إشعاع سنكروترني

هذا المقال هو عن الظاهرة الطبيعية للإشعاع السنكروتروني. إذا كنت تريد تفاصيل حول انتاج هذا الإشعاع وتطبيقاته في المعامل، انظر مصدر ضوء سنكروتروني.

الإشعاع السنكروترني (Synchrotron radiation أو إشعاع برمزشترالونگ المغناطيسي magnetobremsstrahlung radiation) هو إشعاع كهرومغناطيسي ينطلق حين تتعرض الجسيمات المشحونة لتسارع قطري، أي حين تتعرض لتسارع عمودي على سرعتها (a ⊥ v). وتُنتَج، على سبيل المثال، في السنكروترونات باستخدام مغناطيسات منحنية، متموجات و/أو متمايلات. إذا كان الجسيم غير نسباوي، فإن الإشعاع يُدعى إشعاع سيكلوتروني. أما إذا كانت الجسيمات نسباوية، فأحياناً يشار إليها بإسم فائق النسباوية، والإشعاع يسمى "إشعاع سنكروتروني". الإشعاع السنكروتروني قد يتحقق اصطناعياً في السنكروترونات أو حلقات التخزين، أو طبيعياً بسبب مرور إلكترونات سريعة خلال مجالات مغناطيسية. الإشعاع الناتج بهذه الطريقة تكون له استقطاب مميز والترددات المتولدة يمكن أن تنتشر على كامل نطاق الطيف الكهرومغناطيسي، ويسمى ذلك الإشعاع أيضاً إشعاع متصل.

إشعاع سنكروتروني من مغناطيس منحني

إشعاع سنكروتروني من متموج.

من خلال إلكترونيات سريعة الحركة أثناء دورانها في مجال مغناطيسي قوى. والإشعاع السنكروتروني هو سبب الضوضاء والتشويش الراديوي الذي يصدر من خلال معظم المنابع الراديوية في الكون.

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 إشعاع سنكروترني في علم الفلك

تُولد الإشعاعات السنكترونية في الطبيعة في النجوم و المجرات من الإلكترونات السريعة التي تنتشر بسرعات مقاربة لسرعة الضوء وتقع تحت تأثير مجالات مغناطيسية شديدة فيكون مسارها على هيئة مسارات حلزونية يمكن مراقبتها من الأرض . وهذه تتسم بخاصتين : الخاصة الأولى ، وهو أن طيفها ليس من نوع الطيف العادي الذي ينشأ عن الحرارة ، وهذه تتميز بتوزيع يعرف في الفيزياء بتوزيع الفوتونات الصادرة من الجسم الأسود وتعادل طيف الشمس . والخاصية الثانية ، الاستقطاب.

 سدم رياح النباضات

سديم السرطان

الوهج المائل للزرقة من المنطقة الوسطى للسديم هو بسبب الإشعاع السنكروتروني.

 التاريخ

معجل سنكروتروني من جنرال إلكتريك، المبني في 1946، أصل اكتشاف الإشعاع السنكروترني.

منفث نشط من M87's.، صورة من HST. الضوء الأزرق من المنفث المنبثق من لب AGN الساطع، باتجاه أسفل اليمين، هو بسبب الإشعاع السنكروتروني.

 الصياغة

 مجال لينار-ڤيكرت

فلنبدأ بتعبيرات عن مجال لينار-ڤيكرت لشحنة نقطية بكتلة $m$ and charge $q$:

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

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

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

لاحظ أن

• The spatial relationship between β← and قالب:Overset 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 into per solid angle during a finite period of acceleration from t′ = T₁ to t′ = T₂ is

${\begin{aligned}{\frac {\mathrm {d} P}{\mathrm {d} {\mathit {\Omega }}}}&=R(t')^{2}\,[\mathbf {S} (t')\mathbf {\cdot } {\hat {\mathbf {n} }}(t')]\,{\frac {\mathrm {d} t}{\mathrm {d} t'}}=R(t')^{2}\,\mathbf {S} (t')\mathbf {\cdot } {\hat {\mathbf {n} }}(t')\,[1-{\vec {\beta }}(t')\mathbf {\cdot } {\hat {\mathbf {n} }}(t')]\\&={\frac {q^{2}}{16\pi ^{2}\varepsilon _{0}c}}\,{\frac {|{\hat {\mathbf {n} }}(t')\times \{[{\hat {\mathbf {n} }}(t')-{\vec {\beta }}(t')]\times {\dot {\vec {\beta }}}(t')\}|^{2}}{[1-{\vec {\beta }}(t')\mathbf {\cdot } {\vec {\mathbf {n} }}(t')]^{5}}}.\qquad \qquad (4)\end{aligned}}$

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

$P={\frac {q^{2}}{6\pi \varepsilon _{0}c}}\gamma ^{6}\left[\left|{\dot {\vec {\beta }}}\right|^{2}-\left|{\vec {\beta }}\times {\dot {\vec {\beta }}}\right|^{2}\right].\qquad (5)$

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

 السرعة العمودية على التسارع (v ⟂ a): الإشعاع السنكروتروني

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 قالب:Overset is perpendicular to its velocity β←. Choosing a coordinate system such that instantaneously β← is in the z direction and قالب:Overset 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

${\frac {\mathrm {d} P}{\mathrm {d} {\mathit {\Omega }}}}={\frac {q^{2}}{16\pi ^{2}\epsilon _{0}c}}{\frac {|{\dot {\vec {\beta }}}|^{2}}{(1-\beta \cos \theta )^{3}}}\left[1-{\frac {\sin ^{2}\theta \cos ^{2}\phi }{\gamma ^{2}(1-\beta \cos \theta )^{2}}}\right].\qquad (6)$

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

${\frac {\mathrm {d} P}{\mathrm {d} {\mathit {\Omega }}}}\simeq {\frac {2}{\pi }}{\frac {q^{2}}{c^{3}}}\gamma ^{6}{\frac {|{\dot {\mathbf {v} }}|^{2}}{(1+\gamma ^{2}\theta ^{2})^{3}}}\left[1-{\frac {4\gamma ^{2}\theta ^{2}\cos ^{2}\phi }{(1+\gamma ^{2}\theta ^{2})^{2}}}\right].\qquad \qquad (7)$

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 angle) from Eq. (4) is

$P={\frac {q^{2}}{6\pi \epsilon _{0}c}}\left|{\dot {\vec {\beta }}}\right|^{2}\gamma ^{4}={\frac {q^{2}c}{6\pi \epsilon _{0}}}{\frac {\beta ^{4}\gamma ^{4}}{\rho ^{2}}}={\frac {q^{4}}{6\pi \epsilon _{0}m^{4}c^{5}}}B^{2}E^{2}\beta ^{2}={\frac {q^{4}}{6\pi \epsilon _{0}m^{4}c^{5}}}B^{2}(E^{2}-m^{2}c^{4}),\qquad (8)$

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.

باستخدام

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

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:

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

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 2000 times the electron mass.

 تكامل الإشعاع

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

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

Using the Fourier transformation we move to the frequency space

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

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

${\frac {d^{3}W}{d\Omega d\omega }}=2c\varepsilon _{0}R^{2}\left|{\vec {E}}(\omega )\right|^{2}={\frac {q^{2}}{4\pi \varepsilon _{0}4\pi ^{2}c}}\left|\int _{-\infty }^{\infty }{\frac {{\hat {n}}\times \left[\left({\hat {n}}-{\vec {\beta }}\right)\times {\dot {\vec {\beta }}}\right]}{\left(1-{\hat {n}}\cdot {\vec {\beta }}\right)^{2}}}e^{i\omega (t-{\hat {n}}\cdot {\vec {r}}(t)/c)}dt\right|^{2}.\qquad (9)$

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, they require Airy function or the modified Bessel functions).

 مثال 1: المغناطيس المنحني

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 التكامل

مسار قوس محيط

مسار قوس محيط هو

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

In the limit of small angles we compute

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

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

Substituting into the radiation integral and introducing

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

${\frac {d^{3}W}{d\Omega d\omega }}={\frac {3q^{2}}{16\pi ^{3}\varepsilon _{0}c}}\left({\frac {2\omega \rho }{3c\gamma ^{2}}}\right)^{2}\left(1+\gamma ^{2}\theta ^{2}\right)^{2}\left[K_{2/3}^{2}(\xi )+{\frac {\gamma ^{2}\theta ^{2}}{1+\gamma ^{2}\theta ^{2}}}K_{1/3}^{2}(\xi )\right]\qquad (10),$

حيث الدالة K is a modified Bessel function of the second kind.

 توزيع تردد الطاقة المنبعثة

Angular distribution of radiated energy

From Eq.(10), we observe that the radiation intensity is negligible for $\xi \gg 1$. Critical frequency is defined as the frequency when ξ = 1/2 and θ = 0. So,

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

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

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

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.

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

Frequency distribution of radiated energy

فإذا عرّفنا

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

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

حيث y = ω/ω_c. فإن

${\frac {dW}{d\omega }}={\frac {2q^{2}\gamma }{9\varepsilon _{0}c}}S(y).\qquad (11)$

Note that ${\frac {dW}{d\omega }}\sim {\frac {q^{2}}{4\pi \varepsilon _{0}c}}\left({\frac {\omega \rho }{c}}\right)^{1/3}$, if $\omega \ll \omega _{\text{c}}$, and ${\frac {dW}{d\omega }}\approx {\sqrt {\frac {3\pi }{2}}}{\frac {q^{2}}{4\pi \varepsilon _{0}c}}\gamma \left({\frac {\omega }{\omega _{\text{c}}}}\right)^{0.5}e^{-\omega /\omega _{\text{c}}}$, if $\omega \gg \omega _{\text{c}}$

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 (see also modified Bessel functions ) by means of the relation:

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

 انبعاث الإشعاع السنكروتروني كدالة في طاقة الشعاع

Relationship between power radiated and the photon energy

First, define the critical photon energy as

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

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 wavelength.

 استقطاب الإشعاع السنكروتروني

In Eq.(10), the first term $K_{2/3}^{2}(\xi )$ is the radiation power with polarization in the orbit plane, and the second term ${\frac {\gamma ^{2}\theta ^{2}}{1+\gamma ^{2}\theta ^{2}}}K_{1/3}^{2}(\xi )$ is the polarization orthogonal to the orbit plane.

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

${\frac {d^{2}W}{d\Omega }}=\int _{0}^{\infty }{\frac {d^{3}W}{d\omega d\Omega }}d\omega ={\frac {7q^{2}\gamma ^{5}}{64\pi \varepsilon _{0}\rho }}{\frac {1}{(1+\gamma ^{2}\theta ^{2})^{5/2}}}\left[1+{\frac {5}{7}}{\frac {\gamma ^{2}\theta ^{2}}{1+\gamma ^{2}\theta ^{2}}}\right].\qquad (12)$

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.

 المثال 2: المتموج undulator

 حل معادلة الحركة ومعادلة المتموج

يتكون المتموج من مصفوفة دورية من المغناطيسات، بحثي يعطوا مجال مغناطيسي جيبي.

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

undulator

حل معادلة الحركة هو

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

حيث

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

و

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

and the parameter $K$ 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

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

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

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

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

$\lambda _{n}={\frac {\lambda _{\text{u}}}{2\gamma ^{2}n}}\left(1+{\frac {K^{2}}{2}}+\gamma ^{2}\theta ^{2}\right)\qquad (13).$

هذه المعادلة تسمى معادلة المتموج undulator equation.

 الإشعاع من المتموج

تكامل الإشعاع هو

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

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

${\frac {d^{3}W}{d\Omega d\omega }}={\frac {q^{2}\omega ^{2}}{4\pi \varepsilon _{0}4\pi ^{2}c}}\left|\int _{-\lambda _{u}/2{\bar {\beta }}c}^{\lambda _{u}/2{\bar {\beta }}c}{\hat {n}}\times \left({\hat {n}}\times {\vec {\beta }}\right)e^{i\omega (t-{\hat {n}}\cdot {\vec {r}}(t)/c)}dt\right|^{2}\left|1+e^{i\delta }+e^{2i\delta }+\cdots +e^{i(N_{u}-1)\delta }\right|^{2},\qquad (14)$

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

Peak frequencies become sharp as the number N increases

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

Only odd harmonics are radiated on-axis

Off-axis radiation contains many harmonics

The radiation integral in an undulator can be written as

${\frac {d^{3}W}{d\Omega d\omega }}={\frac {q^{2}\gamma ^{2}N^{2}}{4\pi \varepsilon _{0}c}}L\left(N{\frac {\Delta \omega _{n}}{\omega _{\text{res}}(\theta )}}\right)F_{n}(K,\theta ,\phi )\qquad (15).$

where $\Delta \omega _{n}=\omega -n\omega _{\text{res}}$ 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

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

and F_n depends on the angles of observations and K

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

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

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

و

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

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

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

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 المصادر

• مؤمن, عبد الأمير (2006). قاموس دار العلم الفلكي. بيروت، لبنان: دار العلم للملايين. {{cite book}}: Cite has empty unknown parameter: |طبعة أولى coauthors= (help)

 انظر ايضا

• منبع رايوي
• إشعاع كهرومغناطيسي
• إشعاع
• Synchrotron for this type of particle accelerator
• Synchrotron light source for laboratory generation and applications of synchrotron radiation
• Cyclotron Radiation
• Relativistic beaming
• Radiation reaction
• Sokolov–Ternov effect

 وصلات خارجية

• Cosmic Magnetobremsstrahlung (synchrotron Radiation), by Ginzburg, V. L., Syrovatskii, S. I., ARAA, 1965
• Developments in the Theory of Synchrotron Radiation and its Reabsorption, by Ginzburg, V. L., Syrovatskii, S. I., ARAA, 1969
• Lightsources.org
• X-Ray Data Booklet