Larmor Formula for Radiated Power

Statement

For a point charge \( q \) undergoing acceleration \( \vec{a} \) with speed \( v \ll c \), the radiation (far-zone) part of the Liénard–Wiechert fields falls off as \( 1/R \); integrating the Poynting flux of this field over a large sphere yields the total instantaneous power radiated, \[ P = \frac{q^2 a^2}{6\pi\varepsilon_0 c^3}, \] the Larmor formula: radiated power is proportional to the square of the charge and the square of its acceleration, and independent of the sphere radius.

Why it matters

The Larmor formula is the bridge between mechanics and light: it says that acceleration, not velocity, is the source of radiation. Every classical account of emission — antennas, Thomson scattering of X-rays, bremsstrahlung in an X-ray tube, cyclotron and (via its relativistic extension) synchrotron radiation — is this one result dressed in different kinematics.

It also exposes the internal crisis of classical physics. Applied to an electron orbiting a proton, it predicts collapse of the atom in about \( 10^{-11}\ \mathrm{s} \). That quantitative failure, computed in Problem 3 below, is one of the sharpest motivations for quantum mechanics.

Assumptions

Derivation

Result

Limiting cases

• \( a \to 0 \): \( P \to 0 \). A charge in uniform motion radiates nothing — consistent with relativity, since one can boost to its rest frame where only a static Coulomb field exists.
• Harmonic motion \( x = x_0\cos\omega t \): \( \langle a^2\rangle = \tfrac{1}{2}x_0^2\omega^4 \), so \( \langle P\rangle = \dfrac{q^2 x_0^2 \omega^4}{12\pi\varepsilon_0 c^3} \) — the \( \omega^4 \) law behind Rayleigh scattering and the blue sky.
• \( \beta \to \) finite: the formula is the leading term of Liénard's result, recovered from it exactly as \( \gamma \to 1 \); for circular relativistic motion power is enhanced by \( \gamma^4 \) (synchrotron).
• Driven free electron in a wave of intensity \( I \): dividing \( \langle P\rangle \) by \( I \) reproduces the Thomson cross-section \( \sigma_T = \tfrac{8\pi}{3}r_e^2 \) (Problem 2), the low-frequency limit of Compton scattering.
• Angular distribution: along \( \vec{a} \) (\( \theta = 0 \)) the flux vanishes identically; a distant observer on the acceleration axis sees no radiation at all.

Breaks when

Failure modes

• Keeping the velocity field in the flux. Students integrate the full \( \vec{S} \) at finite \( R \) and obtain \( R \)-dependent "power". Only the \( 1/R \) fields contribute as \( R\to\infty \); cross terms fall as \( 1/R^3 \) and die.
• Evaluating \( \vec{a} \) at observer time. Everything in the field expression lives at \( t_r = t - R/c \). For pulsed acceleration this misplaces the emission in time by \( R/c \).
• Maximum along the acceleration. The \( \sin^2\theta \) pattern is zero along \( \vec{a} \) and maximal broadside. Drawing the lobe pointing along \( \vec{a} \) inverts the physics.
• \( \int \sin^2\theta\, d\Omega = 4\pi\cdot\frac{1}{2} \). The solid-angle measure carries an extra \( \sin\theta \); the correct value is \( 8\pi/3 \), not \( 2\pi \). This error changes the \( 6\pi \) in the final denominator to \( 8\pi \).
• Applying Larmor to relativistic beams. Using \( P \propto a^2 \) with the lab-frame acceleration of a synchrotron electron misses the \( \gamma^4 \) enhancement; use Liénard's generalization.
• Confusing power and force. \( P \) is energy per unit time carried by fields; it is not \( \vec{F}\cdot\vec{v} \) of any single force on the charge. Radiation reaction is a separate (and subtler) bookkeeping.
• Gaussian/SI hybrid formulas. Mixing \( P = \frac{2q^2a^2}{3c^3} \) (Gaussian) with SI charges in coulombs produces errors of \( \sim 10^{10} \). Fix the unit system before any numbers.

Discussion

The deepest content of the derivation is the survival of the \( 1/R \) field. A static or uniformly moving charge has fields \( \sim 1/R^2 \), whose energy flux through a large sphere vanishes: the energy stays attached to the charge. Acceleration creates a transverse field component that falls only as \( 1/R \), so \( |\vec{S}|R^2 \) tends to a finite limit — energy that has detached from the source and will never return. Radiation is precisely this irreversibly exported \( 1/R \) structure, a kink in the field lines propagating outward at \( c \) (the Thomson kink construction reproduces \( E_\perp \propto a\sin\theta/R \) geometrically).

The formula unifies an enormous range of phenomena through the single scale \( \tau_e = q^2/(6\pi\varepsilon_0 mc^3) \), with \( P = m\tau_e a^2 \). For the electron \( c\tau_e = \tfrac{2}{3}r_e \) involves the classical electron radius \( r_e = 2.818\times10^{-15}\ \mathrm{m} \): Thomson scattering (\( \sigma_T = \frac{8\pi}{3}r_e^2 \)), bremsstrahlung efficiency in X-ray tubes, cyclotron cooling of plasmas, and the antenna \( \omega^4 \) law are all the same statement \( P \propto a^2 \) evaluated on different trajectories. Because \( a = F/m \), radiation at fixed force scales as \( 1/m^2 \) — this is why electrons dominate radiative losses and why proton synchrotrons can reach far higher energies in the same ring.

Relativistically, the Larmor formula is the rest-frame value of a Lorentz invariant. Writing \( P = \frac{q^2}{6\pi\varepsilon_0 m^2 c^3}\,\frac{dp_\mu}{d\tau}\frac{dp^\mu}{d\tau} \) (with proper time \( \tau \) and four-momentum \( p^\mu \)) shows that emitted power — energy over time, both transforming the same way between frames — is a scalar; evaluating the invariant in an arbitrary frame yields Liénard's \( \gamma^6\!\left[\dot{\vec\beta}^2 - (\vec\beta\times\dot{\vec\beta})^2\right] \) structure. A famous subtlety accompanies uniform acceleration: a uniformly accelerated charge radiates per Larmor, yet a co-accelerating (Rindler) observer detects no radiation, because the radiation escapes behind that observer's horizon — radiation content is not local to the charge but a property of the asymptotic field, and no conflict with the equivalence principle survives careful analysis.

Common misconceptions. (i) "Fast charges radiate" — speed alone is irrelevant; a charge coasting at \( 0.99c \) in vacuum emits nothing. (ii) "The radiated power depends on where you measure it" — the \( R^2 \) of the sphere exactly cancels the \( 1/R^2 \) of the flux; every concentric sphere passes the same power (retardation-shifted). (iii) "Larmor gives the force of radiation reaction" — it gives the energy budget only; the reaction force involves \( \dot{\vec{a}} \), not \( \vec{a} \), and time-averages to the Larmor loss only for bounded motion.

Worked examples

Example 1 — Classical electron in the Bohr orbit: how fast does it lose energy? An electron circles a proton at radius \( r = a_0 = 5.29\times10^{-11}\ \mathrm{m} \). Find the radiated power and the timescale to lose its binding energy (13.6 eV).

Example 2 — Radiative loss in an X-ray tube gap. An electron is accelerated from rest through 10 kV across a uniform-field gap of length \( d = 5.0\ \mathrm{cm} \). What power does it radiate, and what fraction of its final kinetic energy is lost to radiation in transit?

Problems

1. A proton experiences a constant acceleration \( a = 1.0\times10^{18}\ \mathrm{m/s^2} \). Compute the radiated power, and state in which direction an observer detects maximum flux.
   Solution

   The prefactor depends only on \( q^2 = e^2 \), so it is the same as for the electron: \( e^2/(6\pi\varepsilon_0 c^3) = 5.71\times10^{-54}\ \mathrm{W\,s^4/m^2} \). Then \( P = 5.71\times10^{-54}\times(1.0\times10^{18})^2 = 5.7\times10^{-18}\ \mathrm{W} \). The angular distribution is \( dP/d\Omega \propto \sin^2\theta \), so the maximum flux is detected in the plane perpendicular to \( \vec{a} \) (\( \theta = 90^\circ \)); along the acceleration axis the flux is zero. Note the mass of the proton never enters: for a given acceleration all charges \( \pm e \) radiate identically. (For a given force, the proton radiates \( (m_e/m_p)^2 \approx 3\times10^{-7} \) times as much as an electron.)
2. A free electron sits in a linearly polarized plane wave \( E = E_0\cos\omega t \) of intensity \( I = \tfrac{1}{2}\varepsilon_0 c E_0^2 \). Treating the motion nonrelativistically, find the time-averaged reradiated power, and show that \( \sigma \equiv \langle P\rangle / I \) is the frequency-independent Thomson cross-section. Evaluate \( \sigma \) numerically.
   Solution

   Equation of motion: \( m_e a = -eE_0\cos\omega t \), so \( a(t) = -(eE_0/m_e)\cos\omega t \) and \( \langle a^2\rangle = \tfrac{1}{2}(eE_0/m_e)^2 \). Larmor averaged: \( \langle P\rangle = \dfrac{e^2}{6\pi\varepsilon_0 c^3}\cdot\dfrac{e^2E_0^2}{2m_e^2} = \dfrac{e^4E_0^2}{12\pi\varepsilon_0 m_e^2 c^3} \). Divide by \( I = \tfrac{1}{2}\varepsilon_0 cE_0^2 \): \( \sigma = \dfrac{e^4}{6\pi\varepsilon_0^2 m_e^2 c^4} = \dfrac{8\pi}{3}\left(\dfrac{e^2}{4\pi\varepsilon_0 m_e c^2}\right)^2 = \dfrac{8\pi}{3}r_e^2 \). Both \( E_0 \) and \( \omega \) cancel — Thomson scattering is achromatic. Numerically \( r_e = 2.818\times10^{-15}\ \mathrm{m} \), so \( \sigma_T = \frac{8\pi}{3}(2.818\times10^{-15})^2 = 6.65\times10^{-29}\ \mathrm{m^2} = 0.665\ \mathrm{barn} \). This cross-section governs the opacity of ionized plasmas from stellar interiors to the pre-recombination universe.
3. Classical collapse of hydrogen. An electron spirals slowly inward from \( r_0 = a_0 = 5.29\times10^{-11}\ \mathrm{m} \), remaining on a quasi-circular orbit (energy \( E(r) = -\,e^2/8\pi\varepsilon_0 r \)). Using \( dE/dt = -P \), derive the collapse time \( t_c = \dfrac{r_0^3\, m_e^2 c^3 (4\pi\varepsilon_0)^2}{4e^4} \) and evaluate it.
   Solution

   With \( k \equiv 1/4\pi\varepsilon_0 \): circular orbit gives \( a = ke^2/m_e r^2 \) and \( E = -ke^2/2r \), so \( \dfrac{dE}{dt} = \dfrac{ke^2}{2r^2}\dfrac{dr}{dt} \). Larmor: \( P = \dfrac{e^2a^2}{6\pi\varepsilon_0 c^3} = \dfrac{2}{3}\dfrac{ke^2}{c^3}\cdot\dfrac{k^2e^4}{m_e^2r^4} \). Setting \( dE/dt = -P \): \( \dfrac{dr}{dt} = -\dfrac{4}{3}\dfrac{k^2e^4}{m_e^2c^3 r^2} \). Separate and integrate from \( r_0 \) to 0: \( \int_0^{r_0} r^2\,dr = \dfrac{r_0^3}{3} = \dfrac{4}{3}\dfrac{k^2e^4}{m_e^2c^3}\,t_c \), hence \( t_c = \dfrac{r_0^3 m_e^2 c^3}{4k^2e^4} = \dfrac{r_0^3 m_e^2c^3(4\pi\varepsilon_0)^2}{4e^4} \). Numbers: \( r_0^3 = 1.48\times10^{-31}\ \mathrm{m^3} \); \( m_e^2 = 8.30\times10^{-61}\ \mathrm{kg^2} \); \( c^3 = 2.69\times10^{25}\ \mathrm{m^3/s^3} \); numerator \( = 3.31\times10^{-66} \). Denominator \( 4k^2e^4 = 4(8.99\times10^9)^2(1.602\times10^{-19})^4 = 2.13\times10^{-55} \). So \( t_c = 1.6\times10^{-11}\ \mathrm{s} \). Classical electrodynamics destroys the atom in 16 picoseconds; the quasi-circular assumption is self-consistent because the energy lost per orbit is tiny until the final moments.
4. Cyclotron cooling. An electron moves at \( v = 0.010\,c \) perpendicular to a uniform field \( B = 1.0\ \mathrm{T} \). (a) Find the radiated power. (b) Show that the kinetic energy decays exponentially, \( E_k(t) = E_k(0)e^{-t/\tau} \), and evaluate \( \tau \).
   Solution

   (a) \( a = \dfrac{evB}{m_e} = \dfrac{(1.602\times10^{-19})(3.0\times10^{6})(1.0)}{9.109\times10^{-31}} = 5.28\times10^{17}\ \mathrm{m/s^2} \). Then \( P = 5.71\times10^{-54}\times(5.28\times10^{17})^2 = 1.6\times10^{-18}\ \mathrm{W} \). (b) Since \( a = evB/m_e \propto v \), we have \( P = \dfrac{e^4B^2}{6\pi\varepsilon_0 m_e^2c^3}v^2 = \dfrac{e^4B^2}{3\pi\varepsilon_0 m_e^3 c^3}E_k \) using \( E_k = \tfrac12 m_ev^2 \). Thus \( \dfrac{dE_k}{dt} = -\dfrac{E_k}{\tau} \) with \( \tau = \dfrac{3\pi\varepsilon_0 m_e^3c^3}{e^4B^2} \), giving exponential decay. Numerically: \( E_k = \tfrac12(9.109\times10^{-31})(3.0\times10^6)^2 = 4.1\times10^{-18}\ \mathrm{J} \) and \( \tau = E_k/P = 4.1\times10^{-18}/1.6\times10^{-18} \approx 2.6\ \mathrm{s} \). (Check with the closed form: \( \tau = \dfrac{3\pi(8.854\times10^{-12})(9.109\times10^{-31})^3(2.998\times10^8)^3}{(1.602\times10^{-19})^4(1.0)^2} \approx 2.6\ \mathrm{s} \).) Note \( \tau \propto 1/B^2 \) and is independent of \( v \): every nonrelativistic electron in a 1 T trap cools with the same 2.6 s time constant — the principle behind measuring single-electron cyclotron states in Penning traps.
5. Oscillating charge as an antenna element. A charge \( q = 1.0\ \mathrm{nC} \) oscillates along a line, \( x(t) = x_0\cos\omega t \), with \( x_0 = 1.0\ \mathrm{mm} \) and frequency \( f = 100\ \mathrm{MHz} \). (a) Derive \( \langle P\rangle = \dfrac{q^2x_0^2\omega^4}{12\pi\varepsilon_0 c^3} \) and evaluate it. (b) Verify the small-amplitude (dipole) assumption \( x_0 \ll \lambda \), and comment on the \( \omega^4 \) scaling.
   Solution

   (a) \( a(t) = -x_0\omega^2\cos\omega t \), so \( \langle a^2\rangle = \tfrac12 x_0^2\omega^4 \) and \( \langle P\rangle = \dfrac{q^2}{6\pi\varepsilon_0c^3}\cdot\dfrac{x_0^2\omega^4}{2} = \dfrac{q^2x_0^2\omega^4}{12\pi\varepsilon_0c^3} \). With \( \omega = 2\pi f = 6.28\times10^{8}\ \mathrm{s^{-1}} \): \( \omega^4 = 1.56\times10^{35}\ \mathrm{s^{-4}} \), \( q^2x_0^2 = (10^{-9})^2(10^{-3})^2 = 10^{-24}\ \mathrm{C^2m^2} \), and \( 12\pi\varepsilon_0c^3 = 8.99\times10^{15} \) SI. So \( \langle P\rangle = \dfrac{10^{-24}\times1.56\times10^{35}}{8.99\times10^{15}} = 1.7\times10^{-5}\ \mathrm{W} \approx 17\ \mu\mathrm{W} \). (b) \( \lambda = c/f = 3.0\ \mathrm{m} \gg x_0 = 1\ \mathrm{mm} \): the dipole approximation is excellent, and the peak speed \( x_0\omega = 6.3\times10^{5}\ \mathrm{m/s} = 0.002c \) is safely nonrelativistic. The \( \omega^4 \) scaling means doubling the frequency at fixed amplitude multiplies the power by 16 — the same law that makes bound electrons in air molecules scatter blue sunlight (\( \omega_{\mathrm{blue}}/\omega_{\mathrm{red}} \approx 1.7 \), so \( \approx 8\times \) more strongly) and paints the sky.