Synchrotron Radiation

11. Synchrotron Radiation#

11.1. History#

Synchrotron radiation was first observed in GE synchrotron on 1946.

../../_images/ge_synchrotron.png — Fig. 11.1 GE synchrotron#

../../_images/sr_paper_1947.png — Fig. 11.2 First paper of SR#

Then it was realized as the major obstacle to achieve higher electron energy in a ring accelerator. Since the radiation power is scaled as:

(11.1)#

P \sim \frac{γ^{4}}{ρ^{2}}

11.2. The Lienard- Wiechert Potential#

We are interested in the E-M field generated by a moving charged particle Suppose the charge particle has determined trajectory $r_{0} (t_{r})$ , the time $t_{r}$ is the time at the charged particle. The observer located at $P$ as position $r = r_{0} (t_{r}) + R (t_{r})$ . The time $t$ is related to $t_{r}$ by:

(11.2)#

\begin{array}{r} t_{r} + \frac{| R (t_{r}) |}{c} = t \end{array}

$t_{r}$ is the retarded time. The time derivative of $t_{r}$ is given by:

(11.3)#

\begin{aligned} d (t_{r} + \frac{| R (t_{r}) |}{c}) & = d t \\ (1 + \frac{d}{d t_{r}} \frac{| R (t_{r}) |}{c}) d t_{r} & = d t \end{aligned}

from $R^{2} = R \cdot R$ , we have

(11.4)#

\begin{array}{r} \frac{d R}{d t_{r}} = - \frac{v \cdot R}{R} \end{array}

So the $t_{r}$ ‘s derivative is :

(11.5)#

\begin{array}{r} \frac{d t_{r}}{d t} = \frac{1}{1 - β \cdot \hat{R}} \end{array}

From the wave form of the Maxwell equation of one particle:

(11.6)#

\begin{aligned} (- \nabla^{2} + \frac{1}{c^{2}} \frac{\partial^{2}}{\partial t^{2}}) ϕ & = \frac{ρ}{ε_{0}} \\ (- \nabla^{2} + \frac{1}{c^{2}} \frac{\partial^{2}}{\partial t^{2}}) A & = μ_{0} J \end{aligned}

Here, the charged density and current density are

(11.7)#

\begin{aligned} ρ & = q δ (r - r_{0} (t_{r})) \\ J & = q r (t_{r}) δ (r - r_{0} (t_{r})) \end{aligned}

The scalar and vector potential are

(11.8)#

\begin{aligned} ϕ (r, t) & = \frac{1}{4 π ϵ_{0}} \iint \frac{q δ^{3} (r^{'} - r_{s} (t^{'}))}{| r - r^{'} |} δ (t^{'} - t_{r}^{'}) d t^{'} d^{3} r^{'} \\ A (r, t) & = \frac{μ_{0}}{4 π} \iint \frac{q v_{s} (t^{'}) δ^{3} (r^{'} - r_{s} (t^{'}))}{| r - r^{'} |} δ (t^{'} - t_{r}^{'}) d t^{'} d^{3} r^{'} \end{aligned}

Then the potentials reduces to

(11.9)#

\begin{aligned} ϕ (r, t) & = \frac{1}{4 π ϵ_{0}} \frac{q}{(1 - β \cdot \hat{R}) | r - r_{0} |} \\ A (r, t) & = \frac{μ_{0}}{4 π} \frac{q v}{(1 - β \cdot \hat{R}) | r - r_{0} |} \end{aligned}

Both are evaluated at time $t_{r}$ .

<matplotlib.collections.PathCollection at 0x10eae47a0>

The fields therefore can be calculated as from

(11.10)#

\begin{aligned} E & = - \nabla ϕ - \frac{\partial A}{\partial t} \\ B & = \nabla \times A \end{aligned}

To evaluate these differential operations, it is important to know the following relations:

(11.11)#

\begin{aligned} \nabla | r - r_{0} | & = \nabla t_{r} \frac{d | r - r_{0} |}{d t_{r}} + \hat{R} \\ = - \nabla t_{r} c β \cdot \hat{R} + \hat{R} \end{aligned}

And we know that

(11.12)#

\begin{aligned} \nabla t_{r} + \frac{1}{c} \nabla | r - r_{0} | & = 0 \\ \nabla t_{r} (1 - β \cdot \hat{R}) & = - \frac{1}{c} \hat{R} \end{aligned}

Therefore we can gather all useful relations below:

(11.13)#

\begin{aligned} \frac{d t_{r}}{d t} & = \frac{1}{1 - β \cdot \hat{R}} \\ \nabla t_{r} & = - \frac{1}{c} \frac{\hat{R}}{1 - β \cdot \hat{R}} \\ \nabla | r - r_{0} | & = \frac{\hat{R}}{(1 - \hat{R} \cdot β)} \end{aligned}

The the electric field gives:

(11.14)#

\begin{array}{r} E = \frac{q}{4 π ϵ_{0} γ^{2}} \frac{\hat{R} - β}{R^{2} {(1 - β \cdot \hat{R})}^{3}} + \frac{q}{4 π ϵ_{0} c} \hat{R} \times \frac{(\hat{R} - β) \times \dot{β}}{R {(1 - β \cdot \hat{R})}^{3}} \end{array}

The magnetic field gives:

(11.15)#

\begin{array}{r} B = \hat{R} \times E / c \end{array}

The first term does not involve acceleration, which described a ‘uniformly’ moving charge (see “The classical Theory of Fields” by Landau and Lifshitz). After a Lorentz transform, tt is just the static electric field from a charged particle. The emitted energy from this term is zero. Therefore only the contribution from the second term ‘radiates’ out, which is consequence of $\dot{β}$ , or acceleration.

11.3. Radiation Field and Flux#

The Poynting vector of the radiation term gives (using the fact $\hat{R} \cdot E = 0$ ):

(11.16)#

\begin{aligned} S & = \frac{1}{μ_{0}} E \times B \\ = \frac{1}{μ_{0}} E \times (\hat{R} \times E / c) \\ = \frac{1}{μ_{0} c} E^{2} \hat{R} \end{aligned}

The power dissipates in the unit solid angle gives, evalutated as source is :

(11.17)#

\begin{aligned} \frac{d P}{d Ω} & = R^{2} S \cdot \hat{R} \frac{d t}{d t_{r}} \\ = (1 - \hat{R} \cdot β_{s}) \frac{R^{2} E^{2}}{μ_{0} c} \\ = \frac{q^{2}}{16 π^{2} ϵ_{0} c} \frac{(\hat{R} \times ((\hat{R} - β) \times \dot{β}))^{2}}{{(1 - β \cdot \hat{R})}^{5}} \end{aligned}

11.3.1. Non-relativistic limit#

In this limit, the approximation $β ≪ 1$ is made, which leads to:

$\hat{R} - β \approx \hat{R}$
$1 - β \cdot \hat{R} \approx 1$

Then the power flux gives:

(11.18)#

\begin{aligned} \frac{d P}{d Ω} & = \frac{q^{2}}{16 π^{2} ϵ_{0} c} (\hat{R} \times (\hat{R} \times \dot{β}))^{2} \\ = \frac{q^{2}}{16 π^{2} ϵ_{0} c^{3}} {\dot{v}}^{2} {(\sin θ)}^{2} \end{aligned}

Here we establish the spherical coordinate system along $\dot{v}$ and the angle between $\dot{v}$ and $\hat{R}$ is $θ$ . Therefore the total power is simply:

(11.19)#

\begin{aligned} P & = \int \frac{d P}{d Ω} d Ω \\ = \frac{q^{2}}{16 π^{2} ϵ_{0} c^{3}} {\dot{v}}^{2} \int_{0}^{2 π} d ϕ \int_{0}^{π} {(\sin θ)}^{2} \sin θ d θ \\ = \frac{1}{4 π ϵ_{0}} \frac{2 q^{2}}{3 c^{3}} {\dot{v}}^{2} \end{aligned}

Which reduces to the Larmor’s formula

11.3.2. General Case#

In accelerator, we are interested in relativistic case. We start with

(11.20)#

\begin{aligned} \frac{d P}{d Ω} & = \frac{q^{2}}{16 π^{2} ϵ_{0} c} \frac{(\hat{R} \times ((\hat{R} - β) \times \dot{β}))^{2}}{{(1 - β \cdot \hat{R})}^{5}} \\ = \frac{q^{2}}{16 π^{2} ϵ_{0} c} \frac{((\hat{R} \cdot \dot{β}) (\hat{R} - β) - (1 - \hat{R} \cdot β) \dot{β})^{2}}{{(1 - β \cdot \hat{R})}^{5}} \end{aligned}

We need to establish a spherical coordinate system to go further. Let’s assume $β = β (0, 0, 1)$ , $\dot{β} = \dot{β} (\sin Θ, 0, \cos Θ)$ and $\hat{R} = (\sin θ \cos ϕ, \sin θ \sin ϕ, \cos θ)$ and the result is calculated below by Sympy package.

../../_images/0b0ed6ec2592e56b4c84ab4e01f3a523f8a140541b79e3c2d4c59c6b8082e071.png

(11.21)#

\begin{aligned} P & = \int \frac{d P}{d Ω} d Ω \\ = \frac{q^{2}}{16 π^{2} ϵ_{0} c} \frac{8 π {\dot{β}}^{2} (β^{2} \sin^{2} (Θ) - 1)}{3 (β^{6} - 3 β^{4} + 3 β^{2} - 1)} \\ = \frac{1}{4 π ϵ_{0}} \frac{2 q^{2}}{3 c} γ^{6} {\dot{β}}^{2} (1 - β^{2} \sin^{2} (Θ)) \\ = \frac{1}{4 π ϵ_{0}} \frac{2 q^{2}}{3 c} γ^{6} {\dot{β}}^{2} (1 - β^{2} \frac{{| β \times \dot{β} |}^{2}}{β^{2} {\dot{β}}^{2}}) \\ = \frac{1}{4 π ϵ_{0}} \frac{2 q^{2}}{3 c} γ^{6} ({\dot{β}}^{2} - {| β \times \dot{β} |}^{2}) \end{aligned}

We have derived the total power that emitted by an accelerated beam. The first term in the parentheses is the contribution from beam acceleration. The second term is the contribution from the bending term.

11.3.2.1. Radiation in linac#

In a linac, the total radiation power is:

(11.22)#

\begin{array}{r} P = \frac{1}{4 π ϵ_{0}} \frac{2 q^{2}}{3 c} γ^{6} {\dot{β}}^{2} \end{array}

It is more convenient to convert the change of velocity to the energy. From $β^{2} γ^{2} = γ^{2} - 1$ , we have

(11.23)#

\begin{array}{r} β \dot{β} = γ^{- 3} \dot{γ} = \frac{1}{m c^{2} γ^{3}} \frac{d E}{d x} \frac{d x}{d t} = \frac{β}{m c γ^{3}} \frac{d E}{d x} \end{array}

Therefore:

(11.24)#

\begin{aligned} P & = \frac{1}{4 π ϵ_{0}} \frac{2 q^{2}}{3 c} γ^{6} \frac{1}{m^{2} c^{2} γ^{6}} {(\frac{d E}{d x})}^{2} \\ = \frac{1}{4 π ϵ_{0}} \frac{2 q^{2} c}{3 (m c^{2})^{2}} {(\frac{d E}{d x})}^{2} \\ \approx 0.288 e V / s \times {(\frac{Energy gradient [MeV/m]}{Beam rest energy [MeV]})}^{2} \end{aligned}

The angular dependence becomes:

(11.25)#

\begin{array}{r} \frac{d P}{d Ω} \sim \frac{{\dot{β}}^{2} \sin^{2} (θ)}{{(- β \cos (θ) + 1)}^{5}} \end{array}

when the beam is ultra-relativistic, $β \approx 1 - \frac{1}{2 γ^{2}}$ , $\sin θ \approx θ$ , and $\cos θ \approx 1 - θ^{2} / 2$ , therefore, the angular distribution becomes:

(11.26)#

\begin{array}{r} \frac{d P}{d Ω} \sim \frac{{\dot{β}}^{2} θ^{2}}{{(θ^{2} + \frac{1}{γ^{2}})}^{5}} \end{array}

The maximum angle locates at $θ \sim \frac{1}{γ}$ .

11.3.2.2. Radiation in dipole#

In the magnetic field, the velocity only changes its direction, $\dot{β} ⊥ β$ , and the amplitude is

(11.27)#

\begin{array}{r} \dot{β} = \frac{c β^{2}}{ρ} \end{array}

Then the total power is:

(11.28)#

\begin{aligned} P & = \frac{1}{4 π ϵ_{0}} \frac{2 q^{2}}{3 c} γ^{6} {\dot{β}}^{2} (1 - β^{2} \sin^{2} (Θ)) \\ = \frac{1}{4 π ϵ_{0}} \frac{2 q^{2}}{3} \frac{c γ^{4} β^{4}}{ρ^{2}} \end{aligned}

We try to re-express the power in terms of the classical radius $r_{0}$ and radiation constant $C_{γ}$ , who are defined as:

(11.29)#

\begin{array}{r} r_{0} = \frac{1}{4 π ϵ_{0}} \frac{q^{2}}{m c^{2}} \end{array}

(11.30)#

\begin{array}{r} C_{γ} = \frac{4 π}{3} \frac{r_{0}}{{(m c^{2})}^{3}} \end{array}

Then

(11.31)#

\begin{array}{r} P = \frac{C_{γ}}{2 π} c β^{4} \frac{E^{4}}{ρ^{2}} \end{array}

For different particle species the constant $C_{γ}$ gives:

(11.32)#

\begin{array}{r} C_{γ} = {\begin{cases} 8.85 \times 10^{- 5} {m/(GeV)}^{3} & Electron \\ 4.84 \times 10^{- 14} {m/(GeV)}^{3} & Muon \\ 7.78 \times 10^{- 18} {m/(GeV)}^{3} & Proton \end{cases} \end{array}

The total energy radiated of the ring will be:

(11.33)#

\begin{aligned} U_{r} & = \int_{0}^{T} P d t \\ \approx \frac{C_{γ}}{2 π} β^{4} E^{4} \oint \frac{1}{ρ^{2}} d s \end{aligned}

The angular distribution can be found as:

(11.34)#

\begin{aligned} \frac{d P}{d Ω} & \sim \frac{{\dot{β}}^{2}}{{(1 - β \cos (θ))}^{5}} (- β^{2} \sin^{2} (ϕ) \sin^{2} (θ) + β^{2} - 2 β \cos (θ) + \sin^{2} (ϕ) \sin^{2} (θ) + \cos^{2} (θ)) \\ \sim \frac{{\dot{β}}^{2}}{{(1 - β \cos (θ))}^{5}} (\frac{\sin^{2} (ϕ) \sin^{2} (θ)}{γ^{2}} + {(β - \cos θ)}^{2}) \end{aligned}

70.71244595191452

In the relativistic limit, we have

(11.35)#

\begin{aligned} \frac{d P}{d Ω} & \sim \frac{{\dot{β}}^{2}}{{(1 - β \cos (θ))}^{5}} (\frac{\sin^{2} (ϕ) \sin^{2} (θ)}{γ^{2}} + {(β - \cos θ)}^{2}) \\ \sim \frac{{\dot{β}}^{2}}{{(\frac{1}{γ^{2}} + θ^{2})}^{5}} (\frac{4 \sin^{2} (ϕ) θ^{2}}{γ^{2}} + {(\frac{1}{γ^{2}} - θ^{2})}^{2}) \end{aligned}

The power will drop significantly when $θ ≫ 1 / γ$ .

11.4. Radiation Spectrum#

11.4.1. Qualitative analysis#

../../_images/sr_ering.png — Fig. 11.3 Cartoon of an ESR#

We are interested in an electron storage ring. The retarded time interval of shining the SR cone to a far field target is

(11.36)#

\begin{array}{r} Δ t_{r} \approx \frac{ρ θ}{β c} \sim \frac{ρ}{γ c} \end{array}

Convert to the observer’s time:

(11.37)#

\begin{array}{r} Δ t = \frac{d t}{d t_{r}} Δ t_{r} \sim \frac{ρ}{γ^{3} c} \end{array}

Therefore the frequency content is approximated with:

(11.38)#

\begin{array}{r} ω \sim 1 / Δ t \sim \frac{γ^{3} c}{ρ} \sim γ^{3} ω_{0} \end{array}

11.4.2. Quantitative analysis#

To study the frequency spectrum of the synchrotron radiation, we return to the flux, evaluated at the observation point gives

(11.39)#

\begin{aligned} \frac{d P}{d Ω} & = R^{2} S \cdot \hat{R} \\ = \frac{1}{μ_{0} c} R^{2} E^{2} \equiv G (t)^{2} \end{aligned}

We define an amplitude radiation function which is the square root of the flux:

(11.40)#

\begin{array}{r} G (t) = \sqrt{\frac{1}{μ_{0} c}} R (t) E \end{array}

Then the frequency content is given by Fourier transform:

(11.41)#

\begin{aligned} G (ω) & = \frac{1}{\sqrt{2 π}} \int G (t) e^{i ω t} d t \\ = \frac{q}{4 π ϵ_{0} c} \sqrt{\frac{1}{2 π μ_{0} c}} \int \hat{R} \times \frac{(\hat{R} - β) \times \dot{β}}{{(1 - β \cdot \hat{R})}^{3}} e^{i ω t} d t \\ = \sqrt{\frac{q^{2}}{32 π^{3} ϵ_{0} c}} \int \hat{R} \times \frac{(\hat{R} - β) \times \dot{β}}{{(1 - β \cdot \hat{R})}^{2}} e^{i ω (t_{r} + R / c)} d t_{r} \end{aligned}

Now we have to use far field approximation to proceed. We know that $R = r - r_{0}$ . The far field approximation implies $| r | ≫ | r_{0} |$ , then

(11.42)#

\begin{aligned} R & = | r - r_{0} | \\ = r | \hat{r} - r_{0} / r | \\ \approx r (1 - \hat{r} \cdot r_{0} / r) \\ \approx r - \hat{R} \cdot r_{0} \end{aligned}

Then the frequency spectrum becomes

(11.43)#

\begin{aligned} G (ω) & = \sqrt{\frac{q^{2}}{32 π^{3} ϵ_{0} c}} e^{i ω r / c} \int \hat{R} \times \frac{(\hat{R} - β) \times \dot{β}}{{(1 - β \cdot \hat{R})}^{2}} e^{i ω (t_{r} - \hat{R} \cdot r_{0} / c)} d t_{r} \\ = - i ω \sqrt{\frac{q^{2}}{32 π^{3} ϵ_{0} c}} e^{i ω r / c} \int \hat{R} \times (\hat{R} \times β) e^{i ω (t_{r} - \hat{R} \cdot r_{0} / c)} d t_{r} \end{aligned}

Here we use the relation below and integration by part:

(11.44)#

\begin{array}{r} \frac{d}{d t_{r}} (\frac{\hat{R} \times (\hat{R} \times β)}{1 - β \cdot \hat{R}}) = \hat{R} \times \frac{(\hat{R} - β) \times \dot{β}}{{(1 - β \cdot \hat{R})}^{2}} \end{array}

Here we found the frequency spectrum of the radiation amplitude. However, what we can measure is the total radiation energy per solid angle:

(11.45)#

\begin{aligned} \frac{d U}{d Ω} & = \int_{- \infty}^{\infty} \frac{d P}{d Ω} d t \\ = \int_{- \infty}^{\infty} {| G (t) |}^{2} d t = \int_{- \infty}^{\infty} {| G (ω) |}^{2} d ω (Parseval's theorem) \\ = 2 \int_{0}^{\infty} {| G (ω) |}^{2} d ω \equiv \int_{0}^{\infty} \frac{d I}{d ω} d ω (Real signal) \end{aligned}

Therefore we will then focus on the spectrum of the energy:

(11.46)#

\begin{aligned} \frac{d I}{d ω} & = 2 {| G (ω) |}^{2} \\ = \frac{q^{2} ω^{2}}{16 π^{3} ϵ_{0} c} {| \int \hat{R} \times (\hat{R} \times β) e^{i ω (t_{r} - \hat{R} \cdot r_{0} / c)} d t_{r} |}^{2} \end{aligned}

../../_images/sr_coordinate.png — Fig. 11.4 Coordinates for calculation#

Now we concentrate on the spectrum of an electron storage ring with coordinate as shown.
Assume $r_{0}$ is along a ring with radius $ρ$ , therefore:

(11.47)#

\begin{array}{r} r_{0} = ρ (1 - \cos (ω_{0} t_{r}), 0, \sin (ω_{0} t_{r})) \end{array}

Here, $ω_{0}$ is the angular revolution frequency. The velocity therefore gives

(11.48)#

\begin{array}{r} β = β (\sin (ω_{0} t_{r}), 0, \cos (ω_{0} t_{r})) \end{array}

And the unit vector $\hat{R}$ in spherical coordinate is written as:

(11.49)#

\begin{array}{r} \hat{R} = (\sin θ \cos ϕ, \sin θ \sin ϕ, \cos θ) \end{array}

We may only consider the tangential plane ( $y$ - $z$ ), without losing generality, viz. $ϕ = π / 2$ . Therefore

(11.50)#

\begin{array}{r} \hat{R} = (0, \sin θ, \cos θ) \end{array}

Also we use the following assumption:

small cone size $θ \sim 1 / γ ≪ 1$
short retarded time interval $ω_{0} t_{r} ≪ 1$

We have:

(11.51)#

\begin{aligned} \hat{R} \times (\hat{R} \times β) & = \hat{R} (\hat{R} \cdot β) - β \\ = β (- ω_{0} t_{r}, θ, 0) \end{aligned}

and

(11.52)#

\begin{aligned} ω (t_{r} - \hat{R} \cdot r_{0} / c) & = ω (t_{r} - \frac{ρ}{c} \sin (ω_{0} t_{r}) \cos θ) \\ = ω (t_{r} - \frac{β}{ω_{0}} \sin (ω_{0} t_{r}) \cos θ) \\ = ω (t_{r} - \frac{1}{ω_{0}} (1 - \frac{1}{2 γ^{2}}) (ω_{0} t_{r} - \frac{ω_{0}^{3} t_{r}^{3}}{6}) (1 - \frac{θ^{2}}{2})) \\ = ω (t_{r} - (1 - \frac{1}{2 γ^{2}}) (t_{r} - \frac{ω_{0}^{2} t_{r}^{3}}{6}) (1 - \frac{θ^{2}}{2})) \\ = \frac{ω}{2} ((θ^{2} + \frac{1}{γ^{2}}) t_{r} + \frac{ω_{0}^{2} t_{r}^{3}}{3}) \end{aligned}

Now, we redefine the variable $a \equiv γ θ \sim 1$ , and further define:

(11.53)#

\begin{aligned} x & \equiv \sqrt{\frac{ω_{0}^{2} t_{r}^{2}}{θ^{2} + \frac{1}{γ^{2}}}} = \frac{γ ω_{0} t_{r}}{{(1 + a^{2})}^{1 / 2}} \\ ξ & \equiv \frac{ω}{3} \frac{{(1 + a^{2})}^{3 / 2}}{γ^{3} ω_{0}} = \frac{ω}{2 ω_{c}} {(1 + a^{2})}^{3 / 2} \end{aligned}

Here we define the critical frequency of synchrotron radiation:

(11.54)#

\begin{array}{r} ω_{c} \equiv \frac{3}{2} γ^{3} ω_{0} \approx \frac{3 γ^{3} c}{2 ρ} \end{array}

Then, we have:

(11.55)#

\begin{array}{r} ω (t_{r} - \hat{R} \cdot r_{0} / c) = \frac{3}{2} ξ (x + \frac{x^{3}}{3}) \end{array}

The spectrum of the energy flux finally reduces to:

(11.56)#

\begin{aligned} \frac{d I}{d ω} & = \frac{q^{2} ω^{2}}{16 π^{3} ϵ_{0} c} {| \int \hat{R} \times (\hat{R} \times β) e^{i ω (t_{r} - \hat{R} \cdot r_{0} / c)} d t_{r} |}^{2} \\ = \frac{q^{2} ω^{2}}{16 π^{3} ϵ_{0} c} {| \int β (- ω_{0} t_{r} \hat{x} + θ \hat{y}) \exp (i \frac{3}{2} ξ (x + \frac{x^{3}}{3})) d t_{r} |}^{2} \\ \equiv \frac{β^{2} q^{2} ω^{2}}{16 π^{3} ϵ_{0} c} {| (- G_{x} \hat{x} + G_{y} \hat{y})) |}^{2} \end{aligned}

Clearly, the total frequency spectrum of the radiation intensity consists the contribution from two transverse components $G_{x}$ and $G_{y}$ . The integral yields the modified Bessel functions of the second kind:

(11.57)#

\begin{aligned} G_{x} & = \frac{1 + a^{2}}{ω_{0} γ^{2}} \int x \exp (i \frac{3}{2} ξ (x + \frac{x^{3}}{3})) d x = \frac{2}{\sqrt{3}} \frac{1 + a^{2}}{ω_{0} γ^{2}} K_{2 / 3} (ξ) \\ G_{y} & = \frac{a {(1 + a^{2})}^{1 / 2}}{ω_{0} γ^{2}} \int \exp (i \frac{3}{2} ξ (x + \frac{x^{3}}{3})) d x = \frac{2}{\sqrt{3}} \frac{a {(1 + a^{2})}^{1 / 2}}{ω_{0} γ^{2}} K_{1 / 3} (ξ) \end{aligned}

Then the frequency spectrum becomes

(11.58)#

\begin{aligned} \frac{d I}{d ω} & = \frac{β^{2} q^{2} ω^{2}}{16 π^{3} ϵ_{0} c} {| (- G_{x} \hat{x} + G_{y} \hat{y})) |}^{2} \\ = \frac{β^{2} q^{2} ω^{2}}{16 π^{3} ϵ_{0} c} \frac{4}{3} \frac{1 + a^{2}}{ω_{0}^{2} γ^{4}} ((1 + a^{2}) K_{2 / 3}^{2} (ξ) + a^{2} K_{1 / 3}^{2} (ξ)) \end{aligned}

We are interest in the frequency range of the characteristic frequency of the SR. Using the definition of critical frequency, $ω_{c} = 3 γ^{3} ω_{0} / 2$ :

(11.59)#

\begin{aligned} \frac{d I}{d ω} & = \frac{3 q^{2}}{16 π^{3} ϵ_{0} c} \frac{ω^{2}}{ω_{c}^{2}} γ^{2} {(1 + a^{2})}^{2} (K_{2 / 3}^{2} (ξ) + \frac{a^{2}}{1 + a^{2}} K_{1 / 3}^{2} (ξ)) \\ = \frac{3 q^{2}}{4 π^{3} ϵ_{0} c} ξ^{2} γ^{2} {(1 + a^{2})}^{- 1} (K_{2 / 3}^{2} (ξ) + \frac{a^{2}}{1 + a^{2}} K_{1 / 3}^{2} (ξ)) \end{aligned}

The two terms of modified Bessel function corresponds to the contribution from the horizontal and vertical polarized radiation field respectively.
We see that in the bending plane, $a = γ θ = 0$ , the radiation is purely horizontal polarized.

We see that the spectrum is broad band and sudden drops when the the frequency exceeds the critical frequency.

Recall that:

(11.60)#

\begin{array}{r} \frac{d U}{d Ω} = \int_{0}^{\infty} \frac{d I}{d ω} d ω \end{array}

Therefore, the energy spectrum can be achieved by integrating all solid angles:

(11.61)#

\begin{aligned} \frac{d U}{d ω} & = \int \frac{d I}{d ω} d Ω \\ = \frac{\sqrt{3} q^{2}}{4 π ϵ_{0} c} γ \frac{ω}{ω_{c}} \int_{ω / ω_{c}}^{\infty} K_{5 / 3} (ξ) d ξ \end{aligned}

By defining

(11.62)#

\begin{array}{r} S (ξ) = \frac{9 \sqrt{3}}{8 π} ξ \int_{ξ}^{\infty} K_{5 / 3} (ξ^{'}) d ξ^{'} \end{array}

which satisfy $\int_{0}^{\infty} S (ξ) d ξ = 1$ , we will rewrite the energy spectrum as:

(11.63)#

\begin{aligned} \frac{d U}{d ω} & = \frac{2 q^{2}}{9 ϵ_{0} c} γ S (\frac{ω}{ω_{c}}) \end{aligned}

The integral of S function is: 0.9976699617994194

The total energy can be also attained from the frequency domain integration.

(11.64)#

\begin{aligned} U & = \int_{0}^{\infty} \frac{d U}{d ω} d ω \\ = \frac{2 q^{2}}{9 ϵ_{0} c} γ \int_{0}^{\infty} S (\frac{ω}{ω_{c}}) d ω \\ = \frac{2 q^{2}}{9 ϵ_{0} c} γ ω_{c} \\ = \frac{q^{2}}{3 ϵ_{0} ρ} γ^{4} = \frac{2 π ρ}{c} P \end{aligned}

We return to the earlier SR power from time domain integral.

11.5. Quantum Fluctuation#

The photon emission in the SR process is a quantum effect. The energy $u$ of the photon is related to its frequency by $u = ℏ ω$ The photon number is found by:

(11.65)#

\begin{array}{r} u n (u) d u = \frac{d U}{d ω} d ω \end{array}

Therefore: the photon number density gives:

(11.66)#

\begin{aligned} n (u) & = \frac{1}{ℏ^{2} ω} \frac{d U}{d ω} \\ = \frac{1}{ℏ^{2} ω_{c}^{2}} \frac{2 q^{2} ω_{c}}{9 ϵ_{0} c} γ \frac{ω_{c}}{ω} S (\frac{ω}{ω_{c}}) \\ = \frac{U_{S R}}{u_{c}^{2}} \frac{ω_{c}}{ω} S (\frac{ω}{ω_{c}}) \end{aligned}

here $u_{c} = ℏ ω_{c}$ is the critical photon energy. And using the integration fact:

(11.67)#

\begin{array}{r} \int_{0}^{\infty} \frac{S (ξ)}{ξ} d ξ = \frac{15 \sqrt{3}}{8} \end{array}

We get the total emitted photon in one revolution gives:

(11.68)#

\begin{aligned} N & = \int n (u) d u \\ = \frac{15 \sqrt{3}}{8} \frac{U_{S R}}{u_{c}} \\ = \frac{15 \sqrt{3}}{8} \frac{q^{2}}{3 ϵ_{0} ρ ℏ} γ \frac{2 ρ}{3 c} \\ = \frac{5 π}{\sqrt{3}} \frac{q^{2}}{4 π ϵ_{0} ℏ c} γ = \frac{5 π}{\sqrt{3}} α γ \end{aligned}

Here $α$ is the fine structure constant. These photons emits randomly around one revolution.

The average emitted photon gives energy by

(11.69)#

\begin{array}{r} \bar{u} = \frac{1}{N} \int_{0}^{\infty} u n (u) d u = \frac{8}{15 \sqrt{3}} u_{c} \end{array}

And the variation gives by

(11.70)#

\begin{array}{r} \bar{u^{2}} = \frac{1}{N} \int_{0}^{\infty} u^{2} n (u) d u = \frac{11}{27} u_{c}^{2} \end{array}

We can summarized the synchrotron radiation parameters’s dependence on energy:

Quantities	Power of energy
Radiation Power (fix $ρ$ )	$P \sim γ^{4} / ρ^{2}$
Radiation Power (fix field)	$P \sim γ^{2} B^{2}$
Number of Photon	$N \sim γ$
Average photon energy	$\bar{u} \sim γ^{3}$
Photon energy variation	$\bar{u^{2}} \sim γ^{6}$
Accumulated photon energy variation	$\bar{N u^{2}} \sim γ^{7}$