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For constant buoyancy frequency $N$, the log-pressure quasi-geostrophic potential vorticity equation has modes $\psi =e^{z/(2H)}{e}^{i(kx+mz-\omega t)}$ withThe prefactor compensates the density decrease in the energy weight. For $\beta ,k>0$, vertical propagation occurs for $-\beta k/[k^2+f_0^2/(4N^2{H}^2)]<\omega <0$. An outgoing radiation condition selects the root with positive vertical group velocity; bounded weighted amplitude alone does not select between two propagating roots.

The Rossi X-ray Timing Explorer was a space observatory designed to measure rapid X-ray variability, operating from its 1995 launch until 2012. Its time-resolved observations probed accretion and rotation around compact objects such as neutron stars and black holes.

Fast event timing complements an energy spectrum. A narrow feature in a temporal power spectrum identifies an oscillation or rotation frequency of the source, not the frequency of an individual X-ray photon. Comparing such timing features with changes in spectral shape helps distinguish physical states of an accreting system.^[ref]

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.

Consider a two-level system with a real coupling energy $g$ and constant angular drive frequency $\Omega$,The opposite phases and real $g$ make this Hamiltonian Hermitian. Write the state amplitudes as $a=U(t)b$, with the unitary matrixSubstitution into the Schrödinger equation gives the constant quantum rotating frame HamiltonianConsequently $b(t)=[I\cos (\Lambda t/\hslash )-i{H}_{\mathrm{eff}}\sin (\Lambda t/\hslash )/\Lambda ]b(0)$. Starting in the first basis state gives transition probabilityThe diagonal transformation changes phases but not populations. This is an exact solution of the specified rotating-coupling model, without a rotating-wave approximation. If the input phase is instead written $e^{i\omega t/\hslash }$, then $\omega =\hslash \Omega$ has units of energy, rather than angular frequency.

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.

Rough elastic collisions can exchange rotational and translational kinetic energy. If all three translational and all three rotational degrees of freedom of a rigid ball reach classical thermal equilibrium, their six quadratic terms contributeper ball, by the equipartition theorem.

In a constant downward gravitational field above a floor at $z=0$, with no upper boundary, the equilibrium height distribution is exponential:Its mean gravitational potential energy is $⟨mgz⟩=k_{B}T$, so the total mechanical mean energy is $4k_{B}T$. The field is uniform; the height distribution is not.

This result assumes the rough collisions actually thermalize rotation, sufficient room for the exponential height tail and no energy loss. Smooth elastic spheres need not exchange rotational energy with translation. A finite ceiling changes the height distribution and its mean potential energy.

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]

The quadratic rotational Stark effect is a small-field expansion, rather than a response law valid at every field. For a polar linear rigid rotor with rotational constant $B$ in energy units and permanent electric dipole moment $p$, a static field $\mathcal{E}$ gives the Hamiltonian operatorThe dimensionless ratio $p\mathcal{E}/B$ measures competition between rotation and orientation. Once it is not small, the field mixes many field-free angular momentum eigenstates, so diagonalizing the fixed-magnetic quantum number matrix replaces a truncated second-order correction. Increasing the basis cutoff tests numerical convergence.

For a differentiable nondegenerate eigenvalue $E_{\alpha }(\mathcal{E})$, the Hellmann–Feynman theorem gives the mean dipole component $-d{E}_{\alpha }/d\mathcal{E}=p⟨\cos \theta ⟩$. Its magnitude cannot exceed $p$. Continuing the weak-field linear induced-dipole formula indefinitely would violate this bound. Strongly oriented low-lying pendular states instead describe restricted angular motion about the field direction.^[ref]

Write $U=A{U}_0{A}^{-1}$, $A\in SU(2)$, and $A^{-1}\dot{A}=i\mathbf{\omega }\cdot \mathbf{\sigma }/2$. The rotational kinetic energy is $\Lambda |\mathbf{\omega }{|}^2/2$. Since $A$ and $-A$ give the same classical field, the physical orientation space is $SO(3)$, covered by $SU(2)$. For fermionic unit baryons the Finkelstein-Rubinstein constraints require $\Psi (-A)=-\Psi (A)$. The SU(2) representations then give half-integer $j$, with spin angular momentum and isospin both of magnitude $j$. The rigid-rotor energies are $M+j(j+1){\hslash }^2/(2\Lambda )$; the lowest multiplets model the nucleon and Delta baryon. Large rotational energies can excite deformation and radiation, beyond the rigid approximation.

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.

Rudolf Clausius established a relation between average kinetic energy and the forces acting on a bounded mechanical system. The virial theorem follows by differentiating the sum of position-momentum products and averaging when the boundary term remains bounded. For inverse-square gravity it gives $2⟨K⟩+⟨U⟩=0$ under the appropriate equilibrium assumptions. The result connects microscopic motion with binding energy and supplies a bridge between mechanics, gases and gravitating systems.^[ref]

Rudolf Mössbauer discovered that some gamma ray emission and absorption events in a solid occur without changing the crystal's phonon state. The momentum can then be transferred to the whole lattice, with an effective recoil energy far smaller than for a free nucleus. This permits extremely narrow nuclear emission and absorption lines to overlap, producing the Mössbauer effect. The narrow resonance makes very small energy shifts observable, including shifts due to the local electronic and magnetic environment of a nucleus.^[ref]

A physicist.^[ref][ref]