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For $V=m^2{\varphi }^2/2+\lambda ({\varphi }^2{)}^2/4$, $m^2<0$, the vacuum manifold is a sphere of radius $v=\sqrt{-m^2/\lambda }$. At a chosen vacuum the Hessian matrix is $2\lambda v^{2}n{n}^T$, where $n$ is its unit direction. The radial scalar has mass squared $2\lambda v^2$; the $n-1$ angular modes are physical Goldstone bosons when the symmetry is global.

For the two-component case, write $\mathbf{\varphi }=(v+\rho )(\cos \theta ,\sin \theta )$. The kinetic term becomes $\frac{1}{2}({\partial }_{\mu }\rho {)}^2+\frac{1}{2}(v+\rho {)}^2({\partial }_{\mu }\theta {)}^2$. At quadratic order the canonically normalized angular field is $\pi =v\theta$, with no mass term. Its Euler-Lagrange equation is the wave equation, so ${\omega }^2=|\mathbf{k}{|}^2$ in units where the propagation speed is one. The radial field instead obeys ${\omega }^2=|\mathbf{k}{|}^2+2\lambda v^2$. Thus a common rotation costs no energy, whereas a slowly varying rotation propagates as a collective wave. With the selecting source of continuous vacuum selection by an external source, use $\pi =w\theta$ and obtain ${\omega }^2=|\mathbf{k}{|}^2+h/w$. This elementary dispersion relation separates a Goldstone boson, a pseudo-Goldstone boson and a massive radial excitation within one example.

A particle constrained to a stationary smooth surface of revolution $z=f(r)$ under uniform gravity conserves $h=r^2\dot{\theta }$ and mechanical energy per unit mass. Substitute $\dot{z}=f^{\prime }(r)\dot{r}$ and $\dot{\theta }=h/r^2$ into that energy to obtainThis proves the radial reduction. Simple roots of $E-U_h$ are turning points; a stationary point $U_h^{\prime }(r_0)=0$ supports an exact circular orbit. At a local maximum of $U_h$, another trajectory at the same energy can approach the orbit asymptotically as a separatrix. Existence of roots elsewhere is insufficient for confinement: their dynamically accessible ordering relative to the launch radius matters.

The induced electric field opposes the increasing axial current, while the magnetic field circles it. Their Poynting vector points radially outward and transports energy into the growing magnetic-field region.

For an axial magnetic field $\mathbf{B}=(B_0/r){\mathbf{e}}_z$, let $L=m{r}^2\dot{\theta }$. The Lorentz force gives $\dot{L}=-q{B}_0\dot{r}$, so $\ell =(L+q{B}_{0}r)/m$ is constant. Since magnetic forces do no work and $\dot{z}$ is constant, the transverse speed $v_{\perp }$ is constant and ${\dot{r}}^2=v_{\perp }^2-(\ell /r-c{)}^2$. If $c\ell >0$ and $0<v_{\perp }<|c|$, two finite radial turning points confine the motion. If $v_{\perp }>|c|$, an outward branch escapes to infinite radius. At equality a marginal escaping branch exists when $c\ell >0$, but $\ell =0$ gives a circular orbit instead. The transverse escape velocity threshold is $|q{B}_0|/m$; motion parallel to the magnetic field cannot provide transverse escape energy.

For a central-force radial energy equation, choose a monotone radius $q=q(r)$ and clock $\tau$ such that $d\tau /dt=(q/r)dq/dr$. Then $q^{\prime }=(r/q)\dot{r}$, and multiplying the old energy equation by $r^2/q^2$ preserves the centrifugal form $L^2/(2q^2)$. A split of the remaining terms into constant energy and a new Newtonian gravitational potential produces a transformed radial orbit.

In static coordinates on anti-de Sitter spacetime with curvature radius $L$ and $c=1$, put $f=1+r^2/L^2$. The conserved energy per unit rest mass is $E=fdt/d\tau$. Timelike normalization for a radial geodesic gives $(dr/d\tau {)}^2=E^2-1-r^2/L^2$, hence $d^{2}r/d{\tau }^2=-r/L^2$. An outward launch from the origin at proper time ${\tau }_0$ gives the displayed harmonic solution until its first return, after proper time $\pi L$, independently of $E>1$. Beyond that return, use a signed diameter coordinate or change the radial direction to keep the areal radius nonnegative. This is a coordinate realization of Anti-de Sitter timelike geodesic refocusing.

Radiant energy is the amount of energy emitted, transferred or received as electromagnetic radiation. It is measured in joules. For the specified beam or collection of rays crossing a boundary,where radiant flux ${\Phi }_{\mathrm{e}}$ is radiation power. The receiving geometry and time interval are part of the statement; dividing by a receiving area produces radiant exposure, rather than another total energy.^[ref]

For an original passive-optic example, stipulate $5.0\mathrm{mJ}$ incident during an interval, with no remaining change of stored energy afterward. Suppose $3.0\mathrm{mJ}$ is transmitted, $1.25\mathrm{mJ}$ reflected and $0.75\mathrm{mJ}$ absorbed. Their sum is the incoming $5.0\mathrm{mJ}$. A detector collecting only the transmitted path receives $3.0\mathrm{mJ}$, even though the source supplied more. This radiometric energy balance specifies the collection boundary instead of equating every energy readout in the apparatus.

A thermal detector converts absorbed radiation into an observable temperature response; a calibrated model relates that response to received energy. A documented NIST arrangement uses a slow thermal reference and calibrated beam splitting to establish pulse-energy measurements.^[ref] Absorbed energy, incident radiant energy and energy in an uncollected branch must remain distinct in that calibration.

Two sources can deliver the same radiant energy with different radiant flux histories or spectra. Their total energies agree only for the specified integration and collection; this alone does not determine peak power, radiant exposure at a point or detector response.

Radiant exposure measures the radiant energy received per unit surface area during a specified interval. At a fixed surface point $\mathbf{x}$, it is the time integral of incident irradiance $E_{\mathrm{e}}$:The area belongs to the receiving surface: its orientation and location matter. An irradiance in $\mathrm{W}{\mathrm{m}}^{-2}$ specifies a rate, while radiant exposure in $\mathrm{J}{\mathrm{m}}^{-2}$ specifies the accumulated arrival of energy.^[ref]

For a uniform rectangular pulse, $H_{\mathrm{e}}=E_0\tau$, with constant irradiance $E_0$ and duration $\tau$. Stipulate one pulse at $20\mathrm{W}{\mathrm{m}}^{-2}$ for $1\mathrm{s}$, and another at $80\mathrm{W}{\mathrm{m}}^{-2}$ for $0.25\mathrm{s}$. Both deliver $20\mathrm{J}{\mathrm{m}}^{-2}$, but their peak irradiances differ by a factor of four. The two radiant exposures agree only after each complete pulse has arrived. A shorter observation window can intercept different fractions of them, and a nonlinear receiver can respond differently even to their complete pulses.

Integrating over a receiving patch gives its incident radiant energy:Here ${\overline{H}}_A$ is the spatially averaged radiant exposure, rather than necessarily the local maximum. Uniform illumination is the special case where this spatially averaged radiant exposure equals the local value everywhere on the patch. In the rectangular-pulse example, a uniformly illuminated $1\mathrm{c}{\mathrm{m}}^2=10^{-4}{\mathrm{m}}^2$ patch receives $20\times 10^{-4}=2\mathrm{mJ}$.

For an original spatial example, consider a circular Gaussian beam at a fixed transverse plane, with constant Gaussian beam radius $w$ throughout the pulse. Let $P(t)$ be its total radiant power and $U=\int P(t)dt$ its laser pulse energy. Integrating its transverse irradiance in time gives the radiant exposure of a Gaussian beam:The normalization follows by integrating $H(r)2\pi rdr$ from $0$ to $\infty$. The disk of radius $w$ contains only $U(1-e^{-2})$, so its actual spatially averaged radiant exposure is $U(1-e^{-2})/(\pi w^2)$, not $U/(\pi w^2)$. With stipulated $U=1\mu \mathrm{J}$ and $w=1\mathrm{mm}$, the peak is $0.6366\mathrm{J}{\mathrm{m}}^{-2}$, while the average inside that disk is $0.2752\mathrm{J}{\mathrm{m}}^{-2}$. State the Gaussian beam radius convention and the receiving area with any reported value.

An energy readout alone therefore does not specify a surface's radiant exposure. In an ideal linear model for pulse energy measurement with a photodiode, dark-subtracted photocurrent $i_{\mathrm{ph}}(t)=R{P}_{\det }(t)$ has constant spectral responsivity $R$ at the chosen wavelength. Integrating the electric current gives signal electric charge $Q_{\mathrm{sig}}=R{U}_{\det }$. If a known patch of area $A$ is uniformly illuminated and all of its incident energy is collected, thenAs a separate stipulated example, $Q_{\mathrm{sig}}=0.40\mathrm{nC}$ and $R=0.40\mathrm{A}{\mathrm{W}}^{-1}$ correspond to $U_{\det }=1.0\mathrm{nJ}$. Uniform illumination over $A=1.0\mathrm{m}{\mathrm{m}}^2$ gives $H_{\mathrm{e}}=1.0\mathrm{mJ}{\mathrm{m}}^{-2}$. Background subtraction, collection geometry, detector linearity and an integration window covering the full response are part of this measurement model. Continuous-wave responsivity alone does not establish linearity for a short pulse: silicon-photodiode experiments have observed pulse-dependent saturation.^[ref]

A historical NIST measurement calibration separates these operations experimentally. A modulator produces pulses from a continuous-wave laser; a calibrated beam splitter sends one beam to a slow thermal reference and another to the pulse detector. The reference records the accumulated energy, while the pulse repetition rate and injection duration determine the mean energy per pulse. The detector's integrating electronics provide a pulse-energy readout. Converting such a readout to radiant exposure additionally requires the illuminated area and its spatial distribution.^[ref]

The beam-use term fluence denotes the same time integral for a directed beam normal to the chosen surface. For light arriving from many directions, specify the angular geometry: planar irradiance and spherical irradiance are different quantities, so a value defined using one cannot simply be substituted for the other.^[ref]

Radiant flux is the power carried by electromagnetic radiation: radiation energy transferred per unit time, measured in watts, $\mathrm{W}=\mathrm{J}{\mathrm{s}}^{-1}$. Specify what is counted: all radiation emitted by a source, radiation within one beam, or radiation intercepted by a photodetector. Writing $Q(t)$ for the cumulative energy crossing the specified boundary givesThe first expression is an instantaneous derivative; the second is an average over the stated time interval. If $0.020\mathrm{J}$ arrives during $0.010\mathrm{s}$, the mean radiant flux is $2\mathrm{W}$. The photodetector's absorbed power equals this incident radiant flux only if it absorbs all of the incident radiation.

Keeping radiant flux fixed while widening a beam reduces its irradiance, the incident power per receiving area. Radiance additionally resolves the propagation direction and has units $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$; an optical camera can collect different radiant fluxes from sources with the same radiance but different apparent solid angles. Luminous flux, measured in lumens, weights the optical spectrum by the human eye's standardized response instead of counting energy equally at every wavelength. Equal radiant fluxes need not give equal luminous fluxes.^[ref]

Radiant power also differs from the optical power of an optical lens, which is reciprocal focal length in dioptres. Likewise, astronomical radiative flux is power per area. Stating the units prevents these shared words from hiding different physical quantities.

Radiation transports energy away from a source through propagating fields or emitted particles. Electromagnetic radiation includes radio waves, visible light, X-rays and gamma rays; alpha particles, beta particles and neutrons provide particle-radiation examples. The word alone does not specify an energy, a hazard, or a detection method.

An experiment must distinguish the incident field from its interaction with the detector. A photodetector measures a light-induced response, while an ionization chamber collects charge created in a gas. Neither instrument responds equally to every kind of radiation. Ionizing radiation is classified by its ability to remove bound electrons under the relevant interaction conditions, rather than simply by whether the source is radioactive.

The energy entering and leaving a system through electromagnetic radiation. A top-of-atmosphere radiation budget compares absorbed sunlight with outgoing thermal radiation. A latitude-dependent imbalance can be balanced by atmospheric and ocean heat fluxes even when the global mean is zero.

For a hydrostatic monatomic ideal gas with negligible external pressure, $2U+{\Omega }_f=0$, where gravitational potential energy is negative. The final total energy is therefore ${\Omega }_f/2$, giving the displayed radiated energy when no mass or other energy carrier escapes. The common $-{\Omega }_f/2$ result additionally takes the initial total energy to be negligible. Initial rest does not remove initial self-gravitational binding energy.

A charge $q$ undergoing simple harmonic motion of angular frequency $\omega$ and slowly varying amplitude $A$ radiates the averaged powerIts mechanical energy $E=m{A}^2{\omega }^2/2$ therefore has exponential decay rate $\gamma ={\mu }_0{q}^2{\omega }^2/(6\pi mc)$ and half-life

Radiation detection makes interactions of ionizing radiation with matter into observable signals. An ionization chamber collects charge created in a gas; a scintillation detector converts deposited energy into light; a semiconductor detector measures charge carriers created in a solid. The instrument detects interactions, so its response depends on radiation type, energy and geometry.

For an ideal counting model, expected detected events satisfy $N=\epsilon Rt$, where $R$ is the relevant incident event rate, $t$ is observation time and $\epsilon$ is detection efficiency. Background adds counts, while dead time can lose them. Estimating a source's radioactive activity therefore requires calibration and a model of emissions and transport, rather than equating the displayed count rate with decay rate.

Time-ordering the canonical transverse photon field gives a propagator with numerator equal to the spatial transverse projector and denominator $k^2+i0$. Its two energy poles encode forward propagation of the positive-frequency term and backward propagation of the negative-frequency term. This is the reduced free-field propagator; it is distinct from the covariant four-potential propagator.

The characteristic thickness over which a high-energy electron loses all but a fraction $1/e$ of its energy through bremsstrahlung. At fixed thickness $x$, smaller $X_0$ increases the radiative loss; the relevant dimensionless measure is $x/X_0$. Low-energy charged particles can instead be dominated by collisional stopping power.^[ref]

Radiation pressure is the normal momentum flux transferred by electromagnetic radiation to a surface. A photon travelling in vacuum carries momentum $p_{\gamma }=E_{\gamma }/c=h\nu /c$, where $E_{\gamma }$ is its energy, $\nu$ its frequency, $h$ the Planck constant and $c$ the speed of light. Photon absorption transfers that momentum; specular reflection reverses its normal component. This lets light exert a force even when almost no energy becomes heat in the receiver.

Two useful formulas describe different preparations. A directed beam normally incident on a stationary perfect absorber gives pressure $I/c$, and on a stationary perfect mirror gives $2I/c$. Here $I$ is light intensity, the incident power per beam cross-sectional area perpendicular to propagation, in $\mathrm{W}{\mathrm{m}}^{-2}$. An isotropic photon gas has pressure $u/3$, where $u=U/V$ is its radiation internal energy per volume, in $\mathrm{J}{\mathrm{m}}^{-3}$. Stored energy density and travelling energy flux must be distinguished: an isotropic field can have zero net energy flux and positive radiation pressure.

The physical appeal is that the same momentum conservation connects a single photon collision, a mirror's measurable force, and a radiation-filled cavity's equation of state. The directional calculation below complements radiation pressure on a perfectly absorbing surface; the equilibrium spectrum and its temperature dependence belong to blackbody radiation and the photon gas.

For normal incidence in vacuum, a uniform beam delivers energy $IA\Delta t$ to an illuminated surface area $A$ during time $\Delta t$. A perfect absorber receives momentum $IA\Delta t/c$, so force divided by surface area gives $p_{\mathrm{abs}}=I/c$. A perfect stationary mirror sends the photons back with reversed momentum, so momentum conservation gives twice the normal transfer, $p_{\mathrm{refl}}=2I/c$. These are time-averaged pressures; the microscopic arrivals need not be uniform.

For a stipulated model light intensity $I=1000\mathrm{W}{\mathrm{m}}^{-2}$ and the exact SI speed of light $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$,These are computed predictions, not measured data or a universal sunlight value. If the beam fully covers a normally oriented surface area $A=0.010{\mathrm{m}}^2$, the intercepted radiant flux is $\Phi =IA=10\mathrm{W}$ and the absorber's force is $\Phi /c\simeq 33.356\mathrm{nN}$. Using the total source power would be wrong if only part of its beam reaches the receiver. The absorption prediction assumes complete photon absorption, while the reflection prediction assumes complete specular reflection; transmission, scattering of light and directed thermal photon emission require their own outgoing momentum accounting.

Measure the angle of incidence $\theta$ between the incoming path and the surface's normal line, with $0\le \theta <\pi /2$. There are two separate cosine factors. A fixed actual surface area $A$ intercepts only its projected area $A\cos \theta$, so $\Phi =IA\cos \theta$. Each incoming photon supplies normal momentum $(E_{\gamma }/c)\cos \theta$ on absorption, or twice that on specular reflection. Thus, for radiometric absorptance $a$ and reflected power fraction $R$,This model assumes transmitted light continues straight in the same medium and carries away its original momentum. It omits thermal recoil and assumes a stationary receiver. At $\theta =60^{\circ }$, fixed beam light intensity therefore gives one quarter of the normal-incidence pressure. For fixed intercepted radiant flux $\Phi$, however, the normal force has only one cosine factor. Equivalently, if the given quantity is receiving-surface irradiance $E_{\mathrm{inc}}=I\cos \theta$, then $p_{\perp }=(a+2R)E_{\mathrm{inc}}\cos \theta /c$. The distinction matters when interpreting an experiment's angle scan. ^[ref]

Absorption also transfers a tangential momentum component at oblique incidence. It contributes a tangential force, rather than normal pressure. Ideal specular reflection preserves the tangential momentum component, so it transfers only normal momentum.

For radiation in vacuum with isotropy, energy is distributed equally per solid angle: write $u_{\Omega }=u/(4\pi )$ for energy density per unit solid angle. Let $\theta$ be the angle between a ray's path and the wall normal. Its normal velocity $c\cos \theta$ determines the arrival rate, while normal momentum per energy is $\cos \theta /c$. Their product is ${\cos }^2\theta$. For a perfectly reflecting cavity wall,The leading two is the normal momentum reversal; the incoming hemisphere contains half of the field's directional energy density. This is also the normal component of the bulk pressure tensor, $P_{nn}={\int }_{4\pi }{u}_{\Omega }{\cos }^2\theta d\Omega =u/3$. Isotropy makes all three diagonal components equal, while their sum is $u$ because the squared direction components sum to one. The factor one third therefore depends on three-dimensional angular geometry, not on the shape of the blackbody radiation spectrum.

A perfectly absorbing wall exposed to this incoming hemisphere receives only $u/6$ from absorption. If it is in thermal equilibrium with the isotropic radiation, its outgoing emission supplies another $u/6$ of recoil, recovering $u/3$. A cold absorber before equilibrium is a different boundary condition. Similarly, the one-way incident energy flux is $cu/4$, whereas the net energy flux across a plane in the undisturbed isotropic field is zero. Dividing $cu/4$ by $c$ misses the extra normal-momentum projection and does not give the cavity pressure.

For a stipulated isotropic radiation energy density $u=6\mu \mathrm{J}{\mathrm{m}}^{-3}$, the bulk pressure is $2\mu \mathrm{Pa}$ regardless of its spectral shape. In thermal equilibrium, blackbody radiation supplies $u=a_{\mathrm{rad}}{T}^4$ with radiation constant $a_{\mathrm{rad}}=4\sigma /c$, so $p_{\mathrm{rad}}=a_{\mathrm{rad}}{T}^4/3$; Stefan–Boltzmann law defines $\sigma$. The equilibrium photon gas derives this same equation of state from its free energy.^[ref]

Radiation reaction is the influence on a source's motion of its own electromagnetic field, including the balance of radiated energy and momentum with changes of the near field. For sufficiently weak radiation loss, an averaged power balance can give the leading damping law without solving the full self-field dynamics. Replacing a constant-frequency radiation formula by its instantaneous slowly varying value requires an additional time-scale assumption; it is not an exact treatment of arbitrarily rapid motion changes.