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For a nonrelativistic point charge, let $C_{ij}=⟨a_i{a}_j⟩$ be the time-averaged second moment of acceleration, without subtracting its mean. The outgoing radiation zone fields satisfy $\mathbf{B}=\mathbf{n}\times \mathbf{E}/c$ and $\mathbf{E}=q\mathbf{n}\times (\mathbf{n}\times \mathbf{a})/(4\pi {ϵ}_0{c}^{2}r)$. Their Poynting vector gives the displayed power per solid angle. Its integral is $⟨P⟩=q^2\mathrm{tr}C/(6\pi {ϵ}_0{c}^3)$, the averaged Larmor formula.

For periodic motion the mean acceleration vanishes and $C$ is also its covariance matrix. More generally the second moment, rather than the centered covariance matrix, is the relevant quantity. The angular formula is valid to leading nonrelativistic order and evaluates source motion at retarded time.

The Bohr model assumes circular electron orbits with quantized orbital angular momentum $N\hslash$. It reproduces the hydrogen energy levels, but wavefunctions in quantum mechanics describe spatial probability distributions rather than definite trajectories. In particular, spherically symmetric Coulomb bound states have zero orbital angular momentum.

The puzzle is the coexistence of reproducible spectral lines and stable chemical atoms. A classical circular orbit in a Coulomb potential can have any negative energy: Coulomb's law alone selects no special radius or emitted frequency. Moreover, its electron has centripetal acceleration. The Larmor formula then predicts positive radiated power, so conservation of energy makes the orbit shrink. Simply drawing concentric circles explains neither stability nor a discrete spectrum.

Niels Bohr postulated nonradiating stationary states and emission with $h\nu =E_i-E_f$. His circular-orbit condition is $L=n\hslash$. These assumptions override classical radiation; the emitted frequency need not equal an orbital frequency.^[ref]

An accelerating classical charge radiates according to the Larmor formula. For a slowly evolving nonrelativistic circular electron orbit, energy balance gives $d(r^3)/dt=-4c{r}_0^2$, with $r_0$ the classical electron radius. The finite formal collapse time exposes classical atomic instability; it is not a model of a stable quantum atom.

Electromagnetic radiation is a propagating disturbance of the electric field and magnetic field that carries energy and momentum away from its source. It can be a sustained electromagnetic wave or a finite wave packet: periodic oscillation is not required. In vacuum the disturbance travels at the speed of light. A receiver responds to fields arriving now from an earlier source event, as expressed by retarded time. This is how electrical activity at one place can deliver energy and information elsewhere without a connecting wire.

Stored electromagnetic energy alone does not establish outgoing electromagnetic radiation. A stationary electric charge in electrostatics, or a steady localized electric current in magnetostatics, can store electromagnetic energy with zero Poynting vector. Static combinations can have circulating energy flux without net outward power. Near a changing source, reactive energy can return later. The near and far field distinction separates these contributions from outgoing radiation zone fields.^[ref]

A single traveling vacuum plane electromagnetic wave obeying the Maxwell equations has $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$. Its Poynting vector gives the energy density and flux of a plane electromagnetic wave:Here $I$ is wave intensity; ${ϵ}_0$ and ${\mu }_0$ are vacuum permittivity and vacuum permeability. For a monochromatic wave with linear polarization and peak amplitude $E_0$, squared cosine averages to $1/2$. Pulses require the instantaneous Poynting vector; arbitrary incident-and-reflected superpositions need not obey this relation.^[ref]

The electromagnetic momentum density is $\mathbf{g}=\mathbf{S}/c^2$. Completely absorbing a unidirectional vacuum pulse of energy $U$ transfers momentum $U/c$; see radiation pressure on a perfectly absorbing surface. A calculated $1.0\mathrm{mJ}$ pulse transfers $3.34\times 10^{-12}\mathrm{kg}m/s$. Symmetric emission can lose energy without net recoil because opposite momenta cancel.^[ref]

The direction of emission matters as much as its total power. For an oscillating electric dipole moment along $z$, the angular distribution of electric dipole radiation is strongest across the axis and vanishes along it. Its origin is geometric: the electric dipole radiation field is the transverse projection of the second time derivative of the electric dipole moment, so its magnitude contains $\sin \theta$. Squaring this electric field to obtain wave intensity gives ${\sin }^2\theta$. The electric-dipole approximation for radiation requires a source much smaller than its wavelength and observation in the radiation zone. ^[ref]

For example, the power per solid angle at $\theta =30^{\circ }$ is one quarter of its equatorial value, while the electric field amplitude is one half. To normalize the angular distribution to total power, use spherical coordinates with $d\Omega =\sin \theta d\theta d\varphi$:The integral follows by substituting $x=\cos \theta$, giving ${\int }_{-1}^1(1-x^2)dx=4/3$. This is a probability-like normalization over direction, with units of inverse steradian; it does not make the dipole an isotropic radiator. The same angular integral underlies the Larmor formula for nonrelativistic accelerated electric charge.

Detection requires the incident electromagnetic field to do something observable. An antenna develops an electrical response; a thermopile converts absorbed energy into a thermal and then electrical signal; a photodetector can respond through photon absorption. Hertz spark transmission and detection experiment makes this conversion visible as a tiny spark. Detector orientation, sensitivity, and surrounding reflections all matter: failure to see a signal does not prove that no electromagnetic radiation reached the receiver.

Field electron emission releases electrons from matter through electric-field-assisted quantum tunnelling at the surface. The emission mechanism and its rate are distinct from the subsequent acceleration and Larmor formula radiation of an electron already travelling through an approximately uniform electric field.

The relativistic generalization of the Larmor formula uses the invariant proper-acceleration magnitude: $P=-{\mu }_0{q}^2{A}^a{A}_a/(6\pi )$ in signature $(+---)$. In coordinate time this is the displayed expression. For perpendicular magnetic circular motion, $\gamma m|a|=|q|vB$, giving $P={\mu }_0{q}^4({\gamma }^2-1)B^2/(6\pi m^2)$. The statement concerns total instantaneous emitted power; directional radiation flux is not itself invariant.

The Liénard-Wiechert field is the electric field and magnetic field obtained from the retarded potentials of a moving point charge. In the nonrelativistic radiation zone, the electric radiation field is ${\mathbf{E}}_{\mathrm{rad}}=q\hat{\mathbf{n}}\times (\hat{\mathbf{n}}\times \mathbf{a})/(4\pi {ϵ}_0{c}^{2}R)$ at retarded time. Its outgoing Poynting vector integrates to the Larmor formula. Coulomb and induction fields must also be retained when a near-field calculation is required.

Choose polar angle $\theta$ from the acceleration direction. Then $|n\times q\dot{v}{|}^2=q^2|\dot{v}{|}^2{\sin }^2\theta$. Integrating over the solid angle gives $2\pi {\int }_0^{\pi }{\sin }^3\theta d\theta =8\pi /3$, and hence the specified normalization gives Larmor formulaThis uses the source's convention for radiated power; standard SI normalization has an additional factor $1/c$ in both given power formulas.

To leading order when $W\ll E$, neglect the loss while computing the head-on trajectory. Mechanical energy conservation gives $\frac{1}{2}m{\dot{r}}^2+V(r)=E=V(r_0)$ and acceleration magnitude $|V^{\prime }(r)|/m$. On each leg,The incoming and outgoing legs contribute equally. Multiplying the power by this time element, adding both legs, and assuming the integral converges givesUsing the unperturbed turning point is justified at this order in the small radiative loss, for a regular turning trajectory.

In the instantaneous rest frame the total radiation has zero net three-momentum, so its emitted four-momentum is proportional to the four-velocity. A Lorentz boost multiplies its energy by $\gamma$, while the elapsed coordinate time for the source is also multiplied by $\gamma$. Their ratio, total instantaneous radiated power, is therefore invariant. This concerns total emitted power, not the directional flux or an observer's photon arrival rate.

Use the metric of signature $(+---)$ and dots for proper-time derivatives. From $\gamma =(1-v^2{)}^{-1/2}$ obtain $\dot{\gamma }={\gamma }^3\alpha$, where $\alpha =\mathbf{v}\cdot \dot{\mathbf{v}}$. Differentiating $U=(\gamma ,\gamma \mathbf{v})$ givesSquaring in the Minkowski metric, using $1-v^2={\gamma }^{-2}$, gives $\dot{U}\cdot \dot{U}={\gamma }^6{\alpha }^2(1-v^2)-2{\gamma }^4{\alpha }^2-{\gamma }^2|\dot{\mathbf{v}}{|}^2$, henceIn the rest frame, $-{\dot{U}}^2$ is the squared ordinary acceleration, so invariance of power and the low-speed Larmor formula give $P=-({\mu }_0{q}^2/6\pi ){\dot{U}}^2$. In terms of coordinate-time acceleration $\mathbf{a}=d\mathbf{v}/dt$, this is the Lienard formulaFor circular motion perpendicular to $\mathbf{B}$, $\mathbf{a}\perp \mathbf{v}$ and the Lorentz force gives $\gamma m|\mathbf{a}|=|q|v|\mathbf{B}|$. ThusAt low speed ${\gamma }^2-1\sim v^2$, reproducing ordinary Larmor radiation with magnetic acceleration $|q|vB/m$.

Use the Minkowski metric with signature $(-,+,+,+)$. The source point is the intersection of the worldline with the observation event's past light cone: $R^{\mu }{R}_{\mu }=0$ and $R^0>0$. In components this is $c(t-t_r)=|x-y(t_r)|$ with $t_r<t$. For a timelike worldline the retarded solution is unique in the usual complete-source setting.

Writing $R=|x-y(t_r)|$, $n=(x-y)/R$ and $u^{\mu }=\gamma (c,v)$ gives $R\cdot u=-\gamma cR(1-n\cdot v/c)$. Thus the Lienard-Wiechert potentials are $\varphi =q{\mu }_0{c}^2/[4\pi R(1-n\cdot v/c)]$ and $A=q{\mu }_{0}v/[4\pi R(1-n\cdot v/c)]$. At leading order in $1/r$ and first order in $v/c$, set $R\simeq r,n\simeq \hat{r}$ and expand the denominator:

Now use $t_{-}=t-r/c$ and let $a=\dot{v}(t_{-})$. Derivatives of $1/r$ and $\hat{r}$ give near-field terms of order $r^{-2}$; the radiation terms come from ${\partial }_t{t}_{-}=1$ and $\nabla t_{-}=-\hat{r}/c$. The electric field and magnetic field are thereforeThey are transverse to the observation direction. The Poynting vector isIntegrating over a sphere uses $\int {\sin }^2\theta d\Omega =8\pi /3$, yielding the nonrelativistic Larmor formula $P=q^2{\mu }_0|a{|}^2/(6\pi c)$. For simple harmonic motion, $|a{|}^2=A^2{\omega }^4{\cos }^2(\omega t_{-})$, whose period average is $A^2{\omega }^4/2$. Multiplication by the period $2\pi /\omega$ givesThis is the leading radiation energy in the nonrelativistic and far-field approximation.

An accelerating point charge has a transverse radiation field in addition to its near field. At nonrelativistic speed and large distance $R$,The acceleration is evaluated at retarded time. Radiation fields decay as $1/R$, so their outward energy flux integrated over a sphere is independent of radius; the near electric field instead decays as $1/R^2$. The radiation vanishes along the acceleration and is strongest perpendicular to it.^[ref]

Using the Poynting vector givesThe integral uses $\int {\sin }^2\vartheta d\Omega =8\pi /3$. This is the Larmor formula. A harmonically accelerated charge with peak acceleration $a_0$ has mean power $q^2{a}_0^2/(12\pi {ϵ}_0{c}^3)$; external work supplies this radiated energy in steady operation. Uniform inertial motion does not radiate.

At relativistic speed the angular pattern becomes directionally concentrated and retarded-source factors matter. The total instantaneous radiated power generalizes to $q^2{\gamma }^6[a^2-|\mathbf{\beta }\times \mathbf{a}{|}^2]/(6\pi {ϵ}_0{c}^3)$, with $\mathbf{\beta }=\mathbf{v}/c$. The displayed simple ${\sin }^2\vartheta$ pattern is restricted to the nonrelativistic limit.

Write $C=e^2/(4\pi {ϵ}_0)=m{c}^2{r}_0$. For a slowly evolving circular Coulomb orbit, force balance and the mechanical energy giveThe supplied Larmor formula therefore yieldsSet $dE/dt=-P$. Since $dE/dt=C\dot{r}/(2r^2)$,Thus the classical radiation collapse of a Coulomb orbit obeysThe decreasing mechanical energy means a more tightly bound orbit, not a smaller speed. The formal collapse time is $a_0^3/(4r_0^{2}c)$. The nonrelativistic, small-loss-per-orbit approximation eventually fails; the result demonstrates the classical instability rather than describing an actual hydrogen atom.

Since $\mathbf{E}\cdot \mathbf{n}=0$, the Poynting vector isThe area on a distant sphere corresponding to solid angle $d\Omega$ is $r^{2}d\Omega$. ConsequentlyThe average is over the source motion at retarded time. At leading nonrelativistic order, differences between source-time and reception-time averaging are beyond the approximation. The units are power per solid angle, $\mathrm{W}\mathrm{s}{\mathrm{r}}^{-1}$.

The acceleration covariance in nonrelativistic radiation is another way to organize this result: with $C_{ij}=⟨a_i{a}_j⟩$, the angular factor is $\mathrm{tr}C-{\mathbf{n}}^{T}C\mathbf{n}$. Integrating over all observation directions uses $\int [|\mathbf{a}{|}^2-(\mathbf{n}\cdot \mathbf{a}{)}^2]d\Omega =(8\pi /3)|\mathbf{a}{|}^2$ and gives the averaged Larmor formula, $⟨P⟩=q^2⟨a^2⟩/(6\pi {ϵ}_0{c}^3)$.

Thomson scattering describes the low-recoil scattering of electromagnetic radiation by a free charge. For charge $q$ and mass $m$, the weak incident field gives peak acceleration $|q|E_0/m$. The mean Larmor formula power is thereforeThere is no binding force to introduce a characteristic frequency. Dividing by the same incident light intensity gives the Thomson cross-section:For an electron, the length inside parentheses is the classical electron radius. The cross section has units ${\mathrm{m}}^2$ and is independent of wavelength in this classical limit. The driven velocity must remain small compared with $c$, and $\hslash \omega \ll m{c}^2$ must make Compton recoil kinematics negligible.

After field electron emission, let $h$ denote the plate separation of the parallel-plate capacitor, which the question does not specify. The uniform capacitor electric field has magnitude $E_c=|V|/h$, and a nonrelativistic electron of mass $m_e$ has constant acceleration magnitude $a=e{E}_c/m_e$. The sign of its charge affects the direction of the radiation field but not its radiated power.

In the supplied nonrelativistic Liénard-Wiechert field, $\dot{\mathbf{\beta }}=\mathbf{a}/c$. At polar angle $\theta$ to the acceleration, $|\hat{\mathbf{n}}\times (\hat{\mathbf{n}}\times \mathbf{a})|=a\sin \theta$, henceThe outgoing Poynting vector has magnitude $S={ϵ}_{0}c{E}_{\mathrm{rad}}^2$. Its radiated power per solid angle is thereforeUsing $\int {\sin }^2\theta d\Omega =8\pi /3$ gives the Larmor formula and the requested constant power during the uniform-field part of the flight:The brackets in the given field require the acceleration at retarded time; the radiation reaches the far observer after the propagation delay. This derivation assumes that radiation reaction is a small correction to the acceleration and that the flight is nonrelativistic, for example $e|V|\ll m_e{c}^2$ for release from rest. Entry, emission and impact transients are not described by the uniform-field model. A result solely in terms of $V$ cannot be obtained without a plate spacing or a specified internal field.

Thermal bremsstrahlung is radiation from the acceleration of electrons in the Coulomb force of ions. At fixed temperature, the encounter rate per electron is proportional to ion number density $n_i$, and the number of radiating electrons per unit volume is $n_e$. Therefore $j_{\nu }\propto n_e{n}_i$ in a dilute, optically thin plasma. This is a binary-encounter rate, not a one-particle emission process.

For a fixed encounter geometry, the ionic electric field and hence the electron acceleration are proportional to ionic charge $Ze$. The Larmor formula is quadratic in acceleration, giving $j_{\nu }\propto Z^2$ at fixed species number densities. The dimensionless Gaunt factor $g(\nu ,T)$ corrects the simple classical result for the detailed quantum scattering and radiation calculation, after averaging over the thermal electron distribution. It depends on frequency and temperature and, in a more complete notation, ionic charge.

A classical orbiting electron accelerates and therefore radiates according to the Larmor formula. Its mechanical energy becomes more negative, its radius decreases, and the orbit spirals inward rather than remaining a stable atom. The Bohr model replaces this prediction with postulated nonradiating stationary states and quantized orbital angular momentum, together with photon emission or absorption only between states, $h\nu =|E_n-E_m|$.

For a heavy nucleus, $m_{e}vr=n\hslash$. Since the de Broglie wavelength is $h/(m_{e}v)$, this impliesThe electronic wave fits an integer number of wavelengths around the orbit. This is a standing-wave consistency interpretation of the quantization rule, not a classical mechanism that prevents an accelerating charge from radiating. For finite nuclear mass, the two-body relative coordinate uses reduced mass $\mu =m_{e}M/(m_e+M)$; the next derivation retains that correction. In the limit $M\gg m_e$, it reduces to the fixed-nucleus electron description above.

Interpret $3$ GeV as the electron's total energy; the alternative kinetic-energy reading changes the estimate only slightly at this scale. The Lorentz factor and orbital acceleration areUsing the supplied Larmor formula, ${\mu }_0=1/({ϵ}_0{c}^2)$ and the revolution time $2\pi R/v$,For $R=561$ m this gives $P=1.7405\times 10^{-10}$ W per electron and $T_{\mathrm{turn}}=1.17577\times 10^{-5}$ s. Their product isThis is the energy loss of one electron on a full circular orbit, not its energy loss per metre or the total loss of the whole bunch. The fractional loss is only about $4.26\times 10^{-6}$ of the $3$ GeV energy each revolution. Maintaining a steady stored energy requires replacing that energy through the accelerator's rf system.

For uniform circular motion, the transverse Lorentz force supplies the change in direction of three-momentum:The magnetic rigidity relation gives $pc=0.299792458B[\mathrm{T}]r[\mathrm{m}]\mathrm{GeV}=2.99792458\mathrm{GeV}$. Using $m_e{c}^2=0.511\mathrm{MeV}$,The coordinate centripetal acceleration is $a=u^2/r$. For transverse acceleration, the magnitude of the four-acceleration, or proper acceleration, is $a_0={\gamma }^{2}a$. The supplied Larmor formula, in its proper acceleration form, therefore givesThe emitted power here is also the laboratory total emitted power: in the instantaneous rest inertial frame the dipole radiation pattern has zero net radiated three-momentum, so both emitted energy and elapsed coordinate time gain the same Lorentz factor on transformation. This does not assert that a directional detector flux is invariant.

Numerically $\mathcal{T}\simeq 1.048\times 10^{-5}\mathrm{s}$, $P\simeq 2.18\times 10^{-10}\mathrm{W}$ per electron, andThe synchrotron's radiofrequency accelerating system must replace this lost energy to keep $r$ and magnetic rigidity fixed over many revolutions. The fraction lost per turn is about $4.77\times 10^{-6}$, so a fixed-orbit calculation over one turn is self-consistent.^[ref]

Metric-sign qualification. With this paper's $(-,+,+,+)$ metric signature, b givesThe minus sign in the printed Larmor formula would therefore make a positive radiative loss negative. A concrete counterexample is an instant of rest with nonzero acceleration: $F^{\alpha }=(0,m\mathbf{a})$, so the printed expression is $-q^2{a}^2/(6\pi {ϵ}_0{c}^3)$. That sign is appropriate for the opposite metric signature. With the convention already fixed in question 1, positive emitted power instead isAlternatively the printed negative quantity can be interpreted as the particle's signed radiative energy change, rather than the positive loss magnitude.

For fixed rest mass, resolve the acceleration parallel and perpendicular to the velocity, as in b. This givesWriting $C_q=q^2/(6\pi {ϵ}_0{c}^3)$, the Liénard radiated-power formula becomes $P_{\mathrm{rad}}=C_q({\gamma }^6{a}_{\parallel }^2+{\gamma }^4{a}_{\perp }^2)$. For uniform circular motion, the acceleration is perpendicular to the velocity; for straight-line motion it is parallel. Comparing the same particle at equal relativistic energy and equal laboratory acceleration magnitude $a$ therefore givesThis particular comparison makes the linear loss larger. The familiar larger loss from bending applies to a different comparison: at equal applied laboratory force magnitude $F$, $a_{\perp }=F/(m\gamma )$ and $a_{\parallel }=F/(m{\gamma }^3)$, giving $P_{\mathrm{circular}}=C_q{\gamma }^2{F}^2/m^2$ and $P_{\mathrm{linear}}=C_q{F}^2/m^2$. Stating which quantity is held fixed is essential.