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For quadratic three-dimensional dispersion below condensation, the excited-state thermal energy is proportional to $T^{5/2}$. Differentiating at fixed particle number and volume gives the displayed heat capacity, with limiting value approximately $1.926N{k}_B$ at $T_c^{-}$. The macroscopic ground occupation carries no excitation energy in the shifted reference. The exponent differs from the linear-phonon result of an interacting gas.

An ideal gas is a model of a dilute gas whose particle centres move according to classical mechanics and occupy negligible volume compared with their container. Treating the particles as point masses removes the molecular-size correction; neglecting interactions between separated particles removes the intermolecular contribution to potential energy. The usual kinetic theory model retains brief elastic collisions with the walls and, optionally, with other particles: these transfer momentum while conserving the total kinetic energy of the colliding bodies. There is no appreciable interaction energy stored between collisions. These assumptions lead to the ideal gas law, $PV=N{k}_{B}T$. ^{[ref] [ref]}

The model's attraction is its universality: at the same temperature and number density, different classical ideal gases have the same pressure, even when their particle masses differ. In ideal gas pressure from wall collisions, heavier particles move more slowly at a given temperature, and the two effects cancel. For example, one mole of any ideal gas at $300\mathrm{K}$ in $0.025{\mathrm{m}}^3$ has pressure approximately $99.774\mathrm{kPa}$, as computed under the ideal gas law.

The ideal gas law describes the translational contribution to pressure; it does not specify all of the internal energy or the heat capacity. A monatomic gas has three classical translational degrees of freedom, whereas a gas of molecules can also store rotational kinetic energy and energy in vibrational normal modes. An equation of state alone therefore does not determine how much heat a chosen temperature increase requires. It also does not select a process: isothermal expansion and adiabatic expansion need different additional constraints. ^[ref]

A real gas departs from this approximation when particle size or interactions become important, especially at high number density or low temperature near a phase transition. The Van der Waals equation adds such corrections. Independently, at sufficiently low temperature or high number density, quantum statistics can become essential: even particles with negligible interactions need not obey the classical ideal gas law. The classical statistical assumption is the nondegenerate gas regime, which is distinct from neglecting interactions. ^{[ref] [ref]}

For a fixed Newtonian gravitational potential, total energy density is $\mathcal{E}=\rho (u^2/2+\Phi )+p/(\gamma -1)+B^2/(2{\mu }_0)$. Its flux combines advected kinetic/potential energy, specific enthalpy and the Poynting vector. Dotting the ideal magnetohydrodynamic momentum equation with velocity, adding the adiabatic internal energy equation and the magnetic energy equation cancels the magnetic work. The pressure terms combine into $-\nabla \cdot (p\mathbf{u})$, proving the conservation law. A time-dependent imposed potential instead contributes $\rho {\partial }_t\Phi$.

The smooth ideal-MHD system couples mass conservation, momentum, adiabatic pressure transport and the ideal magnetohydrodynamic induction equation, subject to the solenoidal magnetic-field constraint. Without gravity it is a quasilinear partial differential equation system. Across shocks, conservative total energy replaces smooth adiabatic pressure transport.

The ideal magnetohydrodynamic shock conditions are the Rankine-Hugoniot jump conditions for mass, momentum, energy, and magnetic flux. For a dynamically weak magnetic field, the normal field is continuous while the tangential field is multiplied by the gas compression ratio.

If an energy per site $C$ multiplies the identity operator, blocking sends $C$ to $b^{D}C+c(u)$. For free energy per site $F=C+f(u)$, exact partition function invariance gives the displayed inhomogeneous equation with $g=b^{-D}c$. It records eliminated-mode entropy, vacuum contributions and measure normalization. Removing a suitable regular background leaves homogeneous singular scaling; additive logarithmic resonances can obstruct a strictly analytic subtraction.

An idler photon is the partner of a signal photon in a specified nonlinear optical process. In spontaneous parametric down-conversion, one pump photon supplies the energy of the pair, $\hslash {\omega }_p=\hslash {\omega }_s+\hslash {\omega }_i$.

The idler can be measured to herald a signal photon or to condition a correlation measurement. It is not a physically expendable photon: losses in its channel reduce heralding rates and can bias an uncorrected coincidence sample. When the two output modes have the same frequency, the signal/idler distinction comes from their assigned spatial, polarization or experimental roles rather than from energy ordering.

Physical imaginary time runs over $[0,\beta \hslash ]$ with an action divided by $\hslash$. Rescaling it by $\hslash$ gives a variable in $[0,\beta ]$ and the operator ${\partial }_{\tau }+\epsilon$. Matsubara eigenvalues in the latter variable have energy units.

A voltage gain determines a power gain only when the electrical conditions at both ports are included. The familiar equality between $20{\log }_{10}|A_v|$ and $10{\log }_{10}{G}_P$ requires a particular impedance relation; it is not a conversion between arbitrary voltage and power ratios.

Use phasors whose moduli are root mean square amplitudes, at one sinusoidal frequency. At a load with electrical impedance $Z$ and electrical admittance $Y=1/Z$, the current is $I=YV$. Its complex electrical power is $V{I}^{\ast }$, with the complex conjugate arising from the relative phase of the two sinusoids. Averaging $v(t)i(t)$ over a cycle gives the active powerTo see the averaging step, write $v(t)=\sqrt{2}|V|\cos (\omega t+{\varphi }_V)$ and similarly for $i(t)$. The product's cycle average is $|V||I|\cos ({\varphi }_V-{\varphi }_I)$, the real part of $V{I}^{\ast }$. Peak-amplitude phasors instead require an extra factor $1/2$.

Let $Y_{\mathrm{in}}$ describe the actual input impedance and $Y_L$ the actual load impedance. Take $\mathrm{Re}{Y}_{\mathrm{in}}>0$ and $\mathrm{Re}{Y}_L>0$ with nonzero terminal signals, and define currents into the amplifier at the input and into the load at the output. With the measured loaded voltage gain $A_v=V_L/V_{\mathrm{in}}$,Thus $G_P=|A_v{|}^2$ requires equality of the relevant electrical conductances, not merely equality of impedance moduli. For example, input $Z_{\mathrm{in}}=50\Omega$ and load $Z_L=(30+40i)\Omega$ both have modulus $50\Omega$, but their electrical admittances have real parts $0.020$ and $0.012\mathrm{S}$. The multiplier is $0.6$, so $G_P=0.6|A_v{|}^2$.

For real resistive ports this becomesTake an original example with measured $|A_v|=0.5$, $R_{\mathrm{in}}=10\mathrm{k\Omega }$ and $R_L=50\Omega$. The voltage gain is $-6.02\mathrm{dB}$, while the power gain is $50$, or $16.99\mathrm{dB}$. An active buffer can reduce the signal's electric potential difference while delivering more signal power than it accepts at its input; the supply provides the additional energy.

The amplifier's output impedance is distinct from the connected load impedance. To show why that distinction matters, model its output by a Thévenin equivalent with open-circuit voltage $A_0{V}_{\mathrm{in}}$ and series resistance $R_o$. The voltage divider givesWith $A_0=0.6$ and $R_o=10\Omega$, the $50\Omega$ load produces the preceding $A_v=0.5$ example. Differentiating $R_L/(R_o+R_L{)}^2$ gives $(R_o-R_L)/(R_o+R_L{)}^3$, so delivered power is largest at $R_L=R_o$, as in the maximum power transfer theorem. Here its power gain is $90$. Increasing $R_L$ much further makes $A_v$ approach $A_0$ but reduces delivered power. That high-load-impedance regime is impedance bridging, useful for sensing the source's voltage with little loading.

A bench demonstration uses a function generator, a known termination and an oscilloscope. The Keysight guide explains that an expected-load setting changes how amplitude is displayed for generators with a fixed series output impedance; a high-impedance load can receive twice the voltage expected for a matched $50\Omega$ termination.^[ref] As an independent low-frequency resistive calculation, an open-circuit source of $1\mathrm{V}$ root mean square with $50\Omega$ series resistance delivers $0.5\mathrm{V}$ and $5\mathrm{mW}$ to $50\Omega$. Into $1\mathrm{M\Omega }$ it delivers about $0.99995\mathrm{V}$ but only $1.00\mu \mathrm{W}$. The larger terminal reading accompanies much less delivered power. Specify the actual load, frequency and amplitude convention when comparing gain measurements.

The imperfect soft Duffing-van der Pol oscillatorcombines a softening restoring force, velocity-dependent damping and the symmetry-breaking term $2\epsilon v^2$. The equilibrium points are $v=0$ and $v=-\epsilon \pm \sqrt{{\epsilon }^2+\lambda }$, whenever the latter are real. The associated Jacobian matrix has trace $\kappa -v_{\ast }^2$ and determinant $\lambda -4\epsilon v_{\ast }-3v_{\ast }^2$.

At $\epsilon =0$, the equilibrium equation has the reflection symmetry $v\mapsto -v$. For $\epsilon \ne 0$, the branch near zero crosses the zero branch at $\lambda =0$, giving the generic transcritical bifurcation when $\kappa \ne 0$. The two nonzero roots meet at $\lambda =-{\epsilon }^2$, $v=-\epsilon$, giving a generic saddle-node bifurcation when $\kappa \ne {\epsilon }^2$. The excluded trace-zero cases have an additional linear degeneracy and require further analysis.

The mechanical energyhas the exact derivative $\dot{H}=(\kappa -v^2){\dot{v}}^2$. Thus the sign of energy input can change with displacement even at a fixed value of $\kappa$.

For $\lambda >0$, the two outer equilibria are saddle equilibria. Their potential barriers are generally unequal when $\epsilon \ne 0$, so a symmetric saddle connection can split into distinct directed connections under an imperfection. The weak-damping heteroclinic splitting of a tilted quartic oscillator calculates their leading parameter curves. Local equilibrium formulas alone do not guarantee a limit cycle or determine its global termination.

An implicit solution describes a solution of a differential equation through a relation $F(x,y)=C$, instead of isolating $y$ in an explicit formula. The relation describes a level set; an actual scalar solution of a differential equation must be a differentiable function $y(x)$ on an interval whose function graph lies in the equation's domain. A whole level set need not be one such function graph.

For example, considerImplicit differentiation of $x^2+y^2=C$ gives $2x+2y{y}^{\prime }=0$. For $C>0$, the circle contains two solutions of a differential equation,Both satisfy the equation on the stated open interval, and their function graphs stay in $D$. An initial condition $(x_0,y_0)\in D$ fixes $C=x_0^2+y_0^2$ and selects the sign of $y_0$. In particular, $y(0)=2$ gives $C=4$ and the upper branch on $(-2,2)$. The constant alone does not select the branch.^[ref]

The precise local criterion uses the implicit function theorem. Let $F$ be continuously differentiable on an open subset of the plane, with $F(x_0,y_0)=C$. If its partial derivative $F_y(x_0,y_0)\ne 0$, the nearby level set is a unique continuously differentiable function graph $y=g(x)$ through that point. The chain rule givesThis proves that the local function graph solves $y^{\prime }=-F_x/F_y$ there. For a different prescribed equation $y^{\prime }=f(x,y)$, one must also check $F_x+F_{y}f=0$ along that branch and that it stays in the domain of $f$. A level set by itself does not verify an arbitrary ordinary differential equation.^[ref]

For $F=x^2+y^2$, the upper branch through $(1,\sqrt{3})$ has $F_y=2\sqrt{3}$ and tangent line slope $-1/\sqrt{3}$. At $(2,0)$, however, $F_y=0$ and $F_x=4$. The circle remains a smooth regular level set with a vertical tangent line: it can be written locally as $x=\sqrt{4-y^2}$, but not as a single function graph $y=g(x)$ over an open interval containing $2$. Moreover, $-x/y$ is undefined there, so the scalar solution of a differential equation cannot include this point in its domain. Changing the parameter draws the curve through the point but does not extend that scalar initial value problem.

When both partial derivatives vanish, even the geometry needs separate inspection. The level $x^2+y^2=0$ is an isolated point, not a function graph on any nonempty open interval. The zero level set of $G(x,y)=y^2-x^2$ instead consists of the two lines $y=x$ and $y=-x$ crossing at the origin, where $\nabla G=0$. Each line satisfies the implicit ordinary differential equation $y{y}^{\prime }=x$, including at the origin; the explicit form $y^{\prime }=x/y$ excludes the origin. Thus algebraically dividing by a vanishing coefficient changes the domain and can change which solutions of a differential equation are admissible. Vanishing $F_y$ alone is only failure of a sufficient test: $F(x,y)=y^3$ has the smooth zero function graph $y=0$ despite $F_y=0$ everywhere on it.

Implicit solutions are especially useful for exact first-order ordinary differential equations and for finding a first integral of an ordinary differential equation: a conserved relation can reveal allowed states without an explicit formula. For the normalized harmonic oscillator written as $\dot{x}=y$, $\dot{y}=-x$, the energy $H=(x^2+y^2)/2$ has $\dot{H}=x\dot{x}+y\dot{y}=0$. Its positive level sets are the same circles. Where $y\ne 0$, eliminating time gives $dy/dx=-x/y$ and the local implicit solution above. The harmonic oscillator's parametric curve continues through $y=0$ because its vector field governing evolution in time is defined there; treating $x$ as the independent variable fails at that turn. The origin is an equilibrium point of the harmonic oscillator, although it is excluded from the scalar equation $y^{\prime }=-x/y$.

For a complete entry and exit of a small square loop, integrating the drag over two distances $a$ gives impulse magnitude $2B^2{a}^3/R$. This is independent of speed in the ideal sharp-edge, zero-inductance model, though dissipated energy is not.

Take an initially isolated virialized self-gravitating stellar population with kinetic energy $T=|W|/2$ and Newtonian gravitational potential energy $W<0$. Instantaneous spatially uniform removal of mass, without kicks to surviving material, leaves fraction $f$ with $T^{\prime }=fT$ and $W^{\prime }=f^{2}W$. Hence $E^{\prime }=|W|f(1/2-f)$: the homologous whole-system energy is negative only for $f>1/2$. Positive whole-system energy does not forbid a bound core after selective escape, and slow mass loss follows the adiabatic mass-loss expansion law instead.

An apolar axis model with pair energy $-({\mathbf{s}}_i\cdot {\mathbf{s}}_j{)}^2$ is invariant when each molecular representative is independently reversed. The vector average changes under these replacements, but the nematic order parameter made from second moments does not. In an unbiased statistical ensemble the vector average vanishes; an individual finite representative need not have zero average. For independently chosen signs of a perfectly aligned axis, $⟨|N^{-1}{\sum }_i{\mathbf{s}}_i{|}^2⟩=1/N$. This distinguishes exact head-tail invariance of an observable from vanishing only after an average.^[ref]

Consider a stipulated dilute collection of independent two-level systems, each with longitudinal magnetic moment $s\mu$, $s=\pm 1$, and energy $E_s=-s\mu B_0$. The prescribed vacuum/applied magnetic flux density is $B_0={\mu }_0{H}_{\mathrm{app}}$. Ignore interactions, orbital magnetic response, and the small sample-field correction. This is the magnetic version of the existing independent spin-one-half two-level system, with alignment probabilities rather than an interacting Curie–Weiss model.

For temperature $T>0$, the canonical ensemble gives the one-moment partition function $Z_1=2\cosh x$, with $x=\mu B_0/(k_{B}T)$. Taking the expected value over its two Boltzmann weights givesHere $n$ is the number density, $k_B$ the Boltzmann constant, and $\tanh$ the hyperbolic tangent. For $|x|\ll 1$, its small-argument expansion of the hyperbolic tangent gives $M\simeq {\mu }_{0}n{\mu }^2{H}_{\mathrm{app}}/(k_{B}T)$, a Curie's law response in the applied-$H$ convention. For large $|x|$, $M$ approaches the saturation magnetization $n\mu$ with sign set by the field, and its differential magnetic susceptibility falls towards zero.

Both plots use stipulated $n=10^{26}{\mathrm{m}}^{-3}$, $\mu =9.2740101\times 10^{-24}\mathrm{J}{\mathrm{T}}^{-1}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, and ${\mu }_0=1.25663706\times 10^{-6}\mathrm{T}\mathrm{m}{\mathrm{A}}^{-1}$. The moment is approximately one Bohr magneton; the chosen vacuum permeability is a rounded model input. At $T=2\mathrm{K}$, $M_s=927.401\mathrm{A}{\mathrm{m}}^{-1}$ and ${\chi }_{\mathrm{model}}(0)=3.91410\times 10^{-4}$. The linear approximation overestimates $M$ by less than $3\%$ for $|x|\le 0.3$, corresponding to $|{\mu }_0{H}_{\mathrm{app}}|\le 0.89324\mathrm{T}$ or $|H_{\mathrm{app}}|\le 0.71082\mathrm{MA}{\mathrm{m}}^{-1}$. These parameters keep the modeled response weak; neglected magnetic feedback is of relative order ${\chi }_{\mathrm{model}}(0)$, rather than an explanation of a material's ordered phase.

At $B_0=0$, cooling increases the zero-field slope as $1/T$. At fixed $B_0\ne 0$, cooling instead gives ${\chi }_{\mathrm{model}}\sim 4{\mu }_{0}n{\mu }^2{e}^{-2\mu |B_0|/(k_{B}T)}/(k_{B}T)$: a saturated moment can hardly change further. The order of the zero-field and zero-temperature limits matters. This independent model has no finite-temperature collective phase transition; the diverging zero-field slope alone does not establish one.

Before local expulsion, a linear star-formation law $\psi =\nu {\rho }_g$ without return gives ${\rho }_g={\rho }_0{e}^{-\nu t}$. If $q=\gamma E_{\mathrm{SN}}$ is the energy supplied per unit stellar mass, accumulated energy is $q{\rho }_0(1-e^{-\nu t})$. A fixed binding energy per unit remaining gas mass $B(r)$ gives the threshold $q(e^{\nu t}-1)=B(r)$. This assumes a specified retained gravitational profile and neglects exchange between volumes.

Magnetic moments can interact through intermediate orbitals or itinerant carriers rather than direct overlap alone. Superexchange uses virtual charge transfer, the RKKY interaction uses conduction-electron spin response, and double exchange gains carrier kinetic energy through spin-dependent hopping. More than one mechanism can operate in the same solid, with different signs and ranges.

For a linear isotropic unsaturated effective susceptibility contrast $\Delta \chi$, the induced-moment energy is $-V\Delta \chi B^2/(2{\mu }_0)$, giving $\mathbf{F}=V\Delta \chi \nabla B^2/(2{\mu }_0)$. The one-half differs from the fixed permanent-dipole energy because the applied field creates the moment. Demagnetization and saturation are absorbed into or replace the assumed response model.

Inductance relates electric currents to the magnetic flux linkage they produce. For a single linear coil with fixed geometry and magnetic response, its self-inductance $L$ is the constant inHere $\Lambda$ is magnetic flux linkage from the coil's own electric current $I$. The SI unit of inductance is the henry. Knowing inductance lets us predict the induced electromotive force when electric current changes, and the magnetic energy stored by an inductor. ^[ref].

For example, a coil of self-inductance $0.5\mathrm{H}$ carrying electric current $3\mathrm{A}$ has magnetic flux linkage $1.5\mathrm{Wb}$. If its $100$ turns link the same magnetic flux, each turn has magnetic flux $0.015\mathrm{Wb}$. Keeping the geometry and magnetic response fixed while doubling electric current doubles magnetic flux linkage and quadruples the inductor's magnetic energy.

The comparison with capacitance is useful: a linear capacitor has $Q=CV$ and stores capacitor energy $C{V}^2/2$ in its electric field, whereas a linear inductor has $\Lambda =LI$ and stores magnetic energy $L{I}^2/2$ in its magnetic field. The RLC circuit brings both forms of energy storage together.

An inductively coupled plasma receives power through electromagnetic induction. An alternating coil electric current produces a changing magnetic field, which induces an electric field in the gas. Electrons accelerated by that field transfer energy through collisions, sustaining ionization and driving atomic transitions.

A typical analytical source surrounds a gas-carrying dielectric torch with a radiofrequency coil. Excited atoms emit spectral lines useful for atomic spectroscopy, while extracted ions can feed mass spectrometry. The coil supplies energy without requiring it to touch the plasma. Coupling, gas composition and particle losses determine the discharge response. ^[ref]