Synchrotron radiation

This type of radiation was first detected in a jet emitted by Messier 87 in 1956 by Geoffrey R. Burbidge,[11] who saw it as confirmation of a prediction by Iosif S. Shklovsky in 1953. However, it had been predicted earlier (1950) by Hannes Alfvén and Nicolai Herlofson.[12] Solar flares accelerate particles that emit in this way, as suggested by R. Giovanelli in 1948 and described by J.H. Piddington in 1952.[13]

T. K. Breus noted that questions of priority on the history of astrophysical synchrotron radiation are complicated, writing:

In particular, the Russian physicist V.L. Ginzburg broke his relationships with I.S. Shklovsky and did not speak with him for 18 years. In the West, Thomas Gold and Sir Fred Hoyle were in dispute with H. Alfven and N. Herlofson, while K.O. Kiepenheuer and G. Hutchinson were ignored by them.[14]

Crab Nebula. The bluish glow from the central region of the nebula is due to synchrotron radiation.

Supermassive black holes have been suggested for producing synchrotron radiation, by ejection of jets produced by gravitationally accelerating ions through the super contorted 'tubular' polar areas of magnetic fields. Such jets, the nearest being in Messier 87, have been confirmed by the Hubble telescope as apparently superluminal, travelling at 6 × c (six times the speed of light) from our planetary frame. This phenomenon is caused because the jets are travelling very near the speed of light and at a very small angle towards the observer. Because at every point of their path the high-velocity jets are emitting light, the light they emit does not approach the observer much more quickly than the jet itself. Light emitted over hundreds of years of travel thus arrives at the observer over a much smaller time period (ten or twenty years) giving the illusion of faster than light travel, however there is no violation of special relativity.[15]

A class of astronomical sources where synchrotron emission is important is the pulsar wind nebulae, a.k.a. plerions, of which the Crab nebula and its associated pulsar are archetypal. Pulsed emission gamma-ray radiation from the Crab has recently been observed up to ≥25 GeV,[16] probably due to synchrotron emission by electrons trapped in the strong magnetic field around the pulsar. Polarization in the Crab nebula[17] at energies from 0.1 to 1.0 MeV illustrates a typical synchrotron radiation.

Much of what is known about the magnetic environment of the interstellar medium and intergalactic medium is derived from observations of synchrotron radiation. Cosmic ray electrons moving through the medium interact with relativistic plasma and emit synchrotron radiation which is detected on Earth. The properties of the radiation allow astronomers to make inferences about the magnetic field strength and orientation in these regions, however accurate calculations of field strength cannot be made without knowing the relativistic electron density.[10]

We start with the expressions for the Liénard–Wiechert field of a point charge of mass ${\displaystyle m}$ and charge ${\displaystyle q}$:

${\displaystyle \mathbf {B} (\mathbf {r} ,t)=-{\frac {\mu _{0}q}{4\pi }}\left[{\frac {c\,{\hat {\mathbf {n} }}\times {\vec {\beta }}}{\gamma ^{2}R^{2}\,(1-{\vec {\beta }}\mathbf {\cdot } {\hat {\mathbf {n} }})^{3}}}+{\frac {{\hat {\mathbf {n} }}\times [\,{\dot {\vec {\beta }}}+{\hat {\mathbf {n} }}\times ({\vec {\beta }}\times {\dot {\vec {\beta }}})]}{R\,(1-{\vec {\beta }}\mathbf {\cdot } {\hat {\mathbf {n} }})^{3}}}\right]_{\mathrm {retarded} },}$

(1)

${\displaystyle \mathbf {E} (\mathbf {r} ,t)={\frac {q}{4\pi \varepsilon _{0}}}\left[{\frac {{\hat {\mathbf {n} }}-{\vec {\beta }}}{\gamma ^{2}R^{2}\,(1-{\vec {\beta }}\mathbf {\cdot } {\hat {\mathbf {n} }})^{3}}}+{\frac {{\hat {\mathbf {n} }}\times [({\hat {\mathbf {n} }}-{\vec {\beta }})\times {\dot {\vec {\beta }}}\,]}{c\,R\,(1-{\vec {\beta }}\mathbf {\cdot } {\hat {\mathbf {n} }})^{3}}}\right]_{\mathrm {retarded} }.}$

(2)

where R(t′) = r − r₀(t′), R(t′) = |R(t′)|, and n(t′) = R(t′)/R(t′), which is the unit vector between the observation point and the position of the charge at the retarded time, and t′ is the retarded time.

In equation (1), and (2), the first terms for B and E resulting from the particle fall off as the inverse square of the distance from the particle, and this first term is called the generalized Coulomb field or velocity field. These terms represents the particle static field effect, which is a function of the component of its motion that has zero or constant velocity, as seen by a distant observer at r. By contrast, the second terms fall off as the inverse first power of the distance from the source, and these second terms are called the acceleration field or radiation field because they represent components of field due to the charge's acceleration (changing velocity), and they represent E and B which are emitted as electromagnetic radiation from the particle to an observer at r.

If we ignore the velocity field in order to find the power of emitted EM radiation only, the radial component of Poynting's vector resulting from the Liénard–Wiechert fields can be calculated to be

${\displaystyle [\mathbf {S\cdot } {\hat {\mathbf {n} }}]={\frac {q^{2}}{16\pi ^{2}\varepsilon _{0}c}}\left\{{\frac {1}{R^{2}}}\left|{\frac {{\hat {\mathbf {n} }}\times [({\hat {\mathbf {n} }}-{\vec {\beta }})\times {\dot {\vec {\beta }}}]}{(1-{\vec {\beta }}\mathbf {\cdot } {\hat {\mathbf {n} }})^{3}}}\right|^{2}\right\}_{\text{retarded}}.}$

(3)

Note that

• The spatial relationship between β→ and .→β determines the detailed angular power distribution.
• The relativistic effect of transforming from the rest frame of the particle to the observer's frame manifests itself by the presence of the factors (1 − β→⋅n̂) in the denominator of Eq. (3).
• For ultrarelativistic particles the latter effect dominates the whole angular distribution.

The energy radiated per solid angle during a finite period of acceleration from t′ = T₁ to t′ = T₂ is

${\displaystyle {\begin{aligned}{\frac {\mathrm {d} P}{\mathrm {d} {\mathit {\Omega }}}}&=R(t')^{2}\,[\mathbf {S} (t')\mathbf {\cdot } {\hat {\mathbf {n} }}(t')]\,{\frac {\mathrm {d} t}{\mathrm {d} t'}}=R(t')^{2}\,\mathbf {S} (t')\mathbf {\cdot } {\hat {\mathbf {n} }}(t')\,[1-{\vec {\beta }}(t')\mathbf {\cdot } {\hat {\mathbf {n} }}(t')]\\&={\frac {q^{2}}{16\pi ^{2}\varepsilon _{0}c}}\,{\frac {|{\hat {\mathbf {n} }}(t')\times \{[{\hat {\mathbf {n} }}(t')-{\vec {\beta }}(t')]\times {\dot {\vec {\beta }}}(t')\}|^{2}}{[1-{\vec {\beta }}(t')\mathbf {\cdot } {\vec {\mathbf {n} }}(t')]^{5}}}.\end{aligned}}}$

(4)

Integrating Eq. (4) over the all solid angles, we get the relativistic generalization of Larmor's formula

|${\displaystyle P={\frac {q^{2}}{6\pi \varepsilon _{0}c}}\gamma ^{6}\left[\left|{\dot {\vec {\beta }}}\right|^{2}-\left|{\vec {\beta }}\times {\dot {\vec {\beta }}}\right|^{2}\right].}$

However, this also can be derived by relativistic transformation of the 4-acceleration in Larmor's formula.

When the electron velocity approaches the speed of light, the emission pattern is sharply collimated forward.

When the charge is in instantaneous circular motion, its acceleration .→β is perpendicular to its velocity β→. Choosing a coordinate system such that instantaneously β→ is in the z direction and .→β is in the x direction, with the polar and azimuth angles θ and φ defining the direction of observation, the general formula Eq. (4) reduces to

${\displaystyle {\frac {\mathrm {d} P}{\mathrm {d} {\mathit {\Omega }}}}={\frac {q^{2}}{16\pi ^{2}\epsilon _{0}c}}{\frac {|{\dot {\vec {\beta }}}|^{2}}{(1-\beta \cos \theta )^{3}}}\left[1-{\frac {\sin ^{2}\theta \cos ^{2}\phi }{\gamma ^{2}(1-\beta \cos \theta )^{2}}}\right].}$

In the relativistic limit ${\displaystyle (\gamma \gg 1)}$, the angular distribution can be written approximately as

${\displaystyle {\frac {\mathrm {d} P}{\mathrm {d} {\mathit {\Omega }}}}\simeq {\frac {2}{\pi }}{\frac {q^{2}}{c^{3}}}\gamma ^{6}{\frac {|{\dot {\mathbf {v} }}|^{2}}{(1+\gamma ^{2}\theta ^{2})^{3}}}\left[1-{\frac {4\gamma ^{2}\theta ^{2}\cos ^{2}\phi }{(1+\gamma ^{2}\theta ^{2})^{2}}}\right].}$

The factors (1 − βcosθ) in the denominators tip the angular distribution forward into a narrow cone like the beam of a headlight pointing ahead of the particle. A plot of the angular distribution (dP/dΩ vs. γθ) shows a sharp peak around θ = 0.

If we neglect any electric force on the particle, the total power radiated (over all solid angles) from Eq. (4) is

${\displaystyle P={\frac {q^{2}}{6\pi \epsilon _{0}c}}\left|{\dot {\vec {\beta }}}\right|^{2}\gamma ^{4}={\frac {q^{2}c}{6\pi \epsilon _{0}}}{\frac {\beta ^{4}\gamma ^{4}}{\rho ^{2}}}={\frac {q^{4}}{6\pi \epsilon _{0}m^{4}c^{5}}}B^{2}E^{2}\beta ^{2}={\frac {q^{4}}{6\pi \epsilon _{0}m^{4}c^{5}}}B^{2}(E^{2}-m^{2}c^{4}),}$

where E is the particle's total (kinetic plus rest) energy, B is the magnetic field, and ρ is the radius of curvature of the track in the field. Note that the radiated power is proportional to 1/m⁴, 1/ρ², and B². In some cases the surfaces of vacuum chambers hit by synchrotron radiation have to be cooled because of the high power of the radiation.

Using

${\displaystyle B={\frac {E\beta }{q\,r\sin(\alpha )}},}$

where α is the angle between the velocity and the magnetic field and r is the radius of the circular acceleration, the power emitted is:

${\displaystyle P={\frac {q^{2}}{6\pi \epsilon _{0}m^{4}c^{5}r^{2}\sin ^{2}(\alpha )}}E^{4}\beta ^{4}={\frac {q^{2}}{6\pi \epsilon _{0}m^{4}c^{5}r^{2}\sin ^{2}(\alpha )}}(E^{2}-m^{2}c^{4})^{2}.}$

Thus the power emitted scales as energy to the fourth, and decreases with the square of the radius and the fourth power of particle mass. This radiation is what limits the energy of an electron-positron circular collider. Generally, proton-proton colliders are instead limited by the maximum magnetic field; this is why, for example, the LHC has a center-of-mass energy 70 times higher than the LEP even though the proton mass is some 2000 times greater than the electron mass.

The energy received by an observer (per unit solid angle at the source) is

${\displaystyle {\frac {d^{2}W}{d\Omega }}=\int _{-\infty }^{\infty }{\frac {d^{2}P}{d\Omega }}dt=c\varepsilon _{0}\int _{-\infty }^{\infty }\left|R{\vec {E}}(t)\right|^{2}dt.}$

Using the Fourier transformation we move to the frequency space

${\displaystyle {\frac {d^{2}W}{d\Omega }}=2c\varepsilon _{0}\int _{0}^{\infty }\left|R{\vec {E}}(\omega )\right|^{2}d\omega .}$

Angular and frequency distribution of the energy received by an observer (consider only the radiation field)

${\displaystyle {\frac {d^{3}W}{d\Omega d\omega }}=2c\varepsilon _{0}R^{2}\left|{\vec {E}}(\omega )\right|^{2}={\frac {q^{2}}{4\pi \varepsilon _{0}4\pi ^{2}c}}\left|\int _{-\infty }^{\infty }{\frac {{\hat {n}}\times \left[\left({\hat {n}}-{\vec {\beta }}\right)\times {\dot {\vec {\beta }}}\right]}{\left(1-{\hat {n}}\cdot {\vec {\beta }}\right)^{2}}}e^{i\omega (t-{\hat {n}}\cdot {\vec {r}}(t)/c)}dt\right|^{2}}$

Therefore, if we know the particle's motion, cross products term, and phase factor, we could calculate the radiation integral. However, calculations are generally quite lengthy (even for simple cases as for the radiation emitted by an electron in a bending magnet, which require the Airy function or the modified Bessel functions).

Trajectory of the arc of circumference

Trajectory of the arc of circumference is

${\displaystyle {\vec {r}}(t)=\left(\rho \sin {\frac {\beta c}{\rho }}t,\rho \left(1-\cos {\frac {\beta c}{\rho }}t\right),0\right).}$

In the limit of small angles we compute

${\displaystyle {\hat {n}}\times \left({\hat {n}}\times {\vec {\beta }}\right)=\beta \left[-{\vec {\varepsilon }}_{\parallel }\sin \left({\frac {\beta ct}{\rho }}\right)+{\vec {\varepsilon }}_{\perp }\cos \left({\frac {\beta ct}{\rho }}\right)\sin \theta \right]}$

${\displaystyle \omega \left(t-{\frac {{\hat {n}}\cdot {\vec {r}}(t)}{c}}\right)=\omega \left[t-{\frac {\rho }{c}}\sin \left({\frac {\beta ct}{\rho }}\right)\cos \theta \right]}$

Substituting into the radiation integral and introducing

${\displaystyle \xi ={\frac {\rho \omega }{3c\gamma ^{3}}}\left(1+\gamma ^{2}\theta ^{2}\right)^{3/2}}$

${\displaystyle {\frac {d^{3}W}{d\Omega d\omega }}={\frac {3q^{2}}{16\pi ^{3}\varepsilon _{0}c}}\left({\frac {2\omega \rho }{3c\gamma ^{2}}}\right)^{2}\left(1+\gamma ^{2}\theta ^{2}\right)^{2}\left[K_{2/3}^{2}(\xi )+{\frac {\gamma ^{2}\theta ^{2}}{1+\gamma ^{2}\theta ^{2}}}K_{1/3}^{2}(\xi )\right],}$

(5)

where the function K is a modified Bessel function of the second kind.

Angular distribution of radiated energy

From Eq. (5), we observe that the radiation intensity is negligible for ${\displaystyle \xi \gg 1}$. Critical frequency is defined as the frequency when ξ = 1/2 and θ = 0. So,

${\displaystyle \omega _{\text{c}}={\frac {3}{2}}{\frac {c}{\rho }}\gamma ^{3},}$

and critical angle is defined as the angle for which ${\displaystyle \xi (\theta _{\text{c}})\simeq \xi (0)+1}$ and is approximately

${\displaystyle \theta _{\text{c}}\simeq {\frac {1}{\gamma }}\left({\frac {2\omega _{\text{c}}}{\omega }}\right)^{1/3}.}$[18]

For frequencies much larger than the critical frequency and angles much larger than the critical angle, the synchrotron radiation emission is negligible.

Integrating on all angles, we get the frequency distribution of the energy radiated.

${\displaystyle {\frac {dW}{d\omega }}=\oint {\frac {d^{3}W}{d\omega d\Omega }}d\Omega ={\frac {{\sqrt {3}}e^{2}}{4\pi \varepsilon _{0}c}}\gamma {\frac {\omega }{\omega _{\text{c}}}}\int _{\omega /\omega _{\text{c}}}^{\infty }K_{5/3}(x)dx}$

Frequency distribution of radiated energy

If we define

${\displaystyle S(y)\equiv {\frac {9{\sqrt {3}}}{8\pi }}y\int _{y}^{\infty }K_{5/3}(x)dx}$

${\displaystyle \int _{0}^{\infty }S(y)dy=1,}$

where y = ω/ω_c. Then

${\displaystyle {\frac {dW}{d\omega }}={\frac {2q^{2}\gamma }{9\varepsilon _{0}c}}S(y)}$

Note that ${\displaystyle {\frac {dW}{d\omega }}\sim {\frac {q^{2}}{4\pi \varepsilon _{0}c}}\left({\frac {\omega \rho }{c}}\right)^{1/3}}$, if ${\displaystyle \omega \ll \omega _{\text{c}}}$, and ${\displaystyle {\frac {dW}{d\omega }}\approx {\sqrt {\frac {3\pi }{2}}}{\frac {q^{2}}{4\pi \varepsilon _{0}c}}\gamma \left({\frac {\omega }{\omega _{\text{c}}}}\right)^{1/2}e^{-\omega /\omega _{\text{c}}}}$, if ${\displaystyle \omega \gg \omega _{\text{c}}}$

The formula for spectral distribution of synchrotron radiation, given above, can be expressed in terms of a rapidly converging integral with no special functions involved[19] (see also modified Bessel functions) by means of the relation:

${\displaystyle \int _{\xi }^{\infty }K_{5/3}(x)dx={\frac {1}{\sqrt {3}}}\,\int _{0}^{\infty }\,{\frac {9+36x^{2}+16x^{4}}{(3+4x^{2}){\sqrt {1+x^{2}/3}}}}\exp \left[-\xi \left(1+{\frac {4x^{2}}{3}}\right){\sqrt {1+{\frac {x^{2}}{3}}}}\right]\ dx}$

Relationship between power radiated and the photon energy

First, define the critical photon energy as

${\displaystyle \varepsilon _{c}=\hbar \omega _{\text{c}}={\frac {3}{2}}{\frac {\hbar c}{\rho }}\gamma ^{3}.}$

Then, the relationship between radiated power and photon energy is shown in the graph on the right side. The higher the critical energy, the more photons with high energies are generated. Note that, there is no dependence on the energy at longer wavelengths.

In Eq. (5), the first term ${\displaystyle K_{2/3}^{2}(\xi )}$ is the radiation power with polarization in the orbit plane, and the second term ${\displaystyle {\frac {\gamma ^{2}\theta ^{2}}{1+\gamma ^{2}\theta ^{2}}}K_{1/3}^{2}(\xi )}$ is the polarization orthogonal to the orbit plane.

In the orbit plane ${\displaystyle \theta =0}$, the polarization is purely horizontal. Integrating on all frequencies, we get the angular distribution of the energy radiated

${\displaystyle {\frac {d^{2}W}{d\Omega }}=\int _{0}^{\infty }{\frac {d^{3}W}{d\omega d\Omega }}d\omega ={\frac {7q^{2}\gamma ^{5}}{64\pi \varepsilon _{0}\rho }}{\frac {1}{(1+\gamma ^{2}\theta ^{2})^{5/2}}}\left[1+{\frac {5}{7}}{\frac {\gamma ^{2}\theta ^{2}}{1+\gamma ^{2}\theta ^{2}}}\right]}$

Integrating on all the angles, we find that seven times as much energy is radiated with parallel polarization as with perpendicular polarization. The radiation from a relativistically moving charge is very strongly, but not completely, polarized in the plane of motion.

An undulator consists of a periodic array of magnets, so that they provide a sinusoidal magnetic field.

${\displaystyle {\vec {B}}=\left(0,B_{0}\sin(k_{\text{u}}z),0\right)}$

undulator

Solution of equation of motion:

${\displaystyle {\vec {r}}(t)={\frac {\lambda _{\text{u}}K}{2\pi \gamma }}\sin \omega _{\text{u}}t\cdot {\hat {x}}+\left({\bar {\beta _{z}}}ct+{\frac {\lambda _{\text{u}}K^{2}}{16\pi \gamma ^{2}}}\cos(2\omega _{\text{u}}t)\right)\cdot {\hat {z}}}$

where

${\displaystyle K={\frac {qB_{0}\lambda _{\text{u}}}{2\pi mc}},}$

and

${\displaystyle {\bar {\beta _{z}}}=1-{\frac {1}{2\gamma ^{2}}}\left(1+{\frac {K^{2}}{2}}\right),}$

and the parameter ${\displaystyle K}$ is called the undulator parameter.

Constructive interference of the beam in the undulator

Condition for the constructive interference of radiation emitted at different poles is

${\displaystyle d={\frac {\lambda _{\text{u}}}{\bar {\beta }}}-\lambda _{\text{u}}\cos \theta =n\lambda }$

Expanding ${\displaystyle \cos \theta \approx 1-{\frac {\theta ^{2}}{2}}}$ and neglecting the terms ${\displaystyle O(\theta ^{2})}$ in the resulting equation, one obtains

${\displaystyle \lambda _{n}={\frac {\lambda _{\text{u}}}{2\gamma ^{2}n}}\left({\frac {1+{\frac {K^{2}}{2}}+\gamma ^{2}\theta ^{2}}{\bar {\beta }}}\right)}$

For ${\displaystyle {\bar {\beta }}\rightarrow 1}$, one finally gets

${\displaystyle \lambda _{n}={\frac {\lambda _{\text{u}}}{2\gamma ^{2}n}}\left(1+{\frac {K^{2}}{2}}+\gamma ^{2}\theta ^{2}\right)}$

This equation is called the undulator equation.

Radiation integral is

${\displaystyle {\frac {d^{3}W}{d\Omega d\omega }}={\frac {q^{2}}{4\pi \varepsilon _{0}4\pi ^{2}c}}\left|\int _{-\infty }^{\infty }{\frac {{\hat {n}}\times \left[\left({\hat {n}}-{\vec {\beta }}\right)\times {\dot {\vec {\beta }}}\right]}{\left(1-{\hat {n}}\cdot {\vec {\beta }}\right)^{2}}}e^{i\omega (t-{\hat {n}}\cdot {\vec {r}}(t)/c)}dt\right|^{2}}$

Using the periodicity of the trajectory, we can split the radiation integral into a sum over ${\displaystyle N}$ terms, where ${\displaystyle 2N}$ is the total number of bending magnets of the undulator.

${\displaystyle {\frac {d^{3}W}{d\Omega d\omega }}={\frac {q^{2}\omega ^{2}}{4\pi \varepsilon _{0}4\pi ^{2}c}}\left|\int _{-\lambda _{u}/2{\bar {\beta }}c}^{\lambda _{u}/2{\bar {\beta }}c}{\hat {n}}\times \left({\hat {n}}\times {\vec {\beta }}\right)e^{i\omega (t-{\hat {n}}\cdot {\vec {r}}(t)/c)}dt\right|^{2}\left|1+e^{i\delta }+e^{2i\delta }+\cdots +e^{i(N_{u}-1)\delta }\right|^{2}}$

Peak frequencies become sharp as the number N increases

where 　${\displaystyle {\bar {\beta }}=\beta \left(1-{\frac {K^{2}}{4\gamma ^{2}}}\right)}$

and ${\displaystyle \delta ={\frac {2\pi \omega }{\omega _{\text{res}}(\theta )}}}$,　${\displaystyle \omega _{\text{res}}(\theta )={\frac {2\pi c}{\lambda _{\text{res}}(\theta )}}}$,　and　${\displaystyle \lambda _{\text{res}}(\theta )={\frac {\lambda _{u}}{2\gamma ^{2}}}\left(1+{\frac {K^{2}}{2}}+\gamma ^{2}\theta ^{2}\right)}$

Only odd harmonics are radiated on-axis

Off-axis radiation contains many harmonics

The radiation integral in an undulator can be written as

${\displaystyle {\frac {d^{3}W}{d\Omega d\omega }}={\frac {q^{2}\gamma ^{2}N^{2}}{4\pi \varepsilon _{0}c}}L\left(N{\frac {\Delta \omega _{n}}{\omega _{\text{res}}(\theta )}}\right)F_{n}(K,\theta ,\phi ).}$

where ${\displaystyle \Delta \omega _{n}=\omega -n\omega _{\text{res}}}$ is the frequency difference to the n-th harmonic. The sum of δ generates a series of sharp peaks in the frequency spectrum harmonics of fundamental wavelength

${\displaystyle L\left(N{\frac {\Delta \omega _{n}}{\omega _{\text{res}}(\theta )}}\right)={\frac {\sin ^{2}\left(N\pi \Delta \omega _{n}/\omega _{\text{res}}(\theta )\right)}{N^{2}\left(\pi \Delta \omega _{n}/\omega _{\text{res}}(\theta )\right)^{2}}},}$

and F_n depends on the angles of observations and K

${\displaystyle F_{n}(K,\theta ,\phi )\propto \left|\int _{-\lambda _{u}/2{\bar {\beta }}c}^{\lambda _{u}/2{\bar {\beta }}c}{\hat {n}}\times \left({\hat {n}}\times {\vec {\beta }}\right)e^{i\omega (t-{\hat {n}}\cdot {\vec {r}}(t)/c)}dt\right|^{2}.}$

On the axis (θ = 0, φ = 0), the radiation integral becomes

${\displaystyle {\frac {d^{3}W}{d\Omega d\omega }}={\frac {e^{2}\gamma ^{2}N^{2}}{4\pi \varepsilon _{0}c}}L\left(N{\frac {\Delta \omega _{n}}{\omega _{\text{res}}(0)}}\right)F_{n}(K,0,0)}$

and

${\displaystyle F_{n}(K,0,0)={\frac {n^{2}K^{2}}{1+K^{2}/2}}\left[J_{\frac {n+1}{2}}(Z)-J_{\frac {n-1}{2}}(Z)\right]^{2},}$

where ${\displaystyle Z={\frac {nK^{2}}{4(1+K^{2}/2)}}}$

Note that only odd harmonics are radiated on-axis, and as K increases higher harmonic becomes stronger.