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Accelerating charges radiate $P = q^2a^2/6\pi\epsilon_0c^3$; relativistic beaming, synchrotron losses per turn, bremsstrahlung and Thomson scattering.
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By the end of this lesson you will be able to compute the power radiated by an accelerating charge at low and high speed, find synchrotron losses in storage rings, and apply the Thomson cross section.
You know the retarded potentials, the dipole radiation formula, the Poynting vector and the relativistic factor $\gamma$. You saw that an oscillating dipole radiates because its charges accelerate. This lesson goes to the root: a single point charge, accelerating in any way, and how much it radiates at low and at high speed.
| Term | What it means |
|---|---|
| Larmor formula | The power $P = q^2a^2/6\pi\epsilon_0c^3$ radiated by a charge accelerating at low speed. |
| Liénard formula | The relativistic generalization: $\gamma^4$ times Larmor for circular motion, $\gamma^6$ for linear. |
| Synchrotron radiation | Radiation from relativistic charges bent by magnetic fields, beamed into a cone of angle $1/\gamma$. |
| Bremsstrahlung | "Braking radiation" from charges slowed or deflected by other charges, as in an X-ray tube. |
| Thomson scattering | Scattering of light by a free electron driven by the wave's field, with cross section $6.65 \times 10^{-29}$ m². |
| Classical electron radius | $r_e = e^2/4\pi\epsilon_0m_ec^2 = 2.82 \times 10^{-15}$ m, the length that sets Thomson scattering. |
| Radiation reaction | The recoil force on a radiating charge that carries off the energy it emits. |
A charge at rest has a Coulomb field; a charge at constant velocity carries a flattened version of it along, and neither sends energy to infinity. When a charge accelerates, the retarded potentials give an extra field that falls as $1/r$:
$$E_{\text{rad}} = \frac{q\,a\sin\theta}{4\pi\epsilon_0c^2r},$$
where $\theta$ is measured from the direction of the acceleration. Its Poynting flux falls as $1/r^2$, so a constant power crosses every large sphere. Integrating $\sin^2\theta$ over the sphere gives the Larmor formula,
$$P = \frac{q^2a^2}{6\pi\epsilon_0c^3},$$
valid when the charge moves slowly compared with light. For an electron the constants combine to $5.695 \times 10^{-54}$ W s⁴/m². At high speed, Liénard's formula applies: for acceleration perpendicular to the velocity, as in circular motion, the power is $\gamma^4$ times Larmor's, and for acceleration along the velocity it is $\gamma^6$ times. An electron in a ring loses $\Delta E = 88.5\,E^4/R$ keV per turn, with $E$ in GeV and $R$ in m.
Another way: picture
Picture a charge that suddenly jolts sideways. Its field lines, far away, still point to where it used to be, because news of the jolt travels at $c$. A kink in the lines races outward at light speed, connecting the old pattern to the new. That kink is the radiation: transverse, falling as $1/r$, and strongest perpendicular to the jolt. A steadily moving charge has no kink, so it does not radiate.
Another way: steps
Checks. The power must be positive and independent of the observer's distance. It must scale as $q^2$, so an electron and a positron radiate equally. At fixed force, acceleration goes as $1/m$ and power as $1/m^2$, so a proton radiates $1836^2 \approx 3.4 \times 10^6$ times less than an electron. And in the relativistic case, the fourth power of energy should make you pause before every upgrade.
The Liénard–Wiechert potentials, the retarded potentials of a single moving charge, give a field with two parts. The velocity field falls as $1/r^2$ and is just the Coulomb field adjusted for motion. The acceleration field falls as $1/r$ and is proportional to $a$. Only the second carries energy away.
For a slow charge the acceleration field has magnitude $qa\sin\theta/4\pi\epsilon_0c^2r$, and its magnetic partner is $E/c$. The Poynting vector is $E^2/\mu_0c$, and the power per solid angle is $q^2a^2\sin^2\theta/16\pi^2\epsilon_0c^3$. Integrating $\sin^2\theta$ over all directions gives $8\pi/3$, and the Larmor formula follows. The dipole radiation of the last lesson is the same result: an oscillating dipole is charge whose acceleration is $-\omega^2$ times its displacement, and averaging $a^2$ over a cycle gives the factor of one half that turns $12\pi$ into the dipole formula.
At high speed the radiation pattern changes dramatically. In the charge's own frame it is the familiar doughnut, but transforming to the lab frame crowds it forward into a narrow cone of half-angle about $1/\gamma$ around the velocity. A $3$ GeV electron has $\gamma \approx 5900$, so its radiation fills a cone only $0.17$ milliradians wide.
The power also grows. For circular motion, the power is $\gamma^4$ times Larmor's. For linear acceleration, where the same force produces much less acceleration in the lab, the power in terms of acceleration is $\gamma^6$ times Larmor's; in terms of the applied force it is simply Larmor's with $a = F/m$. That is why linear accelerators lose almost nothing to radiation while circular ones, where magnets supply huge transverse forces, lose a great deal.
An electron circling at energy $E$ with bending radius $R$ radiates, per turn,
$$\Delta E = \frac{e^2\gamma^4}{3\epsilon_0R} = 88.5\,\frac{E^4}{R}\ \text{keV}$$
with $E$ in GeV and $R$ in meters. The fourth power is merciless. The Large Electron–Positron collider at CERN, with a $27$ km tunnel and bending radius $3.1$ km, lost about $3.4$ GeV per electron per turn at its top energy of $104.5$ GeV; its radio-frequency cavities had to restore that on every lap. The Large Hadron Collider reuses the tunnel for protons, whose $\gamma$ at the same energy is $1836$ times smaller, so their losses are $1836^4 \approx 10^{13}$ times smaller at equal energy.
What is a nuisance for particle physics is a gift for everyone else. Synchrotron light sources are rings built to radiate: the beamed light is intense, tunable from infrared to hard X-rays, and pulsed, and it serves thousands of researchers in biology, materials science and chemistry.
An electron striking a metal target is deflected and slowed by nuclei, radiating as it does. This braking radiation forms the smooth continuum of every X-ray tube's spectrum. The maximum photon energy equals the electron's kinetic energy, so a tube run at $100$ kV produces X-rays up to $100$ keV and no higher, the Duane–Hunt limit.
The Larmor formula explains two practical facts. Heavy nuclei deflect electrons most violently, so tungsten targets radiate well; the efficiency is roughly proportional to atomic number times voltage. And the efficiency is small: at $100$ kV only about one percent of the beam power becomes X-rays, the rest heat, which is why medical tubes have rotating anodes and heavy cooling.
A free electron in a light wave is driven by the wave's electric field, $a = eE_0/m_e$, and radiates by Larmor's formula. The ratio of scattered power to incident intensity is a cross section, independent of frequency:
$$\sigma_T = \frac{8\pi}{3}r_e^2 = 6.65 \times 10^{-29}\ \text{m}^2,$$
with $r_e = 2.82 \times 10^{-15}$ m the classical electron radius. Thomson scattering is why the Sun is opaque: in its core, with about $6 \times 10^{31}$ electrons per cubic meter, a photon travels only about a quarter of a millimeter before scattering, and energy takes tens of thousands of years or more to random-walk to the surface. It is also why the early universe was opaque until electrons and protons combined into atoms, releasing the cosmic microwave background.
An electron orbiting a proton at the Bohr radius accelerates at about $9.0 \times 10^{22}$ m/s². Larmor's formula says it radiates about $4.7 \times 10^{-8}$ W. Its binding energy is only $13.6$ eV, $2.18 \times 10^{-18}$ J, so it would lose it in about $10^{-11}$ s, spiraling into the nucleus while broadcasting a chirp of rising frequency.
Atoms plainly do not collapse, and this was one of the sharpest failures of classical physics. Bohr's postulate that certain orbits do not radiate, and later quantum mechanics, resolved it: a stationary state has no oscillating charge distribution and so emits nothing. Radiation happens only in transitions, and the quantum rates of the earlier lesson on emission and absorption reproduce the Larmor formula in the limit of large quantum numbers.
A radiating charge must lose energy, so a force must act on it. The Abraham–Lorentz force, $F = m\tau\dot{a}$ with $\tau = q^2/6\pi\epsilon_0mc^3$, is the classical answer. For an electron $\tau = 6.26 \times 10^{-24}$ s, the time light takes to cross a distance comparable to the classical radius. On any ordinary time scale the force is tiny, which is why the Larmor formula can be applied to a trajectory computed without it.
The force has famous pathologies: solutions that run away exponentially with no applied force, and pre-acceleration that begins before a force is applied. They show classical electrodynamics reaching its limits at the scale of $r_e$, where quantum electrodynamics takes over. In the most intense laser fields now available, radiation reaction on electrons has been measured, testing exactly this frontier.
The National Synchrotron Light Source II at Brookhaven National Laboratory in New York stores $3$ GeV electrons in a ring whose dipole magnets bend them with a radius of about $25$ m. Each electron radiates about $287$ keV per turn from the dipoles alone, and with a $500$ mA beam that is more than $140$ kW of X-rays and ultraviolet light, replaced continuously by radio-frequency cavities.
The light is the point. Because the electrons are relativistic, it comes out in a cone only a fraction of a milliradian wide, far brighter than any laboratory X-ray tube, and undulators, rows of alternating magnets, make it brighter still by adding the radiation from many wiggles coherently. Researchers use it to solve protein structures for drug design, watch batteries charge atom by atom and map strain in turbine blades. The Advanced Photon Source at Argonne National Laboratory and similar rings worldwide are built on the physics of this lesson.
Physicists planning the next electron–positron collider face the $E^4/R$ law directly. A circular machine at a few hundred GeV must be enormous to keep synchrotron losses manageable: proposals for a Future Circular Collider at CERN call for a tunnel of about $91$ km, and even then the radio-frequency system must replace tens of megawatts of radiation.
A linear collider avoids the problem because its acceleration lies along the velocity, where for a given accelerating force the radiated power is negligible. The trade-off is that each bunch gets only one chance to collide, instead of passing through the detectors millions of times. Proton machines escape both problems, because $1836^4$ suppresses their losses, which is why the Large Hadron Collider could reach $6.8$ TeV per beam in the tunnel that capped the electron machine at about $105$ GeV.
It is tempting to think any moving charge sends out waves, since a current makes a magnetic field and changing fields make waves. But a charge gliding at constant velocity simply carries its field along with it; in its own rest frame it is a static Coulomb charge, and a static charge radiates nothing. Only acceleration produces the $1/r$ field that carries energy away. A steady current in a straight wire does not radiate; an alternating one does, because its charges accelerate.
A second misconception is that heavier charges radiate more because they carry more momentum. Radiation depends on charge and acceleration. At the same force, a proton accelerates $1836$ times less than an electron and radiates about three million times less power, which is why the Large Hadron Collider can accelerate protons in a tunnel where electrons of the same energy would radiate away nearly everything.
Find the acceleration of an electron orbiting at the Bohr radius $5.29 \times 10^{-11}$ m.
$a = \dfrac{e^2}{4\pi\epsilon_0m_er^2} = \dfrac{8.988 \times 10^9 \times (1.602 \times 10^{-19})^2}{9.109 \times 10^{-31} \times (5.29 \times 10^{-11})^2} = 9.05 \times 10^{22}\ \text{m/s}^2$
Coulomb force over mass.
Square the acceleration.
$a^2 = 8.19 \times 10^{45}\ \text{m}^2/\text{s}^4$
Larmor needs its square.
Evaluate the Larmor power.
$P = 5.695 \times 10^{-54} \times 8.19 \times 10^{45} = 4.66 \times 10^{-8}\ \text{W}$
Tiny in watts, but huge for one atom.
Write the binding energy in joules.
$13.6\ \text{eV} = 2.18 \times 10^{-18}\ \text{J}$
The energy the electron could lose before reaching the nucleus, roughly.
Estimate the time to radiate it.
$t \sim \dfrac{2.18 \times 10^{-18}}{4.66 \times 10^{-8}} = 4.7 \times 10^{-11}\ \text{s}$
A careful integration along the spiral gives $1.6 \times 10^{-11}$ s.
Find $\gamma$ for $104.5$ GeV electrons.
$\gamma = \dfrac{104.5}{0.000511} = 2.045 \times 10^5$
Energy over rest energy.
Raise it to the fourth power.
$\gamma^4 = 1.749 \times 10^{21}$
The factor Liénard adds to Larmor.
Evaluate the prefactor.
$\dfrac{e^2}{3\epsilon_0} = \dfrac{(1.602 \times 10^{-19})^2}{3 \times 8.854 \times 10^{-12}} = 9.66 \times 10^{-28}\ \text{J m}$
From $\Delta E = e^2\gamma^4/3\epsilon_0R$.
Divide by the bending radius of $3096$ m.
$\Delta E = \dfrac{9.66 \times 10^{-28} \times 1.749 \times 10^{21}}{3096} = 5.46 \times 10^{-10}\ \text{J}$
The loss per turn in joules.
Convert to GeV.
$\Delta E = \dfrac{5.46 \times 10^{-10}}{1.602 \times 10^{-10}} = 3.41\ \text{GeV}$
More than three percent of the beam energy on every lap.
Check with the practical formula.
$88.5 \times \dfrac{104.5^4}{3096}\ \text{keV} = 3.41 \times 10^6\ \text{keV}$
The two routes agree.
Drive a free electron with a wave of amplitude $E_0$.
$a = \dfrac{eE_0}{m_e}\cos\omega t$
The wave's magnetic force is negligible at low intensity.
Average the Larmor power over a cycle.
$\langle P\rangle = \dfrac{e^2}{6\pi\epsilon_0c^3}\cdot\dfrac{e^2E_0^2}{2m_e^2}$
The average of $\cos^2$ is one half.
Write the incident intensity.
$I = \tfrac{1}{2}c\epsilon_0E_0^2$
The time-averaged Poynting flux of the wave.
Divide to get a cross section.
$\sigma_T = \dfrac{\langle P\rangle}{I} = \dfrac{8\pi}{3}\left(\dfrac{e^2}{4\pi\epsilon_0m_ec^2}\right)^2 = \dfrac{8\pi}{3}r_e^2$
The field amplitude cancels.
Evaluate the cross section.
$\sigma_T = \dfrac{8\pi}{3} \times (2.818 \times 10^{-15})^2 = 6.65 \times 10^{-29}\ \text{m}^2$
Independent of frequency.
Find the mean free path in the solar core.
$\ell = \dfrac{1}{n_e\sigma_T} = \dfrac{1}{6 \times 10^{31} \times 6.65 \times 10^{-29}} = 2.5 \times 10^{-4}\ \text{m}$
A quarter of a millimeter between scatterings.
Write the practical formula.
$\Delta E = 88.5\,\dfrac{E^4}{R}\ \text{keV}$
Electrons, $E$ in GeV, $R$ in m.
Raise the energy to the fourth power.
$3^4 = 81$
The steep dependence.
Evaluate the loss.
A slow electron radiates $43$ fW while it accelerates. If its acceleration triples, how much does it radiate?
Complete the worked solution: electrons lose $3$ keV per turn in a ring. Find the loss per turn at twice the energy in the same ring, then at twice the energy in a ring of twice the radius, and the extra loss the first upgrade adds, in keV.
Scale the energy by two.
$\Delta E(2E, R) = 2^4 \times 3 =$ a
The fourth power of energy.
Double the bending radius too.
$\Delta E(2E, 2R) = \dfrac{2^4 \times 3}{2} =$ b
A gentler bend radiates less.
Subtract the original loss.
$\Delta E(2E, R) - 3 =$ c
Fifteen times the original loss.
Match each result about radiation from point charges to its expression.
| $q^2a^2/6\pi\epsilon_0c^3$ | $\propto\sin^2\theta$ | $\gamma^4$ times Larmor | $\gamma^6$ times Larmor | |
|---|---|---|---|---|
| slow-charge power | ||||
| low-speed pattern | ||||
| fast circular motion | ||||
| fast linear motion |
An electron ring bends its beam with radius $100$ m. Fill in the energy each electron radiates per turn, in keV, at beam energies of $1$, $2$, $3$ and $5$ GeV.
| $\Delta E$ (keV) | |
|---|---|
| $1$ GeV | |
| $2$ GeV | |
| $3$ GeV | |
| $5$ GeV |
An electron moving slowly has an acceleration of $7 \times 10^{20}$ m/s². How much power does it radiate, in femtowatts? Use $\dfrac{e^2}{6\pi\epsilon_0c^3} = 5.695 \times 10^{-54}$ W s⁴/m².
Answer: fW
Electrons of energy $2.5$ GeV circulate in a storage ring whose bending magnets curve them with radius $8$ m. How much energy does each electron radiate per turn, in MeV?
Answer: MeV
At a synchrotron light source, each electron radiates $236$ keV per turn and the stored beam current is $400$ mA. How much power must the radio-frequency cavities supply to the beam, in kW?
Answer: kW
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Complete the worked solution: electrons lose $17$ keV per turn in a ring. Find the loss per turn at twice the energy in the same ring, then at twice the energy in a ring of twice the radius, and the extra loss the first upgrade adds, in keV.
Scale the energy by two.
$\Delta E(2E, R) = 2^4 \times 17 =$ a
The fourth power of energy.
Double the bending radius too.
$\Delta E(2E, 2R) = \dfrac{2^4 \times 17}{2} =$ b
A gentler bend radiates less.
Subtract the original loss.
$\Delta E(2E, R) - 17 =$ c
Fifteen times the original loss.
You can compute radiation from accelerating charges. Explain to someone why atoms should collapse classically and why they do not.
19. Your turn: a light source stores $3$ GeV electrons with a bending radius of $25$ m. How much energy does each electron radiate per turn, in keV?, step 3
$\Delta E = \dfrac{88.5 \times 81}{25} = 286.7\ \text{keV}$
Restored by the radio-frequency cavities every turn.