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The phenomenon of synchrotron radiation has been introduced in a conceptual way and a number of basic relations have been derived. In this chapter we will approach the physics of synchrotron radiation in a more formal way to exhibit detailed characteristics. Specifically, we will derive expressions for the spatial and spectral distribution of photon emission in a way which is applicable later for special insertion devices.

The theory of synchrotron radiation is intimately related to the electromagnetic fields generated by moving charged particles. Wave equations can be derived from Maxwell’s equations and we will find that any charged particle under the influence of external forces can emit radiation. We will formulate the characteristics of this radiation and apply the results to highly relativistic particles.

1 Radiation Field

The electromagnetic fields for a single moving point charge will be derived first and then applied to a large number of particles. Fields are determined by Maxwell’s equations (1.4) for moving charges in vacuum where \(\epsilon _{\text{r}} =\mu _{\text{r}} = 1\). The magnetic field can be derived from a vector potential\(\boldsymbol{A}\) defined by

$$\displaystyle{ \boldsymbol{B} = \nabla \times A\,. }$$
(25.1)

Inserting the vector potential into Faraday’s law (1.4) we have\(\boldsymbol{\nabla }\times \left (\boldsymbol{E} + \frac{\partial \boldsymbol{A}} {\partial t} \right ) = 0\), or after integration

$$\displaystyle{ \boldsymbol{E} = -\frac{\partial \boldsymbol{A}} {\partial t} -\boldsymbol{\nabla }\varphi \,, }$$
(25.2)

where \(\varphi\) is the scalar potential. We choose the scalar potential such that \(c\boldsymbol{\nabla }\boldsymbol{A} + \frac{1} {c} \frac{\partial \boldsymbol{\varphi }} {\partial t} = 0,\) a condition known as the Lorentz gauge. With (A.21) applied to \(\boldsymbol{A}\) the expression for the electric field together with Ampere’s law (1.4) results in the wave equation

$$\displaystyle{ \nabla ^{2}\boldsymbol{A} - \frac{1} {c^{2}} \frac{\partial ^{2}\boldsymbol{A}} {\partial t^{2}} = \frac{1} {\epsilon _{0}} \rho \boldsymbol{\beta \,}\mathbf{.} }$$
(25.3)

Similarly, we derive the wave equation for the scalar potential

$$\displaystyle{ \nabla ^{2}\varphi - \frac{1} {c^{2}} \frac{\partial ^{2}\varphi } {\partial t^{2}} = -\frac{1} {\epsilon _{0}} \rho \,. }$$
(25.4)

These are the well-known wave equations with the solutions

$$\displaystyle{ \boldsymbol{A}(t) = \frac{\mu _{0}} {4\pi }\int \left.\frac{\boldsymbol{v}\rho (x,y,z)} {R} \right \vert _{t_{\text{r}}}\text{d}x\,\text{d}y\,\text{d}z }$$
(25.5)

and

$$\displaystyle{ \varphi (t) = \frac{1} {4\pi \epsilon _{0}}\int \left.\frac{\rho (x,y,z)} {R} \right \vert _{t_{\text{r}}}\,\text{d}x\,\text{d}y\,\text{d}z\,. }$$
(25.6)

Because of the finite velocity of light, all quantities under the integrals must be evaluated at the retarded time

$$\displaystyle{ t_{\text{r}} = t -\frac{1} {c}R(t_{\text{r}}) }$$
(25.7)

when the radiation was emitted by the moving charge, in contrast to the time t when the radiation is observed at a distant point. The quantity R is the distance between the observation point P(x, y, z) and the location of the charge element \(\rho (x_{\text{r}},y_{\text{r}},z_{\text{r}})\) dxrdyrdzr at the retarded time tr. The vector

$$\displaystyle{ \boldsymbol{R\,}\mathbf{=}(x_{\text{r}} - x,y_{\text{r}} - y,z_{\text{r}} - z) }$$
(25.8)

points away from the observation point to the charge element at the retarded time as shown in Fig. 25.1.

Fig. 25.1
figure 1figure 1

Retarded position of a moving charge distribution

Special care must be exercised in performing the integrations. Although we consider only a point charge q, the integral in (25.6) cannot be replaced by qR but must be integrated over a finite volume followed by a transition to a point charge. As we will see this is a consequence of the fact that the velocity of light is finite and therefore the movement of charge elements must be taken into account.

To define the quantities involved in the integration we use Fig. 25.1. The combined field at the observation point P at time t comes from all charges located at a distance R away from P. We consider the contribution from all charges contained within a spherical shell centered at P with a radius R and thickness dr to the radiation field at P and time t. Radiation emitted at time tr will reach P at the time t. If dσ is a surface element of the spherical shell, the volume element of charge is dx dy dz = dσdr. The retarded time for the radiation from the outer surface of the shell is tr and the retarded time for the radiation from the charge element on the inner surface of the shell is \(t_{\text{r}} -\frac{\text{d}r} {c}\). From Fig. 25.1 we find the electromagnetic field observed at P at time t to originate from the fractional charges within the volume element d σd r or from the charge element dq = ρ dσdr.

The radiation observed at point P and time t is the sum of all radiation arriving simultaneously at P. Elements of this radiation field may have been emitted by different charge elements and at different times and locations. In case of only one electrical charge moving with velocity \(\boldsymbol{v}\), we have to include in the integration those charge elements that move across the inner shell surface into the volume dσdr during the time drc. For a uniform charge distribution this additional charge is \(\delta q =\rho \boldsymbol{ vn}\mathit{\,}\) dt dσ where \(\boldsymbol{n}\) is the vector normal to the surface of the shell and pointing away from the observer

$$\displaystyle{ \boldsymbol{n} = \frac{\boldsymbol{R}} {R}\,. }$$
(25.9)

With d t = drc and \(\boldsymbol{\beta }=\boldsymbol{ v}\mathbf{/}c\), we get then for both contributions to the charge element

$$\displaystyle{ \text{d}q =\rho (1 +\boldsymbol{ n\beta })\,\text{d}r\,\text{d}\sigma \,. }$$
(25.10)

Depending on the direction of the velocity vector \(\boldsymbol{\beta }\), we find an increase or a reduction in the radiation field from moving charges. We solve (25.10) for ρ dr dσ and insert into the integrals (25.5), (25.6). Now we may use the assumption that the electrical charge is a point charge and get for the retarded potentials of a moving point charge q at time t and observation point P

$$\displaystyle{ \boldsymbol{A}(P,t) = \frac{1} {4\pi c\epsilon _{0}}\left. \frac{q} {R} \frac{\boldsymbol{\beta }} {1 +\boldsymbol{ n\beta }}\right \vert _{t_{\text{r}}} }$$
(25.11)

and

$$\displaystyle{ \varphi (P,t) = \frac{1} {4\pi \epsilon _{0}}\left. \frac{q} {R} \frac{1} {1 +\boldsymbol{ n\beta }}\right \vert _{t_{\text{r}}}\,. }$$
(25.12)

These equations are known as the Liénard–Wiechert potentials and express the field potentials of a static or moving charge as functions of the charge parameters at the retarded time. To obtain the electric and magnetic fields we insert the retarded potentials into (25.1), (25.2) noting that the differentiation must be performed with respect to the time t and location P of the observer while the potentials are expressed at the retarded time tr.

In both equations for the vector and scalar potential we have the same denominator

$$\displaystyle{ r = R(1 +\boldsymbol{ n\beta }\mathbf{)\,.} }$$
(25.13)

It will become necessary to calculate the derivative of the retarded time with respect to the time t and since \(t_{\text{r}} = t - R/c\) the time derivative of tr is

$$\displaystyle{ \frac{\text{d}t_{\text{r}}} {\text{d}t} = 1 -\frac{1} {c} \frac{\text{d}R} {\text{d}t_{\text{r}}} \frac{\text{d}t_{\text{r}}} {\text{d}t} }$$
(25.14)

The variation of the distance R with the retarded time depends on the velocity \(\boldsymbol{v}\) of the moving charge and is the projection of the vector \(\boldsymbol{v}\,\) dtr onto the unity vector \(\boldsymbol{n}\). Therefore,

$$\displaystyle{ \text{d}R =\boldsymbol{ vn}\mathit{\,}\text{d}t_{\text{r}} }$$
(25.15)

and (25.14) becomes with (25.14) and (25.13)

$$\displaystyle{ \frac{\text{d}t_{\text{r}}} {\text{d}t} = \frac{1} {1 +\boldsymbol{ n\beta }} = \frac{R} {r}. }$$
(25.16)

The electric field (25.2) is with (25.11), (25.12) and (25.16) after a few manipulations expressed by

$$\displaystyle{ 4\pi \epsilon _{0}\frac{\boldsymbol{E}} {q} = -\frac{1} {c} \frac{R} {r^{2}} \frac{\partial \boldsymbol{\beta }} {\partial t_{\text{r}}} + \frac{\beta \boldsymbol{R}} {cr^{3}} \frac{\partial r} {\partial t_{\text{r}}} + \frac{1} {r^{2}}\boldsymbol{\nabla }_{\text{r}}r\,. }$$
(25.17)

In evaluating the nabla operator and other differentials we remember that all parameters on the r.h.s. must be taken at the retarded time (25.7) which itself depends on the location of the observation point P. To distinguish between the ordinary nabla operator and the case where the dependence of the retarded time on the position P(x, y, z) must be considered, we add to the nabla symbol the indexr like \(\boldsymbol{\nabla }_{\text{r}}\). The components of this operator are then \(\left. \frac{\partial } {\partial x}\right \vert _{\text{r}} = \frac{\partial } {\partial x} + \frac{\partial t_{\text{r}}} {\partial x} \frac{\partial } {\partial t_{\text{r}}}\), and similar for the other components. We evaluate first

$$\displaystyle{ \boldsymbol{\nabla }_{\text{r}}r =\boldsymbol{ \nabla }_{\text{r}}R +\boldsymbol{ \nabla }_{\text{r }}(\beta R) }$$
(25.18)

and with \(\boldsymbol{\nabla }R = -\boldsymbol{n}\) from (25.8)

$$\displaystyle{ \boldsymbol{\nabla }_{\text{r}}R = -\boldsymbol{n} + \frac{\partial R} {\partial t_{\text{r}}}\boldsymbol{\nabla }t_{\text{r}}\,. }$$
(25.19)

For the gradient of the retarded time, we get

$$\displaystyle{ \boldsymbol{\nabla }t_{\text{r}} =\boldsymbol{ \nabla }\left [t -\frac{1} {c}R(t_{\text{r}})\right ] = -\frac{1} {c}\boldsymbol{\nabla }_{\text{r}}R = -\frac{1} {c}\left (-\boldsymbol{n} + \frac{\partial R} {\partial t_{\text{r}}}\boldsymbol{\nabla }t_{\text{r}}\right ) }$$
(25.20)

and performing the differentiation we get with \(\frac{\partial x_{\text{r}}} {\partial t_{\text{r}}} = v_{x},\ldots\)

$$\displaystyle{ \frac{\partial R} {\partial t_{\text{r}}} = \frac{\partial R} {\partial x_{\text{r}}} \frac{\partial x_{\text{r}}} {\partial t_{\text{r}}} + \frac{\partial R} {\partial y_{\text{r}}} \frac{\partial y_{\text{r}}} {\partial t_{\text{r}}} + \frac{\partial R} {\partial z_{\text{r}}} \frac{\partial z_{\text{r}}} {\partial t_{\text{r}}} =\boldsymbol{ nv}\,. }$$
(25.21)

Solving (25.20) for \(\boldsymbol{\nabla }t_{\text{r}}\) we get

$$\displaystyle{ \boldsymbol{\nabla }t_{\text{r}} = \frac{\boldsymbol{R}} {cr} }$$
(25.22)

and (25.19) becomes finally

$$\displaystyle{ \boldsymbol{\nabla }_{\text{r}}R = -\boldsymbol{n} + \frac{\boldsymbol{R}} {r} (\boldsymbol{\beta \,n})\,. }$$
(25.23)

For the second term in (25.18) we note that the velocity \(\boldsymbol{v}\mathit{\,}\) does not depend on the location of the observer and with \(\boldsymbol{\nabla }_{\text{r}}\boldsymbol{R}\mathit{-}1\), (25.22) and

$$\displaystyle{ \frac{\text{d}\boldsymbol{R}} {\text{d}t_{\text{r}}} =\boldsymbol{ v}\mathit{\,} }$$
(25.24)

we get for the second term in (25.18)

$$\displaystyle{ \boldsymbol{\nabla }_{\text{r }}\mathit{(}\boldsymbol{\beta \,R}\mathbf{)} = -\boldsymbol{\beta } + \frac{\partial (\boldsymbol{\beta \,R}\mathbf{)}} {\partial t_{\text{r}}} \boldsymbol{\nabla }t_{\text{r}} = -\boldsymbol{\beta \,} + \left (\boldsymbol{R} \frac{\partial \boldsymbol{\beta \,}} {\partial t_{\text{r}}}\right ) \frac{\boldsymbol{R}} {cr} +\beta ^{2}\frac{\boldsymbol{R}} {r} \,. }$$
(25.25)

To complete the evaluation of the electric field in (25.17), we express the derivative \(\frac{\partial r} {\partial t_{\text{r}}}\) with

$$\displaystyle{ \frac{\partial r} {\partial t_{\text{r}}} = \frac{\partial R} {\partial t_{\text{r}}} + \frac{\partial (\boldsymbol{\beta \,R}\mathbf{)}} {\partial t_{\text{r}}} = cn\beta \, + c\beta ^{2} +\boldsymbol{ R} \frac{\partial \boldsymbol{\beta \,}} {\partial t_{\text{r}}}\,, }$$
(25.26)

where we made use of (25.21). Collecting all differential expressions required in (25.17) we get with (25.18), (25.23), (25.25), (25.26)

$$\displaystyle\begin{array}{rcl} 4\pi \epsilon _{0}\frac{\boldsymbol{E}} {q} & =& \frac{1} {r^{2}}\left [-\boldsymbol{n} -\boldsymbol{\beta \,} +\frac{\boldsymbol{R}} {r} \left (\boldsymbol{n\beta \,}\boldsymbol{\,}+\beta ^{2} + \tfrac{1} {c}\boldsymbol{\dot{\beta }\,R}\right )\right ]_{\text{r}} \\ & -& \frac{R} {cr^{2}}\boldsymbol{\dot{\beta }\,} +\boldsymbol{\beta \,} \frac{R} {r^{3}}\left (\boldsymbol{n}\boldsymbol{\beta \,}+\beta ^{2} + \tfrac{1} {c}\boldsymbol{\dot{\beta }\,R}\right )_{\text{r}}\,,{}\end{array}$$
(25.27)

where \(\boldsymbol{\dot{\beta }\,}=\) d\(\boldsymbol{\beta \,}/\) d tr. After some manipulation and using (A.10), the equation for the electrical field of a charge q moving with velocity\(\boldsymbol{v}\,\) becomes

$$\displaystyle{ 4\pi \epsilon _{0}\frac{\boldsymbol{E}} {q} = \frac{1 -\beta ^{2}} {r^{3}} \left (\boldsymbol{R} + R\boldsymbol{\beta \,}\right )_{\text{r}} + \frac{1} {cr^{3}}\left.\left \{\boldsymbol{R} \times \left [\left (\boldsymbol{R} + R\boldsymbol{\beta \,}\right )_{\text{r}} \times \frac{\text{d}\boldsymbol{\beta \,}} {\text{d}t_{\text{r}}}\right ]\right \}\right \vert _{\text{r}}\,\,, }$$
(25.28)

where we have added the indexr as a reminder that all quantities on the r.h.s. of (25.28) must be taken at the retarded time tr.

This equation for the electric field of a moving charge has two distinct parts. The first part is inversely proportional to the square of the distance between radiation source and observer and depends only on the velocity of the charge. For a charge at rest \(\beta \mathbf{=}0\) this term reduces to the Coulomb field of a point charge q. The area close to the radiating charge where this term is dominant is called the Coulomb regime. The field is directed toward the observer for a positive charge at rest and tilts into the direction of propagation as the velocity of the charge increases. For highly relativistic particles we note the Coulomb field becomes very small.

We will not further consider this regime since we are interested only in the radiation field far away from the moving charge. The second term in (25.28) depends on the velocity as well as on the acceleration of the charge. This term scales linear with the distance r falling off much slower than the Coulomb term and therefore reaches out to large distances from the radiation source. We call this regime the radiation regime and the remainder of this chapter will focus on the discussion of the radiation from moving charges. The electrical field in the radiation regime is

$$\displaystyle{ 4\pi \epsilon _{0}\left.\frac{\boldsymbol{E}\mathbf{(}t\mathbf{)}} {q} \right \vert _{\text{rad}} = \frac{1} {cr^{3}}\left.\left \{\boldsymbol{R} \times \left [\left (\boldsymbol{R} + R\boldsymbol{\beta \,}\boldsymbol{\,}\right )_{\text{r}} \times \frac{\text{d}\boldsymbol{\beta \,}} {\text{d}t_{\text{r}}}\right ]\right \}\right \vert _{\text{r}}\,. }$$
(25.29)

The polarization of the electric field at the location of the observer is purely orthogonal to the direction of observation R. Similar to the derivation of the electric field, we can derive the expression for the magnetic field and get from (25.1) with (25.11)

$$\displaystyle{ \boldsymbol{B}\boldsymbol{ = \nabla }_{\text{r}}\boldsymbol{ \times }\boldsymbol{ A} = q\left [\boldsymbol{\nabla }_{\text{r}} \times \frac{\boldsymbol{\beta \,}} {r}\right ] = \frac{q} {r}\left [\boldsymbol{\nabla }_{\text{r}}\times \boldsymbol{\beta \,}\right ] - \frac{q} {r^{2}}\left [\boldsymbol{\nabla }_{\text{r}}r\times \boldsymbol{\beta \,}\right ]\,, }$$
(25.30)

where again all parameters on the r.h.s. must be evaluated at the retarded time. The evaluation of the “retarded” curl operation \(\boldsymbol{\nabla }_{\text{r}}\times \boldsymbol{\beta }\) becomes obvious if we evaluate one component only, for example, the x component

$$\displaystyle{ \left ( \frac{\partial } {\partial y} + \frac{\partial t_{\text{r}}} {\partial y} \frac{\partial } {\partial t_{\text{r}}}\right )\beta _{z} -\left ( \frac{\partial } {\partial z} + \frac{\partial t_{\text{r}}} {\partial z} \frac{\partial } {\partial t_{\text{r}}}\right )\beta _{y} = \left [\boldsymbol{\nabla }\times \boldsymbol{\beta \,}\right ]_{x} + \left [\boldsymbol{\nabla }t_{\text{r}} \times \frac{\text{d}\boldsymbol{\beta \,}} {\text{d}t_{\text{r}}}\right ]_{x}\,. }$$
(25.31)

In a similar way, we get the other components and find with (25.22) and the fact that the particle velocity \(\boldsymbol{\beta \,}\boldsymbol{\,}\) does not depend on the coordinates of the observation point \(\left (\boldsymbol{\nabla }\times \boldsymbol{\beta \,}\boldsymbol{\,} = 0\right )\),

$$\displaystyle{ \left [\boldsymbol{\nabla }_{\text{r}}r\times \boldsymbol{\beta \,}\right ] = \left [\boldsymbol{\nabla }\times \boldsymbol{\beta \,}\right ] + \left [\boldsymbol{\nabla }t_{\text{r}} \times \frac{\text{d}\boldsymbol{\beta \,}} {\text{d}t_{\text{r}}}\right ] = \frac{1} {cr}\left [\boldsymbol{R} \times \frac{\text{d}\boldsymbol{\beta \,}} {\text{d}t_{\text{r}}}\right ]\,, }$$

The gradient \(\boldsymbol{\nabla }_{\text{r}}r\) has been derived earlier in (25.18) and inserting this into (25.30) we find the magnetic field of an electrical charge moving with velocity v

$$\displaystyle\begin{array}{rcl} 4\pi c\epsilon _{0}\frac{\boldsymbol{B}} {q} & =& -\frac{1} {r^{2}}\left (\boldsymbol{\beta \,}\times \,\boldsymbol{n}\right ) - \frac{R} {cr^{2}}\left.\left [ \frac{\text{d}\boldsymbol{\beta \,}} {\text{d}t}\boldsymbol{ \times \,}\boldsymbol{ n}\right ]\right \vert _{\text{r}} \\ & +& \frac{R} {r^{3}}\left.\left (\boldsymbol{\beta \,}\boldsymbol{n}+\beta ^{2} + \frac{1} {c} \frac{\text{d}\boldsymbol{\beta \,}} {\text{d}t}\boldsymbol{R}\right )\left [\boldsymbol{\beta \,}\times \boldsymbol{n}\right ]\right \vert _{\text{r}}.{}\end{array}$$
(25.32)

Again, there are two distinct groups of field terms. In case of the electrical field the terms that fall off like the square of the distance are the Coulomb fields. For magnetic fields such terms appear only if the charge is moving \(\beta \boldsymbol{\neq }0\) and are identical to the Biot–Savart fields. Here we concentrate only on the far fields or radiation fields which decay inversely proportional to the distance from the source. The magnetic radiation field is then given by

$$\displaystyle{ 4\pi c\epsilon _{0}\left.\frac{\boldsymbol{B}\mathbf{(}t\mathbf{)}} {q} \right \vert _{\text{rad}} = - \frac{R} {cr^{2}}\left [ \frac{\text{d}\boldsymbol{\beta \,}} {\text{d}t} \times \boldsymbol{ n}\right ]_{\text{r}} + \frac{R} {cr^{3}}\left ( \frac{\text{d}\boldsymbol{\beta \,}} {\text{d}t}\boldsymbol{R}\right )\left [\boldsymbol{\beta \,}\times \boldsymbol{n}\right ]_{\text{r}} }$$
(25.33)

Comparing the magnetic field (25.33) with the electrical field (25.28) reveals a very simple correlation between both fields. The magnetic field can be obtained from the electric field, and vice versa, by mere vector multiplication with the unit vector \(\boldsymbol{n}\)

$$\displaystyle{ \boldsymbol{B}\mathbf{=}\tfrac{1} {c}\left [\boldsymbol{E} \times \boldsymbol{ n}\right ]_{\text{r}}\,. }$$
(25.34)

From this equation we can deduce special properties for the field directions by noting that the electric and magnetic fields are orthogonal to each other and both are orthogonal to the direction of observation \(\boldsymbol{n}\). The existence of electric and magnetic fields can give rise to radiation for which the Poynting vector is

$$\displaystyle{ \boldsymbol{S} = \frac{1} {c\mu _{0}}\left [\boldsymbol{E} \times \boldsymbol{ B}\right ]_{\text{r}} =\epsilon _{0}c\left [\boldsymbol{E} \times \mathbf{(}\boldsymbol{E} \times \boldsymbol{ n}\mathbf{)}\right ]_{\text{r}}\,. }$$
(25.35)

Using again the vector relation (A.10) and noting that the electric field is normal to \(\mathit{\,}\boldsymbol{n}\), we get for the Poynting vector or the radiation flux in the direction to the observer

$$\displaystyle{ \boldsymbol{S} = \mathbf{-}\epsilon _{0}c\left.\boldsymbol{E}_{\text{r}}^{2}\mathit{\,}\boldsymbol{n}\right \vert _{\text{r}}\mathit{\,.} }$$
(25.36)

Equation (25.36) defines the energy flux density measured at the observation point P and time t in form of synchrotron radiation per unit cross section and parallel to the direction of observation\(\mathit{\,}\boldsymbol{n}\). All quantities expressing this energy flux are still to be taken at the retarded time. For practical reasons it becomes desirable to express the Poynting vector at the retarded time as well. The energy flux at the observation point in terms of the retarded time is then dW∕d\(t_{\text{r}} = \left (\text{d}W/\text{d}t\right )\left (\text{d}t/\text{d}t_{\text{r}}\right )\) and instead of (25.36) we express the Poynting vector with (25.16) like

$$\displaystyle{ \boldsymbol{S}_{\text{r}} =\boldsymbol{ S} \frac{\text{d}t} {\text{d}t_{\text{r}}} = -\epsilon _{0}c\boldsymbol{E}^{2}\left [\left (1 +\boldsymbol{\beta \, n}\right )\boldsymbol{n}\mathit{\,}\right ]_{\text{r}}\,. }$$
(25.37)

The Poynting vector in this form can be readily used for calculations like those determining the spatial distribution of the radiation power.

2 Total Radiation Power and Energy Loss

So far, no particular choice of the reference system has been assumed, but a particularly simple reference frame \(\mathcal{L}^{{\ast}}\) is the one which moves uniformly with the charge before acceleration. From now on, we use a single particle with a charge e. To an observer in this reference system, the charge moves due to acceleration and the electric field in the radiation regime is from (25.29)

$$\displaystyle{ \mathit{\,}\boldsymbol{E}^{{\ast}}(t) = \frac{1} {4\pi \epsilon _{0}}\left. \frac{e} {cR}\left [\mathit{\,}\boldsymbol{n} \times \,\left (n \times \frac{\text{d}\boldsymbol{\beta \,}^{{\ast}}} {\text{d}t}\right )\mathit{\,}\right ]\right \vert _{\text{r}}\,. }$$
(25.38)

The synchrotron radiation power per unit solid angle and at distance R from the source is from (25.37) with \(\boldsymbol{v} = 0\)

$$\displaystyle{ \frac{\text{d}P^{{\ast}}} {\text{d}\varOmega } = -\boldsymbol{nS}^{{\ast}}R_{\text{r }}^{2} =\epsilon _{ 0}c\left.\boldsymbol{E}^{{\ast}2}R^{2}\right \vert _{ \text{r}}\,. }$$
(25.39)

Introducing the classical particle radius by \(e^{2} = 4\pi \epsilon _{0}r_{\text{c}}mc^{2}\) we obtain expressions which are independent of electromagnetic units and with (25.38)

$$\displaystyle{ \frac{\text{d}P^{{\ast}}} {\text{d}\varOmega } = \frac{r_{\text{c}}mc^{2}} {4\pi c} \left \vert \mathit{\,}\boldsymbol{n} \times \left (\mathit{\,}\boldsymbol{n} \times \frac{\text{d}\boldsymbol{\beta }\,^{{\ast}}} {\text{d}t}\mathit{\,}\right )\right \vert _{\text{r}}^{2} = \frac{r_{\text{c}}mc^{2}} {4\pi c} \left.\frac{\text{d}\boldsymbol{\beta \,}^{{\ast}}} {\text{d}t}\right \vert _{\text{r}}^{2}\sin ^{2}\vartheta _{ \text{r}}\,, }$$
(25.40)

where \(\vartheta _{\text{r}}\) is the retarded angle between the direction of acceleration and the direction of observation \(\boldsymbol{n}\). Integration over all solid angles gives the total radiated power. With d\(\varOmega =\sin \vartheta _{\text{r}}\) d\(\vartheta _{\text{r}}\) dϕ, where ϕ is the azimuthal angle with respect to the direction of acceleration, the total radiation power is in agreement with (24.24)

$$\displaystyle{ P^{{\ast}} = \frac{2} {3}r_{\text{c}}mc\left \vert \frac{\text{d}\boldsymbol{\beta }^{{\ast}}} {\text{d}t}\right \vert _{\text{r}}^{2}. }$$
(25.41)

This equation has been derived first by Larmor [1] within the realm of classical electrodynamics. The emission of a quantized photon, however, exerts a recoil on the electron varying its energy slightly. Schwinger [2] investigated this effect and derived a correction to the radiation power like

$$\displaystyle{ P^{{\ast}} = P_{ \text{classical}}^{{\ast}}\left (1 - \tfrac{55} {16\sqrt{3}} \frac{\epsilon _{\text{c}}} {E}\right )\,, }$$
(25.42)

where εc is the critical photon energy and E the electron energy. The correction is generally very small and we ignore therefore this quantum mechanical effect in our discussions.

Equation (25.41) must be transformed from the particle system to the laboratory frame of reference. This has been done already in Sect. 24.2

2.1 Transition Radiation

Digressing slightly from the discussion of synchrotron radiation we turn our attention to the solution of (25.39). Generally, we do not know the fields E and to solve (25.40) we need to know more about the particular trajectory of the particle motion. In the case of transition radiation, we have, however, all information to formulate a solution. Transition radiation is emitted when a charged particle passes through the boundary of two media with different dielectric constant. We will not go into the detailed general theory of transition radiation but concentrate on the case where a charged particle passes through a thin metallic foil in vacuum. As the particle passes through the foil backward transition radiation is emitted when the particle enters the foil and forward radiation is emitted when it appears on the other side. The emitted radiation energy can be derived directly from (25.39). First, we replace the electric radiation field by the magnetic field component and (25.39) becomes simply

$$\displaystyle{ \frac{\text{d}\varepsilon (t)} {\text{d}t} =\epsilon _{0}c\left.\boldsymbol{B}^{{\ast}2}(t)\,R^{2}\right \vert _{ \text{r}}\text{d}\varOmega. }$$
(25.43)

From Parseval’s theorem (A.42) we know that 

$$\displaystyle{ \int _{-\infty }^{\infty }B^{2}(t)\,\text{d}t = \frac{1} {2\pi }\int _{-\infty }^{\infty }B^{2}(\omega )\text{d}\omega. }$$
(25.44)

The emission of transition radiation occurs in a very short time \(\tau \approx \omega _{\text{p}}^{-1}\), where ωp is the plasma frequency. For this reason, the transition radiation frequency reaches into the x-ray regime. We limit ourselves here to frequencies ω, which are much lower such that τ ≪ ω−1. The magnetic field is nonzero only during the emission process and we can therefore set

$$\displaystyle{ B(\omega ) =\int _{ -\infty }^{\infty }B(t)\,e^{\text{i}\omega t}\,\text{d}t \approx \int _{ -\tau /2}^{\tau /2}B(t)\,\text{d}t\,. }$$
(25.45)

To solve this integral we recall the definition of the vector potential \(\boldsymbol{B}(t) = \nabla \times \boldsymbol{ A}_{\text{r}}\) and keep in mind that all quantities are to be taken at the retarded time. Expressing in component form\(\nabla \times \boldsymbol{ A}_{\text{r}} = \left \{\frac{\partial A_{z}} {\partial y} -\frac{\partial A_{y}} {\partial z},\, \frac{\partial A_{x}} {\partial z} -\frac{\partial A_{z}} {\partial x},\, \frac{\partial A_{x}} {\partial z} -\frac{\partial A_{z}} {\partial x} \right \}_{t_{\text{r}=t-\frac{1} {c}R(t)}}\) the derivatives are \(\frac{\partial A_{z}} {\partial y} = \frac{\partial A_{z}} {\partial t_{\text{r}}} \frac{\partial t_{\text{r}}} {\partial y} \,\) etc. With \(\frac{\partial t_{\text{r}}} {\partial y} = \frac{1} {c} \frac{y_{\text{r}}-y} {R} = \frac{n_{y}} {c}\) we get \(\frac{\partial A_{z}} {\partial y} -\frac{\partial A_{y}} {\partial z} = \frac{1} {c} \frac{\partial A_{z}} {\partial t_{\text{r}}} n_{y} -\frac{1} {c} \frac{\partial A_{y}} {\partial t_{\text{r}}} n_{z}\) or finally

$$\displaystyle{ \boldsymbol{B}(t) = \nabla \times \boldsymbol{ A}_{\text{r}} = \frac{1} {c}\boldsymbol{n}_{\text{r}} \times \frac{\partial } {\partial t_{\text{r}}}\boldsymbol{A}_{\text{r}} = \frac{1} {c} \frac{\partial } {\partial t_{\text{r}}}\left [\boldsymbol{n} \times \boldsymbol{ A}\right ]_{\text{r}}. }$$
(25.46)

The magnetic field spectrum (25.45) becomes then simply

$$\displaystyle{ B(\omega ) =\int _{ -\tau /2}^{\tau /2}B(t)\,\text{d}t = \frac{1} {c}\left.\left [\boldsymbol{n} \times \boldsymbol{ A}\right ]_{\text{r}}\right \vert _{\text{initial}}^{\text{final}}. }$$
(25.47)

Initially, while the electron has not yet vanished into the metallic foil, the vector potential is made up of the Liènard–Wiechert potentials of a free electron and its image charge (a positron) moving in the opposite direction. The vector potential is therefore

$$\displaystyle{ \boldsymbol{A} =\mathop{\underbrace{\mathop{ 4\pi c\epsilon _{0} \frac{e\boldsymbol{\beta \,}} {R(1+\boldsymbol{\beta \,n})}}}\limits }\limits_{ \text{electron}} +\mathop{\underbrace{\mathop{ 4\pi c\epsilon _{0} \frac{e\boldsymbol{\beta \,}} {R(1-\boldsymbol{\beta \,n})}}}\limits.}\limits_{\text{positron}} }$$
(25.48)

Instead of (25.43) we use the spectral radiation energy d\(\varepsilon (\omega ) = \frac{1} {\mu _{0}} R^{2}\) d\(\varOmega \,\frac{1} {2\pi }\boldsymbol{B}_{\text{r}}^{{\ast}2}(t)\) dω 2, where the extra factor of two comes from using only positive frequencies ω > 0, and get with (25.48) and \(e^{2} = r_{\text{c}}mc^{2}4\pi \epsilon _{0}\)

$$\displaystyle\begin{array}{rcl} \frac{\text{d}^{2}\varepsilon } {\text{d}\omega \text{d}\varOmega }& =& \frac{1} {4\pi ^{2}} \frac{r_{\text{c}}mc^{2}} {c} \left \{ \frac{\boldsymbol{n}\times \boldsymbol{\beta \,}} {1 +\boldsymbol{\beta \, n}} + \frac{\boldsymbol{n}\times \boldsymbol{\beta \,}} {1 -\boldsymbol{\beta \, n}}\right \}^{2} {}\\ & =& \frac{r_{\text{c}}mc^{2}} {\pi ^{2}c} \left \vert \boldsymbol{n} \times \boldsymbol{ z}\right \vert ^{2}\left ( \frac{\beta } {1 -\beta ^{2}\left (\boldsymbol{nz}\right )^{2}}\right )^{2}, {}\\ \end{array}$$

where we used \(\boldsymbol{\beta \,}\approx \beta \boldsymbol{ z}\) and where \(\boldsymbol{z}\) is the unit vector along the z-axis. The emission angle \(\vartheta\) is taken with respect to the z-axis. The spectral and spatial transition radiation distribution from a single electron is finally with \(\boldsymbol{nz =}\cos \vartheta\) and \(\boldsymbol{n} \times \boldsymbol{ z =}\sin \vartheta\)

$$\displaystyle{ \frac{\text{d}^{2}\varepsilon } {\text{d}\omega \,\text{d}\varOmega } = \frac{r_{\text{c}}mc^{2}} {\pi ^{2}c} \frac{\beta ^{2}\sin ^{2}\vartheta } {\left (1 -\beta ^{2}\cos ^{2}\vartheta \right )^{2}}. }$$
(25.49)

The spatial radiation distribution of transition radiation is shown in Fig. 25.2. No radiation is emitted along the axis \(\vartheta = 0\) while the radiation intensity reaches a maximum at an emission angle of 1∕γ. Equation (25.49) does not exhibit any frequency dependence, which is due to the fact that the emission process occurs in a very short time generating a uniform spectrum. Very high frequencies in the x-ray regime, where the spectral intensity is expected to drop, have been excluded in this derivation.

Fig. 25.2
figure 2figure 2

Intensity distribution \(\frac{\text{d}^{2}\varepsilon } {\text{d}\omega \text{d}\varOmega } \frac{\pi ^{2}c} {r_{\text{c}}mc^{2}}\) of transition radiation

Integrating (25.49) over a half space, we get

$$\displaystyle\begin{array}{rcl} \frac{\text{d}\varepsilon } {\text{d}\omega }& =& \frac{2r_{\text{c}}mc^{2}} {\pi c} \int _{0}^{\pi /2} \frac{\beta ^{2}\sin ^{2}\vartheta } {\left (1 -\beta ^{2}\cos ^{2}\vartheta \right )^{2}}\sin \vartheta \,\text{d}\vartheta \\ & =& \frac{2r_{\text{c}}mc^{2}} {\pi c} \frac{1} {4\beta }\left [\left (1 +\beta ^{2}\right )\ln \frac{1+\beta } {1-\beta }- 2\beta \right ],{}\end{array}$$
(25.50)

which is for relativistic particles \(\gamma \gg 1\)

$$\displaystyle{ \frac{\text{d}\varepsilon (\omega )} {\text{d}\omega } \approx \frac{2r_{\text{c}}mc^{2}} {\pi c} \ln \gamma \,. }$$
(25.51)

The spectral energy emitted into one half space by a single electron in form of transition radiation is uniform for all frequencies reaching up into the x-ray regime and depends only logarithmically on the particle energy.

3 Spatial Radiation Distribution

Coming back to synchrotron radiation we must define the electron motion in great detail. It is this motion which determines many of the photon beam characteristics. The radiation power and spatial distribution of synchrotron radiation in the electron frame of reference is identical to that from a linear microwave antenna being emitted normal to the direction of acceleration with a sin2-distribution.

Expressions for the radiation fields and Poynting vector exhibit strong vectorial dependencies on the directions of motion and acceleration of the charged particles and on the direction of observation. These vectorial dependencies indicate that the radiation may not be emitted isotropic but rather into specific directions forming characteristic radiation patterns. Similarly, we note a strong dependence on the photon frequency. In the following paragraphs, we will investigate theses dependencies closer.

3.1 Radiation Lobes

In this section we will derive these spatial radiation characteristics and determine the direction of preferred radiation emission.

In (25.40) the radiation power per unit solid angle is expressed in the reference frame of the particle

$$\displaystyle{ \frac{\text{d}P} {\text{d}\varOmega } = \frac{r_{\text{c}}mc} {4\pi } \dot{\beta }_{\text{r}}^{{\ast}2}\sin ^{2}\varTheta }$$
(25.52)

showing a particular directionality of the radiation as shown in Fig. 25.3. The radiation power is mainly concentrated in the x, y-plane and is proportional to sin2Θ where Θ is the angle between the direction of acceleration, in this case the z-axis, and the direction of observation \(\boldsymbol{n}\mathbf{.}\) The radiation pattern in Fig. 25.3 is formed by the end points of vectors with the length dP∕dΩ and angles Θ with respect to the z-axis. Because of symmetry, the radiation is isotropic with respect to the polar angle \(\varphi\) and therefore the radiation pattern is rotation symmetric about the direction of acceleration or in this case about the z-axis.

Fig. 25.3
figure 3figure 3

Radiation pattern in the particle frame of reference or for nonrelativistic particles in the laboratory system

This pattern is the correct representation of the radiation for the reference frame of the radiating particle. We may, however, also consider this pattern as the radiation pattern from non relativistic particles like that from a linear radio antenna. For relativistic particles the radiation pattern differs significantly from the non relativistic case. The Poynting vector in the form of (25.37) can be used to calculate the radiation power per unit solid angle in the direction to the observer \(\mathbf{-}\boldsymbol{n}\)

$$\displaystyle{ \frac{\text{d}P} {\text{d}\varOmega } = -\left.\boldsymbol{nS}\,R^{2}\right \vert _{ \text{r }} =\epsilon _{0}c\left.\boldsymbol{E}^{2}\left (1 +\boldsymbol{\beta \, n}\right )R^{2}\right \vert _{ \text{r}}\,. }$$
(25.53)

We calculate the spatial distribution of the synchrotron radiation for the case of acceleration orthogonal to the propagation of the particle as it happens in beam transport systems where the particles are deflected by a transverse magnetic fields. The particle is assumed to be located at the origin of a right-handed coordinate system as shown in Fig. 25.4 propagating in the z-direction and the orthogonal acceleration in this coordinate system occurs along the x-axis.

Fig. 25.4
figure 4figure 4

Radiation geometry in the laboratory frame of reference for highly relativistic particles

With the expression (25.29) for the electric fields in the radiation regime the spatial radiation power distribution (25.53) becomes

$$\displaystyle{ \frac{\text{d}P} {\text{d}\varOmega } = \frac{c} {4\pi }r_{\text{c}}mc^{2} \frac{R^{5}} {c^{3}r^{5}}\left \{\boldsymbol{n} \times \left [(\boldsymbol{n} +\boldsymbol{\beta } \boldsymbol{\,\,})\times \boldsymbol{\dot{\beta }\,}\right ]\right \}^{2}\,. }$$
(25.54)

We will now replace all vectors by their components to obtain the directional dependency of the synchrotron radiation. The vector \(\boldsymbol{n}\) pointing from the observation point to the source point of the radiation has from Fig. 25.4 the components

$$\displaystyle{ \boldsymbol{n} = \mathbf{(-}\sin \theta \,\cos \varphi,\mathbf{-}\sin \theta \,\sin \varphi,\cos \theta )\,\,, }$$
(25.55)

where the angle θ is the angle between the direction of particle propagation and the direction of emission of the synchrotron light \(-\boldsymbol{n}\). The x-component of the acceleration can be derived from the Lorentz equation

$$\displaystyle{ \gamma m\,\boldsymbol{\dot{v}}_{x} = \frac{\text{d}p_{x}} {\text{d}t} = c\,e\beta _{z}B_{y}\,. }$$
(25.56)

With vz ≈ v we have \(1/\rho = c\,eB_{y}/cp = ceB_{y}/(\gamma mcv)\) and the acceleration vector is

$$\displaystyle{ \boldsymbol{\dot{v}}_{\perp } = (\dot{v},0,0) = \left (\frac{v^{2}} {\rho },0,0\right )\,. }$$
(25.57)

The velocity vector is

$$\displaystyle{ \boldsymbol{v} = (0,0,v) }$$
(25.58)

and after replacing the double vector product in (25.54) by a single vector sum

$$\displaystyle{ \boldsymbol{n} \times \left [(\boldsymbol{n} +\boldsymbol{\beta \,} \mathbf{)}\times \boldsymbol{\beta \,}\right ] = (\boldsymbol{n} +\boldsymbol{\beta } \boldsymbol{\,})\left (\boldsymbol{n}\,\boldsymbol{\beta \,}\right )-\boldsymbol{\beta \,}\mathbf{(1} + \boldsymbol{n}\,\boldsymbol{\beta }\boldsymbol{\,})\,, }$$
(25.59)

we may now square the r.h.s. of (25.54) and replace all vectors by their components. The denominator in (25.54) then becomes

$$\displaystyle{ r^{5} = R^{5}(1 +\boldsymbol{ n\beta }\boldsymbol{\,})^{5} = R^{5}(1-\beta \cos \theta )^{5}\,, }$$
(25.60)

and the full expression for the radiation power exhibiting the spatial distribution is finally

$$\displaystyle{ \frac{\text{d}P} {\text{d}\varOmega } = \frac{r_{\text{c}}mc^{2}c} {4\pi } \frac{\beta ^{4}} {\rho ^{2}} \frac{(1-\beta \cos \theta )^{2} - (1 -\beta ^{2})\sin ^{2}\theta \cos ^{2}\varphi } {(1-\beta \cos \theta )^{5}} \,. }$$
(25.61)

This equation describes the instantaneous synchrotron radiation power per unit solid angle from charged particles moving with velocity v and being accelerated normal to the propagation by a magnetic field. Integration over all angles results again in the total synchrotron radiation power (24.34).

In Fig. 25.5 the radiation power distribution is shown in real space as derived from (25.61). We note that the radiation is highly collimated in the forward direction along the z-axis which is also the direction of particle propagation. Synchrotron radiation in particle accelerators or beam lines is emitted whenever there is a deflecting electromagnetic field and emerges mostly tangentially from the particle trajectory. An estimate of the typical opening angle can be derived from (25.61). We set \(\varphi = 0\) and expand the cosine function for small angles \(\cos \theta \approx 1 -\frac{1} {2}\,\theta ^{2}\). With \(\beta \approx 1 -\frac{1} {2}\,\gamma ^{-2}\) we find the radiation power to scale like (\(\gamma ^{-2} +\theta ^{2})^{-3}\). The radiation power therefore is reduced to about one eighth the peak intensity at an emission angle of \(\theta _{\gamma } = 1/\gamma\) or virtually all synchrotron radiation is emitted within an angle of

$$\displaystyle{ \theta _{\gamma } = \pm \frac{1} {\gamma } }$$
(25.62)

with respect to the direction of the particle propagation.

Fig. 25.5
figure 5figure 5

Spatial synchrotron radiation distribution

From Fig. 25.5 we observe a slightly faster fall off for an azimuthal angle of \(\varphi = 0\) which is in the plane of particle acceleration and propagation. Although the synchrotron radiation is emitted symmetrically within a small angle of the order of \(\pm \frac{1} {\gamma }\) with respect to the direction of particle propagation, the radiation pattern from a relativistic particle as observed in the laboratory is very different in the deflecting plane from that in the nondeflecting plane. While the particle radiates from every point along its path, the direction of this path changes in the deflecting plane but does not in the nondeflecting plane. The synchrotron radiation pattern from a bending magnet therefore resembles the form of a swath where the radiation is emitted evenly and tangentially from every point of the particle trajectory as shown in Fig. 25.6.

Fig. 25.6
figure 6figure 6

Synchrotron radiation from a circular particle accelerator

The extreme collimation of the synchrotron radiation and its high intensity in high energy electron accelerators can cause significant heating problems as well as desorption of gas molecules from the surface of the vacuum chamber. In addition, the high density of thermal energy deposition on the vacuum chamber walls can cause significant mechanical stresses causing cracks in the material. A careful design of the radiation absorbing surfaces to avoid damage to the integrity of the material is required. On the other hand, this same radiation is a valuable source of photons for a wide variety of research applications where, specifically, the collimation of the radiation together with the small source dimensions are highly desired features of the radiation.

4 Radiation Field in the Frequency Domain

Synchrotron radiation is emitted within a wide range of frequencies. As we have seen in the previous paragraph, a particle orbiting in a circular accelerator emits light flashes at the revolution frequency. We expect therefore in the radiation frequency spectrum all harmonics of the revolution frequency up to very high frequencies limited only by the very short duration of the radiation pulse being sent into a particular direction toward the observer. The number of harmonics increases with beam energy and reaches at the critical frequency the order of γ3.

The frequency spectrum of synchrotron radiation has been derived by many authors. In this text, we will stay closer to the derivation by Jackson [3] than others. The general method to derive the frequency spectrum is to transform the electric field from the time domain to the frequency domain by the use of Fourier transforms. Applying this method, we will determine the radiation characteristics of the light emitted by a single pass of a particle in a circular accelerator at the location of the observer. The electric field at the observation point has a strong time dependence and is given by (25.29) while the total radiation energy for one pass is from (25.38)

$$\displaystyle{ \frac{\text{d}W} {\text{d}\varOmega } = -\int \nolimits _{-\infty }^{\infty }\frac{\text{d}P} {\text{d}\varOmega } \,\text{d}t =\int \nolimits _{ -\infty }^{\infty }\boldsymbol{S}_{\text{r}}\boldsymbol{n}\,R^{2}\text{d}t =\epsilon _{ 0}cR^{2}\int \nolimits _{ -\infty }^{\infty }\boldsymbol{E}_{\text{r}}^{2}(t)\,\text{d}t\,. }$$
(25.63)

The transformation from the time domain to the frequency domain is performed by a Fourier transform or an expansion into Fourier harmonics. This is the point where the particular characteristics of the transverse acceleration depend on the magnetic field distribution and are, for example, different in a single bending magnet as compared to an oscillatory wiggler magnet. We use here the method of Fourier transforms to describe the electric field of a single particle passing only once through a homogeneous bending magnet. In case of a circular accelerator the particle will appear periodically with the period of the revolution time and we expect a correlation of the frequency spectrum with the revolution frequency. This is indeed the case and we will later discuss the nature of this correlation. Expressing the electrical field Er(t) by its Fourier transform, we set

$$\displaystyle{ \boldsymbol{E}_{\text{r}}(\omega )\, =\int _{ -\infty }^{\infty }\boldsymbol{E}_{\text{r}}(t)\,\text{e}^{-\text{i}\omega t}\text{d}t\,, }$$
(25.64)

where − < ω < . Applying Parseval’s theorem (A.42) the total absorbed radiation energy from a single pass of a particle is therefore

$$\displaystyle{ \frac{\text{d}W} {\text{d}\varOmega } =\epsilon _{0}\,c\frac{R^{2}} {2\pi } \int _{-\infty }^{\infty }\left \vert \boldsymbol{E}_{\text{r}}(\omega )\,\right \vert ^{2}\text{d}\omega \,. }$$
(25.65)

Evaluating the electrical field by its Fourier components, we derive an expression for the spectral distribution of the radiation energy

$$\displaystyle{ \frac{\text{d}^{2}W} {\text{d}\varOmega \text{d}\omega } =\epsilon _{0}\,\frac{c} {\pi } \left \vert \boldsymbol{E}_{\text{r}}(\omega )\,\right \vert ^{2}\,R_{\text{r}}^{2}\,, }$$
(25.66)

where we have implicitly used the fact that \(\boldsymbol{E}_{\text{r}}(\omega )\, =\boldsymbol{ E}_{\text{r}}(-\omega )\,\) since \(\boldsymbol{E}_{\text{r}}(t)\) is real. To calculate the Fourier transform, we use (25.29) and note that the electrical field is expressed in terms of quantities at the retarded time. The calculation is simplified if we express the whole integrand in (25.64) at the retarded time and get with \(t_{\text{r}} = t -\frac{1} {c}R(t_{\text{r}})\) and d\(t_{\text{r}} = \frac{R(t_{\text{r}})} {r}\) dt instead of (25.64)

$$\displaystyle{ \boldsymbol{E}_{\text{r}}(\omega ) = \frac{1} {4\pi \epsilon _{0}} \frac{e} {c}\int \nolimits _{-\infty }^{\infty }\left.\frac{\boldsymbol{R} \times \left [\left (\boldsymbol{R} +\boldsymbol{\beta \,} R\right )\times \boldsymbol{\dot{\beta }\,}\right ]} {r^{2}R} \right \vert _{\text{r}}\text{e}^{-\text{i}\omega \left (t_{\text{r}}+\frac{R_{\text{r}}} { c} \right )}\,\text{d}t_{\text{r}}\,. }$$
(25.67)

We require now that the radiation be observed at a point sufficiently far away from the source that during the time of emission the vector \(\boldsymbol{R}(t_{\text{r}})\) does not change appreciably in direction. This assumption is generally justified since the duration of the photon emission is of the order of 1∕(ωLγ), where \(\omega _{ \text{L}} = c/\rho\) is the Larmor frequency. The observer therefore should be at a distance from the source large compared to ργ. Equation (25.67) together with (25.14) may then be written like

$$\displaystyle{ \boldsymbol{E}_{\text{r}}(\omega ) = \frac{1} {4\pi \epsilon _{0}} \frac{e} {cR}\int \nolimits _{-\infty }^{\infty }\left.\frac{\boldsymbol{n} \times \left [\left (\boldsymbol{n}\mathbf{+}\boldsymbol{\beta \,}\right )\times \boldsymbol{\dot{\beta }\,}\right ]} {(1 +\boldsymbol{ n}\,\boldsymbol{\beta \,})^{2}} \right \vert _{\text{r}}\text{e}^{-\text{i}\omega \left (t_{\text{r}}+\frac{R_{\text{r}}} { c} \right )}\,\text{d}t_{\text{r}}\,. }$$
(25.68)

With

$$\displaystyle{ \frac{\boldsymbol{n} \times \left [\left (\boldsymbol{n}+\boldsymbol{\beta \,}\right )\times \boldsymbol{\dot{\beta }\,}\right ]} {(1 +\boldsymbol{ n}\,\boldsymbol{\beta \,})^{2}} = \frac{\text{d}} {\text{d}t_{\text{r}}} \frac{\boldsymbol{n} \times \left (\boldsymbol{n}\times \boldsymbol{\beta \,}\right )} {1 +\boldsymbol{ n}\,\boldsymbol{\beta \,}} \,, }$$
(25.69)

we integrate (25.68) by parts while noting that the integrals vanish at the boundaries and get

$$\displaystyle{ \boldsymbol{E}_{\text{r}}(\omega ) = \frac{1} {4\pi \epsilon _{0}} \frac{-\,\text{i}e\omega } {cR} \int \nolimits _{-\infty \text{ }}^{\infty }\left [\boldsymbol{n} \times \left (\boldsymbol{n}\times \boldsymbol{\beta \,}\right )\right ]_{\text{r}}\text{e}^{-\text{i}\omega \left (t_{\text{r}}+\frac{R_{\text{r}}} { c} \right )}\,\text{d}t_{\text{r}}\,. }$$
(25.70)

After insertion into (25.66) the spectral and spatial intensity distribution is

$$\displaystyle{ \frac{\text{d}^{2}W} {\text{d}\varOmega \,\text{d}\omega } = \frac{r_{\text{c}}mc^{2}} {4\pi c} \omega ^{2}\left \vert \int \nolimits _{ -\infty \text{ }}^{\infty }\left [\boldsymbol{n} \times \left (\boldsymbol{n}\times \boldsymbol{\beta \,}\right )\right ]\text{e}^{-\text{i}\omega \left (t_{\text{r}}+\frac{R_{\text{r}}} { c} \right )}\,\text{d}t_{\text{r}}\right \vert _{\text{r}}^{2}\,. }$$
(25.71)

The spectral and spatial radiation distribution depends on the Fourier transform of the particle trajectory which itself is a function of the magnetic field distribution. The trajectory in a uniform dipole field is different from say the step function of real lumped bending magnets or oscillating deflecting fields from wiggler magnets and the radiation characteristics may therefore be different. In this chapter, we will concentrate only on a uniform dipole field and postpone the discussion of specific radiation characteristics for insertion devices to Chap. 26

The integrand in (25.71) can be expressed in component form to simplify integration. For that we consider a fixed coordinate system (x, y, z) as shown in Fig. 25.7. The observation point is far away from the source point and we focus on the radiation that is centered about the tangent to the orbit at the source point. The observation point P and the vectors \(\boldsymbol{R}\) and \(\boldsymbol{n}\) are therefore within the (y, z)-plane and radiation is emitted at angles θ with respect to the z-axis.

Fig. 25.7
figure 7figure 7

Radiation geometry

The vector from the origin of the coordinate system P0 to the observation point P is \(\boldsymbol{r}\), the vector \(\boldsymbol{R}\) is the vector from P to the particle at Pp and \(\boldsymbol{r}_{\text{p}}\) is the vector from the origin to Pp. With this we have

$$\displaystyle{ \boldsymbol{r} = \,\boldsymbol{r}_{\text{p}} -\boldsymbol{ R}\left (t_{\text{r}}\right )\,, }$$
(25.72)

where \(\boldsymbol{r}_{\text{p}}\) and \(\boldsymbol{R}_{\text{r}}\) are taken at the retarded time. The exponent in (25.71) is then

$$\displaystyle{ \omega (t_{\text{r}} + R_{\text{r}}/c) =\omega (t_{\text{r}} +\boldsymbol{ nR}_{\text{r}}/c) = \frac{\omega } {c}\left (ct_{\text{r}} +\boldsymbol{ nr}_{\text{p}} -\boldsymbol{ nr}\right ) }$$
(25.73)

and the term \(-\frac{\omega }{c}\boldsymbol{nr}\) is independent of the time generating only a constant phase factor which is completely irrelevant for the spectral distribution and may therefore be ignored.

Following the above discussion the azimuthal angle is constant and set to \(\varphi = \frac{1} {2}\,\pi\) because we are interested only in the vertical radiation distribution. The horizontal distribution is uniform by virtue of the tangential emission along the orbit. With these assumptions, we get the vector components for the vector \(\boldsymbol{n}\) from (25.55)

$$\displaystyle{ \boldsymbol{n} = \mathbf{(}0\mathbf{,-}\sin \theta,-\cos \theta )\,. }$$
(25.74)

The vector rp is defined by Fig. 25.7 and depends on the exact variation of the deflecting magnetic field along the path of the particles. Here we assume a constant bending radius ρ and have

$$\displaystyle{ \boldsymbol{r}_{\text{p}} = \left [-\rho \cos (\omega _{\text{L}}t_{\text{r}}),0,\rho \sin (\omega _{\text{L}}t_{\text{r}})\right ]\,, }$$
(25.75)

where \(\omega _{\text{L}} =\beta c/\rho\) is the Larmor frequency. From these component representations the vector product

$$\displaystyle{ \boldsymbol{nr}_{\text{p}} = -\rho \sin (\omega _{\text{L}}t_{\text{r}})\cos \theta }$$
(25.76)

Noting that both arguments of the trigonometric functions in (25.76) are very small, we may expand the r.h.s. of (25.76) up to third order in tr and the factor \(t_{\text{r}} +\boldsymbol{ nr}_{\text{p}}/c\) in (25.73) becomes

$$\displaystyle{ ct_{\text{r}} +\boldsymbol{ nr}_{\text{p}} = ct_{\text{r}} -\rho \left [\omega _{\text{L}}t_{\text{r}} -\tfrac{1} {6}(\omega _{\text{L}}t_{\text{r}})^{3}\left (1 -\tfrac{1} {2}\theta ^{2}\right )\right ]\,. }$$
(25.77)

With \(\omega _{\text{L}} =\beta c/\rho\) we get \(t_{\text{r}}(1 -\rho \,\omega _{\text{L}}/c) = (1-\beta )\,t_{\text{r}} \approx t_{\text{r}}/(2\gamma ^{2})\). Keeping only up to third order terms in \(\omega _{\text{L}}t_{\text{r}}\) and θ we have finally for high energetic particles β ≈ 1

$$\displaystyle{ t_{\text{r}} + \frac{\boldsymbol{nr}_{\text{p}}} {c} = \tfrac{1} {2}\left (\gamma ^{-2} +\theta ^{2}\right )\,t_{\text{r}} + \tfrac{1} {6}\omega _{\text{L}}^{2}t_{\text{r}}^{3}\,. }$$
(25.78)

The triple vector product in (25.71) can be evaluated in a similar way. For the velocity vector we derive from Fig. 25.7

$$\displaystyle{ \boldsymbol{\beta }=\beta \, \left [-\text{sign}(1/\rho )\sin (\omega _{\text{L}}t_{\text{r}}),0,\cos (\omega _{\text{L}}t_{\text{r}})\right ]\,. }$$
(25.79)

Consistent with the definition of the curvature, the sign of the curvature sign(1∕ρ) is positive for a positive charge and a positive magnetic field vector By. The vector relation (A.10) and (25.74), (25.79) can be used to express the triple vector product in terms of its components

$$\displaystyle\begin{array}{rcl} \boldsymbol{n} \times \left (\boldsymbol{n}\times \boldsymbol{\beta \,}\right ) =\beta \left [\text{sign}(1/\rho )\sin (\omega _{\text{L}}t_{\text{r}}), \tfrac{1} {2}\sin 2\theta \cos (\omega _{\text{L}}t_{\text{r}}),-\sin ^{2}\theta \cos (\omega _{ \text{L}}t_{\text{r}})\right ]\,.& &{}\end{array}$$
(25.80)

Splitting this three-dimensional vector into two parts will allow us to characterize the polarization states of the radiation. To do this, we take the unit vector \(\boldsymbol{u}_{\perp }\) in the x-direction and \(\boldsymbol{u}_{\Vert }\) a unit vector normal to \(\boldsymbol{u}_{\perp }\) and normal to \(\boldsymbol{r}\). The y and z − components of (25.80) are then also the components of \(\boldsymbol{u}_{\Vert }\) and we may express the vector (25.80) by

$$\displaystyle{ \boldsymbol{n} \times \left (\boldsymbol{n}\times \boldsymbol{\beta }\right ) =\beta \, \text{sign}(1/\rho )\sin (\omega _{\text{L}}t_{\text{r}})\,\boldsymbol{u}_{\perp } +\beta \sin \theta \cos (\omega _{\text{L}}t_{\text{r}})\boldsymbol{u}_{\Vert }\,, }$$
(25.81)

Inserting (25.78) and (25.81) into the integrand (25.70) we get with β ≈ 1

$$\displaystyle\begin{array}{rcl} \boldsymbol{E}_{\text{r}}(\omega ) = -\frac{1} {4\pi \epsilon _{0}} \frac{e} {R} \frac{\omega } {c}\int _{-\infty }^{\infty }\left [\text{sign}(1/\rho )\sin (\omega _{ \text{L}}t_{\text{r}})\boldsymbol{u}_{\perp } +\sin \theta \,\cos (\omega _{\text{L}}t_{\text{r}})\boldsymbol{u}_{\Vert }\right ]\text{e}^{X}\text{d}t_{\text{r}}\,,& &{}\end{array}$$
(25.82)

where

$$\displaystyle{ X = -\text{i} \frac{\omega } {2\gamma ^{2}}\left [(1 +\gamma ^{2}\theta ^{2})\,t_{\text{r}} + \tfrac{1} {3}\gamma ^{2}\omega _{ \text{L}}^{2}t_{\text{r}}^{3}\right ]\,. }$$

Two polarization directions have been defined for the electric radiation field. One of which \(\left (\boldsymbol{u}_{\perp }\right )\) is in the plane of the particle path being perpendicular to the particle velocity and to the deflecting magnetic field. Following Sokolov and Ternov [4] we call this the σ-mode (\(\boldsymbol{u}_{\perp } =\boldsymbol{ u}_{\sigma })\). The other polarization direction in the plane containing the deflecting magnetic field and the observation point is perpendicular to \(\boldsymbol{n}\) and is called the π-mode (\(\boldsymbol{u}_{\Vert } =\boldsymbol{ u}_{\pi }\)). Since the emission angle θ is very small, we find this polarization direction to be mostly parallel to the magnetic field. Noting that most accelerators or beam lines are constructed in the horizontal plane, the polarizations are also often referred to as the horizontal polarization for the σ-mode and as the vertical polarization for the π-mode.

4.1 Spectral Distribution in Space and Polarization

As was pointed out by Jackson [3], the mathematical need to extend the integration over infinite times does not invalidate our expansion of the trigonometric functions where we assumed the argument ωLtr to be small. Although the integral (25.82) extends over all past and future times, the integrand oscillates rapidly for all but the lowest frequencies and therefore only times of the order \(ct_{\text{r}} = \pm \,\rho /\gamma\) centered about tr contribute to the integral. This is a direct consequence of the fact that the radiation is emitted in the forward direction and therefore only photons from a very small segment of the particle trajectory reach the observation point. For very low frequencies of the order of the Larmor frequency, however, we must expect considerable deviations from our results. In practical circumstances such low harmonics will, however, not propagate in the vacuum chamber [5] and the observed photon spectrum therefore is described accurately for all practical purposes.

The integral in (25.82) can be expressed by modified Bessel’s functions in the form of Airy’s integrals as has been pointed out by Schwinger [6]. Since the deflection angle ωLtr is very small, we may use linear expansions\(\,\sin (\omega _{\text{L}}t_{\text{r}}) \approx \omega _{\text{L}}t_{\text{r}}\) and \(\cos (\omega _{\text{L}}t_{\text{r}}) \approx 1\). Inserting the expression for the electric field (25.82) into (25.65) we note that cross terms of both polarizations vanish \(\boldsymbol{u}_{\perp }\boldsymbol{u}_{\Vert } = 0\;\) and the radiation intensity can therefore be expressed by two separate orthogonal polarization components. Introducing in (25.82) the substitutions [6]

$$\displaystyle\begin{array}{rcl} \omega _{\text{L}}t_{\text{r}}& = \sqrt{\frac{1} {\gamma ^{2}} +\theta ^{2}}\,x\,,&{}\end{array}$$
(25.83)
$$\displaystyle\begin{array}{rcl} \xi & = \tfrac{1} {3} \frac{\omega } {\omega _{\text{L}}} \frac{1} {\gamma ^{3}} (1 +\gamma ^{2}\theta ^{2})^{3/2} = \tfrac{1} {2} \frac{\omega } {\omega _{\text{c}}}(1 +\gamma ^{2}\theta ^{2})^{3/2}\,,&{}\end{array}$$
(25.84)

where \(\hslash \omega _{\text{c}}\;\) is the critical photon energy, the argument in the exponential factor of (25.82) becomes

$$\displaystyle{ \frac{\omega } {2\gamma ^{2}}\left [(1 +\gamma ^{2}\theta ^{2})\,t_{\text{r}} + \tfrac{1} {3}\gamma ^{2}\omega _{ \text{L}}^{2}t_{\text{r}}^{3}\right ] = \tfrac{1} {2}\xi (3x + x^{3})\,. }$$
(25.85)

With these substitutions, (25.82) can be evaluated noting that only even terms contribute to the integral. With ωLtr and θ being small quantities we get integrals of the form [7]

$$\displaystyle{ \begin{array}{l} \int _{0}^{\infty }\cos \left [\tfrac{1} {2}\xi (3x + x^{3})\right ]\,\text{d}x = \tfrac{1} {\sqrt{3}}K_{1/3}(\xi )\,, \\ \int _{0}^{\infty }\sin \left [\tfrac{1} {2}\xi (3x + x^{3})\right ]\,\text{d}x = \tfrac{1} {\sqrt{3}}K_{2/3}(\xi )\,,\end{array} \ \ }$$
(25.86)

where the functions Kν are modified Bessels’s functions of the second kind. These functions assume finite values for small arguments but vanish exponentially for large arguments as shown in Fig. 24.12 Fast converging series for these modified Bessels’s functions with fractional index have been derived by Kostroun [8]. The Fourier transform of the electrical field (25.82) finally becomes

$$\displaystyle\begin{array}{rcl} \boldsymbol{E}_{\text{r}}(\omega ) = \frac{-1} {4\pi \epsilon _{0}} \frac{\sqrt{3}e} {cR} \frac{\omega } {\omega _{\text{c}}}\gamma (1 +\gamma ^{2}\theta ^{2})\,\left [\text{sign}\left (\frac{1} {\rho } \right )K_{2/3}(\xi )\,\boldsymbol{u}_{\sigma } -\text{i} \frac{\gamma \theta K_{1/3}(\xi )} {\sqrt{1 +\gamma ^{2 } \theta ^{2}}}\boldsymbol{u}_{\pi }\right ],& &{}\end{array}$$
(25.87)

describing the spectral radiation field far from the source for particles traveling through a uniform magnetic dipole field. Later, we will modify this expression to make it suitable for particle motion in undulators or other nonuniform fields.

The spectral synchrotron radiation energy emitted by one electron per pass is proportional to the square of the electrical field (25.87) and is from (25.66)

$$\displaystyle{ \frac{\text{d}^{2}W} {\text{d}\varOmega \text{d}\omega } = \frac{3\,r_{\text{c}}mc} {4\pi ^{2}} \gamma ^{2}\left ( \frac{\omega } {\omega _{\text{c}}}\right )^{2}(1 +\gamma ^{2}\theta ^{2})^{2}\left [K_{ 2/3}^{2}(\xi )\boldsymbol{u}_{\sigma }^{2} + \frac{\gamma ^{2}\theta ^{2}K_{ 1/3}^{2}(\xi )} {1 +\gamma ^{2}\theta ^{2}} \boldsymbol{u}_{\pi }^{2}\right ]. }$$
(25.88)

The radiation spectrum has two components of orthogonal polarization, one in the plane of the particle trajectory and the other almost parallel to the deflecting magnetic field. In (25.87) both polarizations appear explicitly through the orthogonal unit vectors. Forming the square of the electrical field to get the radiation intensity, cross terms disappear because of the orthogonality of the unit vectors \(\boldsymbol{u}_{\sigma }\) and \(\boldsymbol{u}_{\pi }\). The expression for the radiation intensity therefore preserves separately the two polarization modes in the square brackets of (25.88) representing the σ-mode and π-mode of polarization, respectively.

It is interesting to study the spatial distribution for the two polarization modes in more detail. Not only are the intensities very different but the spatial distribution is different too. The spatial distribution of the σ-mode is directed mainly in the forward direction while the π-mode radiation is emitted into two lobes at finite angles and zero intensity in the forward direction θ = 0. In Fig. 25.8 the instantaneous radiation lobes are shown for both the σ- and the π-mode at the critical photon energy and being emitted tangentially from the orbit at the origin of the coordinate system.

Fig. 25.8
figure 8figure 8

Radiation lobes for σ- and π-mode polarization

4.2 Spectral and Spatial Photon Flux

The radiation intensity W from a single electron and for a single pass may not always be the most useful parameter. A more useful parameter is the spectral photon flux per unit solid angle into a frequency bin Δ ωω and for a circulating beam current I

$$\displaystyle{ \frac{\text{d}^{2}\dot{N}_{\text{ph}}\left (\omega \right )} {\text{d}\theta \,\text{d}\psi } = \frac{\text{d}^{2}W\left (\omega \right )} {\text{d}\omega \,\text{d}\varOmega } \frac{1} {\hslash } \frac{I} {e} \frac{\varDelta \omega } {\omega }. }$$
(25.89)

Here we have replaced the solid angle by its components, the vertical angle θ and the bending angle ψ. In more practical units the differential photon flux is

$$\displaystyle{ \frac{\text{d}^{2}\dot{N}_{\text{ph}}\left (\omega \right )} {\text{d}\theta \,\text{d}\psi } = C_{\varOmega }E^{2}I \frac{\varDelta \omega } {\omega }\left ( \frac{\omega } {\omega _{\text{c}}}\right )^{2}K\,_{ 2/3}^{2}(\xi )F(\xi,\theta )\,, }$$
(25.90)

where

$$\displaystyle{ C_{\varOmega } = \frac{3\alpha } {4\pi ^{2}e(mc^{2})^{2}} = 1.3273 \times 10^{16}\; \frac{\text{photons}} {\text{s mrad}^{2}\text{ GeV}^{2}\text{A}}\,, }$$
(25.91)

α the fine structure constant and

$$\displaystyle{ F(\xi,\theta ) = (1 +\gamma ^{2}\theta ^{2})^{2}\left [1 + \frac{\gamma ^{2}\theta ^{2}} {1 +\gamma ^{2}\theta ^{2}} \frac{K\,_{1/3}^{2}(\xi )\,} {K\,_{2/3}^{2}(\xi )\,}\right ]. }$$
(25.92)

For approximate numerical calculations of photon fluxes, we may use the graphic representation in Fig. 24.12 for the modified \(\left.\text{Bessel}\right.^{\text{'}}\) s function.

The spatial radiation pattern varies with the frequency of the radiation. Specifically, the angular distribution concentrates more and more in the forward direction as the radiation frequency increases. The radiation distribution in frequency and angular space is shown for both the σ- (Fig. 25.9) and the π-mode (Fig. 25.10) at the fundamental frequency. The high collimation of synchrotron radiation in the forward direction makes it a prime research tool to probe materials and its atomic and molecular properties.

Fig. 25.9
figure 9figure 9

Distribution in frequency and angular space for σ-mode radiation

Fig. 25.10
figure 10figure 10

Distribution in frequency and angular space for π-mode radiation

4.3 Harmonic Representation

Expression (25.88) can be transformed into a different formulation emphasizing the harmonic structure of the radiation spectrum. The equivalence between both formulations has been shown by Sokolov and Ternov [4] expressing the modified Bessel’s functions K1∕3 and K2∕3 by regular Bessel’s functions of high order. With \(\nu = \frac{\omega } {\omega _{ \text{L}}}\) the asymptotic formulas for \(\nu \gg 1\) are

$$\displaystyle\begin{array}{rcl} K_{1/3}(\xi )& = \frac{\sqrt{3}\pi } {\sqrt{1-\beta ^{2 } \cos ^{2}\theta }} J_{\nu }(\nu \beta \cos \theta ),&{}\end{array}$$
(25.93)
$$\displaystyle\begin{array}{rcl} K_{2/3}(\xi )& = \frac{\sqrt{3}\pi } {1-\beta ^{2}\cos ^{2}\theta } J\,_{\nu }^{{\prime}}(\nu \beta \cos \theta ),&{}\end{array}$$
(25.94)

where \(\xi = \frac{\nu } {3}\left (1 -\beta ^{2}\cos ^{2}\theta \right )^{3/2} \approx \frac{\nu } {3}\left (\gamma ^{-2} +\beta ^{2}\theta ^{2}\right )^{3/2}\) for small angles. These approximations are justified since we are only interested in very large harmonics of the revolution frequency. The harmonic number ν for the critical photon frequency, for example, is given by \(\nu _{\text{c}} =\omega _{\text{c}}/\omega _{\text{L}} = \frac{3} {2}\gamma ^{3}\) which for practical cases is generally a very large number. Inserting these approximations into (25.88) gives the formulation that has been derived first by Schott [911] in 1907 long before synchrotron radiation was discovered in an attempt to calculate the radiation intensity of atomic spectral lines

$$\displaystyle{ \frac{\text{d}^{2}P} {\text{d}\varOmega \text{d}\nu } = \frac{r_{\text{c}}mc^{3}} {2\pi \rho ^{2}} \nu ^{2}\left [J_{\nu }^{{\prime}2}(\nu \cos \theta ) +\theta ^{2}J_{\nu }^{2}(\nu \cos \theta )\right ], }$$
(25.95)

where we have introduced the radiation power \(P = W \frac{c} {2\pi \rho }\). This form still exhibits the separation of the radiation into the two polarization modes.

4.4 Spatial Radiation Power Distribution

Integrating over all frequencies we obtain the angular distribution of the synchrotron radiation. From (25.88) we note the need to perform integrals of the form \(\int _{-\infty }^{\infty }\omega ^{2}K_{\mu }^{2}(a\omega )\,\) d ω, where a ω = ξ. The solution can be found in the integral tables of Gradshteyn and Ryzhik [12] as solution number GR(6.576.4)Footnote 1

$$\displaystyle{ \int _{0}^{\infty }\omega ^{2}K_{\mu }^{2}(a\omega )\,d\omega = \frac{\pi ^{2}} {32a^{3}} \frac{1 - 4\mu ^{2}} {\cos \pi \mu } \,, }$$
(25.96)

for a > 0, and − 1. 5 < μ < 1. 5. Applying this solution to (25.88) and integrating over all frequencies, we get for the angular energy distribution of the synchrotron radiation per electron

$$\displaystyle{ \frac{\text{d}W} {\text{d}\varOmega } = \frac{7} {16} \frac{r_{\text{c}}mc^{2}} {\rho } \frac{\gamma ^{5}} {(1 +\gamma ^{2}\theta ^{2})^{5/2}}\left (1 + \frac{5} {7} \frac{\gamma ^{2}\theta ^{2}} {1 +\gamma ^{2}\theta ^{2}}\right )\,. }$$
(25.97)

This result is consistent with the angular radiation power distribution (25.61) where we found that the radiation is collimated very much in the forward direction with most of the radiation energy being emitted within an angle of ± 1∕γ. There are two contributions to the total radiation intensity, the σ-mode and the π-mode. The σ-mode has a maximum intensity in the forward direction, while the maximum intensity for the π-mode occurs at an angle of \(\theta _{\pi } = 1/(\sqrt{5/2}\ \gamma )\). The quantity dW∕dΩ is the radiation energy per unit solid angle from a single electron and a single pass and the average radiation power is therefore \(P_{\gamma } = W\,/\,T_{\text{rev}}\) or (25.97) becomes

$$\displaystyle{ \frac{\text{d}P_{\gamma }} {\text{d}\varOmega } = \frac{7\,r_{\text{c}}mc^{3}} {32\,\pi \rho ^{2}} \frac{\gamma ^{5}} {(1 +\gamma ^{2}\theta ^{2})^{5/2}}\left (1 + \frac{5} {7} \frac{\gamma ^{2}\theta ^{2}} {1 +\gamma ^{2}\theta ^{2}}\right ). }$$
(25.98)

Integrating (25.98) over all angles θ, we find the synchrotron radiation power into both polarization modes. In doing so, we note first that (25.98) can be simplified with (24.34) and β = 1

$$\displaystyle{ \frac{\text{d}P_{\gamma }} {\text{d}\varOmega } = \frac{21} {32} \frac{P_{\gamma }} {2\pi } \frac{\gamma } {(1 +\gamma ^{2}\theta ^{2})^{5/2}}\left (1 + \frac{5} {7} \frac{\gamma ^{2}\theta ^{2}} {1 +\gamma ^{2}\theta ^{2}}\right ). }$$
(25.99)

This result is consistent with (25.61) although it should be noted that (25.99) gives the average radiation power from a circular accelerator with uniform intensity in ψ, while (25.61) is the instantaneous power into the forward lobe. Equation (25.99) exhibits the power into each polarization mode for which the total power can be obtained by integration over all angles. First, we integrate over all points along the circular orbit and get a factor 2π since the observed radiation power does not depend on the location along the orbit. Continuing the integration over all angles of θ, we find the contributions to the integral to become quickly negligible for angles larger than 1∕γ. If it were not so, we could not have used (25.99) where the trigonometric functions have been replaced by their small arguments. Both terms in (25.99) can be integrated readily and the first term becomes with GR(2.271.6) [12]

$$\displaystyle{ \int _{\theta _{\text{max}}\gamma \ll 1}^{\theta _{\text{max}}\gamma \gg 1} \frac{\gamma \text{d}\theta } {(1 +\gamma ^{2}\theta ^{2})^{5/2}} = \frac{4} {3}. }$$
(25.100)

The second term is with GR[2.272.7] [12]

$$\displaystyle{ \int _{\theta _{\text{max}}\gamma \ll 1}^{\theta _{\text{max}}\gamma \gg 1} \frac{\gamma ^{3}\theta ^{2}\text{d}\theta } {(1 +\gamma ^{2}\theta ^{2})^{7/2}} = \frac{4} {15}. }$$
(25.101)

With these integrals and (25.99) the radiation power into the σ- and π-mode with Pγ from (24.34) is

$$\displaystyle{ \begin{array}{l} P_{\sigma } = \tfrac{7} {8}P_{\gamma }, \\ P_{\pi } = \tfrac{1} {8}P_{\gamma }. \end{array} }$$
(25.102)

The horizontally polarized component of synchrotron radiation greatly dominates the photon beam characteristics and only 12. 5 % of the total intensity is polarized in the vertical plane. In the forward direction the σ-polarization even approaches 100 %. Obviously, the sum of both components is equal to the total radiation power. This high polarization of the radiation provides a valuable characteristic for experimentation with synchrotron radiation. In addition, the emission of polarized light generates a slow polarizing reaction on the particle beam orbiting in a circular accelerator like in a storage ring [13].

5 Asymptotic Solutions

Expressions for the radiation distribution can be greatly simplified if we restrict the discussion to very small or very large arguments of the modified Bessel’s functions for which approximate expressions exist [14]. Knowledge of the radiation distribution at very low photon frequencies becomes important for experiments using such radiation or for beam diagnostics where the beam cross section is being imaged to a TV camera using the visible part of the radiation spectrum. To describe this visible part of the spectrum, we may in most cases assume that the photon frequency is much lower than the critical photon frequency.

5.1 Low Frequencies and Small Observation Angles

For very small arguments or low frequencies and small angles, we find the following approximations AS(9.6.9) [14]

$$\displaystyle\begin{array}{rcl} K_{1/3}^{2}(\xi & \longrightarrow 0) \approx \frac{\varGamma ^{2}(1/3)} {2^{2/3}} \left ( \frac{\omega }{\omega _{\text{c}}}\right )^{-2/3} \frac{1} {1+\gamma ^{2}\theta ^{2}},&{}\end{array}$$
(25.103a)
$$\displaystyle\begin{array}{rcl} K_{2/3}^{2}(\xi & \longrightarrow 0) \approx 2^{2/3}\varGamma ^{2}(2/3)\left ( \frac{\omega } {\omega _{\text{c}}}\right )^{-4/3} \frac{1} {(1+\gamma ^{2}\theta ^{2})^{2}},&{}\end{array}$$
(25.103b)

where the Gamma functions \(\varGamma (1/3) = 2.6789385\) and \(\varGamma (2/3) = 1.351179\) and from (25.85)

$$\displaystyle{ \xi = \frac{1} {2}\, \frac{\omega } {\omega _{\text{c}}}(1 +\gamma ^{2}\theta ^{2})^{3/2}\,. }$$
(25.104)

Inserting this into (25.90) the photon flux spectrum in the forward direction becomes for θ = 0 and \(\frac{\omega }{\omega _{\text{c}}} \ll 1\)

$$\displaystyle{ \frac{\text{d}^{2}\dot{N}_{\text{ph}}} {\text{d}\theta \,\text{d}\psi } \approx C_{\varOmega }\,E^{2}I\,\,\varGamma ^{2}(2/3)\left (\frac{2\omega } {\omega _{\text{c}}}\right )^{2/3}\frac{\varDelta \omega } {\omega }\,. }$$
(25.105)

The photon spectrum at very low frequencies is independent of the particle energy since \(\omega _{\text{c}} \propto E^{3}\). Clearly, in this approximation there is no angular dependence for the σ-mode radiation and the intensity increases with frequency. The π-mode radiation on the other hand is zero for θ = 0 and increases in intensity with the square of θ as long as the approximation is valid.

5.2 High Frequencies or Large Observation Angles

For large arguments of the modified Bessel’s functions or for high frequencies and large emission angles different approximations hold. In this case, the approximate expressions are actually the same for both Bessel’s functions indicating the same exponential drop off for high energetic photons AS(9.7.2) [14]

$$\displaystyle\begin{array}{rcl} K_{1/3}^{2}(\xi & \longrightarrow \infty ) \approx \frac{\pi } {2} \frac{\text{e}^{-2\xi }} {\xi } \,,&{}\end{array}$$
(25.106a)
$$\displaystyle\begin{array}{rcl} K_{2/3}^{2}(\xi & \longrightarrow \infty ) \approx \frac{\pi } {2} \frac{\text{e}^{-2\xi }} {\xi } \,.&{}\end{array}$$
(25.106b)

The photon flux distribution in this approximation becomes from (25.90)

$$\displaystyle{ \frac{\text{d}^{2}N_{\text{ph}}} {\text{d}\theta \text{d}\psi } \approx \frac{3r_{\text{c}}mc^{2}} {4\pi \hslash \,c} \gamma ^{2} \frac{\omega } {\omega _{\text{c}}}\text{e}^{-2\xi }\sqrt{1 +\gamma ^{2 } \theta ^{2}}\frac{\varDelta \omega } {\omega } \frac{I} {e}\left (1 + \frac{\gamma ^{2}\theta ^{2}} {1 +\gamma ^{2}\theta ^{2}}\right )\,, }$$
(25.107)

where Nph is the number of photons emitted per pass. The spatial radiation distribution is greatly determined by the exponential factor and the relative amplitude with respect to the forward direction scales therefore like

$$\displaystyle{ \exp \left \{-\frac{\omega } {\omega _{\text{c}}}\left [\left (1 +\gamma ^{2}\theta ^{2}\right )^{3/2} - 1\right ]\right \}\,. }$$
(25.108)

We look now for the specific angle for which the intensity has fallen to 1∕e. Since \(\omega \gg \omega _{\text{c}},\) this angle must be very small γ θ ≪ 1 and we can ignore other θ-dependent factors. The exponential factor becomes equal to 1∕e for

$$\displaystyle{ \tfrac{3} {2} \frac{\omega } {\omega _{\text{c}}}\gamma ^{2}\theta _{ 1/e}^{2} \approx 1\, }$$
(25.109)

and solving for θ1∕e we get finally

$$\displaystyle{ \theta _{1/e} = \sqrt{\tfrac{2} {3}} \frac{1} {\gamma } \frac{\omega _{\text{c}}} {\omega } \qquad \text{for }\omega \gg \omega _{\text{c}}\,. }$$
(25.110)

The high energy end of the synchrotron radiation spectrum is more and more collimated into the forward direction. The angular distribution is graphically illustrated for both polarization modes in Figs. 25.9 and 25.10.

6 Angle-Integrated Spectrum

Synchrotron radiation is emitted over a wide range of frequencies and it is of great interest to know the exact frequency distribution of the radiation. Since the radiation is very much collimated in the forward direction, it is useful to integrate over all angles of emission to obtain the total spectral photon flux that might be accepted by a beam line with proper aperture. To that goal, (25.88) will be integrated with respect to the emission angles to obtain the frequency spectrum of the radiation. The emission angle θ appears in (25.88) in a rather complicated way which makes it difficult to perform the integration directly. We replace therefore the modified Bessel’s functions by Airy’s functions defined by AS(10.4.14) and AS(10.4.31) [14]

$$\displaystyle\begin{array}{rcl} \mathcal{A}\text{i}(z)& = \frac{\sqrt{z}} {\sqrt{3}\pi }K_{1/3}(\xi )\,,&{}\end{array}$$
(25.111a)
$$\displaystyle\begin{array}{rcl} \mathcal{A}\text{i}^{{\prime}}(z)& = - \frac{z} {\sqrt{3}\pi }K_{2/3}(\xi )\,.&{}\end{array}$$
(25.111b)

With the definition

$$\displaystyle{ \eta = \tfrac{3} {4} \frac{\omega } {\omega _{\text{c}}} }$$
(25.112)

we get from (25.85)

$$\displaystyle{ z = \left (\tfrac{3} {2}\xi \right )^{2/3} =\eta ^{2/3}\left (1 +\gamma ^{2}\theta ^{2}\right )\,. }$$
(25.113)

We apply this to the periodic motion of particles orbiting in a circular accelerator. In this case the spectral distribution of the radiation power can be obtained by noting that the differential radiation energy (25.88) is emitted every time the particle passes by the source point. A short pulse of radiation is sent towards the observation point at periodic time intervals equal to the revolution time. The spectral power distribution (25.88) expressed by Airy functions is

$$\displaystyle{ \frac{\text{d}^{2}P_{\gamma }} {\text{d}\omega \,\text{d}\varOmega } = \frac{9P_{\gamma }} {2\pi } \frac{\gamma } {\omega _{\text{c}}}\left [\eta ^{2/3}\left.\mathcal{A}\text{i}^{{\prime}}\right.^{2}\text{ }(z) +\eta ^{4/3}\gamma ^{2}\theta ^{2}\mathcal{A}\text{i}^{2}\text{ }(z)\right ]\,. }$$
(25.114)

To obtain the photon frequency spectrum, we integrate over all angles of emission which is accomplished by integrating along the orbit contributing a mere factor of 2π and over the angle θ. Although this latter integration is to be performed between -π∕2 and +π∕2, we choose the mathematically easier integration from − to + because the Airy functions fall off very fast for large arguments. In fact, we have seen already that most of the radiation is emitted within a very small angle of ± 1∕γ. The integrals to be solved are of the form \(\int _{0}^{\infty }\theta ^{n}\mathcal{A}\)i2\(\left [\eta ^{2/3}(1 +\gamma ^{2}\theta ^{2}\right ]\,\) dθ where n = 0 or 2. We concentrate first on the second term in (25.114) and form with (25.86) and (25.111a) the square of the Airy function

$$\displaystyle\begin{array}{rcl} \theta ^{2}\,\mathcal{A}\text{i}^{2}(z) = \frac{1} {\pi ^{2}} \int _{0}^{\infty }\theta ^{2}\cos \left [\tfrac{1} {3}x^{3} + zx\right ]\,\text{d}x\int _{ 0}^{\infty }\theta ^{2}\cos \left [\tfrac{1} {3}\,y^{3} + z\,y\right ]\,\text{d}y\,.& &{}\end{array}$$
(25.115)

We solve these integrals by making use of the trigonometric relation

$$\displaystyle{ \cos (\alpha +\tfrac{1} {2}\beta )\,\cos (\alpha -\tfrac{1} {2}\beta ) =\cos \alpha \,\cos \beta \,. }$$
(25.116)

After introducing the substitutions \(x + y = s\) and \(x - y = t\), we obtain integrals over two terms which are symmetric in s and t and therefore can be set equal to get

$$\displaystyle{ \theta ^{2}\mathcal{A}\text{i}^{2}(z) = \frac{1} {2\pi ^{2}}\int _{0}^{\infty }\int _{ 0}^{\infty }\theta ^{2}\cos \left [ \tfrac{1} {12}s^{3} + 3\,s\,t^{2} + z\,s\right ]\text{d}s\,\text{d}y\,, }$$
(25.117)

where the factor \(\frac{1} {2}\) comes from the transformation of the area element ds d\(y = \frac{\text{d}s\,} {\sqrt{2}} \frac{\text{d}t} {\sqrt{2}}\). In our problem we replace the argument z by the expression \(z =\eta ^{2/3}\left (1 +\gamma ^{2}\theta ^{2}\right )\) and integrate over the angle θ

$$\displaystyle\begin{array}{rcl} \pi ^{2}\int \limits _{ -\infty }^{\infty }\theta ^{2}\mathcal{A}\text{i}^{2}(z)\,\text{d}\theta =\iiint \limits _{ -\infty }^{\infty }\theta ^{2}\cos \left [ \tfrac{1} {12}s^{3} + 3st^{2} + s\eta ^{2/3}\left (1 +\gamma ^{2}\theta ^{2}\right )\right ]\text{d}s\,\text{d}y\,\text{d}\theta.& &{}\end{array}$$
(25.118)

The integrand is symmetric with respect to θ and the integration therefore needs to be performed only from 0 to with the result being doubled. We also note that the integration is taken over only one quadrant of the (s, t)-space. Further simplifying the integration, the number of variables in the argument of the cosine function can be reduced in the following way. We note the coefficient \(\frac{1} {4}\,t^{2} +\eta ^{2/3}\gamma ^{2}\theta ^{2}\) which is the sum of squares. Setting \(\frac{1} {2}t = r\cos \varphi\) and \(\eta ^{1/3}\gamma \theta = r\sin \varphi\) this term becomes simply r2. The area element transforms like dt d\(\theta = 2/(\eta ^{1/3}\gamma )\)r dr d\(\varphi\) and integrating over \(\varphi\) from 0 to π∕2, since we need integrate only over one quarter plane, (25.118) becomes finally

$$\displaystyle\begin{array}{rcl} \int \limits _{-\infty }^{\infty }\theta ^{2}\mathcal{A}\text{i}^{2}\,\,(z)\,\text{d}\theta = \frac{1} {2\pi \eta \gamma ^{3}}\iint \nolimits _{0}^{\infty }r^{2}\cos \left [ \tfrac{1} {12}s^{3} + s\,\eta ^{2/3} + r^{2}\right ]\,r\,\text{d}r\,\,\text{d}s\,\,.& &{}\end{array}$$
(25.119)

The integrand of (25.119) has now a form close to that of an Airy integral and we will try to complete that similarity. With \(q = (3\xi /2)^{1/3}x\) the definition of the Airy functions AS(10.4.31) [14] are consistent with (25.111)

$$\displaystyle{ \mathcal{A}\text{i}(z) = \frac{1} {\pi } \int _{0}^{\infty }\cos \left [\tfrac{1} {3}q^{3} + z\,q\right ]\text{d}q\,. }$$
(25.120)

Equation (25.119) can be modified into a similar form by setting

$$\displaystyle{ w^{3} = \tfrac{1} {4}s^{3}\qquad \text{and }\qquad s\,(\eta ^{2/3} + r^{2}\text{)} = y\,w\,. }$$
(25.121)

Solving for w we get \(w = s/2^{2/3}\) and with \(y = 2^{2/3}(\eta ^{2/3} + r^{2}\)), d\(s = 2^{2/3}\) dw and d\(y = 2^{5/3}r\,\) dr Eq. (25.119) becomes

$$\displaystyle{ \int _{-\infty }^{\infty }\theta ^{2}\mathcal{A}\text{i}^{2}(z)\,\text{d}\theta = \frac{1} {4\eta \gamma ^{3}}\int \nolimits _{y_{0}}^{\infty }\left ( \frac{y} {2^{2/3}} -\eta ^{2/3}\right )\mathcal{A}\text{i}(y)\,\text{d}y\,\,, }$$
(25.122)

where we have used the definition of Airy’s function and where the integration starts at

$$\displaystyle{ y_{0} = (2\eta )^{2/3} = \left (\tfrac{3} {2} \frac{\omega } {\omega _{\text{c}}}\right )^{2/3} }$$
(25.123)

corresponding to r = 0.

We may separate this integral into two parts and get a term \(y\mathcal{A}\) i(y) under one of the integrals. This term is by the definition of Airy’s functions AS(10.4.1) [14] equal to \(\mathcal{A}\) i′ ′. Integration of this second derivative gives

$$\displaystyle{ \int _{y_{0}}^{\infty }\mathcal{A}\text{i}^{{\prime\prime}}(y)\,\text{d}y = -\mathcal{A}\text{i}^{{\prime}}(y_{ 0}) }$$
(25.124)

and collecting all terms in (25.122) we have finally

$$\displaystyle{ \int _{-\infty }^{\infty }\theta ^{2}\mathcal{A}\text{i}^{2}(z)\,\text{d}\theta = - \frac{1} {4\eta ^{1/3}\gamma ^{3}}\left [\frac{\mathcal{A}\text{i}^{{\prime}}(y_{0})} {y_{0}} +\int _{ y_{0}}^{\infty }\mathcal{A}\text{i}(y)\,\text{d}y\right ]\,. }$$
(25.125)

The derivation of the complete spectral radiation power distribution (25.114) requires also the evaluation of the integral \(\int \mathcal{A}\) i\(^{{\prime\prime}}(z)\,\) d θ. This can be done with the help of the integral \(\int \mathcal{A}\) i′ ′(z) dθ and (25.125). We follow a similar derivation that led us just from (25.118) to (25.119) and get instead of (25.125)

$$\displaystyle{ \int _{-\infty }^{\infty }\,\,\mathcal{A}\text{i}^{2}(z)\,\text{d}\theta = - \frac{1} {2\eta ^{1/3}\gamma }\int _{y_{0}}^{\infty }\mathcal{A}\text{i}(y)\,\text{d}y\,. }$$
(25.126)

Recalling the definition of the argument \(y =\eta ^{2/3}\left (1 +\gamma ^{2}\theta ^{2}\right )\), we differentiate (25.126) twice with respect to η2∕3 to get

$$\displaystyle{ 2\int _{-\infty }^{\infty }\left [\mathcal{A}\text{i}^{{\prime\prime}}(z) + \mathcal{A}\text{i}^{{\prime}2}(z)\right ]\,\text{d}\theta = -\frac{2^{1/3}} {\eta ^{1/3}\gamma } \mathcal{A}\text{i}\,^{{\prime}}(y_{ 0})\,\,. }$$
(25.127)

Using the relation \(\mathcal{A}\) i\(^{{\prime\prime}}(z) = z\mathcal{A}\) i (z) and the results (25.124), (25.125) in (25.127) we get

$$\displaystyle{ \int _{-\infty }^{\infty }\mathcal{A}\text{i}{\prime}^{2}(z)\,\text{d}\theta = -\frac{\eta ^{1/3}} {4\gamma } \left [\frac{3\mathcal{A}\text{i}\,^{{\prime}}(y_{0})} {y_{0}} +\int _{ y_{0}}^{\infty }\mathcal{A}\text{i}(y)\,\text{d}y\right ]\,. }$$
(25.128)

At this point, all integrals have been derived that are needed to describe the spectral radiation power separately in both polarization modes and the spectral radiation power from (25.114) becomes

$$\displaystyle\begin{array}{rcl} \frac{\text{d}P_{\gamma }} {\text{d}\omega \,} & =& \frac{27P_{\gamma }\omega } {16\,\omega _{\text{c}}^{2}}\left [\left (-\frac{3\mathcal{A}\text{i}\,^{{\prime}}\,(y_{0})} {y_{0}} -\int \nolimits _{y_{0}}^{\infty }\mathcal{A}\text{i}(y)\,\text{d}y\right )\right. \\ & & \qquad \qquad \quad \qquad \qquad \left.-\left (\frac{\mathcal{A}\text{i}^{{\prime}}\,(y_{0})} {y_{0}} +\int \nolimits _{ y_{0}}^{\infty }\mathcal{A}\text{i}(y)\,\text{d}y\right )\right ].{}\end{array}$$
(25.129)

The first term describes the σ-mode of polarization and the second term the π-mode. Combining both polarization modes, we may derive a comparatively simple expression for the spectral radiation power. To this goal, we replace the Airy’s functions by modified Bessel’s functions

$$\displaystyle{ \frac{\mathcal{A}\text{i}^{{\prime}}\,(y_{0})} {y_{0}} = - \tfrac{1} {\sqrt{3}\pi }K_{2/3}(x_{0})\,, }$$
(25.130)

where from (25.111), (25.112), and (25.122) \(x_{0} =\omega /\omega _{\text{c}}\). With \(\sqrt{y}\,\) dy = dx, the recurrence formula \(2K_{2/3}^{{\prime}} = -K_{1/3} + K_{5/3}\) and (25.111) the Airy integral is

$$\displaystyle\begin{array}{rcl} \int _{y_{0}}^{\infty }\mathcal{A}\text{i}(y)\,\text{d}y& =& - \tfrac{2} {\sqrt{3}\pi }\int _{x_{0}}^{\infty }K_{ 2/3}^{{\prime}}(x)\,\,\text{d}x - \tfrac{1} {\sqrt{3}\pi }\int _{x_{0}}^{\infty }K_{ 5/3}(x)\,\text{d}x \\ & =& \tfrac{2} {\sqrt{3}\pi }K_{2/3}(x_{0}) - \tfrac{1} {\sqrt{3}\pi }\int _{x_{0}}^{\infty }K_{ 5/3}(x)\,\text{d}x\,. {}\end{array}$$
(25.131)

We use (25.130) and (25.131) in (25.129) and get the simple expression for the synchrotron radiation spectrum

$$\displaystyle{ \frac{\text{d}P_{\gamma }} {\text{d}\omega \,} = \frac{P_{\gamma }} {\omega _{\text{c}}} \,\tfrac{9\sqrt{3}} {8\pi } \frac{\omega } {\omega _{\text{c}}}\int _{x_{0}}^{\infty }K_{ 5/3}(x)\,\text{d}x = \frac{P_{\gamma }} {\omega _{\text{c}}} \,S\left ( \frac{\omega } {\omega _{\text{c}}}\right )\,, }$$
(25.132)

where we defined the universal function

$$\displaystyle{ S\left ( \frac{\omega } {\omega _{\text{c}}}\right ) = \tfrac{9\sqrt{3}} {8\pi } \frac{\omega } {\omega _{\text{c}}}\int _{\omega /\omega _{\text{c}}}^{\infty }K_{ 5/3}(x)\text{d}x\,. }$$
(25.133)

The spectral distribution depends only on the critical frequency ωc, the total radiation power and a purely mathematical function. This result has been derived originally by Ivanenko and Sokolov [15] and independently by Schwinger [6]. Specifically, it should be noted that the synchrotron radiation spectrum, if normalized to the critical frequency, does not depend on the particle energy and is represented by the universal function shown in Fig. 25.11. The energy dependence is contained in the cubic dependence of the critical frequency acting as a scaling factor for the real spectral distribution.

Fig. 25.11
figure 11figure 11

Universal function: \(S(\xi ) = \frac{9\sqrt{3}} {8\pi } \xi \int _{\xi }^{\infty }K_{ 5/3}(x)\,\) dx,  with \(\xi =\omega /\omega _{\text{c}}\)

The mathematical function is properly normalized as we can see by integrating over all frequencies.

$$\displaystyle{ \int _{0}^{\infty }\frac{\text{d}P_{\gamma }} {\text{d}\omega \,} \text{d}\omega = \tfrac{9\sqrt{3}} {8\pi } P_{\gamma }\int _{0}^{\infty }x_{ 0}\int _{x_{0}}^{\infty }K_{ 5/3}(x)\text{d}x\,\text{d}x_{0}\,. }$$
(25.134)

After integration by parts, the result can be derived from GR[6.561.16] [12]

$$\displaystyle{ \int \nolimits _{0}^{\infty }\frac{\text{d}P_{\gamma }} {\text{d}\omega \,} \text{d}\omega = \tfrac{9\sqrt{3}} {16\pi } P_{\gamma }\int \nolimits _{0}^{\infty }x_{ 0}^{2}\,K_{ 5/3}(x_{0})\text{d}x_{0} =\varGamma \left (4/3\right )\,\varGamma (2/3)\,. }$$
(25.135)

Using the triplication formula AS(6.1.19) [14] the product of the gamma functions becomes

$$\displaystyle{ \varGamma \left (4/3\right )\varGamma (2/3) = \tfrac{4} {9} \tfrac{2\pi } {\sqrt{3}}. }$$
(25.136)

With this equation the proper normalization of (25.134) is demonstrated

$$\displaystyle{ \int _{0}^{\infty }\frac{\text{d}P_{\gamma }} {\text{d}\omega \,} \text{d}\omega = P_{\gamma }. }$$
(25.137)

Of more practical use is the spectral photon flux per unit angle of deflection in the bending magnet. With the photon flux d\(\dot{N}_{\text{ph}} =\) dPℏ ω we get from (25.132)

$$\displaystyle{ \frac{\text{d}\dot{N}_{\text{ph}}} {\text{d}\psi } = \frac{P_{\gamma }} {2\pi \hslash \omega _{\text{c}}} \frac{\varDelta \omega } {\omega }S\left ( \frac{\omega } {\omega _{\text{c}}}\right ) }$$
(25.138)

and with (24.34) and (24.49)

$$\displaystyle{ \frac{\text{d}\dot{N}_{\text{ph}}} {\text{d}\psi } = \frac{4\alpha } {9}\gamma \frac{I} {e} \frac{\varDelta \omega } {\omega }S\left ( \frac{\omega } {\omega _{\text{c}}}\right ), }$$
(25.139)

where ψ is the deflection angle in the bending magnet and α the fine structure constant. In practical units, this becomes

$$\displaystyle{ \frac{\text{d}\dot{N}_{\text{ph}}} {\text{d}\psi } = C_{\psi }EI \frac{\varDelta \omega } {\omega }S\left ( \frac{\omega } {\omega _{\text{c}}}\right ) }$$
(25.140)

with

$$\displaystyle{ C_{\psi } = \frac{4\alpha } {9e\,mc^{2}} = 3.9614 \times 10^{16} \frac{\text{photons}} {\text{s mrad A GeV}}\,. }$$
(25.141)

The synchrotron radiation spectrum in Fig. 25.11 is rather uniform up to the critical frequency beyond which the intensity falls off rapidly. Equation (25.132) is not well suited for quick calculation of the radiation intensity at a particular frequency. We may, however, express (25.132) in much simpler form for very low and very large frequencies as discussed in Sect. 24.3

7 Statistical Radiation Parameters

The emission of synchrotron radiation is a classical phenomenon. For some applications it is, however, useful to express some parameters in statistical form. Knowing the spectral radiation distribution, we may follow Sands [16] and express some quantities in the photon picture. We have used such to derive expressions for the equilibrium beam size and energy spread. Equilibrium beam parameters are determined by the statistical emission of photons and its recoil on the particle motion. For this purpose, we are mainly interested in an expression for \(\varepsilon _{ \text{ph}}^{2}\) and the photon flux at energy \(\varepsilon _{\text{ph}}\). From these quantities, we may derive an expression for the average photon energy \(\left \langle \varepsilon _{ \text{ph}}^{2}\right \rangle _{ z}\) emitted along the circumference of the storage ring. With \(\varPi \left (\varepsilon _{\text{ph}}\right )\) being the probability to emit a photon with energy \(\varepsilon _{ \text{ph}}\) we have

$$\displaystyle{ \left \langle \varepsilon _{\text{ph}}^{2}\right \rangle _{ z} =\int _{ 0}^{\infty }\varepsilon _{ \text{ph}}^{2}\varPi \left (\varepsilon _{ \text{ph}}\right )\,\text{d}\varepsilon _{\text{ph}}\,. }$$
(25.142)

The probability \(\varPi \left (\varepsilon _{\text{ph}}\right )\) is defined by the ratio of the photon flux \(\dot{n}(\varepsilon _{\text{ph}})\) emitted at energy \(\varepsilon _{\text{ph}}\) to the total photon flux \(\dot{N}_{\text{ph}}\)

$$\displaystyle{ \varPi \left (\varepsilon _{\text{ph}}\right ) = \frac{\dot{n}(\varepsilon _{\text{ph}})} {\dot{N}_{\text{ph}}} \,, }$$
(25.143)

where

$$\displaystyle{ \dot{n}(\varepsilon _{\text{ph}}) = \frac{P_{\gamma }} {\varepsilon _{\text{c}}^{2}} \frac{S\left (x\right )} {x},\qquad \text{with } \qquad x = \frac{\varepsilon _{\text{ph}}} {\varepsilon _{\text{c}}}. }$$
(25.144)

The photon flux at \(\varepsilon _{\text{ph}}\) is related to the spectral photon power by \(\varepsilon _{ \text{ph}}\dot{n}(\varepsilon _{\text{ph}}) = P\left (\varepsilon _{\text{ph}}\right )\). Integrating (25.138) over all angles ψ and multiplying by \(\hslash \omega =\varepsilon _{\text{ph}}\) we get for the spectral radiation power

$$\displaystyle{ P\left (\varepsilon _{\text{ph}}\right )\,\text{d}\varepsilon _{\text{ph}} =\varepsilon _{\text{ph}} \frac{\text{d}\dot{N}} {\text{d}\varepsilon _{\text{ph}}}\text{d}\varepsilon _{\text{ph}} = \frac{P_{\gamma }} {\varepsilon _{\text{c}}} \,S\left (\frac{\varepsilon _{\text{ph}}} {\varepsilon _{\text{c}}} \right )\text{d}\varepsilon _{\text{ph}}\,. }$$
(25.145)

The total number of emitted photons per unit time is just the integral

$$\displaystyle{ \dot{N}_{\text{ph}} =\int _{ 0}^{\infty }\dot{n}(\varepsilon _{ \text{ph}})\,\text{d}\varepsilon _{\text{ph}} = \frac{P_{\gamma }} {\varepsilon _{\text{c}}} \,\int _{0}^{\infty }\frac{S\left (x\right )} {x} \text{d}x = \frac{15\sqrt{3}} {8} \frac{P_{\gamma }} {\varepsilon _{\text{c}}} \,. }$$
(25.146)

With this, the probability to emit a photon of energy \(\varepsilon _{ \text{ph}}\) is finally

$$\displaystyle{ \varPi \left (\varepsilon _{\text{ph}}\right ) = \frac{8} {15\sqrt{3}} \frac{1} {\varepsilon _{\text{c}}} \frac{S\left (x\right )} {x} \,, }$$
(25.147)

and

$$\displaystyle{ \left \langle \varepsilon _{\text{ph}}^{2}\right \rangle _{ z} = \frac{8\varepsilon _{\text{c}}^{2}} {15\sqrt{3}}\int _{0}^{\infty }x\,S(x)\,\text{d}x = \frac{11} {27}\varepsilon _{\text{c}}^{2}. }$$
(25.148)

To calculate equilibrium beam parameters in Chaps. 11.3 and 11.4, for example, we need to know the quantity \(\left \langle \dot{N}_{\mathrm{ph}}\langle \varepsilon _{ \text{ph}}^{2}\rangle \right \rangle _{ z}\) which is now from (25.146), (25.148)

$$\displaystyle{ \left \langle \dot{N}_{\mathrm{ph}}\langle \varepsilon ^{2}\rangle \right \rangle _{ z} = \frac{55} {24\sqrt{3}}\left \langle \varepsilon _{\text{c}}P_{\gamma }\right \rangle _{z}\,, }$$
(25.149)

where the average is to be taken along the orbit and around the storage ring through all magnets. Expressing the critical photon energy by (24.49) and the radiation power by (24.34) and we get finally

$$\displaystyle{ \left \langle \dot{N}_{\mathrm{ph}}\langle \varepsilon ^{2}\rangle \right \rangle _{ z} = \frac{55} {24\sqrt{3}}r_{\text{c}}cmc^{2}\hslash c\gamma ^{7}\left \langle \frac{1} {\rho ^{3}} \right \rangle _{z}\,. }$$
(25.150)

Problems

25.1 (S). Integrate the radiation power distribution (25.61) over all solid angles and prove that the total radiation power is equal to (24.34).

25.2 (S). In the ESRF (European Synchrotron Radiation Facility) synchrotron radiation source in Grenoble (France) an electron beam of 200 mA circulates at an energy of 6 GeV. The bending magnet field is 1.0 T. Derive and sketch the spectral photon flux into a band width of 1 % and an acceptance angle of 10 mrad as a function of photon energy.

25.3 (S). Derive an expression identifying the angle at which the spectral intensity has dropped to p % from the maximum intensity. Derive approximate expressions for very low or very large photon energies. Find the angle at which the total radiation intensity has dropped to 10 %. 

25.4. Derive the wave equations (25.3) and (25.4).

25.5. Derive (25.17).

25.6. Derive (25.28) from (25.27). Show that the electrical field in the radiation regime is purely orthogonal to the direction of observation. Is the field also parallel to the acceleration?

25.7. Design a synchrotron radiation source for a critical photon energy of your choice. Use a simple FODO lattice and specify the minimum beam energy, beam current, and bending radius which will produce a bending magnet photon flux of 1014 photons/s/mrad at the desired photon energy and into a band width of \(\varDelta \omega /\omega = 1\,\%\). What is the minimum and maximum photon energy for which the photon flux is at least 1011 photons/s/mrad? How big is your ring assuming a 30 % fill factor for bending magnets?