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Fundamentals of Beam Physics$

J. B. Rosenzweig

Print publication date: 2003

Print ISBN-13: 9780198525547

Published to Oxford Scholarship Online: January 2010

DOI: 10.1093/acprof:oso/9780198525547.001.0001

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(p.241) Appendix A: selected problem solutions

(p.241) Appendix A: selected problem solutions

Source:
Fundamentals of Beam Physics
Publisher:
Oxford University Press

The following solutions to selected problems have been primarily generated by the author’s assistants, who include a number of students who have taken the UCLA Department of Physics and Astronomy beam physics course, Physics 150. The varying styles of solution reflect the diverse approaches to the material taken by different contributors.

Chapter 1

Exercise 1.1

In a three-dimensional simple harmonic oscillator where T = 1 2 m x ˙ and V = 1 2 k x 2 : (a) The Lagrangian is

Appendix A: selected problem solutions
In Cartesian coordinates, x = x x ^ + y y ^ + z z ^ Therefore
Appendix A: selected problem solutions

(b) Canonical momenta are given by p c , i = L / x ˙ c , i (subscript c denotes canonical variable). For this system

Appendix A: selected problem solutions
The Hamiltonian is
Appendix A: selected problem solutions
The total energy of the system is conserved if ∂H/∂t = 0. Since this condition is true for the Hamiltonian given above, the Hamiltonian is a constant of the motion. Notice also that the portion of the Hamiltonian governing horizontal motion alone is constant, ∂H x/∂t = 0 (and similarly for H y and H z), and the (p.242) energy in each phase plane (e.g. (x c, p c,x)) is conserved. Therefore, we can write,
Appendix A: selected problem solutions
where H x = ( p c , x 2 / 2 m ) + 1 2 k x 2 etc.

(c) In cylindrical coordinates

Appendix A: selected problem solutions
where x 2 = x 2 + y 2 + z 2 = ρ 2 cos φ + ρ 2 sin 2 φ + z 2 = ρ 2 + z 2 and
Appendix A: selected problem solutions
so the Lagrangian is
Appendix A: selected problem solutions
or
Appendix A: selected problem solutions

(d) The canonical momenta, p c , i = L / x ˙ c , i , in cylindrical coordinates are

Appendix A: selected problem solutions
so the Hamilton, H ( x c , p c ) where x c = ρ ρ ^ + φ φ ^ + z z ^ and p c = p c , ρ ρ ^ + p c , φ φ ^ + p c , z z ^ is
Appendix A: selected problem solutions
or in terms of canonical variables,
Appendix A: selected problem solutions
From inspection of the Hamiltonian,
Appendix A: selected problem solutions
is a constant of motion.

(p.243) (e) In spherical coordinates,

Appendix A: selected problem solutions
where
Appendix A: selected problem solutions
and the Lagrangian is
Appendix A: selected problem solutions
or
Appendix A: selected problem solutions

(f) The canonical momenta, p c , i = L / x ˙ c , i , in spherical coordinates are

Appendix A: selected problem solutions
so the Hamilton, H ( x c , p c ) , where x c = r r ^ + θ θ ^ + φ φ p c = p c , r r ^ + p c , θ θ ^ + p c , φ φ ^ is
Appendix A: selected problem solutions
From inspection of the Hamiltonian,
Appendix A: selected problem solutions
is a constant of motion.

Exercise 1.3

In a sextupole magnet with a force, F x = –ax 2 (where a is constant);

(a) the Hamiltonian associated with one-dimensional motion is obtained by first solving for the Lagrangian—the difference between the kinetic and (p.244) potential energy of the system, L = T – V, where

Appendix A: selected problem solutions
The Hamiltonian is
Appendix A: selected problem solutions

(b) In order to plot representative constant H curves, first solve for the parameter p x, and set H, m, and a to arbitrary constants,

Appendix A: selected problem solutions
Below, H = 1, 4, and 8, respectively, while m = 1 and a = 1. The curves are not closed, indicating that the particle motion is unbounded.

Appendix A: selected problem solutions

In an octupole field, the force is F x = −ax 3. Following the above method, L = TV, where

Appendix A: selected problem solutions
The Hamiltonian is
Appendix A: selected problem solutions
In an octupole-like field, the equation for plotting the phase space becomes
Appendix A: selected problem solutions
We consider the sign of the constant a, to find if the particle motion is affected.

For a = −1 (i.e. a < 0), m = 1 and values of H = 0.25, 1, and 4, respectively, the motion is unbounded.

(p.245)

Appendix A: selected problem solutions

For a = 1 (i.e. a > 0) and the same values assumed for m and H, the motion is bounded. The closed curves are not ellipses, but have a “race-track” appearance.

Appendix A: selected problem solutions

Exercise 1.5

The Taylor series expansion for a small x is

Appendix A: selected problem solutions
The equation U = γm 0 c 2= m 0 c 2(1–(v 2/c 2)) –1/2 can be expanded using v/c, since v/c ≪ 1. Therefore
Appendix A: selected problem solutions
and
Appendix A: selected problem solutions
The first term in the mechanical energy represents the rest energy and the second term represents the familiar non-relativistic kinetic energy.

(p.246) Chapter 2

Exercise 2.1

Given

Appendix A: selected problem solutions

(a) Show × A = B given A = ( B 0 / 2 ) ρ φ ^ .

In cylindrical coordinates,

Appendix A: selected problem solutions

(b) Write the Lagrangian in cylindrical coordinates:

Appendix A: selected problem solutions
where
Appendix A: selected problem solutions

(c) Lagrange–Euler equations,

Appendix A: selected problem solutions
Since we are considering a pure magnetic field, the total energy of this problem is conserved, and γ = constant, or dγ/dt = 0.

(p.247) The Lagrange–Euler equations become, after use of the constants of the motion:

Appendix A: selected problem solutions
and
Appendix A: selected problem solutions
For circular motion, the equilibrium solution requires that ρ ¨ = ρ ˙ = 0 ; therefore, ρ must be a constant.

Thus, we have φ ˙ = q B 0 / γ m 0 = ω c .

Now, using the fact that the total energy is conserved, U=γm 0 c 2= constant, and 1 / γ 2 = 1 ( ρ 2 φ ˙ 2 / c 2 ) is also constant.

Using φ ˙ = q B 0 / γ m 0  and  ρ = R (radius of curvature), we have

Appendix A: selected problem solutions
so P = qB 0 R, which is the same as given in Eq. (2.7).

Exercise 2.2

Starting with p 0(t) = eB 0(t)R

(a) Differentiate the above equation with respect to time

Appendix A: selected problem solutions

(a) Using Maxwell’s equations we have that ∇ × E = −∂B/∂t. Let us apply Stokes Theorem

Appendix A: selected problem solutions
Using the definition of the average magnetic field given in the problem
Appendix A: selected problem solutions
we obtain that B 0 ( t ) = ( B ¯ ( t ) / 2 ) +  constant . The constant of integration can be chosen to be non-zero. It, in practice, often is made non-zero to avoid saturation (p.248) effects in the iron. This is called a reverse-biased betatron. For a description of saturation see discussion in Chapter 6.

Exercise 2.9

(a) Starting with Eq. (2.29) (for motion in a static, constant electric field)

Appendix A: selected problem solutions
Take the time derivative
Appendix A: selected problem solutions
Using the paraxial approximation:
Appendix A: selected problem solutions
With this expression we obtain
Appendix A: selected problem solutions

(a) Integrate this equation:

Appendix A: selected problem solutions
to find
Appendix A: selected problem solutions
Here, C 1 is a constant of integration, but since γ(0) = 1, we have C 1 = 0. Solving for γ, we obtain
Appendix A: selected problem solutions

(b) Now, let us directly integrate Eq. (2.25)

Appendix A: selected problem solutions
Here, C 2 is a constant of integration but since γ(0) = 1 and v z(0) = 0, then C 2 = 0, and
Appendix A: selected problem solutions
Using the paraxial approximation,
Appendix A: selected problem solutions
(p.249) we have
Appendix A: selected problem solutions
Solving for γ we obtain
Appendix A: selected problem solutions
This matches the previous result.

Exercise 2.11

First let us compare the equations of motion for a particle traveling in z ^ and its antiparticle traveling in the opposite direction z ^ .

The particle has charge q, mass m 0, and momentum p z = γ m 0 ν z z ^ . The antiparticle has charge – q, mass m 0, and momentum p z = γ m 0 ν z ( z ^ ) .

For magnetic focusing channel:

Appendix A: selected problem solutions
In this case we need to omit the skew term (a 2 = 0) and keep the normal term (b 2 > 0)
Appendix A: selected problem solutions
The force is given by F = q ( v z × B )
Appendix A: selected problem solutions
and the equations of motion are the same.

Now let us do the same thing for the electric quadrupole fields:

Appendix A: selected problem solutions
In this case we set the skew term b 2 = 0, and keep the normal term (a 2 > 0)
Appendix A: selected problem solutions
The force is given by F ¯ = q E .
Appendix A: selected problem solutions
In this case the dimensions of focusing and defocusing are exchanged for the particle and antiparticle.

(p.250) Exercise 2.14

(a) Consider a cylindrically symmetric beam in a solenoid. Using Gauss’ Law:

Appendix A: selected problem solutions
where λ(ρ) is the enclosed charge per unit length, related to the beam density n b by λ(ρ) = qn bπρ 2. Thus,
Appendix A: selected problem solutions
The radial space charge electric force is
Appendix A: selected problem solutions
where
Appendix A: selected problem solutions

(a) Due to length contraction of the moving cylinder (l = l′/γ) the density in the rest frame is given by

Appendix A: selected problem solutions
In the rest frame
Appendix A: selected problem solutions
Transform to the lab frame:
Appendix A: selected problem solutions
In the lab frame
Appendix A: selected problem solutions
Then
Appendix A: selected problem solutions
(p.251) Now let us look at the forces
Appendix A: selected problem solutions
where v zβc.

We have the compact result that the total electromagnetic force is given by

Appendix A: selected problem solutions
This result could also have been obtained by direct calculation of the radial electric and azimuthal magnetic field in the lab frame. (b) We also have two other forces: the solenoid force
Appendix A: selected problem solutions
and the centripetal force
Appendix A: selected problem solutions
The total force is
Appendix A: selected problem solutions
At equilibrium:
Appendix A: selected problem solutions
or
Appendix A: selected problem solutions
Here we have used
Appendix A: selected problem solutions
In more standard form, we have
Appendix A: selected problem solutions

(p.252)

Appendix A: selected problem solutions

(c) Plot ωpc versus ωrc.

The rotation frequency that maximizes the equilibrium density of the beam is ωr = ωc/2.

Exercise 2.16

The helical undulator has on on-axis field given by

Appendix A: selected problem solutions
The vector potential in cylindrical coordinates is
Appendix A: selected problem solutions
where I v (x) are the modified Bessel functions which obey he following relations:
Appendix A: selected problem solutions
Let us use some abbreviations to simplify notation
Appendix A: selected problem solutions
(a) Show that the 2 A = 0 Since there are no magnetic source terms, × B = × ( × A ) = ( . A ) 2 A = 0 Therefore, ( . A ) = 2 A (p.253) First let us calculate . A (in cylindrical coordinates):
Appendix A: selected problem solutions
So from ( . A ) = 2 A we can conclude that 2 A = 0 . (b) Next we calculate the field components in cylindrical coordinates:
Appendix A: selected problem solutions
Let us calculate each term individually. The ρ-component is:
Appendix A: selected problem solutions
The ɸ-component is:
Appendix A: selected problem solutions
The z-component is:
Appendix A: selected problem solutions
(c) First we can use small angle approximations for the modified Bessel functions since k uρ ≪ 1.
Appendix A: selected problem solutions
In the first-order approximation these become
Appendix A: selected problem solutions
(p.254) We need to transform the vector potential from cylindrical coordinates to Cartesian coordinates:
Appendix A: selected problem solutions
The Hamiltonian for this system is given by
Appendix A: selected problem solutions
The equations of motion can be derived from:
Appendix A: selected problem solutions
The canonical momenta are:
Appendix A: selected problem solutions
The Hamiltonian is constant and H= γmc 2. since p ˙ z = 0 , we have p z=γmβ z c and formally
Appendix A: selected problem solutions
A particle entering a helical undulator’s magnetic field is assumed to have no transverse velocity components. This means that p x = p y = 0. Introducing the undulator parameter K = qB/mkc, we now have:
Appendix A: selected problem solutions
The equations for x(t) and y(t) describe the motion of a helix with radius ρ = cK/kγ = constant.

(p.255) From the equations of motion we can see that the mechanical angular momentum p φ , mech = γ m ρ φ ˙ is constant. Let us return to our coordinate transformations and differentiate with respect to time realizing that ρ is a constant.

Appendix A: selected problem solutions
Combining this and the previous expressions for x ˙ and y ˙ , we obtain two expressions for φ ˙ .
Appendix A: selected problem solutions
The left-hand side of these equations can be written as total derivatives of sine and cosine functions:
Appendix A: selected problem solutions
Recall that in the paraxial approximation, we have
Appendix A: selected problem solutions
Using this we obtain,
Appendix A: selected problem solutions
which can be integrated to find
Appendix A: selected problem solutions
Combining these expressions we have,
Appendix A: selected problem solutions
or φ ˙ = c k β z = constant . Recall that p ɸ,mech = constant. Throughout this exercise we have kept terms only up to first order in k uρ.

(p.256) Chapter 3

Exercise 3.2

Using the equipartition theorem,

Appendix A: selected problem solutions
and assuming that
Appendix A: selected problem solutions
To calculate the field index, n, required to guarantee that the betatron beam obeys x rms = 2y rms, start with:
Appendix A: selected problem solutions
The tunes are v x = 1 n   and ν y = n .

Following the given requirements with the above relations:

Appendix A: selected problem solutions
and
Appendix A: selected problem solutions
gives 2v x = v y, which can be written as 2 1 n = n , or 1-n=n/4 so the required field index is n = 4/5.

Exercise 3.4

(a) x = M T x 0 = M O M F x 0 The total transformation matrix for one period is M T = M o. M F

Appendix A: selected problem solutions
(b) To ensure linear stability, |TrM T| ≤ 2 giving in this case
Appendix A: selected problem solutions
which is simply
Appendix A: selected problem solutions

(p.257) Exercise 3.12

Consider the FOFO focusing system: (a) The transformation matrix of the FOFO system is given by

Appendix A: selected problem solutions
The phase advance, μ, is a function of κ0 l:
Appendix A: selected problem solutions
(b) In the limit of κ0 l ≪, use the following expansions up to fourth order:
Appendix A: selected problem solutions
and
Appendix A: selected problem solutions
Substituting the above relations into cos μ = cos(κ0 l) – (κ0 l/2) sin(κ0 l) yields
Appendix A: selected problem solutions
Simplifying, we have
Appendix A: selected problem solutions
which yields
Appendix A: selected problem solutions
Now, evaluate this expression for k sec 2 :
Appendix A: selected problem solutions
(c) Now define κ 0 2 as a harmonic series
Appendix A: selected problem solutions

The Fourier series is useful:

Appendix A: selected problem solutions
(p.258) The coefficients are:
Appendix A: selected problem solutions
and so
Appendix A: selected problem solutions
(d) The Fourier series solution is a superposition of a uniform focusing system of strength κ 0 2 / 2 (first term on the RHS) and an FD system of the same strength (the second term on the LHS). The averaged focusing strength is
Appendix A: selected problem solutions
Using bn = 2/nπ and kp = 2π/Lp = π/l,
Appendix A: selected problem solutions
or
Appendix A: selected problem solutions
(e) Solutions to (b) and (d) are identical.

Exercise 3.15

In the smooth approximation,

Appendix A: selected problem solutions
Referring to the evolution of horizontal dispersion, Eq. (3.65),
Appendix A: selected problem solutions
Solutions are given by Eq. (3.66),
Appendix A: selected problem solutions
The solution is given in two parts. The homogeneous solution is given by the first two terms with coefficients A and B. The final term on the RHS is the (p.259) particular solution,
Appendix A: selected problem solutions
The given Tevatron values are v x = 19.4 and R 0 = 1 km. The approximate constant value of the dispersion can be estimated for the Tevatron using only the particular solution
Appendix A: selected problem solutions

Exercise 3.18

In Fig. 3.13, the transport line translates the beam to the side. The dispersion and its derivative vanish at the second bend exit. The dispersion also vanishes at the mid-point between the magnets. (a) The given conditions are

Appendix A: selected problem solutions
The focal length is derived using
Appendix A: selected problem solutions
Since the system is completely symmetrical about the midpoint, this expression can be simplified by rewriting the expression for the focal length,
Appendix A: selected problem solutions
with
Appendix A: selected problem solutions
In calculating the midpoint expression, it is only necessary to solve M 13 =0,
Appendix A: selected problem solutions
Appendix A: selected problem solutions
(p.260) To simplify notation, normalize all terms to R:
Appendix A: selected problem solutions
Solving for the focal length f′ in terms of (normalized) a and b,
Appendix A: selected problem solutions
(b) The expression for the momentum compaction parameter is
Appendix A: selected problem solutions
or
Appendix A: selected problem solutions

Chapter 4

Exercise 4.1

The drift tube linac has an electric field profile given by

Appendix A: selected problem solutions
(a) Since the system is periodic,
Appendix A: selected problem solutions
The coefficients a n are given by
Appendix A: selected problem solutions
Thus,
Appendix A: selected problem solutions
(b) Assuming the particle is synchronous with the first harmonic (n = 1), the particle velocity can be written as
Appendix A: selected problem solutions
(p.261) since k z,n = 2nπ/d. The relationship between the period and the frequency is,
Appendix A: selected problem solutions
(c) For a partially relativistic ion beam, the periodicity must change to keep particles in resonance. The relationship between kinetic energy and velocity is derived from the energy expression u = T + m 0 c 2 = γm 0 c 2.

Solving for γ, we have γ = (T/m 0 c 2) + 1.

We were given T 0 = 5 MeV and T f = 250 MeV. Recall that the rest mass of a proton is m p c 2 = 939 MeV, and thus

Appendix A: selected problem solutions
Then the velocity can be found from
Appendix A: selected problem solutions
Using the results from part (b) we obtain:
Appendix A: selected problem solutions

Exercise 4.2

Evaluate αrf:

Appendix A: selected problem solutions
(a) For a high gradient electron linac: f = ω/2π = 2856 MHz, E 0 = 50 MV/m, qE 0 = 50 MeV/m, U = 1 GeV, m e c 2 = 0.511 MeV.

Determine the value of k z: Begin with

Appendix A: selected problem solutions
and using the paraxial approximation vv z:
Appendix A: selected problem solutions
We can express γ in terms of the energy:
Appendix A: selected problem solutions
Now evaluate αrf:
Appendix A: selected problem solutions
(b) For a moderate gradient proton linac: f = ω/2π = 805 MHz, qE 0 = 8 MeV/m, T = 200 MeV (here T is the kinetic energy), and m p c 2 = 939 MeV. Recall that the total energy U = γm 0 c 2 = T + m 0 c 2 or γ = 1 + (T/m 0 c 2).

(p.262) Plugging this value into

Appendix A: selected problem solutions
we find
Appendix A: selected problem solutions
For a proton linac αrf:
Appendix A: selected problem solutions

Exercise 4.3

The Hamiltonian for this problem is given by

Appendix A: selected problem solutions
where
Appendix A: selected problem solutions
(a) For optimum capture, in the final state ɸf = π/2, and v zc, so χ−1 ⇒ 0, and
Appendix A: selected problem solutions
The Hamiltonian is a constant of motion, so in the initial state we have
Appendix A: selected problem solutions
The minimum value of αrf which allows solution of this expression is clearly 1. (b) From the above final state Hamiltonian, we have
Appendix A: selected problem solutions
(c) The initial phase for the electron linac: f = ω/2π = 2856 MHz, with E 0 = 50MV/m, or qE 0 = 50MeV/m, and m e c 2 = 0.511 MeV, we have αrf = 1.63. So ɸ0 = 127.57°. (p.263) (d) Hamiltonian method: Since the Hamiltonian is a constant, the initial and final values are the same,
Appendix A: selected problem solutions

Using ϕi = 127.57° - 1° = 126.57° (one degree early):

Appendix A: selected problem solutions

First-order analysis

Appendix A: selected problem solutions
The parameters given are
Appendix A: selected problem solutions
and
Appendix A: selected problem solutions
Both of these methods yield the same result.

Exercise 4.4

The parameters given in Exercise 4.2(b) are:

Appendix A: selected problem solutions
The momentum is p 0 = γ0 m 0β0 c. (a) The bucket height is given by Eq. (4.32):
Appendix A: selected problem solutions
with our parameters this becomes:
Appendix A: selected problem solutions
(p.264) (b) The area of the bucket is given by Eq. (4.33):
Appendix A: selected problem solutions
We want to calculate the momentum in units of MeV/c.
Appendix A: selected problem solutions
Combining the above equations we obtain:
Appendix A: selected problem solutions
(c) The synchrotron frequency is given by Eq. (4.36):
Appendix A: selected problem solutions
The normalized synchrotron frequency is:
Appendix A: selected problem solutions

Exercise 4.5

Appendix A: selected problem solutions
The equation of motion is
Appendix A: selected problem solutions
so the Hamiltionian can be written as
Appendix A: selected problem solutions
Solving for ζ ˙ , we have
Appendix A: selected problem solutions
To obtain a dependence on amplitude, we need to evaluate H at a turning point Hmax, 0)
Appendix A: selected problem solutions
or
Appendix A: selected problem solutions
(p.265) which can be written
Appendix A: selected problem solutions

Let us now set

Appendix A: selected problem solutions
Integrating, we obtain
Appendix A: selected problem solutions
or
Appendix A: selected problem solutions
Thus,
Appendix A: selected problem solutions
is an elliptical integral of the first order.

You can find the solution to this in a table. See Gradshteyn, I.S., Table of Integrals, Series, and Products, pp. 904–909.

The frequency is then given by

Appendix A: selected problem solutions

Chapter 5

Exercise 5.1

Equation (5.9) is

Appendix A: selected problem solutions
Equation (5.11) is
Appendix A: selected problem solutions

(p.266) Normalization constant for Eq. (5.9): In terms of σPx we have

Appendix A: selected problem solutions
or
Appendix A: selected problem solutions
In terms of T x
Appendix A: selected problem solutions

Normalization constant for Eq. (5.11): in terms of σx (same form, same solution)

Appendix A: selected problem solutions
In terms of T x
Appendix A: selected problem solutions

Exercise 5.6

An initial correlated bi-Gaussian distribution function is written in standard form as

Appendix A: selected problem solutions
Given the transformation
Appendix A: selected problem solutions
we can also write the inverse transformation
Appendix A: selected problem solutions
The final distribution function can be written as
Appendix A: selected problem solutions
which is a correlated bi-Gaussian.

(p.267) The projection integrals in such correlated bi-Gaussians can be done by simply completing the squares in the exponential arguments:

Appendix A: selected problem solutions
with
Appendix A: selected problem solutions
with
Appendix A: selected problem solutions
For more information on such Twiss parameter transformations, see the solution to Exercise 5.9.

Exercise 5.9

Equation (5.32) gives a relationship between the Twiss parameters,

Appendix A: selected problem solutions
Given the trace space vector transformation
Appendix A: selected problem solutions
we can also write the inverse transformation
Appendix A: selected problem solutions
We can thus write out our first equation in terms of initial Twiss parameters and final phase space variables as
Appendix A: selected problem solutions
We next need to collect terms proportional to x f 2 , x f x f and x f 2 , to give
Appendix A: selected problem solutions
Identifying the coefficients of the terms with the appropriate Twiss parameters, we can write
Appendix A: selected problem solutions
as desired.

(p.268) Exercise 5.12

Equation (5.48) is

Appendix A: selected problem solutions
The envelope equation is
Appendix A: selected problem solutions
The following is a general method for the solution for the force-free case:
Appendix A: selected problem solutions
and the first integral is
Appendix A: selected problem solutions
where we have substituted the waist condition. The final integration is
Appendix A: selected problem solutions
which can be written as
Appendix A: selected problem solutions
as expected from the matrix treatment.

Exercise 5.14

Equation (5.67) is

Appendix A: selected problem solutions
where εn = const., if conserved
Appendix A: selected problem solutions

For linear transformations, x = k x 2 x , and the right-hand side of the equation is

Appendix A: selected problem solutions
so
Appendix A: selected problem solutions

(p.269) Exercise 5.17

We have

Appendix A: selected problem solutions
where
Appendix A: selected problem solutions

The following relation must be fulfilled

Appendix A: selected problem solutions
and so
Appendix A: selected problem solutions
and
Appendix A: selected problem solutions

Exercise 5.17

(a) The path length through all four magnets in the chicane is

Appendix A: selected problem solutions
As an explicit function of the radius of curvature R, the path length is
Appendix A: selected problem solutions
(b) The differential of the path length with respect to the radius of curvature is
Appendix A: selected problem solutions
Therefore, we have
Appendix A: selected problem solutions
and
Appendix A: selected problem solutions
(p.270) or
Appendix A: selected problem solutions

Exercise 5.20

For a tightly bunched beam, the distribution is bi-Gaussian, and the emittance is given by

Appendix A: selected problem solutions
where
Appendix A: selected problem solutions
Thus, we have
Appendix A: selected problem solutions

Exercise 5.23

The given conditions are σζ = 1 mm and λrf = 0.105, so σφ = k zσζ = 2π. 0.001/0.105 = 0.060, and φ0 = 13π/36. (a) The longitudinal Twiss parameters are

Appendix A: selected problem solutions
(b) The needed value of the matrix element in question is calculated as
Appendix A: selected problem solutions
The radius of curvature of the 20MeV beam in a 0.3 T magnetic field is R = 22.2 cm. From Exercise 3.19, we have derived that
Appendix A: selected problem solutions
Solving for the bend angle, we have θ = 0.65 = 37°.

(p.271) Exercise 5.27

(a) The square of the J matrix is given by

Appendix A: selected problem solutions
(b) The quantity
Appendix A: selected problem solutions
(c) From part (b), the mapping of a matrix applied n times can be written compactly as
Appendix A: selected problem solutions
This should not be surprising, given that μ is the phase advance per period.

Chapter 6

Exercise 6.2

(a) We know that the trajectory is

Appendix A: selected problem solutions
and the tangent unit vector is
Appendix A: selected problem solutions
Thus, the trajectory tangent vector unit vector is
Appendix A: selected problem solutions
Using Eq. (4.16)
Appendix A: selected problem solutions
(p.272) and
Appendix A: selected problem solutions
since sin is an odd function and ( a 2 + z 2 ) 3 is an even function. The integral of an odd function over a symmetric integral is 0.
Appendix A: selected problem solutions
where K −2(k u a) is the modified Bessel function.
Appendix A: selected problem solutions
Putting these expressions together, we obtain
Appendix A: selected problem solutions
is the longitudinal field.

(b) The return winding (of the opposite polarity) serves to cancel the net axial current in the device. If the helical undulator consisted of only one winding it would have, like a single wire, a field gradient that points away from the wire. This gradient serves either repel or attract the beam (like current will attract), and so the design trajectory is not stable.

(p.273) Exercise 6.4

(a) In general the superposition of multipoles in the magnetic field is give by

Appendix A: selected problem solutions
with the additional restriction Δ ˙ B ˙ = 0 . Therefore we have
Appendix A: selected problem solutions
and
Appendix A: selected problem solutions
The induced voltage is
Appendix A: selected problem solutions
Using Φ = ω
Appendix A: selected problem solutions
and the voltage
Appendix A: selected problem solutions
The signal due to the j-th multipole contributes a component at j times the rotation frequency.

(b) If one was to place the coil at some position offset from the axis of a device one would measure the fields produced by that device and also the lower order fields. For example, if one placed the coil correctly inside of a pure quadrupole, one would only see the quadrupole field. If you didn’t place the coil correctly inside the quadrupole you would also measure a dipole field.

Exercise 6.5

We know from Eq. (2.39)

Appendix A: selected problem solutions
We are given that b 3 = 10T/m2 and the pole tip radius a = 5 cm, and so ψ = b 3 a 3 sin(3ɸ).

For a sextupole ɸ = 30° and therefore ψ = b 3 a 3.

From Eq. (6.20) Δψ = I encμ0 we get 2ψ = 2I poleμ0 or ψ/μ 0 = I pole, and

Appendix A: selected problem solutions

(p.274) Exercise 6.6

  1. (a) Since the H-magnet has a winding around each leg of the yolk and magnetic fields are vector fields that add linearly, the H-magnet gives twice as much field as the C-magnet.

  2. (b) If one of the windings is turned off the magnet does not behave like a C-magnet. The flux produced by the remaining winding that is still on is free to travel through the low reluctance iron in the opposite yoke. Therefore one expects very little the magnetic field produced in the gap in this case.

Exercise 6.8

Given that B varies sinusoidally from 0.1 to 1 T

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(a)
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(b) For ohmic losses,
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From the information given above we know B gap = 0.55 T + 0.45 T sin(ωt). We also know that μ/μ0 = 104 in the iron and so

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We want to find E induction :
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and
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Time averaging, <cos2(ωt)> = 0.5, and using σc = 1.03 × 103 m/Ω we have
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(p.275) Chapter 7

Exercise 7.2

Each scalar component of the electromagnetic wave (e.g. u = E x), obeys the wave equation,

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For the spherically symmetric case, the Laplacian operator takes a simple form that can be substituted to obtain,
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The function u = u 0 exp[ik(rct)]/r has first derivative
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and thus
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and with ω2 = k 2 c 2,
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The condition r ^ × E = 0 can be obtained from Gauss’ law by noting that
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But, since the transverse derivatives of the field components vanish, E = 0 , and E r=0, and r ^ × E = 0 . Note that this is a general condition, applying to any case (such as plane waves) with uniform phase fronts.

Exercise 7.3

(a) In S-band, we have

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(b) For the λ = 10.6 μm case (4 orders of magnitude smaller than S-band)
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(c) ωε/σ c = 1 for λ = 0.4 Å, which is in the X-ray regime. For this case, the Drude model is completely invalid, and the radiation frequency is in fact well above the metal’s plasma frequency.

(p.276) Exercise 7.4

If one begins with, νɸνg = c 2, it helps to first write it explicitly as

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Integrating once, we obtain

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which is the form we desire.

Exercise 7.7

(a) With a line charge charge density λe, the radial electric field is easily found by Gauss’ law with cylindrical symmetry,

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The total line charge density on the outside conductor’s inner surface (ρ = b) must be equal to –λe in order to cancel the field for ρ > b.

(b) The surface currents that are found on the conductors are

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The magnetic field that is associated with these currents is contained between the conductors and is, from Ampere’s law,
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The electric and magnetic fields have the same functional form, and their ratio is a constant.

(c) The wave impedance is defined to be this constant, which is

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as expected.

Exercise 7.8

For 500 MHz waves in copper, one has the attenuation constant of the coaxial line (2 mm, 5 mm inner/outer radius, κe = 2.25)

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In a 100-m run of this coaxial line, the attenuation of the power is exp(–0.44) = 0.65. The voltage on the line is only attenuated by the square root of this factor, or 0.80.

(p.277) Exercise 7.11

We begin with the stored energy, in the rectangular cavity, for fixed L z

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We have an additional constraint, which is that the frequency be held constant as the geometry is varied,
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Without loss of generality, we may choose n = 0 (this is the obvious choice for an accelerating cavity), and l = m = 1,
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We now introduce the aspect ratio g = L y/L x and write the stored energy explicitly including the constraint of constant frequency
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We can vary the stored energy with respect to choice of g to find the minimum at
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This expression has as a solution g = 1. One should also check the second derivative to make sure that it is a minimum condition, not maximum.

Exercise 7.14

(a) Inclusion the azimuthal derivatives in the Laplacian operator in Eq. (7.1) (the wave equation in vacuum), gives

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(b) Separation of variables by the substitution E z = R(ρ) Φ (ɸ)Z(z) gives three equations:
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(p.278) and
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The solutions to the first equation are simple harmonic exp(ik z,n z), with k z,n determined from conducting boundaries in z.

The second equation has solutions of the form exp(i), and are periodic in ɸ, which gives the eigenvalue requirement that m is an integer.

The final equation is a Bessel equation, which when m is not equal to zero admits solutions that do not necessarily have vanishing derivative (maximum or minimum) at ρ = 0. The value of m gives the form of the radial solution, as the Bessel function J m(k r r). The order of the Bessel function is specified by m, but the argument is given by the factor multiplying the final term in the operator,

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In practice, the radial wavenumber is determined by the transverse boundary conditions—metallic pipe radius in the simplest case, where the /th zero of the Bessel function must occur at the boundary R w, and k r,lm = j 0l/R w. Thus, the frequency of a given mode is given by
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(c) In Eq. (4.74), we are considering only speed-of-light modes, in which case k r 2 = ( ω 2 / c 2 ) k z , n 2 0 . Thus, the Bessel functions all take on a limiting form as their arguments approach zero, i.e. J 0 ⇒ 1, J 1 ∝ ρ, and the takes on the form of powers in ρ.

Exercise 7.16

The dispersion relation for such a cavity would be given by

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where both k r 2 and k z 2 are non-vanishing, and positive (the radial wavenumber must always be positive to satisfy the boundary condition at cavity edge). Therefore, the phase velocity of the forward wave in such a cavity could not be made to equal the speed of light—it in fact must exceed the speed of light. Lining a simple cavity such as this with dielectric can slow the wave down (see Ex. 7.25).

Exercise 7.18

In a single cavity with are many resonances having distinct resonant frequencies ω0, corresponding different values of lumped capacitance C and inductance L arise from different patterns of surface current and charge density associated with each mode. Thus each mode has a distinct “circuit” based on a different physical configuration of the sources of electric and magnetic fields.

(p.279) Exercise 7.21

The VSWR is defined as

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The voltage at a given point in a waveguide is given by
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where V F and V r are complex voltage amplitudes.

The rms power voltage is obtained by taking the square average of this function, and averaging in time.

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This is a sinusoidal function of z. The time-averaged power as a function of z will be proportional to the rms voltage squared. (b) From part (a), the rms voltage has minimum value of
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and a maximum value of
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Thus, the VSWR is given by the ratio of the maximum to minimum rms voltage

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Determination of maximum and minimum voltage points is sufficient to determine the VSWR.

Exercise 7.24

The frequency of the nth mode is given by

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Thus, the 12th, or π-mode, frequency is
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while the nearest neighbor mode (11th) is given by
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We require a mode separation five times the resonance bandwidth, 5ωπ/Q = ωπ/1200, so the minimum allowable coupling constant κc is given by the condition
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(p.280) Chapter 8

Exercise 8.2

Using the convention

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the transformation becomes:
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At point z:
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The matrix for the in-plane bend is
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Now lets look at the out-of-plane bend. We need to flip s p , by employing

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The bend matrix (M ipb) is the same. (a) Using this we obtain the matrix for the out of plane bend
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(b) All drifts are given by the matrix
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where
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Using the above definitions the total transformation is
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(p.281) Exercise 8.5

Given R 0 = 50 × 10−6m, n 2 = 20m −2, n 0 = 1.5. Start with Eq. (8.20):

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with
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This has the solution
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In the paraxial approximation:
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r max is achieved when sin ( κ f z ) = 1 ; r max = κ f R 0 = 1.825 × 10 4 rad, or ψ ≅ 0.02°.

Exercise 8.6

The transformation matrix is MT = M f,2 M 0 M f,1

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The stability condition is |Tr(MT)| ≤ 2
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Now let y = L/R 1 and x = R 2/R 1. The stability condition becomes:
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Now let us plot for both cases y(x)/f(x) versus x: Case I: defocusing
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(p.282) Case 2: focusing

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In both cases the stable region is found above the x-axis and below the curve.

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Exercise 8.8

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First let us rewrite u(x, y, z) in terms of q(z)
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but
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and with ρ2 = x 2 + y 2
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Then, writing this in cylindrical coordinates
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or
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(p.283) With
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or
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we obtain
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Combination of these expressions yields
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Thus, u(x, y, z) is a solution to the paraxial wave equation.

Exercise 8.12

(a) We want to find the ratio of the beam size at the mirrors to that at the waist.

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(b) We are given: R = 50 cm =. 5 m; λ = 1 × 10−6 m; R M = 1 in. = 2.54 cm = 0.0254 m; ω(L/2) = (1/4)d M = (l/2)R M = 0.0127 m.

(p.284) We know:

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So we may deduce that
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Rearranging this expression we obtain
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Let us define
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for simplicity
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So either l = 0 (trivial case) or, more relevantly, l = R/(1 + C 2).

Finally, we may calculate

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Exercise 8.13

We know:

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(p.285) (a) Power:
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Let v = p 2 and dv = 2p dp:
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So, we have
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or simply
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(b)
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Q grows very quickly with the mirror size.

Exercise 8.18

A two-level quantum system will not work as a laser. Since there are only two levels you cannot maintain population inversion. By the same considerations, a four-level quantum system works better than a three-level quantum system. The four-level quantum system has the extra level above ground state. Since the rate from the third level to the second level is very slow (this is the lasing transition) and the rate from the second level to ground is very fast the population of the second level is essentially zero. This helps when considering the single pass gain for the laser. Since the population of the level is approximately zero you do not have to consider it when calculating the rate equations.

Exercise 8.23

(a) The four-vector is

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We are given ω = 0, k z = k u.

(p.286) So the four-vector becomes

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The Lorentz transformation is
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In the rest frame of the electron
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Therefore,
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(b) The magnetic field is B = B 0 sin ( k u z ) y . The vector potential is
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The four-vector potential is
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A is a constant under Lorentz transformations.

But z transforms as z = γ(z′ + β0 ct′). This is a wave traveling in the z ^ direction. The field is then

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If β0 ≈ 1, these two equations define an EM wave. (p.287) (c) The four-vector is
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As in (a) the transformation to the rest frame is k ˜ = L k ˜ :
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The energy of the Compton scattering is conserved
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So β0 k u = k(1 – β0 cosθ). In this case β0 ≈ 1.

Therefore, to first order,

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Finally, we have
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Exercise 8.24

We start with the Hamiltonian H = ( p e A ) 2 c 2 + ω 2 c 2 , where

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As
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we can set p x = 0, otherwise there would be off-axis motion and as p ˙ y = 0 and y ˙ = 0 , again we can set p y = 0, y = 0.

Energy is conserved, so we have

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or
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Resonance then requires
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be equal to v z, which gives us
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(p.288) Exercise 8.25

(a)

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(b)
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since in this case σx = σy. (c)
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The growth in the beam size is given by
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