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The Science and Technology of Undulators and Wigglers$

James A. Clarke

Print publication date: 2004

Print ISBN-13: 9780198508557

Published to Oxford Scholarship Online: September 2007

DOI: 10.1093/acprof:oso/9780198508557.001.0001

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Fundamentals of Synchrotron Radiation Emission

Fundamentals of Synchrotron Radiation Emission

Chapter:
(p.7) 2 Fundamentals of Synchrotron Radiation Emission
Source:
The Science and Technology of Undulators and Wigglers
Author(s):

James A. Clarke

Publisher:
Oxford University Press
DOI:10.1093/acprof:oso/9780198508557.003.0002

Abstract and Keywords

This chapter derives the general solution for the electric and magnetic fields emitted by an accelerated relativistic charged particle using Maxwell’s equations as the starting point. This general solution is then applied to the case of a bending magnet and the photon flux and brightness levels found. Finally, it is shown that by using a special magnet called a wavelength shifter, it is possible to alter the output spectrum significantly to enhance the output at X-ray wavelengths.

Keywords:   electric field, bending magnet, Maxwell’s equations, flux, brightness, wavelength shifter, X-ray

Although this book is primarily concerned with the subject of Insertion Devices (IDs) it is important that we cover the basics first before rushing into the exact output characteristics from undulators and wigglers.

The concept involved in the production of Synchrotron Radiation (SR) is essentially straightforward. The radiation is produced when a relativistic charged particle is accelerated, usually in a magnetic field. Any charged particle will produce SR but the electron is the particle most usually associated with the phenomenon since, as will become apparent, the low rest mass of the electron means that it emits by far the most radiation for a given particle energy (˜ 1013 times more than a proton). It is only in the very latest generation of proton accelerators (such as the Large Hadron Collider at CERN) that the emission of SR from protons is of any consequence. The vast majority of synchrotron light sources around the world use electrons as their emitter of SR. The remainder rely on the positron which, although it does have some subtle accelerator physics advantages over the electron, produces exactly the same SR output. Throughout this book I have consistently referred to the electron as the emitting particle but I could equally well have used a positron instead without any loss of accuracy.

This chapter covers the very basics of SR from first principles. An arbitrary electron trajectory is studied to give us some fundamental relationships linking the emitting particle to the observer. The general electric field due to the electron at the observer is then derived and this is applied to the scenario of an electron travelling on a circular path through a uniform magnetic field. This is the extremely important case of bending magnet or dipole radiation. Finally, equations are derived for the actual number of photons arriving at the observer at particular photon wavelengths and also the exact angular distribution of these photons.

2.1 Emission and Observation

First we consider an electron moving with relativistic velocity (i.e., close to the velocity of light) on an arbitrary path (Fig. 2.1). If the electron emits a photon in the direction of an observer then by the time it arrives at the observer the electron will no longer be in the same position. So, if we want to consider the light that is arriving at the observer now we must appreciate that this was emitted by the moving electron some time ago.

(p.8)

                   Fundamentals of Synchrotron Radiation Emission

FIG. 2.1. Geometry of the electron and the observer.

The electron emits a photon at a time t′, which arrives at the observer at the later time of t. The time t′ generally goes by the peculiar name of the retarded time but perhaps a better, more intuitive, description is the emission time. The time t will be referred to as the observation time. The photon travels at the speed of light, c, towards the observer along the path R(t′) (note that vector quantities are shown in bold font) and so the observation time is related to the emission time by

(2.1)
t = t + R ( t ) c .
The relationship between the emission and observation times is important and so we will explore how one changes with respect to the other
(2.2)
d t d t = 1 + 1 c d R ( t ) d t .
And since
(2.3)
R ( t ) = x r ( t ) ,
it is clear that
(2.4)
d R ( t ) d t = d r ( t ) d t = v ( t ) ,
where υ is the velocity of the electron, as x does not vary with time.

Recall the standard result that by differentiating the dot product a(t) · a(t) with respect to time we get

a ( t ) d a ( t ) d t = a ( t ) d a ( t ) d t .
Hence (p.9)
R ( t ) d R ( t ) d t = R ( t ) d R ( t ) d t = R ( t ) υ ( t )
and so
(2.5)
d R ( t ) d t = R ( t ) R ( t ) υ ( t ) = n ( t ) υ ( t ) ,
where n(t′) is the unit vector pointing towards the observer along R(t′). Substituting this result into (2.2) gives the important result
(2.6)
d t d t = 1 n ( t ) β ( t ) ,
where β = υ/c, is the electron velocity relative to the speed of light. At this point it is also worth noting that
β = 1 1 γ 2 ,
where γ is the relativistic Lorentz factor that is the ratio of the electron's energy, E, to its rest mass energy, E 0
γ = E E 0 .
Since E 0 for an electron is 0.511 MeV and the energy of the electrons in typical synchrotron light sources is a few GeV, the factor γ is usually of the order of several thousands.

2.2 Electric Field at the Observer

A widely used approach to reducing the number of unknowns in Maxwell's equations is to introduce auxiliary variables, which are known as vector and scalar potentials [12]. From the standard vector analysis result [13] · (∇ × a) = 0 it is clear that for Maxwell's first equation, ∇ · B = 0, to hold, it is sufficient to have

(2.7)
B = × A ,
where A(t) is termed the vector potential. Similarly, by differentiating (2.7), Maxwell's second equation (Faraday's law) (p.10)
× E = B t
can be rewritten as
× E = × ( A t )
and so inspection of another vector analysis result [13], × ( U) = 0, implies that a scalar potential, Φ(t), can be introduced such that
× E = × ( A t + Φ )
and so
(2.8)
E = Φ A t .
Clearly there is some arbitrariness in the choice of A and Φ so far. By applying a third condition to this pair of equations it is possible to produce a consistent set of potentials. In particular, if the Lorentz condition is applied [12],
A + 1 c 2 Φ t = 0
then the resulting potentials are termed the Liénard–Wiechert potentials. We will now use these to derive the electric and magnetic fields received by the observer. The scalar potential in this case is given by
(2.9)
Φ ( t ) = e 4 π 0 ( 1 R ( t ) ( 1 n ( t ) β ( t ) ) ) = e 4 π 0 ( 1 R ( 1 n β ) ) t
where the t′ subscript implies that the variables are all calculated at the emission time. Also note that e is the electronic charge and є0 is the permittivity of free space. The vector potential is given by
(2.10)
A ( t ) = e 4 π c 0 ( β R ( 1 n β ) ) t .
The derivation of the electric field, E, from A and Φ is somewhat longwinded but it is such an important quantity in synchrotron radiation emission that it is worth the effort. Further background on this derivation can be found in [14, 15].

The first term in eqn (2.8) is the gradient of Φ, (p.11)

                   Fundamentals of Synchrotron Radiation Emission

FIG. 2.2. Geometry of the electron as seen by two observers.

Φ = e 4 π 0 ( 1 R R β ) t ,
which, using the standard result for the differential of a fraction, gives
(2.11)
Φ = e 4 π 0 ( ( R R β ) ( R R β ) ) t = e 4 π 0 ( ( R R β ) R 2 ( n β ) 2 ) t .
So, to find Φ we will need to calculate R(1 −n·β). Consider two observers A and B that are separated by a small amount Δx (Fig. 2.2), two photons emitted from A′ and B′ arrive at the observers at exactly the same time t. The emission times must differ by an amount Δt′. It is clear then that
v Δ t + R + Δ R = R + Δ x
and so
Δ R = Δ x c Δ t β ,
hence
(2.12)
Δ R = n Δ R = n Δ x c Δ t n β .
Since both photons arrive at the same time at both observers we have (p.12)
t = t + R c = ( t + Δ t ) + ( R + Δ R ) c
and so
Δ R = c Δ t .
Equating this result with (2.12) gives
(2.13)
Δ t = n Δ x c ( 1 n β ) .
Furthermore
Δ ( R ( 1 n β ) ) = Δ ( R R β ) = Δ R Δ R β R Δ β .
Substituting for ΔR and ΔR gives
Δ ( R R β ) = c Δ t ( Δ x c Δ t β ) β R β t Δ t .
Substituting Δt′ from (2.13) and using the standard notation of βad; = ∂β/∂t′, gives
Δ ( R R β ) = Δ x ( n ( 1 n β ) β β 2 n ( 1 n β ) + ( R β . ) n c ( 1 n β ) ) .
Since
R ( 1 n β ) = Δ ( R R β ) Δ x
in the limit as Δx → 0, we have
R ( 1 n β ) = n ( 1 n β ) β- β 2 n ( 1 n β ) + ( R β . ) n c ( 1 n β ) = c ( 1 β 2 ) n c ( 1 n β ) β+R ( n β . ) n c ( 1 n β ) .
Substituting this result into (2.11) gives
(2.14)
Φ = e 4 π 0 ( c ( 1 β 2 ) n c ( 1 n β ) β + R ( n β . ) n c R 2 ( 1 n β ) 3 ) t .

The second term in (2.8) contains the differential of the magnetic vector potential with respect to the observer time, t, (p.13)

(2.15)
A t = d t d t A t = 1 ( 1 n β ) υ A t ,

where we have used the result of (2.6). Differentiating the magnetic vector potential with respect to the emission time, t′, we get

A t = e 4 π c 0 ( R ( 1 n β ) β . β ( R ( 1 n β ) ) t R 2 ( 1 n β ) 2 ) t .
Note that the derivative in the second term of the numerator is
( R ( 1 n β ) ) t = ( R R β ) t = R t R t β-R β . = n υ + υ β-R β . = c ( n β ) + c β 2 R β , .
where we have used the results of (2.4) and (2.5). Therefore,
A t = e 4 π c 0 ( R ( 1 n β ) β . β ( c ( n β ) + c β 2 R β . ) R 2 ( 1 n β ) 2 ) t = e 4 π c 0 ( β . R ( 1 n β ) β ( c ( n β ) + c β 2 R β . ) R 2 ( 1 n β ) 2 ) t .
Putting this result into (2.15) gives
A t = e 4 π c 0 ( 1 ( 1 n β ) ( β . R ( 1 n β ) β ( c ( n β ) + c β 2 R β . ) R 2 ( 1 n β ) 2 ) ) t = e 4 π c 0 ( 1 ( 1 n β ) 3 ( β . ( 1 n β ) + β ( n β . ) R + c ( β ( n β ) ββ 2 ) R 2 ) ) t .
Finally, this result can be combined with that from (2.14) to solve (2.8), (p.14)
E ( t ) = e 4 π 0 ( c ( 1 β 2 ) n c ( 1 n β ) β + R ( n β . ) n c R 2 ( 1 n β ) 3 ) t e 4 π 0 ( 1 ( 1 n β ) 3 ( β . ( 1 n β ) + β ( n β . ) R + c ( β ( n β ) ββ 2 ) R 2 ) ) t .
Gathering terms together with denominators of order R and R 2 gives
(2.16)
E ( t ) = e 4 π ε 0 ( c ( 1 β 2 ) n c ( 1 β 2 ) β R 2 ( 1 n β ) 3 + ( n β . ) ( n β ) ( 1 n β ) β . R ( 1 n β ) 3 ) t = e 4 π ε 0 ( c ( 1 β 2 ) ( n β ) R 2 ( 1 n β ) 3 + n × ( ( n β ) × β . ) R ( 1 n β ) 3 ) t ,
where we have used the triple vector product result
A × ( B × C ) = ( A C ) B ( A B ) C .
In a similar manner, we can derive the result for the magnetic field experienced by the observer as
(2.17)
B ( t ) = n × E ( t ) c .

Now that we have derived the electric and magnetic field experienced by the observer we can make several remarks. First, the magnetic field, B, must be perpendicular to both the electric field, E, and the unit vector pointing towards the observer, n since by definition the result a × b is perpendicular to both a and b. Second, if the electron is stationary (β = βad; = 0), then the electric field is given by

E = e n 4 π 0 R 2 ,
which is simply Coulomb's law. Note also that the electric field has two terms, the first is proportional to 1/R 2 and the second to 1/R. As R increases the first term becomes less and less significant and it is very common for (2.16) to be written in the simplified form of
(2.18)
E ( t ) = e 4 π c 0 ( n × ( ( n β ) × β . ) R ( 1 n β ) 3 ) t .
It is very important to remember, however, that this equation only holds in the so-called far field when R is large. Exactly how large R needs to be for this (p.15) approximation to be valid will be covered later. It will be sufficient to state now that for most situations of interest encountered in a synchrotron light source the far field approximation is valid. A simple consequence of this far field case is that, by virtue of the cross product, the electric field is also perpendicular to n.

2.3 Fourier Transform of the Electric Field

Although we now have a means of calculating the electric field as a function of time experienced by an observer due to an electron on an arbitrary path, it is not altogether clear how to interpret this result. It would be much nicer if we could analyse the electric field to determine what frequency content, and so what electromagnetic radiation, it represented. Of course we can convert from time to frequency with a Fourier Transform and this is the next step. We will assume the far field case is valid as this simplifies things somewhat. The Fourier Transform is given by

E ( ω ) = 1 2 π E ( t ) e i ω t d t .
Obviously the integral is performed with respect to t because we are interested in the frequency content experienced by the observer. However, the integral is actually more straightforward if we transform the integration variable to t′ using (2.1) and (2.6)
E ( ω ) = 1 2 π E ( t ) e i ω ( t + R ( t ) c ) ( 1 n β ) t d t = e 4 π 2 π c ε 0 ( n × ( ( n β ) × β . ) R ( 1 n β ) 2 ) t e i ω ( t + R ( t ) c ) d t .
The integral can be simplified if the assumption is made that R does not vary with time (dn/dt =0), which is reasonable for the relativistic far field case. After integrating by parts (one part being the exponential term and the other being the cross product terms) the expression then becomes
E ( ω ) = e 4 π 2 π c ε 0 R [ [ ( n × ( n × β ) ) ( 1 n β ) e i ω ( t + R ( t ) c ) ] + i ω ( n × ( n × β ) ) e i ω ( t + R ( t ) c ) d t ] .
The first term on the right-hand side can be conveniently dropped since it should be evaluated at t′ = ±∞ and so will not therefore be observed. This gives us the final expression for the far field case of an electron moving on an arbitrary path of (p.16)
                   Fundamentals of Synchrotron Radiation Emission

FIG. 2.3. Coordinate frame of an electron on a circular orbit.

2.19
E ( ω ) = i e ω 4 π 2 π c 0 R ( n × ( n × β ) ) e i ω ( t + R ( t ) c ) d t .

Again, the magnetic field is related to the electric field by

2.20
B ( ω ) = n × E ( ω ) c .

2.4 Synchrotron Radiation in a Bending Magnet

We will use the equations that we have just derived to look at the special case of a single relativistic electron moving on a circular orbit. Passing electrons through a uniform magnetic field (commonly referred to as a bending magnet or more correctly, dipole magnet) is the most basic method of generating synchrotron radiation in particle accelerators.

The angular velocity, ω 0, of the electron is given by

ω 0 = βc ρ ,
where ρ is the bending radius of the electron. Assuming the geometry of Fig. 2.3 and with the time coordinate chosen so that t′ = 0 occurs at the origin, we can determine the position, normalized velocity, and normalized acceleration of the electron to be [14]
r ( t ) = ( ρ ( 1 cos ( ω 0 t ) ) , 0 , ρ sin ( ω 0 t ) ) β ( t ) = ( β sin ( ω 0 t ) , 0 , β cos  ( ω 0 t ) ) β . ( t ) = ( β ω 0  cos ( ω 0 t ) , 0 , β ω 0 sin ( ω 0 t ) ) .
(p.17) Again we assume the far field condition so R is large and we can consider R to be constant over the short time during which the emitted radiation can be received by the observer. The unit vector is given by
n = ( 0 , sin ψ , cos ψ )
and the vector triple product from (2.19) is
( n × ( n × β ) ) = ( n . β ) n β = β cos ψ cos ( ω 0 t ) n β = ( β sin ( ω 0 t ) , β sin ψ cos ψ cos ( ω 0 t ) , β sin 2 ψ cos ( ω 0 t ) ) .
Since we are only considering relativistic particles, with the relativistic Lorentz factor γ= ( 1-β 2 ) 1 / 2 1 , the vertical opening angle of the radiation is very small, ψ ˜ 1/γ, and only a small part of the trajectory contributes to the radiation seen by our observer (Δø = ω 0 t′ « 1). We can therefore approximate the vector product to
(2.21)
( n × ( n × β ) ) β ( ω 0 t , ψ , 0 ) .
Also, the term in the exponent of (2.19) can be expressed as follows
(2.22)
t + R ( t ) c = t + n . x c n . r c
from (2.3), and so, using small angle approximations again we get
t + R ( t ) c t + x c ρ c sin ( ω 0 t ) cos ψ t + x c ρ c ( ω 0 t ω 0 3 t 3 6 ) ( 1 ψ 2 2 ) x c + t ( 1 ρ ω 0 c + ρ ω 0 ψ 2 2 c ) + t 3 ω 0 3 ρ 6 c ( 1 ψ 2 2 )
substituting ω 0 =βc/ρ and noting that β= 1-1/γ 2 1 1 / 2 γ 2 we get
(2.23)
t + R ( t ) c x c + t ( 1 β + βψ 2 2 ) + c 2 β 3 t 3 6 ρ 2 ( 1 ψ 2 2 ) x c + t ( 1 2 γ 2 + βψ 2 2 ) + c 2 β 3 t 3 6 ρ 2 ( 1 ψ 2 2 ) x c + t 2 γ 2 ( 1 + γ 2 ψ 2 ) + c 2 t 3 6 ρ 2 .
(p.18) An alternative but equivalent approach to deriving the above equation is to apply the small angle approximation to dt/dt′ (eqn (2.6)) and then to integrate this with respect to dt′ [16].

Introducing our results from (2.21) and (2.23) into the electric field as a function of frequency (2.19) we obtain

E ( ω ) = i e ω 4 π 2 π c 0 R × β ( ω 0 t , ψ , 0 ) exp ( i ω ( x c + t 2 γ 2 ( 1 + γ 2 ψ 2 ) + c 2 t 3 6 ρ 2 ) ) d t .
We can now express the exponential part in terms of sine and cosine and again use the approximation that β ≍ 1. We can also remove the term x/c in the exponential as this is a constant (the time taken to travel between the origin and the observer) and so will just give a fixed phase shift, which is of no importance to us. We are then left with
E ( ω ) = i e ω 4 π 2 π c 0 R ( ω 0 t , ψ , 0 ) ( cos U + i sin U ) d t ,
where
U = ω ( t 2 γ 2 ( 1 + γ 2 ψ 2 ) + c 2 t 3 6 ρ 2 ) .
If we now look at the x, y, and s terms separately we will get an interesting insight into the emitted radiation. The first remark to make is that there is no electric field in the direction of s (i.e. E s(ω) = 0). In fact, we had already established this earlier when we found that the electric field was perpendicular to n in Section 2.2. We will now look at the electric field seen by the observer that is aligned in the x direction, E x(ω).
E x ( ω ) = i e ω ω 0 4 π 2 π c 0 R [ t cos U d t + i t sin U d t ] .
A function of the form f(x) = x cos g(x) is a so-called odd function, which means that f(x) = −f(−x). The significance of this is that the integral of such a function between −∞ and ∞ will always equal zero. Hence, the above equation can be reduced to
E x ( ω ) = e ω ω 0 4 π 2 π c 0 R t sin ( ω ( t 2 γ 2 ( 1 + γ 2 ψ 2 ) + c 2 t 3 6 ρ 2 ) ) d t .
This can be made a little more easy to digest by substituting the variable u = t′ (ωc 2/2ρ 2)1/3 and also by introducing the critical frequency, ω c = 3 3/2ρ. (p.19)
(2.24)
E x ( ω ) = e ω ω 0 4 π 2 π c 0 R × ( 2 ρ 2 ω c 2 ) 2 / 3 u sin ( ω 2 γ 2 ( 2 ρ 2 ω c 2 ) 1 / 3 ( 1 + γ 2 ψ 2 ) u + u 3 3 ) d u = e ω ω 0 4 π 2 π c 0 R ( 2 ρ 2 ω c 2 ) 2 / 3 × u sin ( ( 3 ω 4 ω c ) 2 / 3 ( 1 + γ 2 ψ 2 ) u + u 3 3 ) d u .

Similarly for the electric field in the vertical plane

E y ( ω ) = i e ω ψ 4 π 2 π c 0 R [ cos U d t + i sin U d t ] .
Note again that f(x) = sin x is an odd function and so the second integral in the above equation can be neglected, leaving us with
(2.25)
E y ( ω ) = i e ω ψ 4 π 2 π c 0 R cos ( ω ( t 2 γ 2 ( 1 + γ 2 ψ 2 ) + c 2 t 3 6 ρ 2 ) ) d t = i e ω ψ 4 π 2 π c 0 R ( 2 ρ 2 ω c 2 ) 1 / 3 × cos ( ( 3 ω 4 ω c ) 2 / 3 ( 1 + γ 2 ψ 2 ) u + u 3 3 ) d u .

We can present (2.24) and (2.25) in a more useful form but first we need to make a brief diversion into Airy functions and modified Bessel functions. A standard result from math texts for the Airy function, Ai(x), is [17]

A i ( x ) = 1 π 0 cos ( x t + t 3 3 ) d t
and so the differential of this with respect to x would be
d A i ( x ) d x = A i ( x ) = 1 π 0 t sin ( x t + t 3 3 ) d t .
Two more standard results express the Airy functions in terms of modified Bessel functions [17], K 1/3 and K 2/3
A i ( x ) = 1 π x 3 K 1 / 3 ( 2 x 3 / 2 3 ) A i ( x ) = 1 π x 3 K 2 / 3 ( 2 x 3 / 2 3 ) .
(p.20) We will now use these functions to express our results for E x(ω) and E y(ω) in terms of either Airy functions or modified Bessel functions. First, expressed as Airy functions we have
E x ( ω ) = e γ 2 π c 0 R ( 3 ω 4 ω c ) 1 / 3 A i ( ( 3 ω 4 ω c ) 2 / 3 ( 1 + γ 2 ψ 2 ) ) E y ( ω ) = i e ψ γ 2 2 π c 0 R ( 3 ω 4 ω c ) 2 / 3 A i ( ( 3 ω 4 ω c ) 2 / 3 ( 1 + γ 2 ψ 2 ) ) ,
and then expressed as Bessel functions this becomes
(2.26)
E x ( ω ) = 3 e γ 4 π 2 π c 0 R ( ω ω c ) ( 1 + γ 2 ψ 2 ) K 2 / 3 ( G )
(2.27)
E y ( ω ) = i 3 e ψ γ 2 4 π 2 π c 0 R ( ω ω c ) ( 1 + γ 2 ψ 2 ) 1 / 2 K 1 / 3 ( G ) ,
where
G = ( ω 2 ω c ) ( 1 + γ 2 ψ 2 ) 3 / 2 .

Some example graphs of E x(ω) and E y(ω) are given in Figs. 2.4 and 2.5. The result for E x(ω) looks very much like the classic bending magnet spectrum, as we shall see later in this section. Note that the lower frequencies become more and more important in both the planes as the vertical angle (γψ) increases. This is also a standard bending magnet behaviour where long wavelength radiation extends much further out in vertical angle than shorter wavelength radiation. It is also important to note that the vertical electric field vanishes when ψ = 0, or in other words, the radiation is polarized entirely in the horizontal plane when observed on-axis.

Whether one chooses to work with either the Airy functions or the Bessel functions is purely a matter of personal preference. In the literature, the Bessel function form appears more often though some find the Airy function easier to work with analytically. I have found the Bessel function to be more readily available in commercial software such as spreadsheets and programming libraries and so I generally prefer this form. Table 2.1 gives the value of K 1/3(u) and K 2/3(u) for a wide range of u, as well as two other useful Bessel functions that we will meet in the following section.

We will need the expressions for the electric fields in the next part where we will look at how the synchrotron radiation power emitted by the electron is distributed in angle.

(p.21)

                   Fundamentals of Synchrotron Radiation Emission

FIG. 2.4. Horizontal electric field as a function of frequency for different vertical angles.

                   Fundamentals of Synchrotron Radiation Emission

FIG. 2.5. Vertical electric field as a function of frequency for different vertical angles.

2.4.1 Angular Power Distribution

A standard result in electromagnetic theory is that the Poynting vector, S, gives the energy flow per unit area per unit time [18].

S = 1 μ 0 ( E × B ) = 0 c 2 ( E × B ) ,
(p.22)

Table 2.1 Numerical values for modified Bessel functions that are encountered in synchrotron radiation calculations

u

K 1/3(u)

K 2/3(u)

K 5/3(u)

u K 5 / 3 ( u ) d u

0.0001

36.28

498.9

6.651E+06

995.9

0.0002

28.76

314.3

2.095E+06

626.7

0.0004

22.79

198.0

6.599E+05

394.1

0.0005

21.13

170.6

4.550E+05

339.4

0.0007

18.86

136.3

2.597E+05

270.8

0.0008

18.03

124.7

2.079E+05

247.6

0.001

16.72

107.5

1.433E+05

213.1

0.002

13.19

67.69

4.514E+04

133.6

0.004

10.38

42.62

1.422E+04

83.49

0.005

9.594

36.72

9.802E+03

71.70

0.007

8.514

29.33

5.594E+03

56.93

0.008

8.116

26.82

4.478E+03

51.93

0.01

7.486

23.10

3.087E+03

44.50

0.02

5.781

14.50

972.3

27.36

0.04

4.386

9.052

306.1

16.57

0.05

3.991

7.762

211.0

14.03

0.07

3.437

6.136

120.3

10.85

0.08

3.231

5.581

96.25

9.777

0.1

2.900

4.753

66.27

8.182

0.2

1.979

2.802

20.66

4.517

0.4

1.206

1.517

6.263

2.255

0.5

0.9890

1.206

4.205

1.742

0.7

0.6965

0.8148

2.248

1.126

0.8

0.5932

0.6839

1.733

0.9280

1

0.4384

0.4945

1.098

0.6514

2

0.1165

0.1248

0.1998

0.1508

4

0.01130

0.01173

0.01521

0.01321

5

0.003729

0.003844

0.004754

0.004250

7

4.280E-04

4.376E-04

5.113E-04

4.725E-04

8

1.474E-04

1.504E-04

1.725E-04

1.611E-04

10

1.787E-05

1.816E-05

2.030E-05

1.922E-05

where μ 0 is the permeability of free space. Using our result for B found earlier (2.17) we have that
S = 0 c ( E × ( n × E ) ) = 0 c ( E 2 n ( n E ) E )
but n · E = 0 since they are orthogonal from (2.18) and so
S = 0 c E 2 n .
(p.23) If we define an area at the observer in terms of a solid angle ΔΩ then we can say that the energy radiated by the electron through this area (R 2ΔΩ) in the time Δt is
W = ( n S ) Δ t R 2 Δ Ω = 0 c E 2 R 2 Δ t Δ Ω .
The power observed will be
P = d W d t = 0 c E 2 R 2 d Ω ,
and the power observed per solid angle will be
d P d Ω = d 2 W d t d Ω = 0 c E 2 R 2.
The total energy received by the observer in a single turn of the electron is
W = 0 4 π 0 c E 2 ( t ) R 2 d t d Ω
and the energy received per unit solid angle is
d W d Ω = 0 c E 2 ( t ) R 2 d t
We now express E(t) in terms of its Fourier Transform
E ( t ) = 1 2 π E ( ω ) e i ω t d ω ,
and we get
d W d Ω = 0 c R 2 2 π E ( ω ) e i ω t E ( ω ) e i ω t E ( ω ) e i ω t d ω d ω d t .
Noting at this point that the Dirac delta function has integral form [19]
δ ( x ) = 1 2 π e i x t d t
and so
d W d Ω = 0 c R 2 E ( ω ) E ( ω ) δ ( ω ω ) d ω d ω .
Since E(t) is a real function, its conjugate E *(t) = E(t) (this is easy to see if one expresses the exponential term in trigonometric form). This then has the (p.24) consequence that E * (ω) = E(−ω). Remembering that, by definition, the integral of the delta function is equal to unity, the energy per solid angle then becomes
d W d Ω = 0 c R 2 E ( ω ) E ( ω ) d ω = 0 c R 2 E ( ω ) E ( ω ) d ω = 0 c R 2 | E ( ω ) | 2 d ω = 2 0 c R 2 | E ( ω ) | 2 d ω ,
and the spectral angular distribution radiated by an electron on a single revolution is
(2.28)
d 2 W d Ω d ω = 2 0 c R 2 | E ( ω ) | 2 ,
where ‘spectral’ implies that it is into a particular frequency band. Since the electron is on a circular path it will make c/2πρ revolutions per second. And so the spectral power angular density will be
d 2 P d Ω d ω = c 2 π ρ d 2 W d Ω d ω = R 2 π μ 0 ρ | E ( ω ) | 2 .
We can insert | E ( ω ) | 2 = E x 2 ( ω ) + E y 2 ( ω ) into the above equation and use our results from (2.26) and (2.27)
(2.29)
d 2 P d Ω d ω = 3 e 2 γ 2 32 π 4 0 ρ ( ω ω c ) 2 ( 1 + γ 2 ψ 2 ) 2 [ K 2 / 3 2 ( G ) + γ 2 ψ 2 ( 1 + γ 2 ψ 2 ) K 1 / 3 2 ( G ) ]
Integrating this over all angles will give the spectral power, remembering that Ω is the solid angle [14].

(2.30)
d P d ω = d 2 P d Ω d ω d Ω = 2 π d 2 P d ψ d ω d ψ = P 0 ω C S ( ω ω C ) = P 0 ω C ( S x ( ω ω c ) + S y ( ω ω c ) ) ,
where P 0 is the total power radiated by one electron [12]
P 0 = c e 2 γ 4 6 π 0 ρ 2
and (p.25)
S x ( ω ω c ) = 9 3 ω 8 π ω c [ ω / ω c K 5 / 3 ( u ) d u + K 2 / 3 ( ω ω c ) ] S y ( ω ω c ) = 9 3 ω 16 π ω c [ ω / ω c K 5 / 3 ( u ) d u K 2 / 3 ( ω ω c ) ] S ( ω ω c ) = 9 3 ω 16 π ω c ω / ω c K 5 / 3 ( u ) d u .

The functions S, S x, and S y are plotted in Fig. 2.6 and values for the integral of K 5/3 are given in Table 2.1. It is clear from this graph that the majority of the power is horizontally polarized. We can integrate dP/dω over frequency to find some remarkable results. First if we consider all frequencies then

0 S ( ω ω c ) d ( ω / ω c ) = 1 0 S x ( ω ω c ) d ( ω / ω c ) = 7 8 0 S y ( ω ω c ) d ( ω / ω c ) = 1 8
Therefore, of all of the power radiated, exactly 7/8 is horizontally polarized and exactly 1/8 is vertically polarized. If we only integrate the frequencies up to ω = ω c then we find
0 1 S ( ω ω c ) d ( ω / ω c ) = 1 2 ,
which demonstrates that the critical frequency, ω c, actually divides the power spectrum into two equal parts. This fact is often used in the literature to define the critical frequency.

Having integrated (2.29) over all angles to examine the power emitted as a function of frequency, we will now do the reverse and integrate over frequency to determine how the power varies with angle [14].

(2.31)
d P d Ω = d 2 P d Ω d ω d ω = 2 π d 2 P d ψ d ω d ω d P d ψ = 21 P 0 γ 32 ( 1 + γ 2 ψ 2 ) 5 / 2 [ 1 + 5 γ 2 ψ 2 7 ( 1 + γ 2 ψ 2 ) ] .

The first and second terms in the square brackets again correspond to the horizontal and vertical polarizations, respectively. This function is plotted in (p.26)

                   Fundamentals of Synchrotron Radiation Emission

FIG. 2.6. The power spectral density and the horizontal and vertical polarization components.

                   Fundamentals of Synchrotron Radiation Emission

FIG. 2.7. The angular power distribution and the horizontal and vertical polarization components.

Fig.2.7, again demonstrating that there is no vertical power on-axis. We could integrate this result further over ψ to show that of the total power emitted 7/8 is horizontally polarized and 1/8 vertically as seen earlier.

2.4.2 Photon Flux

It is quite straightforward to convert the power results from the previous section into a more commonly used result in terms of the number of photons. The energy (p.27)

Table 2.2 Example values for the critical photon energy and angular frequency for three light sources

Ring

Energy

γ

B

ρ

ω c

є c

λc

(GeV)

(T)

(m)

(×1018s−1)

(keV)

(nm)

SRS

2

3914

1.2

5.56

4.9

3.19

0.39

DIAMOND

3

5871

1.4

7.15

13

8.38

0.15

ESRF

6

11742

0.8

25.0

29

19.2

0.06

of each photon comes from the well known result є = /2π = ħω, where h is the Planck constant. Note at this point that the critical photon energy is, as expected, given by λ c = ħω c. Similarly the critical wavelength is given by λc = 2πc/ω c. Typical values for ω c, є c, and other common parameters are given in Table 2.2.

If the number of photons emitted per second with energy є is then the power emitted at that photon energy is simply N˙є. The number of photons emitted per second per solid angle by one electron into a relative photon energy bandwidth Δє/є is

d 2 N . d Ω d / = d 2 N . d Ω d = d 2 P d Ω d = d 2 P d Ω d ω = 3 e 2 γ 2 32 π 4 0 ρ ( ω ω c ) 2 ( 1 + γ 2 ψ 2 ) 2 [ K 2 / 3 2 ( G ) + γ 2 ψ 2 ( 1 + γ 2 ψ 2 ) K 1 / 3 2 ( G ) ] .
Remembering that the fine structure constant, α, is given by
α = e 2 2 c h 0 1 137
this becomes
(2.32)
d N . d Ω = 3 α γ 2 4 π 2 c 2 π ρ ( Δ ) ( ω ω c ) 2 × ( 1 + γ 2 ψ 2 ) 2 [ K 2 / 3 2 ( G ) + γ 2 ψ 2 ( 1 + γ 2 ψ 2 ) K 1 / 3 2 ( G ) ] .
Let us consider for a moment how many electrons, N e, are passing a given point per second for a given beam current of electrons, I b. The electron beam current, measured in amperes, is simply the amount of charge flowing in unit time. For a beam of electrons this is simple to calculate as the charge carried by each electron, e, is well known. However, we must remember that as the electrons (p.28) are travelling on a circular path they will pass the same point many times in one second. Putting all this together we find that the beam current is given by the product of the number of electrons, the charge per electron, and the number of revolutions per second
(2.33)
I b = N e e c 2 π ρ .
Rearranging this we find that the number of electrons is given by
N e = I b e 2 π ρ c .
So the number of photons emitted per second per solid angle into a bandwidth Δє/є by a beam current, I b, is
(2.34)
d N . d Ω = 3 α γ 2 4 π 2 I b e ( Δ ) ( ω ω c ) 2 ( 1 + γ 2 ψ 2 ) 2 × [ K 2 / 3 2 ( G ) + γ 2 ψ 2 ( 1 + γ 2 ψ 2 ) K 1 / 3 2 ( G ) ] .
The quantity d—/dΩ is referred to as the spectral angular flux density or more correctly the spectral intensity. We can write this quantity in more useful terms, with units of photons per second per milliradian2 per 0.1% bandwidth
(2.35)
d N . d Ω = 1.33 × 10 13 E 2 I b ( ω ω c ) 2 × ( 1 + γ 2 ψ 2 ) 2 [ K 2 / 3 2 ( G ) + γ 2 ψ 2 ( 1 + γ 2 ψ 2 ) K 1 / 3 2 ( G ) ] ,
where E is the electron energy in GeV and I b is given in A. If we consider only the radiation that is on-axis (ψ = 0) then this simplifies to
(2.36)
d N . d Ω | ψ = 0 = 1.33 × 10 13 E 2 I b ( ω ω c ) 2 K 2 / 3 2 ( ω 2 ω c ) .

Three examples for the spectral intensity are given in Fig. 2.8, each for a different photon angular frequency. Note that at lower frequency and longer wavelength, the radiation extends further out in the vertical angle and also that the vertical polarization contribution becomes quite significant. Since the vertical electric field is 90° out of phase with the horizontal component (by inspection of (2.26) and (2.27)), the result is circular polarization. We can also see that at frequencies close to ω c the approximation that the radiation is emitted with a (p.29)

                   Fundamentals of Synchrotron Radiation Emission

FIG. 2.8. Spectral intensity at ω c, 0.1ω c, and 10ω c for a 300 mA, 3 GeV electron beam passing through a 1.4 T bending magnet.

vertical angle of ±1/γ is a good one. Finally we can see that at high frequencies the angular distribution is significantly less and the beam is more tightly collimated.

We can use the same approach to look at the number of photons emitted per electron per second into all angles (p.30)

d N = d P d ω = P 0 ω c S ( ω ω c ) = c e 2 γ 4 6 π 0 ρ 2 c 9 3 ω 8 π ω c ω / ω c K 5/3 ( u ) d u .
Substituting in for the critical photon energy
(2.37)
c = ω c = 3 h c γ 3 4 π ρ
and noting that ω/ω c =є/єc we get
d N . d / = 3 γ e 2 2 c h 0 c 2 π ρ ( c ) / c K 5 / 3 ( u ) d u = 3 α γ c 2 π ρ ( c ) / c K 5 / 3 ( u ) d u .

Again, multiplying this result by the number of electrons will give us the number of photons emitted per second into all angles for a beam current, I b, into a relative energy bandwidth Δє/є

(2.38)
N . = 3 α γ I b e ( Δ ) ( c ) / c K 5 / 3 ( u ) d u .

The parameter is referred to as the spectral photon flux or the vertically integrated spectral flux, this latter name conveys the message that the photon emission has been summed over all angles (as the electron travels on a circle of 2π radians it is automatically integrated horizontally). Again, in practical units this reduces to

(2.39)
N . = 2.46 × 10 13 E I b ( c ) / c K 5 / 3 ( u ) d u
in units of photons per second per milliradian horizontally per 0.1% bandwidth.

A plot of versus є/єc is known as the universal curve (Fig. 2.9). Absolute flux levels for a particular electron energy and beam current can be quickly scaled off the curve for any photon energy once the critical photon energy is calculated. It should be clear from this that all bending magnet sources have the same characteristic spectrum. The spectral flux always increases slowly from the low photon energies, peaking at approximately є/єc = 0.25. The spectrum then falls off sharply, with a typical consideration being that the flux is useful up to about є/єc ≍ 5. The low photon energies extend down until the emitted radiation reaches a wavelength of the order of the vacuum chamber dimensions (p.31)

                   Fundamentals of Synchrotron Radiation Emission

FIG. 2.9. The universal flux curve. Note that both axes have a logarithmic scale.

                   Fundamentals of Synchrotron Radiation Emission

FIG. 2.10. Spectral output from the bending magnets of three well known electron storage rings. A beam current of 200 mA has been assumed. Note the logarithmic scale along the x-axis.

(usually tens of mm) and so becomes cut-off from the observer. Typical bending magnet spectral curves are shown in Fig. 2.10 for the three storage rings whose parameters are given in Table 2.2.

In summary then, the spectral range covered by a bending magnet is fixed by the critical photon energy, which is a function of the electron energy and the bending radius (2.37). The bending radius itself depends upon the electron (p.32) energy and the magnetic field, B

ρ = E 0.3 B ,
where E is expressed in GeV and B is in Tesla and so in more practical units the critical energy is given (in eV) by
c = 665 E 2 B
In an electron storage ring with a fixed energy the SR spectrum can be shifted sideways along the photon energy axis if a different єc can be generated locally. It is possible to change єc in this way by having a different magnetic field in a local part of the storage ring. The insertion devices which can produce this local change in єc and so shift the SR spectrum, go by the name of wavelength shifters and these are the subject of Section 2.6.

2.4.3 Vertical Opening Angle

We have already seen that different photon energies emit SR over quite different vertical angular distributions. It is useful to be able to estimate what a ‘typical’ angle might be for each particular photon energy. This ‘typical’ angle is the so-called vertical opening angle of the radiation, denoted as σr′. The most common method for estimating this opening angle is to assume that the angular distribution follows a Gaussian or Normal distribution. This is not a particularly good assumption, especially when є « єc where the vertical polarization component can become quite significant (see Fig. 2.8, for example) so the results should always be treated with some caution.

Let's just remind ourselves of the main features of a Gaussian distribution. First it has a functional form

f ( x ) = 1 2 π σ exp ( ( x μ ) 2 2 σ 2 ) ,
where μ is the mean and σ is the standard deviation. Second, the integral of ƒ(x) over all x is equal to unity and finally, the full width at half maximum (FWHM) for a Gaussian distribution is equal to 2.355 σ.

Now, we are assuming that the vertical angular distribution is of Gaussian form and symmetric about ψ = 0, so

(2.40)
d N . d Ω = d N . d Ω | ψ = 0 exp ( ψ 2 2 σ r 2 ) .
Integrating this over all vertical angles we get
N . 2 π = 2 π σ r d N . d Ω | ψ = 0
therefore (p.33)
σ r 1 = N . 2 π 2 π d N . d Ω | ψ = 0
Substituting our results from (2.38) and (2.34) gives
(2.41)
σ r = 3 α γ ( I b / e ) ( Δ / ) ( / c ) / c K 5 / 3 ( u ) d u 2 π 2 π ( 3 αγ 2 / 4 π 2 ) ( I b / e ) ( Δ / ) ( / c ) 2 K 2 / 3 2 ( / 2 c ) = 2 π 3 1 γ ( c ) / c K 5 / 3 ( u ) d u K 2 / 3 2 ( / 2 c )
So, for instance, at є = єc we find that
σ r = 0.65 γ

A plot of the output from the above result for σ r′ is given in Fig. 2.11 for a 3 GeV electron beam. Note that the vertical opening angle changes by two orders of magnitude between low and high photon energies. The examples that were used for Fig. 2.8 are plotted again in Fig. 2.12 but this time with the superimposed ideal Gaussian distribution using the calculated value of σ r′ as well. It is clear from this that the Gaussian assumption is only correct at the highest photon energies, the approximation is already struggling at є = єc. Since this treatment is only an approximation, if the exact vertical angular distribution is of particular interest the distribution dN˙/dΩ should be examined directly.

2.4.4 Bending Magnet Brightness

If we take a perpendicular slice through the SR travelling towards the observer then we will intercept millions of photons. Each of these photons in this slice will have a particular position and angular direction. Some will have a position close to the axis but a relatively large divergence and others will be far from the axis but have a very shallow trajectory. This concept of particles having a position and an angle that evolves with time as they travel towards the observer is known as the phase space. It is an important concept that also crops up in many other areas of physics. In particular, it is often used in accelerator physics to describe the distribution of the charged particles travelling around an accelerator.

The brightness of a source is the phase space density of the photon flux (i.e. the photons per unit solid angle per unit solid area) and it is a figure of merit that takes into account not only the number of photons emitted but also their concentration. It is often encountered in geometric optics where it is widely used because it is a quantity which is invariant in an ideal optical beam transport system (a result from thermodynamics known as Liouville's theorem [20]), unlike angular flux density, for instance.

(p.34)

                   Fundamentals of Synchrotron Radiation Emission

FIG. 2.11. A plot of σ r vs. ∈/∈cfor a 3 GeV electron beam.

To calculate the bending magnet brightness we need to consider the effective phase space area from which the photon flux is being emitted taking account of both the finite electron and photon beam sizes and divergences. First, there is no need to consider any horizontal angle effects as the light is emitted smoothly over the full horizontal 2π radians. The effective vertical angle, ∑y′, will be a combination of the electron vertical beam divergence, σ y′, and the photon beam opening angle, σ r′. Since these are both from Gaussian distributions they are added in quadrature

y = σ y 2 + σ r 2 .
For the effective source sizes we have to combine the electron beam sizes, σ x and σ y, with the photon beam size, σ r, in a similar manner
x = σ x 2 + σ r 2 y = σ y 2 + σ r 2 .
The photon beam size is found by approximating the single electron photon source to the fundamental mode of an optical resonator (see Section 4.4)
σ r = λ σ r .
The bending magnet brightness, B, is then given by
B = N ( 2 π ) 3 / 2 x y y ,
(p.35)
                   Fundamentals of Synchrotron Radiation Emission

FIG. 2.12. Spectral intensity at ω c, 0.1ω c, and 10ω c for a 300 mA, 3 GeV electron beam passing through a 1.4 T bending magnet. Also plotted is the Gaussian distribution for the appropriate vertical opening angle σ r′.

where is the photon flux per unit horizontal angle (2.38). Note that each Σ term introduces a 2 π because the rectangular function of equal area to a Gaussian has a width of 2 π σ (see Section 2.4.3).

In general, σ r « σ x,y and the brightness equation can be simplified to (p.36)

B = N ˙ ( 2 π ) 3 / 2 σ x σ y ( σ y 2 + σ r 2 ) 1 / 2 .
The bending magnet brightness has units of photons per second per solid area per solid angle per spectral bandwidth.

2.5 Power and Power Density from a Bending Magnet

A considerable amount of power can be generated in the form of synchrotron radiation in a storage ring. This is not only important for the users of the radiation but also those who have to design and operate the accelerator. The power and power density levels are often high enough in a synchrotron light source to cause damage to the accelerator itself. A number of accelerators around the world have melted components inside the vacuum chamber and the vacuum chamber itself in some cases! For this reason all light sources have to water cool many items inside the vacuum system that the synchrotron radiation impinges upon. In some accelerators sophisticated monitoring of the electron beam position is necessary to ensure that it is operating safely and if the beam moves outside of certain prescribed limits it is quickly dumped to prevent any possible thermal damage.

2.5.1 Total Power

The instantaneous total power emitted by a single relativistic electron is [12]

P 0 = c e 2 γ 4 6 π 0 ρ 2 ,
where ρ is the radius of curvature of the electron path. So, in a storage ring of circumference, C, with fixed bending radius in the dipoles, ρ 0, the energy radiated by each electron per turn is
(2.42)
Δ E = e 2 γ 4 6 π 0 0 C d s ρ 0 2
(the c is cancelled because the integral is expressed with respect to s and not t). And since the integral equals 2π/ρ 0 (there is no emission in the straight sections) this becomes
Δ E = e 2 γ 4 3 0 ρ 0 .
So, the total power emitted by a beam of electrons of current, I b, using (2.33) is
P total = e γ 4 I b 3 0 ρ 0 ,
which in practical units is
(2.43)
P total = 88.46 E 4 I b ρ 0 ,
where the power is in kW and the electron energy, E, is in GeV.

(p.37)

Table 2.3 Example values for the power, power per horizontal angle and power density on-axis for three light sources

Ring

Energy (GeV)

ρ (m)

I b (mA)

P total (kW)

dP/dθ (W/mrad)

dP/dΩ (W/mrad2)

SRS

2

5.56

200

50.9

8.1

20.8

DIAMOND

3

7.15

300

300.7

47.9

184.4

ESRF

6

25.0

200

916.5

145.9

1124.0

Another useful number to know is the power per horizontal angle. Again, expressed in practical units this is given by

d P d θ = 14.08 E 4 I b ρ 0 ,
where the result is in W/mrad.

2.5.2 Power Density

The bending magnet power distribution in the vertical plane has already been derived (2.31) and plotted in Fig. 2.7. We can use this result to find the power density. In particular, the power density on-axis (ψ = 0) is given by

d P d Ω | ψ = 0 = 21 γ 64 π P total ,
which in practical units of W/mrad2 is
(2.44)
d P d Ω | ψ = 0 = 18.08 E 5 I b ρ 0 .
Some actual values for the power levels that are experienced in modern light sources are given in Table 2.3.

2.6 Wavelength Shifters

Wavelength shifters are a type of insertion device, which essentially produces bending magnet style radiation. The advantage that they have over the storage ring bend magnets is that their magnetic field, and so the critical photon energy, can be tailored to a specific beamline requirement. Generally, wavelength shifters are used to shift the spectrum towards the higher photon energy end. In the SRS, for instance, the 6.0 T magnetic field in one of the wavelength shifters produces a critical photon energy five times higher than the 1.2 T bending magnets [21]. This gives a larger flux at the high photon energy end of the spectrum than is normally available (Fig. 2.13). From this plot it becomes apparent why the term wavelength shifter is used since the spectrum literally shifts along the photon energy x-axis. (p.38)

                   Fundamentals of Synchrotron Radiation Emission

FIG. 2.13. A comparison between the spectral photon flux emitted by a 1.2 T SRS bending magnet and the 6.0 T wavelength shifter. The beam current is 200 mA and the electron energy is 2 GeV.

There is no absolute change to the total number of photons emitted, just the wavelength at which they are emitted changes.

A wavelength shifter usually consists of one high magnetic field central pole surrounded by two weaker side poles of opposite polarity. A typical magnetic field profile along a wavelength shifter and the electron trajectory through such a magnet is shown in Fig. 2.14. The magnetic field strengths are arranged so that the total integrated field strength along the longitudinal axis is equal to zero. This ensures that the overall angular deflection to the electron is zero and that the electron exits the insertion device on the same axis that it entered on.

Since the magnetic field along the length of the wavelength shifter is not constant, the critical photon energy also varies along its length (see Fig. 2.14). This means that although the SR produced has the same characteristics of bending magnet radiation, the exact characteristics observed depend on which part of the electron trajectory the observer is looking at. Furthermore, the observer may simultaneously also see SR produced by the side poles, which may enhance the flux but also give a light source with more than one source point.

A wavelength shifter will typically deflect the electron beam by the order of 10 mm at the peak of the bump, which is of course the optimum source point (highest magnetic field strength). Clearly if the magnetic field strength of the wavelength shifter is altered the size of this deflection will alter, in turn changing the source point position. It is possible to produce a wavelength shifter, which always has the main source point on-axis by the inclusion of two further side poles. Examples of this are the 7 T wavelength shifters at CAMD [23] and BESSY II [24].

(p.39)

                   Fundamentals of Synchrotron Radiation Emission

FIG. 2.14. The magnetic field profile for the SRS 5 T wavelength shifter [22], the critical energy and the position and angle of a 2 GeV electron through the magnet.

2.7 Extension to Multipole Wigglers

We have just seen that a wavelength shifter, which is simply a single, large bump on the electron trajectory can produce synchrotron radiation that is essentially bending magnet radiation. Imagine putting several identical wavelength shifters one after the other in a straight section of a storage ring. The electron would simply travel through each wavelength shifter in turn emitting synchrotron radiation in the forward direction. Each wavelength shifter is independent of the other and the electron returns back to the beam axis after passing through each one. Although there is no fundamental relationship between each wavelength shifter (p.40)

                   Fundamentals of Synchrotron Radiation Emission

FIG. 2.15. Magnetic field and electron trajectory for three consecutive wavelength shifters in a straight section. The observer will receive three times more flux than from one wavelength shifter.

in terms of the emission of light, an observer will see an enhancement in the flux received simply because there are now several sources emitting radiation in his direction. This point is illustrated in Fig. 2.15. This concept of multiple sources, separated longitudinally and each emitting independently, forms the basis for the multipole wiggler. Note that the term wiggler on its own is used somewhat carelessly in the literature and refers sometimes to a wavelength shifter and sometimes to a multipole wiggler.

Of course, putting several wavelength shifters in a straight section is not the most efficient way of creating a multiple source. A better arrangement of alternating magnetic fields is shown in Fig. 2.16. In fact a typical multipole wiggler has a magnetic field, which closely resembles a sinusoidal profile. A detailed derivation of the synchrotron radiation emission from a multipole wiggler is given in Chapter 3 and the design of magnets to create the desired magnetic field is covered in Chapter 7 for permanent magnet based solutions and Chapter 8 for electromagnet based solutions.

(p.41)

                   Fundamentals of Synchrotron Radiation Emission

FIG. 2.16. Magnetic field arrangement that utilizes the available straight space more economically. This concept of alternating magnetic fields to generate a flux enhancement is the basis of the multipole wiggler.