Advanced Topics

Abstract

In this chapter, some more-advanced aspects of longitudinal beam manipulations in synchrotrons are discussed.

These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

In this chapter, some more-advanced aspects of longitudinal beam manipulations in synchrotrons are discussed.

5.1 Different Phase Space Descriptions

In this section, we discuss alternatives to the longitudinal phase space $(\Delta t,\Delta W)$.

5.1.1 Phase Space $(\varphi,\delta )$

In Chap. 3, we carefully derived the tracking equations (3.8),

$$\displaystyle{\Delta \gamma _{n} = \Delta \gamma _{n-1} + \frac{Q} {m_{0}c_{0}^{2}}\Delta V _{n-1},}$$

and (3.14),

$$\displaystyle{\Delta t_{n} = \Delta t_{n-1} + \frac{l_{\mathrm{R}}\eta _{\mathrm{R},n}} {\beta _{n}\beta _{\mathrm{R},n}^{2}c_{0}} \frac{\Delta \gamma _{n}} {\gamma _{\mathrm{R},n}}.}$$

This allowed us to show that the phase space area measured in eVs is invariant. These tracking equations are, of course, not the only possible ones. For example, we may use the approximation

$$\displaystyle{ \delta:= \frac{\Delta p} {p_{\mathrm{R}}} \approx \frac{1} {\beta _{\mathrm{R}}^{2}} \frac{\Delta \gamma } {\gamma _{\mathrm{R}}} }$$

(5.1)

to convert the first equation into

$$\displaystyle{\delta _{n} =\delta _{n-1} + \frac{Q} {\beta _{\mathrm{R}}^{2}W_{\mathrm{R}}}\Delta V _{n-1}.}$$

For harmonic gap voltages, we get

$$\displaystyle{ \delta _{n} =\delta _{n-1} + \frac{Q\hat{V }} {\beta _{\mathrm{R}}^{2}W_{\mathrm{R}}}\left (\sin \varphi _{\mathrm{RF},n-1} -\sin \varphi _{\mathrm{R}}\right ). }$$

(5.2)

Here we assumed that $\beta _{\mathrm{R}}^{2}\gamma _{\mathrm{R}}$ does not change significantly from revolution n − 1 to revolution n. We may also multiply the second equation by

$$\displaystyle{\omega _{\mathrm{RF}} = 2\pi hf_{R} = 2\pi h\frac{c_{0}\beta _{\mathrm{R}}} {l_{\mathrm{R}}} }$$

to get

$$\displaystyle{\Delta \varphi _{\mathrm{RF},n} = \Delta \varphi _{\mathrm{RF},n-1} + 2\pi h \frac{\eta _{\mathrm{R},n}} {\beta _{n}\beta _{\mathrm{R},n}} \frac{\Delta \gamma _{n}} {\gamma _{\mathrm{R},n}}.}$$

Here we assumed that the RF frequency does not change significantly from revolution n − 1 to revolution n. If one also assumes that $\varphi _{\mathrm{R},n}$ does not differ significantly from $\varphi _{\mathrm{R},n-1}$, one gets

$$\displaystyle{\varphi _{\mathrm{RF},n} =\varphi _{\mathrm{RF},n-1} + 2\pi h \frac{\eta _{\mathrm{R},n}} {\beta _{n}\beta _{\mathrm{R},n}} \frac{\Delta \gamma _{n}} {\gamma _{\mathrm{R},n}},}$$

since

$$\displaystyle{\varphi _{\mathrm{RF},n} =\varphi _{\mathrm{R},n} + \Delta \varphi _{\mathrm{RF},n}}$$

holds. If we furthermore make use of the approximation (5.1), we get

$$\displaystyle{ \varphi _{\mathrm{RF},n} =\varphi _{\mathrm{RF},n-1} + 2\pi h\eta _{\mathrm{R}}\delta _{n}. }$$

(5.3)

Equations (5.2) and (5.3) may also be found, for example, as Eq. (3.28) in the textbook by Lee [1]. They obviously use the phase space $(\varphi _{\mathrm{RF}},\delta )$ instead of our original phase space $(\Delta t,\Delta W)$. If one calculates the Jacobian

$$\displaystyle{ \frac{\partial (\delta _{n},\varphi _{\mathrm{RF},n})} {\partial (\delta _{n-1},\varphi _{\mathrm{RF},n-1})},}$$

one finds that it equals 1. Therefore, the modified tracking equations preserve the phase space area. It should be clear, however, that this is an artifact caused by our sloppy derivation of the modified tracking equations. We know that the phase space $(\Delta t,\Delta W)$ leads to area invariance on a long-term basis. Due to the approximations we used to derive the tracking equations for the phase space $(\varphi _{\mathrm{RF}},\delta )$, we cannot be sure that these equations are still exact. Although the approximations are valid with a certain precision in each tracking step, they may lead to large deviations on a long-term basis. This should not astonish the reader, since we already pointed out previously that making the tracking equations more symmetric also destroys area preservation, although this modification is negligible in each step. There is no canonical transformation that converts the phase space coordinates $(\Delta t,\Delta W)$ into the phase space coordinates $(\varphi _{\mathrm{RF}},\delta )$ or vice versa.

Of course, one may analyze the phase space $(\varphi _{\mathrm{RF}},\delta )$ in the same way as we did for the phase space $(\Delta t,\Delta W)$. A Hamiltonian may be derived, and the bucket height or the bucket area can be calculated.

As an example, we consider the ratio of the principal axes. According to Eq. (3.28), we have

$$\displaystyle{\frac{\Delta \hat{W}} {\Delta \hat{t}} = f_{\mathrm{S,0}}\;\frac{2\pi W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} {\vert \eta _{\mathrm{R}}\vert }.}$$

If we use the momentum spread

$$\displaystyle{\delta = \frac{\Delta p} {p_{\mathrm{R}}} \approx \frac{1} {\beta _{\mathrm{R}}^{2}} \frac{\Delta W} {W_{\mathrm{R}}} }$$

instead of $\Delta W$ and the RF phase deviation

$$\displaystyle{\Delta \varphi _{\mathrm{RF}} =\omega _{\mathrm{RF}}\Delta t}$$

instead of $\Delta t$, we obtain

$$\displaystyle{ \frac{\hat{\delta }} {\Delta \hat{\varphi }_{\mathrm{RF}}} = f_{\mathrm{S,0}}\; \frac{2\pi } {\vert \eta _{\mathrm{R}}\vert \;\omega _{\mathrm{RF}}} = \frac{1} {\vert \eta _{\mathrm{R}}\vert h} \frac{f_{\mathrm{S,0}}} {f_{\mathrm{R}}}.}$$

This corresponds to Eq. (3.55) in Lee [1].

Since the phase space coordinates $(\varphi _{\mathrm{RF}},\delta )$ lead to an area preservation that is not justified from a physical point of view, we will not use this phase space in most parts of this book.

5.1.2 Relation to Phase Space ($\Delta t,\Delta W$)

Now we determine the bucket area for the phase space ($\Delta \varphi _{\mathrm{RF}},\delta$). For this purpose, we consider the variables

$$\displaystyle\begin{array}{rcl} Q = \Delta t,& \mbox{ }& P = \Delta W, {}\\ q = \Delta \varphi _{\mathrm{RF}},& \mbox{ }& p =\delta = \frac{\Delta p} {p_{\mathrm{R}}}. {}\\ \end{array}$$

This leads to

$$\displaystyle{ \frac{q} {Q} = \frac{\Delta \varphi _{\mathrm{RF}}} {\Delta t} =\omega _{\mathrm{RF}} = \frac{2\pi h} {T_{\mathrm{R}}} }$$

(5.4)

and

$$\displaystyle{ \frac{p} {P} = \frac{\delta } {\Delta W} = \frac{1} {\beta _{\mathrm{R}}^{2}m_{0}c_{0}^{2}\gamma _{\mathrm{R}}}. }$$

(5.5)

For the first of these two equations, one has to take into account that $\Delta \varphi _{\mathrm{RF}}$ is the RF phase. If this varies from −π to π, the variable $\Delta t$ will move through only one bucket with the time span T_R∕h, not through the whole circumference equivalent to T_R.

For deriving the second equation, we made use of the relation

$$\displaystyle{\frac{\Delta W} {W_{\mathrm{R}}} = \frac{\Delta \gamma } {\gamma _{\mathrm{R}}} \approx \beta _{\mathrm{R}}^{2}\frac{\Delta p} {p_{\mathrm{R}}}.}$$

For the bucket area, we have

$$\displaystyle{A_{\mathrm{B}}^{\Delta \varphi _{\mathrm{RF}},\delta } =\int \int \mathrm{ d}q\;\mathrm{d}p}$$

and

$$\displaystyle{A_{\mathrm{B}}^{\Delta t,\Delta W} =\int \int \mathrm{ d}Q\;\mathrm{d}P,}$$

respectively. The transformation law is

$$\displaystyle{A_{\mathrm{B}}^{\Delta \varphi _{\mathrm{RF}},\delta } =\int \int \left \vert \begin{array}{cc} \frac{\partial q} {\partial Q} & \frac{\partial q} {\partial P} \\ \frac{\partial p} {\partial Q} & \frac{\partial p} {\partial P} \end{array} \right \vert \;\mathrm{d}Q\;\mathrm{d}P,}$$

the Jacobian equals

$$\displaystyle{\xi = \left \vert \begin{array}{cc} \frac{\partial q} {\partial Q} & \frac{\partial q} {\partial P} \\ \frac{\partial p} {\partial Q} & \frac{\partial p} {\partial P} \end{array} \right \vert = \frac{2\pi h} {T_{\mathrm{R}}\beta _{\mathrm{R}}^{2}m_{0}c_{0}^{2}\gamma _{\mathrm{R}}} = \frac{2\pi h} {T_{\mathrm{R}}\beta _{\mathrm{R}}^{2}W_{\mathrm{R}}},}$$

as Eqs. (5.4) and (5.5) show. Please note that this factor ξ changes if an acceleration takes place. The transformation from q, p to Q, P does not correspond to a canonical transformation. The factor ξ does not depend on the integration variables, so that we may take it out of the integral:

$$\displaystyle{A_{\mathrm{B}}^{\Delta \varphi _{\mathrm{RF}},\delta } =\xi \; A_{\mathrm{ B}}^{\Delta t,\Delta W}.}$$

Therefore, Eq. (3.41),

$$\displaystyle{A_{\mathrm{B}}^{\Delta t,\Delta W} = \frac{4\sqrt{2}} {\pi h} T_{\mathrm{R}}\;\sqrt{\frac{W_{\mathrm{R} } \beta _{\mathrm{R} }^{2 }\vert Q\vert \hat{V }} {\pi h\vert \eta _{\mathrm{R}}\vert }} \;\alpha (\varphi _{\mathrm{R}}),}$$

leads to

$$\displaystyle{A_{\mathrm{B}}^{\Delta \varphi _{\mathrm{RF}},\delta } = 8\sqrt{2}\;\sqrt{ \frac{\vert Q\vert \hat{V }} {\pi h\vert \eta _{\mathrm{R}}\vert W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}}}\;\alpha (\varphi _{\mathrm{R}})}$$

for the bucket area of a single bucket. By means of Eq. (5.5), we may convert the bucket height in Eq. (3.32),

$$\displaystyle{\Delta W_{\mathrm{max,stat}} = \sqrt{\frac{2\;W_{\mathrm{R} } \beta _{\mathrm{R} }^{2 }\;\vert Q\vert \;\hat{V }} {\pi h\;\vert \eta _{\mathrm{R}}\vert }},}$$

into the bucket height

$$\displaystyle{\delta _{\mathrm{max,stat}} = \Delta W_{\mathrm{max,stat}}\; \frac{1} {W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} = \sqrt{ \frac{2\;\vert Q\vert \hat{V }} {\pi h\vert \eta _{\mathrm{R}}\vert W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}}}}$$

for the momentum spread. For an accelerated bucket, we have

$$\displaystyle{\delta _{\mathrm{max}} =\delta _{\mathrm{max,stat}}Y (\varphi _{\mathrm{R}}).}$$

5.1.3 Scale Transformation with Invariant Bucket Area

If we transform the original coordinate/momentum pair (q, p) into a new one (Q, P), and if Q and P depend on q and p only via constant factors, then the bucket area changes by the factor

$$\displaystyle{\xi = \left \vert \begin{array}{cc} \frac{\partial Q} {\partial q} &\frac{\partial Q} {\partial p} \\ \frac{\partial P} {\partial q} &\frac{\partial P} {\partial p} \end{array} \right \vert,}$$

as shown above. If we furthermore assume that Q depends only on q, and P only on p, then

$$\displaystyle{\xi = \frac{\partial Q} {\partial q} \frac{\partial P} {\partial p} }$$

holds. If the bucket area is to remain unchanged under the transformation, the condition ξ = 1 is required. Therefore, if one variable p is multiplied by a factor f (P = fp), the other variable q must be divided by that factor f ($Q = q/f$) in order to preserve the bucket area.

Let us, for example, begin with the phase space $(\Delta t,\Delta W)$ and multiply the first variable $\Delta t$ by ω_RF in order to obtain $\Delta \varphi _{\mathrm{RF}}$. Then $\Delta W$ must be divided by ω_RF. Hence, the coordinates $(\Delta \varphi _{\mathrm{RF}},\Delta W/\omega _{\mathrm{RF}})$ lead to the same bucket area as the coordinates ($\Delta t,\Delta W$). If instead of the RF phase $\Delta \varphi _{\mathrm{RF}}$, one considers the angle θ of the whole accelerator ring, i.e.,

$$\displaystyle{\Delta \theta = \Delta \varphi _{\mathrm{RF}}/h,}$$

then the pair $(\Delta \theta,\Delta W/\omega _{\mathrm{R}})$ is obtained as another alternative.

The transition to the new coordinates may be regarded as a canonical transformation. Even though we did not introduce generating functions in the scope of this book, we now use F₃(p, Q). The reader may consult, for example, the book of Goldstein [2] for the definition and the properties of generating functions. We begin with

$$\displaystyle{q = \Delta t\mbox{, }p = \Delta W\mbox{, }Q = \Delta \varphi _{\mathrm{RF}},}$$

and due to

$$\displaystyle{q = -\frac{\partial F_{3}(p,Q)} {\partial p},}$$

we obtain

$$\displaystyle{\Delta t = -\frac{\partial F_{3}(\Delta W,\Delta \varphi _{\mathrm{RF}})} {\partial \Delta W}.}$$

This relation is obviously satisfied for

$$\displaystyle{F_{3} = -\Delta W\;\Delta \varphi _{\mathrm{RF}}\; \frac{1} {\omega _{\mathrm{RF}}},}$$

since $\Delta \varphi _{\mathrm{RF}} =\omega _{\mathrm{RF}}\Delta t$ holds. This allows us to determine the new generalized momentum variable P:

$$\displaystyle{P = -\frac{\partial F_{3}} {\partial Q} = - \frac{\partial F_{3}} {\partial \Delta \varphi _{\mathrm{RF}}} = \frac{\Delta W} {\omega _{\mathrm{RF}}}.}$$

As a result, we obtain the same pair of variables ($\Delta \varphi _{\mathrm{RF}},\Delta W/\omega _{\mathrm{RF}}$) as above.

Now we consider the case that the factor used for the scale transformation (ω_RF) may vary with time. Just in order to use a different generating function this time, we choose F₂(q, P, t). We set

$$\displaystyle{F_{2} = q\;P\;\lambda (t).}$$

Therefore, we obtain

$$\displaystyle{p = \frac{\partial F_{2}} {\partial q} = P \cdot \lambda (t)\mbox{ } \Rightarrow P = \frac{p} {\lambda (t)},}$$

$$\displaystyle{Q = \frac{\partial F_{2}} {\partial P} = q \cdot \lambda (t)\mbox{ } \Rightarrow Q = q \cdot \lambda (t).}$$

The new Hamiltonian is

$$\displaystyle{K = H + \frac{\partial F_{2}} {\partial t} = H + q\;P\; \frac{\mathrm{d}\lambda } {\mathrm{d}t} = H + QP\;\frac{1} {\lambda } \frac{\mathrm{d}\lambda } {\mathrm{d}t}.}$$

We see that only for constant λ is the same Hamiltonian K = H obtained.

In Schmutzer [3, volume I, p. 417], it is shown that for all canonical transformations, i.e., time-dependent ones, the Jacobian equals 1 in general. Therefore, the phase space area remains the same even though the Hamiltonian becomes time-dependent.

We finally check that the new function K actually is a Hamiltonian:

$$\displaystyle\begin{array}{rcl} \frac{\partial K} {\partial Q}& =& \frac{\partial H} {\partial q} \; \frac{\partial q} {\partial Q} + \frac{\partial H} {\partial p} \; \frac{\partial p} {\partial Q} + P \frac{\dot{\lambda }} {\lambda } = {}\\ & =& -\dot{p}\;\frac{1} {\lambda } +\dot{ q} \cdot 0 + P \frac{\dot{\lambda }} {\lambda } = {}\\ & =& -(\dot{\lambda }P +\lambda \dot{ P})\frac{1} {\lambda } + P \frac{\dot{\lambda }} {\lambda } = {}\\ & =& -\dot{P}, {}\\ \end{array}$$

$$\displaystyle\begin{array}{rcl} \frac{\partial K} {\partial P}& =& \frac{\partial H} {\partial q} \; \frac{\partial q} {\partial P} + \frac{\partial H} {\partial p} \; \frac{\partial p} {\partial P} + Q\frac{\dot{\lambda }} {\lambda } = {}\\ & =& -\dot{p} \cdot 0 +\dot{ q}\;\lambda + Q\frac{\dot{\lambda }} {\lambda } = {}\\ & =& \frac{\dot{Q}\lambda - Q\dot{\lambda }} {\lambda ^{2}} \;\lambda + Q\frac{\dot{\lambda }} {\lambda } = {}\\ & =& \dot{Q}. {}\\ \end{array}$$

5.2 Special Remarks on Linear ODEs of Second Order

Linear ODEs of second order with variable coefficients occur very often in mathematical physics. In the following, we will therefore discuss some of their properties.

5.2.1 Removing the Attenuation Term

As mentioned in Kamke [4, volume 1, Sect. 16.3], the second term of a linear ODE may be removed by means of a suitable transformation. For the special case

$$\displaystyle{ a_{2}(t)\ddot{y} + a_{1}(t)\;\dot{y} + a_{0}(t)\;y = 0, }$$

(5.6)

the ansatz

$$\displaystyle\begin{array}{rcl} y& =& uv, {}\\ \dot{y}& =& \dot{u}v + u\dot{v}, {}\\ \ddot{y}& =& \ddot{u}v + 2\dot{u}\dot{v} + u\ddot{v}, {}\\ \end{array}$$

leads to the ODE

$$\displaystyle{\ddot{u}(a_{2}v) +\dot{ u}(2a_{2}\dot{v} + a_{1}v) + u(a_{2}\ddot{v} + a_{1}\dot{v} + a_{0}v) = 0}$$

$$\displaystyle{\Rightarrow \ddot{ u} +\dot{ u}\left (2\frac{\dot{v}} {v} + \frac{a_{1}} {a_{2}}\right ) + u\left (\frac{\ddot{v}} {v} + \frac{a_{1}} {a_{2}} \frac{\dot{v}} {v} + \frac{a_{0}} {a_{2}}\right ) = 0.}$$

Here we make the second term vanish by setting

$$\displaystyle{\frac{\dot{v}} {v} = -\frac{a_{1}} {2a_{2}},}$$

which leads to

$$\displaystyle{\int \frac{\mathrm{d}v} {v} = -\frac{1} {2}\int \frac{a_{1}} {a_{2}}\mathrm{d}t}$$

$$\displaystyle{\Rightarrow \ln \vert v\vert = -\frac{1} {2}\int \frac{a_{1}} {a_{2}}\mathrm{d}t +\mathrm{ const}}$$

$$\displaystyle{\Rightarrow v = v_{0}\;\exp \left (-\frac{1} {2}\int \frac{a_{1}} {a_{2}}\mathrm{d}t\right ).}$$

This choice leads to the ODE

$$\displaystyle{\ddot{u} + u\left (\frac{\ddot{v}} {v} + \frac{a_{1}} {a_{2}} \frac{\dot{v}} {v} + \frac{a_{0}} {a_{2}}\right ) = 0,}$$

without attenuation term. Due to

$$\displaystyle\begin{array}{rcl} \dot{v}& =& -\frac{1} {2} \frac{a_{1}} {a_{2}}v, {}\\ \ddot{v}& =& -\frac{1} {2} \frac{\dot{a}_{1}a_{2} -\dot{ a}_{2}a_{1}} {a_{2}^{2}} v -\frac{1} {2} \frac{a_{1}} {a_{2}}\dot{v} = \left (-\frac{1} {2} \frac{\dot{a}_{1}} {a_{2}} + \frac{1} {2} \frac{\dot{a}_{2}} {a_{2}} \frac{a_{1}} {a_{2}} + \frac{1} {4} \frac{a_{1}^{2}} {a_{2}^{2}}\right )v, {}\\ \end{array}$$

one obtains

$$\displaystyle{ \ddot{u} + u\left (\frac{a_{0}} {a_{2}} -\frac{1} {2} \frac{\dot{a}_{1}} {a_{2}} + \frac{1} {2} \frac{\dot{a}_{2}} {a_{2}} \frac{a_{1}} {a_{2}} -\frac{1} {4} \frac{a_{1}^{2}} {a_{2}^{2}}\right ) = 0. }$$

(5.7)

In principle, the trick that was presented here that made the first-order derivative vanish cannot change the physical behavior of the system under consideration. The attenuation that was obviously present in the original ODE was just (partly) shifted into the function v(t). Furthermore, it is clear that the transformation will usually lead to a very complicated ODE of the type (5.7), which is no easier to solve analytically than the original ODE (5.6).

However, the mathematical trick showed us that it is very useful to analyze ODEs of the type

$$\displaystyle{\ddot{u} + K(t)\;u = 0,}$$

because it also allows statements about the original ODE (5.6).

5.2.2 Solution by Integration of the Phase

We begin with the homogeneous linear ODE

$$\displaystyle{ \ddot{u} + \Omega ^{2}(t)u = 0 }$$

(5.8)

of second order.¹ If $\Omega$ did not depend on time, we could use sine, cosine, or exponential functions as an ansatz for the solution. For a time-dependent $\Omega$, it is therefore straightforward to test the ansatz

$$\displaystyle{u(t) = u_{0}(t)\exp \left (j\int \Omega _{0}(t)\;\mathrm{d}t\right ),}$$

which leads to

$$\displaystyle\begin{array}{rcl} \dot{u}& =& (\dot{u}_{0} + j\Omega _{0}u_{0})e^{j\varphi } {}\\ \ddot{u}& =& (\ddot{u}_{0} + j\dot{\Omega }_{0}u_{0} + j\Omega _{0}\dot{u}_{0} + j\Omega _{0}(\dot{u}_{0} + j\Omega _{0}u_{0}))e^{j\varphi } = {}\\ & =& (\ddot{u}_{0} + j\dot{\Omega }_{0}u_{0} + 2j\Omega _{0}\dot{u}_{0} - \Omega _{0}^{2}u_{ 0})e^{j\varphi }. {}\\ \end{array}$$

Here we used

$$\displaystyle{\varphi (t) =\int \Omega _{0}(t)\;\mathrm{d}t.}$$

If we insert this into the ODE, we obtain

$$\displaystyle{\ddot{u}_{0} + (\Omega ^{2} - \Omega _{ 0}^{2})u_{ 0} = 0}$$

for the real part and

$$\displaystyle{\dot{\Omega }_{0}u_{0} + 2\Omega _{0}\dot{u}_{0} = 0}$$

for the imaginary part. According to

$$\displaystyle{\int \frac{\mathrm{d}u_{0}} {u_{0}} = -\frac{1} {2}\int \frac{\mathrm{d}\Omega _{0}} {\Omega _{0}},}$$

the latter may be separated, so that

$$\displaystyle{\ln \vert u_{0}\vert = -\frac{1} {2}\ln \vert \Omega _{0}\vert +\mathrm{ const}}$$

$$\displaystyle{\Rightarrow u_{0} = \frac{M} {\sqrt{\Omega _{0}}}}$$

is obtained. If $\Omega _{0}(t)$ is known, then u₀(t) is determined as well, and a solution u(t) of the ODE is found.

We obviously have

$$\displaystyle\begin{array}{rcl} \dot{u}_{0}& =& -\frac{M} {2} \Omega _{0}^{-3/2}\dot{\Omega }_{ 0}, {}\\ \ddot{u}_{0}& =& \frac{3M} {4} \Omega _{0}^{-5/2}\dot{\Omega }_{ 0}^{2} -\frac{M} {2} \Omega _{0}^{-3/2}\ddot{\Omega }_{ 0}, {}\\ \end{array}$$

which may be inserted into the ODE for u₀(t):

$$\displaystyle{\frac{3} {4}\Omega _{0}^{-5/2}\dot{\Omega }_{ 0}^{2} -\frac{1} {2}\Omega _{0}^{-3/2}\ddot{\Omega }_{ 0} + (\Omega ^{2} - \Omega _{ 0}^{2})\Omega _{ 0}^{-1/2} = 0}$$

$$\displaystyle{ \Rightarrow \Omega ^{2} = \Omega _{ 0}^{2} + \frac{1} {2} \frac{\ddot{\Omega }_{0}} {\Omega _{0}} -\frac{3} {4} \frac{\dot{\Omega }_{0}^{2}} {\Omega _{0}^{2}}. }$$

(5.9)

In general, it is, of course, very difficult to solve this nonlinear ODE if $\Omega (t)$ is given. In Appendix A.13, however, we use this result to construct a test scenario that can be solved analytically.

5.2.3 Discussion of a Sample Solution

Now we consider some concrete numbers for the test scenario defined in Appendix A.13. In the synchrotron SIS18 at GSI, U⁷³⁺ is stored at a kinetic energy of 11. 4 MeV∕u. According to Eq. (A.95), this leads to

$$\displaystyle{a = \frac{-\eta _{\mathrm{R}}} {\beta _{\mathrm{R}}^{2}\gamma _{\mathrm{R}}} = 38.732.}$$

At the time t = 0, the total gap voltage is chosen in such a way that a synchrotron frequency of 1640 Hz is obtained:

$$\displaystyle{\hat{\Omega }_{0} = 10,304\,\mathrm{s}^{-1}.}$$

We fix

$$\displaystyle{k = 10\,\mathrm{s}^{-1}}$$

and consider the time span from t = 0 to t = 100 ms (see Eq. (A.90)). Finally, we analyze an asynchronous particle with an initial time deviation

$$\displaystyle{\Delta t_{0} = 300\,\mathrm{ns}.}$$

For this example, Fig. 5.1 shows a phase space plot. In order to keep the plot simple, only three time intervals with a length of 10 ms each are displayed. It is remarkable that the revolutions are almost closed. Therefore, one may calculate the area approximately that is enclosed by the trajectory. According to the figure, this area obviously does not change significantly.

Fig. 5.1

Phase space plot for $k = 10\,\mathrm{s}^{-1}$

Figures 5.2 and 5.3 show how the situation changes if k has a larger value. In all cases, kt sweeps from 0 to 1, so that $\Omega _{0}$ decreases to about 30% of its initial value.

Fig. 5.2

Phase space plot for $k = 100\,\mathrm{s}^{-1}$

Fig. 5.3

Phase space plot for $k = 1000\,\mathrm{s}^{-1}$

For values of k that are too large, the curves are strongly deformed, and we may therefore no longer regard them as closed. This makes it more difficult to define the area that is enclosed in a unique way.

Figures 5.4, 5.5, and 5.6 show that for sufficiently small k, the values of $\Omega$ and $\Omega _{0}$ are almost equal.

Fig. 5.4

$\Omega _{0}$ (solid line) and $\Omega$ (dotted line) for $k = 10\,\mathrm{s}^{-1}$

Fig. 5.5

$\Omega _{0}$ (solid line) and $\Omega$ (dotted line) for $k = 100\,\mathrm{s}^{-1}$

Fig. 5.6

$\Omega _{0}$ (solid line) and $\Omega$ (dotted line) for $k = 1000\,\mathrm{s}^{-1}$

If we consider Eq. (5.9),

$$\displaystyle{\Omega ^{2} = \Omega _{ 0}^{2} + \frac{1} {2} \frac{\ddot{\Omega }_{0}} {\Omega _{0}} -\frac{3} {4} \frac{\dot{\Omega }_{0}^{2}} {\Omega _{0}^{2}},}$$

we immediately see that the two requirements

$$\displaystyle\begin{array}{rcl} \left \vert \frac{\dot{\Omega }_{0}} {\Omega _{0}}\right \vert & \ll & \frac{1} {T_{0}},{}\end{array}$$

(5.10)

$$\displaystyle\begin{array}{rcl} \left \vert \frac{\ddot{\Omega }_{0}} {\Omega _{0}}\right \vert & \ll & \frac{1} {T_{0}^{2}},{}\end{array}$$

(5.11)

lead to

$$\displaystyle{\Omega \approx \Omega _{0} = \frac{2\pi } {T_{0}}.}$$

Therefore, the behavior that is observed in our example is valid in general.

Now we check under what conditions the curves in phase space are almost closed. Equations (A.97) and (A.98) are abbreviated according to

$$\displaystyle{\Delta t = \Delta t_{0}c^{-1}\;\cos \varphi }$$

and

$$\displaystyle{\Delta \gamma = \frac{\Delta t_{0}} {a} \left [-k\;c^{-2}\;s\;\cos \varphi +\hat{ \Omega }_{ 0}\;c\;\sin \varphi \right ].}$$

For the sake of simplicity, we consider one revolution from $\varphi = 2\pi m +\pi /2$ to $\varphi = 2\pi (m + 1) +\pi /2$. In this case, at the starting point and at the end point, we have

$$\displaystyle{\Delta t = 0,}$$

and $\Delta \gamma$ changes only according to the cosine function

$$\displaystyle{c =\cos (\mathit{kt}).}$$

Hence, it is obvious that during one revolution, $\Delta \gamma$ changes by the factor

$$\displaystyle{\frac{\cos (k(t + T_{0})) -\cos (\mathit{kt})} {\cos (\mathit{kt})} =\cos (kT_{0}) -\tan (\mathit{kt})\;\sin (kT_{0}) -\mathrm{1}.}$$

For our choice

$$\displaystyle{\Omega _{0}(t) =\hat{ \Omega }_{0}\cos ^{2}(\mathit{kt}),}$$

the requirements above lead to

$$\displaystyle{(kT_{0})^{2} \ll 1,}$$

so that

$$\displaystyle{\frac{\cos (k(t + T_{0})) -\cos (\mathit{kt})} {\cos (\mathit{kt})} \approx -\frac{(kT_{0})^{2}} {2} - kT_{0}\;\tan (\mathit{kt})}$$

is obtained. Due to

$$\displaystyle{kT_{0}\;\tan (\mathit{kt}) \ll \frac{1} {2},}$$

the expression on the right-hand side is much less than 1. Hence, with the required inequalities (5.10) and (5.11), the revolutions are almost closed.

Since the curves are almost closed, we may answer the question how large the enclosed area is. For this purpose, we may use Leibniz’s sector formula

$$\displaystyle{\fbox{$A_{0} =\int _{ 0}^{T_{0} } \frac{\dot{y}\;x -\dot{ x}\;y} {2} \;\mathrm{d}t,$}}$$

which gives the area inside a closed curve (x(t), y(t)) parameterized by the parameter t ∈ [0, T₀]. In order to define this area in a unique way, we demand that for a given t, the expression kt will be kept constant during the following revolution. Due to kT₀ ≪ 1, the value of kt will not change significantly. The expression $\varphi \approx \Omega _{0}t$, however, increases by 2π during one revolution. For a given revolution, $\varphi$ will then be the only variable quantity on the right-hand side of the following expressions:

$$\displaystyle\begin{array}{rcl} \Delta t& =& \Delta t_{0}\;c^{-1}\;\cos \varphi, {}\\ \Delta \gamma & =& \frac{\Delta t_{0}} {a} \left [-k\;c^{-2}\;s\;\cos \varphi +\hat{ \Omega }_{ 0}\;c\;\sin \varphi \right ]. {}\\ \end{array}$$

Due to

$$\displaystyle\begin{array}{rcl} \Delta \dot{t}& =& -a\;\Delta \gamma, {}\\ \Delta \dot{\gamma }& =& b\;\Delta t, {}\\ \end{array}$$

and

$$\displaystyle{b = \frac{\Omega ^{2}} {a},}$$

we obtain

$$\displaystyle\begin{array}{rcl} A_{0}& =& \int _{0}^{T_{0} } \frac{\Delta \dot{\gamma }\;\Delta t - \Delta \dot{t}\;\Delta \gamma } {2} \;\mathrm{d}t = \frac{1} {2}\int _{0}^{T_{0} }\left (b\;\Delta t^{2}\,+\,a\Delta \gamma ^{2}\right )\;\mathrm{d}t = {}\\ & =& \frac{1} {2}\int _{0}^{T_{0} }\left[\frac{\Omega ^{2}} {a} \Delta t_{0}^{2}\;c^{-2}\;\cos ^{2}\varphi \,+\,\frac{\Delta t_{0}^{2}} {a} \left (-k\;c^{-2}\;s\;\cos \varphi \,+\,\hat{\Omega }_{ 0}\;c\;\sin \varphi \right )^{2}\right]\;\mathrm{d}t = {}\\ & =& \frac{\Delta t_{0}^{2}} {2a} \int _{0}^{T_{0} }\left[\left (\Omega ^{2}\;c^{-2} + k^{2}\;c^{-4}\;s^{2}\right )\;\cos ^{2}\varphi - 2k\;c^{-1}\;s\;\hat{\Omega }_{ 0}\;\sin \varphi \;\cos \varphi \right. {}\\ & & +\left.\hat{\Omega }_{0}^{2}\;c^{2}\;\sin ^{2}\varphi \right]\;\mathrm{d}t. {}\\ \end{array}$$

With

$$\displaystyle{\varphi =\int \Omega _{0}\;\mathrm{d}t,\mbox{ } \frac{\mathrm{d}\varphi } {\mathrm{d}t} = \Omega _{0} =\hat{ \Omega }_{0}\;c^{2},}$$

it follows that

$$\displaystyle\begin{array}{rcl} A_{0}& =& \frac{\Delta t_{0}^{2}} {2a} \int _{0}^{2\pi }\left[\left (\frac{\Omega ^{2}} {\hat{\Omega }_{0}}\;c^{-4} + \frac{k^{2}} {\hat{\Omega }_{0}}\;c^{-6}\;s^{2}\right )\;\cos ^{2}\varphi - 2k\;c^{-3}\;s\;\sin \varphi \;\cos \varphi \right. {}\\ & & +\left.\hat{\Omega }_{0}\;\sin ^{2}\varphi \right]\;\mathrm{d}\varphi. {}\\ \end{array}$$

Now we have

$$\displaystyle{\int _{0}^{2\pi }\sin ^{2}\varphi \;\mathrm{d}\varphi =\int _{ 0}^{2\pi }\cos ^{2}\varphi \;\mathrm{d}\varphi =\pi \mbox{ and }\int _{ 0}^{2\pi }\sin \varphi \;\cos \varphi \;\mathrm{d}\varphi = 0,}$$

so that

$$\displaystyle{A_{0} = \frac{\pi \Delta t_{0}^{2}} {2a\hat{\Omega }_{0}}\left (\Omega ^{2}\;c^{-4} + k^{2}\;c^{-6}\;s^{2} +\hat{ \Omega }_{ 0}^{2}\right )}$$

is the result. We may now substitute $\Delta t_{0}$ and a if we define the oscillation amplitudes

$$\displaystyle{\Delta \hat{\gamma } = \frac{\Delta t_{0}\hat{\Omega }_{0}c} {a},\mbox{ }\Delta \hat{t} = \Delta t_{0}c^{-1}\mbox{ } \Rightarrow \Delta \hat{t}\;\Delta \hat{\gamma } = \frac{\Delta t_{0}^{2}\hat{\Omega }_{ 0}} {a}.}$$

One should note that the product $\Delta \hat{t}\;\Delta \hat{\gamma }$ is constant, even though the factors are changing with time:

$$\displaystyle{A_{0} = \frac{\pi } {2}\;\Delta \hat{t}\;\Delta \hat{\gamma }\left ( \frac{\Omega ^{2}} {\hat{\Omega }_{0}^{2}}\;c^{-4} + \frac{k^{2}} {\hat{\Omega }_{0}^{2}}\;c^{-6}\;s^{2} + 1\right ).}$$

Due to

$$\displaystyle{\Omega \approx \Omega _{0} =\hat{ \Omega }_{0}c^{2}\mbox{ and }T_{ 0} = \frac{2\pi } {\Omega _{0}} = \frac{2\pi c^{-2}} {\hat{\Omega }_{0}},}$$

this may be written approximately as

$$\displaystyle{ A_{0} \approx \frac{\pi } {2}\;\Delta \hat{t}\;\Delta \hat{\gamma }\left (2 + \left [\frac{kT_{0}} {2\pi } \tan (kt)\right ]^{2}\right ). }$$

(5.12)

Finally, with the help of

$$\displaystyle{kT_{0}\;\tan (kt) \ll \frac{1} {2},}$$

we find that

$$\displaystyle{A_{0} \approx \pi \; \Delta \hat{t}\;\Delta \hat{\gamma }}$$

is almost constant.

We analyze this phenomenon more thoroughly. According to Eq. (A.90),

$$\displaystyle{\Omega _{0}(t) =\hat{ \Omega }_{0}\;\cos ^{2}(kt),}$$

the angular frequency $\Omega _{0} =\hat{ \Omega }_{0}$ is obtained for t = 0. This value will be reduced to a specific final value $\Omega _{\mathrm{final}}$. This leads to a certain requirement for kt. By choosing k, we may now select whether this final value is reached in a shorter or a longer time t = t_final. Under these conditions, we have

$$\displaystyle{A_{1} = A_{0}(t = 0) =\pi \; \Delta \hat{t}\;\Delta \hat{\gamma }}$$

and

$$\displaystyle{A_{2} = A_{0}(t = t_{\mathrm{final}}) \approx \pi \; \Delta \hat{t}\;\Delta \hat{\gamma }\left (1 + \frac{1} {8}\left [\frac{kT_{0}} {\pi } \tan (kt_{\mathrm{final}})\right ]^{2}\right ).}$$

For the same final value $\Omega _{\mathrm{final}}$, the tangent function always has the same value, since kt_final is thereby determined in a unique way. However, we see that the difference between A₁ and A₂ can be made as small as desired by choosing k. If k is very small, i.e., if the frequency change is slow and the final frequency $\Omega _{\mathrm{final}}$ is reached for large t_final, this difference will be very small. In fact, it can be made as small as one likes by increasing t_final accordingly.

5.3 Adiabaticity

The example that was presented in the previous section was a very specific one. Nevertheless, we can use it to make some assumptions about phenomena of a general nature.

We summarize the observations we made based in the example:

• The frequency was chosen in such a way that it does not change significantly during one revolution in phase space.

• Therefore, the curves in phase space were almost closed.

• From a mathematical point of view, we required that the inequalities (5.10),

  $$\displaystyle{\left \vert \frac{\dot{\Omega }_{0}} {\Omega _{0}}\right \vert \ll \frac{1} {T_{0}},}$$

  and (5.11)

  $$\displaystyle{\left \vert \frac{\ddot{\Omega }_{0}} {\Omega _{0}}\right \vert \ll \frac{1} {T_{0}^{2}}}$$

  be valid.

• The area that is enclosed by the orbits remains almost constant (its shape changes, however)—even during long time intervals. The slower the frequency is changed, the smaller is the change in the area.

The observation described in the second bullet point motivates the following definition:

Definition 5.1.

Consider a Hamiltonian system with the Hamiltonian

$$\displaystyle{H(q,p,\lambda )}$$

that depends on a time-dependent control parameter λ(t). For constant λ, suppose that the orbits in phase space are closed. Under these conditions, the function

$$\displaystyle{J(t) = \frac{1} {2\pi }\int _{A(t)}\mathrm{d}q\;\mathrm{d}p}$$

is called an action variable of the system. The domain A(t) is defined in such a way that the boundary curve ∂ A(t) is a closed orbit $(q(\tilde{t}),p(\tilde{t}))$ of the system with λ = λ(t) kept fixed while the time parameter $\tilde{t}$ changes. Furthermore, the point (q(t), p(t)) must belong to ∂ A(t).

This means that for each time t, the actual position (q, p) in phase space is determined. Based on this position, a specific orbit is considered that would be obtained if λ were constant (even though λ varies a bit in reality). This trick leads to a closed orbit that is determined in a unique way for each time t. Please note that this closed orbit is not an orbit that the system actually traverses, since the real system orbit is not closed. Therefore, we cannot use area preservation in phase space for time-dependent Hamiltonians as an argument for the case described in this definition. The action J(t) may in principle be time-dependent.

Please note that Definition 5.1 is a generalization of the definition of J in Sect. 2.11.7.2, which was applicable only to closed orbits. With Definition 5.1, one may now also consider orbits that are not closed or almost closed.

A quantity for which the phenomenon applies that was described in the last bullet point above is called an (Ehrenfest) adiabatic invariant.

According to Vladimir Arnold (see reprint of the article “Small denominators and problems of stability of motion in classical and celestial mechanics,” Russ. Math. Surveys 18:6 85–191(1963) in [5]), one makes the following definition.

Definition 5.2.

Consider a Hamiltonian system with one degree of freedom specified by a Hamiltonian

$$\displaystyle{H(q,p,\lambda )}$$

that depends on a time-dependent control parameter λ(t). Suppose that for constant λ, the orbits in phase space are closed and that the control parameter varies according to λ(t) = f(μ t), where f(x) is a smooth function.

A function J is called an adiabatic invariant of this system H(q, p, λ) if for every κ > 0, there exists μ₀ > 0 such that for all μ satisfying 0 < μ < μ₀, the relation

$$\displaystyle{\left \vert J(t) - J(0)\right \vert <\kappa \mbox{ for all }t\mbox{ with }0 <\mu t < 1}$$

holds.

This definition becomes transparent if we have a look at our example. Let us assume that kt sweeps from 0 to 1. 5. We may then set

$$\displaystyle{\mu = \frac{k} {1.5},}$$

so that

$$\displaystyle{0 < kt < 1.5}$$

and

$$\displaystyle{0 <\mu t < 1}$$

are equivalent. Therefore, the normalization that is performed by the transition from k to μ ensures that the final value is obtained for μ t = 1. Our example showed that for every given deviation κ, a maximum value for k can be specified that corresponds to the maximum value μ₀ for μ in the definition.

If we define $t_{\mathrm{final}} = 1/\mu$ and $t_{\mathrm{final,min}} = 1/\mu _{0}$, we may rewrite Definition 5.2 in the following form:

Definition 5.3.

Consider a Hamiltonian system with one degree of freedom specified by the Hamiltonian

$$\displaystyle{H(q,p,\lambda )}$$

that depends on a time-dependent control parameter λ(t). Suppose that for constant λ, the orbits in phase space are closed and that the control parameter varies according to $\lambda (t) = f(t/t_{\mathrm{final}})$, where f(x) is a smooth function.

A function J is called an adiabatic invariant of this system H(q, p, λ) if for every κ > 0, there exists t_final,min > 0 such that for all t_final satisfying t_final > t_final,min, the relation

$$\displaystyle{\left \vert J(t) - J(0)\right \vert <\kappa \mbox{ for all }t\mbox{ with }0 < t < t_{\mathrm{final}}}$$

holds.

This means that the tolerated increase in the adiabatic invariant can be fixed a priori as small as desired. By fixing the function f(x), it is clear how the control parameter will change in principle, but since t_final is not fixed, the speed of changing the control parameter—and hence also the time span for this process—is open. If J is actually an adiabatic invariant, it is possible to specify a minimum time that is needed for the process in order to satisfy the requirement concerning the tolerated increase of J.

The ODE in our example is a simple case for a Hamiltonian system with one degree of freedom given by the Hamiltonian

$$\displaystyle{H(q,p,\lambda ),}$$

which depends on a time-dependent parameter λ(t). For this type of system, the following theorem holds (cf. [6, 7]):

Theorem 5.4.

The action variable J of a Hamiltonian system with one degree of freedom is an adiabatic invariant, provided that the revolution frequency in phase space is not equal to zero.

This shows that it is theoretically possible to limit the increase in the longitudinal emittance to an arbitrarily small value by just making the beam manipulation process slow enough. Considering longitudinal beam manipulations at constant reference energy, it becomes clear that the change in the voltage amplitude $\hat{V }$ has to be adiabatic.

Theorem 5.4 does not make any statement about how the slowness of the changes of the parameter may be determined in order for the area deviation to remain below a certain limit.

For this practical purpose, in the case of longitudinal beam manipulations, one usually defines the adiabaticity parameter

$$\displaystyle{ \epsilon = \frac{1} {\omega _{\mathrm{S}}^{2}}\left \vert \frac{\mathrm{d}\omega _{\mathrm{S}}} {\mathrm{d}t}\right \vert = \frac{T_{\mathrm{S}}} {2\pi } \left \vert \frac{\dot{\omega }_{\mathrm{S}}} {\omega _{\mathrm{S}}}\right \vert }$$

(5.13)

(cf. [8, p. 316]). This definition becomes transparent if one looks at requirement (5.10), which states that 2π ε ≪ 1.

5.3.1 Pendulum with Variable Length

A famous example using an adiabatic invariant is a pendulum similar to the one shown in Fig. 2.18 on p. 105. However, it does not have a fixed suspension point where the massless thread is fixed. Instead, the thread passes through a hole in a metal plate at the top. Hence, the length R of the pendulum may be varied by moving the thread up and down through the hole. This type of pendulum is called a Rayleigh pendulum or Lorentz–Einstein pendulum.

If the length R is shortened very slowly by pulling the thread upward, the action variable will be an adiabatic invariant.

Before we consider a variable length R, we keep R fixed, and we analyze the case of small oscillation amplitudes with α ≪ 1. Due to

$$\displaystyle{\cos \;\alpha =\cos ^{2} \frac{\alpha } {2} -\sin ^{2} \frac{\alpha } {2} = 1 - 2\;\sin ^{2} \frac{\alpha } {2},}$$

we obtain

$$\displaystyle{1-\cos \;\alpha = 2\;\sin ^{2} \frac{\alpha } {2} \approx \frac{\alpha ^{2}} {2},}$$

so that the potential energy is

$$\displaystyle{W_{\mathrm{pot}} = mgx = mgR(1-\cos \;\alpha ) \approx mgR\frac{\alpha ^{2}} {2}.}$$

Due to $v = R\dot{\alpha }$, the kinetic energy is

$$\displaystyle{W_{\mathrm{kin}} = \frac{1} {2}\;mv^{2} = \frac{1} {2}\;mR^{2}\dot{\alpha }^{2}.}$$

The sum W_pot + W_kin remains constant. Hence, the time derivative is zero:

$$\displaystyle{mgR\alpha \dot{\alpha } + mR^{2}\dot{\alpha }\ddot{\alpha } = 0.}$$

Therefore, we obtain the well-known ODE

$$\displaystyle{\ddot{\alpha }+ \frac{g} {R}\alpha = 0}$$

with angular frequency

$$\displaystyle{\omega = \sqrt{ \frac{g} {R}}.}$$

We now use the length coordinate

$$\displaystyle{q = R\alpha }$$

and the momentum

$$\displaystyle{p = mR\dot{\alpha }}$$

as the pair of coordinates to define the Hamiltonian

$$\displaystyle{H(q,p) = \frac{p^{2}} {2m} + \frac{m} {2} \frac{g} {R}q^{2}}$$

as the sum W_pot + W_kin. We verify that H is actually a Hamiltonian:

$$\displaystyle{\frac{\partial H} {\partial p} = \frac{p} {m} =\dot{ q},}$$

$$\displaystyle{\frac{\partial H} {\partial q} = m \frac{g} {R}q = mg\alpha.}$$

Due to

$$\displaystyle{\dot{p} = mR\ddot{\alpha } = -mR \frac{g} {R}\alpha = -mg\alpha,}$$

we actually obtain

$$\displaystyle{\frac{\partial H} {\partial q} = -\dot{p}.}$$

For small oscillation amplitudes, a solution of the ODE is

$$\displaystyle\begin{array}{rcl} \alpha & =& \hat{\alpha }\;\cos (\omega t), {}\\ \dot{\alpha }& =& -\omega \;\hat{\alpha }\;\sin (\omega t), {}\\ q& =& \hat{q}\;\cos (\omega t), {}\\ p& =& -\hat{p}\;\sin (\omega t), {}\\ \end{array}$$

with

$$\displaystyle{\hat{q} = R\hat{\alpha }\mbox{, }\hat{p} = mR\omega \;\hat{\alpha },}$$

leading to the following value of the Hamiltonian:

$$\displaystyle{H\,=\,\frac{1} {2}\;mR^{2}\omega ^{2}\;\hat{\alpha }^{2}\;\sin ^{2}(\omega t)\,+\,\frac{1} {2}\;m \frac{g} {R}R^{2}\hat{\alpha }^{2}\;\cos ^{2}(\omega t)\,=\,\frac{1} {2}\;mR^{2}\omega ^{2}\;\hat{\alpha }^{2}\,=\,\frac{1} {2}\;m\omega ^{2}\;\hat{q}^{2}.}$$

The action may easily be determined as the area of the ellipse divided by 2π:

$$\displaystyle{ J = \frac{1} {2\pi }\int \int \mathrm{d}q\;\mathrm{d}p = \frac{1} {2\pi }\pi \hat{q}\hat{p} = \frac{1} {2}mR^{2}\omega \hat{\alpha }^{2} = \frac{1} {2}\;m\omega \;\hat{q}^{2}. }$$

(5.14)

The last two equations show that

$$\displaystyle{ \omega = \frac{H} {J} = \sqrt{ \frac{g} {R}} }$$

(5.15)

is valid.

Now we allow slow changes of the control parameter R with time so that the Hamiltonian is modified² according to

$$\displaystyle{H(q,p,R(t)) = \frac{p^{2}} {2m} + \frac{m} {2} \frac{g} {R(t)}q^{2}.}$$

We know that J is an adiabatic invariant, i.e., its value will remain almost constant if R(t) is modified slowly. Therefore, according to Eq. (5.15), we obtain

$$\displaystyle{H \sim \omega \sim R^{-1/2}.}$$

Due to Eq. (5.14),

$$\displaystyle{\hat{q} \sim \omega ^{-1/2} \sim R^{1/4}}$$

holds. This leads to

$$\displaystyle{\hat{\alpha }= \frac{\hat{q}} {R} \sim R^{-3/4} \sim \omega ^{3/2}.}$$

Please note that H is not constant, since energy is exchanged with the pendulum if the thread is moved up or down through the hole. When R is reduced, ω and the amplitude $\hat{\alpha }$ of the angle increase. However, the amplitude $\hat{q} = R\hat{\alpha }$ decreases.

A more detailed analysis of the Rayleigh pendulum with the same results can be found in [9, 10].

5.3.2 Iso-Adiabatic Ramps

Consider a coasting beam that is stored in a synchrotron at constant energy. This beam will be captured, i.e., bunched, by means of an adiabatic increase of the voltage amplitude $\hat{V }$. This procedure is called an adiabatic capture process. The voltage amplitude $\hat{V }$ plays the role of the control parameter λ discussed earlier. As we will see, the voltage begins at a small but positive value, so that the synchrotron frequency does not vanish.

We assume that the adiabaticity parameter

$$\displaystyle{\alpha _{\mathrm{adiab}} = T_{\mathrm{S}}\frac{\dot{\omega }_{\mathrm{S}}} {\omega _{\mathrm{S}}} = 2\pi \frac{\dot{\omega }_{\mathrm{S}}} {\omega _{\mathrm{S}}^{2}}}$$

remains constant (so-called isoadiabatic ramp). Please note that in the literature, the adiabaticity parameter is sometimes defined without the additional factor 2π, and sometimes the absolute value is used as well, so that the coefficient is always nonnegative (cf. Garoby in [8, Sect. 4.8.1]); see Eq. (5.13).

Due to Eq. (3.25), we know that in the stationary case,

$$\displaystyle{\omega _{\mathrm{S,0,stat}} = k\sqrt{\hat{V}}\mbox{ with }k = f_{\mathrm{R}}\sqrt{\frac{2\pi \vert \eta _{\mathrm{R} } \;Q\vert \;h} {W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}}} > 0}$$

holds, so that one obtains

$$\displaystyle{\alpha _{\mathrm{adiab}} = \frac{2\pi } {k^{2}\hat{V }} \frac{k} {2\sqrt{\hat{V}}}\dot{\hat{V }} = \frac{\pi \;\dot{\hat{V }}} {k\;\hat{V }^{3/2}}.}$$

Separation of this ODE yields

$$\displaystyle{\int \alpha _{\mathrm{adiab}}\;\mathrm{d}t = \frac{\pi } {k}\int \hat{V }^{-3/2}\mathrm{d}\hat{V }}$$

$$\displaystyle{\Rightarrow -2\hat{V }^{-1/2} =\beta _{\mathrm{ adiab}}t + c_{\mathrm{adiab}}\mbox{ with }\beta _{\mathrm{adiab}} = \frac{k\alpha _{\mathrm{adiab}}} {\pi } }$$

$$\displaystyle{ \Rightarrow \hat{ V } = \frac{4} {\left (\beta _{\mathrm{adiab}}t + c_{\mathrm{adiab}}\right )^{2}},\mbox{ }\beta _{\mathrm{adiab}}t = - \frac{2} {\sqrt{\hat{V}}} - c_{\mathrm{adiab}}. }$$

(5.16)

This solution is now written depending on the initial value $\hat{V }(t = 0) =\hat{ V }_{1}$ and the final value $\hat{V }(t = T) =\hat{ V }_{2}$:

$$\displaystyle{\hat{V }_{1} = \frac{4} {c_{\mathrm{adiab}}^{2}},\mbox{ }c_{\mathrm{adiab}} = - \frac{2} {\sqrt{\hat{V }_{1}}},}$$

$$\displaystyle{\hat{V }_{2}\,=\, \frac{4} {\left (\beta _{\mathrm{adiab}}T\,+\,c_{\mathrm{adiab}}\right )^{2}},\mbox{ }\beta _{\mathrm{adiab}}\,=\, \frac{1} {T}\left (\,-\, \frac{2} {\sqrt{\hat{V }_{2}}}\,-\,c_{\mathrm{adiab}}\right )\,=\, \frac{2} {T}\left ( \frac{1} {\sqrt{\hat{V }_{1}}}\,-\, \frac{1} {\sqrt{\hat{V }_{2}}}\right ).}$$

For increasing synchrotron frequencies ω_S, the amplitude $\hat{V }$ will increase, and α_adiab is positive, so that β_adiab is positive as well. For decreasing synchrotron frequencies ω_S, the amplitude $\hat{V }$ will decrease, and α_adiab is negative, so that β_adiab is negative as well. It follows that

$$\displaystyle{\hat{V } = \frac{4} {\left ( \frac{t} {T}\left ( \frac{2} {\sqrt{\hat{V }_{1}}} - \frac{2} {\sqrt{\hat{V }_{2}}} \right ) - \frac{2} {\sqrt{\hat{V }_{1}}} \right )^{2}}}$$

$$\displaystyle{\Rightarrow \fbox{$\hat{V } = \frac{\hat{V }_{1}} {\left ( \frac{t} {T}\left (\sqrt{\frac{\hat{V }_{1 } } {\hat{V }_{2}}} - 1\right ) + 1\right )^{2}}.$}}$$

The total time T must, of course, be chosen sufficiently large that

$$\displaystyle{\vert \alpha _{\mathrm{adiab}}\vert \ll 1}$$

holds, where α_adiab can be calculated as follows:

$$\displaystyle{\alpha _{\mathrm{adiab}} = \frac{\pi } {k}\beta _{\mathrm{adiab}} = \frac{2\pi } {kT}\left ( \frac{1} {\sqrt{\hat{V }_{1}}} - \frac{1} {\sqrt{\hat{V }_{2}}}\right ) = \frac{1} {f_{\mathrm{R}}T}\sqrt{\frac{2\pi \;W_{\mathrm{R} } \beta _{\mathrm{R} }^{2 }} {\vert \eta _{\mathrm{R}}\;Q\vert \;h}} \left ( \frac{1} {\sqrt{\hat{V }_{1}}} - \frac{1} {\sqrt{\hat{V }_{2}}}\right ).}$$

An example of the amplitude ramp is shown in Fig. 5.7. Let us briefly discuss the bunching case with $\hat{V }_{2} >\hat{ V }_{1}$. Since the voltage $\hat{V }_{1}$ is switched on instantaneously at the very beginning, a certain increase in the longitudinal emittance cannot be avoided. One therefore tries to keep $\hat{V }_{1}$ as small as technically achievable. For a required value of α_adiab, however, the time T will then be large. Therefore, a compromise has to be found.

Fig. 5.7

Example of isoadiabatic ramps

5.4 Bunch Compression and Unmatched Bunches

In the following, we will discuss some beam manipulations such as bunch compression and barrier bucket operation. These schemes belong to advanced scenarios that are sometimes called “RF gymnastics” [11].

If only the innermost part of the bucket is filled with particles, one may describe the motion in phase space by small values for the momentum spread and the time deviation. In this case, we may linearize the equations of motion. Both the time deviation and the energy deviation will then be harmonic functions. If the first one is proportional to cos(ω_St), the latter one will be proportional to sin(ω_St). Assuming suitable scaling, the orbits in phase space will be circles. If the phase space occupation looks like a horizontal bar at the beginning, this bar will be oriented in an upright way after one quarter of the synchrotron period (similar to the first three pictures in Fig. 3.2 on p. 125). Hence, a long bunch with a comparatively small momentum spread will be transformed into a short bunch with a larger momentum spread. Therefore, this process, which corresponds to a 90^∘ rotation in phase space, is called a fast bunch compression.

It is easy to see what happens in the time domain if one analyzes how many particles are located inside a certain time interval. At the beginning (horizontal bar), one will find a comparatively small number of particles in a certain time interval. This corresponds to a long bunch with small beam current. After one-quarter revolution, we will find many particles in the same time interval, provided that this time interval is close enough to the bunch center (or we will find zero particles outside). This corresponds to a short bunch with large beam current.

If we observe this process depending on time, we will identify an amplitude and length modulation of the bunch. In reality, the bucket will also be filled in the vicinity of the separatrix, so that nonlinearities will be present. The angular velocity of the particles in phase space will no longer be constant, but it will decrease by an amount that increases with the distance of the particles from the bucket center. A bar-shaped bunch that is oriented horizontally at the beginning will therefore be deformed in an S-shaped way after one-quarter of a synchrotron period.

We now analyze the simple case that the bunch is located in the linear region of the bucket and that it fills an elliptical area. The ellipse will be matched to the bucket so that the ratio of the principal axes is given by Eq. (3.28),

$$\displaystyle{\frac{\Delta \hat{W}_{1}} {\Delta \hat{t}_{1}} = f_{\mathrm{S,1}}\;\frac{2\pi W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} {\vert \eta _{\mathrm{R}}\vert }.}$$

The index 1 denotes the situation before the voltage is increased at t = 0. Therefore, the total length is $2\Delta \hat{t}_{1}$. We now consider a particle at $(\Delta t,\Delta W) = (0,-\Delta \hat{W}_{1})$. Due to the voltage increase, it will move to the point $(\Delta t,\Delta W) = (\Delta \hat{t}_{2},0)$ after one-quarter of the synchrotron period. The new ratio of the principal axes is given by

$$\displaystyle{\frac{\Delta \hat{W}_{1}} {\Delta \hat{t}_{2}} = f_{\mathrm{S,2}}\;\frac{2\pi W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} {\vert \eta _{\mathrm{R}}\vert },}$$

since the synchrotron frequency has changed (in the numerator, we replaced $\Delta \hat{W}_{2}$ by $\Delta \hat{W}_{1}$, since this corresponds to the starting point of the trajectory). We have to take into account that the energy W_R of the reference particle does not change when the voltage increases.

We calculate the quotient of the last two equations:

$$\displaystyle{\frac{\Delta \hat{t}_{2}} {\Delta \hat{t}_{1}} = \frac{f_{\mathrm{S,1}}} {f_{\mathrm{S,2}}}.}$$

According to Eq. (3.25), the synchrotron frequency is proportional to the square root of the voltage amplitude, so that

$$\displaystyle{\frac{\Delta \hat{t}_{2}} {\Delta \hat{t}_{1}} = \sqrt{\frac{\hat{V }_{1 } } {\hat{V }_{2}}}}$$

is obtained. Hence, we have found an approximation for the bunch length reduction as a result of the voltage increase. The above-mentioned S-shaped deformation is, of course, not taken into account.

As described before, the bunch compression is intentionally based on the phenomenon that the bunch contour does not fit to the trajectories. If the beam is extracted in time, the short bunch that is generated may be used for an experiment.

The same phenomenon is also observed in a weaker form if the bunch is unmatched by accident (unmatched bunch). In this case, the particles will smear over the whole phase space area that is determined by the particles that correspond to the outermost trajectory. As a result, this outermost trajectory will be filled with particles in a homogeneous way after several revolutions in phase space. As mentioned in Sect. 3.3, this effect is called filamentation and phase space dilution. The particles finally cover more phase space area than in the beginning.

This observation seems to contradict Liouville’s theorem (here we are even confronted with the case that the Hamiltonian does not explicitly depend on time, so that Liouville’s theorem is valid in its simplest form). Especially during the first revolution, we may easily see, however, that the S-shaped area filled by the particles has, in fact, the same area as the initial ellipse. Even though this observation is more complicated to verify in the following revolutions, one will always come to the conclusion that area preservation is still valid. Hence, Liouville’s theorem is still satisfied.

After some revolutions, however, a lot of “air” will be enclosed between the particles, so that the impression of a completely filled phase space area is induced. Since the number of particles is large but still finite, it is difficult to determine the border of the area that is filled with particles. Therefore, the contradiction is resolved: due to the finite number of particles, the “air” that is curled up leads to an effective increase in the phase space area that cannot be withdrawn. The filamentation process thus leads to an emittance increase.

Please note that Liouville’s theorem was derived for a continuous distribution of particles in phase space. When the spiral arms have become longer and longer so that they contain only a few individual particles, this assumption of a continuous distribution is no longer justified, and Liouville’s theorem is no longer applicable. In other words, the seeming contradiction may be resolved by the fact that on the one hand, the bunch is described as a continuum, and on the other hand, as a cloud of discrete particles.

The increase in the longitudinal emittance that is caused by unmatched bunches is of course undesirable. The larger emittance usually leads to a larger bucket area that is required to keep all particles bunched. According to Eq. (3.38), this also leads to larger RF voltage (and hence RF power installation) requirements. This is clearly a negative effect of the filamentation process. There are, however, also positive aspects of the momentum spread of a bunch that may also be explained based on our unmatched bunch example. The unmatched bunch as a whole obviously performs coherent oscillations, which may primarily be measured as amplitude oscillations of the beam current. After filamentation has taken effect, these oscillations will have stopped. Therefore, the filamentation process may also be regarded as a damping effect. The momentum spread of the bunch suppresses a collective motion of the particles. This effect is called Landau damping, and it may prevent beam instabilities from growing, provided that Landau damping is stronger than the growth of the specific instability (cf. [12, 13, Sect. 9.5]).

5.5 Dual-Harmonic Operation and Barrier Buckets

According to Sect. 2.1.3 (see Eq. (2.9)) and Sect. 2.1.6, for a strongly bunched beam, the beam current amplitude at the fundamental harmonic is twice as large as the DC component, i.e., the average beam current.

In general, the interaction of the beam with the environment or with itself may lead to instabilities. Some of these instabilities become stronger as the ratio of maximum beam current to average beam current increases. Hence, the bunching factor

$$\displaystyle{B_{\mathrm{f}} = \frac{\bar{I}_{\mathrm{beam}}} {I_{\mathrm{beam,max}}}}$$

defined in Eq. (2.37) should be as large as possible. Please note that the wording “bunching factor” is misleading because this factor decreases when the bunching gets stronger.

Increasing the bunching factor means to fill a larger fraction of the synchrotron circumference l_R with bunches and/or to distribute the particles more homogeneously in the longitudinal direction. One method to achieve this is to add a second RF voltage with a higher frequency to the normal accelerating voltage (dual-harmonic operation). For ions with positive charge below transition, the bunches are located on the rising slope of the RF voltage. If the higher-harmonic RF that is added has a negative slope at this point, the phase focusing effect will be reduced locally. Hence, the bunches will show a flat profile.

A different method that also increases the bunching factor is barrier bucket operation. Let us assume that h bunches are present in the synchrotron ring and that they are captured as usual by a sinusoidal RF voltage. If some of these sine waves are now omitted, the adjacent bunches will now merge across the missing sine slope. Hence, a longer bunch with a sausage-like shape is created. In other words, two single-sine pulses may be used to keep the beam bunched between them. It is even possible to shift the phase of one of these sine pulses with respect to the revolution frequency. This allows one to adiabatically squeeze the bunch together, thereby increasing its momentum spread. One speaks of moving barriers in this case.

5.5.1 Barrier Bucket Signal Generation

Let us assume that a standard cavity system with quality factor Q_p (loaded Q) is used to generate single-sine pulses for barrier bucket operation. The cavity system can be described by the standard equivalent circuit shown in Fig. 4.5 on p. 181, so that the transfer function is given by Eq. (4.26)

$$\displaystyle{Z_{\mathrm{tot}}(s) = \frac{R_{\mathrm{p}} \frac{\omega _{\mathrm{res}}} {Q_{\mathrm{p}}} s} {s \frac{\omega _{\mathrm{res}}} {Q_{\mathrm{p}}} + s^{2} +\omega _{ \mathrm{res}}^{2}}}$$

$$\displaystyle{\Rightarrow Y _{\mathrm{tot}}(s) = \frac{1} {Z_{\mathrm{tot}}(s)} = s \frac{Q_{\mathrm{p}}} {R_{\mathrm{p}}\omega _{\mathrm{res}}} + \frac{1} {R_{\mathrm{p}}} + \frac{\omega _{\mathrm{res}}Q_{\mathrm{p}}} {R_{\mathrm{p}}} \frac{1} {s}.}$$

We want to generate a single-sine pulse

$$\displaystyle\begin{array}{rcl} V _{\mathrm{gap}}(t)& =& \hat{V }_{\mathrm{gap}}\;\sin (\omega _{1}t)\left [\Theta (t) - \Theta (t - T_{1})\right ] = {}\\ & =& \hat{V }_{\mathrm{gap}}\;\sin (\omega _{1}t)\;\Theta (t) -\hat{ V }_{\mathrm{gap}}\;\sin (\omega _{1}(t - T_{1}))\;\Theta (t - T_{1}) {}\\ \end{array}$$

with $\omega _{1} = 2\pi /T_{1}$. According to Table A.4 (p. 418) and to the time shifting property of the Laplace transform (Sect. 2.2), this corresponds to

Now we want to determine the required generator current for negligible beam current. In the Laplace domain, we get

$$\displaystyle{ I_{\mathrm{gen}}(s) = Y _{\mathrm{tot}}(s)\;V _{\mathrm{gap}}(s) = \left (1 - e^{-T_{1}s}\right )\;I_{ x}(s) }$$

(5.17)

with

$$\displaystyle{I_{x}(s) = \frac{\hat{V }_{\mathrm{gap}}} {R_{\mathrm{p}}} \left [\omega _{1}\frac{Q_{\mathrm{p}}} {\omega _{\mathrm{res}}} \; \frac{s} {s^{2} +\omega _{ 1}^{2}} + \frac{\omega _{1}} {s^{2} +\omega _{ 1}^{2}} +\omega _{\mathrm{res}}Q_{\mathrm{p}}\omega _{1}\; \frac{1} {s(s^{2} +\omega _{ 1}^{2})}\right ].}$$

Now we need a transformation back to the time domain. The first two expressions in the square brackets are directly available from Table A.4. For the last term, a partial fraction decomposition leads to

$$\displaystyle{ \frac{1} {s(s^{2} +\omega _{ 1}^{2})} = \frac{1} {\omega _{1}^{2}}\left (\frac{1} {s} - \frac{s} {s^{2} +\omega _{ 1}^{2}}\right ).}$$

Now the transformation back to the time domain can be accomplished with the help of Table A.4:

$$\displaystyle{I_{x}(t) = \Theta (t)\;\frac{\hat{V }_{\mathrm{gap}}} {R_{\mathrm{p}}} \;\left [\left (\omega _{1}\frac{Q_{\mathrm{p}}} {\omega _{\mathrm{res}}} -\omega _{\mathrm{res}}\frac{Q_{\mathrm{p}}} {\omega _{1}} \right )\;\cos (\omega _{1}t) +\sin (\omega _{1}t) +\omega _{\mathrm{res}}\frac{Q_{\mathrm{p}}} {\omega _{1}} \right ].}$$

The required generator current is obtained using the time shifting property based on Eq. (5.17):

$$\displaystyle{I_{\mathrm{gen}}(t) = I_{x}(t) - I_{x}(t - T_{1}).}$$

It is obvious that the solution can be simplified significantly if the cavity is tuned to the angular frequency ω₁ that corresponds to the duration T₁ of the single-sine pulse, i.e., if ω_res = ω₁:

$$\displaystyle{I_{x}(t) = \Theta (t)\;\frac{\hat{V }_{\mathrm{gap}}} {R_{\mathrm{p}}} \;\left [Q_{\mathrm{p}} +\sin (\omega _{1}t)\right ]}$$

$$\displaystyle{\Rightarrow I_{\mathrm{gen}}(t) = \Theta (t)\;\frac{\hat{V }_{\mathrm{gap}}} {R_{\mathrm{p}}} \;\left [Q_{\mathrm{p}} +\sin (\omega _{1}t)\right ] - \Theta (t - T_{1})\;\frac{\hat{V }_{\mathrm{gap}}} {R_{\mathrm{p}}} \;\left [Q_{\mathrm{p}} +\sin (\omega _{1}(t - T_{1}))\right ]}$$

$$\displaystyle{ \Rightarrow I_{\mathrm{gen}}(t) = \frac{\hat{V }_{\mathrm{gap}}} {R_{\mathrm{p}}} \;\left [Q_{\mathrm{p}} +\sin (\omega _{1}t)\right ]\left [\Theta (t) - \Theta (t - T_{1})\right ]. }$$

(5.18)

We see that even though the cavity is tuned to the frequency of the single-sine pulse, a DC offset pulse is needed that is Q_p times higher than the peak value of the sinusoidal current component. This is not very efficient if a narrowband cavity is used. Hence, the quality factor (loaded Q) of the barrier bucket cavity system should be kept as low as possible (broadband cavity system). Nevertheless, it is possible to produce single-sine voltage pulses by means of the superposition of a DC current pulse and a sinusoidal current pulse even if Q_p is on the order of ten [14].

Of course, in reality, there are always deviations from the ideal behavior. Therefore, it may be necessary to correct the theoretical generator current given by Eq. (5.18) slightly in order to avoid microbunching effects. One way to optimize the shape of the gap voltage is to perform a Fourier analysis of the measured gap voltage, which makes it possible to calculate a predistorted control signal [15].

5.5.2 Phase and Amplitude Relations for Dual-Harmonic Operation

As in most cases, we assume ions with positive charge below transition. Let us consider the case that a higher-harmonic component k ω_RF is added to the fundamental harmonic ω_RF. This leads to the total voltage

$$\displaystyle{V (t) =\hat{ V }_{1}\;\sin (\omega _{\mathrm{RF}}t +\varphi _{1}) +\hat{ V }_{2}\;\sin (k\omega _{\mathrm{RF}}t +\varphi _{2}).}$$

We now require that a saddle point be created at the location of the bunch and that the slope be positive to the left and to the right of this saddle point (as it was before the second harmonic component was added). For the bunch center at t = 0, we therefore require

$$\displaystyle{V (t = 0)\,=\,V _{0}\mbox{, }V '(t = 0)\,=\,0\mbox{, }V ''(t = 0)\,=\,0,\mbox{ and }V '''(t\,=\,0) > 0.}$$

Due to

$$\displaystyle\begin{array}{rcl} V '(t)& =& \omega _{\mathrm{RF}}\;\hat{V }_{1}\;\cos (\omega _{\mathrm{RF}}t +\varphi _{1}) + k\omega _{\mathrm{RF}}\;\hat{V }_{2}\;\cos (k\omega _{\mathrm{RF}}t +\varphi _{2}), {}\\ V ''(t)& =& -\omega _{\mathrm{RF}}^{2}\;\hat{V }_{ 1}\;\sin (\omega _{\mathrm{RF}}t +\varphi _{1}) - k^{2}\omega _{ \mathrm{RF}}^{2}\;\hat{V }_{ 2}\;\sin (k\omega _{\mathrm{RF}}t +\varphi _{2}), {}\\ V '''(t)& =& -\omega _{\mathrm{RF}}^{3}\;\hat{V }_{ 1}\;\cos (\omega _{\mathrm{RF}}t +\varphi _{1}) - k^{3}\omega _{ \mathrm{RF}}^{3}\;\hat{V }_{ 2}\;\cos (k\omega _{\mathrm{RF}}t +\varphi _{2}), {}\\ \end{array}$$

this leads to:

$$\displaystyle\begin{array}{rcl} V _{0}& =& \hat{V }_{1}\;\sin \varphi _{1} +\hat{ V }_{2}\;\sin \varphi _{2},{}\end{array}$$

(5.19)

$$\displaystyle\begin{array}{rcl} \hat{V }_{1}\;\cos \varphi _{1}& =& -k\;\hat{V }_{2}\;\cos \varphi _{2},{}\end{array}$$

(5.20)

$$\displaystyle\begin{array}{rcl} \hat{V }_{1}\;\sin \varphi _{1}& =& -k^{2}\;\hat{V }_{ 2}\;\sin \varphi _{2},{}\end{array}$$

(5.21)

$$\displaystyle\begin{array}{rcl} -\hat{V }_{1}\;\cos \varphi _{1}& -& k^{3}\;\hat{V }_{ 2}\;\cos \varphi _{2} > 0.{}\end{array}$$

(5.22)

The last equation shows that the substitution

$$\displaystyle{\varphi '_{2} =\varphi _{2}-\pi }$$

is suitable, since we have

$$\displaystyle{\cos \varphi _{2} = -\cos \varphi '_{2}}$$

and

$$\displaystyle{\sin \varphi _{2} = -\sin \varphi '_{2}.}$$

Equations (5.20) and (5.21) then lead to

$$\displaystyle{ \tan \varphi _{1} = k\;\tan \varphi '_{2}. }$$

(5.23)

Equation (5.20) provides

$$\displaystyle{ \frac{\hat{V }_{2}} {\hat{V }_{1}} = \frac{\cos \varphi _{1}} {k\;\cos \varphi '_{2}}. }$$

(5.24)

If we insert Eq. (5.21) into Eq. (5.19), we obtain

$$\displaystyle{V _{0} =\hat{ V }_{1}\;\sin \varphi _{1}\left (1 - \frac{1} {k^{2}}\right ),}$$

or

$$\displaystyle{V _{0} =\hat{ V }_{2}\;\sin \varphi '_{2}\left (k^{2} - 1\right ).}$$

From k > 1 and V₀ ≥ 0, it follows that

$$\displaystyle{0 \leq \varphi _{1},\varphi '_{2} \leq \pi.}$$

Phases outside the interval $\left]-\pi,+\pi \right]$ do not lead to signal forms that differ from those within the interval and are therefore excluded.

Inserting Eq. (5.20) into the inequality (5.22) leads to

$$\displaystyle{\hat{V }_{1}\;\cos \varphi _{1}\left (k^{2} - 1\right ) > 0}$$

and

$$\displaystyle{\hat{V }_{2}\;\cos \varphi '_{2}\left (k^{3} - k\right ) > 0.}$$

This reduces the selection to

$$\displaystyle{0 \leq \varphi _{1},\varphi '_{2} < \frac{\pi } {2}.}$$

The phase relation given by Eq. (5.23) and the amplitude relation given by Eq. (5.24) are both displayed in Fig. 5.8 for the simplest case k = 2.

Fig. 5.8

Phase and amplitude relations between first and second harmonics for k = 2

The signal forms V (t) that are obtained for different values of $\varphi _{1}$ are shown in Fig. 5.9. It is obvious that phases $\varphi _{1} > 45^{\circ }$ are usually not of interest, because the corresponding bucket³ becomes too small.

Fig. 5.9

Dual-harmonic signal forms V (t) for k = 2

5.6 Bunch Description by Means of Moments

We will now analyze the influence of gap voltage modulations. The results presented in this section and in the next one are based on the article [16] and were extended in [17].

Based on the unmodulated harmonic gap voltage

$$\displaystyle{V (t) =\hat{ V }_{0}(t)\;\sin (\varphi _{\mathrm{RF}}(t))}$$

with

$$\displaystyle{\varphi _{\mathrm{RF}}(t) =\int \omega _{\mathrm{RF}}(t)\;\mathrm{d}t,}$$

we now introduce a phase modulation $\Delta \varphi _{\mathrm{gap}}(t)$ and an amplitude modulation ε(t) such that

$$\displaystyle{ V (t) =\hat{ V }_{0}(t)(1 +\epsilon (t))\;\sin (\varphi _{\mathrm{RF}}(t) - \Delta \varphi _{\mathrm{gap}}(t)) }$$

(5.25)

is obtained.

The reference particle is defined in such a way that it arrives at the accelerating gap when $\varphi _{\mathrm{RF}}(t) =\varphi _{\mathrm{R}}(t) + 2\pi k$ is satisfied. Here, the integer k is the bunch repetition number. For an asynchronous particle, the arrival time is defined by $\varphi _{\mathrm{RF}}(t) =\varphi _{\mathrm{R}}(t) + 2\pi k + \Delta \varphi _{\mathrm{RF}}(t)$. The magnetic field B in the bending dipoles and the quantities $\hat{V }_{0}$, ω_RF, and $\varphi _{\mathrm{R}}$ are chosen in such a way that the reference particle follows the reference path. All these quantities vary slowly with time in comparison with the synchrotron oscillation. The modulation functions ε(t) and $\Delta \varphi _{\mathrm{gap}}(t)$, however, may vary faster.

According to Eqs. (3.18) and (3.19), the nonlinear differential equations are

$$\displaystyle\begin{array}{rcl} \Delta \dot{\varphi }_{\mathrm{RF}}& =& \frac{2\pi h\eta _{\mathrm{R}}} {T_{\mathrm{R}}\beta _{\mathrm{R}}^{2}W_{\mathrm{R}}}\Delta W,{}\end{array}$$

(5.26)

$$\displaystyle\begin{array}{rcl} \Delta \dot{W}& =& \frac{Q\hat{V }_{0}} {T_{\mathrm{R}}} [(1+\epsilon ) \cdot \sin (\varphi _{\mathrm{R}} + \Delta \varphi _{\mathrm{RF}} - \Delta \varphi _{\mathrm{gap}}) -\sin \varphi _{\mathrm{R}}].{}\end{array}$$

(5.27)

For small values of $\vert \Delta \varphi _{\mathrm{RF}} - \Delta \varphi _{\mathrm{gap}}\vert \ll 1$, we have

$$\displaystyle{\Delta \dot{W} \approx \frac{Q\hat{V }_{0}\;\cos \varphi _{\mathrm{R}}} {T_{\mathrm{R}}} (1+\epsilon )\left (\Delta \varphi _{\mathrm{RF}} - \Delta \varphi _{\mathrm{gap}} + \frac{\epsilon \;\tan \varphi _{\mathrm{R}}} {1+\epsilon }\right ).}$$

Defining

$$\displaystyle{ \Delta \tilde{\varphi }_{\mathrm{gap}} = \Delta \varphi _{\mathrm{gap}} - \frac{\epsilon } {1+\epsilon }\;\tan \varphi _{\mathrm{R}} }$$

(5.28)

leads to

$$\displaystyle{ \Delta \dot{W} \approx \frac{Q\hat{V }_{0}\;\cos \varphi _{\mathrm{R}}} {T_{\mathrm{R}}} (1+\epsilon )(\Delta \varphi _{\mathrm{RF}} - \Delta \tilde{\varphi }_{\mathrm{gap}}). }$$

(5.29)

By a combination of Eqs. (5.26) and (5.29), we get

$$\displaystyle{\Delta \ddot{\varphi }_{\mathrm{RF}} = \frac{2\pi h\eta _{\mathrm{R}}Q\hat{V }_{0}\;\cos \varphi _{\mathrm{R}}} {T_{\mathrm{R}}^{2}\beta _{\mathrm{R}}^{2}W_{\mathrm{R}}}(1+\epsilon )(\Delta \varphi _{\mathrm{RF}} - \Delta \tilde{\varphi }_{\mathrm{gap}}).}$$

According to Eq. (3.25), the synchrotron frequency is defined by

$$\displaystyle{\omega _{\mathrm{S}} =\omega _{\mathrm{S,0}} = 2\pi f_{\mathrm{S,0}} = \sqrt{\frac{2\pi h\hat{V }_{0 } (-\eta _{\mathrm{R} } \;Q\;\cos \varphi _{\mathrm{R} } )} {T_{\mathrm{R}}^{2}\beta _{\mathrm{R}}^{2}W_{\mathrm{R}}}},}$$

which yields

$$\displaystyle{ \Delta \ddot{\varphi }_{\mathrm{RF}} +\omega _{ \mathrm{S}}^{2}(1+\epsilon )\Delta \varphi _{\mathrm{ RF}} =\omega _{ \mathrm{S}}^{2}(1+\epsilon )\Delta \tilde{\varphi }_{\mathrm{ gap}}. }$$

(5.30)

Using the new variables

$$\displaystyle{x = \Delta \varphi _{\mathrm{RF}},\mbox{ }y = C\Delta \dot{\varphi }_{\mathrm{RF}},}$$

we obtain

$$\displaystyle\begin{array}{rcl} \dot{x}& =& \frac{1} {C}\;y, {}\\ \dot{y}& =& -C\omega _{\mathrm{S}}^{2}(1+\epsilon )(x - \Delta \tilde{\varphi }_{\mathrm{ gap}}). {}\\ \end{array}$$

Here the factor C was introduced because we want the trajectories to be circles if no excitations are present (ε = 0 and $\Delta \tilde{\varphi }_{\mathrm{gap}} = 0$). Based on this requirement, we can easily determine C:

$$\displaystyle{x =\cos (\omega _{\mathrm{S}}t)\mbox{ } \Rightarrow y = C\dot{x} = -C\omega _{\mathrm{S}}\;\sin (\omega _{\mathrm{S}}t)\mbox{ } \Rightarrow C = -\frac{1} {\omega _{\mathrm{S}}}.}$$

Thus, we obtain (note that C and ω_S vary slowly, and therefore we neglect the time derivative)

$$\displaystyle\begin{array}{rcl} \dot{x}& =& -\omega _{\mathrm{S}}\;y,{}\end{array}$$

(5.31)

$$\displaystyle\begin{array}{rcl} \dot{y}& =& \omega _{\mathrm{S}}(1+\epsilon )(x - \Delta \tilde{\varphi }_{\mathrm{gap}}).{}\end{array}$$

(5.32)

5.6.1 Phase Oscillations

Whereas Eqs. (5.31) and (5.32) are valid for individual particles, we now consider bunches with N_b particles. The mean values are defined as

$$\displaystyle{\bar{x} = \frac{1} {N_{\mathrm{b}}}\sum _{k=1}^{N_{\mathrm{b}} }x_{k},\mbox{ }\bar{y} = \frac{1} {N_{\mathrm{b}}}\sum _{k=1}^{N_{\mathrm{b}} }y_{k}.}$$

This leads to

$$\displaystyle{ \dot{\bar{x}} = \frac{1} {N_{\mathrm{b}}}\sum _{k=1}^{N_{\mathrm{b}} }\dot{x}_{k} = -\omega _{\mathrm{S}} \frac{1} {N_{\mathrm{b}}}\sum _{k=1}^{N_{\mathrm{b}} }y_{k} = -\omega _{\mathrm{S}}\bar{y} }$$

(5.33)

and

$$\displaystyle{\dot{\bar{y}} = \frac{1} {N_{\mathrm{b}}}\sum _{k=1}^{N_{\mathrm{b}} }\dot{y}_{k} =\omega _{\mathrm{S}}(1+\epsilon ) \frac{1} {N_{\mathrm{b}}}\sum _{k=1}^{N_{\mathrm{b}} }(x_{k} - \Delta \tilde{\varphi }_{\mathrm{gap}})}$$

$$\displaystyle{ \Rightarrow \dot{\bar{ y}} =\omega _{\mathrm{S}}(1+\epsilon )(\bar{x} - \Delta \tilde{\varphi }_{\mathrm{gap}}). }$$

(5.34)

If we combine the results (5.33) and (5.34) for slowly varying ω_S, we obtain

$$\displaystyle{ \ddot{\bar{x}} = -\omega _{\mathrm{S}}\dot{\bar{y}} = -\omega _{\mathrm{S}}^{2}(1+\epsilon )(\bar{x} - \Delta \tilde{\varphi }_{\mathrm{ gap}}). }$$

(5.35)

We may interpret $\bar{x}$ as the bunch center. Hence, we see that the equation for the bunch center has the same form as Eq. (5.30) for the individual particles. This means that the whole bunch may oscillate with the synchrotron frequency. One should note, however, that we allowed only small oscillation amplitudes to get to this result. Such an oscillation of a whole bunch is called a coherent dipole oscillation. It may, for instance, be generated by placing a matched bunch off-center into the bucket. Since all particles rotate in phase space with the synchrotron frequency, this will, in this case, also be true for the whole bunch. However, it is clear that for a realistic bunch size, such an oscillation will soon lead to filamentation. Hence, our linearization will describe only the coherent dipole oscillation in the very beginning and only for small oscillation amplitudes.

5.6.2 Amplitude Oscillations

We define the following quantities:

$$\displaystyle{ a_{x} = \frac{1} {N_{\mathrm{b}}}\sum _{k=1}^{N_{\mathrm{b}} }x_{k}^{2},\mbox{ }a_{ y} = \frac{1} {N_{\mathrm{b}}}\sum _{k=1}^{N_{\mathrm{b}} }y_{k}^{2},\mbox{ }\xi = \frac{1} {N_{\mathrm{b}}}\sum _{k=1}^{N_{\mathrm{b}} }x_{k}y_{k}, }$$

(5.36)

$$\displaystyle{v_{x} = \frac{1} {N_{\mathrm{b}}}\sum _{k=1}^{N_{\mathrm{b}} }(x_{k} -\bar{ x})^{2} = \frac{1} {N_{\mathrm{b}}}\sum _{k=1}^{N_{\mathrm{b}} }(x_{k}^{2} - 2x_{ k}\bar{x} +\bar{ x}^{2}) = a_{ x} -\bar{ x}^{2},}$$

$$\displaystyle{v_{y} = a_{y} -\bar{ y}^{2}.}$$

Please note that v_x corresponds to the variance of the quantities x_k if a division by N_b − 1 is used instead of division by N_b. Since we are interested only in large (particle) numbers N_b, this difference is negligible.

The quantity $\sqrt{v_{x}}$ represents the (rms) bunch length, whereas $\sqrt{v_{ y}}$ represents the (rms) height of the bunch in the phase space (x, y). We get

$$\displaystyle{\dot{a}_{x} = \frac{1} {N_{\mathrm{b}}}\sum _{k=1}^{N_{\mathrm{b}} }2x_{k}\dot{x}_{k} = -2\omega _{\mathrm{S}}\xi,}$$

$$\displaystyle{\dot{a}_{y} = \frac{1} {N_{\mathrm{b}}}\sum _{k=1}^{N_{\mathrm{b}} }2y_{k}\dot{y}_{k} = 2\omega _{\mathrm{S}}(1+\epsilon ) \frac{1} {N_{\mathrm{b}}}\sum _{k=1}^{N_{\mathrm{b}} }y_{k}(x_{k} - \Delta \tilde{\varphi }_{\mathrm{gap}})}$$

$$\displaystyle{\Rightarrow \dot{ a}_{y} = 2\omega _{\mathrm{S}}(1+\epsilon )(\xi -\bar{y}\Delta \tilde{\varphi }_{\mathrm{gap}})}$$

$$\displaystyle\begin{array}{rcl} \dot{\xi }& =& \frac{1} {N_{\mathrm{b}}}\sum _{k=1}^{N_{\mathrm{b}} }(\dot{x}_{k}y_{k} + x_{k}\dot{y}_{k}) = {}\\ & =& -\omega _{\mathrm{S}}a_{y} +\omega _{\mathrm{S}}(1+\epsilon )(a_{x} -\bar{ x}\Delta \tilde{\varphi }_{\mathrm{gap}}), {}\\ \end{array}$$

$$\displaystyle{ \dot{v}_{x} =\dot{ a}_{x} - 2\bar{x}\dot{\bar{x}} = -2\omega _{\mathrm{S}}\xi + 2\omega _{\mathrm{S}}\bar{x}\bar{y} = -2\omega _{\mathrm{S}}\alpha. }$$

(5.37)

Here we defined $\alpha =\xi -\bar{x}\bar{y}$ in order to have the same form in the expressions for a_x and for v_x. We obtain

$$\displaystyle\begin{array}{rcl} \dot{v}_{y} =\dot{ a}_{y} - 2\bar{y}\dot{\bar{y}}& =& 2\omega _{\mathrm{S}}(1+\epsilon )(\xi -\bar{y}\Delta \tilde{\varphi }_{\mathrm{gap}}) - {}\\ &-& 2\omega _{\mathrm{S}}(1+\epsilon )\bar{y}(\bar{x} - \Delta \tilde{\varphi }_{\mathrm{gap}}) {}\\ \end{array}$$

$$\displaystyle{ \Rightarrow \dot{ v}_{y} = 2\omega _{\mathrm{S}}(1+\epsilon )\alpha, }$$

(5.38)

$$\displaystyle\begin{array}{rcl} \dot{\alpha }& =& \dot{\xi }-\dot{\bar{x}}\bar{y} -\bar{ x}\dot{\bar{y}} = {}\\ & =& -\omega _{\mathrm{S}}a_{y} +\omega _{\mathrm{S}}(1+\epsilon )(a_{x} -\bar{ x}\Delta \tilde{\varphi }_{\mathrm{gap}}) + {}\\ & +& \omega _{\mathrm{S}}\bar{y}^{2} -\omega _{\mathrm{ S}}(1+\epsilon )\bar{x}(\bar{x} - \Delta \tilde{\varphi }_{\mathrm{gap}}) {}\\ \end{array}$$

$$\displaystyle{ \Rightarrow \dot{\alpha }= -\omega _{\mathrm{S}}v_{y} +\omega _{\mathrm{S}}(1+\epsilon )v_{x}. }$$

(5.39)

Now we are able to derive a differential equation for v_x, i.e., for the bunch length oscillation.

Combining Eqs. (5.37) and (5.38) yields

$$\displaystyle{ \dot{v}_{y} = -(1+\epsilon )\dot{v}_{x}. }$$

(5.40)

We now combine Eq. (5.37) with Eq. (5.39):

$$\displaystyle{ \ddot{v}_{x} = 2\omega _{\mathrm{S}}^{2}v_{ y} - 2\omega _{\mathrm{S}}^{2}(1+\epsilon )v_{ x} }$$

(5.41)

$$\displaystyle{\Rightarrow \stackrel{\ldots }{v}_{x} = 2\omega _{\mathrm{S}}^{2}\dot{v}_{ y} - 2\omega _{\mathrm{S}}^{2}(1+\epsilon )\dot{v}_{ x} - 2\omega _{\mathrm{S}}^{2}\dot{\epsilon }v_{ x}.}$$

Using Eq. (5.40), we finally get

$$\displaystyle{ \stackrel{\ldots }{v}_{x} = -4\omega _{\mathrm{S}}^{2}(1+\epsilon )\dot{v}_{ x} - 2\omega _{\mathrm{S}}^{2}\dot{\epsilon }v_{ x}. }$$

(5.42)

Please note that for ε = 0, the standard differential equation

$$\displaystyle{ \ddot{v}_{x} + (2\omega _{\mathrm{S}})^{2}v_{ x} =\mathrm{ const} }$$

(5.43)

is obtained, which corresponds to an oscillation with frequency 2ω_S, a so-called quadrupole oscillation. Due to the linearization, an initial quadrupole oscillation will continue forever.

We saw above that a dipole oscillation is generated if a matched bunch is placed off-center in the bucket. In order to generate a quadrupole oscillation, we do not place the bunch off-center in the bucket, but we consider an unmatched bunch. Since a matched bunch has a circular shape in our phase space coordinates (x, y), this means that a slightly elliptical bunch has to be considered. Let us assume that the major axis of the ellipse in phase space is oriented in the x direction at the beginning. Since the individual particles rotate in phase space with the synchrotron frequency, it is then clear that after a quarter of a synchrotron period, the major axis will be directed in the y direction, i.e., the ellipse is then standing upright. After half a synchrotron period, the major of the ellipse will again be oriented horizontally. This explains the oscillation frequency 2ω_S of the quadrupole oscillation. The following points have to be emphasized:

• The derivations presented in this section imply that only small deviations from the matched bunch are allowed.

• The larger the bunch is, the more time will the particles on the bunch contour need for one revolution in phase space. Therefore, the oscillation is more accurately described if an effective synchrotron frequency ω_S is considered instead of the synchrotron frequency ω_S,0, which is valid for particles close to the bucket center. In this case, ω_S describes the coherent motion of the particles.

• For realistic bunch sizes, filamentation will occur after a few oscillation periods, so that that a pure quadrupole oscillation will be visible only at the very beginning.

Now we derive the differential equation for v_y, i.e., for the amplitude oscillation. Equation (5.38) yields

$$\displaystyle{ \frac{\dot{v}_{y}} {1+\epsilon } = 2\omega _{\mathrm{S}}\alpha.}$$

The time derivative is

$$\displaystyle{\frac{\ddot{v}_{y}(1+\epsilon ) -\dot{\epsilon }\dot{ v}_{y}} {(1+\epsilon )^{2}} = -2\omega _{\mathrm{S}}^{2}v_{ y} + 2\omega _{\mathrm{S}}^{2}(1+\epsilon )v_{ x},}$$

where we used Eq. (5.39) on the right-hand side. We divide by (1 +ε):

$$\displaystyle{ \frac{\ddot{v}_{y}} {(1+\epsilon )^{2}} - \frac{\dot{\epsilon }\dot{v}_{y}} {(1+\epsilon )^{3}} = -2\omega _{\mathrm{S}}^{2} \frac{v_{y}} {1+\epsilon } + 2\omega _{\mathrm{S}}^{2}v_{ x}.}$$

Now another time derivative leads to $\dot{v}_{x}$ on the right-hand side, so that we can use Eq. (5.40) to eliminate v_x completely. After some steps, we obtain

$$\displaystyle\begin{array}{rcl} & & \stackrel{\ldots }{v}_{y} -\frac{3\ddot{v}_{y}\dot{\epsilon }} {1+\epsilon } +\dot{ v}_{y}\left (4\omega _{\mathrm{S}}^{2}(1+\epsilon ) - \frac{\ddot{\epsilon }} {1+\epsilon } + \frac{3\dot{\epsilon }^{2}} {(1+\epsilon )^{2}}\right ) = 2\omega _{\mathrm{S}}^{2}\dot{\epsilon }v_{ y}.{}\end{array}$$

(5.44)

This differential equation for v_y differs from Eq. (5.42) for v_x only by terms that are of higher order with respect to ε. Furthermore, the sign of the excitation term $2\omega _{\mathrm{S}}^{2}\dot{\epsilon }v_{y}$ is different for v_x and v_y, which matches our expectation, since the bunch is short when its amplitude is high, whereas the bunch is long when its amplitude is small.

Please note that we have not introduced any approximations to derive the differential equations (5.42) and (5.44) from Eqs. (5.31) and (5.32).

According to [4], these differential equations have the following solution:

$$\displaystyle\begin{array}{rcl} v_{x} = C_{x1}w_{x1}^{2} + C_{ x2}w_{x1}w_{x2} + C_{x3}w_{x2}^{2},& &{}\end{array}$$

(5.45)

$$\displaystyle\begin{array}{rcl} v_{y} = C_{y1}w_{y1}^{2} + C_{ y2}w_{y1}w_{y2} + C_{y3}w_{y2}^{2}.& &{}\end{array}$$

(5.46)

The functions w_x1 and w_x2 are the linearly independent solutions of

$$\displaystyle{\ddot{w}_{x} +\omega _{ \mathrm{S}}^{2}(1+\epsilon )w_{ x} = 0,}$$

whereas the functions w_y1 and w_y2 are the linearly independent solutions of

$$\displaystyle{\ddot{w}_{y} - \frac{\dot{\epsilon }} {1+\epsilon }\dot{w}_{y} +\omega _{ \mathrm{S}}^{2}(1+\epsilon )w_{ y} = 0.}$$

In the trivial case ε = 0, we may choose

$$\displaystyle{w_{x1} = w_{y1} =\cos (\omega _{\mathrm{S}}t)\mbox{, }w_{x2} = w_{y2} =\sin (\omega _{\mathrm{S}}t)}$$

as a solution. Due to Eqs. (5.45) and (5.46), v_x and v_y will oscillate with twice the frequency, in accord with Eq. (5.43).

5.6.3 Linearization

The derived equations (5.33), (5.34), (5.41), and (5.40),

$$\displaystyle\begin{array}{rcl} \dot{\bar{x}}& =& -\omega _{\mathrm{S}}\bar{y}, {}\\ \dot{\bar{y}}& =& \omega _{\mathrm{S}}(1+\epsilon )(\bar{x} - \Delta \tilde{\varphi }_{\mathrm{gap}}), {}\\ \ddot{v}_{x}& =& 2\omega _{\mathrm{S}}^{2}v_{ y} - 2\omega _{\mathrm{S}}^{2}(1+\epsilon )v_{ x}, {}\\ \dot{v}_{y}& =& -(1+\epsilon )\dot{v}_{x}, {}\\ \end{array}$$

can be written as a state-space model⁴

$$\displaystyle{\dot{\vec{x}} = \vec{f}(\vec{x},\epsilon,\Delta \tilde{\varphi }_{\mathrm{gap}})}$$

with state vector

$$\displaystyle{\vec{x} = \left (\begin{array}{*{10}c} x_{1} & x_{2} & x_{3} & x_{4} & x_{5} \end{array} \right )^{\mathrm{T}} = \left (\begin{array}{*{10}c} \bar{x}&\bar{y}&v_{x}&\dot{v}_{x}&v_{y} \end{array} \right )^{\mathrm{T}}}$$

and the nonlinear function

$$\displaystyle{\vec{f}(\vec{x},\epsilon,\Delta \tilde{\varphi }_{\mathrm{gap}}) = \left (\begin{array}{*{10}c} -\omega _{\mathrm{S}}x_{2} \\ \omega _{\mathrm{S}}(1+\epsilon )(x_{1} - \Delta \tilde{\varphi }_{\mathrm{gap}})\\ x_{ 4} \\ 2\omega _{\mathrm{S}}^{2}x_{5} - 2\omega _{\mathrm{S}}^{2}(1+\epsilon )x_{3} \\ -(1+\epsilon )x_{4} \end{array} \right ).}$$

In the following, a linearization with $\Delta \vec{x} = \vec{x} -\vec{x}_{\mathrm{op}}$ around the operating point

$$\displaystyle{\vec{x}_{\mathrm{op}} = \left (\begin{array}{*{10}c} 0&0&v_{0} & 0&v_{0} \end{array} \right )^{\mathrm{T}},\quad \epsilon _{\mathrm{ op}} = 0,\quad \Delta \tilde{\varphi }_{\mathrm{gap,op}} = 0}$$

is performed, which corresponds to the matched circle-shaped bunch. This linearization (see Sect. 7.1.3, cf. [18]) leads to the linear system

$$\displaystyle\begin{array}{rcl} \Delta \dot{\vec{x}}(t) = A \cdot \Delta \vec{x}(t) + \vec{b}_{1}\epsilon (t) + \vec{b}_{2}\Delta \tilde{\varphi }_{\mathrm{gap}}(t)& &{}\end{array}$$

(5.47)

with the system matrix (Jacobian matrix)

$$\displaystyle{A = \left.\frac{\partial \vec{f}} {\partial \vec{x}}\right\vert _{\mathrm{op}} = \left(\begin{array}{*{10}c} 0&-\omega _{\mathrm{S}} & 0 & 0 & 0 \\ \omega _{\mathrm{S}} & 0 & 0 & 0 & 0 \\ 0& 0 & 0 & 1 & 0 \\ 0& 0 &-2\omega _{\mathrm{S}}^{2} & 0 &2\omega _{\mathrm{S}}^{2} \\ 0& 0 & 0 &-1& 0 \end{array} \right)}$$

and the input matrices

$$\displaystyle{\vec{b}_{1} = \left.\frac{\partial \vec{f}} {\partial \epsilon } \right\vert _{\mathrm{op}} = \left(\begin{array}{*{10}c} 0&0&0&-2\omega _{\mathrm{S}}^{2}v_{0} & 0 \end{array} \right)^{\mathrm{T}}}$$

and

$$\displaystyle{\vec{b}_{2} = \left.\frac{\partial \vec{f}} {\partial \Delta \tilde{\varphi }_{\mathrm{gap}}}\right\vert _{\mathrm{op}} = \left(\begin{array}{*{10}c} 0&-\omega _{\mathrm{S}} & 0&0&0 \end{array} \right)^{\mathrm{T}}.}$$

Please note that the matrix A has a block diagonal structure with one block corresponding to the dynamics of the bunch center $\bar{x}$ and one to the dynamics of the bunch variance v_x. In addition, the bunch center is influenced only by $\Delta \tilde{\varphi }_{\mathrm{gap}}$, and the bunch variance only by ε.

Comparing the equations for $\Delta x_{3}$ and $\Delta x_{5}$ in (5.47) yields $\Delta \dot{x}_{3}(t) = -\Delta \dot{x}_{5}(t)$ and thus

$$\displaystyle{ \Delta x_{3}(t) + \Delta x_{5}(t) = v_{x}(t) + v_{y}(t) - 2v_{0} =\mathrm{ const}, }$$

(5.48)

which implies that the bunch variances are connected by an algebraic equation and cannot be controlled independently. It must be possible in principle that the solution $\vec{x}$ of the differential equation reaches the operating point $\vec{x}_{\mathrm{op}}$ (e.g., as an initial condition). For the operating point,

$$\displaystyle{ \Delta x_{3} + \Delta x_{5} = 0 }$$

(5.49)

is valid, which therefore holds in general for every t, due to Eq. (5.48).

With (5.47) and (5.49), linear differential equations of second order can be derived for the bunch center using $\Delta x_{1} =\bar{ x}$ and for the bunch variance using $\Delta x_{3} = v_{x} - v_{0}$:

$$\displaystyle{\ddot{\bar{x}} +\omega _{ \mathrm{S}}^{2}\bar{x} =\omega _{ \mathrm{ S}}^{2}\Delta \tilde{\varphi }_{\mathrm{ gap}},}$$

$$\displaystyle{\ddot{v}_{x} + 4\omega _{\mathrm{S}}^{2}(v_{ x} - v_{0}) = -2\omega _{\mathrm{S}}^{2}v_{ 0}\epsilon.}$$

A phase modulation mainly influences the dipole oscillation whereas an amplitude modulation primarily affects the quadrupole oscillation.

5.6.4 RMS Emittance

In the previous sections, we defined the longitudinal emittance of the beam as the area in phase space that is filled by the particles. This is a very transparent definition from a geometric point of view. For a real particle distribution whose density differs from point to point in phase space, however, this definition is not satisfactory. In practice, one cannot decide easily where the boundaries of a given bunch are.

Therefore, one sometimes defines the emittance as the area of a contour that contains 95% of all particles. On the one hand, this percentage is chosen arbitrarily, and on the other hand, it is not easy to define this contour in a unique way. Therefore, an emittance definition is needed that can be determined for every cloud of particles. The motivation for this definition of the RMS emittance [19, 20], which is presented in the following, is based on [21].

Let us assume that particle number i is located in a Cartesian coordinate system (which represents our phase space) at the position $\vec{r}_{i} = x_{i}\;\vec{e}_{x} + y_{i}\;\vec{e}_{y}$. The area of the triangle formed by particle number i, particle number k, and the origin is then given by the vector product

$$\displaystyle{\frac{1} {2}\;\vec{r}_{i} \times \vec{ r}_{k} = \frac{1} {2}\;(x_{i}\;\vec{e}_{x} + y_{i}\;\vec{e}_{y}) \times (x_{k}\;\vec{e}_{x} + y_{k}\;\vec{e}_{y}) = \frac{\vec{e}_{z}} {2} \;(x_{i}\;y_{k} - x_{k}y_{i}).}$$

$$\displaystyle{\Rightarrow A_{ik} = \frac{1} {2}\;\vert x_{i}\;y_{k} - x_{k}y_{i}\vert.}$$

For the next step, we abandon the idea of calculating an exact area for the particle cloud. We just need an expression that has some similarity to an area. Therefore, we simply sum up the squares of these areas for all possible particle pairs:

$$\displaystyle{A^{2} = \frac{1} {N_{\mathrm{b}}^{2} - N_{\mathrm{b}}}\sum _{i=1}^{N_{\mathrm{b}} }\sum _{k=1}^{N_{\mathrm{b}} }A_{ik}^{2}.}$$

The terms where i = k holds do not contribute anything to the sum. Therefore, we divided by N_b(N_b − 1) to get the average. We obtain

$$\displaystyle{A^{2} = \frac{1} {4(N_{\mathrm{b}}^{2} - N_{\mathrm{b}})}\sum _{i=1}^{N_{\mathrm{b}} }\sum _{k=1}^{N_{\mathrm{b}} }(x_{i}\;y_{k}-x_{k}y_{i})^{2} = \frac{1} {4(N_{\mathrm{b}}^{2} - N_{\mathrm{b}})}\sum _{i=1}^{N_{\mathrm{b}} }\sum _{k=1}^{N_{\mathrm{b}} }(x_{i}^{2}\;y_{ k}^{2}+x_{ k}^{2}\;y_{ i}^{2}-2x_{ i}\;x_{k}\;y_{i}\;y_{k}).}$$

The first two terms lead to the same sum, since only the roles of i and k are interchanged:

$$\displaystyle{ A^{2} = \frac{1} {4(N_{\mathrm{b}}^{2} - N_{\mathrm{b}})}\sum _{i=1}^{N_{\mathrm{b}} }\sum _{k=1}^{N_{\mathrm{b}} }(2x_{i}^{2}\;y_{ k}^{2}-2x_{ i}\;x_{k}\;y_{i}\;y_{k}) = \frac{1} {2(N_{\mathrm{b}}^{2} - N_{\mathrm{b}})}\sum _{i=1}^{N_{\mathrm{b}} }\sum _{k=1}^{N_{\mathrm{b}} }(x_{i}^{2}\;y_{ k}^{2}-x_{ i}\;x_{k}\;y_{i}\;y_{k}). }$$

(5.50)

We now return to our definitions (5.36),

$$\displaystyle{a_{x}:= \overline{x^{2}} = \frac{1} {N_{\mathrm{b}}}\sum _{k=1}^{N_{\mathrm{b}} }x_{k}^{2}}$$

and

$$\displaystyle{a_{y}:= \overline{y^{2}} = \frac{1} {N_{\mathrm{b}}}\sum _{k=1}^{N_{\mathrm{b}} }y_{k}^{2},}$$

which lead to

$$\displaystyle{a_{x}a_{y} = \overline{x^{2}}\;\overline{y^{2}} = \frac{1} {N_{\mathrm{b}}^{2}}\left (\sum _{i=1}^{N_{\mathrm{b}} }x_{i}^{2}\right )\left (\sum _{ k=1}^{N_{\mathrm{b}} }y_{k}^{2}\right ) = \frac{1} {N_{\mathrm{b}}^{2}}\sum _{i=1}^{N_{\mathrm{b}} }\sum _{k=1}^{N_{\mathrm{b}} }(x_{i}^{2}y_{ k}^{2}).}$$

This obviously reproduces the first term in Eq. (5.50). In a similar way, the definition

$$\displaystyle{\xi:= \overline{xy} = \frac{1} {N_{\mathrm{b}}}\sum _{k=1}^{N_{\mathrm{b}} }x_{k}y_{k}}$$

leads to

$$\displaystyle{\xi ^{2} = (\overline{xy})^{2} = \frac{1} {N_{\mathrm{b}}^{2}}\sum _{i=1}^{N_{\mathrm{b}} }\sum _{k=1}^{N_{\mathrm{b}} }x_{i}x_{k}y_{i}y_{k},}$$

which reproduces the second term in Eq. (5.50). We therefore get

$$\displaystyle{2A^{2}\left (1 - \frac{1} {N_{\mathrm{b}}}\right ) = a_{x}a_{y} -\xi ^{2} = \overline{x^{2}}\;\overline{y^{2}} - (\overline{xy})^{2}.}$$

Please note that the term in parentheses is close to 1 for sufficiently large numbers N_b. Since we summed up the squares of the areas of the triangles, we have to apply the square root to get a quantity with the correct dimension. This defines the RMS emittance

$$\displaystyle{\varepsilon = C_{\varepsilon }\sqrt{\overline{x^{2 } } \;\overline{y^{2 } } - (\overline{xy } )^{2}},}$$

for which different constant factors $C_{\varepsilon }$ are used in the literature (e.g., $C_{\varepsilon } = 4$ in [19]).

Due to the construction of the emittance (based on triangles that have the origin as one vertex), it is obvious that the value $\varepsilon$ is not invariant under translations of the origin. Points that are far from the origin contribute with large areas, whereas points close to the origin contribute with small areas. Therefore, the origin should correspond to the center of the particle cloud:

$$\displaystyle{\bar{x} = 0,\mbox{ }\bar{y} = 0.}$$

Nevertheless, the RMS emittance has the disadvantage that particles at the boundaries of the bunch have more influence than they deserve and that it is not always a measure for the area (especially for deformed bunches). For bunches with small distortions, however, the RMS emittance may often be used successfully.

We may regard the RMS definition as a special case for a finite number of particles. For a continuous distribution, one would use expected values instead:

$$\displaystyle{\left ( \frac{\varepsilon } {C_{\varepsilon }}\right )^{2} = E(X^{2})\;E(Y ^{2}) - E^{2}(XY ),\mbox{ }E(X) = 0,\mbox{ }E(Y ) = 0.}$$

If one now assumes that the random variables X and Y are independent, one sees that

$$\displaystyle{\left ( \frac{\varepsilon } {C_{\varepsilon }}\right )^{2} = E(X^{2})\;E(Y ^{2})}$$

holds. In this case,

$$\displaystyle{ \frac{\varepsilon } {C_{\varepsilon }} =\sigma _{X}\;\sigma _{Y }}$$

is obtained. For a bunch that has an elliptical shape in phase space, this result is expected, because the area of an ellipse with the two semiaxes r_X, r_Y is π r_Xr_Y. If the distribution of the particles is Gaussian, it is obvious that $C_{\varepsilon }$ can be used to define an elliptical contour that contains a certain percentage of particles.

For purposes of illustration, we conclude this section with a simple example. Let us consider only N_b = 3 particles with the following positions:

$$\displaystyle\begin{array}{rcl} \vec{r}_{1}& =& (+10,+10), {}\\ \vec{r}_{2}& =& (0,+10), {}\\ \vec{r}_{3}& =& (-10,+10). {}\\ \end{array}$$

This leads to $N_{\mathrm{b}}(N_{\mathrm{b}} - 1)/2 = 3$ triangles with areas 50, 50, and 100, respectively. The average of the squares is $15000/3 = 5000$, so that

$$\displaystyle{A = \sqrt{5000} \approx 70.71}$$

is obtained. This can also be calculated formally based on

$$\displaystyle{a_{x} = \frac{200} {3},\mbox{ }a_{y} = \frac{300} {3} = 100,\mbox{ }\xi = 0.}$$

In this example, the origin does not correspond to the center of the particle cloud. If we shifted the origin to this center so that

$$\displaystyle{\bar{x} = 0,\mbox{ }\bar{y} = 0}$$

holds, we would get an area of zero (A = 0). This is due to the fact that the three particles are located on a straight line.

5.7 Longitudinal Bunch Oscillations

We now analyze a few specific oscillations of bunched beams in longitudinal phase space. The undisturbed bunch, i.e., a bunch that is matched to the bucket, is the starting point. As in the previous section, a scaling of the phase space coordinates will be performed in such a way that the contour of the matched bunch is a circle. Instead of physical phase space variables, we again use simple coordinates (x, y) for which the undisturbed bunch is a unit circle:

$$\displaystyle{x =\cos \varphi,\mbox{ }y =\sin \varphi.}$$

The physical phase space representation is obtained if both coordinates are multiplied by the corresponding factors.

5.7.1 Coherent Dipole Mode

As we already discussed in the previous section, the coherent dipole mode of oscillation is obtained if the bunch as a whole is shifted along one coordinate so that it is located off-center in the bucket afterward:

$$\displaystyle{x =\epsilon +\cos \varphi,\mbox{ }y =\sin \varphi.}$$

For the radius r, we obtain

$$\displaystyle{r^{2} = x^{2} + y^{2} = 1 +\epsilon ^{2} + 2\epsilon \;\cos \varphi.}$$

With the help of

$$\displaystyle{\sqrt{1 + a} \approx 1 + \frac{a} {2},}$$

we find, for sufficiently small ε ≪ 1,

$$\displaystyle{r \approx 1 +\epsilon \;\cos \varphi.}$$

5.7.2 Quadrupole Mode

A quadrupole oscillation is obtained if the ratio of the principal axes is slightly modified. For our circular bunch, this means that it becomes elliptical:

$$\displaystyle{x = a\;\cos \varphi,\mbox{ }y = b\;\sin \varphi,}$$

$$\displaystyle{r^{2} = a^{2}\cos ^{2}\varphi + b^{2}\sin ^{2}\varphi = a^{2} + (b^{2} - a^{2})\sin ^{2}\varphi.}$$

Due to

$$\displaystyle{\sin ^{2}\varphi = \frac{1} {2} -\frac{1} {2}\cos (2\varphi ),}$$

we obtain

$$\displaystyle{r^{2} = \frac{a^{2} + b^{2}} {2} + \frac{a^{2} - b^{2}} {2} \;\cos (2\varphi ) = \frac{a^{2} + b^{2}} {2} \left (1 + \frac{a^{2} - b^{2}} {a^{2} + b^{2}}\;\cos (2\varphi )\right ).}$$

For

$$\displaystyle{a = 1 + \frac{\epsilon } {2},\mbox{ }b = 1 - \frac{\epsilon } {2},}$$

we obtain

$$\displaystyle{r \approx 1 +\epsilon \;\cos (2\varphi ).}$$

The quadrupole mode leads to the effect that the bunch is elongated in phase space. The rotation in phase space then leads to an oscillation between short bunches with large peak current and long bunches with small peak current. Hence, a bunch length and bunch amplitude modulation is present.

5.7.3 Generalization

We are now able to see that the two cases discussed before may be generalized to the formula

$$\displaystyle{r \approx 1 +\epsilon \;\cos (m\varphi ).}$$

The dipole mode is obtained for m = 1, the quadrupole oscillation for m = 2. For m = 3, the sextupole oscillation is obtained; for m = 4, the octupole mode, etc.

The mode number m also specifies the eigenfrequency of the oscillation. As the diagrams in Fig. 5.10 show, the bunch in phase space is a polygon with m rounded corners. The rotation of the bunch with frequency f_S results in a projection onto the time axis whose frequency is m times higher. Together with the revolution period, one obtains spectral components at kf_R ± mf_S.

Fig. 5.10

Longitudinal modes of oscillation

In the case of coupled-bunch oscillations, the h bunches in the ring interact with each other. In general, not only do the bunches oscillate as a whole (as would be the case for m = 1), but the individual bunches may be deformed, as described by the bunch shape mode number m.

If we consider coupled oscillations for a specific m, we find that there are h modes altogether, which are characterized by the coupled-bunch mode number n with 0 ≤ n ≤ h − 1, since h bunches may oscillate in h ways with respect to one another. In the simplest case, m = 1 (dipole oscillation of individual bunches) and h = 2, the two bunches may oscillate either in phase (n = 0) or out of phase (n = 1). In general, the phase advance from bunch to bunch is

$$\displaystyle{2\pi \frac{n} {h}.}$$

Hence, coupled-bunch oscillations may be characterized using the two mode numbers m and n (cf. [22] and [23, Sect. 5.6]).

The modes of oscillation may alternatively be defined on the basis of the spectral lines observed in the beam signal. If, for example, strong spectral components are observed at kf_R ± 2f_S, this would indicate a quadrupole oscillation by definition.

The modes of oscillation that were introduced here may be excited if the initial conditions do not correspond to those of a matched bunch. If, for example, the bunch is not centered inside the bucket, coherent dipole oscillations will be the result.

A different cause of longitudinal oscillations is (longitudinal) impedances that act on the beam. If these impedances lead to an unstable situation, specific modes of oscillation will be excited; they have certain growth rates. With respect to the modes of oscillation defined above with mode numbers m and n, one then speaks of coupled-bunch instabilities. Instabilities may be damped using feedback systems (cf. [24]). So-called longitudinal feedback systems are used to reduce undesired longitudinal beam oscillations, whereas transverse feedback systems damp transverse beam oscillations.

5.7.4 Spectrum of the Dipole Oscillation

Consider the time function

$$\displaystyle{ f(t) =\sum _{ k=-\infty }^{+\infty }\delta (t - kT_{\mathrm{ R}} -\tau _{k}), }$$

(5.51)

where τ_k is periodic with period T_S of the synchrotron oscillation. Here T_R is the revolution time,⁵ and T_S is an integer multiple of T_R, so that f(t) is strictly periodic with period T_S. The function f(t) obviously represents a strongly bunched beam that performs coherent dipole oscillations. The corresponding Fourier coefficients are

$$\displaystyle{c_{n} = \frac{1} {T_{\mathrm{S}}}\int _{-T_{\mathrm{S}}/2}^{+T_{\mathrm{S}}/2}f(t)\;e^{-jn\omega _{\mathrm{S}}t}\;\mathrm{d}t.}$$

Hence we have

$$\displaystyle{c_{n} = \frac{1} {T_{\mathrm{S}}}\sum _{k\in M}e^{-jn\omega _{\mathrm{S}}[kT_{\mathrm{R}}+\tau _{k}]}\;\mathrm{d}t.}$$

Here M denotes the set of all indices for which kT_R +τ_k runs from $-T_{\mathrm{S}}/2$ to $+T_{\mathrm{S}}/2$. In order to determine this set in a unique way, we want to satisfy

$$\displaystyle{\tau _{k} \approx \epsilon \;\sin (\omega _{\mathrm{S}}t)}$$

approximately. Since ω_S ≪ ω_R holds, we may insert

$$\displaystyle{t = kT_{\mathrm{R}}}$$

approximately, so that

$$\displaystyle{ \tau _{k} =\epsilon \;\sin (k\;\omega _{\mathrm{S}}T_{\mathrm{R}}) =\epsilon \;\sin (k\;2\pi T_{\mathrm{R}}/T_{\mathrm{S}}) }$$

(5.52)

is now considered as an exact definition. The limits of integration will be obtained for

$$\displaystyle{kT_{\mathrm{R}} = \pm T_{\mathrm{S}}/2,}$$

i.e., for

$$\displaystyle{k = \pm \frac{T_{\mathrm{S}}} {2T_{\mathrm{R}}}.}$$

At this point, we require that $T_{\mathrm{S}}/T_{\mathrm{R}} =\omega _{\mathrm{R}}/\omega _{\mathrm{S}}$ be even in order to have an integer k at the integration limits. For these values of k, the quantity τ_k vanishes. Therefore, the integration limits are located exactly on two Dirac pulses. Since we have to integrate only one period, only one of these two Dirac pulses must be taken into account in the summation. Therefore, we have

$$\displaystyle{M = \left \{-\left ( \frac{T_{\mathrm{S}}} {2T_{\mathrm{R}}} - 1\right ),\ldots, \frac{T_{\mathrm{S}}} {2T_{\mathrm{R}}}\right \}.}$$

Therefore, we obtain

$$\displaystyle{c_{n} = \frac{1} {T_{\mathrm{S}}} + \frac{1} {T_{\mathrm{S}}}e^{-jn\omega _{\mathrm{S}}\left [\frac{T_{\mathrm{S}}} {2} +\epsilon \sin \pi \right ]} + \frac{2} {T_{\mathrm{S}}}\sum _{k=1}^{ \frac{T_{\mathrm{S}}} {2T_{\mathrm{R}}} -1}\cos \left (n\omega _{\mathrm{S}}\left [kT_{\mathrm{R}} +\epsilon \;\sin \left (2\pi kT_{\mathrm{R}}/T_{\mathrm{S}}\right )\right ]\right ).}$$

The first term is obtained for k = 0, the second one for $k = \frac{T_{\mathrm{S}}} {2T_{\mathrm{R}}}$. It follows that

$$\displaystyle{ \fbox{$c_{n} = \frac{1} {T_{\mathrm{S}}}\left [1 + (-1)^{n} + 2\sum _{ k=1}^{ \frac{T_{\mathrm{S}}} {2T_{\mathrm{R}}} -1}\cos \left (2\pi n \frac{1} {T_{\mathrm{S}}}\left [kT_{\mathrm{R}} +\epsilon \;\sin \left (2\pi kT_{\mathrm{R}}/T_{\mathrm{S}}\right )\right ]\right )\right ].$} }$$

(5.53)

We are now looking for a specific expression for the coefficients c_n at the revolution harmonics.

The coefficients c_n correspond to the frequencies n ω_S. Hence the revolution harmonics are located at

$$\displaystyle{p\omega _{\mathrm{R}} = p\frac{\omega _{\mathrm{R}}} {\omega _{\mathrm{S}}} \omega _{\mathrm{S}} = p\frac{T_{\mathrm{S}}} {T_{\mathrm{R}}}\omega _{\mathrm{S}}\mbox{ } \Rightarrow n = p\frac{T_{\mathrm{S}}} {T_{\mathrm{R}}}.}$$

Since T_S∕T_R is an even number, n is also even, and it follows that

$$\displaystyle{c_{n} = \frac{2} {T_{\mathrm{S}}} + \frac{2} {T_{\mathrm{S}}}\sum _{k=1}^{ \frac{T_{\mathrm{S}}} {2T_{\mathrm{R}}} -1}\cos \left (2\pi p \frac{\epsilon } {T_{\mathrm{R}}}\;\sin \left (2\pi kT_{\mathrm{R}}/T_{\mathrm{S}}\right )\right ).}$$

For ε = 0 (no coherent dipole oscillation), we obtain

$$\displaystyle{c_{n} = \frac{2} {T_{\mathrm{S}}} + \frac{2} {T_{\mathrm{S}}}\left ( \frac{T_{\mathrm{S}}} {2T_{\mathrm{R}}} - 1\right ) = \frac{1} {T_{\mathrm{R}}},}$$

which is what one expects for a simple Dirac comb without phase modulation. Now we would have to show that the coefficients c_n that do not correspond to revolution harmonics vanish for ε = 0. At this point, however, we omit the calculation (for odd ratios T_S∕T_R, the calculation is presented in the next section).

Now let T_S∕T_R be an odd integer. Then the summation limits are defined by

$$\displaystyle{k = \pm \frac{1} {2}\left ( \frac{T_{\mathrm{S}}} {T_{\mathrm{R}}} - 1\right ).}$$

It follows that

$$\displaystyle{ \fbox{$c_{n} = \frac{1} {T_{\mathrm{S}}}\left [1 + 2\sum _{k=1}^{\frac{1} {2} \left ( \frac{T_{\mathrm{S}}} {T_{\mathrm{R}}} -1\right )}\cos \left (2\pi n \frac{1} {T_{\mathrm{S}}}\left [kT_{\mathrm{R}} +\epsilon \;\sin \left (2\pi kT_{\mathrm{R}}/T_{\mathrm{S}}\right )\right ]\right )\right ].$} }$$

(5.54)

Also here we seek a specific expression for the coefficients c_n at the revolution harmonics. To this end, we make the substitution

$$\displaystyle{n = p\frac{T_{\mathrm{S}}} {T_{\mathrm{R}}}}$$

and obtain

$$\displaystyle{c_{n} = \frac{1} {T_{\mathrm{S}}}\left [1 + 2\sum _{k=1}^{\frac{1} {2} \left ( \frac{T_{\mathrm{S}}} {T_{\mathrm{R}}} -1\right )}\cos \left (2\pi p \frac{\epsilon } {T_{\mathrm{R}}}\;\sin \left (2\pi kT_{\mathrm{R}}/T_{\mathrm{S}}\right )\right )\right ].}$$

As expected, we get for ε = 0,

$$\displaystyle{c_{n} = \frac{1} {T_{\mathrm{S}}}\left [1 + \left ( \frac{T_{\mathrm{S}}} {T_{\mathrm{R}}} - 1\right )\right ] = \frac{1} {T_{\mathrm{R}}}.}$$

Now we show that all other coefficients that do not belong to the revolution harmonics equal zero. As one finds in Gradshteyn [25] (formula 1.342,2),

$$\displaystyle{\sum _{k=1}^{m}\cos (kx) =\cos \left (\frac{m + 1} {2} x\right )\;\sin \frac{mx} {2} \;\mathrm{cosec}\frac{x} {2}}$$

holds. From Eq. (5.54), we conclude for ε = 0 with the help of this formula that

$$\displaystyle\begin{array}{rcl} c_{n}& =& \frac{1} {T_{\mathrm{S}}}\left(1 + 2\;\cos \left[\frac{1} {4}\left( \frac{T_{\mathrm{S}}} {T_{\mathrm{R}}} + 1\right)\;2\pi n\frac{T_{\mathrm{R}}} {T_{\mathrm{S}}} \right]\;\right. \cdot {}\\ & &\left.\cdot \sin \left[\frac{1} {4}\left( \frac{T_{\mathrm{S}}} {T_{\mathrm{R}}} - 1\right)\;2\pi n\frac{T_{\mathrm{R}}} {T_{\mathrm{S}}} \right]\;\mathrm{cosec}\left[\pi n\frac{T_{\mathrm{R}}} {T_{\mathrm{S}}} \right]\right) {}\\ \end{array}$$

$$\displaystyle{\Rightarrow c_{n}\,=\, \frac{1} {T_{\mathrm{S}}}\left (1\,+\,2\;\cos \left [ \frac{\pi } {2}n\left (1\,+\,\frac{T_{\mathrm{R}}} {T_{\mathrm{S}}} \right )\right ]\;\sin \left [ \frac{\pi } {2}n\left (1 -\frac{T_{\mathrm{R}}} {T_{\mathrm{S}}} \right )\right ]\;\mathrm{cosec}\left [\pi n\frac{T_{\mathrm{R}}} {T_{\mathrm{S}}} \right ]\right ).}$$

From

$$\displaystyle{a = \frac{\pi } {2}n\mbox{ and }b = \frac{\pi } {2}n\frac{T_{\mathrm{R}}} {T_{\mathrm{S}}},}$$

it follows that

$$\displaystyle{c_{n} = \frac{1} {T_{\mathrm{S}}}\left [1 + 2\frac{\cos (a + b)\;\sin (a - b)} {\sin (2b)} \right ].}$$

With the help of trigonometric identities, one easily shows that

$$\displaystyle{\frac{\cos (a + b)\;\sin (a - b)} {\sin (2b)} = \frac{1} {2}\left (\frac{\sin (2a)} {\sin (2b)} - 1\right )}$$

is valid. This leads to

$$\displaystyle{c_{n} = \frac{1} {T_{\mathrm{S}}} \frac{\sin (2a)} {\sin (2b)} = \frac{1} {T_{\mathrm{S}}} \frac{\sin (\pi n)} {\sin \left (\pi n\frac{T_{\mathrm{R}}} {T_{\mathrm{S}}} \right )}.}$$

The numerator is always equal to zero. The coefficient c_n can equal 1∕T_S only if the denominator is zero as well, which is the case if nT_R∕T_S is an integer, i.e., if

$$\displaystyle{n\frac{T_{\mathrm{R}}} {T_{\mathrm{S}}} = p}$$

holds. This, however, is just the condition derived above for the beam harmonics. Hence, the spectrum for ε = 0 corresponds to a Dirac comb, as expected for a Dirac comb in the time domain.

The Fourier coefficients derived here (Eq. (5.53) for even ratios T_S∕T_R and Eq. (5.54) for odd ratios T_S∕T_R) are exact formulas without approximations that are valid for the coherent dipole oscillation defined by Eqs. (5.51) and (5.52). These formulas can also be proven in the scope of distribution theory [16]. It is easily possible to consider realistic bunches instead of the Dirac pulses if one performs a convolution as shown in Sect. 2.1.6.

5.8 A Simple Space Charge Model

We have heretofore assumed that each charged particle with charge Q that belongs to a bunch experiences the same voltage V (t) (depending, of course, on its arrival time) produced by a cavity. If the density of particles becomes larger and larger, this is no longer the case. The charge distribution of the whole particle cloud will influence an individual particle. Such phenomena are called space charge effects.

In this section, a simple space charge model is analyzed. For this purpose, we consider a reference frame in which the particle bunches are at rest. Therefore, a pure electrostatic problem with $\vec{B} = 0$ has to be solved. In this case, Maxwell’s equations reduce to

$$\displaystyle\begin{array}{rcl} \mathrm{div}\;\vec{D}& =& \rho _{q},{}\end{array}$$

(5.55)

$$\displaystyle\begin{array}{rcl} \mathrm{curl}\;\vec{E}& =& 0.{}\end{array}$$

(5.56)

5.8.1 Field in the Rest Frame of the Bunch

We now assume that the beam pipe is perfectly conducting and that it has a cylindrical shape (radius r_bp). The longitudinal axis of the beam pipe defines the z-axis of a cylindrical coordinate system (coordinates $\rho,\varphi,z$), and the beam pipe is assumed to be infinitely long. This means that the curvature of the synchrotron is neglected. The beam itself is modeled as a charge distribution with nonzero charge density inside a cylinder of radius r_beam < r_bp:

$$\displaystyle{\rho _{q}(\rho,z) = \left\{\begin{array}{ll} \rho _{q,0}(z)&\mbox{ for }0 \leq \rho < r_{\mathrm{beam}}\mbox{ (region A),} \\ 0 &\mbox{ for }r_{\mathrm{beam}} \leq \rho < r_{\mathrm{bp}}\mbox{(region B).} \end{array} \right. }$$

In each cross section at constant z, the space charge density is constant for ρ < r_beam. According to⁶

$$\displaystyle{\lambda _{q}(z) =\pi r_{\mathrm{beam}}^{2}\rho _{ q,0}(z),}$$

this space charge density ρ_q may be converted into a line charge density λ_q.

Due to the special setup, we may require $D_{\varphi } = 0$ and no $\varphi$-dependence of the fields. In cylindrical coordinates, Eqs. (5.55) and (5.56) may be written as

$$\displaystyle\begin{array}{rcl} \frac{\partial D_{\rho }} {\partial \rho } + \frac{1} {\rho } \;D_{\rho } + \frac{\partial D_{z}} {\partial z} & =& \rho _{q}(\rho,z),{}\end{array}$$

(5.57)

$$\displaystyle\begin{array}{rcl} \frac{\partial D_{\rho }} {\partial z} -\frac{\partial D_{z}} {\partial \rho } & =& 0.{}\end{array}$$

(5.58)

We calculate the derivative of the first equation with respect to ρ and insert the second equation:

$$\displaystyle{ \frac{\partial ^{2}D_{\rho }} {\partial \rho ^{2}} -\frac{1} {\rho ^{2}} \;D_{\rho } + \frac{1} {\rho } \;\frac{\partial D_{\rho }} {\partial \rho } + \frac{\partial ^{2}D_{\rho }} {\partial z^{2}} = 0. }$$

(5.59)

If one uses the derivative of Eq. (5.57) with respect to z instead, one obtains by inserting Eq. (5.58),

$$\displaystyle{ \frac{\partial ^{2}D_{z}} {\partial \rho ^{2}} + \frac{1} {\rho } \;\frac{\partial D_{z}} {\partial \rho } + \frac{\partial ^{2}D_{z}} {\partial z^{2}} = \frac{\partial \rho _{q}} {\partial z}. }$$

(5.60)

We now attempt to solve Eq. (5.59) by means of a separation ansatz:

$$\displaystyle{D_{\rho } = f(\rho )\;g(z)}$$

$$\displaystyle{\Rightarrow g\;\frac{\partial ^{2}f} {\partial \rho ^{2}} -\frac{1} {\rho ^{2}} \;f\;g + \frac{1} {\rho } \;g\;\frac{\partial f} {\partial \rho } + f \frac{\partial ^{2}g} {\partial z^{2}} = 0}$$

$$\displaystyle{\Rightarrow \frac{1} {f}\;\frac{\partial ^{2}f} {\partial \rho ^{2}} -\frac{1} {\rho ^{2}} + \frac{1} {\rho f}\;\frac{\partial f} {\partial \rho } + \frac{1} {g}\;\frac{\partial ^{2}g} {\partial z^{2}} = 0.}$$

The last term on the left-hand side may depend only on z, whereas all the other terms may depend only on ρ. Therefore, these terms must be constant:

$$\displaystyle{ \frac{1} {f}\;\frac{\partial ^{2}f} {\partial \rho ^{2}} -\frac{1} {\rho ^{2}} + \frac{1} {\rho f}\;\frac{\partial f} {\partial \rho } = C_{0},}$$

$$\displaystyle{\frac{1} {g}\;\frac{\partial ^{2}g} {\partial z^{2}} = -C_{0}.}$$

We need solutions that are periodic in the z direction, because the fields must repeat themselves after one revolution in the synchrotron (we assume that this requirement in combination with the straight cylindrical beam pipe leads to solutions that are similar to a closed, bent beam pipe). Therefore, C₀ > 0 will be valid, and we set $C_{0} = k_{z}^{2}$:

$$\displaystyle{ \rho ^{2}\frac{\mathrm{d}^{2}f} {\mathrm{d}\rho ^{2}} +\rho \; \frac{\mathrm{d}f} {\mathrm{d}\rho } - (\rho ^{2}k_{ z}^{2} + 1)\;f = 0, }$$

(5.61)

$$\displaystyle{\frac{\mathrm{d}^{2}g} {\mathrm{d}z^{2}} + k_{z}^{2}g = 0.}$$

The second equation obviously has the solution

$$\displaystyle{g(z) = g_{1}\;\cos (k_{z}z) + g_{2}\;\sin (k_{z}z)}$$

with constants g₁ and g₂. Due to the periodicity of the solutions after one synchrotron revolution,

$$\displaystyle{k_{z}l = 2\pi p\mbox{ } \Rightarrow k_{z} = p\frac{2\pi } {l} \mbox{ with }p \in \{ 0,1,2,\ldots \}}$$

must be valid.

By means of the substitution u = k_zρ, the first equation (5.61) may be transformed into the modified Bessel’s differential equation

$$\displaystyle{\fbox{$u^{2}\frac{\mathrm{d}^{2}f} {\mathrm{d}u^{2}} + u\;\frac{\mathrm{d}f} {\mathrm{d}u} - (u^{2} + m^{2})\;f = 0$}}$$

with m = 1. Two independent solutions are the modified Bessel functions I_m(u) and K_m(u); see Table A.2 on p. 415 and Fig. A.19. Hence, we get

$$\displaystyle{f(\rho ) = f_{1}\;\mathrm{I}_{1}(k_{z}\rho ) + f_{2}\;\mathrm{K}_{1}(k_{z}\rho )}$$

$$\displaystyle{\Rightarrow D_{\rho } = \left (f_{1}\;\mathrm{I}_{1}(k_{z}\rho ) + f_{2}\;\mathrm{K}_{1}(k_{z}\rho )\right )\left (g_{1}\;\cos (k_{z}z) + g_{2}\;\sin (k_{z}z)\right )}$$

with constants f₁ and f₂. For the derivative with respect to z, we obtain

$$\displaystyle{\frac{\partial D_{\rho }} {\partial z} = k_{z}\left (f_{1}\;\mathrm{I}_{1}(k_{z}\rho ) + f_{2}\;\mathrm{K}_{1}(k_{z}\rho )\right )\left (-g_{1}\;\sin (k_{z}z) + g_{2}\;\cos (k_{z}z)\right ).}$$

According to Eq. (5.58), we get D_z by means of an integration with respect to ρ:

$$\displaystyle{D_{z} = k_{z}\left (-g_{1}\;\sin (k_{z}z) + g_{2}\;\cos (k_{z}z)\right )\int \left (f_{1}\;\mathrm{I}_{1}(k_{z}\rho ) + f_{2}\;\mathrm{K}_{1}(k_{z}\rho )\right )\;\mathrm{d}\rho.}$$

Since the functions K_m(u) have poles at u = 0 for m ∈ { 0, 1, 2, …}, the function K₁(u) cannot be used in region A (0 ≤ ρ < r_beam), because there is no singular charge density at ρ = 0. Hence, the solution in region A is

$$\displaystyle\begin{array}{rcl} D_{\rho }& =& \mathrm{I}_{1}(k_{z}\rho )\;\left (g_{1}^{A}\;\cos (k_{ z}z) + g_{2}^{A}\;\sin (k_{ z}z)\right ),{}\end{array}$$

(5.62)

$$\displaystyle\begin{array}{rcl} D_{z}& =& \zeta (\rho )\;\left (-g_{1}^{A}\;\sin (k_{ z}z) + g_{2}^{A}\;\cos (k_{ z}z)\right ),{}\end{array}$$

(5.63)

where

$$\displaystyle{ \zeta (\rho ) =\int _{ 0}^{\rho }k_{ z}\mathrm{I}_{1}(k_{z}\rho )\;\mathrm{d}\rho +\zeta (0) =\mathrm{ I}_{0}(k_{z}\rho ) - 1 +\zeta (0). }$$

(5.64)

In the last step, we used Table A.2 on p. 415, formula (A.78),

$$\displaystyle{ \fbox{$\mathrm{I}'_{0}(u) =\mathrm{ I}_{1}(u),$} }$$

(5.65)

which implies

$$\displaystyle{\fbox{$\int k_{z}\mathrm{I}_{1}(k_{z}\rho )\;\mathrm{d}\rho =\mathrm{ I}_{0}(k_{z}\rho ) + \mathrm{const}$}}$$

and the function value I₀(0) = 1. In the following, we will also make use of the general formula (A.77),

$$\displaystyle{ \fbox{$\mathrm{K}'_{0}(u) = -\mathrm{K}_{1}(u),$} }$$

(5.66)

which implies

$$\displaystyle{\fbox{$\int k_{z}\mathrm{K}_{1}(k_{z}\rho )\;\mathrm{d}\rho = -\mathrm{K}_{0}(k_{z}\rho ) + \mathrm{const}.$}}$$

In region B (r_beam ≤ ρ < r_bp), we have

$$\displaystyle\begin{array}{rcl} D_{\rho }& =& \left (f_{1}^{B}\;\mathrm{I}_{ 1}(k_{z}\rho ) + f_{2}^{B}\;\mathrm{K}_{ 1}(k_{z}\rho )\right )\left (g_{1}^{B}\;\cos (k_{ z}z) + g_{2}^{B}\;\sin (k_{ z}z)\right ),{}\end{array}$$

(5.67)

$$\displaystyle\begin{array}{rcl} D_{z}& =& \left (f_{1}^{B}\;\int _{ r_{\mathrm{beam}}}^{\rho }k_{z}\mathrm{I}_{1}(k_{z}\rho )\;\mathrm{d}\rho + f_{2}^{B}\;\int _{ r_{\mathrm{beam}}}^{\rho }k_{z}\mathrm{K}_{1}(k_{z}\rho )\;\mathrm{d}\rho +\zeta (r_{\mathrm{beam}})\right ) \cdot \\ & \cdot & \left (-g_{1}^{B}\;\sin (k_{ z}z) + g_{2}^{B}\;\cos (k_{ z}z)\right ) = \\ & =& \left (f_{1}^{B}\left [\mathrm{I}_{ 0}(k_{z}\rho ) -\mathrm{ I}_{0}(k_{z}r_{\mathrm{beam}})\right ] + f_{2}^{B}\left [\mathrm{K}_{ 0}(k_{z}r_{\mathrm{beam}}) -\mathrm{ K}_{0}(k_{z}\rho )\right ] +\zeta (r_{\mathrm{beam}})\right ) \cdot \\ & \cdot & \left (-g_{1}^{B}\;\sin (k_{ z}z) + g_{2}^{B}\;\cos (k_{ z}z)\right ). {}\end{array}$$

(5.68)

The field continuity between the two regions A and B at ρ = r_beam leads to the integration constant ζ(r_beam) in the last equation, and it also implies

$$\displaystyle{g_{1}^{A} = g_{ 1}^{B} = g_{ 1},\mbox{ }g_{2}^{A} = g_{ 2}^{B} = g_{ 2}.}$$

Furthermore, we get

$$\displaystyle{ \mathrm{I}_{1}(k_{z}r_{\mathrm{beam}}) = f_{1}^{B}\;\mathrm{I}_{ 1}(k_{z}r_{\mathrm{beam}}) + f_{2}^{B}\;\mathrm{K}_{ 1}(k_{z}r_{\mathrm{beam}}). }$$

(5.69)

The field continuity is also the reason why we did not use different symbols for k_z in the two regions.

At ρ = r_bp, the ideally conducting beam pipe leads to the condition E_z = 0 (the longitudinal component of the electric field remains unchanged by the Lorentz transformation, so that E_z = 0 in the laboratory frame corresponds to E_z = 0 in the rest frame of the beam). This leads to

$$\displaystyle{ f_{1}^{B}\left [\mathrm{I}_{ 0}(k_{z}r_{\mathrm{bp}}) -\mathrm{ I}_{0}(k_{z}r_{\mathrm{beam}})\right ]+f_{2}^{B}\left [\mathrm{K}_{ 0}(k_{z}r_{\mathrm{beam}}) -\mathrm{ K}_{0}(k_{z}r_{\mathrm{bp}})\right ]+\zeta (r_{\mathrm{beam}}) = 0. }$$

(5.70)

The last two equations can be used to determine the constants $f_{1}^{B}$ and $f_{2}^{B}$. However, ζ(0) still has to be calculated in order to get ζ(r_beam) by means of Eq. (5.64).

Now we determine the space charge density in region A. Due to

$$\displaystyle\begin{array}{rcl} \frac{\partial D_{\rho }} {\partial \rho } & =& k_{z}\mathrm{I}'_{1}(k_{z}\rho )\;\left (g_{1}^{A}\;\cos (k_{ z}z) + g_{2}^{A}\;\sin (k_{ z}z)\right ), {}\\ \frac{\partial D_{z}} {\partial z} & =& k_{z}\;\zeta (\rho )\;\left (-g_{1}^{A}\;\cos (k_{ z}z) - g_{2}^{A}\;\sin (k_{ z}z)\right ), {}\\ \end{array}$$

we get

$$\displaystyle\begin{array}{rcl} \rho _{q}(z)& =& \frac{\partial D_{\rho }} {\partial \rho } + \frac{1} {\rho } \;D_{\rho } + \frac{\partial D_{z}} {\partial z} = {}\\ & =& \left (g_{1}^{A}\;\cos (k_{ z}z) + g_{2}^{A}\;\sin (k_{ z}z)\right )\left (k_{z}\mathrm{I}'_{1}(k_{z}\rho ) + \frac{1} {\rho } \;\mathrm{I}_{1}(k_{z}\rho ) - k_{z}\zeta (\rho )\right ). {}\\ \end{array}$$

According to [26, p. 376, formula (9.6.28)], we have

$$\displaystyle{\frac{1} {u} \frac{\mathrm{d}} {\mathrm{d}u}\left (u\mathrm{I}_{1}(u)\right ) =\mathrm{ I}_{0}(u)}$$

$$\displaystyle{\Rightarrow \fbox{$\frac{1} {u}\;\mathrm{I}_{1}(u) +\mathrm{ I}'_{1}(u) =\mathrm{ I}_{0}(u)$}}$$

$$\displaystyle{\Rightarrow \frac{1} {k_{z}\rho }\;\mathrm{I}_{1}(k_{z}\rho ) +\mathrm{ I}'_{1}(k_{z}\rho ) =\mathrm{ I}_{0}(k_{z}\rho ),}$$

so that

$$\displaystyle{\rho _{q}(z) = \left (g_{1}^{A}\;\cos (k_{ z}z) + g_{2}^{A}\;\sin (k_{ z}z)\right )\left (k_{z}\mathrm{I}_{0}(k_{z}\rho ) - k_{z}\zeta (\rho )\right ),}$$

or

$$\displaystyle{ \rho _{q}(z) =\xi \; \left (g_{1}^{A}\;\cos (k_{ z}z) + g_{2}^{A}\;\sin (k_{ z}z)\right ), }$$

(5.71)

with

$$\displaystyle{\xi = k_{z}\mathrm{I}_{0}(k_{z}\rho ) - k_{z}\zeta (\rho ) = k_{z}\left (\mathrm{I}_{0}(k_{z}\rho ) -\zeta (\rho )\right )}$$

is obtained. The derivative with respect to ρ is

$$\displaystyle{\frac{\mathrm{d}\xi } {\mathrm{d}\rho } = k_{z}^{2}\mathrm{I}'_{ 0}(k_{z}\rho ) - k_{z}^{2}\mathrm{I}'_{ 0}(k_{z}\rho ) = 0.}$$

Here we used Eq. (5.64). With Eq. (5.71),

$$\displaystyle{\frac{\partial \rho _{q}} {\partial \rho } = 0}$$

is also valid, which means that in each region, the space charge density depends only on z, as required. Please note that the quantity ξ introduced above does not depend on ρ, even though the individual terms do depend on ρ. One may therefore evaluate ξ for all values of ρ. For ρ = 0, we get

$$\displaystyle{ \xi = k_{z}\left (1 -\zeta (0)\right ), }$$

(5.72)

since I₀(0) = 1 holds.

As a last step, we now have to satisfy the boundary condition for the normal component of the electric field at ρ = r_bp:

$$\displaystyle{D_{n} =\sigma _{q}.}$$

In general, σ_q denotes the surface charge density defined by

$$\displaystyle{\int \int \int \rho _{q}\;\mathrm{d}V =\int \int \sigma _{q}\;\rho \;\mathrm{d}\varphi \;\mathrm{d}z =\int \lambda _{q}\;\mathrm{d}z.}$$

In our special case, the radius is ρ = r_bp, $D_{n} = -D_{\rho }$ holds, and the charge on the beam pipe must be the same as the charge of the beam with negative sign (image charge in the rest frame of the beam):

$$\displaystyle{\sigma _{q}\;2\pi r_{\mathrm{bp}} = -\lambda _{q} = -\pi r_{\mathrm{beam}}^{2}\rho _{ q}}$$

$$\displaystyle{\Rightarrow D_{\rho }\vert _{\rho =r_{\mathrm{bp}}} = \frac{\lambda _{q}} {2\pi r_{\mathrm{bp}}} = \frac{r_{\mathrm{beam}}^{2}\rho _{q}} {2r_{\mathrm{bp}}}.}$$

According to Eqs. (5.67) and (5.71) this leads to the condition

$$\displaystyle{ f_{1}^{B}\;\mathrm{I}_{ 1}(k_{z}r_{\mathrm{bp}}) + f_{2}^{B}\;\mathrm{K}_{ 1}(k_{z}r_{\mathrm{bp}}) = \frac{r_{\mathrm{beam}}^{2}\xi } {2r_{\mathrm{bp}}}. }$$

(5.73)

In combination with Eqs. (5.69), (5.70), and (5.72), this defines how ζ(ρ) and ξ depend on each other.

The solution discussed above does not include the case that the charge density and the fields are constant in the longitudinal direction. For p = 0, all field components vanish.

Therefore, we now consider Eq. (5.59) for the case that D_ρ does not depend on z:

$$\displaystyle{\frac{\mathrm{d}^{2}D_{\rho }} {\mathrm{d}\rho ^{2}} -\frac{1} {\rho ^{2}} \;D_{\rho } + \frac{1} {\rho } \;\frac{\mathrm{d}D_{\rho }} {\mathrm{d}\rho } = 0}$$

$$\displaystyle{\Rightarrow \rho ^{2}\;\frac{\mathrm{d}^{2}D_{\rho }} {\mathrm{d}\rho ^{2}} +\rho \; \frac{\mathrm{d}D_{\rho }} {\mathrm{d}\rho } - D_{\rho } = 0.}$$

This is a homogeneous Euler–Cauchy ODE, which can be solved by the substitution

$$\displaystyle{\rho = e^{u},}$$

so that we have

$$\displaystyle{\frac{\mathrm{d}D_{\rho }} {\mathrm{d}u} = \frac{\mathrm{d}D_{\rho }} {\mathrm{d}\rho } \frac{\mathrm{d}\rho } {\mathrm{d}u} =\rho \; \frac{\mathrm{d}D_{\rho }} {\mathrm{d}\rho } }$$

and

$$\displaystyle{\frac{\mathrm{d}^{2}D_{\rho }} {\mathrm{d}u^{2}} = \frac{\mathrm{d}} {\mathrm{d}\rho }\left (\rho \;\frac{\mathrm{d}D_{\rho }} {\mathrm{d}\rho } \right ) \frac{\mathrm{d}\rho } {\mathrm{d}u} = \left (\frac{\mathrm{d}D_{\rho }} {\mathrm{d}\rho } +\rho \frac{\mathrm{d}^{2}D_{\rho }} {\mathrm{d}\rho ^{2}} \right )\;\rho =\rho \; \frac{\mathrm{d}D_{\rho }} {\mathrm{d}\rho } +\rho ^{2}\frac{\mathrm{d}^{2}D_{\rho }} {\mathrm{d}\rho ^{2}}.}$$

We thereby obtain

$$\displaystyle{\frac{\mathrm{d}^{2}D_{\rho }} {\mathrm{d}u^{2}} - D_{\rho } = 0.}$$

The ansatz

$$\displaystyle{D_{\rho } \sim e^{ku}}$$

leads to the characteristic equation

$$\displaystyle{k^{2} - 1 = 0\mbox{ } \Rightarrow k = \pm 1,}$$

which yields the solutions

$$\displaystyle{D_{\rho } \sim \rho \mbox{ and }D_{\rho } \sim 1/\rho.}$$

In region A (0 ≤ ρ < r_beam), the second solution would lead to a singularity at ρ = 0, although the charge distribution will not be singular. Therefore, we have

$$\displaystyle{D_{\rho }^{A} = h_{ 1}^{A}\rho }$$

and

$$\displaystyle{D_{\rho }^{B} = h_{ 1}^{B}\rho + \frac{h_{2}^{B}} {\rho }.}$$

We evaluate Eq. (5.57),

$$\displaystyle{\frac{\partial D_{\rho }} {\partial \rho } + \frac{1} {\rho } \;D_{\rho } + \frac{\partial D_{z}} {\partial z} =\rho _{q}(z),}$$

for the two regions A (0 < ρ < r_beam) and B (r_beam < ρ < r_bp):

$$\displaystyle{2h_{1}^{A} + \frac{\partial D_{z}^{A}} {\partial z} =\rho _{q,0}(z) =\rho _{q,0,\mathrm{DC}},}$$

$$\displaystyle{h_{1}^{B} -\frac{h_{2}^{B}} {\rho ^{2}} + h_{1}^{B} + \frac{h_{2}^{B}} {\rho ^{2}} + \frac{\partial D_{z}^{B}} {\partial z} = 0\mbox{ } \Rightarrow 2h_{1}^{B} + \frac{\partial D_{z}^{B}} {\partial z} = 0.}$$

From a physical point of view, solutions for D_z that increase or decrease linearly with z may be excluded, since the fields must be periodic with respect to the synchrotron circumference. Therefore, and because of $\frac{\partial D_{z}} {\partial \rho } = \frac{\partial D_{\rho }} {\partial z} = 0$, only a constant D_z may be considered. However, D_z must vanish for ρ = r_bp. Due to the field continuity at ρ = r_beam, D_z must then be zero everywhere. Therefore, we have

$$\displaystyle{h_{1}^{A} = \frac{\rho _{q,0,\mathrm{DC}}} {2},\mbox{ }h_{1}^{B} = 0.}$$

For ρ = r_beam, we obtain

$$\displaystyle{ \frac{h_{2}^{B}} {r_{\mathrm{beam}}} = h_{1}^{A}r_{\mathrm{ beam}} = \frac{\rho _{q,0,\mathrm{DC}}} {2} r_{\mathrm{beam}}.}$$

Therefore, we get the following solution:

$$\displaystyle\begin{array}{rcl} D_{\rho }^{A}& =& \frac{\rho _{q,0,\mathrm{DC}}} {2} \;\rho, {}\\ D_{\rho }^{B}& =& \frac{\rho _{q,0,\mathrm{DC}}r_{\mathrm{beam}}^{2}} {2\rho } {}\\ \end{array}$$

Due to $\lambda _{q,0,\mathrm{DC}} =\pi r_{\mathrm{beam}}^{2}\;\rho _{q,0,\mathrm{DC}}$, we get

$$\displaystyle\begin{array}{rcl} D_{\rho }^{A}& =& \frac{\lambda _{q,0,\mathrm{DC}}} {2\pi r_{\mathrm{beam}}^{2}}\;\rho,{}\end{array}$$

(5.74)

$$\displaystyle\begin{array}{rcl} D_{\rho }^{B}& =& \frac{\lambda _{q,0,\mathrm{DC}}} {2\pi \rho }.{}\end{array}$$

(5.75)

The general solution is the sum of the DC charge distribution result (Eqs. (5.74) and (5.75)) and the harmonic solutions (Eqs. (5.62), (5.63), (5.67), and (5.68)). Before we write down the general solution, we now assume that it belongs to a reference frame $\bar{S}$ that is the rest frame of the beam. Later, we will analyze a Lorentz transformation to the frame S that is the rest frame of the synchrotron, i.e., the laboratory frame. In the frame $\bar{S}$, we now have⁷

$$\displaystyle\begin{array}{rcl} \bar{D}_{\rho }^{A}& =& \frac{\bar{\lambda }_{q,0,\mathrm{DC}}} {2\pi r_{\mathrm{beam}}^{2}}\;\bar{\rho } +\sum _{ k=1}^{\infty }\mathrm{I}_{ 1}(\bar{k}_{zk}\bar{\rho })\left (g_{1k}\;\cos (\bar{k}_{zk}\bar{z}) + g_{2k}\;\sin (\bar{k}_{zk}\bar{z})\right ),{}\end{array}$$

(5.76)

$$\displaystyle\begin{array}{rcl} \bar{D}_{z}^{A}& =& \sum _{ k=1}^{\infty }\zeta _{ k}(\bar{\rho })\left (-g_{1k}\;\sin (\bar{k}_{zk}\bar{z}) + g_{2k}\;\cos (\bar{k}_{zk}\bar{z})\right ),{}\end{array}$$

(5.77)

$$\displaystyle\begin{array}{rcl} \bar{D}_{\rho }^{B}& =& \frac{\bar{\lambda }_{q,0,\mathrm{DC}}} {2\pi \bar{\rho }} +\sum _{ k=1}^{\infty }\left (f_{ 1k}\;\mathrm{I}_{1}(\bar{k}_{zk}\bar{\rho }) + f_{2k}\;\mathrm{K}_{1}(\bar{k}_{zk}\bar{\rho })\right ) \cdot \\ & &\cdot \left (g_{1k}\;\cos (\bar{k}_{zk}\bar{z}) + g_{2k}\;\sin (\bar{k}_{zk}\bar{z})\right ), {}\end{array}$$

(5.78)

$$\displaystyle\begin{array}{rcl} \bar{D}_{z}^{B}& =& \sum _{ k=1}^{\infty }\left(f_{ 1k}\left[\mathrm{I}_{0}(\bar{k}_{zk}\bar{\rho }) -\mathrm{ I}_{0}(\bar{k}_{zk}r_{\mathrm{beam}})\right]\right. \\ & & +\,\left.f_{2k}\left[\mathrm{K}_{0}(\bar{k}_{zk}r_{\mathrm{beam}}) -\mathrm{ K}_{0}(\bar{k}_{zk}\bar{\rho })\right] +\zeta _{k}(r_{\mathrm{beam}})\right) \cdot \\ & \cdot & \left(-g_{1k}\;\sin (\bar{k}_{zk}\bar{z}) + g_{2k}\;\cos (\bar{k}_{zk}\bar{z})\right). {}\end{array}$$

(5.79)

Taking the Lorentz transformation into account, we see that the constant $\bar{k}_{zk}$ equals

$$\displaystyle{\bar{k}_{zk} = k \frac{2\pi } {\bar{l}_{\mathrm{R}}} = k \frac{2\pi \gamma } {l_{\mathrm{R}}}.}$$

It is assumed that the charge density distribution is given by Eq. (5.83) below, so that the constants g_1k, g_2k are known.

For each k, the three constants f_1k, f_2k, ζ_k(r_beam) that were introduced above and two further constants ζ_k(0), ξ_k can then be determined by solving the linear system of equations that consists of Eqs. (5.80), (5.81), (5.82), (5.84), and finally Eq. (5.85) for the specific value $\bar{\rho }= r_{\mathrm{beam}}$.

The first of these five equations is a rewritten form of Eq. (5.69),

$$\displaystyle{ \mathrm{I}_{1}(\bar{k}_{zk}r_{\mathrm{beam}}) = f_{1k}\;\mathrm{I}_{1}(\bar{k}_{zk}r_{\mathrm{beam}}) + f_{2k}\;\mathrm{K}_{1}(\bar{k}_{zk}r_{\mathrm{beam}}), }$$

(5.80)

the second of Eq. (5.70),

$$\displaystyle{ f_{1k}\left [\mathrm{I}_{0}(\bar{k}_{zk}r_{\mathrm{bp}}) -\mathrm{ I}_{0}(\bar{k}_{zk}r_{\mathrm{beam}})\right ]+f_{2k}\left [\mathrm{K}_{0}(\bar{k}_{zk}r_{\mathrm{beam}}) -\mathrm{ K}_{0}(\bar{k}_{zk}r_{\mathrm{bp}})\right ]+\zeta _{k}(r_{\mathrm{beam}}) = 0, }$$

(5.81)

and the third of Eq. (5.73),

$$\displaystyle{ f_{1k}\;\mathrm{I}_{1}(\bar{k}_{zk}r_{\mathrm{bp}}) + f_{2k}\;\mathrm{K}_{1}(\bar{k}_{zk}r_{\mathrm{bp}}) = \frac{r_{\mathrm{beam}}^{2}\xi _{k}} {2r_{\mathrm{bp}}}. }$$

(5.82)

From Eq. (5.71), we obtain, for the charge density in region A,

$$\displaystyle{ \bar{\rho }_{q} =\bar{\rho } _{q,0,\mathrm{DC}} +\sum _{ k=1}^{\infty }\xi _{ k}\left (g_{1k}\;\cos (\bar{k}_{zk}\bar{z}) + g_{2k}\;\sin (\bar{k}_{zk}\bar{z})\right ), }$$

(5.83)

with

$$\displaystyle{\xi _{k} =\bar{ k}_{zk}\left (\mathrm{I}_{0}(\bar{k}_{zk}\bar{\rho }) -\zeta _{k}(\bar{\rho })\right )}$$

constant, so that this expression may be evaluated for different values of $\bar{\rho }$, e.g., $\bar{\rho }= 0$:

$$\displaystyle{ \xi _{k} =\bar{ k}_{zk}\left (1 -\zeta _{k}(0)\right ). }$$

(5.84)

Equation (5.64) now reads

$$\displaystyle{ \zeta _{k}(\bar{\rho }) =\mathrm{ I}_{0}(\bar{k}_{zk}\bar{\rho }) - 1 +\zeta _{k}(0). }$$

(5.85)

5.8.2 Transformation to the Rest Frame of the Synchrotron

For the Lorentz transformation, the following formulas are valid in our specific case (cf. Eqs. (2.56)–(2.59), (2.60)–(2.63), (2.69)–(2.74)):

$$\displaystyle{\rho =\bar{\rho },\mbox{ }\bar{z} =\gamma (z - vt),\mbox{ }\bar{k}_{zk}\bar{z} =\gamma \bar{ k}_{zk}z -\gamma \bar{ k}_{zk}vt =:\phi (z,t),}$$

$$\displaystyle{D_{\rho }^{A} =\gamma \bar{ D}_{\rho }^{A},\mbox{ }D_{\rho }^{B} =\gamma \bar{ D}_{\rho }^{B},}$$

$$\displaystyle{D_{z}^{A} =\bar{ D}_{ z}^{A},\mbox{ }D_{ z}^{B} =\bar{ D}_{ z}^{B},}$$

$$\displaystyle{H_{\varphi }^{A} =\gamma v\bar{D}_{\rho }^{A},\mbox{ }H_{\varphi }^{B} =\gamma v\bar{D}_{\rho }^{B},}$$

$$\displaystyle{J_{z} =\gamma v\;\bar{\rho }_{q},}$$

$$\displaystyle{\rho _{q} =\gamma \;\bar{\rho } _{q},}$$

$$\displaystyle{\int \lambda _{q}\;\mathrm{d}z =\int \bar{\lambda } _{q}\;\mathrm{d}\bar{z}\mbox{ } \Rightarrow \bar{\lambda }_{q} = \frac{1} {\gamma } \;\lambda _{q}.}$$

Therefore, we obtain

$$\displaystyle\begin{array}{rcl} D_{\rho }^{A}& =& \frac{\lambda _{q,0,\mathrm{DC}}} {2\pi r_{\mathrm{beam}}^{2}}\;\rho +\gamma \sum _{ k=1}^{\infty }\mathrm{I}_{ 1}(\bar{k}_{zk}\rho )\left (g_{1k}\;\cos \phi + g_{2k}\;\sin \phi \right ),{}\end{array}$$

(5.86)

$$\displaystyle\begin{array}{rcl} D_{z}^{A}& =& \sum _{ k=1}^{\infty }\zeta _{ k}(\rho )\left (-g_{1k}\;\sin \phi + g_{2k}\;\cos \phi \right ),{}\end{array}$$

(5.87)

$$\displaystyle\begin{array}{rcl} D_{\rho }^{B}& =& \frac{\lambda _{q,0,\mathrm{DC}}} {2\pi \rho } +\gamma \sum _{ k=1}^{\infty }\left (f_{ 1k}\;\mathrm{I}_{1}(\bar{k}_{zk}\rho ) + f_{2k}\;\mathrm{K}_{1}(\bar{k}_{zk}\rho )\right ) \cdot \\ & &\cdot \left (g_{1k}\;\cos \phi + g_{2k}\;\sin \phi \right ), {}\end{array}$$

(5.88)

$$\displaystyle\begin{array}{rcl} D_{z}^{B}& =& \sum _{ k=1}^{\infty }\left(f_{ 1k}\left[\mathrm{I}_{0}(\bar{k}_{zk}\rho ) -\mathrm{ I}_{0}(\bar{k}_{zk}r_{\mathrm{beam}})\right]\right. \\ & & +\left.f_{2k}\left[\mathrm{K}_{0}(\bar{k}_{zk}r_{\mathrm{beam}}) -\mathrm{ K}_{0}(\bar{k}_{zk}\rho )\right] +\zeta _{k}(r_{\mathrm{beam}})\right) \cdot \\ & \cdot & \left(-g_{1k}\;\sin \phi + g_{2k}\;\cos \phi \right), {}\end{array}$$

(5.89)

$$\displaystyle{\rho _{q} =\rho _{q,0,\mathrm{DC}} +\gamma \sum _{ k=1}^{\infty }\xi _{ k}\left (g_{1k}\;\cos \phi + g_{2k}\;\sin \phi \right ),}$$

$$\displaystyle{\dot{\rho }_{q} = -\gamma ^{2}v\;\sum _{ k=1}^{\infty }\bar{k}_{ zk}\xi _{k}\left (-g_{1k}\;\sin \phi + g_{2k}\;\cos \phi \right ).}$$

Figure 5.11 presents an example for the field $\vec{D}$ that is obtained if a line charge density according to Fig. 5.12 is assumed.

Fig. 5.11

Electric displacement field $\vec{D}$ for the line charge density given in Fig. 5.12 (parameters: l_R = 216 m, r_bp = 7 cm, r_beam = 1 cm)

Fig. 5.12

Line charge density $\lambda _{q} =\pi r_{\mathrm{beam}}^{2}\rho _{q}$ with an average beam current of $\bar{I}_{\mathrm{beam}} = 13.6\,\mathrm{mA}$ for β = 0. 15583, and $\hat{I}_{\mathrm{beam}} = 20.0\,\mathrm{mA}$ at h = 4

5.8.3 Longitudinal Electric Field

Now we introduce some approximations. We assume that only a few Fourier components, say 10, are needed to describe the beam current. Furthermore, we assume that the maximum energy is below γ = 6. Therefore, the maximum $\bar{k}_{zk}$ is equal to

$$\displaystyle{\bar{k}_{zk} = k \frac{2\pi } {\bar{l}_{\mathrm{R}}} = k \frac{2\pi \gamma } {l_{\mathrm{R}}},}$$

with k = 10 and γ = 6. Here, l_R is the circumference of the synchrotron, for which a length contraction

$$\displaystyle{\bar{l}_{\mathrm{R}} = \frac{l_{\mathrm{R}}} {\gamma } }$$

has to be taken into account in the rest frame $\bar{S}$ of the beam. If we now assume that the beam pipe radius r_bp is smaller than 10 cm whereas the synchrotron circumference l_R is larger than 200 m, we get

$$\displaystyle{\bar{k}_{zk}r_{\mathrm{bp}} < 0.1885.}$$

Therefore, we will assume

$$\displaystyle{\bar{k}_{zk}r_{\mathrm{bp}} < 0.2}$$

in the following. The specific numbers may, of course, be modified for different cases.

For small arguments z, one obtains⁸

$$\displaystyle{\fbox{$\mathrm{I}_{0}(z) \approx 1 + \frac{z^{2}} {4},\mbox{ }\mathrm{K}_{0}(z) \approx -\ln z,\mbox{ }\mathrm{I}_{1}(z) \approx \frac{z} {2},\mbox{ }\mathrm{K}_{1}(z) \approx \frac{1} {z}.$}}$$

We use these approximations to simplify Eqs. (5.80)–(5.82):

$$\displaystyle\begin{array}{rcl} \frac{\bar{k}_{zk}r_{\mathrm{beam}}} {2} & =& f_{1k}\;\frac{\bar{k}_{zk}r_{\mathrm{beam}}} {2} + f_{2k}\; \frac{1} {\bar{k}_{zk}r_{\mathrm{beam}}},{}\end{array}$$

(5.90)

$$\displaystyle\begin{array}{rcl} & & f_{1k}\;\frac{\bar{k}_{zk}^{2}(r_{\mathrm{bp}}^{2} - r_{\mathrm{beam}}^{2})} {4} + f_{2k}\;\ln \frac{r_{\mathrm{bp}}} {r_{\mathrm{beam}}} + \frac{\bar{k}_{zk}^{2}r_{\mathrm{beam}}^{2}} {4} +\zeta _{k}(0) = 0,\quad {}\end{array}$$

(5.91)

$$\displaystyle\begin{array}{rcl} & & f_{1k}\;\frac{\bar{k}_{zk}r_{\mathrm{bp}}} {2} + f_{2k}\; \frac{1} {\bar{k}_{zk}r_{\mathrm{bp}}} = \frac{r_{\mathrm{beam}}^{2}\xi _{k}} {2r_{\mathrm{bp}}} = \frac{r_{\mathrm{beam}}^{2}} {2r_{\mathrm{bp}}} \;\bar{k}_{zk}\left (1 -\zeta _{k}(0)\right ).\quad {}\end{array}$$

(5.92)

In the last two equations, we used Eqs. (5.84) and (5.85). The first of these three equations leads to

$$\displaystyle{ f_{1k} = 1 - f_{2k} \frac{2} {(\bar{k}_{zk}r_{\mathrm{beam}})^{2}}. }$$

(5.93)

If we insert this into Eq. (5.91), we get

$$\displaystyle{ \frac{\bar{k}_{zk}^{2}r_{\mathrm{bp}}^{2}} {4} + f_{2k}\left (\ln \frac{r_{\mathrm{bp}}} {r_{\mathrm{beam}}} -\frac{1} {2} \frac{r_{\mathrm{bp}}^{2}} {r_{\mathrm{beam}}^{2}} + \frac{1} {2}\right ) +\zeta _{k}(0) = 0. }$$

(5.94)

If one inserts Eq. (5.93) into Eq. (5.92), one obtains

$$\displaystyle{\frac{\bar{k}_{zk}r_{\mathrm{bp}}} {2} + \frac{f_{2k}} {\bar{k}_{zk}}\left ( \frac{1} {r_{\mathrm{bp}}} - \frac{r_{\mathrm{bp}}} {r_{\mathrm{beam}}^{2}}\right ) = \frac{r_{\mathrm{beam}}^{2}} {2r_{\mathrm{bp}}} \;\bar{k}_{zk}\left (1 -\zeta _{k}(0)\right )}$$

$$\displaystyle{\Rightarrow f_{2k} = - \frac{\bar{k}_{zk}^{2}r_{\mathrm{beam}}^{2}r_{\mathrm{bp}}^{2}} {2(r_{\mathrm{beam}}^{2} - r_{\mathrm{bp}}^{2})} + \frac{r_{\mathrm{beam}}^{4}} {2(r_{\mathrm{beam}}^{2} - r_{\mathrm{bp}}^{2})}\;\bar{k}_{zk}^{2}\left (1 -\zeta _{ k}(0)\right ).}$$

This may be inserted into Eq. (5.94):

$$\displaystyle{\frac{\bar{k}_{zk}^{2}r_{\mathrm{bp}}^{2}} {4} +\left [ \frac{\bar{k}_{zk}^{2}r_{\mathrm{beam}}^{2}r_{\mathrm{bp}}^{2}} {2(r_{\mathrm{bp}}^{2} - r_{\mathrm{beam}}^{2})} - \frac{\bar{k}_{zk}^{2}r_{\mathrm{beam}}^{4}} {2(r_{\mathrm{bp}}^{2} - r_{\mathrm{beam}}^{2})}\;\left (1 -\zeta _{k}(0)\right )\right ]\left (\ln \frac{r_{\mathrm{bp}}} {r_{\mathrm{beam}}} -\frac{1} {2} \frac{r_{\mathrm{bp}}^{2}} {r_{\mathrm{beam}}^{2}} + \frac{1} {2}\right )+\zeta _{k}(0) = 0}$$

$$\displaystyle\begin{array}{rcl} & \Rightarrow & \zeta _{k}(0)\left [1 + \frac{\bar{k}_{zk}^{2}r_{\mathrm{beam}}^{4}} {2(r_{\mathrm{bp}}^{2} - r_{\mathrm{beam}}^{2})}\left (\ln \frac{r_{\mathrm{bp}}} {r_{\mathrm{beam}}} -\frac{1} {2} \frac{r_{\mathrm{bp}}^{2}} {r_{\mathrm{beam}}^{2}} + \frac{1} {2}\right )\right ] = {}\\ & =& \bar{k}_{zk}^{2}\left [-\frac{r_{\mathrm{bp}}^{2}} {4} + \frac{r_{\mathrm{beam}}^{4} - r_{\mathrm{beam}}^{2}r_{\mathrm{bp}}^{2}} {2(r_{\mathrm{bp}}^{2} - r_{\mathrm{beam}}^{2})} \left (\ln \frac{r_{\mathrm{bp}}} {r_{\mathrm{beam}}} -\frac{1} {2} \frac{r_{\mathrm{bp}}^{2}} {r_{\mathrm{beam}}^{2}} + \frac{1} {2}\right )\right ] {}\\ & =& \bar{k}_{zk}^{2}\left [-\frac{r_{\mathrm{bp}}^{2}} {4} -\frac{r_{\mathrm{beam}}^{2}} {2} \left (\ln \frac{r_{\mathrm{bp}}} {r_{\mathrm{beam}}} -\frac{1} {2} \frac{r_{\mathrm{bp}}^{2}} {r_{\mathrm{beam}}^{2}} + \frac{1} {2}\right )\right ] {}\\ & =& \bar{k}_{zk}^{2}\left [-\frac{r_{\mathrm{beam}}^{2}} {2} \left (\ln \frac{r_{\mathrm{bp}}} {r_{\mathrm{beam}}} + \frac{1} {2}\right )\right ] {}\\ \end{array}$$

$$\displaystyle{\Rightarrow \zeta _{k}(0) = -\bar{k}_{zk}^{2}\; \frac{r_{\mathrm{beam}}^{2}\left (\ln \frac{r_{\mathrm{bp}}} {r_{\mathrm{beam}}} + \frac{1} {2}\right )} {2 + (\bar{k}_{zk}r_{\mathrm{beam}})^{2}\frac{\ln \frac{r_{\mathrm{bp}}} {r_{\mathrm{beam}}} -\frac{1} {2} \frac{r_{\mathrm{bp}}^{2}} {r_{\mathrm{beam}}^{2}} +\frac{1} {2} } { \frac{r_{\mathrm{bp}}^{2}} {r_{\mathrm{beam}}^{2}} -1} }.}$$

For $\frac{r_{\mathrm{bp}}} {r_{\mathrm{beam}}} > 1$, the fraction in the denominator is in the range $\left]-0.5,0\right[$. If we also take into account that $\bar{k}_{zk}r_{\mathrm{beam}} <\bar{ k}_{zk}r_{\mathrm{bp}} < 0.2$ holds, we see that this part in the denominator can be neglected in comparison with 2:

$$\displaystyle{ \zeta _{k}(0) = -\frac{\bar{k}_{zk}^{2}r_{\mathrm{beam}}^{2}} {2} \left (\frac{1} {2} +\ln \frac{r_{\mathrm{bp}}} {r_{\mathrm{beam}}}\right ) = -\frac{\bar{k}_{zk}^{2}r_{\mathrm{bp}}^{2}} {2} \left [\frac{r_{\mathrm{beam}}^{2}} {r_{\mathrm{bp}}^{2}} \left (\frac{1} {2} +\ln \frac{r_{\mathrm{bp}}} {r_{\mathrm{beam}}}\right )\right ].\quad }$$

(5.95)

Therefore, according to Eq. (5.87), the longitudinal field on the beam axis is

$$\displaystyle{ D_{z}\vert _{\rho =0} = -\frac{r_{\mathrm{beam}}^{2}} {2} \left (\frac{1} {2} +\ln \frac{r_{\mathrm{bp}}} {r_{\mathrm{beam}}}\right )\sum _{k=1}^{\infty }\bar{k}_{ zk}^{2}\left (-g_{ 1k}\;\sin \phi + g_{2k}\;\cos \phi \right ). }$$

(5.96)

By means of Eq. (5.83), we obtain

$$\displaystyle{ \frac{\mathrm{d}\bar{\lambda }_{q}} {\mathrm{d}\bar{z}} =\pi r_{\mathrm{beam}}^{2}\; \frac{\mathrm{d}\bar{\rho }_{q}} {\mathrm{d}\bar{z}} =\pi r_{\mathrm{beam}}^{2}\;\sum _{ k=1}^{\infty }\bar{k}_{ zk}\;\xi _{k}\left (-g_{1k}\;\sin (\bar{k}_{zk}\bar{z}) + g_{2k}\;\cos (\bar{k}_{zk}\bar{z})\right ). }$$

(5.97)

Due to $\frac{r_{\mathrm{bp}}} {r_{\mathrm{beam}}} > 1$, the expression in Eq. (5.95) in square brackets is in the range $\left]0,0.5\right[$. With $\bar{k}_{zk}r_{\mathrm{bp}} < 0.2$, one sees that

$$\displaystyle{\left \vert \zeta _{k}(0)\right \vert \ll 1,}$$

so that

$$\displaystyle{\xi _{k} \approx \bar{ k}_{zk}}$$

is valid according to Eq. (5.84). Therefore, the sums in the two equations (5.96) and (5.97) are identical. Hence, we obtain

$$\displaystyle{D_{z}\vert _{\rho =0} = -\frac{1} {2\pi }\left (\frac{1} {2} +\ln \frac{r_{\mathrm{bp}}} {r_{\mathrm{beam}}}\right )\;\frac{\mathrm{d}\bar{\lambda }_{q}} {\mathrm{d}\bar{z}}.}$$

According to the Lorentz transformation,

$$\displaystyle{\frac{\mathrm{d}\bar{\lambda }_{q}} {\mathrm{d}\bar{z}} = \frac{1} {\gamma ^{2}} \frac{\partial \lambda _{q}} {\partial z}}$$

is valid, and we finally obtain

$$\displaystyle{D_{z}\vert _{\rho =0} = -\frac{1} {4\pi \gamma ^{2}}\left (1 + 2\;\ln \frac{r_{\mathrm{bp}}} {r_{\mathrm{beam}}}\right )\;\frac{\partial \lambda _{q}} {\partial z} = -\frac{g_{0}} {4\pi \gamma ^{2}}\; \frac{\partial \lambda _{q}} {\partial z}}$$

with the geometry factor

$$\displaystyle{\fbox{$g_{0} = 1 + 2\;\ln \frac{r_{\mathrm{bp}}} {r_{\mathrm{beam}}}.$}}$$

The result

$$\displaystyle{ \fbox{$E_{z}\vert _{\rho =0} = -\frac{g_{0}} {4\pi \epsilon _{0}\gamma ^{2}}\; \frac{\partial \lambda _{q}} {\partial z}$} }$$

(5.98)

may also be found in Edwards/Syphers [27] as formula (6.33) and in Ng [28] as formula (2.42). In these references, a much simpler derivation is offered, which, however, requires some advance knowledge about the field. Please note that in our derivation and in [27], λ_q denotes the charge density. Sometimes, λ_q, norm is defined as a normalized density function for the bunch, so that the total bunch charge N_bQ has to be added explicitly as a factor:

$$\displaystyle{\lambda _{q} = N_{b}Q\lambda _{q,\mathrm{norm}}.}$$

Finally, it has to be emphasized that the derivation presented here is based on the simplest model with transversally constant charge density. If different models are used, one also obtains different expressions for the geometry factor. This is discussed, for example, in Reiser [29, Sect. 6.3.2], and in Zotter [23, Sect. 12.1.1].

5.8.4 Space Charge Impedance

We determine the current at a constant position z:

$$\displaystyle{I_{\mathrm{beam}} =\int \vec{ J} \cdot \mathrm{ d}\vec{A} =\pi r_{\mathrm{beam}}^{2}J_{ z} =\pi r_{\mathrm{beam}}^{2}\gamma v\bar{\rho }_{ q} =\gamma v\bar{\lambda }_{q}.}$$

In $\bar{S}$, the line charge density $\bar{\lambda }_{q}$ depends only on $\bar{z}$ and not on time, since it determines the static solution. By means of the Lorentz transformation, however, $\bar{z} =\gamma (z - vt)$ depends on z and on t in the laboratory frame S. Therefore, the time derivative is

$$\displaystyle{ \frac{\partial I_{\mathrm{beam}}} {\partial t} =\gamma v\frac{\mathrm{d}\bar{\lambda }_{q}} {\mathrm{d}\bar{z}} \frac{\partial \bar{z}} {\partial t} = -\gamma ^{2}v^{2}\left.\frac{\mathrm{d}\bar{\lambda }_{q}} {\mathrm{d}\bar{z}}\right\vert _{\bar{z}=\gamma (z-vt)}. }$$

(5.99)

Now we calculate the voltage that is seen by a specific particle of the bunch. In the rest frame $\bar{S}$ of the bunch, we have

$$\displaystyle{\bar{E}_{z}\vert _{\rho =0} = E_{z}\vert _{\rho =0} = -\frac{g_{0}} {4\pi \epsilon _{0}}\; \frac{\mathrm{d}\bar{\lambda }_{q}} {\mathrm{d}\bar{z}}.}$$

Since the particle is moving with the bunch, the derivative $\frac{\mathrm{d}\bar{\lambda }_{q}} {\mathrm{d}\bar{z}}$ is always constant.⁹ Therefore, the integration is simple. We have only to multiply this expression by l_R to get the voltage in S:

$$\displaystyle{ V =\int \vec{ E} \cdot \mathrm{ d}\vec{r} =\int E_{z}\;\mathrm{d}z = -\frac{g_{0}l_{\mathrm{R}}} {4\pi \epsilon _{0}} \; \frac{\mathrm{d}\bar{\lambda }_{q}} {\mathrm{d}\bar{z}}. }$$

(5.100)

The expression $\frac{\mathrm{d}\bar{\lambda }_{q}} {\mathrm{d}\bar{z}}$ has to be evaluated at the position of the specific particle. In the frame S, it will be given by $z = z_{0} + vt$, so that

$$\displaystyle{\bar{z}_{0} =\bar{ z} =\gamma z_{0}}$$

is obtained. If the beam current is also measured at $z = z_{0} + vt$, we may combine Eqs. (5.99) and (5.100) and obtain

$$\displaystyle{V = -\frac{g_{0}l_{\mathrm{R}}} {4\pi \epsilon _{0}} \; \frac{1} {-\gamma ^{2}v^{2}}\;\frac{\partial I_{\mathrm{beam}}} {\partial t} = \frac{g_{0}l_{\mathrm{R}}} {4\pi \epsilon _{0}\gamma ^{2}v^{2}}\;\frac{\partial I_{\mathrm{beam}}} {\partial t}.}$$

If this equation is transformed into the frequency domain, the time derivative will be converted into a multiplication by j ω:

$$\displaystyle{\hat{\underline{V }}(\omega ) = j\omega \frac{g_{0}l_{\mathrm{R}}} {4\pi \epsilon _{0}\gamma ^{2}v^{2}}\;\hat{\underline{I}}_{\mathrm{beam}}(\omega ).}$$

According to our standard equivalent circuit in Fig. 4.5 on p. 181, we have to define the longitudinal space charge impedance as

$$\displaystyle{Z_{\mathrm{sc}} = - \frac{\hat{\underline{V }}(\omega )} {\hat{\underline{I}}_{\mathrm{beam}}(\omega )}}$$

if we want to treat it similarly to the impedance of a cavity. Hence we get

$$\displaystyle{ Z_{\mathrm{sc}} = -j\omega \frac{g_{0}l_{\mathrm{R}}} {4\pi \epsilon _{0}\gamma ^{2}v^{2}}. }$$

(5.101)

If we evaluate this expression at the nth harmonic of the revolution frequency, i.e., at

$$\displaystyle{\omega = n\omega _{\mathrm{R}} = \frac{2\pi n} {T_{\mathrm{R}}} = \frac{2\pi nv} {l_{\mathrm{R}}},}$$

we obtain

$$\displaystyle{ \fbox{$\frac{Z_{\mathrm{sc}}} {n} = -j \frac{g_{0}} {2\epsilon _{0}\gamma ^{2}v} = -j\frac{g_{0}Z_{0}} {2\gamma ^{2}\beta }.$} }$$

(5.102)

Here we used

$$\displaystyle{c_{0} = \frac{1} {\sqrt{\epsilon _{0 } \mu _{0}}}\mbox{ and }Z_{0} = \sqrt{\frac{\mu _{0 } } {\epsilon _{0}}} \approx 376.73\;\Omega,}$$

respectively the velocity of light and the impedance of free space. According to the sign in Eq. (5.102), the space charge impedance is capacitive, although the frequency dependence corresponds to an inductance. Formula (5.102) can be found in many publications and textbooks on accelerator physics (cf. Hofmann/Pedersen [30, eqn. (8)]; Edwards/Syphers [27, eqn. (6.50)]; Zotter [23, eqn. (12.1)]; Ng [28, eqn. (2.45)]; Chao/Tigner [8, Sect. 2.5.3.1, p. 128, formula (1)]; Reiser [29, Sect. 6.3.3, eqn. (6.114)]). In many cases, however, the definition of the sign is reversed. Instead of the impedance itself, one may also consider the impedance per length. In this sense, Eq. (5.101), e.g., corresponds to equation (6.94b) in Reiser [29].

According to Eq. (5.98), the electric field induced by the space charge is proportional to $-\frac{\partial \lambda _{q}} {\partial z}$. In terms of time, it is therefore proportional to $+\frac{\partial \lambda _{q}} {\partial t}$. This is due to the fact that we have

$$\displaystyle{\frac{\partial \bar{\lambda }_{q}} {\partial z} = \frac{\mathrm{d}\bar{\lambda }_{q}} {\mathrm{d}\bar{z}} \frac{\partial \bar{z}} {\partial z} =\gamma \; \frac{\mathrm{d}\bar{\lambda }_{q}} {\mathrm{d}\bar{z}}}$$

and

$$\displaystyle{\frac{\partial \bar{\lambda }_{q}} {\partial t} = \frac{\mathrm{d}\bar{\lambda }_{q}} {\mathrm{d}\bar{z}} \frac{\partial \bar{z}} {\partial t} = -\gamma v\;\frac{\mathrm{d}\bar{\lambda }_{q}} {\mathrm{d}\bar{z}}.}$$

From a physical point of view, this change of sign is also obvious, since the head of the bunch is located at larger z than the tail of the bunch. With respect to time, the head of the bunch will reach a certain point earlier than the tail of the bunch. We now have a look at the four cases shown in Fig. 3.3 on p. 133. If we draw the electric field that is generated by the space charge for each of these four cases, we get rising slopes in the bunch center for the right-hand diagrams and falling slopes in the left-hand diagrams. This corresponds to a defocusing effect in the upper two diagrams, since the slope is opposite to that of V (t), and to a focusing effect in the lower two diagrams, since the slope is the same as that of V (t). Therefore, the longitudinal space charge leads to a defocusing effect below transition. Above transition, it may lead to a beam instability (cf. Edwards [27, Sect. 6.2]; Reiser [29, Sect. 6.3.3]).

We now take GSI’s synchrotron SIS18 at injection energy (positive charges below transition) as an example. Let us consider the stationary case. Under these conditions, we have

$$\displaystyle{\beta = 0.15503,\mbox{ }\gamma = 1.012238,}$$

and we are interested in the Fourier component of the space charge voltage at $n = h = 4$. We assume

$$\displaystyle{ \frac{r_{\mathrm{bp}}} {r_{\mathrm{beam}}} = 7\mbox{ } \Rightarrow g_{0} = 4.89,}$$

which leads to a space charge impedance of

$$\displaystyle{Z_{\mathrm{sc}} = -j \cdot 23\;k\Omega }$$

if Eq. (5.102) is evaluated. If we now assume a beam current amplitude of

$$\displaystyle{\hat{\underline{I}}_{\mathrm{beam}} = 20\,\mathrm{mA}}$$

at the RF frequency (h = 4), this results in a space charge voltage

$$\displaystyle{\hat{\underline{V }} = -Z_{\mathrm{sc}}\;\hat{\underline{I}}_{\mathrm{beam}} = +j \cdot 460\,\mathrm{V}.}$$

A beam current

$$\displaystyle{I_{\mathrm{beam}}(t) = 20\,\mathrm{mA} \cdot \cos (\omega _{\mathrm{RF}}t)}$$

therefore leads to a voltage

$$\displaystyle{V (t) = 460\,\mathrm{V} \cdot \cos (\omega _{\mathrm{RF}}t +\pi /2) = -460\,\mathrm{V} \cdot \sin (\omega _{\mathrm{RF}}t).}$$

This is actually defocusing, because the RF voltage must be proportional to + sin(ω_RFt) to keep the particles bunched.