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Otto Robert Frisch and Lise Meitner interpreted the barium found after uranium bombardment as evidence that a heavy atomic nucleus could divide into two large fragments. Their 1939 paper related the available energy to nuclear mass differences and the electrical repulsion of the fragments. This supplied a physical account of nuclear fission, connecting a chemical identification of the products with a new type of nuclear transformation and its large energy release.^[ref]

A physicist who worked with Lise Meitner on the interpretation of nuclear fission.^[ref]

A Green function for stationary scattering is chosen to contain waves travelling away from its source. In one dimension at positive energy it is proportional to $e^{ik|x-x^{\prime }|}$; the condition selects the $E+i0$ resolvent boundary value.

An outgoing radiation pulse carries a finite amount of electromagnetic energy through successively larger enclosing spheres. In vacuum, far from a localized source and without absorption or additional reflections, let $P_{\mathrm{emit}}(t)$ be the emitted power. The retarded time at radius $r$ is $s=t-r/c$. Define ${\overline{S}}_r$ as the angular average of the radial Poynting vector over the sphere, not a cycle average:Here $\overline{u}$ is the angularly averaged energy density in the leading radiation zone approximation. Multiplying by shell area removes geometric dilution: the energy per unit radial thickness translates outward without losing its shape. These angular averages do not assume that an actual electric dipole emits equally in all directions. Nor do they specify an exact spherically symmetric transverse solution of the Maxwell equations.

For an explicitly stipulated model, emitAfter emission ends, $t>\tau$, the pulse lies between $r=c(t-\tau )$ and $r=ct$, provided this shell is in the radiation zone. Its radial thickness remains $c\tau$. Substituting $s=t-r/c$ in the shell integral $\int 4\pi r^2\overline{u}dr$ gives $U=P_0\tau /2$ at every such time. The same substitution in the time integral of $4\pi r^2{\overline{S}}_r$ at any enclosing sphere gives that same total crossing energy. This is conservation of energy in two different views.

Along any fixed direction with nonzero outgoing power, compare the same part of the pulse at two radii, hence at different arrival times. The leading electric field amplitude falls as $1/r$, while its squared wave intensity falls as $1/r^2$. At $2r$ the field is one half and the energy flux one quarter; the signal arrives $r/c$ later. A small fixed area receiver therefore intercepts less energy at greater distance even though every enclosing sphere receives the same total pulse energy. The inverse-square law expresses spreading over area, rather than destruction of energy. Close to the source, the neglected near and far field terms prevent extrapolating these formulas down to $r=0$.

Consider a noninteracting band model in which all states of one energy band would be occupied and the next energy band empty. Define the global indirect band gapIf ${\Delta }_{\mathrm{ind}}>0$, this filling can have a positive bulk excitation gap. If ${\Delta }_{\mathrm{ind}}<0$, transferring an electron from the highest occupied states of the lower energy band to sufficiently low states of the upper energy band lowers the total energy. The nominal complete-band filling is not the equilibrium state: electron and electron hole populations can occupy the overlapping bands, precluding a gapped band insulator at that filling.

The equality case requires care. At ${\Delta }_{\mathrm{ind}}=0$, the excitation gap closes, but contact only at isolated wavevectors need not create finite-volume carrier pockets or a finite carrier density at zero temperature. It is therefore incorrect to infer strictly energy-lowering redistribution or partially occupied pockets merely from a band touching. Electron interactions and symmetry breaking can alter this single-particle description.

Choose the water, vessel and immersed paddle as the closed system, with fixed outer volume and negligible endpoint changes in bulk mechanical energy. Between common equilibrium temperatures $T_1$ and $T_2$,The effective heat capacity $C_{\mathrm{eff}}$ includes every component that participates in the common temperature change. For nearly constant $C_{\mathrm{eff}}$, the readout is $C_{\mathrm{eff}}\Delta T$. Heat transferred between water and paddle stays inside this boundary; it must not be added again as external input. The drive supplies shaft work even though the vessel has no expansion work.

For a simplified modern model, let $M$ be the total descending mass, let each of $N$ descents cover height $h$, and let $W_{\mathrm{loss},\mathrm{out}}$ be energy diverted by the external drive. With negligible remaining drive kinetic energy and elastic storage,The signed $Q_{\mathrm{room},\mathrm{out}}=-Q_{\mathrm{room},\mathrm{in}}$ may be negative if the room warms the assembly. This model omits the small endpoint and elastic corrections explicitly considered in the historical analysis. It does not equate all of the weights' lost gravitational potential energy with the water's gained internal energy.

As a proposed modern protocol, measure the falling masses and distances, calibrate the thermometer and $C_{\mathrm{eff}}$ independently, record room temperature and resting drift before and after each run, and wait for fluid motion to decay before the final equilibrium reading. Repeat with different fall budgets and run durations. A stable ratio of corrected input to corrected thermal storage supports the first law of thermodynamics over the tested range. Resting controls only estimate leakage if their thermal conditions represent the driven run; sensor lag, nonuniform temperature, unmeasured drive losses, evaporation or changing heat capacity can spoil the inference. Those are limits of the measurement, rather than evidence that heat was stored as a separate substance.

For Hartree direct integrals $J_{ij}$ and orbital eigenvalues ${ϵ}_i$, the product-state energy is $E={\sum }_i{ϵ}_i-{\sum }_{i<j}{J}_{ij}$. Each pair enters two orbital mean fields but only once in the total interaction energy. Orbital energies are therefore not independent additive particle energies.

A pairing gap is an energy scale for fermionic excitations above a paired superfluid or superconductor ground state. In a simple isotropic paired system the quasiparticle dispersion has the form $E_{\mathbf{k}}=\sqrt{{\xi }_{\mathbf{k}}^2+|\Delta {|}^2}$, where ${\xi }_{\mathbf{k}}$ measures the unpaired single-particle energy relative to the chemical potential. A nonzero minimum gap gives exponentially small fermionic thermal occupations when $k_{B}T\ll |\Delta |$. If the gap varies with direction and has nodes, gapless directions can instead produce power-law thermodynamics. Collective excitations must be counted separately from the gapped fermionic quasiparticles.

For a reconstructed electron–Positron pair, four-momentum conservation givesNeglecting the electron mass compared with the daughter energies, and calling the opening angle $\theta$, this becomesAs an original example, choose $E_{+}=2\mathrm{GeV}$, $E_{-}=3\mathrm{GeV}$ and $\theta =60^{\circ }$. Then $M_{ee}{c}^2=\sqrt{6}\mathrm{GeV}\simeq 2.449\mathrm{GeV}$, although the pair's laboratory energy is $5\mathrm{GeV}$. The pair carries net momentum. Neither its laboratory energy nor the original proton beam energy is its invariant mass; unobserved products carry additional four-momentum.

For equal, oppositely directed incoming beams with equal total energy $E_b$, their total momentum vanishes. Thus the incoming center-of-mass energy isStipulated $1.50\mathrm{GeV}$ beams therefore provide $3.00\mathrm{GeV}$ of incoming center-of-mass energy. This identity is exact for the stated equal-beam geometry. Radiation before annihilation can reduce the energy forming the final state; the apparatus also has a distribution of beam energies. A nominal machine setting is therefore not an exact event-by-event final-state invariant mass.

Pair production creates a particle and its antiparticle while conserving energy, momentum and relevant charges. For an electron–Positron pair, the available invariant energy must at least supply their combined rest mass, with additional kinetic energy and other final-state requirements included.

A single free photon in empty flat spacetime cannot produce just that pair: its four-momentum has zero invariant mass, while the massive pair has a positive invariant mass. A second photon or another participant, such as a nucleus taking recoil, can satisfy the full kinematics. The pair's charges cancel globally, but each outgoing particle can leave its own detector track. A laboratory-frame photon energy threshold therefore depends on the complete reaction, not only on writing twice the electron rest energy.

A vertical direct transition at a common wavevector puts an electron above the conduction edge and leaves a hole below the valence edge. Their parabolic kinetic energies are inversely proportional to their effective masses. The excess photon energy is therefore split as $K_e/K_h=m_h^{\ast }/m_e^{\ast }$, rather than equally unless the masses match. The initial valence electron energy is $-K_h$ relative to its band maximum; the final conduction energy is $E_g+K_e$.

In a dry galaxy merger, zero orbital energy at infinite separation and retention of all mass and energy give $E_F=E_I+E_A$. If the progenitors and remnant satisfy the virial theorem, their gravitational radii obey $R_F/R_I=(1+\eta {)}^2/(1+ϵ\eta )$, where $\eta =M_A/M_I$ and $ϵ=⟨v_A^2⟩/⟨v_I^2⟩$. The latter average is weighted by satellite mass. Nonzero orbital energy, escaping stars, or dissipative losses change the energy accounting.

For a slab of thickness $2a$ with parallel applied field, the local London equation gives $B(x)=B_a\cosh (x/\lambda )/\cosh (a/\lambda )$. Averaging gives $M=-B_a[1-\tanh (a/\lambda )/(a/\lambda )]/{\mu }_0$. Integrating $-Md{B}_a$ and comparing with the bulk condensation energy yields the displayed thin-slab enhancement. It assumes field-independent condensate density and can be limited by pair breaking or nonlocal electrodynamics.

If $g_2=-g_1$ in a shared magneton convention, the Zeeman energy of both parallel-spin states cancels. Their energy stays $\Delta /4$, with exact degeneracy two for every field. Opposite charges alone do not enforce this equality because the masses and magnetic factors may differ.

For a parity-symmetric one-dimensional potential, the even and odd parity states diagonalize the One-dimensional S-matrix:For a real potential, unitarity further gives $|S_{++}|=|S_{--}|=1$ at real positive energy.

When a quadratic fermion spectrum was derived in a fixed fermion parity sector, filling all negative-energy modes is permitted only if the resulting particle number has that parity. If it does not, toggle the occupation of a mode with smallest absolute energy. Compare the minimum energies of the physical sectors. Zero-energy modes can create degeneracy without energy cost.

An even one-dimensional confining potential preserves even and odd parity. Restricting a normalized trial state to the odd sector makes its Rayleigh quotient an upper bound on the lowest odd energy, which is the first excited energy for the usual nonsingular one-dimensional bound-state problem. This avoids contamination by the even ground state. For an adjustable-width odd Gaussian, optimize its own width rather than borrowing the ground-state optimum.

The contribution of a specified decay channel to the total decay width. In an exponential-decay description, its probability rate is ${\lambda }_i={\Gamma }_i/\hslash$, the total width is $\Gamma ={\sum }_i{\Gamma }_i=\hslash /\tau$, and its branching fraction is ${\Gamma }_i/\Gamma$. Channels must be specified consistently, including any inclusive treatment of unresolved radiation. A partial decay width has energy units rather than inverse-time units.

For an eigenvector normalized by ${\sum }_{n=1}^N|{\psi }_n{|}^2=1$, the participation ratio in its site basis isIt estimates how many sites carry appreciable probability. Equal weight on $K$ sites and zero elsewhere gives $P=K$, irrespective of their wave phases. A single occupied site gives $P=1$; uniform probability on all sites gives $P=N$. These are exact constructed examples, not claims about particular Hamiltonian operators.

To prove the bounds, write $p_n=|{\psi }_n{|}^2$. Since $0\le p_n\le 1$, ${\sum }_n{p}_n^2\le {\sum }_n{p}_n=1$, giving $P\ge 1$. The Cauchy-Schwarz inequality gives $1=({\sum }_n{p}_n{)}^2\le N{\sum }_n{p}_n^2$, giving $P\le N$. Thus the measure has useful limits without assuming an exponential shape. Comparing $P$ as $N$ increases helps distinguish a state filling the sample from one occupying a fixed region, provided energy selection and disorder statistics are controlled.

A particle accelerator increases the kinetic energy of charged particles through the work of an electric field. A single electric potential difference $V$ gives an energy gain of magnitude $|q|V$ to a particle of electric charge $q$ crossing in the accelerating direction. Reusing a smaller electric potential difference many times can give a much larger final kinetic energy without requiring the whole apparatus to withstand the equivalent single voltage. A magnetic field can return the particles to the accelerating region: magnetic forces do no work, but they make repeated energy gains possible. The cyclotron turns this separation between bending and acceleration into a particularly elegant machine.

A particle beam is a directed stream of particles prepared for transport, interaction or detection. A molecular beam can contain neutral chemical atoms or molecules; an accelerator beam can contain charged electrons, protons or ions. Its direction, energy spread, spatial distribution and particle flux describe different aspects of the preparation.

A source, direction-selecting apertures or focusing elements, a transport region and a detector perform separate tasks. For a charged beam, electric fields can change its energy and magnetic fields can bend its trajectory through the Lorentz force. A neutral atomic beam requires interactions appropriate to its neutral constituents. A vacuum reduces unwanted collisions along either transport path, but it does not itself make every particle's speed equal.