressure heating givesHere $c_p$ is specific heat capacity at constant pressure; fixed-volume heating instead uses specific heat capacity at constant volume $c_V$. A cooling step has negative $Q_{\mathrm{sens}}$ with the sign convention in heat. A temperature difference has the same numerical value in units of the kelvin or degree Celsius. The integral follows by adding the small transfers $\delta Q=m{c}_p(T)dT$, rather than assuming a constant heat capacity.

As a stylized ice-to-water example for a fixed-composition closed system at fixed pressure $1\mathrm{atm}$, take $m=0.5\mathrm{kg}$, an initial temperature of $-10{}^{\circ }\mathrm{C}$, melting at $0{}^{\circ }\mathrm{C}$, and a final temperature of $20{}^{\circ }\mathrm{C}$. Assume constant specific heat capacity at constant pressure values $c_{\mathrm{ice}}=2.1\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $c_{\mathrm{water}}=4.18\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and latent heat of fusion $L_f=334\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$. These are stipulated, rounded model values; ignore the container and heat losses, retain all the material, and assume only pressure-volume work under the quasistatic process conditions above. The three steps areOnly the first and third terms are sensible heat; the middle term is latent heat. The plateau begins after $10.5\mathrm{kJ}$ and ends after $177.5\mathrm{kJ}$ of total supplied heat. At $94\mathrm{kJ}$ total input, half the mass is liquid water and the temperature remains $0{}^{\circ }\mathrm{C}$. The equilibrium phase transition temperature depends on pressure; this example stipulates a single fixed pressure. ^[ref]

This split makes heating budgets easy to interpret: melting consumes $167/219.3\approx 76.2\%$ of the supplied heat in this model despite causing no temperature rise. Within either single thermodynamic phase, the heating curve slope is $dT/dQ=1/(m{c}_p)$; a larger heat capacity makes the slope shallower. A flat interval represents changing phase fraction, so it cannot be modeled by applying $Q=m{c}_p\Delta T$ across the phase transition. The complementary constructed example in latent heat tracks that phase fraction explicitly.

In a series RLC circuit, the same electric current $i=q^{\prime }$ flows through the resistor with electrical resistance $R$, the inductor, and the capacitor. Choose $q$ as the signed electric charge on the capacitor plate into which positive $i$ flows. With terminal drops measured in that direction, Kirchhoff's voltage law givesHere $V(t)$ is the applied electric potential difference; assume constant $L>0$, $C>0$, and $R>0$. With the applied source set to zero and the loop kept closed, the free response satisfiesThe angular frequency ${\omega }_0$ is the value for lossless free oscillation; $\alpha$ is the decay parameter. The three possibilities are underdamped RLC response for $\alpha <{\omega }_0$, critically damped RLC response for $\alpha ={\omega }_0$, and overdamped RLC response for $\alpha >{\omega }_0$. ^[ref].

For a fair comparison, start each circuit with the same electric charge $q(0)=q_0$ and zero electric current $q^{\prime }(0)=0$. Set $\tau ={\omega }_{0}t$ and $a=\alpha /{\omega }_0$; then $x=q/q_0$ obeys $d^{2}x/d{\tau }^2+2adx/d\tau +x=0$. The plot compares $a=0.5,1,2$. These dimensionless curves apply to any $L,C$ after rescaling time and electric charge.

The stored energy is the sum of magnetic energy and capacitor energy:Differentiating and using the circuit equation gives the complete power balance,In the free response, $E^{\prime }=-R{i}^2\le 0$: Joule heating accounts exactly for the lost energy. For the plotted initial conditions, $E(0)=q_0^2/(2C)$ and $E/E(0)=x^2+(dx/d\tau {)}^2$. Since every free solution tends to zero for $R>0$,In the lossless limit $R=0$, energy stays constant and shuttles between the capacitor and inductor. Thus an oscillation of electric charge and a decrease of total energy are compatible statements.

The analogy with a damped harmonic oscillator is exact within these ideal models: electric charge corresponds to displacement, electric current to velocity, inductance to mass, $1/C$ to spring constant, and $R$ to the coefficient of a resisting force proportional to velocity. The two stored energy terms play the roles of kinetic energy and elastic potential energy. ^[ref].

Differentiating the driven equation also gives $L{i}^{\prime \prime }+R{i}^{\prime }+i/C=V^{\prime }$. When using this equation to solve for electric current, retain the original constraint $L{i}^{\prime }(0)+Ri(0)+q(0)/C=V(0)$: the initial electric charge fixes the initial rate of change of electric current.

For a well of depth $V_0$ and full width $2a$, measured from zero exterior potential energy, even-state matching gives $q=k\tan ka$. If $a\sqrt{m{V}_0}/\hslash \ll 1$, then $ka\ll 1$ and $q\simeq 2m{V}_{0}a/{\hslash }^2$. Using $E_0=-{\hslash }^2{q}^2/(2m)$ gives the displayed leading binding energy. It agrees with the attractive delta-potential result for integrated strength $2a{V}_0$.

Simple harmonic motion is motion in which the acceleration is proportional to the displacement from equilibrium and points back towards it:Here $x$ is measured from equilibrium, and $\omega$ is the constant angular frequency. For a point mass attached to an ideal spring, Hooke's law gives the restoring force $F=-kx$, with stiffness $k>0$, and Newton's second law yields $\omega =\sqrt{k/m}$. The minus sign matters: for $x>0$ the force is negative, and for $x<0$ it is positive. A restoring force need not oppose velocity; on the return towards equilibrium it increases speed.

The general solution isThe amplitude $A\ge 0$ and oscillation phase $\phi$ encode the initial conditions. We use $P$ for the oscillation period to distinguish it from kinetic energy $T$. For $A>0$, the oscillation period is independent of amplitude, the property of isochronous harmonic motion. At $A=0$ the body remains at equilibrium. ^[ref]

Choose $m=0.5\mathrm{kg}$, $k=2N/m$, and release the mass from rest at displacement $A=0.3\mathrm{m}$. Then $\omega =2{\mathrm{s}}^{-1}$, $P=\pi \mathrm{s}$, and $\phi =0$. Writing time $t$ in seconds,At $t=0$ the velocity vanishes, but the acceleration is $-1.2m/{\mathrm{s}}^2$: a turning point is not a state of rest that persists. At $t=\pi /4\mathrm{s}$, the body crosses equilibrium with velocity $-0.6m/s$ and zero acceleration. At $t=\pi /2\mathrm{s}$ it reaches the opposite turning point, $x=-0.3\mathrm{m}$.

Taking potential energy zero at equilibrium, the same solution givesAt each turning point all mechanical energy is potential energy; at each equilibrium crossing it is kinetic energy. Both repeat twice during one complete position cycle because squaring removes the sign. At $t=\pi /8\mathrm{s}$, for example, each contributes $0.045\mathrm{J}$.

The physical position trace covers a line segment, $-0.3\le x\le 0.3$ in metres. In the phase plane, the paired position and velocity instead trace the ellipseThis is precisely $m{v}^2/2+k{x}^2/2=E$, rewritten: the single conserved mechanical energy fixes the amplitude and the whole ellipse. It does not fix the starting oscillation phase. At $x=0$ both $v=-0.6m/s$ and $v=+0.6m/s$ lie on it; position alone cannot distinguish them. Replacing $v$ by momentum $p=mv$ gives the equivalent phase-space trajectory. The arrows show increasing time, clockwise in $(x,v)$ coordinates.

This closed ellipse assumes a time-independent spring and no damping or applied drive. Adding a resisting force $-bv$, $b>0$, gives $m\ddot{x}+b\dot{x}+kx=0$ and $dE/dt=-b{v}^2\le 0$. An underdamped damped harmonic oscillator spirals inward in the phase plane, crossing successively smaller energy contours. The ideal harmonic oscillator is also useful beyond springs: near a smooth potential energy minimum with $V^{\prime \prime }(0)>0$, its Taylor expansion begins with $V(0)+V^{\prime \prime }(0)x^2/2$. For sufficiently small displacements, this gives the same restoring force to leading order, explaining why small oscillations share one solvable pattern.

Finite optical pumping time lets an atom climb a light-shift hill before transfer into a lower magnetic sublevel. Spontaneous emission dissipates the gained potential energy; climbing alone is conservative.^[ref]

To make the geometry and signs explicit, take an original choice of phase origin for equal opposing waves:The real electric field is $\mathrm{Re}[\mathcal{E}(z)e^{-i{\omega }_{L}t}]$; $\lambda$ is wavelength. The fractions $w_{\pm }$ are squared projections onto the defined circular basis vectors, divided by total intensity. Their sum is one: interference changes electromagnetic polarization despite constant total intensity. At $z=0,\lambda /4,\lambda /2$ the field is respectively ${\mathbf{ϵ}}_{+},{\mathbf{ϵ}}_{-},{\mathbf{ϵ}}_{+}$; halfway between these positions it is linearly polarized. Specifying the vectors avoids an observer-dependent verbal handedness convention.

For an illustrative $J_g=1/2\to J_e=3/2$ angular momentum transition, use the squared Clebsch-Gordan coefficients $1$ and $1/3$ for the strong and weak circular couplings from each lower magnetic sublevel. Interchanging circular components interchanges these strengths. The weighted sums are $w_{+}+w_{-}/3=(2+\cos (2kz))/3$ and $w_{-}+w_{+}/3=(2-\cos (2kz))/3$. Rescaling their common negative shift givesBoth potential energies are negative relative to the unilluminated level. Each hill has height $U_0$, and opposite hills are displaced by $\lambda /4$. An atom starting at the $U_{+}$ minimum at $z=0$ loses $U_0$ of kinetic energy if it reaches $z=\lambda /4$ on that branch. A pump transfer there lowers its potential energy by $U_0$, leaving it at a $U_{-}$ minimum. The next uphill segment ends at $z=\lambda /2$, where the favoured transfer is reversed. Actual transfers are stochastic and need not occur exactly at these extrema.

For one completed climb and pump, conservation of energy gives $\Delta K=-U_0$, while the ground-state potential energy returns to its starting value. Neglecting recoil and treating the scattered photon as the sole outgoing radiation change, $\hslash ({\omega }_{\mathrm{out}}-{\omega }_L)=U_0$. A potential drop is an internal-state reset, not an automatic equal upward kick in kinetic energy. Random emission nevertheless adds momentum fluctuations and recoil energy, so useful cooling requires the dissipated energy to compete with that heating.

The recoil energy $E_R=(\hslash k{)}^2/(2m)$ sets a different scale from $k_B{T}_D=\hslash \Gamma /2$. Conventional Sisyphus cooling reaches the recoil regime when reducing the hill height also weakens cooling; it is not a claim of arbitrarily small temperature at vanishing intensity. Subrecoil techniques need further physics.^[ref]

A magnetic dipole of fixed magnitude $m_B$, free to rotate about an axis with moment of inertia $J$, has potential energy $-m_{B}B\cos \theta$ in a uniform magnetic field. Near stable alignment, $J\ddot{\theta }=-m_{B}B\theta$, so its small-angle period is $2\pi \sqrt{J/(m_{B}B)}$. This neglects damping, bearing friction and any other restoring torque.

A particle constrained to the graph $z=h(x,y)$ has kinetic energy $m[|\dot{\mathbf{r}}{|}^2+(\nabla h\cdot \dot{\mathbf{r}}{)}^2]/2$ and potential energy $mgh$. For a gently varying graph and velocities for which its geometric kinetic correction is higher order, the leading Euler-Lagrange equations are the displayed planar gradient-force equations. Small slope alone at arbitrarily large speed or curvature does not justify discarding the normal inertial acceleration. For a radial height, the leading planar angular momentum is conserved, allowing an effective potential reduction.

For overdamped Brownian motion at uniform temperature in a potential energy $U$, constant diffusion coefficient $D$ gives the displayed Fokker-Planck equation, with $\beta =(k_{B}T{)}^{-1}$. The probability current is $\mathbf{J}=-D(\nabla p+\beta p\nabla U)$. Zero current recovers the Boltzmann distribution; absorbing boundaries instead permit transport and mean first-passage times. This diffusion equation should be distinguished from the similarly named coagulation equation.

For locally linear electron confinement potential energy $U(y)$, two Zeeman splitting branches cross the same Fermi energy at positions differing by the energy splitting divided by $|U^{\prime }|$. The expression neglects self-consistent electrostatic screening and exchange interaction enhancement. If the quoted slope is an electrostatic voltage gradient instead of an electron-energy gradient, its magnitude must be multiplied by the elementary charge. A screened compressible quantum Hall edge strip is not a single sharp crossing with an imposed bare slope.

Attach the table-end mass to a linear spring of spring constant k. Adding spring potential energy $k(e_0+x{)}^2/2$ gives restoring curvature $k-\rho g$ about equilibrium. The positive gravitational feedback from increasing hanging chain length subtracts from the spring stiffness. Stable oscillations require $k>\rho g$; equality is neutral in this ideal linear model, while $k<\rho g$ gives exponential departure. The frequency is independent of the preload, although the equilibrium position is not.

A pendulum bob and a horizontally moving support of equal mass $m$, with support spring spring constant $k$, have the quadratic potential energy $k{x}_1^2/2+mg(x_2-x_1{)}^2/(2l)$. With $\kappa =k/m$, $q=g/l$, the normal mode eigenvectors can be normalized to $(1,r_{\pm })$, where $r_{\pm }=(\kappa /2\mp \sqrt{{\kappa }^2/4+q^2})/q$. The lower mode is in phase and the upper mode in opposite phase. The positive eigenvalues of the symmetric stiffness-per-mass matrix give the displayed squared angular frequencies.

For real scalar fields with quartic potential energy density $V_4=\lambda ({\varphi }_1^4+{\varphi }_2^4)+2\mu {\varphi }_1^2{\varphi }_2^2$, nonnegativity in every direction is equivalent to the displayed conditions. Necessity follows on an axis and on ${\varphi }_1={\varphi }_2$. Sufficiency follows by writing $V_4=\lambda ({\varphi }_1^2-{\varphi }_2^2{)}^2+2(\lambda +\mu ){\varphi }_1^2{\varphi }_2^2$. With a positive quadratic mass term, equality cases still give a stable isolated classical vacuum at the origin. The quartic alone is strictly positive away from the origin precisely when $\lambda >0$ and $\mu >-\lambda$. Rotation symmetry additionally requires $\mu =\lambda$.

The sum of specific enthalpy and macroscopic kinetic energy per unit mass. In a steady adiabatic flow without shaft work, its change is determined by the change in external potential energy. At essentially constant external potential energy, it is conserved through a stationary nonradiating shock wave even though entropy increases.