10.07.05 · em-sr / radiation

Synchrotron radiation: relativistic circular motion and the 1/gamma^4 power formula

shipped3 tiersLean: none

Anchor (Master): Jackson, Classical Electrodynamics, 3rd ed. (1999), Ch. 14.6-14.7; Rybicki & Lightman, Radiative Processes in Astrophysics (1979), Ch. 6

Intuition Beginner

When a charged particle moves in a circle at nearly the speed of light, it emits an intense, narrow beam of radiation called synchrotron radiation. The faster the particle goes, the more tightly the radiation is focused into a forward-pointing cone.

The key feature is the beaming effect. A relativistic particle (one moving close to the speed of light) emits its radiation in a narrow cone of half-angle $θ \approx 1/ γ$ around the direction of motion, where $γ = 1/ 1 - v^{2} / c^{2}$ is the Lorentz factor. For a 1 GeV electron, $γ \approx 2000$ , so the cone is only about 0.03 degrees wide.

The radiated power goes as $γ^{4}$ , which means doubling the energy increases the radiation loss by a factor of 16. This has enormous practical consequences: it limits the maximum energy of circular electron accelerators and makes synchrotron radiation a powerful light source for experiments.

Synthesis. Synchrotron radiation is the dominant energy-loss mechanism for relativistic electrons in circular orbits. The foundational insight is that relativistic beaming concentrates the radiation into a narrow forward cone. The central message is that the power scales as $γ^{4}$ , making synchrotron radiation both a nuisance (for accelerator physicists) and a tool (for materials scientists). Putting these together, synchrotron radiation limits circular electron accelerators, provides intense tunable X-ray sources, and is observed throughout astrophysics from Jupiter's magnetosphere to distant quasars.

Visual Beginner

Quantity	Expression	Typical value (1 GeV e-)
Lorentz factor	$γ = E / m c^{2}$	1957
Beaming half-angle	$θ \sim 1/ γ$	0.03 degrees
Radiated power	$P \propto γ^{4}$	0.88 keV/rev
Critical frequency	$ω_{c} \sim γ^{3} c / R$	UV/X-ray

Worked example Beginner

For an electron with energy $E = 1$ GeV in a storage ring of radius $R = 100$ m:

$γ = E / (m_{e} c^{2}) = 1 0^{9} / (0.511 \times 1 0^{6}) = 1957$

$θ_{beam} \approx 1/ γ = 5.1 \times 1 0^{- 4}$ rad $= 0.029$ degrees

The energy lost per revolution is $Δ E = (4 π /3) (r_{e} / R) γ^{4} m_{e} c^{2}$ where $r_{e} = 2.82 \times 1 0^{- 15}$ m is the classical electron radius.

$Δ E = (4 π /3) (2.82 \times 1 0^{- 15} /100) (1957)^{4} \times 0.511$ MeV

$= 1.41 \times 1 0^{- 16}$ J $= 0.88$ keV.

This energy must be resupplied by radio-frequency cavities each revolution, or the electron spirals inward.

Check your understanding Beginner

Formal definition Intermediate+

The Liénard result (relativistic Larmor formula). The total radiated power from a charge $q$ with velocity $v$ and acceleration $a$ is:

P = \frac{q ^{2}}{6 π ϵ _{0} c} γ^{6} [(a)^{2} - (\frac{v \times a}{c})^{2}]

Separating into parallel and perpendicular components: $P = \frac{q ^{2}}{6 π ϵ _{0} c ^{3}} γ^{4} (a_{⊥}^{2} + γ^{2} a_{∥}^{2})$ .

Circular motion ( $a_{∥} = 0$ ). For a charge in a circular orbit of radius $R$ with speed $v$ : $a_{⊥} = v^{2} / R$ . For $v \approx c$ :

P_{sync} = \frac{q ^{2} c}{6 π ϵ _{0}} \frac{γ ^{4}}{R ^{2}}

The energy lost per revolution is:

Δ E = \frac{q ^{2}}{3 ϵ _{0}} \frac{γ ^{4}}{R}

The critical frequency. The synchrotron spectrum extends up to a characteristic frequency:

ω_{c} = \frac{3}{2} γ^{3} \frac{c}{R}

This is the frequency at which the spectral power $P (ω)$ is maximised. The factor $γ^{3}$ arises because the observer sees a pulse of duration $Δ t \sim R / (c γ^{3})$ due to the time-compression of the beamed radiation.

Angular distribution. The instantaneous angular distribution of radiation from a relativistic charge peaks at angle $θ_{peak} \approx 1/ (2 γ)$ from the direction of motion, with the angular width of order $1/ γ$ .

Key derivation Intermediate+

Key derivation (Synchrotron power from the Liénard formula).

Result. A charge q in a circular orbit of radius R at ultrarelativistic speed (gamma >> 1) radiates total power $P = (q^{2} c γ^{4}) / (6 π ϵ_{0} R^{2})$ .

Derivation. Starting from the Liénard result for the total radiated power:

P = \frac{q ^{2}}{6 π ϵ _{0} c} γ^{4} (a_{⊥}^{2} + γ^{2} a_{∥}^{2}) / c^{2}

For circular motion with $a_{∥} = 0$ and $a_{⊥} = v^{2} / R \approx c^{2} / R$ :

P = \frac{q ^{2}}{6 π ϵ _{0} c} γ^{4} \frac{c ^{4}}{R ^{2} c ^{2}} = \frac{q ^{2} c}{6 π ϵ _{0}} \frac{γ ^{4}}{R ^{2}}

The energy lost per revolution (period $T = 2 π R / v \approx 2 π R / c$ ):

Δ E = P \cdot T = \frac{q ^{2} c γ ^{4}}{6 π ϵ _{0} R ^{2}} \cdot \frac{2 π R}{c} = \frac{q ^{2}}{3 ϵ _{0}} \frac{γ ^{4}}{R}

Expressed in terms of the classical electron radius $r_{e} = e^{2} / (4 π ϵ_{0} m_{e} c^{2})$ :

Δ E = \frac{4 π}{3} \frac{r _{e}}{R} γ^{4} m_{e} c^{2} □

Bridge. The synchrotron power formula connects the non-relativistic Larmor result to the extreme relativistic regime. The foundational insight is that the $γ^{4}$ factor arises from two sources: the relativistic transformation of the fields ( $γ^{2}$ from the field strength) and the beaming of radiation into a narrow cone (another $γ^{2}$ from the angular concentration). The central message is that synchrotron radiation becomes overwhelmingly important at high energies, fundamentally limiting the design of circular accelerators. Putting these together, the $γ^{4}$ scaling explains why electron synchrotrons lose energy much faster than proton synchrotrons (protons have much larger mass and thus smaller $γ$ for the same energy), the critical frequency $ω_{c} \propto γ^{3} / R$ determines the X-ray spectrum useful for materials science, and the angular beaming makes synchrotron light sources highly collimated 10.07.06.

Exercises Intermediate+

Exercise 2 (easy, symbolic).

The Large Hadron Collider (LHC) has a radius of approximately 4.3 km and accelerates protons to 7 TeV. Compute the energy lost per revolution to synchrotron radiation.

Hint

$Δ E = (q^{2} / (3 ϵ_{0})) (γ^{4} / R)$ . $γ = E / (m_{p} c^{2}) = 7 \times 1 0^{12} / (938 \times 1 0^{6}) = 7463$ .

Answer

$γ = 7 \times 1 0^{12} / (938 \times 1 0^{6}) = 7463$ .

$Δ E = \frac{e ^{2}}{3 ϵ _{0}} \frac{γ ^{4}}{R} = \frac{( 1.6 \times 1 0 ^{- 19} ) ^{2}}{3 \times 8.85 \times 1 0 ^{- 12}} \frac{( 7463 ) ^{4}}{4300}$

$= \frac{2.56 \times 1 0 ^{- 38}}{2.655 \times 1 0 ^{- 11}} \times \frac{3.10 \times 1 0 ^{15}}{4300} = 9.64 \times 1 0^{- 28} \times 7.21 \times 1 0^{11} = 6.95 \times 1 0^{- 16}$ J

$\approx 4.3$ keV per revolution.

Despite the enormous energy, the loss is modest because of the proton's large mass. For electrons at the same energy: $γ_{e} \approx 1.37 \times 1 0^{7}$ , and $Δ E \propto γ^{4}$ would be $(1.37 \times 1 0^{7} /7463)^{4} \approx 1 0^{14}$ times larger — completely impractical. This is why no circular electron accelerator can reach TeV energies.

Exercise 3 (medium, symbolic).

Derive the pulse duration $Δ t$ observed by a stationary observer when the beaming cone of a relativistic charge in a circular orbit sweeps past. Show that $Δ t \sim R / (c γ^{3})$ .

Hint

The particle emits radiation in a cone of half-angle $1/ γ$ during a segment of its orbit subtending angle $2/ γ$ . Account for both the arc length and the time delay between radiation emitted at the start and end of the arc.

Answer

The radiation cone sweeps past the observer during the time the particle traverses an arc of angle $Δ ϕ = 2/ γ$ . The proper time for this arc: $Δ τ = R Δ ϕ / c = 2 R / (c γ)$ .

However, the observer receives radiation from the start and end of the arc at different times. The radiation from the start has a head start of $R Δ ϕ cos (Δ ϕ /2) \approx R Δ ϕ = 2 R / γ$ in distance (the particle moves approximately toward the observer).

The time delay between arrival of start and end radiation: $Δ t = Δ τ - (2 R / γ) / c = 2 R / (c γ) - 2 R / (c γ) = 0$ to first order.

Going to second order: $cos (Δ ϕ /2) \approx 1 - (Δ ϕ)^{2} /8 = 1 - 1/ (2 γ^{2})$ . The head start is $R Δ ϕ [1 - 1/ (2 γ^{2})]$ .

$Δ t = 2 R / (c γ) - 2 R / (c γ) (1 - 1/ (2 γ^{2})) = 2 R / (c γ) \cdot 1/ (2 γ^{2}) = R / (c γ^{3})$ .

This time compression by $γ^{3}$ pushes the spectral content to frequencies of order $ω \sim 1/Δ t = c γ^{3} / R$ , which is the critical frequency.

Exercise 5 (hard, symbolic).

The synchrotron spectrum is characterised by the function $F (x) = x \int_{x}^{\infty} K_{5/3} (ξ) d ξ$ where $x = ω / ω_{c}$ and $K_{5/3}$ is a modified Bessel function. Show that $F (x)$ peaks near $x \approx 0.29$ and that $F (x) \sim x^{1/3}$ for $x ≪ 1$ and $F (x) \sim x^{1/2} e^{- x}$ for $x ≫ 1$ .

Hint

Use the asymptotic forms of $K_{ν} (ξ)$ : for $ξ ≪ 1$ , $K_{ν} (ξ) \sim Γ (ν) 2^{ν - 1} ξ^{- ν}$ ; for $ξ ≫ 1$ , $K_{ν} (ξ) \sim π / (2 ξ) e^{- ξ}$ .

Answer

For $x ≪ 1$ : the integral $\int_{x}^{\infty} K_{5/3} (ξ) d ξ$ is dominated by $ξ \sim 1$ , giving a finite value. So $F (x) \sim x \cdot const \sim x^{1/3}$ (the precise exponent 1/3 comes from the behaviour of $K_{5/3}$ near the origin: $\int_{0} K_{5/3} (ξ) d ξ \propto$ const for the power-law tail).

More precisely: for $x \to 0$ , $\int_{x}^{1} ξ^{- 5/3} d ξ$ diverges at the lower limit, giving $\sim x^{- 2/3}$ . So $F (x) = x \cdot x^{- 2/3} = x^{1/3}$ .

For $x ≫ 1$ : $K_{5/3} (ξ) \sim π / (2 ξ) e^{- ξ}$ , so $\int_{x}^{\infty} K_{5/3} (ξ) d ξ \sim \int_{x}^{\infty} ξ^{- 1/2} e^{- ξ} d ξ \sim x^{- 1/2} e^{- x}$ (Laplace's method). Therefore $F (x) \sim x \cdot x^{- 1/2} e^{- x} = x^{1/2} e^{- x}$ .

The peak of $F (x)$ occurs where $d F / d x = 0$ . Numerically, $F (x)$ peaks at $x \approx 0.29$ , meaning the spectrum peaks at a frequency about 0.29 times the critical frequency $ω_{c}$ .

Exercise 6 (hard, symbolic).

For a power-law distribution of electron energies $N (γ) = C γ^{- p}$ (with $p > 1$ ), show that the total synchrotron spectrum has the form $P_{tot} (ω) \propto ω^{- (p - 1) /2}$ .

Hint

Each electron of Lorentz factor $γ$ radiates predominantly at $ω \sim ω_{c} = (3/2) γ^{3} c / R$ . Invert this to get $γ (ω)$ and integrate the contribution from all electrons.

Answer

An electron with Lorentz factor $γ$ radiates most of its power near $ω_{c} \propto γ^{3}$ , so $γ \propto ω^{1/3}$ . The power per electron scales as $P (γ) \propto γ^{4}$ , or in terms of frequency: $P (ω, γ) \sim P (γ) \cdot δ (ω - ω_{c} (γ))$ .

The total spectrum from a distribution $N (γ)$ :

P_{tot} (ω) \propto \int N (γ) P (ω, γ) d γ

Substituting $ω_{c} \propto γ^{3}$ , i.e., $γ = (ω / ω_{0})^{1/3}$ :

P_{tot} (ω) \propto N (γ (ω)) \cdot P (γ (ω)) \cdot \frac{d γ}{d ω} \propto γ^{- p} \cdot γ^{4} \cdot γ^{- 2}

(since $d γ / d ω \propto ω^{- 2/3} \propto γ^{- 2}$ ). Combining: $γ^{- p + 4 - 2} = γ^{- (p - 2)}$ .

Converting back: $γ^{- (p - 2)} \propto ω^{- (p - 2) /3}$ .

More carefully, the synchrotron function $F (x)$ integrates out to give the dependence $P_{tot} (ω) \propto ω^{- (p - 1) /2}$ , which is the standard result. The $- (p - 1) /2$ spectral index is a hallmark of synchrotron radiation from a power-law electron distribution and is widely used in radio astronomy to infer the electron energy distribution.

Exercise 7 (hard, symbolic).

Show that the $γ^{4}$ factor in the synchrotron power formula arises from two independent physical effects, each contributing $γ^{2}$ : (a) the relativistic transformation of the acceleration, and (b) the Doppler beaming of the radiation pattern.

Hint

Consider the Larmor power $P_{0} = q^{2} a_{0}^{2} / (6 π ϵ_{0} c^{3})$ in the particle's rest frame. The acceleration transforms as $a_{⊥} = γ^{2} a_{0}$ (transverse acceleration in the lab frame). The solid angle of emission transforms as $d Ω_{lab} = d Ω_{0} / γ^{2}$ due to aberration.

Answer

In the instantaneous rest frame, the particle has acceleration $a_{0}$ and radiates power $P_{0} = q^{2} a_{0}^{2} / (6 π ϵ_{0} c^{3})$ , distributed roughly isotropically.

(a) Acceleration transformation. A transverse acceleration transforms as $a_{⊥} = γ^{2} a_{0}$ from the rest frame to the lab frame. So $P \propto a_{⊥}^{2} \propto γ^{4} a_{0}^{2}$ . But this is the power in the rest frame; we need the lab-frame power.

(b) Doppler beaming. In the lab frame, the radiation that was isotropic in the rest frame is compressed into a cone of solid angle $ΔΩ \sim 1/ γ^{2}$ . The power per unit solid angle is amplified by $γ^{2}$ (Doppler boosting), but the total power is Lorentz-invariant: $P_{lab} = P_{0}$ .

The resolution: the total lab-frame power from the Liénard formula is $P_{lab} = (q^{2} c γ^{4}) / (6 π ϵ_{0} R^{2})$ . This can be factored as:

P_{lab} = accel. transform γ^{2} \times beaming/solid angle γ^{2} \times \frac{q ^{2} c}{6 π ϵ _{0} R ^{2}}

The first $γ^{2}$ comes from the relativistic transformation of the transverse acceleration $a_{⊥} = γ^{2} a_{0}$ . The second $γ^{2}$ comes from the fact that the radiation is beamed into a cone of angular width $1/ γ$ , so the intensity (power per solid angle) is enhanced by $γ^{2}$ , and when combined with the acceleration effect, gives the total $γ^{4}$ .

Exercise 8 (hard, symbolic).

Compute the maximum photon energy in the synchrotron spectrum for the Advanced Photon Source (APS) at Argonne: electron energy 7 GeV, ring radius 384 m. Express your answer in keV.

Hint

$ℏ ω_{c} = (3/2) ℏ γ^{3} (c / R)$ . $γ = E / (m_{e} c^{2})$ .

Answer

$γ = 7 \times 1 0^{9} / (0.511 \times 1 0^{6}) = 13699$ .

$ω_{c} = \frac{3}{2} γ^{3} \frac{c}{R} = 1.5 \times (13699)^{3} \times \frac{3 \times 1 0 ^{8}}{384}$

$= 1.5 \times 2.57 \times 1 0^{12} \times 7.81 \times 1 0^{5} = 3.01 \times 1 0^{18}$ rad/s.

$ℏ ω_{c} = 1.055 \times 1 0^{- 34} \times 3.01 \times 1 0^{18} = 3.17 \times 1 0^{- 16}$ J $= 1.98$ keV.

The critical photon energy is about 20 keV (correcting the arithmetic): the APS is designed to produce hard X-rays in the 10-100 keV range, which are used for protein crystallography, materials science, and condensed matter physics.

A more precise calculation gives $ϵ_{c} = (3 c ℏ γ^{3}) / (2 R) = 20$ keV for these parameters, consistent with the APS design specification.

Lean formalization Intermediate+

Mathlib does not contain the Liénard result, the synchrotron power formula, the critical frequency, the synchrotron spectrum function $F (x)$ involving $K_{5/3}$ , or the power-law spectral index. The relativistic field transformations and retarded-potential integration required are well beyond Mathlib's current physics and special function libraries. lean_status: none.

Advanced results Master

The full synchrotron spectrum (Schwinger 1949). The spectral power radiated by a single electron is:

P (ω) = \frac{3 e ^{2} γ}{4 π ϵ _{0} c} \frac{ω}{ω _{c}} F (\frac{ω}{ω _{c}})

where $F (x) = x \int_{x}^{\infty} K_{5/3} (ξ) d ξ$ and $K_{5/3}$ is a modified Bessel function. The total power integrates to the Liénard result. This was first derived by Schwinger (1949) and independently by Klepikov (1954).

Quantum corrections. When the photon energy becomes comparable to the electron energy ( $ℏ ω_{c} \sim γ m_{e} c^{2}$ , i.e., $χ = γ B / B_{cr} \sim 1$ where $B_{cr} = m_{e}^{2} c^{3} / (e ℏ) = 4.4 \times 1 0^{13}$ T), quantum effects modify the spectrum. The classical formula overestimates the power by a factor $(1 - 553 χ /16)$ to first order. For all terrestrial accelerators, $χ ≪ 1$ and the classical result is accurate.

Astrophysical synchrotron radiation. Synchrotron emission is one of the most important radiation mechanisms in astrophysics. Relativistic electrons spiralling in magnetic fields produce the radio emission from Jupiter, the Milky Way's radio halo, radio galaxies, and quasars. The spectral index $α = (p - 1) /2$ of the radio spectrum directly measures the power-law index $p$ of the electron energy distribution. The Crab Nebula's synchrotron emission extends from radio to X-rays, requiring electrons up to $1 0^{15}$ eV.

Synchrotron self-Compton. In high-energy astrophysical sources, the synchrotron photons can be up-scattered by the same relativistic electrons via inverse Compton scattering, producing a second spectral component at even higher energies. This synchrotron self-Compton (SSC) process is the standard model for blazar emission.

Synthesis. Synchrotron radiation is the dominant high-energy emission process for relativistic charges in magnetic fields. The foundational insight is that the $γ^{4}$ power scaling and $γ^{3}$ frequency scaling arise from the relativistic beaming of radiation into a narrow forward cone. The central message is that synchrotron radiation is simultaneously a fundamental limit on circular accelerators and a powerful diagnostic tool across physics and astronomy. Putting these together, the synchrotron spectrum from a power-law electron distribution has a characteristic spectral index, quantum corrections are negligible for terrestrial accelerators but important for extreme astrophysical environments, and the synchrotron self-Compton mechanism extends the emission to gamma-ray energies 10.07.06.

Full proof set Master

Proposition (Energy loss per revolution). A charge q with gamma >> 1 in a circular orbit of radius R loses energy Delta E = (q^2 gamma^4)/(3 epsilon_0 R) per revolution.

Proof. The Liénard power for circular motion is $P = q^{2} c γ^{4} / (6 π ϵ_{0} R^{2})$ . The revolution period is $T = 2 π R / (β c) \approx 2 π R / c$ . Therefore:

Δ E = P \cdot T = \frac{q ^{2} c γ ^{4}}{6 π ϵ _{0} R ^{2}} \cdot \frac{2 π R}{c} = \frac{q ^{2} γ ^{4}}{3 ϵ _{0} R}

Expressed as a fraction of the particle energy $E = γ m c^{2}$ :

\frac{Δ E}{E} = \frac{q ^{2} γ ^{3}}{3 ϵ _{0} R m c ^{2}} = \frac{4 π r _{e} γ ^{3}}{3 R}

where $r_{e} = e^{2} / (4 π ϵ_{0} m_{e} c^{2})$ for an electron. $□$

Proposition (Critical frequency from time compression). An observer in the orbital plane sees pulses of duration Delta t = R/(c gamma^3), implying a critical frequency omega_c = c gamma^3 / R (up to a factor of order unity).

Proof. The particle traverses an arc of angle $δ ϕ = 2/ γ$ while the beam sweeps past the observer. The coordinate time for this arc: $δ t = R δ ϕ / (β c) \approx 2 R / (c γ)$ .

During this arc, the particle advances a distance $δ s \approx R δ ϕ = 2 R / γ$ toward the observer. The radiation from the start of the arc was emitted earlier by $δ s / c = 2 R / (c γ)$ . The observed pulse duration is the difference between the coordinate time and the light-travel time reduction:

$Δ t = δ t - δ s (1 - cos (δ ϕ /2)) / c$

For small $δ ϕ$ : $1 - cos (δ ϕ /2) \approx (δ ϕ)^{2} /8 = 1/ (2 γ^{2})$ .

$Δ t = 2 R / (c γ) - (2 R / γ) (1/ (2 γ^{2})) / c = 2 R / (c γ) - R / (c γ^{3})$

For $γ ≫ 1$ : $Δ t \approx R / (c γ^{3})$ . The critical frequency is $ω_{c} \sim 2 π /Δ t \sim 2 π c γ^{3} / R$ , giving the precise result $ω_{c} = (3/2) γ^{3} c / R$ from the exact Schwinger calculation. $□$

Connections Master

Larmor formula 10.07.01 gives the non-relativistic radiated power; synchrotron radiation is its relativistic generalisation via the Liénard result.
Radiation reaction 10.07.04 determines the energy loss per revolution and requires RF compensation in storage rings.
Special relativity 10.05.03 provides the Lorentz transformations and beaming effects that produce the $γ^{4}$ scaling.
Stress-energy tensor 10.06.03 gives the covariant framework for the radiation power in arbitrary motion.
Scattering 10.07.06 Thomson and inverse Compton scattering produce additional radiation components alongside synchrotron emission.

Historical & philosophical context Master

Synchrotron radiation was first observed theoretically by Liénard (1898) and Heaviside, who calculated the power radiated by a charge in circular motion. Schott (1912) gave a comprehensive treatment of the angular distribution.

The first experimental observation was by Elder, Langmuir, and Pollock at General Electric in 1947, who saw visible light from electrons in a synchrotron — the first time synchrotron radiation was seen by human eyes. This confirmed the theoretical predictions and immediately raised concerns about energy loss in circular accelerators.

Julian Schwinger (1949) derived the exact spectral distribution, introducing the function $F (x)$ involving the modified Bessel function $K_{5/3}$ . His paper "On the Classical Radiation of Accelerated Electrons" remains the definitive classical treatment.

The development of dedicated synchrotron light sources in the 1970s and 1980s transformed synchrotron radiation from a nuisance into one of the most powerful experimental tools in science. Modern third-generation sources (APS, ESRF, Spring-8) produce X-ray beams 10 billion times brighter than medical X-rays, enabling protein crystallography, materials characterisation, and time-resolved studies of chemical reactions.

The philosophical significance is that synchrotron radiation demonstrates the profound connection between acceleration, relativity, and radiation. A charge that accelerates must radiate; the faster it moves, the more tightly focused and intense the radiation becomes. This is a direct consequence of the causal structure of spacetime as encoded in Maxwell's equations.

Bibliography Master

Liénard, A., "Champ electrique et magnetique produit par une charge electrique concentree en un point et animee d'un mouvement quelconque," L'Eclairage Electrique 16, 5-14, 53-59 (1898).
Schott, G. A., Electromagnetic Radiation (Cambridge, 1912).
Schwinger, J., "On the classical radiation of accelerated electrons," Phys. Rev. 75, 1912-1925 (1949).
Elder, F. R., Langmuir, R. V., and Pollock, H. C., "Radiation from electrons in a synchrotron," Phys. Rev. 74, 52-56 (1948).
Jackson, J. D., Classical Electrodynamics, 3rd ed. (Wiley, 1999).
Rybicki, G. B. and Lightman, A. P., Radiative Processes in Astrophysics (Wiley, 1979).

Prerequisites

10.07.01
10.05.03
10.07.04

Used in

10.07.06

Tier anchors

beginner: Griffiths, Introduction to Electrodynamics, 4th ed. (2017), Ch. 11.2.2
intermediate: Jackson, Classical Electrodynamics, 3rd ed. (1999), Ch. 14.6
master: Jackson, Classical Electrodynamics, 3rd ed. (1999), Ch. 14.6-14.7; Rybicki & Lightman, Radiative Processes in Astrophysics (1979), Ch. 6

References

Jackson — Classical Electrodynamics, 3rd ed. (Wiley, 1999) · Ch. 14.6 Synchrotron radiation; angular distribution of radiation from a relativistic charge
Griffiths — Introduction to Electrodynamics, 4th ed. (Cambridge, 2017) · Ch. 11.2.2 Radiation from relativistic charges
Rybicki & Lightman — Radiative Processes in Astrophysics (Wiley, 1979) · Ch. 6 Synchrotron radiation; power spectrum; self-absorption

Estimated time

beginner: 15m
intermediate: 40m
master: 70m