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For a uniform ideal gas sphere, set $\rho =mn$, $p=n{k}_{B}T$ and $M=4\pi \rho R^3/3$. Building the sphere by concentric shells gives gravitational potential energy $W=-3G{M}^2/(5R)$. During a small homologous contraction, released gravitational energy balances the work against gas pressure when $3G{M}^2|dR|/(5R^2)=4\pi R^{2}p|dR|$, or $GMm/R=5k_{B}T$. Eliminating $R$ gives the displayed mass. At the same initial mass density and temperature, larger masses have stronger gravitational contraction than pressure support in this estimate. Its numerical coefficient belongs to the uniform-sphere pressure-work model; the infinite-medium Jeans instability and a pressure-confined isothermal sphere use different equilibrium and boundary assumptions.

A time-independent Hamiltonian evolves a state by the unitary operator $U(t)=e^{-iHt/\hslash }$.

For a closed quantum system, unitary time evolution changes a quantum state through a unitary operator, preserving norms and inner products. For a time-independent self-adjoint Hamiltonian operator $H$, choose the initial time as zero:Here $\hslash =h/(2\pi )$ is the reduced Planck constant. The same $U(t)$ acts on every initial quantum state; it depends on the Hamiltonian operator and elapsed time, rather than the chosen preparation.^[ref]

The self-adjointness of $H$ makes the exponential a unitary operator. In finite dimension, conjugating its matrix exponential givesThe two exponents commute because both are multiples of $H$. Hence, for any two initial vectors,This short proof explains why normalization survives, why orthogonality of initial vectors survives, and why the transition probability between two states evolved by the same $U$ is unchanged. It does not require that a fixed measurement in quantum mechanics have time-independent probabilities: its measurement basis need not move with the quantum states.

For an unbounded self-adjoint operator, define the exponential through the spectral theorem, rather than assuming that its power series acts on every vector. It is a unitary operator on the entire Hilbert space; the strong Schrödinger equation holds for initial vectors in the operator domain $D(H)$. For a density operator, unitary time evolution reads $\rho (t)=U(t)\rho (0)U(t{)}^{†}$. It preserves eigenvalues and purity of a density operator, but can change its entries in a fixed basis.

The useful distinction is between conserving total probability and conserving each measurement's probabilities. A fixed energy spectrum supplies different ticking rates for different probability amplitudes. Their relative quantum phases can move the Bloch vector and create quantum interference even while every energy eigenstate population stays fixed. Qubit precession under a constant Hamiltonian makes this visible; Ramsey interferometry converts the accumulated relative quantum phase into a population difference that a simple readout can detect.

In a uniform U-tube containing incompressible liquid of density $\rho$, cross section $A$ and total moving length $l$, let one surface fall by $y$ while the other rises by $y$. The height difference is $2y$, and potential energy relative to equilibrium is $U=\rho gA{y}^2$. The whole moving mass is $\rho Al$, so $K=\rho Al{\dot{y}}^2/2$. Newton's second law or energy conservation gives $\ddot{y}+(2g/l)y=0$ and ${\omega }_0=\sqrt{2g/l}$. This assumes small excursions that keep the free surfaces in the uniform vertical limbs.

The VSEPR model predicts approximate molecular geometry by arranging the electron domains around a central chemical atom to reduce their mutual repulsion. Start with a Lewis structure, count the domains, and then distinguish the resulting electron pair geometry from the positions of the chemical atoms themselves. A lone pair counts as a domain even though it is not another chemical atom.^[ref]

Three worked counts have the same total of eight valence electrons. In methane, carbon's four plus the contributions of four hydrogen chemical atoms give $4+4=8$: four bond pairs, no lone pairs, and tetrahedral molecular geometry. In Ammonia, nitrogen's five plus three from hydrogen give $5+3=8$: three bond pairs and one lone pair, producing trigonal pyramidal molecular geometry. In water, oxygen's six plus two from hydrogen give $6+2=8$: two bond pairs and two lone pairs, producing bent molecular geometry. All three have four electron domains; their molecular geometries differ.

The regular tetrahedral directions can be represented by the unit vectorsEach pair has dot product $-1/3$, so the ideal bond angle is $\theta =\arccos (-1/3)\approx 109.47^{\circ }$. This is a geometric calculation, rather than a prediction that every four-domain molecule has that exact bond angle. Lone pairs and different kinds of chemical bonds distort the ideal arrangement; water's H–O–H angle is about $104.5^{\circ }$.^[ref]

The model is attractive because small-number bookkeeping produces a spatial prediction without first solving for every molecular orbital. Shape then helps explain chemical polarity and which hydrogen bond donor or hydrogen bond acceptor sites are accessible. It is a qualitative model, not a full calculation of the electronic energy or a universal solution for every chemical compound. Gillespie and Nyholm's 1957 review ^[ref] is a historical anchor for its development.

The number of symmetry-related equivalent electronic valleys contributing to a specified low-energy energy band. It multiplies the single-valley density of states, independently of spin degeneracy. Equal electron and hole effective masses alone do not ensure equal total density of states values unless their valley degeneracies and spin degeneracies also agree.

The only zeros of $f$, modulo the reciprocal lattice, are the two Dirac valleys in graphene,At these momenta ${\sum }_j\sin ({\mathbf{K}}_{\tau }\cdot {\mathbf{b}}_j)=\tau 3\sqrt{3}/2$. Thus the two Dirac mass coefficients, measured in energy units, are $m_{\tau }=M-\tau \mu$, with $\mu =3\sqrt{3}{t}_2\sin \varphi$. For $\mathbf{k}={\mathbf{K}}_{\tau }+\mathbf{q}$,The two Dirac cones have opposite orientation, encoded by $\tau$. A scalar sublattice offset $M$ gives equal-sign Dirac masses; imaginary second-neighbour hopping at $M=0$ gives opposite-sign Dirac masses. Both mechanisms open valley gapsAn observed band gap alone therefore cannot establish a Chern insulator. In particular, the gapless nearest-neighbour model of graphene and its sublattice-mass deformation should not be confused with the topologically nontrivial Haldane model.

A band gap closes when all three components of $\mathbf{d}$ vanish. Since $d_x=d_y=0$ only at the two valleys, the phase boundaries are exactly $M=\pm 3\sqrt{3}{t}_2\sin \varphi$. Usually one valley becomes massless; at $M=0$, $\sin \varphi =0$, both do. The occupied-band Chern number is undefined at every such touching, even if a numerical momentum grid misses its location.

A hypothetical Van der Waals equation model illustrates positive and negative Joule-Thomson coefficient regions and a finite throttle that first warms, then cools. Use the molar convention already distinguished in Van der Waals equation:Here $v_{\mathrm{m}}$ is molar volume, $R$ the molar gas constant, $u_{\mathrm{m}}$ internal energy per mole, and $C_{V,\mathrm{m}}$ the stipulated constant molar heat capacity at fixed volume. The second equation is an additional caloric assumption: an equation of state alone does not specify the temperature dependence of internal energy. It satisfies the equilibrium identity $(\partial u_{\mathrm{m}}/\partial v_{\mathrm{m}}{)}_T=T(\partial p/\partial T{)}_{v_{\mathrm{m}}}-p=a_{\mathrm{m}}/v_{\mathrm{m}}^2$. The arbitrary energy reference is set to zero.

Define dimensionless variables $r=v_{\mathrm{m}}/b_{\mathrm{m}}$, $\theta =R{b}_{\mathrm{m}}T/a_{\mathrm{m}}$, $\pi =b_{\mathrm{m}}^{2}p/a_{\mathrm{m}}$ and $\eta =b_{\mathrm{m}}{h}_{\mathrm{m}}/a_{\mathrm{m}}$, where $h_{\mathrm{m}}$ is molar enthalpy. These differ from the critical-value scaling in Van der Waals law of corresponding states. The throttle calculation reduces toThus a prescribed inlet fixes $\eta$, and solving the first and last equations at the outlet pressure gives the final temperature. For an inlet $({\theta }_1,{\pi }_1)=(0.8,0.12)$ and outlet ${\pi }_2=0.025$, the model gives ${\theta }_2\simeq 0.752287$, a decrease of $0.047713$. The inlet $(2.2,0.12)$ instead gives ${\theta }_2\simeq 2.205786$, an increase of $0.005786$. The purple inlet $(1.45,0.30)$ initially warms but ends at ${\theta }_2\simeq 1.437585$ for the same outlet pressure. These are reduced toy values, not kelvin readings or experimental evidence. No physical gas parameters were fitted.

For a short inversion derivation, implicit differentiation of the Van der Waals equation at fixed pressure givesInsert this into $T(\partial v_{\mathrm{m}}/\partial T{)}_p=v_{\mathrm{m}}$ and substitute back into the equation of state. The Joule-Thomson inversion curve isPositive inversion pressure requires $r>3/2$. Its maximum pressure is $\pi =1/3$ at $r=3$, while the upper dilute limit is ${\theta }_{\mathrm{inv}}\to 2$. Using Van der Waals critical point values, the latter is $T_{\mathrm{inv},\max }/T_c=27/4$. The inversion locus depends on the equation of state; the chosen molar heat capacity additionally determines how much temperature changes along each finite throttle.^[ref]

The displayed region has $\theta >8/27$, above the model's critical temperature, and uses only $r>1$, $p>0$, $(\partial p/\partial v_{\mathrm{m}}{)}_T<0$ and positive $C_{p,\mathrm{m}}$. Subcritical calculations require attention to phase coexistence, metastability and the Maxwell construction; an unstable homogeneous continuation is not an equilibrium throttle endpoint. Even on a stable branch, this mean-field theory with constant $C_{V,\mathrm{m}}$ is a qualitative model, not an accurate inversion map for a particular real gas. Phase transitions, changing composition, appreciable heat leakage or bulk kinetic energy changes require a different property calculation or steady-flow energy balance. At phase coexistence, the smooth single-thermodynamic phase coefficient formula cannot replace a balance involving the phase fraction.

The variational method estimates the lowest energy of a Quantum Hamiltonian by choosing a trial wavefunction, evaluating its energy, and optimizing the trial parameters. It uses the quantum variational principle, the Rayleigh-Ritz variational principle expressed for quantum states. Let $H$ be a self-adjoint operator bounded below, and define $E_0=\inf \sigma (H)$, the bottom of its spectrum. Every normalized admissible trial quantum state satisfiesWhen $E_0$ is an energy eigenvalue, it is the ground-state energy; equality holds precisely for a quantum state in the ground-state subspace. Otherwise the bottom of the spectrum need not be attained by any normalizable wavefunction.

Admissible means finite form energy. Choose a real $b\le E_0$ and use the shifted quadratic form of a positive quantum Hamiltonian:The form domain $Q(H)$ is independent of this choice of $b$ and can be larger than the operator domain $D(H)$. For $\psi \in D(H)$, $E_{\psi }=⟨\psi ,H\psi ⟩$ is the usual expectation value; for nonzero unnormalized trials, divide $q_H[\psi ]$ by $\Vert \psi {\Vert }^2$ to get the form version of the Rayleigh quotient. See the quantum variational principle for the general domain justification.

Thus each trial energy is an upper bound on $E_0$, and minimizing within a trial family improves that upper bound. It is not a lower bound on $E_0$. The Gaussian variational energy for the harmonic oscillator reaches $E_0=1/2$ because its family contains an exact ground state. The exact variational solution in a two-level system makes the equality condition visible as vanishing excited-state probability amplitude.

The method is useful when solving the Schrödinger equation exactly is difficult but integrals of a trial wavefunction are tractable. Enlarging a trial family cannot increase its optimal energy, although an optimum need not be attained. The beauty is that even an imperfect trial gives a rigorous bound, provided its form energy is computed exactly: numerical integration errors require separate control. The Gaussian variational estimate for a solvable sextic potential provides an existing example beyond a quantum harmonic oscillator.

A short proof explains the direction. With a complete orthonormal eigenbasis of energy eigenstates, write $\psi ={\sum }_n{c}_n{\varphi }_n$, where $H{\varphi }_n=E_n{\varphi }_n$ and ${\sum }_n|c_n{|}^2=1$. ThenThis is a weighted mean of nonnegative energy differences; equality requires every occupied energy eigenstate to have energy $E_0$. ^[ref]. For continuous spectrum, the spectral theorem replaces the sum by an integral against the scalar spectral measure of $\psi$, with the same nonnegative integrand. This proves the bound without assuming an orthonormal eigenbasis exists.

The historical setting predates quantum mechanics: Rayleigh's The Theory of Sound, volume I, appeared in 1877, and Ritz's paper on solving variational problems of mathematical physics appeared in volume 135 of Crelle's journal, catalogued by its publisher as 1909, pp. 1–61. These are original works associated with the Rayleigh-Ritz method; the dates here identify the publications rather than assert priority. ^[ref]; ^[ref].

Change in vascular volume per change in distending pressure; it describes storage rather than an independent source of energy.

For fixed cloud mass $M$ and isothermal sound speed $c_s$, approximate the binding energy by $-3G{M}^2/(5r)$. The boundary-pressure virial theorem gives $A=3M{c}_s^2/(4\pi )$ and $B=3G{M}^2/(20\pi )$. The required external pressure peaks at $r_c=4B/(3A)$. A homologous radial displacement is restoring on the branch $d{p}_{\mathrm{ext}}/dr<0$, or $r>r_c$; the smaller-radius positive-pressure branch is unstable. This is a one-radius virial model, not the exact hydrostatic isothermal-sphere sequence.

The virial theorem relates the average value of a function along a trajectory to the forces that bind it. For an isolated system of particles interacting through Newtonian gravitational potential energy, it gives $2⟨K⟩+⟨U⟩=0$: the average kinetic energy is half the magnitude of the average negative potential energy. An eccentric Kepler orbit makes the distinction vivid: kinetic energy and potential energy vary throughout the orbit, although their sum is constant. The relation concerns their time averages, rather than their values at every instant. It underlies the virial mass estimate and the negative heat capacity of a virialized gravitational system.

Use constant positive particle masses $m_i$, positions ${\mathbf{r}}_i$, mechanical momenta ${\mathbf{p}}_i=m_i{\dot{\mathbf{r}}}_i$, and total kinetic energy $K={\sum }_i{\mathbf{p}}_i^2/(2m_i)$. The auxiliary quantity $G(t)$ is a sum of dot products, not the gravitational constant, which we denote here by $G_N$:This finite-$T$ identity is exact wherever Newton's second law holds along a regular trajectory; ${\mathbf{F}}_i$ must include every force. The endpoint term is what a finite observation window leaves behind. For the infinite-time theorem, suppose the averages on the left exist and $(G(T)-G(0))/T\to 0$. A sufficient condition is uniformly bounded positions and velocities for finitely many particles. More generally $G(T)=o(T)$ suffices, even without those individual bounds. Bounded positions alone do not supply a bound on momenta, especially near a singular potential energy; nor does a small endpoint term prove thermodynamic equilibrium. Periodic motion with $G(T)=G(0)$ gives the identity without an endpoint term at that finite $T$. For an isolated system, use an inertial centre of mass frame so translational drift does not contribute bulk kinetic energy.^[ref]

For a differentiable potential energy that is a homogeneous function of degree $q$, choose its additive constant so that the scaling law really holds. With conservative forces, the Euler theorem for homogeneous functions givesThus a harmonic oscillator has $q=2$ and $⟨K⟩=⟨U⟩$. An isolated system with Newtonian gravitational potential energy hasEach interacting pair occurs once, and $U=0$ at infinite separation. Here the total energy $E=K+U$ is conserved. An external field or a mixture of different scaling powers requires its own force term; one cannot assign the whole potential energy a single degree $q$. The virial theorem for a homogeneous pair potential states the corresponding general energy relation.

Consider the exact two-body problem with masses $m_1,m_2$. In the inertial centre of mass frame, its internal kinetic energy is $K=m|\dot{\mathbf{r}}{|}^2/2$, where $\mathbf{r}$ is their relative position, $m=m_1{m}_2/(m_1+m_2)$ is the reduced mass, and $U=-\kappa /r$ with $\kappa =G_N{m}_1{m}_2$. Take a noncollision elliptic Kepler orbit with semi-major axis $a$ and orbital eccentricity $0<e<1$. Write $\eta$ for eccentric anomaly, to avoid confusing it with total energy $E$. With mean motion $n=\sqrt{\kappa /(m{a}^3)}$ and orbital period $P=2\pi /n$, Kepler's equation and the orbit geometry giveThe time origin is periapsis. Since $dt=(1-e\cos \eta )d\eta /n$, the time average of $1/r$ over one orbital period is exactly $1/a$. Therefore $⟨K{⟩}_P=\kappa /(2a)$ and $⟨U{⟩}_P=-\kappa /a$, independently of orbital eccentricity. Equal steps in eccentric anomaly are not equal time steps; unweighted averaging of a drawing of the ellipse gives the wrong answer.

The original dimensionless example in both diagrams has $m=a=\kappa =1$ and $e=0.6$. At periapsis, $r=0.4$, $K=2$, $U=-2.5$; at apoapsis, $r=1.6$, $K=0.125$, $U=-0.625$. Both have $E=-0.5$, but $2K+U$ is respectively $1.5$ and $-0.375$. Only the orbital period averages are $⟨K{⟩}_P=0.5$ and $⟨U{⟩}_P=-1$. The instantaneous zero of $2K+U$ occurs at $r=a$, not at every point of the orbit.

The endpoint term can also be evaluated exactly: $G=m\mathbf{r}\cdot \dot{\mathbf{r}}=mn{a}^{2}e\sin \eta$. Starting at periapsis, a window ending at eccentric anomaly $\eta$ hasAt $T=P/4$, Kepler's equation gives $\eta \simeq 2.09133$ and a positive residual $\simeq 0.3314\kappa /a$. At $T=P/2$ or $P$, $G$ returns to zero and the residual vanishes exactly. The lower panel shows the running residual and its endpoint expression coinciding; the shaded quarter-orbital period window explains why a short observation need not satisfy the infinite-time theorem.

A hopping process through an intermediate configuration of different unperturbed energy, contributing to a low-energy effective description through perturbation theory. If an intermediate state costs $\Delta >0$, second-order nondegenerate perturbation theory gives a contribution $-|⟨a|T|0⟩{|}^2/\Delta$. Its amplitude is of order $t/\Delta$ and its energy correction of order $-t^2/\Delta$; a degenerate low-energy sector requires the corresponding degenerate perturbation theory before identifying the ground state.

An intermediate contribution to a second-order transition amplitude need not be an occupied, on-shell stationary atomic state. Both time orderings must be summed with their energy denominators. Large detuning can suppress real intermediate occupation while preserving a coherent two-photon coupling. Near resonance, finite linewidth, saturation and sequential population transfer must be treated explicitly.

A channel supporting ballistic electron transport carries incident populations set by different source and drain chemical potentials, rather than an everywhere-equilibrated local distribution with an Ohmic voltage gradient. Its electrostatic barrier responds self-consistently to contacts, gates and electrostatic screening. A symmetric rigid-barrier model places reservoir energies at $E_F\pm eV/2$ relative to the barrier. Asymmetric partition or bias-dependent transmission changes the weights in the differential electrical conductance, and must be included when extracting an energy scale.

For a Newtonian fluid and a flat wall with $0<b<\infty$, the Navier slip boundary condition is equivalent to a tangential wall-on-fluid tractionThe sign expresses resisting friction: ${\partial }_{n}u$ uses the inward unit normal, whereas the wall-on-fluid traction uses the outward unit normal of the fluid domain. The irreversible loss of mechanical energy per unit wall area and time isFor two tangential directions replace $u_{\mathrm{rel}}^2$ by the squared Euclidean norm of relative tangential velocity. At a moving wall this relative-velocity loss must be distinguished from the power supplied by the wall's motion. Bulk viscous dissipation also remains present. A finite slip length models reduced interfacial resistance, without declaring every boundary ideal or removing dynamic viscosity.

The limiting loss depends on how the flow is driven. In Couette flow with a slipping lower wall,It tends to zero for both $b\to 0$ and $b\to \infty$ at fixed $H$ and $U$. At $b=0$ impose the no-slip boundary condition directly, rather than dividing by zero. In planar Poiseuille flow with equal wall slip at fixed $G$, each wall instead dissipates $G^{2}b{H}^2/(4\mu )$, which grows with $b$ because relative velocity grows. The bulk loss per unit width is $G^2{H}^3/(12\mu )$; adding both wall losses gives $Gq$, the pressure work per unit channel length and width. This energy balance shows why small friction coefficients do not guarantee small losses under every driving constraint.

A Wannier-Mott exciton has an internal wave function extending over many unit cells. Its simplest model assumes a three-dimensional bulk crystal, approximately parabolic valence band and conduction band, positive isotropic scalar effective masses $m_e^{\ast },m_h^{\ast }$, a constant scalar relative permittivity ${ϵ}_r$, and a dilute pair population. Replace microscopic crystal forces by those effective masses and dielectric screening, and neglect exchange corrections and coupling to phonons. These assumptions must hold on the spatial and energy scales of the particular exciton.

With relative coordinate $\mathbf{r}={\mathbf{r}}_e-{\mathbf{r}}_h$, the internal Hamiltonian operator is the three-dimensional hydrogen atom problem with a reduced mass and a screened Coulomb potential:Here $e$ is the elementary charge, ${ϵ}_0$ the vacuum permittivity, and $\hslash$ the reduced Planck constant. The internal energy eigenvalues, radius scale, and exciton binding energies areThe principal quantum number is $n=1,2,\dots$. This reduction and these scales are derived in Excitons in Semiconductor Quantum Dots by Matjaž Humar (§3, printed pp. 5–6, equations (13)–(26)).

To check the scales by dimensional analysis, write $C=e^2/(4\pi {ϵ}_0{ϵ}_r)$, which has units of energy times length. Then $a_B^{\ast }={\hslash }^2/(\mu C)$ is a length and $R^{\ast }=\mu C^2/(2{\hslash }^2)$ is an energy. Rescaling $r=a_B^{\ast }\rho$ makes $H_{\mathrm{rel}}/R^{\ast }=-{\nabla }_{\rho }^2-2/\rho$, the dimensionless hydrogen atom operator. The Bohr radius $a_B^{\ast }$ is a wave function scale, not the radius of a classical orbit; for this model's lowest internal bound state, the mean separation is $3a_B^{\ast }/2$.

For the diagrams' hypothetical inputs $m_e^{\ast }=0.10m_0$, $m_h^{\ast }=0.40m_0$, ${ϵ}_r=10$ and $E_g=1.500\mathrm{eV}$, where $m_0$ is the free electron mass, independent substitution gives $\mu =0.080m_0$, $a_B^{\ast }\simeq 6.615\mathrm{nm}$ and $R^{\ast }\simeq 10.885\mathrm{meV}$. Here electronvolts express energy. The exciton binding energies for $n=1,2,3$ are approximately $10.885$, $2.721$ and $1.209\mathrm{meV}$. These are model calculations, not measured material parameters. The large-radius approximation additionally requires $a_B^{\ast }$ to be much larger than the crystal's unit cell length scale. In a thin layer, with anisotropic effective masses, or with spatially varying dielectric screening, this three-dimensional formula is not generally valid.

Choose approximate atomic weights $A_{\mathrm{r}}(\mathrm{H})=1.008$ and $A_{\mathrm{r}}(\mathrm{O})=15.999$, and use $M_{\mathrm{u}}\approx 1\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. The chemical formula ${\mathrm{H}}_2\mathrm{O}$ then gives water molar massThese are deliberately chosen rounded inputs, not exact atomic weights or a universal value for every water sample. Different isotopic abundances, notably those in heavy water, change the mean molar mass; chemical formula alone does not specify them. Adding atomic weights also neglects the tiny mass change associated with chemical bond energies, immaterial at this rounding precision.

For an illustrative pure water sample of mass $4.50375\mathrm{g}$, this convention givesThe count refers to water molecules. Their hydrogen chemical atom count is approximately $2N=3.01107038\times 10^{23}$, and their oxygen chemical atom count is approximately $N$. The mean mass per water molecule is $M/N_{\mathrm{A}}\approx 2.99146\times 10^{-26}\mathrm{kg}$. Extra displayed digits expose the arithmetic; they do not imply that the sample composition or mass was measured that precisely.

The same conversion makes stoichiometry a mass calculation. For the chemical equation $2{\mathrm{H}}_2+{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}$, complete conversion of $0.250000\mathrm{mol}$ of molecular hydrogen and $0.125000\mathrm{mol}$ of molecular oxygen gives $0.250000\mathrm{mol}$ of water. With the chosen approximate atomic weights, the reactant masses are $0.50400\mathrm{g}$ and $3.99975\mathrm{g}$, summing to $4.50375\mathrm{g}$. The $2:1:2$ ratio belongs to molecule counts and amount of substance, not to masses; this bookkeeping neglects the much smaller reaction energy contribution to mass.

The watt is the SI unit of power, one joule per second. Joule heating at $12.5\mathrm{W}$ transfers $125\mathrm{J}$ of energy over a time of $10\mathrm{s}$ if that power stays constant. ^[ref].

A wave is a propagating or standing disturbance that can carry energy and momentum. Its oscillations are described by quantities such as wavelength, frequency, and wave speed.