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For unstressed central Hookean springs, the quadratic elastic energy is $\frac{1}{2}{\sum }_{ij}{k}_{ij}[{\mathbf{n}}_{ij}\cdot ({\mathbf{u}}_i-{\mathbf{u}}_j){]}^2$, with unit edge direction ${\mathbf{n}}_{ij}$. A common translation gives zero displacement difference. An infinitesimal rotation ${\mathbf{u}}_i=\mathbf{\Omega }\times {\mathbf{r}}_i$ also gives zero extension because a vector is perpendicular to its cross product with $\mathbf{\Omega }$. Thus a noncollinear three-dimensional free framework has three translational and three rotational zero modes. Extra mechanisms can give additional zero modes; six is a lower bound until rigidity is established.

Ringdown is the freely decaying motion of a resonator after an applied drive is removed. For an underdamped damped harmonic oscillator, the amplitude decays as $e^{-\gamma t}$ while the angular frequency is $\nu =\sqrt{{\Omega }^2-{\gamma }^2}$. In the weak-damping limit, the cycle-averaged stored energy decays as $e^{-2\gamma t}$, giving the quality factor from ringdown $Q\simeq \Omega /(2\gamma )$.

For two weakly damped normal modes with initial intensity ratio $R_0>1$, independent exponential energy decay gives $I_1(t)/I_2(t)=R_0{e}^{-t({\omega }_1/Q_1-{\omega }_2/Q_2)}$. An observed reversal to a ratio below one implies the displayed upper bound on $Q_1$, provided the denominator is positive. A strict intensity comparison does not determine a unique quality factor; equality fixes only the crossing threshold.

An RLC circuit combines electrical resistance $R$, inductance $L$, and capacitance $C$. The capacitor stores energy in its electric field; the inductor stores energy in its magnetic field. Their exchange can produce an oscillation, while Joule heating removes energy. A series RLC circuit gives a particularly clear example: changing $R$ turns ringing into a direct return to equilibrium, and changing $L$ or $C$ tunes a resonance.

These circuits connect two useful viewpoints. In time, they describe how stored energy is released after switching. In frequency, they select a range of signals: the resonant series RLC response explains radio tuning and simple filters. The same mathematics describes a damped harmonic oscillator, so insight about mechanical ringing transfers directly to an electrical circuit. ^[ref]; ^[ref].

An RL circuit combines electrical resistance $R$ and self-inductance $L$. In the series model, the same electric current $I$ flows through both components. An ideal battery maintains terminal electric potential difference $V$; the drops in the direction of positive electric current obey Ohm's law and the inductor relation:This is Kirchhoff's voltage law with the inductor's terminal drop, rather than its oppositely signed induced electromotive force. For constant $R>0$, $L>0$, and $V$, the first-order linear differential equation has solutionThe time constant $\tau$ measures exponential decay of the difference from the final electric current. ^[ref].

For a stipulated example, take electrical resistance $R=2\Omega$, self-inductance $L=0.5\mathrm{H}$, terminal electric potential difference $V=6\mathrm{V}$, and initial electric current $I(0)=0$. ThenWith $t$ measured numerically in seconds, this is $I(t)=3(1-e^{-4t})$ in amperes. At one time constant,and the terminal electric potential difference drops areTheir sum is exactly $6\mathrm{V}$. At $t=0^{+}$, the inductor takes the entire terminal electric potential difference while the electric current is still zero. At steady electric current, the inductor's terminal drop is zero and the entire drop is across the electrical resistance.

The final magnetic energy stored by the inductor is $U_{\infty }=L{I}_{\infty }^2/2=2.25\mathrm{J}$. At one time constant, the electric current has reached about $63.21\%$ of its final value, but its magnetic energy has reached only about $39.96\%$:This difference follows from the square of electric current in the inductor's magnetic energy.

For release, take $t=0$ at the start of a separate experiment with initial electric current $3\mathrm{A}$. Remove the battery from the model while maintaining a closed path through the same electrical resistance and inductor. ThenNow $v_L=-v_R$: the induced electromotive force supports the continuing positive electric current as its magnitude decreases. This closed-loop model describes exponential decay, not opening the sole path and imposing an instantaneous jump to zero electric current. The inductor's magnetic energy and the power of Joule heating areTheir time constant is $\tau /2=0.125\mathrm{s}$. The total energy dissipated by Joule heating is exactly the initial magnetic energy:The second diagram in inductor shows this energy transfer. A finite closed path lets us predict both how long the electric current persists and how much energy becomes heat. ^[ref].

Wilhelm Conrad Röntgen's 1895 first communication describes an induction coil driving an evacuated gas-discharge tube, enclosed in black cardboard in darkness. A paper screen coated with barium platinocyanide showed fluorescence even with its uncoated side facing the tube, visible two metres away. The covering excluded ordinary illumination.^[ref]

He interposed materials: a thousand-page book transmitted; $1.5\mathrm{mm}$ lead was practically opaque. A stepped stack of foil produced a photographic plate record. Different materials were not equivalent at equal density and thickness. Photographic plates checked visual observations while remaining covered against ordinary illumination.^[ref]

Section 14 lists photographs of hand bones, wire on a bobbin, boxed weights, an enclosed compass, and a lead-painted door. It does not identify the hand's owner. Magnetic deflection moved the cathode ray impact point and thus the source, without bending the emerging X-rays. Their nature remained unsettled.^[ref]

The observations belong to 1895; the contemporary English translation appeared in Nature on 23 January 1896.^[ref] The experimental logic separates production, transmission and detection. A visibly opaque covering can transmit X-rays; fluorescence then converts deposited energy into visible light. A shadow measures a difference between paths through an object, rather than a direct view of its surface. Neither a screen's brightness nor photographic darkness alone measures an absolute X-ray energy or establishes the X-ray attenuation law below.

For $\alpha =\omega t+{\alpha }_0$ and $|\omega |<m$, the kink is $\varphi =2\arctan e^{\kappa (x-x_0)}$ with $\kappa =\sqrt{m^2-{\omega }^2}$. Its rest energy and mechanical internal charge are $M=r{m}^2/\kappa$ and $Q_0=r\omega /\kappa$. Eliminating the frequency gives the displayed mass relation. The orientation rotates while the energy profile remains stationary.

A rotating ideal half-wave plate flips circular polarization. The spin change has magnitude $2\hslash$ per photon, so its mechanical energy exchange produces a frequency shift of magnitude $2|\Omega |$. For Jones convention $H(\theta )=R(\theta )\mathrm{diag}(1,-1)R(-\theta )$ and $Q=\mathrm{diag}(1,i)$, the combination $QH(\theta )Q=e^{i2\theta {\sigma }_x}$. Acting on $(1,1{)}^T/\sqrt{2}$ gives $e^{i2\theta }$; with time dependence $e^{-i\omega t}$ and $\theta =\Omega t$, the output frequency is $\omega -2\Omega$. The opposite input eigenpolarization reverses the shift.

In the midplane Mestel disk potential $\Phi =v_c^2\log (R/R_0)$, take specific energy $E=(v_R^2+v_{\varphi }^2)/2+\Phi$ and $L_z=R{v}_{\varphi }$. A prograde galactic distribution function $f=A{L}_z^q{e}^{-E/s}$ for $L_z>0$, with $s>0$, has surface density proportional to $R^{q-v_c^2/s}$. Matching $\Sigma ={\Sigma }_0{R}_0/R$ gives $q=v_c^2/s-1>-1$ andThe denominator is the product of the full radial Gaussian integral and the positive-azimuthal half-line moment. Here $s$ is a squared-velocity scale, not a velocity dispersion itself. Changing the potential's additive constant changes $A$. With positive exponent $e^{\epsilon /s}$, $\epsilon$ must instead be the negative of the specific total energy.

A straight rigidly rotating classical string has speed of light at a free endpoint. Its energy and angular momentum satisfy a Regge trajectory. A segment with one anchored and one free end has $J=E^2/(\pi T)$, twice the slope of the string with two free ends.

A frequency change associated with rotation and exchange of angular momentum. If an optical element rotates at angular velocity $\Omega$ about a chosen axis and gains angular momentum $\Delta L$ from a photon, its mechanical energy gain is $\Omega \Delta L$ in the prescribed slow-rotation model. Energy conservation then gives photon frequency change $\Delta \omega =-\Omega \Delta L/\hslash$. The rotating half-wave plate frequency shift is a spin-polarization example.

Rotational kinetic energy is the kinetic energy of a rigid body's rotation. For pure rotation about an axis fixed in an inertial frame,Here $r_{\perp }$ is the perpendicular distance of a mass element $dm$ from that axis, $I$ is the moment of inertia about that same axis, and $\omega$ is the common angular speed, measured in radians per second. The rigid body assumption fixes the relative positions of the mass elements; they share angular velocity but have different speeds $v=\omega r_{\perp }$. Summing their kinetic energies $dm{v}^2/2$ gives the formula. This is a nonrelativistic model; deformation or internal motion can contribute additional kinetic energy.^[ref]

For a computed example, take a uniform disc and an ideal thin ring, each of mass $M=2\mathrm{kg}$ and radius $R=0.30\mathrm{m}$, rotating about its central perpendicular axis at angular speed $10rad/s$. The moment of inertia of a uniform disc is $M{R}^2/2=0.090\mathrm{kg}{\mathrm{m}}^2$, while the ring has moment of inertia $M{R}^2=0.180\mathrm{kg}{\mathrm{m}}^2$. Their rotational kinetic energies are therefore $4.5\mathrm{J}$ and $9.0\mathrm{J}$. The ring stores twice the energy at the same angular speed, because more mass lies farther from the axis. At the same nonzero rotational kinetic energy, its angular speed would instead be $1/\sqrt{2}$ times the disc's. This makes radial placement useful for storing energy in a rotating wheel, although a real design must also withstand the stresses of rotation.

For fixed-axis rotation with constant moment of inertia, write the signed rotation rate as $\Omega =\dot{\theta }$, with angle $\theta$ in radians. The axial component of net external torque obeys ${\tau }_{\mathrm{axis}}=I\dot{\Omega }$, so the work-energy theorem becomesAn opposing torque removes rotational kinetic energy; bearing friction must be included in the net torque. The signed $\Omega$ matters when the direction reverses. Rotation about a fixed axis need not be about a principal axis: the scalar energy formula still holds, but transverse bearing torques can be needed to maintain that axis.

For general rigid body motion, use the inertia tensor ${\mathsf{I}}_G$ about the centre of mass $G$:This is the kinetic-energy decomposition about the centre of mass. In a principal body frame, the rotational term is $\frac{1}{2}(I_1{\omega }_1^2+I_2{\omega }_2^2+I_3{\omega }_3^2)$, using the principal moments of inertia. Away from a principal axis, angular momentum and angular velocity need not be parallel. A freely rotating asymmetric rigid body can change the direction of angular velocity relative to the body even while rotational kinetic energy remains constant; Euler equations for a torque-free rigid body govern that motion. One constant axial moment of inertia is then insufficient.^[ref]

For the spin-free linear rigid rotor in thermal equilibrium at temperature $T>0$, assume all integer $J\ge 0$ are allowed and ignore nuclear-spin restrictions. Each $J$ level has $2J+1$ quantum states sharing one energy eigenvalue, an instance of degenerate energy levels, so the Boltzmann distribution givesHere $Q_{\mathrm{rot}}$ is a rotational partition function, $h$ the Planck constant, $k_{\mathrm{B}}$ the Boltzmann constant, and $B$ a frequency in $\mathrm{Hz}$. Thus $hB/(k_{\mathrm{B}}T)$ is dimensionless. The $2J+1$ factor counts magnetic quantum numbers; it does not mean that a whole level is one quantum state. For the stipulated $B=5\mathrm{GHz}$, the most populated levels are $J=2$ at $3\mathrm{K}$ and $J=7$ at $30\mathrm{K}$.

Population alone is insufficient to predict intensity. In weak, unsaturated absorption, the integrated absorption coefficient includes the transition frequency, the squared electric dipole moment matrix element, and the difference of lower and upper substate populations because stimulated emission returns energy to the field. A pulsed FTMW spectroscopy signal additionally depends on excitation, quantum coherence, cavity tuning, and receiver response.^[ref]

For an isotropic sample with polarization of an electromagnetic wave along $z$, write $d_z=d\cos \theta$. Normalized spherical harmonics giveThe last equality follows from ${\sum }_{M=-J}^J{M}^2=J(J+1)(2J+1)/3$. Combining this sum with the per-substate Boltzmann distribution and subtracting stimulated emission gives relative integrated absorption areasThe common factor includes $d^2/3$ and cancels upon normalization. This formula describes integrated weak absorption in a thermal, spin-free linear rigid rotor, not arbitrary peak heights. The example's strongest integrated lines originate at $J=3$ and $J=13$ for $3$ and $30\mathrm{K}$, respectively: they differ from the most populated levels. A missing line can reflect population, coupling, coverage, or sensitivity; it does not prove that the corresponding quantum state is absent.

For a linear rigid rotor, rotational temperature expresses the rotational energy constant in kelvins: $E_J=k_B{\theta }_{r}J(J+1)$, where $I$ is the perpendicular moment of inertia. With all integer $J\ge 0$ allowed and degeneracies $2J+1$, its rotational canonical partition function is $Z={\sum }_J(2J+1)e^{-{\theta }_{r}J(J+1)/T}$. It approaches $1+3e^{-2{\theta }_r/T}$ at low temperature and $T/{\theta }_r$ at high temperature. The latter limit gives mean rotational energy $k_{B}T$ and heat capacity $k_B$ per rotor.

A roton is an excitation near a local minimum of a superfluid's dispersion relation at nonzero momentum $p_0$. The displayed local approximation defines a positive energy gap $\Delta$ and positive curvature effective mass $m_r$; it is not a free helium atom or a macroscopic quantum vortex.^[ref] In the stipulated plot, these correspond to $p_0=2p_{\ast }$, $\Delta =0.6E_{\ast }$ and $m_r=p_{\ast }^2/(4E_{\ast })$. The plot illustrates how a low energy at appreciable momentum can lower the Landau criterion below the speed of sound, without assigning any measured parameter to real helium.

Landau's 1947 revision joined the low-momentum linear part to a spectrum with a finite-momentum minimum, retaining the macroscopic superfluid theory while changing its excitation model.^[ref]

The ratio of useful energy recovered on discharge to the energy supplied during charging for a complete storage cycle. Generation efficiency alone excludes the preceding charging losses.

The ruby pressure scale converts a shift of a ruby fluorescence spectral line into pressure. Excitation light produces a fluorescence emission spectrum; fitting its $R_1$ line gives a wavelength ${\lambda }_{R_1}$. The reference ${\lambda }_{R_1,0}$ must be measured or assigned at specified reference pressure and temperature. The difference $\Delta \lambda$ is an observable; interpreting it as pressure requires a pressure calibration and suitable stress conditions.

The original 1972 report found sharp $R_1,R_2$ emission shifting toward lower photon energy as pressure increased; it tentatively associated broadening with nonhydrostatic conditions.^[ref] In use, excitation rejection, a calibrated wavelength axis and a fitted background matter because leakage or background can move an apparent peak. A brighter trace improves counting statistics but does not alone establish a correct pressure scale.

The ruby R-line shift compares the same emission feature under two conditions. It is distinct from the Stokes shift, which compares absorption and emission features. In ruby, chromium ions in the corundum crystal provide the electronic transitions; compression changes their local environment and transition energies.^[ref]

For consistently defined vacuum wavelengths, the reciprocal relation givesThus a positive wavelength shift means a negative reciprocal shift, and $E=hc/\lambda$ gives lower photon energy; $h$ is the Planck constant and $c$ the speed of light. For spectroscopic wavenumber, use vacuum wavelength and keep that convention consistent. A calibration specified in air wavelength must retain its convention or be converted consistently. The reciprocal relation is nonlinear: a line exactly linear in $\Delta \lambda$ is not exactly linear in $\Delta (1/\lambda )$.

For a purely mathematical check with chosen vacuum wavelengths ${\lambda }_0=700.0\mathrm{nm}$ and $\Delta \lambda =2.0\mathrm{nm}$, the reciprocal shift is $-40.7000\mathrm{c}{\mathrm{m}}^{-1}$. The small-shift approximation $-\Delta \lambda /{\lambda }_0^2$ gives $-40.8163\mathrm{c}{\mathrm{m}}^{-1}$, about $0.286\%$ too large in magnitude. These chosen values illustrate the conversion; they are not measured ruby lines or an additional pressure calibration.

The Rutherford model places a concentrated central electric charge inside a much larger chemical atom. In 1911, Ernest Rutherford argued that rare large deflections were predominantly single close encounters, rather than accumulated small deviations in a diffuse-charge atom. He treated the central electric charge as positive for calculation while noting that the scattering deductions did not establish its sign. He explicitly deferred atomic stability.^[ref]

The physical inference is a compact region generating a strong electric field, not a miniature hard wall. Most impact parameters miss that region, while a small subset give large deflections. The model supplies a way to infer interior structure from outgoing angles, a central use of scattering throughout atomic physics and nuclear physics. This inference does not by itself specify electron quantum states or explain spectral lines.

The repulsive model also sets a probing scale. At closest approach, conservation of energy with zero radial velocity givesFor the numerical model above, even a head-on projectile turns at $r_{\min }=2a=45.504\mathrm{fm}$. This is a calculated closest-approach distance, not a measured atomic nucleus radius. A point-charge fit can constrain how compact the scattering centre must be, but cannot directly resolve structure at distances the projectiles never reach.

The Rutherford differential cross-section must not be extrapolated to zero angle: the unscreened point model has a forward divergence, while real neutral chemical atoms exhibit the screening effect, finite beam acceptance, and accumulated small deflections. At very short separation, finite nuclear size and the nuclear force can change the interaction. Nuclear recoil changes the relation between laboratory and centre-of-mass angles for lighter targets; appreciable energy loss or repeated large deflections invalidates the simple thin-foil scattering counts. These limits delimit the calculation rather than supplying evidence for a nuclear contact collision in the early experiment.

Rutherford scattering is elastic scattering by a Coulomb potential. In the repulsive classical case, a positively charged projectile bends around a positively charged atomic nucleus. A close approach can turn it backwards without a contact collision. The calculation connects an invisible impact parameter with a measurable outgoing angle, then with a differential scattering cross-section and a detector count.

Assume a projectile of mass $m$ and electric charge $ze$, a stationary point atomic nucleus of electric charge $Ze$, and positive $z,Z$. Here $e$ is the elementary charge. Neglect nuclear recoil, electron screening effect, energy loss, and the nuclear force. With vacuum permittivity ${\epsilon }_0$, defineHere $E$ and $v$ are the incident kinetic energy and speed far from the atomic nucleus, $b\ge 0$ is the impact parameter, and $L$ is the magnitude of angular momentum. The nonrelativistic limit requires $v\ll c$, with $c$ the speed of light. The fixed-target approximation requires the target mass to be much greater than $m$. In SI, $\kappa$ has units $\mathrm{J}\mathrm{m}$, $E$ has units $\mathrm{J}$, and $b$ has units $\mathrm{m}$.