## 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

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

In Cartesian coordinates, $x → = x x ^ + y y ^ + z z ^$ Therefore

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

The Hamiltonian is
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,
where $H x = ( p c , x 2 / 2 m ) + 1 2 k x 2$ etc.

(c) In cylindrical coordinates

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
so the Lagrangian is
or

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

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
or in terms of canonical variables,
From inspection of the Hamiltonian,
is a constant of motion.

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

where
and the Lagrangian is
or

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

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
From inspection of the Hamiltonian,
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

The Hamiltonian is

(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,

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.

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

The Hamiltonian is
In an octupole-like field, the equation for plotting the phase space becomes
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)

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.

## Exercise 1.5

The Taylor series expansion for a small x is

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
and
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

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

In cylindrical coordinates,

(b) Write the Lagrangian in cylindrical coordinates:

where

(c) Lagrange–Euler equations,

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:

and
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 (radius of curvature), we have

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

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

Using the definition of the average magnetic field given in the problem
we obtain that . 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)

Take the time derivative
Using the paraxial approximation:
With this expression we obtain

(a) Integrate this equation:

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

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

Here, C 2 is a constant of integration but since γ(0) = 1 and v z(0) = 0, then C 2 = 0, and
Using the paraxial approximation,
(p.249) we have
Solving for γ we obtain
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:

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

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

In this case we set the skew term b 2 = 0, and keep the normal term (a 2 > 0)
The force is given by$F ¯ ⊥ = q E$.
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:

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

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

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

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

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
and the centripetal force
The total force is
At equilibrium:
or
Here we have used
In more standard form, we have

(p.252)

(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

The vector potential in cylindrical coordinates is
where I v (x) are the modified Bessel functions which obey he following relations:
Let us use some abbreviations to simplify notation
(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):
So from $∇ → ( ∇ → . A → ) = ∇ → 2 A →$ we can conclude that $∇ → 2 A → = 0$. (b) Next we calculate the field components in cylindrical coordinates:
Let us calculate each term individually. The ρ-component is:
The ɸ-component is:
The z-component is:
(c) First we can use small angle approximations for the modified Bessel functions since k uρ ≪ 1.
In the first-order approximation these become
(p.254) We need to transform the vector potential from cylindrical coordinates to Cartesian coordinates:
The Hamiltonian for this system is given by
The equations of motion can be derived from:
The canonical momenta are:
The Hamiltonian is constant and H= γmc 2. since $p ˙ z = 0$, we have p z=γmβ z c and formally
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:
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.

Combining this and the previous expressions for $x ˙$ and$y ˙$, we obtain two expressions for $φ ˙$.
The left-hand side of these equations can be written as total derivatives of sine and cosine functions:
Recall that in the paraxial approximation, we have
Using this we obtain,
which can be integrated to find
Combining these expressions we have,
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,

and assuming that
To calculate the field index, n, required to guarantee that the betatron beam obeys x rms = 2y rms, start with:
The tunes are and$ν y = n$.

Following the given requirements with the above relations:

and
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

(b) To ensure linear stability, |TrM T| ≤ 2 giving in this case
which is simply

## (p.257) Exercise 3.12

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

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

The Fourier series is useful:

(p.258) The coefficients are:
and so
(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
Using bn = 2/nπ and kp = 2π/Lp = π/l,
or
(e) Solutions to (b) and (d) are identical.

## Exercise 3.15

In the smooth approximation,

Referring to the evolution of horizontal dispersion, Eq. (3.65),
Solutions are given by Eq. (3.66),
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,
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

## 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

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

# Chapter 4

## Exercise 4.1

The drift tube linac has an electric field profile given by

(a) Since the system is periodic,
The coefficients a n are given by
Thus,
(b) Assuming the particle is synchronous with the first harmonic (n = 1), the particle velocity can be written as
(p.261) since k z,n = 2nπ/d. The relationship between the period and the frequency is,
(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

Then the velocity can be found from
Using the results from part (b) we obtain:

## Exercise 4.2

Evaluate αrf:

(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

and using the paraxial approximation vv z:
We can express γ in terms of the energy:
Now evaluate αrf:
(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

we find
For a proton linac αrf:

## Exercise 4.3

The Hamiltonian for this problem is given by

where
(a) For optimum capture, in the final state ɸf = π/2, and v zc, so χ−1 ⇒ 0, and
The Hamiltonian is a constant of motion, so in the initial state we have
The minimum value of αrf which allows solution of this expression is clearly 1. (b) From the above final state Hamiltonian, we have
(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,

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

First-order analysis

The parameters given are
and
Both of these methods yield the same result.

## Exercise 4.4

The parameters given in Exercise 4.2(b) are:

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

## Exercise 4.5

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

Let us now set

Integrating, we obtain
or
Thus,
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

# Chapter 5

## Exercise 5.1

Equation (5.9) is

Equation (5.11) is

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

or
In terms of T x

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

In terms of T x

## Exercise 5.6

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

Given the transformation
we can also write the inverse transformation
The final distribution function can be written as
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:

with
with
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,

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

## (p.268) Exercise 5.12

Equation (5.48) is

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

## Exercise 5.14

Equation (5.67) is

where εn = const., if conserved

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

so

## (p.269) Exercise 5.17

We have

where

The following relation must be fulfilled

and so
and

## Exercise 5.17

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

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

## Exercise 5.20

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

where
Thus, we have

## 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

(b) The needed value of the matrix element in question is calculated as
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
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

(b) The quantity
(c) From part (b), the mapping of a matrix applied n times can be written compactly as
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

and the tangent unit vector is
Thus, the trajectory tangent vector unit vector is
Using Eq. (4.16)
(p.272) and
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.
where K −2(k u a) is the modified Bessel function.
Putting these expressions together, we obtain
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

with the additional restriction $Δ ˙ ⋅ B ˙ = 0$. Therefore we have
and
The induced voltage is
Using Φ = ω
and the voltage
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)

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

## (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

(a)
(b) For ohmic losses,

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

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

# (p.275) Chapter 7

## Exercise 7.2

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

For the spherically symmetric case, the Laplacian operator takes a simple form that can be substituted to obtain,
The function u = u 0 exp[ik(rct)]/r has first derivative
and thus
and with ω2 = k 2 c 2,
The condition $r ^ × E → = 0$ can be obtained from Gauss’ law by noting that

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

(b) For the λ = 10.6 μm case (4 orders of magnitude smaller than S-band)
(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

Integrating once, we obtain

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,

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

The magnetic field that is associated with these currents is contained between the conductors and is, from Ampere’s law,

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

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)

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

We have an additional constraint, which is that the frequency be held constant as the geometry is varied,
Without loss of generality, we may choose n = 0 (this is the obvious choice for an accelerating cavity), and l = m = 1,
We now introduce the aspect ratio g = L y/L x and write the stored energy explicitly including the constraint of constant frequency
We can vary the stored energy with respect to choice of g to find the minimum at
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

(b) Separation of variables by the substitution E z = R(ρ) Φ (ɸ)Z(z) gives three equations:
(p.278) and
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,

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
(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

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

The voltage at a given point in a waveguide is given by
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.

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
and a maximum value of

Thus, the VSWR is given by the ratio of the maximum to minimum rms voltage

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

Thus, the 12th, or π-mode, frequency is
while the nearest neighbor mode (11th) is given by
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

# (p.280) Chapter 8

## Exercise 8.2

Using the convention

the transformation becomes:
At point z:
The matrix for the in-plane bend is

Now lets look at the out-of-plane bend. We need to flip $s → → p →$, by employing

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

## (p.281) Exercise 8.5

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

with
This has the solution
In the paraxial approximation:
$r ′ max$ 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

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

(p.282) Case 2: focusing

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

## Exercise 8.8

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

(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:

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

Finally, we may calculate

## Exercise 8.13

We know:

(p.285) (a) Power:
Let v = p 2 and dv = 2p dp:
So, we have
or simply
(b)
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

We are given ω = 0, k z = k u.

(p.286) So the four-vector becomes

The Lorentz transformation is
In the rest frame of the electron
Therefore,
(b) The magnetic field is $B → = B 0 sin ⁡ ( k u z ) ⋅ y →$. The vector potential is
The four-vector potential is
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

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

Therefore, to first order,

Finally, we have

## Exercise 8.24

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

As
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

or
Resonance then requires
be equal to v z, which gives us

## (p.288) Exercise 8.25

(a)

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