RF Acceleration

Abstract

This chapter is devoted to the longitudinal motion of charged particles in a synchrotron.

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This chapter is devoted to the longitudinal motion of charged particles in a synchrotron.

3.1 Centripetal Force

For the derivation of the equations of motion in a synchrotron, we need the centripetal force. Therefore, we briefly show that the expression for the centripetal force is the same in special relativity as in classical mechanics.

For the momentum vector, we have

$$\displaystyle{\vec{p} = m\;\vec{u} = m(\dot{x}\;\vec{e}_{x} +\dot{ y}\;\vec{e}_{y} +\dot{ z}\;\vec{e}_{z}),}$$

where

$$\displaystyle{m = m_{0}\gamma }$$

is the velocity-dependent mass. We assume that wherever the centripetal force is active, the absolute value of the velocity will not be changed. Therefore, one obtains

$$\displaystyle{\dot{u} = 0,\qquad \dot{\beta } = 0,\qquad \dot{\gamma } = 0,\qquad \dot{m} = 0.}$$

Hence, only

$$\displaystyle{\vec{F} =\dot{\vec{ p}} =\dot{ m}\;\vec{u} + m\;\dot{\vec{u}} = m(\ddot{x}\;\vec{e}_{x} +\ddot{ y}\;\vec{e}_{y} +\ddot{ z}\;\vec{e}_{z})}$$

has to be evaluated. On a circular orbit, we obtain

$$\displaystyle\begin{array}{rcl} \begin{array}{rcl} x& =&r\;\cos (\omega t), \\ y& =&r\;\sin (\omega t), \\ z& =&\mathrm{const},\end{array} & \qquad \begin{array}{rcl} u_{x} =\dot{ x}& =& -\omega r\;\sin (\omega t), \\ u_{y} =\dot{ y}& =& +\omega r\;\cos (\omega t), \\ u_{z} =\dot{ z}& =&0, \end{array} & \qquad \begin{array}{rcl} \dot{u}_{x} =\ddot{ x}& =& -\omega ^{2}r\;\cos (\omega t), \\ \dot{u}_{y} =\ddot{ y}& =& -\omega ^{2}r\;\sin (\omega t), \\ \dot{u}_{z} =\ddot{ z}& =&0. \end{array} {}\\ \end{array}$$

If we define the unit vector

$$\displaystyle{\vec{e}_{r} =\vec{ e}_{x}\;\cos (\omega t) +\vec{ e}_{y}\;\sin (\omega t)}$$

pointing radially outward, we can see directly the following relations:

$$\displaystyle{\dot{\vec{u}} = -\omega ^{2}r\;\vec{e}_{ r},}$$

$$\displaystyle{u^{2} =\vec{ u} \cdot \vec{ u} =\omega ^{2}r^{2}.}$$

Hence we obtain

$$\displaystyle{\vec{F} = -m\;\omega ^{2}r\;\vec{e}_{ r} = -\frac{mu^{2}} {r} \;\vec{e}_{r}.}$$

This is the well-known formula for the centripetal force, which is now verified in the scope of special relativity, provided that the energy of the particle remains constant (as satisfied in pure magnetic fields). In Appendix A.7.3, it is shown that this result remains true for arbitrary plane curves (e.g., if the magnetic field is no longer constant).

3.2 Simplified Model Synchrotron

In the scope of this book, we are interested only in longitudinal particle motion. Therefore, we significantly decrease the complexity of the problem by employing a model synchrotron that comprises only two straight sections and two dipole magnets. We have to emphasize that such a synchrotron will not work, because quadrupole magnets are essential for transverse focusing. If we keep this in mind, however, we may use the model nevertheless to study longitudinal motion in principle.

For the dipole magnets, we obtain

$$\displaystyle{F = \mathit{QuB} = m\frac{u^{2}} {r_{\mathrm{R}}}}$$

$$\displaystyle{ \Rightarrow \fbox{$p = \mathit{Qr}_{\mathrm{R}}B.$} }$$

(3.1)

The product r_RB is the magnetic rigidity. In one of the straight sections (the upper one in Fig. 3.1), we place a short ceramic gap. Let us assume that at t = 0, the reference particle is located at the gap. By means of the gap voltage, the energy¹ of the particle will increase from γ_R, 0 to γ_R, 1. Therefore, in the nth revolution, the particle has energy γ_R, n. Hereinafter, we will in general use the index R for quantities that are related to the reference particle.

Fig. 3.1

Strongly simplified synchrotron

By definition, the magnetic field is increased in such a way that the reference particle always remains in the same orbit (r_R = const).

For the reference particle, we therefore obtain

$$\displaystyle{ p_{\mathrm{R},n} = \mathit{Qr}_{\mathrm{R}}B_{n}. }$$

(3.2)

Here B_n is the magnetic dipole field, and p_R, n is the momentum of the reference particle in the nth revolution. The reference particle will reach the beginning of the gap at time t_R, n after the nth revolution (t_R, 0 = 0).

For the orbit length, one obtains

$$\displaystyle{l_{\mathrm{R}} = 2\pi r_{\mathrm{R}} + 2d.}$$

From

$$\displaystyle{u_{\mathrm{R},n} = \frac{l_{\mathrm{R}}} {T_{\mathrm{R},n}},}$$

it follows that

$$\displaystyle{ t_{\mathrm{R},n} = t_{\mathrm{R},n-1} + T_{\mathrm{R},n} = t_{\mathrm{R},n-1} + \frac{l_{\mathrm{R}}} {\beta _{\mathrm{R},n}c_{0}}. }$$

(3.3)

Before the nth revolution takes place, the energy of the particle is increased from γ_R, n−1 to γ_R, n. This happens due to the voltage V_R, n−1 at the gap:

$$\displaystyle{(\gamma _{\mathrm{R},n} -\gamma _{\mathrm{R},n-1})m_{0}c_{0}^{2} = \mathit{QV }_{\mathrm{ R},n-1}}$$

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

(3.4)

This completes the analysis of the reference particle.

After the nth revolution, an asynchronous particle, i.e., an off-momentum particle, will be located at the gap at time t_n. For the nth revolution, this asynchronous particle will need time T_n:

$$\displaystyle{u_{n} = \frac{l_{\mathrm{R}} + \Delta l_{n}} {T_{n}},}$$

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

(3.5)

Before the nth revolution, the energy of the asynchronous particle increases from γ_n−1 to γ_n. This energy step is caused by the voltage V_n−1:

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

(3.6)

From now on, we shall use $\Delta$ quantities to specify differences between the asynchronous particle and the reference particle (i.e., the synchronous particle). If we set $\Delta t_{n} = t_{n} - t_{\mathrm{R},n}$ and $\Delta \gamma _{n} =\gamma _{n} -\gamma _{\mathrm{R},n}$, then Eqs. (3.3) and (3.5) lead to

$$\displaystyle\begin{array}{rcl} \Delta t_{n}& =& \Delta t_{n-1} + \frac{l_{\mathrm{R}} + \Delta l_{n}} {\beta _{n}c_{0}} - \frac{l_{\mathrm{R}}} {\beta _{\mathrm{R},n}c_{0}} = {}\\ & =& \Delta t_{n-1} + \frac{\beta _{\mathrm{R},n}l_{\mathrm{R}} +\beta _{\mathrm{R},n}\Delta l_{n} -\beta _{n}l_{\mathrm{R}}} {\beta _{n}\beta _{\mathrm{R},n}c_{0}} {}\\ & =& \Delta t_{n-1} + l_{\mathrm{R}}\frac{\frac{\Delta l_{n}} {l_{\mathrm{R}}} \beta _{\mathrm{R},n} - \Delta \beta _{n}} {\beta _{n}\beta _{\mathrm{R},n}c_{0}} {}\\ \end{array}$$

$$\displaystyle{ \Rightarrow \Delta t_{n} = \Delta t_{n-1} + \frac{l_{\mathrm{R}}} {\beta _{n}c_{0}}\left (\frac{\Delta l_{n}} {l_{\mathrm{R}}} -\frac{\Delta \beta _{n}} {\beta _{\mathrm{R},n}}\right ). }$$

(3.7)

Based on Eqs. (3.4) and (3.6), one obtains, by means of $\Delta V _{n} = V _{n} - V _{\mathrm{R},n}$,

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

(3.8)

We are now interested in the change of the orbit length that is caused by the change of momentum.

This is relevant, because the orbit of the asynchronous particle will have a different radius r_n from that of the reference particle, since its momentum is different:

$$\displaystyle{p_{n} = \mathit{Qr}_{n}B_{n}.}$$

The magnetic field B_n, however, is the same as for the reference particle, since we assume that the asynchronous particle is still inside the straight section when the reference particle is located at the gap.

We obviously have

$$\displaystyle{\frac{r_{n}} {r_{\mathrm{R}}} = \frac{p_{n}} {p_{\mathrm{R},n}},\qquad r_{n} = r_{\mathrm{R}} \frac{p_{n}} {p_{\mathrm{R},n}} = r_{\mathrm{R}} + \left ( \frac{p_{n}} {p_{\mathrm{R},n}} - 1\right )r_{\mathrm{R}}.}$$

It follows that

$$\displaystyle{l_{n} = 2\pi r_{n} + 2d = l_{\mathrm{R}} + 2\pi \left ( \frac{p_{n}} {p_{\mathrm{R},n}} - 1\right )r_{\mathrm{R}}.}$$

Using $\Delta l_{n} = l_{n} - l_{\mathrm{R}}$ and $\Delta p_{n} = p_{n} - p_{\mathrm{R},n}$, one obtains

$$\displaystyle{\Delta l_{n} = 2\pi r_{\mathrm{R}}\frac{\Delta p_{n}} {p_{\mathrm{R},n}}.}$$

Since l_R and r_R are constant, the quantity

$$\displaystyle{ \alpha _{\mathrm{c}} = \frac{2\pi r_{\mathrm{R}}} {l_{\mathrm{R}}} }$$

(3.9)

can be defined, leading to

$$\displaystyle{ \fbox{$\frac{\Delta l_{n}} {l_{\mathrm{R}}} =\alpha _{\mathrm{c}}\frac{\Delta p_{n}} {p_{\mathrm{R},n}}.$} }$$

(3.10)

Due to this equation, α_c is called the momentum compaction factor. The simple expression in Eq. (3.9) is due to our very special simplified model synchrotron. In general, it depends on the synchrotron lattice, i.e., on the special combination of all magnets. In Appendix A.8, a more general equation, namely (A.54), is discussed. In most synchrotrons, the momentum compaction factor is positive (cf. [1, Sect. 2.1.1]):

$$\displaystyle{\alpha _{\mathrm{c}} > 0.}$$

In this book, we will restrict ourselves to this case.

By means of equation (3.10), we may convert Eq. (3.7) as follows:

$$\displaystyle{\Delta t_{n} = \Delta t_{n-1} + \frac{l_{\mathrm{R}}} {\beta _{n}c_{0}}\left (\alpha _{\mathrm{c}}\frac{\Delta p_{n}} {p_{\mathrm{R},n}} -\frac{\Delta \beta _{n}} {\beta _{\mathrm{R},n}}\right ).}$$

With

$$\displaystyle\begin{array}{rcl} \beta _{n}\gamma _{n} -\beta _{\mathrm{R},n}\gamma _{\mathrm{R},n}& =& (\beta _{\mathrm{R},n} + \Delta \beta _{n})(\gamma _{\mathrm{R},n} + \Delta \gamma _{n}) -\beta _{\mathrm{R},n}\gamma _{\mathrm{R},n} = {}\\ & =& \beta _{\mathrm{R},n}\Delta \gamma _{n} + \Delta \beta _{n}\gamma _{\mathrm{R},n} + \Delta \beta _{n}\Delta \gamma _{n}, {}\\ \end{array}$$

the equation

$$\displaystyle{\frac{\Delta p_{n}} {p_{\mathrm{R},n}} = \frac{\beta _{n}\gamma _{n} -\beta _{\mathrm{R},n}\gamma _{\mathrm{R},n}} {\beta _{\mathrm{R},n}\gamma _{\mathrm{R},n}} }$$

leads to

$$\displaystyle{\frac{\Delta p_{n}} {p_{\mathrm{R},n}} = \frac{\Delta \gamma _{n}} {\gamma _{\mathrm{R},n}} + \frac{\Delta \beta _{n}} {\beta _{\mathrm{R},n}} + \frac{\Delta \beta _{n}} {\beta _{\mathrm{R},n}} \frac{\Delta \gamma _{n}} {\gamma _{\mathrm{R},n}}.}$$

We conclude that

$$\displaystyle{ \Delta t_{n} = \Delta t_{n-1} + \frac{l_{\mathrm{R}}} {\beta _{n}c_{0}}\left (\alpha _{\mathrm{c}}\left (\frac{\Delta \gamma _{n}} {\gamma _{\mathrm{R},n}} + \frac{\Delta \beta _{n}} {\beta _{\mathrm{R},n}} + \frac{\Delta \beta _{n}} {\beta _{\mathrm{R},n}} \frac{\Delta \gamma _{n}} {\gamma _{\mathrm{R},n}}\right ) -\frac{\Delta \beta _{n}} {\beta _{\mathrm{R},n}}\right ). }$$

(3.11)

Please note that up to this point, all derivations have been exact in the scope of our simplified synchrotron model. We now make first approximations, assuming

$$\displaystyle{\frac{\Delta \gamma _{n}} {\gamma _{\mathrm{R},n}} \ll 1\quad \mbox{ and}\quad \frac{\Delta \beta _{n}} {\beta _{\mathrm{R},n}} \ll 1.}$$

In Table 2.3 (p. 48), we find the conversion

$$\displaystyle{ \frac{\Delta \beta _{n}} {\beta _{\mathrm{R},n}} \approx \frac{1} {\gamma _{\mathrm{R},n}^{2}\beta _{\mathrm{R},n}^{2}} \frac{\Delta \gamma _{n}} {\gamma _{\mathrm{R},n}}. }$$

(3.12)

This leads to

$$\displaystyle{\Delta t_{n} = \Delta t_{n-1} + \frac{l_{\mathrm{R}}} {\beta _{n}c_{0}} \frac{\Delta \gamma _{n}} {\gamma _{\mathrm{R},n}}\left (\alpha _{\mathrm{c}}\left (1 + \frac{1} {\gamma _{\mathrm{R},n}^{2}\beta _{\mathrm{R},n}^{2}}\right ) - \frac{1} {\gamma _{\mathrm{R},n}^{2}\beta _{\mathrm{R},n}^{2}}\right ),}$$

and due to

$$\displaystyle{1 +\beta ^{2}\gamma ^{2} =\gamma ^{2},}$$

one obtains

$$\displaystyle{\Delta t_{n} = \Delta t_{n-1} + \frac{l_{\mathrm{R}}} {\beta _{n}c_{0}} \frac{\Delta \gamma _{n}} {\gamma _{\mathrm{R},n}}\left (\alpha _{\mathrm{c}} \frac{1} {\beta _{\mathrm{R},n}^{2}} - \frac{1} {\gamma _{\mathrm{R},n}^{2}\beta _{\mathrm{R},n}^{2}}\right ).}$$

By means of

$$\displaystyle{ \eta _{\mathrm{R},n} =\alpha _{\mathrm{c}} - \frac{1} {\gamma _{\mathrm{R},n}^{2}}, }$$

(3.13)

it follows that

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

(3.14)

If we assume that the time T_R, n needed by the particle for one revolution is short in comparison with the times in which $\Delta t_{n}$ changes significantly, we may calculate the following limit:

$$\displaystyle{\frac{\Delta t_{n} - \Delta t_{n-1}} {T_{\mathrm{R},n}} \rightarrow \frac{\mathrm{d}\Delta t} {\mathrm{d}t}.}$$

The quantities without the index R that describe the asynchronous particle and the quantities with the index R that describe the synchronous particle differ only by a small amount. Therefore, we obtain, by the limiting process,

$$\displaystyle{\frac{\mathrm{d}\Delta t} {\mathrm{d}t} = \frac{l_{\mathrm{R}}\eta _{\mathrm{R}}} {T_{\mathrm{R}}\beta _{\mathrm{R}}^{3}\gamma _{\mathrm{R}}c_{0}}\;\Delta \gamma }$$

$$\displaystyle{ \Leftrightarrow \frac{\mathrm{d}\Delta t} {\mathrm{d}t} = \frac{\eta _{\mathrm{R}}} {\beta _{\mathrm{R}}^{2}\gamma _{\mathrm{R}}}\;\Delta \gamma. }$$

(3.15)

In an analogous way, we may perform a limiting process for Eq. (3.8):

$$\displaystyle{\frac{\mathrm{d}\Delta \gamma } {\mathrm{d}t} = \frac{Q} {T_{\mathrm{R}}m_{0}c_{0}^{2}}\;\Delta V.}$$

If we assume that the voltage

$$\displaystyle{V (t) =\hat{ V }(t)\;\sin (\varphi _{\mathrm{RF}}(t))\qquad \mbox{ with}\quad \varphi _{\mathrm{RF}}(t) =\int _{ 0}^{t}\omega _{ \mathrm{RF}}(\tilde{t})\;\mathrm{d}\tilde{t}}$$

is harmonic (see Chap. 1) and that the reference particle experiences the reference phase (also called synchronous phase) $\varphi _{\mathrm{RF}} =\varphi _{\mathrm{R}}$ when it passes the gap, we obtain

$$\displaystyle{ V _{\mathrm{R}} = V (t_{\mathrm{R}}) =\hat{ V }(t_{\mathrm{R}})\;\sin \varphi _{\mathrm{R}} }$$

(3.16)

for the synchronous particle and

$$\displaystyle{V (t = t_{\mathrm{R}} + \Delta t) \approx \hat{ V }(t_{\mathrm{R}})\;\sin (\omega _{\mathrm{RF}}\Delta t +\varphi _{\mathrm{R}})}$$

for an off-momentum particle. Here we have assumed that neither the amplitude $\hat{V }$ nor the frequency ω_RF changes significantly during the time $\Delta t$. It follows that

$$\displaystyle{ \frac{\mathrm{d}\Delta \gamma } {\mathrm{d}t} = \frac{Q\hat{V }} {T_{\mathrm{R}}m_{0}c_{0}^{2}}\left (\sin (\omega _{\mathrm{RF}}\Delta t +\varphi _{\mathrm{R}}) -\sin \varphi _{\mathrm{R}}\right ). }$$

(3.17)

Equations (3.15) and (3.17) may be written in the form

$$\displaystyle{\frac{\mathrm{d}\vec{r}} {\mathrm{d}t} =\vec{ v}(\vec{r}),}$$

which is compatible with Sect. 2.8 if

$$\displaystyle{\vec{r} = \left ({ \Delta t \atop \Delta \gamma } \right )\qquad \mbox{ and}\quad \vec{v} = \left ({ v_{\Delta t} \atop v_{\Delta \gamma }} \right )}$$

are defined. For the divergence of $\vec{v}$, one obtains

$$\displaystyle{\mathrm{div}\;\vec{v} = \frac{\partial v_{\Delta t}} {\partial \Delta t} + \frac{\partial v_{\Delta \gamma }} {\partial \Delta \gamma } = 0.}$$

Thus we have shown that the flow in phase space preserves the phase space area (see Sect. 2.10.3); Liouville’s theorem is satisfied.² Therefore, one usually uses the unit eVs for the phase space area, i.e., a product of time and energy. This unit is obtained if we calculate with $\Delta W$ instead of $\Delta \gamma$. This does not change the invariance of the phase space area, since the required factor $m_{0}c_{0}^{2}$ is constant:

$$\displaystyle{\frac{\mathrm{d}\Delta t} {\mathrm{d}t} = \frac{\eta _{\mathrm{R}}} {m_{0}c_{0}^{2}\beta _{\mathrm{R}}^{2}\gamma _{\mathrm{R}}}\;\Delta W}$$

$$\displaystyle{ \Leftrightarrow \fbox{$\frac{\mathrm{d}\Delta t} {\mathrm{d}t} = \frac{\eta _{\mathrm{R}}} {W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}}\;\Delta W,$} }$$

(3.18)

$$\displaystyle{ \fbox{$\frac{\mathrm{d}\Delta W} {\mathrm{d}t} = \frac{Q\hat{V }} {T_{\mathrm{R}}} \left (\sin (\omega _{\mathrm{RF}}\Delta t +\varphi _{\mathrm{R}}) -\sin \varphi _{\mathrm{R}}\right ).$} }$$

(3.19)

Here we used the total energy

$$\displaystyle{ \fbox{$W_{\mathrm{R}} = \mathit{mc}_{0}^{2} = m_{ 0}c_{0}^{2}\gamma _{ \mathrm{R}}$} }$$

(3.20)

of the synchronous particle.

3.3 Tracking Equations

Equations (3.8) and (3.14) may be used as recurrence steps for a so-called particle tracking program. It is remarkable that the result of Eq. (3.8) has to be inserted on the right-hand side of Eq. (3.14). Therefore, one cannot evaluate both equations at the same time. If one replaced $\Delta \gamma _{n}$ in Eq. (3.14) by $\Delta \gamma _{n-1}$, which does not seem to change things significantly at first sight, then unstable orbits instead of stable orbits would be the result. This may be verified as follows:

Equation (3.14) obviously has the form

$$\displaystyle{x_{k} = x_{k-1} + f(y_{k}).}$$

Equation (3.8), however, has the form

$$\displaystyle{y_{k} = y_{k-1} + g(x_{k-1}).}$$

If we insert the second equation into the first in order to have only quantities on the right-hand side that belong to step k − 1, we obtain

$$\displaystyle\begin{array}{rcl} x_{k}& =& x_{k-1} + f(y_{k-1} + g(x_{k-1})), {}\\ y_{k}& =& y_{k-1} + g(x_{k-1}). {}\\ \end{array}$$

A map

$$\displaystyle{\vec{r}_{k} =\vec{ F}(\vec{r}_{k-1})}$$

is area-preserving if the absolute value of the Jacobian equals 1, as we discussed in Sect. 2.10.2. In the case under consideration here, the Jacobian is

$$\displaystyle{ \frac{\partial (x_{k},y_{k})} {\partial (x_{k-1},y_{k-1})} = \left \vert \begin{array}{cc} 1 + f^{{\prime}}(y_{ k-1} + g(x_{k-1}))\;g^{{\prime}}(x_{ k-1})&f^{{\prime}}(y_{ k-1} + g(x_{k-1})) \\ g^{{\prime}}(x_{k-1}) & 1 \end{array} \right \vert = 1.}$$

Hence, the system that is described by the tracking equations (3.8) and (3.14) is actually area-preserving. The discrete tracking equations therefore are a reasonable discretization of the continuous system; the corresponding iteration algorithm is called a leapfrog scheme (cf. [2, Appendix E]).

If, however, one starts with the symmetric pair of equations

$$\displaystyle\begin{array}{rcl} x_{k}& =& x_{k-1} + f(y_{k-1}), {}\\ y_{k}& =& y_{k-1} + g(x_{k-1}), {}\\ \end{array}$$

one obtains the Jacobian

$$\displaystyle{ \frac{\partial (x_{k},y_{k})} {\partial (x_{k-1},y_{k-1})} = \left \vert \begin{array}{cc} 1 &f^{{\prime}}(y_{ k-1}) \\ g^{{\prime}}(x_{k-1})& 1 \end{array} \right \vert \neq 1.}$$

In contrast to the continuous system, these discrete equations do not describe a conservative system. We may therefore conclude that Eqs. (3.8) and (3.14) are correct, even though they are not symmetric.

By means of the tracking equations, we are now able to simulate the behavior of particle clouds in longitudinal phase space. An example is shown in Fig. 3.2. At the start of the simulation, several particles are randomly distributed in phase space inside an ellipse. A harmonic voltage

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

is assumed. Each particle moves in phase space according to the tracking equations.

Fig. 3.2

Example of a tracking simulation of particles in phase space

The simulation shows some effects that can also be seen in the snapshots displayed in Fig. 3.2:

• The particles perform rotations in longitudinal phase space (counterclockwise). This is the so-called synchrotron oscillation of individual particles that was already mentioned in the introduction. The conclusion that the particles oscillate around the synchronous phase, i.e., that they perform oscillations on the axis $\Delta t$, is now visualized in a different way: since the energy deviation is added in phase space, this oscillation on the time axis is converted into a rotation in phase space. For the sake of clarity, it should again be emphasized (cf. Footnote 4 on p. 5) that the oscillation does not occur around a single slope of the voltage V (t) but that an overlay of several revolutions is necessary to make this oscillation (and also the rotation in phase space) visible. After all, the particles fly in the longitudinal direction with very high speed.

• The combination of all the particles (e.g., in the first diagram of Fig. 3.2) is called a bunch.

• The third picture shows approximately the initial bunch rotated by 90^∘ in phase space (and stretched due to the arbitrary scaling of the axes). However, the bunch shape is not identical to the original elliptical bunch, because the outer particles rotate with a lower oscillation frequency; they lag behind. In the beginning, this leads to an S-shaped bunch, and later, spiral galaxies are formed in phase space.

• After several revolutions, the different spiral arms can no longer be distinguished from each other.

• Based on the first four (or even five) pictures, one may verify that the area occupied by the particles in phase space remains constant, as proven above. In the last picture, the occupied area seems to be larger, but this is due to the effect mentioned earlier that the spiral arms can no longer be identified clearly. One speaks of filamentation and phase space dilution. The reason is the limited number of particles used instead of a continuous particle distribution. Winding up the spiral arms will involve some empty phase space area that is also wound up in between. Strictly speaking, the definition of the area that is occupied by particles is possible only for a continuous particle distribution with clear boundaries, i.e., for an infinite number of particles. For a finite number of particles, the definition may be used only approximately. In the beginning, the number of particles is large enough to identify the boundaries of the bunch. Later, the same number of particles is no longer sufficient to identify the more complex bunch shape with the wound up spiral arms. Therefore, as a result of the lack of a clear definition of area for a finite number of particles, the visible area occupied by the particles in the last diagram is larger than that in the first diagrams. We will return to this problem in Sects. 3.24, 5.4, and 5.6.4.

• The phase space area that is occupied by particles is called the longitudinal emittance. Due to Liouville’s theorem, the longitudinal emittance is constant.³ For the longitudinal emittance, one usually uses eVs as the physical unit. In this book, we are not dealing with transverse beam dynamics, and we will therefore omit the word “longitudinal” from time to time.

• If the simulation runs even longer than displayed here, the macroscopic shape of the bunch will no longer change significantly. The individual particles will, of course, continue to move, but the distribution will no longer change. As we will see later, this is the case of a matched bunch, for which

  $$\displaystyle{ \frac{\partial \rho } {\partial t} = 0}$$

  holds everywhere in phase space.

• All particles that are located inside a small time interval (i.e., inside a narrow vertical strip) will contribute to the beam current during this time interval. In other words, the specific instantaneous energy of these particles does not matter. Therefore, the beam signal is obtained as the projection

  $$\displaystyle{I_{\mathrm{beam}}(t) = I_{\mathrm{beam}}(t_{\mathrm{R}} + \Delta t) =\int \rho _{q}(\Delta t,\Delta W)\;\mathrm{d}(\Delta W)}$$

  of the bunch onto the time axis. This is why both the particle distribution in phase space and the pulses of the beam signal are called bunches.

Of course, the example shown in Fig. 3.2 is a very special one. If, for example, a larger bunch had been assumed as an initial distribution, one would have seen that the outer particles may also move on unstable trajectories. Therefore, we will discuss the phase space behavior in more detail in the following sections. Nevertheless, the individual particles will still show a very transparent behavior that justifies our phase space analysis.

The phenomena discussed here were based on a computer simulation of several particles. These tracking simulations are inevitable for complicated situations. Nevertheless, it is possible to study the basic effects analytically. This will be done in the following sections.

3.4 Phase Slip Factor and Transition Energy

The quantity⁴

$$\displaystyle{ \fbox{$\eta _{\mathrm{R}} =\alpha _{\mathrm{c}} - \frac{1} {\gamma _{\mathrm{R}}^{2}},$} }$$

(3.21)

which was introduced in Eq. (3.13), is called⁵ a slip factor (cf. [4]), phase slip factor (cf. [5]), frequency slip factor (cf. [6]), or slippage factor [7]. This name will be explained in the following.

Based on

$$\displaystyle{u_{\mathrm{R}} = \frac{l_{\mathrm{R}}} {T_{\mathrm{R}}}\qquad \Leftrightarrow T_{\mathrm{R}} = \frac{l_{\mathrm{R}}} {\beta _{\mathrm{R}}c_{0}},}$$

we may obtain the following relation by calculating the derivative with respect to time:

$$\displaystyle{\dot{T}_{\mathrm{R}} = \frac{\dot{l}_{\mathrm{R}}\beta _{\mathrm{R}}c_{0} -\dot{\beta }_{\mathrm{R}}c_{0}l_{\mathrm{R}}} {\beta _{\mathrm{R}}^{2}c_{0}^{2}} }$$

$$\displaystyle{\Rightarrow \frac{\dot{T}_{\mathrm{R}}} {T_{\mathrm{R}}} = \frac{\dot{l}_{\mathrm{R}}} {l_{\mathrm{R}}} -\frac{\dot{\beta }_{\mathrm{R}}} {\beta _{\mathrm{R}}}.}$$

Therefore, for a deviation $\Delta T$ of the revolution time, one obtains approximately

$$\displaystyle{\frac{\Delta T} {T_{\mathrm{R}}} = \frac{\Delta l} {l_{\mathrm{R}}} -\frac{\Delta \beta } {\beta _{\mathrm{R}}}.}$$

On the one hand, we may make use of Eq. (3.10),

$$\displaystyle{\frac{\Delta l} {l_{\mathrm{R}}} =\alpha _{\mathrm{c}}\frac{\Delta p} {p_{\mathrm{R}}},}$$

to calculate the deviation $\Delta l$ of the orbit length. On the other hand, one obtains

$$\displaystyle\begin{array}{rcl} \gamma ^{2} = \frac{1} {1 -\beta ^{2}}\qquad & \Rightarrow & \gamma ^{2} -\beta ^{2}\gamma ^{2} = 1\qquad \Rightarrow \beta \gamma = \sqrt{\gamma ^{2 } - 1}\qquad {}\\ & \Rightarrow & \frac{\mathrm{d}} {\mathrm{d}t}(\beta \gamma ) = \frac{2\gamma \dot{\gamma }} {2\sqrt{\gamma ^{2 } - 1}} = \frac{\dot{\gamma }} {\beta } {}\\ & \Rightarrow & \frac{1} {\beta \gamma } \frac{\mathrm{d}} {\mathrm{d}t}(\beta \gamma ) = \frac{1} {\beta ^{2}} \frac{\dot{\gamma }} {\gamma }\qquad \Rightarrow \frac{\Delta p} {p_{\mathrm{R}}} \approx \frac{1} {\beta _{\mathrm{R}}^{2}} \frac{\Delta \gamma } {\gamma _{\mathrm{R}}}. {}\\ \end{array}$$

Together with Eq. (3.12),

$$\displaystyle{\frac{\Delta \beta } {\beta _{\mathrm{R}}} = \frac{1} {\gamma _{\mathrm{R}}^{2}\beta _{\mathrm{R}}^{2}} \frac{\Delta \gamma } {\gamma _{\mathrm{R}}},}$$

one obtains

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

which may be inserted above:

$$\displaystyle{\frac{\Delta T} {T_{\mathrm{R}}} = \left (\alpha _{\mathrm{c}} - \frac{1} {\gamma _{\mathrm{R}}^{2}}\right )\frac{\Delta p} {p_{\mathrm{R}}}.}$$

Therefore,

$$\displaystyle{\fbox{$\frac{\Delta T} {T_{\mathrm{R}}} =\eta _{\mathrm{R}}\;\frac{\Delta p} {p_{\mathrm{R}}} $}}$$

is valid. In conclusion, the factor η_R describes by what percentage the revolution time changes if the momentum changes by a certain percentage. Every time deviation $\Delta t$ of a particle with respect to the reference particle may be converted into a phase deviation:

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

$$\displaystyle{\Leftrightarrow \frac{\Delta \varphi _{\mathrm{RF}}} {2\pi } = h\; \frac{\Delta t} {T_{\mathrm{R}}}.}$$

This explains the term “phase slip factor.”

For sufficiently low energies, η_R < 0 is valid, so that a larger momentum causes the particles to arrive again earlier at the gap. For

$$\displaystyle{\alpha _{\mathrm{c}} > \frac{1} {\gamma _{\mathrm{R}}^{2}},}$$

this is no longer valid. Then the path that has to be taken by the particles increases so much that the momentum increase cannot compensate for it; the particles will arrive later at the gap. One defines the transition gammaγ_T by

$$\displaystyle{\fbox{$\alpha _{\mathrm{c}} = \frac{1} {\gamma _{\mathrm{T}}^{2}} \gamma _{\mathrm{T}} = \frac{1} {\sqrt{\alpha _{\mathrm{c}}}},$}}$$

so that this transition point is reached for γ_R = γ_T. The corresponding energy is called the transition energy.

3.5 Accelerating Voltage

The magnetic field B_n will be given at each revolution n. This leads to the question how large the voltage V_R, n must be to ensure that the synchronous particle always takes the same reference path of length l_R.

Due to Eq. (3.2), we first of all note that from a knowledge of B_n, the momentum p_R, n of the reference particle is known for each revolution n. Therefore, β_R, n and γ_R, n are also known.

From $\gamma = \sqrt{1 + (\beta \gamma )^{2}}$, Eq. (3.4) leads to

$$\displaystyle{V _{\mathrm{R},n} = \frac{m_{0}c_{0}^{2}} {Q} (\gamma _{\mathrm{R},n+1} -\gamma _{\mathrm{R},n}) = \frac{m_{0}c_{0}^{2}} {Q} \left (\sqrt{1 + \left (\frac{p_{\mathrm{R},n+1 } } {m_{0}c_{0}} \right )^{2}} -\sqrt{1 + \left ( \frac{p_{\mathrm{R},n } } {m_{0}c_{0}}\right )^{2}}\right ).}$$

If we insert

$$\displaystyle{p_{\mathrm{R},n+1} = p_{\mathrm{R},n} +\delta p,}$$

we obtain

$$\displaystyle\begin{array}{rcl} & & \sqrt{1 + \left (\frac{p_{\mathrm{R},n+1 } } {m_{0}c_{0}} \right )^{2}} = \sqrt{1 + \left ( \frac{p_{\mathrm{R},n } } {m_{0}c_{0}}\right )^{2} + 2 \frac{p_{\mathrm{R},n}\delta p} {(m_{0}c_{0})^{2}} + \frac{\delta p^{2}} {(m_{0}c_{0})^{2}}} = {}\\ & & = \sqrt{1 + \left ( \frac{p_{\mathrm{R},n } } {m_{0}c_{0}}\right )^{2}}\sqrt{1 + \frac{2 \frac{p_{\mathrm{R},n } \delta p} {(m_{0}c_{0})^{2}} + \frac{\delta p^{2}} {(m_{0}c_{0})^{2}} } {1 + \left ( \frac{p_{\mathrm{R},n}} {m_{0}c_{0}} \right )^{2}}}. {}\\ \end{array}$$

Using $\sqrt{1 + x} \approx 1 + \frac{x} {2}$ for x ≪ 1 in the second square root and neglecting the quadratic term leads to

$$\displaystyle{\sqrt{1 + \left (\frac{p_{\mathrm{R},n+1 } } {m_{0}c_{0}} \right )^{2}} \approx \sqrt{1 + \left ( \frac{p_{\mathrm{R},n } } {m_{0}c_{0}}\right )^{2}}\left (1 + \frac{ \frac{p_{\mathrm{R},n}\delta p} {(m_{0}c_{0})^{2}} } {1 + \left ( \frac{p_{\mathrm{R},n}} {m_{0}c_{0}} \right )^{2}}\right ).}$$

This yields the approximation

$$\displaystyle{V _{\mathrm{R},n} \approx \frac{m_{0}c_{0}^{2}} {Q} \sqrt{1 + \left ( \frac{p_{\mathrm{R},n } } {m_{0}c_{0}}\right )^{2}}\; \frac{ \frac{p_{\mathrm{R},n}\delta p} {(m_{0}c_{0})^{2}} } {1 + \left ( \frac{p_{\mathrm{R},n}} {m_{0}c_{0}} \right )^{2}} = \frac{1} {m_{0}Q} \frac{p_{\mathrm{R},n}\delta p} {\sqrt{1 + \left ( \frac{p_{\mathrm{R},n } } {m_{0}c_{0}} \right )^{2}}}.}$$

Due to $\frac{p_{\mathrm{R},n}} {m_{0}c_{0}} =\gamma _{\mathrm{R},n}\beta _{\mathrm{R},n}$ and $\sqrt{ 1 + (\gamma _{\mathrm{R},n}\beta _{\mathrm{R},n})^{2}} =\gamma _{\mathrm{R},n}$, it follows that

$$\displaystyle{V _{\mathrm{R},n} \approx \frac{p_{\mathrm{R},n}\delta p} {m_{0}Q\gamma _{\mathrm{R},n}} = \frac{c_{0}\beta _{\mathrm{R},n}\delta p} {Q} = \frac{u_{\mathrm{R},n}\delta p} {Q} = \frac{l_{\mathrm{R}}\delta p} {T_{\mathrm{R},n}Q}.}$$

With the help of Eq. (3.2), one obtains

$$\displaystyle{\delta p = p_{\mathrm{R},n+1} - p_{\mathrm{R},n} = \mathit{Qr}_{\mathrm{R}}(B_{n+1} - B_{n}),}$$

and one finally obtains

$$\displaystyle{V _{\mathrm{R},n} \approx l_{\mathrm{R}}r_{\mathrm{R}}\;\frac{B_{n+1} - B_{n}} {T_{\mathrm{R},n}},}$$

which leads to the continuous counterpart

$$\displaystyle{ \fbox{$V _{\mathrm{R}} \approx l_{\mathrm{R}}r_{\mathrm{R}}\dot{B}.$} }$$

(3.22)

This is the voltage that the reference particle actually experiences, not the required amplitude. The relation between these two quantities is given by Eq. (3.16):

$$\displaystyle{ \fbox{$V _{\mathrm{R}} =\hat{ V }\;\sin \varphi _{\mathrm{R}}.$} }$$

(3.23)

Without loss of generality, the reference phase $\varphi _{\mathrm{R}}$ will always be in the range $\varphi _{\mathrm{R}} \in [-\pi,+\pi ]$. We discuss two cases:

• $\dot{B} = 0$: This means that the magnetic field remains constant, i.e., that the reference particle does not change its momentum or energy, as Eq. (3.1) shows. In this case, V_R = 0 must hold. This can be realized either by $\hat{V } = 0$ or by $\hat{V } > 0$ and $\varphi _{\mathrm{R}} \in \{ 0,\pm \pi \}$. The first case means that none of the particles experiences a voltage, so that a coasting beam will be present. The second case means that—in contrast to the reference particle—the off-momentum particles will experience a voltage. The phase-focusing principle (see Chap. 1) will lead to a bunched beam in this case. The larger the amplitude $\hat{V }$ is, the stronger will be the bunching effect, i.e., the shorter the bunches (lower bunching factor B_f according to Eq. (2.37)). The condition $\dot{B} = 0$ is usually satisfied before acceleration and after acceleration, i.e., at injection energy and at extraction energy. One also speaks of the injection plateau and the extraction plateau. The latter is also called the flat top energy.

• $\dot{B} > 0$: This means that the magnetic field, and hence the momentum and energy of the reference particle, increases. This is the acceleration phase, which is possible only if $\hat{V } > 0$ and $\varphi _{\mathrm{R}} > 0$ for Q > 0 (or $\varphi _{\mathrm{R}} < 0$ for Q < 0) are chosen. Only bunched beams can be accelerated.

We will return to these facts and discuss them in more detail in Sect. 3.22.

3.6 Synchrotron Oscillation

For small time deviations $\Delta t$ with $\omega _{\mathrm{RF}}\Delta t \ll 1$, the approximation

$$\displaystyle\begin{array}{rcl} \sin (\omega _{\mathrm{RF}}\Delta t +\varphi _{\mathrm{R}})& =& \sin (\omega _{\mathrm{RF}}\Delta t)\;\cos \varphi _{\mathrm{R}} +\cos (\omega _{\mathrm{RF}}\Delta t)\;\sin \varphi _{\mathrm{R}} {}\\ & \approx & \omega _{\mathrm{RF}}\Delta t\;\cos \varphi _{\mathrm{R}} +\sin \varphi _{\mathrm{R}} {}\\ \end{array}$$

is valid, so that a linear approximation of Eq. (3.19) leads to the relation

$$\displaystyle{ \frac{\mathrm{d}\Delta W} {\mathrm{d}t} = \frac{Q\hat{V }\;\omega _{\mathrm{RF}}\;\cos \varphi _{\mathrm{R}}} {T_{\mathrm{R}}} \;\Delta t. }$$

(3.24)

Together with Eq. (3.18), it follows that

$$\displaystyle{\frac{\mathrm{d}^{2}\Delta W} {\mathrm{d}t^{2}} = \frac{\eta _{\mathrm{R}}\;Q\hat{V }\;\omega _{\mathrm{RF}}\;\cos \varphi _{\mathrm{R}}} {W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}T_{\mathrm{R}}}\;\Delta W,}$$

if one takes into account that the quantities in the fraction will change slowly. This is an oscillator equation of the form

$$\displaystyle{\frac{\mathrm{d}^{2}\Delta W} {\mathrm{d}t^{2}} +\omega _{ \mathrm{S,0}}^{2}\Delta W = 0}$$

with

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

$$\displaystyle{\Leftrightarrow \omega _{\mathrm{S,0}} = \sqrt{- \frac{\eta _{\mathrm{R} } \;Q\hat{V }\;h2\pi \;\cos \varphi _{\mathrm{R} } } {W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}T_{\mathrm{R}}^{2}}}}$$

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

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

(3.25)

The oscillation frequency f_S,0 is called the synchrotron frequency. The index 0 indicates that it is valid for particles with small oscillation amplitudes with respect to the reference particle. Please note that for positive charges Q > 0 below the transition η_R < 0, one chooses $0 \leq \varphi _{\mathrm{R}} <\pi /2$, so that the argument of the square root is positive. The various cases have to be treated accordingly (see Fig. 3.3), so that

$$\displaystyle{ \eta _{\mathrm{R}}\;Q\;\cos \varphi _{\mathrm{R}} < 0 }$$

(3.26)

holds.

Fig. 3.3

Phase relation between gap voltage and beam signal (solid line: stationary case with $\varphi _{\mathrm{R}} = 0$ for η_RQ < 0 or $\varphi _{\mathrm{R}} = \pm \pi$ for η_RQ > 0; dotted line: acceleration case)

3.7 Principal Axes

Due to the simple oscillator equation for small oscillation amplitudes, we get elliptic trajectories in the $(\Delta t,\Delta W)$ phase space. This is true for all particles inside a bunch that is sufficiently short. Now we try to find the principal axes of the bunch ellipse in phase space for small oscillation amplitudes. For this purpose, we begin with Eq. (3.24),

$$\displaystyle{\frac{\mathrm{d}\Delta W} {\mathrm{d}t} = \frac{Q\hat{V }\;\omega _{\mathrm{RF}}\;\cos \varphi _{\mathrm{R}}} {T_{\mathrm{R}}} \;\Delta t,}$$

and insert the ansatz

$$\displaystyle{\Delta W = \Delta \hat{W}\;\sin \omega _{\mathrm{S,0}}t.}$$

This leads to

$$\displaystyle{\frac{\mathrm{d}\Delta W} {\mathrm{d}t} = \Delta \hat{W}\;\omega _{\mathrm{S,0}}\;\cos \omega _{\mathrm{S,0}}t = \frac{Q\hat{V }\;\omega _{\mathrm{RF}}\;\cos \varphi _{\mathrm{R}}} {T_{\mathrm{R}}} \;\Delta t.}$$

Hence, one obtains

$$\displaystyle{\Delta t = \pm \Delta \hat{t}\;\cos \omega _{\mathrm{S,0}}t}$$

with

$$\displaystyle{\Delta \hat{t} = \pm \Delta \hat{W}\; \frac{\omega _{\mathrm{S,0}}T_{\mathrm{R}}} {Q\hat{V }\;\omega _{\mathrm{RF}}\;\cos \varphi _{\mathrm{R}}}.}$$

Thus, we obtain the ratio of the principal axes

$$\displaystyle{\frac{\Delta \hat{W}} {\Delta \hat{t}} = \pm \frac{Q\hat{V }\;\omega _{\mathrm{RF}}\;\cos \varphi _{\mathrm{R}}} {\omega _{\mathrm{S,0}}T_{\mathrm{R}}} = \pm \frac{Q\hat{V }\;h\;f_{\mathrm{R}}^{2}\;\cos \varphi _{\mathrm{R}}} {f_{\mathrm{S,0}}}.}$$

From Eq. (3.25), it follows that

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

(3.27)

leading to

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

(3.28)

3.8 Hamiltonian

Now we want to analyze the longitudinal phase space dynamics without restricting ourselves to small oscillation amplitudes. For this purpose, we will determine a Hamiltonian. Equations (3.18) and (3.19) are converted into Hamilton’s equations if we set

$$\displaystyle{ \frac{\partial H} {\partial \Delta t} = \Delta \dot{W},}$$

$$\displaystyle{ \frac{\partial H} {\partial \Delta W} = -\Delta \dot{t}.}$$

By integration of the first equation with respect to $\Delta t$, one obtains

$$\displaystyle{H = \frac{Q\hat{V }} {T_{\mathrm{R}}} \left (- \frac{1} {\omega _{\mathrm{RF}}}\cos (\omega _{\mathrm{RF}}\Delta t +\varphi _{\mathrm{R}}) - \Delta t\;\sin \varphi _{\mathrm{R}}\right ) + C_{1}(\Delta W).}$$

Here we assumed that $\hat{V }$, T_R, ω_RF, $\varphi _{\mathrm{R}}$ are constant or change sufficiently slowly. An integration of the second equation with respect to $\Delta W$ leads to

$$\displaystyle{H = - \frac{\eta _{\mathrm{R}}} {W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}}\;\frac{\Delta W^{2}} {2} + C_{2}(\Delta t).}$$

We combine both results to obtain

$$\displaystyle{H = - \frac{\eta _{\mathrm{R}}} {W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}}\;\frac{\Delta W^{2}} {2} -\frac{Q\hat{V }} {T_{\mathrm{R}}} \left ( \frac{1} {\omega _{\mathrm{RF}}}\cos (\omega _{\mathrm{RF}}\Delta t +\varphi _{\mathrm{R}}) + \Delta t\;\sin \varphi _{\mathrm{R}}\right ) +\mathrm{ const}.}$$

The constant is now fixed so that H(0, 0) = 0 holds:

$$\displaystyle{ \fbox{$H = - \frac{\eta _{\mathrm{R}}} {W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}}\;\frac{\Delta W^{2}} {2} -\frac{Q\hat{V }} {T_{\mathrm{R}}} \left ( \frac{1} {\omega _{\mathrm{RF}}}\left [\cos (\omega _{\mathrm{RF}}\Delta t +\varphi _{\mathrm{R}}) -\cos \varphi _{\mathrm{R}}\right ] + \Delta t\;\sin \varphi _{\mathrm{R}}\right ).$} }$$

(3.29)

3.9 Separatrix

We will now analyze the properties of the Hamiltonian $H(\Delta W,\Delta t)$ in Eq. (3.29). First of all, we determine the fixed points of the system. For these fixed points, $\frac{\partial H} {\partial \Delta W} = 0$ and $\frac{\partial H} {\partial \Delta t} = 0$ must be valid (see Sect. 2.11.4). The first of these requirements leads directly to $\Delta W = 0$. The second equation leads to

$$\displaystyle{-\sin (\omega _{\mathrm{RF}}\Delta t +\varphi _{\mathrm{R}}) +\sin \varphi _{\mathrm{R}} = 0.}$$

One sees that this equation is satisfied for $\omega _{\mathrm{RF}}\Delta t +\varphi _{\mathrm{R}} =\varphi _{\mathrm{R}}$ and for⁶$\omega _{\mathrm{RF}}\Delta t +\varphi _{\mathrm{R}} =\pi \;\mathrm{ sign}\,\varphi _{\mathrm{R}} -\varphi _{\mathrm{R}}$. Hence, we obtain the two fixed points $(\Delta W,\Delta t) = (0,0)$ and $(\Delta W,\Delta t) = \left (0, \frac{\pi \;\mathrm{sign}\,\varphi _{\mathrm{R}}-2\varphi _{\mathrm{R}}} {\omega _{\mathrm{RF}}} \right )$, for which, of course, further fixed points exist periodically.

According to Appendix A.9, the first one is the stable fixed point, for which H has a minimum or a maximum, and the second one is a saddle point. For the saddle point,

$$\displaystyle{\Delta \varphi _{\mathrm{saddle}} =\pi \;\mathrm{ sign}\,\varphi _{\mathrm{R}} - 2\varphi _{\mathrm{R}},}$$

H has the value

$$\displaystyle\begin{array}{rcl} H_{s}& =& -\frac{Q\hat{V }} {T_{\mathrm{R}}} \left ( \frac{1} {\omega _{\mathrm{RF}}}\left [\cos (\pi \;\mathrm{sign}\,\varphi _{\mathrm{R}} -\varphi _{\mathrm{R}}) -\cos \varphi _{\mathrm{R}}\right ] + \frac{\pi \;\mathrm{sign}\,\varphi _{\mathrm{R}} - 2\varphi _{\mathrm{R}}} {\omega _{\mathrm{RF}}} \;\sin \varphi _{\mathrm{R}}\right )= \\ & =& - \frac{Q\hat{V }} {T_{\mathrm{R}}\omega _{\mathrm{RF}}}\left (\cos (\pi \;\mathrm{sign}\,\varphi _{\mathrm{R}} -\varphi _{\mathrm{R}}) -\cos \varphi _{\mathrm{R}} + (\pi \;\mathrm{sign}\,\varphi _{\mathrm{R}} - 2\varphi _{\mathrm{R}})\;\sin \varphi _{\mathrm{R}}\right ) = \\ & =& \frac{Q\hat{V }} {T_{\mathrm{R}}\omega _{\mathrm{RF}}}\left (2\;\cos \varphi _{\mathrm{R}} + (2\varphi _{\mathrm{R}} -\pi \;\mathrm{ sign}\,\varphi _{\mathrm{R}})\;\sin \varphi _{\mathrm{R}}\right ). {}\end{array}$$

(3.30)

The Hamiltonian keeps this value H_s on the whole separatrix (see Sect. 2.8.9). Therefore, using Eq. (3.29), one obtains for the separatrix

$$\displaystyle\begin{array}{rcl} - \frac{\eta _{\mathrm{R}}} {W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}}\;\frac{\Delta W^{2}} {2} & =& \frac{Q\hat{V }} {T_{\mathrm{R}}\omega _{\mathrm{RF}}}\left [\cos (\omega _{\mathrm{RF}}\Delta t +\varphi _{\mathrm{R}}) -\cos \varphi _{\mathrm{R}} +\omega _{\mathrm{RF}}\Delta t\;\sin \varphi _{\mathrm{R}}+\right. {}\\ & +& \left.2\;\cos \varphi _{\mathrm{R}} + (2\varphi _{\mathrm{R}} -\pi \;\mathrm{ sign}\,\varphi _{\mathrm{R}})\;\sin \varphi _{\mathrm{R}}\right ], {}\\ \end{array}$$

$$\displaystyle\begin{array}{rcl} \Delta W^{2}& =& \frac{2W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}Q\hat{V }} {-\eta _{\mathrm{R}}T_{\mathrm{R}}\omega _{\mathrm{RF}}} \left [\cos (\omega _{\mathrm{RF}}\Delta t +\varphi _{\mathrm{R}}) +\omega _{\mathrm{RF}}\Delta t\;\sin \varphi _{\mathrm{R}} +\cos \varphi _{\mathrm{R}}\right. \\ & & \qquad \qquad \qquad + \left.(2\varphi _{\mathrm{R}} -\pi \;\mathrm{ sign}\,\varphi _{\mathrm{R}})\;\sin \varphi _{\mathrm{R}}\right ]. {}\end{array}$$

(3.31)

The maximum of this function is characterized by the fact that the derivative with respect to $\Delta t$ vanishes, so that

$$\displaystyle{-\omega _{\mathrm{RF}}\sin (\omega _{\mathrm{RF}}\Delta t +\varphi _{\mathrm{R}}) +\omega _{\mathrm{RF}}\;\sin \varphi _{\mathrm{R}} = 0}$$

holds. This is satisfied for $\Delta t = 0$ and $\omega _{\mathrm{RF}}\Delta t +\varphi _{\mathrm{R}} =\pi \;\mathrm{ sign}\,\varphi _{\mathrm{R}} -\varphi _{\mathrm{R}}$ (further zeros occur periodically). For the latter of these two values, $\Delta W = 0$, so that it does not correspond to the desired maximum. Therefore, the maximum is obtained for $\Delta t = 0$:

$$\displaystyle{\Delta W_{\mathrm{max}} = \left (\frac{2W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}Q\hat{V }} {-\eta _{\mathrm{R}}T_{\mathrm{R}}\omega _{\mathrm{RF}}} \left (2\;\cos \varphi _{\mathrm{R}} + (2\varphi _{\mathrm{R}} -\pi \;\mathrm{ sign}\,\varphi _{\mathrm{R}})\;\sin \varphi _{\mathrm{R}}\right )\right )^{1/2}}$$

$$\displaystyle{\Leftrightarrow \Delta W_{\mathrm{max}} = \left (\frac{W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}Q\hat{V }} {-h\pi \eta _{\mathrm{R}}} \left (2\;\cos \varphi _{\mathrm{R}} + (2\varphi _{\mathrm{R}} -\pi \;\mathrm{ sign}\,\varphi _{\mathrm{R}})\;\sin \varphi _{\mathrm{R}}\right )\right )^{1/2}.}$$

For the stationary case, i.e., for $\varphi _{\mathrm{R}} = 0$ if Q and η_R have opposite signs (this is our standard case if nothing else is stated) or for $\varphi _{\mathrm{R}} = \pm \pi$ if Q and η_R have identical signs, one obtains

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

(3.32)

It follows that

$$\displaystyle{ \fbox{$\Delta W_{\mathrm{max}} = \Delta W_{\mathrm{max,stat}}\;Y (\varphi _{\mathrm{R}})$} }$$

(3.33)

with

$$\displaystyle{ \fbox{$Y (\varphi _{\mathrm{R}}) = \sqrt{\left \vert \cos \varphi _{\mathrm{R} } + \left (\varphi _{\mathrm{R} } - \frac{\pi } {2}\;\mathrm{sign}\,\varphi _{\mathrm{R}}\right )\;\sin \varphi _{\mathrm{R}}\right \vert }.$} }$$

(3.34)

Let us now briefly analyze the function $Y (\varphi _{\mathrm{R}})$. An extremum is expected for

$$\displaystyle{-\sin \varphi _{\mathrm{R}} +\sin \varphi _{\mathrm{R}} + \left (\varphi _{\mathrm{R}} - \frac{\pi } {2}\;\mathrm{sign}\,\varphi _{\mathrm{R}}\right )\;\cos \varphi _{\mathrm{R}} = 0,}$$

i.e., for $\varphi _{\mathrm{R}} = \pm \pi /2$. As shown in Fig. 3.4, this is a minimum. In the interval $0 \leq \varphi _{\mathrm{R}} \leq \pi /2$, the function $Y (\varphi _{\mathrm{R}})$ decreases monotonically from 1 to 0. Please note that Eq. (3.34) is valid for all four cases discussed in the next section, i.e., for $-\pi \leq \varphi _{\mathrm{R}} \leq +\pi$; $Y (\varphi _{\mathrm{R}})$ is an even function.

Fig. 3.4

Bucket height reduction factor $Y (\varphi _{\mathrm{R}})$

3.10 Symmetry with Respect to Transition Energy and Sign of Charge

Depending on whether positive or negative charges are accelerated, and depending on whether operation below or above transition energy is chosen, different cases have to be distinguished that are shown in Fig. 3.3 on p. 133 and Table 3.1 on p. 139.

One usually tries to avoid a crossing of the transition energy during acceleration.⁷ However, in principle, transition crossing is also possible (cf. [8, Sect. 4.7] and [4, Sect. 2.2.3]). In this case, a γ_T jump (cf. [9, Sect. 7.7] and [8, Sect. 4.7]) may, for example, be applied in order to cross the transition energy quickly. According to Table 3.1, this includes a shift of the RF phase.

We will now analyze different substitutions in Eq. (3.29):

• If $\varphi _{\mathrm{R}}^{{\prime}} = -\varphi _{\mathrm{R}}$ and $\Delta t' = -\Delta t$ are substituted, the expression in parentheses will still have the same form. If, in addition, the signs of Q and of η_R are reversed, the Hamiltonian will change its sign, which does not modify the trajectories. These steps lead us from case 1 to case 4 or from case 2 to case 3. In total, the separatrix and the trajectories will experience a reflection across the ordinate.

• If $\varphi _{\mathrm{R}}^{{\prime}} =\varphi _{\mathrm{R}} -\pi \,\mathrm{ sign}\,\varphi _{\mathrm{R}}$ is substituted (this corresponds to $\varphi _{\mathrm{R}} =\varphi _{ \mathrm{R}}^{{\prime}}-\pi \,\mathrm{ sign}\,\varphi _{\mathrm{R}}^{{\prime}}$ if the values 0, π, and −π are excluded), the expression in parentheses will still have the same form, but its sign will change. If, in addition, the sign of Q is inverted, the Hamiltonian will not change at all. These steps lead us from case 1 to case 3 or from case 2 to case 4. In total, the separatrix and the trajectories will not change at all.

In sum, all four cases lead to the same trajectories and separatrices. Only a reflection across the ordinate may be necessary.

This allows us to restrict the following analysis of the orbits and the bucket geometry to case 1 without suffering any loss of generality.

Table 3.1 Symmetry of the buckets (acceleration case only)

3.11 Orbits

From Eq. (3.29), one concludes for Q > 0, η_R < 0, and $0 <\varphi _{\mathrm{R}} <\pi /2$, if only the upper half of the orbit is considered, that

$$\displaystyle{\Delta W = \sqrt{\frac{2W_{\mathrm{R} } \beta _{\mathrm{R} }^{2 }} {-\eta _{\mathrm{R}}} \;H + \frac{Q\hat{V }} {T_{\mathrm{R}}} \frac{2W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} {-\eta _{\mathrm{R}}} \left ( \frac{1} {\omega _{\mathrm{RF}}}\left [\cos (\omega _{\mathrm{RF}}\Delta t +\varphi _{\mathrm{R}}) -\cos \varphi _{\mathrm{R}}\right ] + \Delta t\;\sin \varphi _{\mathrm{R}}\right )}.}$$

Using Eq. (3.32) leads to

$$\displaystyle{ \Delta W = \Delta W_{\mathrm{max,stat}}\;\sqrt{ \frac{\pi h} {Q\hat{V }}\;H + \frac{1} {2}\left (\cos (\omega _{\mathrm{RF}}\Delta t +\varphi _{\mathrm{R}}) -\cos \varphi _{\mathrm{R}} +\omega _{\mathrm{RF}}\Delta t\;\sin \varphi _{\mathrm{R}}\right )}. }$$

(3.35)

For H = H_s, the first term under the square root sign equals $\frac{1} {2}\left (2\;\cos \varphi _{\mathrm{R}}\right.+$$(2\varphi _{\mathrm{R}}-\pi )\;\sin \varphi _{\mathrm{R}}\left.\right ) = [Y (\varphi _{\mathrm{R}})]^{2}$, as Eqs. (3.30) and (3.34) show. For $\Delta t = 0$, it follows that $\Delta W = \Delta W_{\mathrm{max,stat}}Y (\varphi _{\mathrm{R}}) = \Delta W_{\mathrm{max}}$, as expected.

If one wants to draw an orbit inside the separatrix, one inserts a value $\tilde{H} < [Y (\varphi _{\mathrm{R}})]^{2}$ for $\tilde{H} = \frac{\pi h} {Q\hat{V }}\;H$. For trajectories outside the separatrix, however, one sets $\tilde{H} > [Y (\varphi _{\mathrm{R}})]^{2}$. The orbits themselves are obtained if $\Delta W$ as a function of $\Delta \varphi _{\mathrm{RF}} =\omega _{\mathrm{RF}}\Delta t$ is plotted according to Eq. (3.35). Of course, one obtains only the upper half of the orbits in this case.

Figure 3.5 shows a typical sketch of the longitudinal phase space. It clearly shows the characteristics that were discussed in Sect. 2.11.4. The stable fixed point in the middle is a center. The smaller the orbits around this center, the more closely they resemble an ellipse. The unstable fixed point is a saddle point. The orbits that approach this saddle point define the separatrix. Inside the separatrix, the orbits are closed. Outside the separatrix, the orbits are unstable. The region inside the separatrix is also called a bucket, since the particles that are inside will remain inside (provided that parameters such as the amplitude $\hat{V }$ are not changed rapidly). Only parts of the bucket are usually filled with particles. The region that is filled with particles is called a bunch.

Fig. 3.5

Orbits in longitudinal phase space, acceleration with $\varphi _{\mathrm{R}} = 42^{\circ }$

The bucket shown in Fig. 3.5 corresponds to a positive reference phase $\varphi _{\mathrm{R}} = 42^{\circ }$, i.e., to the acceleration case; one sometimes calls this a fishlike bucket. In the stationary case, defined by $\varphi _{\mathrm{R}} = 0^{\circ }$ for Q > 0 and η_R < 0, no overall acceleration takes place. The stationary bucket leads to an eye-shaped separatrix (see Fig. 3.6) that is symmetric with respect to both the energy and time axes. The bucket height of the stationary bucket is given by $2\Delta W_{\mathrm{max,stat}}$, according to Eq. (3.32), whereas the bucket height for the acceleration case is $2\Delta W_{\mathrm{max}}$, according to Eq. (3.33).

Figure 3.7 shows the same two cases for a larger range of $\Delta \varphi _{\mathrm{RF}}$ values. It is obvious that the fixed points occur periodically. The buckets for the stationary case are larger than those for the acceleration case if all parameters except $\varphi _{\mathrm{R}}$ are kept constant. This effect will be analyzed in the Sect. 3.12 below. In the acceleration case, the energy deviation $\Delta W$ of a particle on an unstable orbit (outside the separatrix) will grow without bound, as Fig. 3.7 shows. The same is therefore valid for its deviation from the reference orbit; it will sooner or later hit the beam pipe walls, which is, of course, undesirable. Therefore, one always has to minimize the number of particles that are not captured inside a bucket.

Fig. 3.6

Orbits in longitudinal phase space, stationary case ($\varphi _{\mathrm{R}} = 0^{\circ }$)

Fig. 3.7

Orbits in longitudinal phase space (top: acceleration with $\varphi _{\mathrm{R}} = 42^{\circ }$, bottom: stationary case $\varphi _{\mathrm{R}} = 0^{\circ }$)

Strictly speaking, the trajectories shown in Figs. 3.5, 3.6, and 3.7 are obtained only if an autonomous system is assumed, i.e., if no parameters (such as voltage $\hat{V }$) change with time. We will see later, however (cf. Sect. 5.3), that slow changes of control parameters are acceptable.

3.12 Bucket Area

Now we want to determine the area inside the separatrix. First of all, we consider the stationary case with η_RQ < 0 and $\varphi _{\mathrm{R}} = 0$ (or η_RQ > 0 and $\varphi _{\mathrm{R}} = \pm \pi$, respectively). For the separatrix, we then obtain, using Eq. (3.31),

$$\displaystyle{ \Delta W = \left [\frac{2W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}\vert Q\vert \hat{V }} {\vert \eta _{\mathrm{R}}\vert T_{\mathrm{R}}\omega _{\mathrm{RF}}} \left (\cos (\omega _{\mathrm{RF}}\Delta t) + 1\right )\right ]^{1/2}. }$$

(3.36)

The value $\Delta W = 0$ is achieved for $\Delta \varphi _{\mathrm{RF}} = -\pi$ and $\Delta \varphi _{\mathrm{RF}} = +\pi$, so these values are the integration limits. Due to

$$\displaystyle{\sqrt{1 +\cos \Delta \varphi _{\mathrm{RF }}} = \sqrt{1 +\cos ^{2 } \frac{\Delta \varphi _{\mathrm{RF } } } {2} -\sin ^{2}\frac{\Delta \varphi _{\mathrm{RF}}} {2}} = \sqrt{2}\;\cos \frac{\Delta \varphi _{\mathrm{RF}}} {2},}$$

one obtains

$$\displaystyle\begin{array}{rcl} \int _{-\pi }^{+\pi }\sqrt{1 +\cos \Delta \varphi _{\mathrm{ RF}}}\;\mathrm{d}\Delta \varphi _{\mathrm{RF}}& =& 2\;\int _{0}^{+\pi }\sqrt{2}\;\cos \frac{\Delta \varphi _{\mathrm{RF}}} {2} \;\mathrm{d}\Delta \varphi _{\mathrm{RF}} \\ & =& 4\;\sqrt{2}\left [\sin \frac{\Delta \varphi _{\mathrm{RF}}} {2} \right ]_{0}^{\pi } = 4\;\sqrt{2},{}\end{array}$$

(3.37)

and it follows that

$$\displaystyle{A_{\mathrm{B,stat}}^{\Delta \varphi _{\mathrm{RF}},\Delta W}/2 =\int _{ -\pi }^{+\pi }\Delta W\;\mathrm{d}\Delta \varphi _{\mathrm{ RF}} = \sqrt{\frac{2W_{\mathrm{R} } \beta _{\mathrm{R} }^{2 }\vert Q\vert \hat{V }} {\vert \eta _{\mathrm{R}}\vert T_{\mathrm{R}}\omega _{\mathrm{RF}}}} \;4\sqrt{2}}$$

$$\displaystyle{\Rightarrow A_{\mathrm{B,stat}}^{\Delta \varphi _{\mathrm{RF}},\Delta W} = 8\sqrt{2}\;\sqrt{\frac{2W_{\mathrm{R} } \beta _{\mathrm{R} }^{2 }\vert Q\vert \hat{V }} {\vert \eta _{\mathrm{R}}\vert T_{\mathrm{R}}\omega _{\mathrm{RF}}}}.}$$

Since we calculated the integration with respect to $\Delta \varphi _{\mathrm{RF}} =\omega _{\mathrm{RF}}\Delta t$, this is the bucket area in ($\Delta W$,$\Delta \varphi _{\mathrm{RF}}$) coordinates. In order to obtain the bucket area in ($\Delta W$,$\Delta t$) coordinates, we have to divide by ω_RF:

$$\displaystyle{ \fbox{$A_{\mathrm{B,stat}}^{\Delta t,\Delta W} = \frac{4\sqrt{2}} {\pi h} T_{\mathrm{R}}\;\sqrt{\frac{W_{\mathrm{R} } \beta _{\mathrm{R} }^{2 }\vert Q\vert \hat{V }} {\pi h\vert \eta _{\mathrm{R}}\vert }} = \frac{8\sqrt{2}} {\omega _{\mathrm{RF}}} \;\sqrt{\frac{W_{\mathrm{R} } \beta _{\mathrm{R} }^{2 }\vert Q\vert \hat{V }} {\pi h\vert \eta _{\mathrm{R}}\vert }}.$} }$$

(3.38)

This equation corresponds to formula (2.75) in Edwards/Syphers [4].

If $\varphi _{\mathrm{R}}$ is not equal to zero, we have to integrate $\Delta W$ in Eq. (3.31) instead of $\Delta W$ in Eq. (3.36). This integration can be performed for the case Q > 0, η_R < 0, $0 <\varphi _{\mathrm{R}} <\pi /2$ in the following, because the result will not depend on this choice, due to the symmetry as discussed in Sect. 3.10. In order to find the integration limits, we first have to determine the zeros of $\Delta W$ as a function of $\Delta \varphi _{\mathrm{RF}}$. The coefficients of $\sin \varphi _{\mathrm{R}}$ in parentheses in Eq. (3.31) become zero for

$$\displaystyle{\Delta \varphi _{\mathrm{RF}} = \Delta \varphi _{\mathrm{saddle}} =\pi -2\varphi _{\mathrm{R}}.}$$

For this value, the remaining terms in parentheses also vanish; this obviously corresponds to the upper integration limit. In order to find the lower integration limit $\Delta \varphi _{\mathrm{RF}} = \Delta \varphi _{\mathrm{l}}$, the equation

$$\displaystyle{ \sin \varphi _{\mathrm{R}}\;(\sin \Delta \varphi _{\mathrm{l}} +\pi -2\varphi _{\mathrm{R}} - \Delta \varphi _{\mathrm{l}}) =\cos \varphi _{\mathrm{R}}\;(\cos \Delta \varphi _{\mathrm{l}} + 1) }$$

(3.39)

(which is obtained by setting the expression in parentheses equal to zero) has to be solved numerically with $\Delta \varphi _{\mathrm{l}}$ in the range between −π and 0. For this purpose, one may determine the zero of the function

$$\displaystyle{f(\Delta \varphi _{\mathrm{l}}) =\sin \varphi _{\mathrm{R}}\;(\sin \Delta \varphi _{\mathrm{l}} +\pi -2\varphi _{\mathrm{R}} - \Delta \varphi _{\mathrm{l}}) -\cos \varphi _{\mathrm{R}}\;(\cos \Delta \varphi _{\mathrm{l}} + 1)}$$

by means of an interval division method, since for $0 <\varphi _{\mathrm{R}} <\pi /2$, the function is always positive at $\Delta \varphi _{\mathrm{l}} = -\pi$, whereas it is always negative at $\Delta \varphi _{\mathrm{l}} = +\pi$. The resulting function $\Delta \varphi _{\mathrm{l}}(\varphi _{\mathrm{R}})$ is shown in Fig. 3.9. It may be extended easily to the range $-\pi \leq \varphi _{\mathrm{R}} \leq +\pi$ (see Sect. 3.10), because $\Delta \varphi _{\mathrm{l}}(\varphi _{\mathrm{R}})$ is an odd function.

If one now constructs the integral

$$\displaystyle{A_{\mathrm{B}}^{\Delta \varphi _{\mathrm{RF}},\Delta W}/2 =\int _{ \Delta \varphi _{\mathrm{l}}}^{\Delta \varphi _{\mathrm{saddle}} }\Delta W\;\mathrm{d}\Delta \varphi _{\mathrm{RF}} =\int _{ \Delta \varphi _{\mathrm{l}}}^{\pi -2\varphi _{\mathrm{R}} }\Delta W\;\mathrm{d}\Delta \varphi _{\mathrm{RF}},}$$

one sees by comparison with Eqs. (3.31), (3.36), and (3.37) that it is smaller than that for the stationary case by a factor of

$$\displaystyle\begin{array}{rcl} \alpha (\varphi _{\mathrm{R}})& =& \frac{1} {4\;\sqrt{2}}\ \ \cdot \\ & &\quad \ \cdot \int _{\Delta \varphi _{\mathrm{l}}}^{\pi -2\varphi _{\mathrm{R}} }\sqrt{(2\varphi _{\mathrm{R} } + \Delta \varphi _{\mathrm{RF } } -\pi )\sin \varphi _{\mathrm{R} } + \cos \varphi _{\mathrm{R} } + \cos (\varphi _{\mathrm{R} } + \Delta \varphi _{\mathrm{RF } } )}\;\mathrm{d}\Delta \varphi _{\mathrm{RF}}.{}\end{array}$$

(3.40)

The denominator of the first fraction equals the integral for the stationary case with $\varphi _{\mathrm{R}} = 0$, as shown above. We obtain

$$\displaystyle{A_{\mathrm{B}}^{\Delta \varphi _{\mathrm{RF}},\Delta W} = A_{\mathrm{ B,stat}}^{\Delta \varphi _{\mathrm{RF}},\Delta W}\;\alpha (\varphi _{\mathrm{ R}}).}$$

The conversion to ($\Delta W$,$\Delta t$) coordinates leads to the same factor on both sides, and we obtain

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

(3.41)

By means of a numerical calculation of the functions $\Delta \varphi _{\mathrm{l}}(\varphi _{\mathrm{R}})$ and $\alpha (\varphi _{\mathrm{R}})$ and subsequent fitting by suitable functions, the following approximations may be found:

$$\displaystyle{ \Delta \varphi _{\mathrm{l}}(\varphi _{\mathrm{R}}) \approx \pi \left [\left (\frac{2} {\pi } \varphi _{\mathrm{R}}\right )^{0.4373} - 1\right ], }$$

(3.42)

$$\displaystyle{ \alpha (\varphi _{\mathrm{R}}) \approx 1 -\sin \varphi _{\mathrm{R}} - a\left (\frac{2} {\pi } \varphi _{\mathrm{R}}\right ) + 2a\left (\frac{2} {\pi } \varphi _{\mathrm{R}}\right )^{2} - a\left (\frac{2} {\pi } \varphi _{\mathrm{R}}\right )^{3}\qquad \mbox{ with }a = 1.0879. }$$

(3.43)

The accuracy of the approximation for $\Delta \varphi _{\mathrm{l}}$ is better than ± 2^∘. The accuracy of the approximation for α in the range $0 \leq \varphi _{\mathrm{R}} \leq 85^{\circ }$ relevant in practical applications is better than + 5% or − 16%, respectively. Figure 3.8 shows the dependence of α on the synchronous phase $\varphi _{\mathrm{R}}$. Please note that the function $\alpha (\varphi _{\mathrm{R}})$ may easily be extended to the range $-\pi \leq \varphi _{\mathrm{R}} \leq +\pi$, because it is an even function. The RF phase that corresponds to one end of the bucket, $\Delta \varphi _{\mathrm{RF}} = \Delta \varphi _{\mathrm{l}}$, is shown in Fig. 3.9; it is an odd function $\Delta \varphi _{\mathrm{l}}(\varphi _{\mathrm{R}})$. Since the other end of the bucket is given by $\Delta \varphi _{\mathrm{RF}} = \Delta \varphi _{\mathrm{saddle}} =\pi \;\mathrm{ sign}\,\varphi _{\mathrm{R}} - 2\varphi _{\mathrm{R}}$, the bucket length is

$$\displaystyle{\varphi _{\mathrm{RF,B,len}} = \vert \Delta \varphi _{\mathrm{saddle}} - \Delta \varphi _{\mathrm{l}}\vert = \vert \pi \;\mathrm{sign}\,\varphi _{\mathrm{R}} - 2\varphi _{\mathrm{R}} - \Delta \varphi _{\mathrm{l}}\vert.}$$

Fig. 3.8

Bucket area reduction factor $\alpha (\varphi _{\mathrm{R}})$; solid line: Eq. (3.40), dashed line: approximation (3.43)

Fig. 3.9

RF phase $\Delta \varphi _{\mathrm{l}}(\varphi _{\mathrm{R}})$; solid line: solution of Eq. (3.39), dashed line: approximation (3.42)

This function is shown in Fig. 3.10; it is an even function (see Sect. 3.10). In all these diagrams, the left half of the curves ($0 \leq \varphi _{\mathrm{R}} \leq 90^{\circ }$) corresponds to case 1 (stationary case for $\varphi _{\mathrm{R}} = 0$), whereas the right half ($90^{\circ }\leq \varphi _{\mathrm{R}} \leq 180^{\circ }$) corresponds to case 2 (stationary case for $\varphi _{\mathrm{R}} =\pi$) discussed in Sect. 3.10.

Fig. 3.10

Bucket length $\varphi _{\mathrm{RF,B,len}}(\varphi _{\mathrm{R}})$

3.13 Approximation of Bucket Area

If we simply approximate the shape of the stationary bucket by an ellipse, we obtain the formula

$$\displaystyle{A_{\mathrm{B,stat}}^{\Delta t,\Delta W} \approx \pi \Delta t_{\mathrm{ max,stat}}\Delta W_{\mathrm{max,stat}}.}$$

By means of

$$\displaystyle{\Delta t_{\mathrm{max,stat}} = \frac{T_{\mathrm{R}}} {2h} }$$

and the bucket height (3.32),

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

it follows that

$$\displaystyle{A_{\mathrm{B,stat}}^{\Delta t,\Delta W} \approx \frac{\pi } {\sqrt{2}}\;\frac{T_{\mathrm{R}}} {h} \;\sqrt{\frac{W_{\mathrm{R} } \beta _{\mathrm{R} }^{2 }\vert Q\vert \hat{V }} {\pi h\vert \eta _{\mathrm{R}}\vert }}.}$$

The coefficient $\frac{\pi }{\sqrt{2}} \approx 2.22$ does not differ much from that of the exact solution ($\frac{4\sqrt{2}} {\pi } \approx 1.80$; see Eq. (3.38)). The relative deviation is about 23%.

3.14 Ratio of Bucket Height to Bucket Length

The bucket length $2\Delta t_{\mathrm{max,stat}}$ of a stationary bucket is obviously given by the condition

$$\displaystyle{\omega _{\mathrm{RF}}\Delta t_{\mathrm{max,stat}} =\pi.}$$

Together with Eq. (3.32), this leads to

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

We may now use Eq. (3.25) for the stationary case,

$$\displaystyle{f_{\mathrm{S,0,stat}} = f_{\mathrm{R}}\sqrt{\frac{\vert \eta _{\mathrm{R} } \;Q\vert \;\hat{V }\;h} {2\pi W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}}} \qquad \Rightarrow \vert Q\vert \hat{V } = \frac{2\pi W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} {\vert \eta _{\mathrm{R}}\vert \;h} \left (\frac{f_{\mathrm{S,0,stat}}} {f_{\mathrm{R}}} \right )^{2},}$$

and obtain

$$\displaystyle{\frac{\Delta W_{\mathrm{max,stat}}} {\Delta t_{\mathrm{max,stat}}} = \frac{2W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} {\vert \eta _{\mathrm{R}}\vert \;h} \;\frac{f_{\mathrm{S,0,stat}}} {f_{\mathrm{R}}} \;\frac{\omega _{\mathrm{RF}}} {\pi } }$$

$$\displaystyle{ \Rightarrow \fbox{$\frac{\Delta W_{\mathrm{max,stat}}} {\Delta t_{\mathrm{max,stat}}} = f_{\mathrm{S,0,stat}}\;\frac{4W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} {\vert \eta _{\mathrm{R}}\vert }.$} }$$

(3.44)

If we compare this with the ratio of the principal axes for small oscillation amplitudes

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

according to Eq. (3.28), we see that the expressions differ only by the factor 4 vs. 2π. In comparison with small-amplitude orbits, the bucket shape is flattened. This is also visible in Fig. 3.6.

Equation (3.25) for the synchrotron frequency

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

in a stationary bucket ($\varphi _{\mathrm{R}} = 0$ or $\varphi _{\mathrm{R}} = \pm \pi$, respectively) and Eq. (3.38) for the bucket area

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

have similar expressions. Therefore, we get

$$\displaystyle{ \frac{f_{\mathrm{S,0,stat}}} {A_{\mathrm{B,stat}}^{\Delta t,\Delta W}} = \frac{\pi \mathit{hf }_{\mathrm{R}}^{2}} {4\sqrt{2}} \frac{\vert \eta _{\mathrm{R}}\vert \;h} {\sqrt{2}\;W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} = \frac{\pi f_{\mathrm{RF}}^{2}\;\vert \eta _{\mathrm{R}}\vert } {8W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}}. }$$

(3.45)

This allows us to write the principal axes ratio given by Eq. (3.28),

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

in the form

$$\displaystyle{ \fbox{$\frac{\Delta \hat{W}} {\Delta \hat{t}} = A_{\mathrm{B,stat}}^{\Delta t,\Delta W}\;\frac{\pi ^{2}} {4}f_{\mathrm{RF}}^{2} = A_{\mathrm{ B,stat}}^{\Delta t,\Delta W}\frac{\omega _{\mathrm{RF}}^{2}} {16}.$} }$$

(3.46)

This leads us to the following expression for the bunch area if only small bunches are considered:

$$\displaystyle{A_{\mathrm{b,stat}}^{\Delta t,\Delta W} \approx \pi \Delta \hat{t}\Delta \hat{W} =\pi \Delta \hat{t}^{2}A_{\mathrm{ B,stat}}^{\Delta t,\Delta W}\frac{\omega _{\mathrm{RF}}^{2}} {16} = \frac{\pi } {16}\Delta \hat{\varphi }_{\mathrm{RF}}^{2}A_{\mathrm{ B,stat}}^{\Delta t,\Delta W}.}$$

Here $\Delta \hat{t}$ and $\Delta \hat{W}$ are interpreted as the semiaxes of the outermost particle in the bunch. If we define the bucket filling factor η_fill by

$$\displaystyle{\eta _{\mathrm{fill}} = \frac{A_{\mathrm{b,stat}}^{\Delta t,\Delta W}} {A_{\mathrm{B,stat}}^{\Delta t,\Delta W}},}$$

we obtain

$$\displaystyle{ \eta _{\mathrm{fill}} \approx \frac{\pi } {16}\Delta \hat{\varphi }_{\mathrm{RF}}^{2} }$$

(3.47)

for sufficiently small bunches. In Sect. 3.17, an exact expression for arbitrary bunch sizes will be derived.

For later use, we write Eq. (3.45) in the form

$$\displaystyle{ \fbox{$ \frac{\omega _{\mathrm{S,0,stat}}} {A_{\mathrm{B,stat}}^{\Delta t,\Delta W}} = \frac{\omega _{\mathrm{RF}}^{2}\;\vert \eta _{\mathrm{R}}\vert } {16\;W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}}.$} }$$

(3.48)

From now on, we will often discard the superscript specification of the phase space coordinates. If nothing else is stated, we assume

$$\displaystyle{A_{\mathrm{B}}\,:=\,A_{\mathrm{B}}^{\Delta t,\Delta W},\qquad A_{\mathrm{ B,stat}}\,:=\,A_{\mathrm{B,stat}}^{\Delta t,\Delta W},\qquad A_{\mathrm{ b}}\,:=\,A_{\mathrm{b}}^{\Delta t,\Delta W},\qquad A_{\mathrm{ b,stat}}\,:=\,A_{\mathrm{b,stat}}^{\Delta t,\Delta W}.}$$

3.15 Choice of the Harmonic Number

According to Eq. (3.38), the total area of h buckets,

$$\displaystyle{h \cdot A_{\mathrm{B,stat}} = \frac{4\sqrt{2}} {\pi } T_{\mathrm{R}}\;\sqrt{\frac{W_{\mathrm{R} } \beta _{\mathrm{R} }^{2 }\vert Q\vert \hat{V }} {\pi h\vert \eta _{\mathrm{R}}\vert }},}$$

remains constant if both the amplitude $\hat{V }$ and the harmonic number h are increased by the same factor.

If we assume, however, that the ramp rate$\dot{B}$ is the same, the required accelerating voltage V_R will also remain the same, as Eq. (3.22),

$$\displaystyle{ V _{\mathrm{R}} \approx l_{\mathrm{R}}r_{\mathrm{R}}\dot{B}, }$$

(3.49)

shows. Therefore, an increase in $\hat{V }$ and h by the same factor will lead to a reduced reference phase $\varphi _{\mathrm{R}}$, since

$$\displaystyle{ V _{\mathrm{R}} =\hat{ V }\sin \varphi _{\mathrm{R}} }$$

(3.50)

will remain constant. This leads to a larger value of the bucket reduction factor $\alpha (\varphi _{\mathrm{R}})$, which results in a larger area

$$\displaystyle{h \cdot A_{\mathrm{B}} = h \cdot A_{\mathrm{B,stat}}\alpha (\varphi _{\mathrm{R}})}$$

of the accelerated buckets.

In conclusion, for the acceleration case, the simultaneous increase of RF amplitude and harmonic number by the same factor leads to a larger area in phase space, since the reference phase is reduced.

3.16 Revolution Time in the Stationary Bucket

Let us consider the Hamiltonian in Eq. (3.29) for the stationary case with Q > 0, η_R < 0, $\varphi _{\mathrm{R}} = 0$:

$$\displaystyle{H = - \frac{\eta _{\mathrm{R}}} {W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}}\;\frac{\Delta W^{2}} {2} -\frac{Q\hat{V }} {T_{\mathrm{R}}} \frac{1} {\omega _{\mathrm{RF}}}\left [\cos (\omega _{\mathrm{RF}}\Delta t) - 1\right ].}$$

Setting

$$\displaystyle{ A = - \frac{\eta _{\mathrm{R}}} {2\;W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} }$$

(3.51)

leads to

$$\displaystyle{\Delta W^{2} = \frac{1} {A}\left (H + \frac{Q\hat{V }} {T_{\mathrm{R}}} \frac{1} {\omega _{\mathrm{RF}}}\left [\cos (\omega _{\mathrm{RF}}\Delta t) - 1\right ]\right ).}$$

One obtains the action variable

$$\displaystyle\begin{array}{rcl} J& =& \frac{1} {2\pi }\int \int \mathrm{d}(\Delta t)\;\mathrm{d}(\Delta W) \\ & =& \frac{1} {\pi \sqrt{A}}\int _{-\Delta \hat{t}}^{\Delta \hat{t}}\sqrt{ H + \frac{Q\hat{V }} {T_{\mathrm{R}}} \frac{1} {\omega _{\mathrm{RF}}}\left [\cos (\omega _{\mathrm{RF}}\Delta t) - 1\right ]}\;\mathrm{d}(\Delta t).{}\end{array}$$

(3.52)

By means of the substitution

$$\displaystyle{\Delta \varphi _{\mathrm{RF}} =\omega _{\mathrm{RF}}\Delta t,\qquad \frac{\mathrm{d}\Delta \varphi _{\mathrm{RF}}} {\mathrm{d}\Delta t} =\omega _{\mathrm{RF}},}$$

we obtain, since the integrand is an even function,

$$\displaystyle{J = \frac{2} {\pi \omega _{\mathrm{RF}}\sqrt{A}}\int _{0}^{\Delta \hat{\varphi }_{\mathrm{RF}} }\sqrt{H + \frac{Q\hat{V }} {T_{\mathrm{R}}} \frac{1} {\omega _{\mathrm{RF}}}\left [\cos \;\Delta \varphi _{\mathrm{RF}} - 1\right ]}\;\mathrm{d}(\Delta \varphi _{\mathrm{RF}}).}$$

In order to convert this into a well-known elliptic integral, we make use of the trigonometric identity

$$\displaystyle{\cos \;\Delta \varphi _{\mathrm{RF}} =\cos ^{2}\frac{\Delta \varphi _{\mathrm{RF}}} {2} -\sin ^{2}\frac{\Delta \varphi _{\mathrm{RF}}} {2} = 1 - 2\;\sin ^{2}\frac{\Delta \varphi _{\mathrm{RF}}} {2} }$$

and get

$$\displaystyle{J = \frac{2} {\pi \omega _{\mathrm{RF}}\sqrt{A}}\int _{0}^{\Delta \hat{\varphi }_{\mathrm{RF}} }\sqrt{H - 2\; \frac{Q\hat{V }} {T_{\mathrm{R}}\;\omega _{\mathrm{RF}}}\sin ^{2}\frac{\Delta \varphi _{\mathrm{RF}}} {2}} \;\mathrm{d}(\Delta \varphi _{\mathrm{RF}}).}$$

If we now set

$$\displaystyle{ B = 2\; \frac{Q\hat{V }} {T_{\mathrm{R}}\;\omega _{\mathrm{RF}}} = \frac{Q\hat{V }} {h\;\pi } }$$

(3.53)

and

$$\displaystyle{x = \frac{\Delta \varphi _{\mathrm{RF}}} {2},\qquad \frac{\mathrm{d}x} {\mathrm{d}\Delta \varphi _{\mathrm{RF}}} = \frac{1} {2},}$$

we obtain

$$\displaystyle{J = \frac{4} {\pi \omega _{\mathrm{RF}}\sqrt{A}}\int _{0}^{x_{\mathrm{max}} }\sqrt{H - B\;\sin ^{2 } x}\;\mathrm{d}x.}$$

Obviously,

$$\displaystyle{x_{\mathrm{max}} =\arcsin \sqrt{\frac{H} {B}}}$$

is valid, because the integrand must vanish where the orbits cross the x-axis. One obtains

$$\displaystyle{ \frac{\mathrm{d}J} {\mathrm{d}H} = \frac{4} {\pi \omega _{\mathrm{RF}}\sqrt{A}}\left [\frac{\mathrm{d}x_{\mathrm{max}}} {\mathrm{d}H} \sqrt{H - B\;\sin ^{2 } x_{\mathrm{max }}} +\int _{ 0}^{x_{\mathrm{max}} } \frac{\mathrm{d}x} {2\;\sqrt{H - B\;\sin ^{2 } x}}\right ].}$$

The first term equals zero:

$$\displaystyle{ \frac{\mathrm{d}J} {\mathrm{d}H} = \frac{2} {\pi \omega _{\mathrm{RF}}\sqrt{HA}}\int _{0}^{x_{\mathrm{max}} } \frac{\mathrm{d}x} {\sqrt{1 - \frac{B} {H}\;\sin ^{2}x}}.}$$

For

$$\displaystyle{k^{2} = \frac{B} {H},}$$

one obtains an elliptic integral of the first kind:

$$\displaystyle{ \frac{\mathrm{d}J} {\mathrm{d}H} = \frac{2} {\pi \omega _{\mathrm{RF}}\sqrt{HA}}\;\mathrm{F}(x_{\mathrm{max}},k). }$$

(3.54)

Obviously,

$$\displaystyle{x_{\mathrm{max}} =\arcsin \frac{1} {k}}$$

is valid. Now we make use of formula 17.4.15 in Abramowitz/Stegun [10]:

$$\displaystyle{\mathrm{F}(\varphi,k) = \frac{1} {k}\;\mathrm{F}\left (\arcsin (k\!\sin \varphi ), \frac{1} {k}\right )}$$

$$\displaystyle{\Rightarrow \mathrm{ F}(x_{\mathrm{max}},k) = \frac{1} {k}\;\mathrm{F}\left (\arcsin (1), \frac{1} {k}\right ).}$$

Because of

$$\displaystyle{\arcsin (1) = \frac{\pi } {2},}$$

this corresponds to a complete elliptic integral of the first kind:

$$\displaystyle{\mathrm{F}(x_{\mathrm{max}},k) = \frac{1} {k}\;\mathrm{F}\left ( \frac{\pi } {2}, \frac{1} {k}\right ) = \frac{1} {k}\;\mathrm{K}\left (\frac{1} {k}\right ).}$$

Equation (3.54) therefore yields

$$\displaystyle{ \frac{\mathrm{d}J} {\mathrm{d}H} = \frac{2} {\pi \omega _{\mathrm{RF}}\sqrt{BA}}\;\mathrm{K}\left (\sqrt{\frac{H} {B}}\right ).}$$

We see from Eqs. (3.51), (3.53), and (3.25) that

$$\displaystyle{AB = - \frac{\eta _{\mathrm{R}}} {2\;W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} \frac{Q\hat{V }} {h\;\pi } = \left (\frac{\omega _{\mathrm{S,0,stat}}} {\omega _{\mathrm{R}}} \right )^{2} \frac{1} {h^{2}} = \left (\frac{\omega _{\mathrm{S,0,stat}}} {\omega _{\mathrm{RF}}} \right )^{2}}$$

is valid, so that

$$\displaystyle{ \frac{\mathrm{d}J} {\mathrm{d}H} = \frac{2} {\pi \omega _{\mathrm{S,0,stat}}}\;\mathrm{K}\left (\sqrt{\frac{H} {B}}\right )}$$

$$\displaystyle{\Rightarrow \omega _{\mathrm{S,stat}} = \frac{\mathrm{d}H} {\mathrm{d}J} =\omega _{\mathrm{S,0,stat}} \frac{\pi } {2\;\mathrm{K}\left (\sqrt{\frac{H} {B}}\right )}}$$

results. Instead of the dependence on the value of the Hamiltonian, the dependence on the maximum phase deviation of the particles is of course more transparent. Due to

$$\displaystyle{x_{\mathrm{max}} =\arcsin \sqrt{\frac{H} {B}} = \frac{\Delta \hat{\varphi }_{\mathrm{RF}}} {2},}$$

one finally obtains

$$\displaystyle{ \fbox{$\omega _{\mathrm{S,stat}} =\omega _{\mathrm{S,0,stat}} \frac{\pi } {2\;\mathrm{K}\left (\sin \frac{\Delta \hat{\varphi }_{\mathrm{RF}}} {2} \right )}.$} }$$

(3.55)

This formula for the synchrotron frequency of particles with arbitrary oscillation amplitude may also be found in Lee [5, p. 240] or Ng [1, Sect. 2.1.2.1]. The decrease in the oscillation frequency is shown in Fig. 3.11. For zero argument, we have

$$\displaystyle{\mathrm{K}(0) = \frac{\pi } {2},}$$

and

$$\displaystyle{\omega _{\mathrm{S,stat}} =\omega _{\mathrm{S,0,stat}}}$$

results for small phase amplitudes $\Delta \hat{\varphi }_{\mathrm{RF}}$, as expected. By means of a Taylor series expansion of Eq. (3.55), one obtains

$$\displaystyle{ \frac{\omega _{\mathrm{S,stat}}} {\omega _{\mathrm{S,0,stat}}} \approx 1 -\frac{\Delta \hat{\varphi }_{\mathrm{RF}}^{2}} {16} }$$

(3.56)

for sufficiently small amplitudes $\Delta \hat{\varphi }_{\mathrm{RF}}$. This, however, does not restrict the validity significantly, because the error of this formula is below 5% if $\Delta \hat{\varphi }_{\mathrm{RF}} < 164^{\circ }$ holds.

Fig. 3.11

Decrease of synchrotron frequency vs. oscillation amplitude (solid line: exact formula (3.55), dashed line: approximation (3.56))

3.17 Bunch Area

In this section, we calculate the bunch area, i.e., the area that is enclosed by a particle trajectory in phase space. We restrict ourselves to the stationary case with Q > 0, η_R < 0, $\varphi _{\mathrm{R}} = 0$. The bunch area already appeared in Eq. (3.52):

$$\displaystyle{A_{\mathrm{b,stat}} = 2\pi J = \frac{2} {\sqrt{A}}\int _{-\Delta \hat{t}}^{\Delta \hat{t}}\sqrt{H + \frac{Q\hat{V }} {2\pi h} \left [\cos (\omega _{\mathrm{RF}}\Delta t) - 1\right ]}\;\mathrm{d}(\Delta t).}$$

According to Eq. (3.51), we have

$$\displaystyle{A = - \frac{\eta _{\mathrm{R}}} {2\;W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}}.}$$

By means of the substitution

$$\displaystyle{\Delta \varphi _{\mathrm{RF}} =\omega _{\mathrm{RF}}\Delta t\qquad \Rightarrow \frac{\mathrm{d}\Delta \varphi _{\mathrm{RF}}} {\mathrm{d}\Delta t} =\omega _{\mathrm{RF}},}$$

one gets

$$\displaystyle\begin{array}{rcl} A_{\mathrm{b,stat}}& =& 2\pi J = \frac{2} {\omega _{\mathrm{RF}}\sqrt{A}}\int _{-\Delta \hat{\varphi }_{\mathrm{RF}}}^{\Delta \hat{\varphi }_{\mathrm{RF}} }\sqrt{H + \frac{Q\hat{V }} {2\pi h} \left [\cos (\Delta \varphi _{\mathrm{RF}}) - 1\right ]}\;\mathrm{d}(\Delta \varphi _{\mathrm{RF}}) = \\ & =& \frac{4} {\omega _{\mathrm{RF}}\sqrt{A}}\int _{0}^{\Delta \hat{\varphi }_{\mathrm{RF}} }\sqrt{a + b\;\cos \Delta \varphi _{\mathrm{RF }}}\;\mathrm{d}(\Delta \varphi _{\mathrm{RF}}) {}\end{array}$$

(3.57)

with

$$\displaystyle{a = H -\frac{Q\hat{V }} {2\pi h} }$$

and

$$\displaystyle{ b = \frac{Q\hat{V }} {2\pi h}. }$$

(3.58)

The integration limit is defined by the integrand being zero:

$$\displaystyle{H + \frac{Q\hat{V }} {2\pi h} \left [\cos (\Delta \hat{\varphi }_{\mathrm{RF}}) - 1\right ] = 0}$$

$$\displaystyle{\Rightarrow a = -\frac{Q\hat{V }} {2\pi h} \;\cos (\Delta \hat{\varphi }_{\mathrm{RF}}) = -b\;\cos (\Delta \hat{\varphi }_{\mathrm{RF}}),}$$

$$\displaystyle{ H = b + a = b\left (1 -\cos (\Delta \hat{\varphi }_{\mathrm{RF}})\right ). }$$

(3.59)

The integral in Eq. (3.57) is already well known from Sect. 2.11.9, Eq. (2.134). If we take the different definitions of the parameters a and b into account, we may adopt the solution (2.135). We get

$$\displaystyle{A_{\mathrm{b,stat}} = \frac{4} {\omega _{\mathrm{RF}}\sqrt{A}}\sqrt{\frac{2} {b}}\left [(a - b)\mathrm{F}\left (\gamma, \frac{1} {r}\right ) + 2b\mathrm{E}\left (\gamma, \frac{1} {r}\right )\right ]_{0}^{\Delta \hat{\varphi }_{\mathrm{RF}} }}$$

with

$$\displaystyle{r = \sqrt{ \frac{2b} {a + b}}\qquad \mbox{ and}\quad \gamma =\arcsin \sqrt{\frac{b(1 -\cos \Delta \varphi _{\mathrm{RF } } )} {a + b}}.}$$

Here E(γ, k) is an elliptic integral of the second kind (see Footnote 23 on p. 108). For $\Delta \varphi _{\mathrm{RF}}\,=\,0$, we get γ = 0, and the expression in the square brackets disappears. For $\Delta \varphi _{\mathrm{RF}} = \Delta \hat{\varphi }_{\mathrm{RF}}$, we get

$$\displaystyle{\gamma =\arcsin \sqrt{\frac{b(1 -\cos \Delta \hat{\varphi }_{\mathrm{RF } } )} {a + b}}.}$$

Because of Eq. (3.59), we therefore obtain

$$\displaystyle{\gamma =\arcsin \; 1 = \frac{\pi } {2}.}$$

Due to

$$\displaystyle{\mathrm{F}\left ( \frac{\pi } {2},k\right ) =\mathrm{ K}(k),\qquad \mathrm{E}\left ( \frac{\pi } {2},k\right ) =\mathrm{ E}(k),}$$

we obtain

$$\displaystyle{A_{\mathrm{b,stat}} = \frac{4\sqrt{2}} {\omega _{\mathrm{RF}}\sqrt{Ab}}\left [(a - b)\mathrm{K}(k) + 2b\mathrm{E}(k)\right ]}$$

if we set

$$\displaystyle\begin{array}{rcl} & k = \frac{1} {r} = \sqrt{\frac{a+b} {2b}} = \sqrt{\frac{H} {2b}} = \sqrt{\frac{1-\cos \Delta \hat{\varphi }_{\mathrm{RF } } } {2}} & {}\\ & \Rightarrow k =\sin \; \frac{\Delta \hat{\varphi }_{\mathrm{RF}}} {2}. & {}\\ \end{array}$$

Since

$$\displaystyle{a - b = H - 2b = -b(1 +\cos \Delta \hat{\varphi }_{\mathrm{RF}})}$$

is valid, we obtain

$$\displaystyle\begin{array}{rcl} & A_{\mathrm{b,stat}} = \frac{4\sqrt{2}} {\omega _{\mathrm{RF}}\sqrt{Ab}}\;2b\left [\mathrm{E}(k) -\frac{1+\cos \Delta \hat{\varphi }_{\mathrm{RF}}} {2} \;\mathrm{K}(k)\right ] & {}\\ & \Rightarrow A_{\mathrm{b,stat}} = \frac{8\sqrt{2}\;\sqrt{b}} {\omega _{\mathrm{RF}}\sqrt{A}} \left [\mathrm{E}(k) -\cos ^{2}\frac{\Delta \hat{\varphi }_{\mathrm{RF}}} {2} \;\mathrm{K}(k)\right ] & {}\\ & \Rightarrow A_{\mathrm{b,stat}} = \frac{8\sqrt{2}\;} {\omega _{\mathrm{RF}}} \;\sqrt{-\frac{Q\hat{V }\;W_{\mathrm{R} } \beta _{\mathrm{R} }^{2 }} {\pi h\eta _{\mathrm{R}}}} \;\left [\mathrm{E}(k) -\mathrm{ K}(k)\;\cos ^{2}\frac{\Delta \hat{\varphi }_{\mathrm{RF}}} {2} \right ].& {}\\ \end{array}$$

In the last step, we used Eqs. (3.51) and (3.58). A comparison with Eq. (3.38) leads to

$$\displaystyle\begin{array}{rcl} & A_{\mathrm{b,stat}} = A_{\mathrm{B,stat}}\;\left [\mathrm{E}(k) -\mathrm{ K}(k)\;\cos ^{2}\frac{\Delta \hat{\varphi }_{\mathrm{RF}}} {2} \right ]& \\ & \Leftrightarrow \fbox{$A_{\mathrm{b,stat}} = A_{\mathrm{B,stat}}\;\eta _{\mathrm{fill}}$} &{}\end{array}$$

(3.60)

with the bucket filling factor

$$\displaystyle{ \fbox{$\eta _{\mathrm{fill}}(\Delta \hat{\varphi }_{\mathrm{RF}}) =\mathrm{ E}\left (\sin \frac{\Delta \hat{\varphi }_{\mathrm{RF}}} {2} \right ) -\mathrm{ K}\left (\sin \frac{\Delta \hat{\varphi }_{\mathrm{RF}}} {2} \right )\;\cos ^{2}\frac{\Delta \hat{\varphi }_{\mathrm{RF}}} {2}.$} }$$

(3.61)

Please note that $\Delta \hat{\varphi }_{\mathrm{RF}}$ denotes the RF phase amplitude at the border of the bunch, not the bucket limit.

Now we check our result for small values of $\Delta \hat{\varphi }_{\mathrm{RF}}$. By means of the approximations

$$\displaystyle{\mathrm{E}(k) \approx \frac{\pi } {2}\left (1 -\frac{k^{2}} {4} \right ) = \frac{\pi } {2}\left (1 -\frac{\Delta \hat{\varphi }_{\mathrm{RF}}^{2}} {16} \right ),}$$

$$\displaystyle{\mathrm{K}(k) \approx \frac{\pi } {2}\left (1 + \frac{k^{2}} {4} \right ) = \frac{\pi } {2}\left (1 + \frac{\Delta \hat{\varphi }_{\mathrm{RF}}^{2}} {16} \right ),}$$

$$\displaystyle{\cos ^{2}\frac{\Delta \hat{\varphi }_{\mathrm{RF}}} {2} = 1 -\sin ^{2}\frac{\Delta \hat{\varphi }_{\mathrm{RF}}} {2} \approx 1 -\frac{\Delta \hat{\varphi }_{\mathrm{RF}}^{2}} {4},}$$

we obtain

$$\displaystyle\begin{array}{rcl} \eta _{\mathrm{fill}}(\Delta \hat{\varphi }_{\mathrm{RF}})& \approx & \frac{\pi } {2}\left [1 -\frac{\Delta \hat{\varphi }_{\mathrm{RF}}^{2}} {16} -\left (1 + \frac{\Delta \hat{\varphi }_{\mathrm{RF}}^{2}} {16} \right )\left (1 -\frac{\Delta \hat{\varphi }_{\mathrm{RF}}^{2}} {4} \right )\right ] \approx {}\\ &\approx & \frac{\pi } {2}\left [-\frac{\Delta \hat{\varphi }_{\mathrm{RF}}^{2}} {16} -\frac{\Delta \hat{\varphi }_{\mathrm{RF}}^{2}} {16} + \frac{\Delta \hat{\varphi }_{\mathrm{RF}}^{2}} {4} \right ] {}\\ \end{array}$$

$$\displaystyle{ \eta _{\mathrm{fill}}(\Delta \hat{\varphi }_{\mathrm{RF}}) \approx \frac{\pi } {16}\;\Delta \hat{\varphi }_{\mathrm{RF}}^{2}. }$$

(3.62)

As expected, this is equivalent to Eq. (3.47). Figure 3.12 shows the difference between the exact solution and this approximation. For $\Delta \hat{\varphi }_{\mathrm{RF}} \leq 90^{\circ }$, the error is below 15%. An improved approximation will be derived in Sect. 3.18.

Fig. 3.12

Bucket filling factor (solid line: exact solution according to Eq. (3.61), dashed line (uppermost curve): approximation (3.62), dash-dotted line: improved approximation (3.65))

3.18 Ratio of Bunch Height to Bunch Length

We now determine the ratio of bunch height to bunch length for the stationary case ($\varphi _{\mathrm{R}} = 0$ or $\varphi _{\mathrm{R}} = \pm \pi$). In this case, Eq. (3.29) reads

$$\displaystyle{H = - \frac{\eta _{\mathrm{R}}} {W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}}\;\frac{\Delta W^{2}} {2} -\frac{Q\hat{V }} {T_{\mathrm{R}}} \left ( \frac{1} {\omega _{\mathrm{RF}}}\left [\pm \cos (\omega _{\mathrm{RF}}\Delta t) \mp 1\right ]\right ).}$$

If we assume that the value of H is given for a specific closed trajectory that marks the outer limits of the bunch, we may calculate the bunch height $2\Delta W = 2\Delta \hat{W}$ by setting $\Delta t = 0$:

$$\displaystyle{ \Delta \hat{W} = \sqrt{\frac{2W_{\mathrm{R} } \beta _{\mathrm{R} }^{2 }} {-\eta _{\mathrm{R}}} H}. }$$

(3.63)

The bunch length $2\Delta t = 2\Delta \hat{t}$ is obtained if we set $\Delta W = 0$ for the same value of H:

$$\displaystyle{H = -\frac{Q\hat{V }} {T_{\mathrm{R}}} \frac{1} {\omega _{\mathrm{RF}}}\left [\pm \cos (\omega _{\mathrm{RF}}\Delta \hat{t}) \mp 1\right ].}$$

Due to

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

we obtain

$$\displaystyle{H = \pm \frac{Q\hat{V }} {2\pi h} \;2\;\sin ^{2}\frac{\Delta \hat{\varphi }_{\mathrm{RF}}} {2} }$$

with

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

Inserting this into Eq. (3.63) leads to

$$\displaystyle{\Delta \hat{W} = \sqrt{\pm \frac{2W_{\mathrm{R} } \beta _{\mathrm{R} }^{2 }Q\hat{V }} {-\pi h\eta _{\mathrm{R}}}} \;\sin \frac{\Delta \hat{\varphi }_{\mathrm{RF}}} {2} = \sqrt{\frac{W_{\mathrm{R} } \beta _{\mathrm{R} }^{2 }\;\vert Q\vert \;\hat{V }} {2\pi h\;\vert \eta _{\mathrm{R}}\vert }} \;2\;\sin \frac{\Delta \hat{\varphi }_{\mathrm{RF}}} {2}.}$$

Due to Eq. (3.27), we have

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

which leads to

$$\displaystyle{\Delta \hat{W} = \frac{W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} {h\;\vert \eta _{\mathrm{R}}\vert } \frac{f_{\mathrm{S,0,stat}}} {f_{\mathrm{R}}} \;2\;\sin \frac{\Delta \hat{\varphi }_{\mathrm{RF}}} {2} = \frac{W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} {\vert \eta _{\mathrm{R}}\vert } \frac{\omega _{\mathrm{S,0,stat}}} {\omega _{\mathrm{RF}}} \;2\;\sin \frac{\Delta \hat{\varphi }_{\mathrm{RF}}} {2} }$$

$$\displaystyle{ \Rightarrow \fbox{$\frac{\Delta \hat{W}} {\Delta \hat{t}} = \frac{W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} {\vert \eta _{\mathrm{R}}\vert } \;\omega _{\mathrm{S,0,stat}}\;\mathrm{si}\frac{\Delta \hat{\varphi }_{\mathrm{RF}}} {2}.$} }$$

(3.64)

For small values of $\Delta \hat{\varphi }_{\mathrm{RF}}$, i.e., for small bunches, this is equivalent to Eq. (3.28), as expected. If the bunch fills the whole bucket, we have $\Delta \hat{\varphi }_{\mathrm{RF}} =\pi$, which leads to

$$\displaystyle{\mathrm{si}\frac{\Delta \hat{\varphi }_{\mathrm{RF}}} {2} =\mathrm{ si} \frac{\pi } {2} = \frac{2} {\pi },}$$

so that

$$\displaystyle{\frac{\Delta \hat{W}} {\Delta \hat{t}} = \frac{2W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} {\pi \vert \eta _{\mathrm{R}}\vert } \;\omega _{\mathrm{S,0,stat}} = \frac{4W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} {\vert \eta _{\mathrm{R}}\vert } \;f_{\mathrm{S,0,stat}}}$$

is obtained. As expected, this is equivalent to Eq. (3.44).

One may sometimes approximate the bunch area by the formula for the area of an ellipse:

$$\displaystyle{A_{\mathrm{b,stat}} \approx \pi \Delta \hat{t}\Delta \hat{W} =\pi \Delta \hat{t}^{2}\frac{W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} {\vert \eta _{\mathrm{R}}\vert } \;\omega _{\mathrm{S,0,stat}}\;\mathrm{si}\frac{\Delta \hat{\varphi }_{\mathrm{RF}}} {2}.}$$

If we make use of Eq. (3.48),

$$\displaystyle{ \frac{\omega _{\mathrm{S,0,stat}}} {A_{\mathrm{B,stat}}} = \frac{\omega _{\mathrm{RF}}^{2}\;\vert \eta _{\mathrm{R}}\vert } {16\;W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}},}$$

we get

$$\displaystyle{ \eta _{\mathrm{fill}} = \frac{A_{\mathrm{b,stat}}} {A_{\mathrm{B,stat}}} \approx \frac{\pi } {16}\Delta \hat{\varphi }_{\mathrm{RF}}^{2}\;\mathrm{si}\frac{\Delta \hat{\varphi }_{\mathrm{RF}}} {2} }$$

(3.65)

for the bucket filling factor. In comparison with the exact solution (3.61), the maximum error is 23. 4% for $\Delta \hat{\varphi }_{\mathrm{RF}} =\pi$ (180^∘). For $\Delta \hat{\varphi }_{\mathrm{RF}} < 110^{\circ }$, the error is smaller than 5%, while for $\Delta \hat{\varphi }_{\mathrm{RF}} < 140^{\circ }$, the error is smaller than 10%. This shows that it is often justified to simplify the shape of the trajectory by replacing it with an ellipse.

3.19 Frequency and RF Amplitude

Inside the deflection magnets, the magnetic field acts as centripetal force:

$$\displaystyle{m\frac{u_{\mathrm{R}}^{2}} {r_{\mathrm{R}}} = Q\;u_{\mathrm{R}}\;B}$$

$$\displaystyle{ \Rightarrow p_{\mathrm{R}} = Q\;r_{\mathrm{R}}\;B. }$$

(3.66)

By means of this formula, we may calculate the revolution frequency of the particles. One obtains

$$\displaystyle\begin{array}{rcl} & u_{\mathrm{R}} =\beta _{\mathrm{R}}c_{0} = l_{\mathrm{R}}f_{\mathrm{R}}& {}\\ & \Rightarrow f_{\mathrm{R}} = \frac{c_{0}} {l_{\mathrm{R}}} \;\beta _{\mathrm{R}}. & {}\\ \end{array}$$

Since

$$\displaystyle{p_{\mathrm{R}} = m_{0}\gamma _{\mathrm{R}}\beta _{\mathrm{R}}c_{0}\qquad \Rightarrow \gamma _{\mathrm{R}}\beta _{\mathrm{R}} = \frac{\mathit{Qr}_{\mathrm{R}}B} {m_{0}c_{0}} }$$

is valid, we obtain, with $\beta _{\mathrm{R}} = \frac{\gamma _{\mathrm{R}}\beta _{\mathrm{R}}} {\sqrt{1+\gamma _{\mathrm{R} }^{2 }\beta _{\mathrm{R} }^{2}}}$,

$$\displaystyle{ \fbox{$f_{\mathrm{R}} = \frac{c_{0}} {l_{\mathrm{R}}}\; \frac{\frac{\mathit{Qr}_{\mathrm{R}}B} {m_{0}c_{0}} } {\sqrt{1 + \left (\frac{\mathit{Qr } _{\mathrm{R} } B} {m_{0}c_{0}} \right )^{2}}}.$} }$$

(3.67)

For a given accelerator, l_R and r_R are constants. Therefore, we see that the revolution frequency is given by the ratio

$$\displaystyle{ \frac{\mathit{Qr}_{\mathrm{R}}B} {m_{0}c_{0}} =\beta _{\mathrm{R}}\gamma _{\mathrm{R}} }$$

(3.68)

in a unique way. Due to

$$\displaystyle{\gamma _{\mathrm{R}}^{2} = \frac{1} {1 -\beta _{\mathrm{R}}^{2}}\qquad \Leftrightarrow \gamma _{\mathrm{R}}^{2} -\beta _{\mathrm{ R}}^{2}\gamma _{ \mathrm{R}}^{2} = 1\qquad \Leftrightarrow \beta _{\mathrm{ R}}\gamma _{\mathrm{R}} = \sqrt{\gamma _{\mathrm{R} }^{2 } - 1},}$$

the frequency is alternatively determined by a given γ_R. This shows that a normalization of the kinetic energy with respect to the mass according to

$$\displaystyle{\frac{W_{\mathrm{kin}}} {m_{0}} = \frac{m_{0}c_{0}^{2}(\gamma _{\mathrm{R}} - 1)} {m_{0}} = c_{0}^{2}(\gamma _{\mathrm{ R}} - 1)}$$

leads to values that can be uniquely assigned to a certain revolution frequency. This is why energies are often specified using the unit MeV∕u. Independent of the ion species, one obtains the same revolution frequency and the same γ_R for the same value in MeV∕u.

According to Eq. (3.68), however, this does not determine the magnetic field in a unique way, since the mass-to-charge ratio is still relevant.

Now we present a simplified derivation of Eq. (3.22). Based on Eq. (3.66), we determine the force that accelerates the particles in the longitudinal direction:

$$\displaystyle{\Rightarrow F_{\mathrm{R}} = \frac{\mathrm{d}p_{\mathrm{R}}} {\mathrm{d}t} = Q\;r_{\mathrm{R}}\;\dot{B}.}$$

If $\dot{B}$ is constant, which is approximately true during one revolution, this force F_R has to be applied continuously. We therefore imagine that the force F_R is continuously distributed⁸ along the ring accelerator. Then the kinetic energy that the particle gains during one revolution is

$$\displaystyle{W =\int F_{\mathrm{R}}\;\mathrm{d}s = Q\;r_{\mathrm{R}}\;\dot{B}\;l_{\mathrm{R}}.}$$

This energy is delivered by the accelerating voltage in the accelerating gap:

$$\displaystyle\begin{array}{rcl} & W =\int QE\;\mathrm{d}s = Q\;V _{\mathrm{R}}& \\ & \Rightarrow V _{\mathrm{R}} = r_{\mathrm{R}}\;\dot{B}\;l_{\mathrm{R}}. &{}\end{array}$$

(3.69)

If $\dot{B}$ is given also, the required accelerating voltage is known. The acceleration, however, does not take place at the maximum of the accelerating voltage:

$$\displaystyle\begin{array}{rcl} & V _{\mathrm{R}} =\hat{ V }\;\sin \varphi _{\mathrm{R}} & {}\\ & \Rightarrow \hat{ V } = \frac{r_{\mathrm{R}}\;\dot{B}\;l_{\mathrm{R}}} {\sin \varphi _{\mathrm{R}}}.& {}\\ \end{array}$$

The frequency of the accelerating voltage is larger than the revolution frequency by a factor of h:

$$\displaystyle{V (t) =\hat{ V }\;\sin (\omega _{\mathrm{RF}}t +\varphi _{\mathrm{R}}) =\hat{ V }\;\sin (2\pi \mathit{hf }_{\mathrm{R}}t +\varphi _{\mathrm{R}}).}$$

3.20 Voltage Versus Bunch Length

As an application of Eq. (3.28) for the ratio of the principal axes in phase space, we now determine the voltage that is necessary to produce a bunch of a given length.

For this purpose, we assume that the bunch area

$$\displaystyle{A_{\mathrm{b}} =\pi \Delta \hat{t}\Delta \hat{W}}$$

in phase space, i.e., the longitudinal emittance, is given. This formula includes the assumption that the target bunch in phase space is an ellipse. Therefore, the bunch must be sufficiently short in comparison with the bucket. Furthermore, only the stationary case is considered. Together with Eq. (3.28),

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

one obtains

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

If we now insert the synchrotron frequency from Eq. (3.25), it follows that

$$\displaystyle{A_{\mathrm{b,stat}}^{2} =\pi ^{2}\Delta \hat{t}^{4}\;f_{\mathrm{ R}}^{2}\;\frac{\vert \eta _{\mathrm{R}}\vert \;\vert Q\vert \;\hat{V }\;h} {2\pi W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} \;\frac{(2\pi W_{\mathrm{R}}\beta _{\mathrm{R}}^{2})^{2}} {\eta _{\mathrm{R}}^{2}} =\pi ^{2}\Delta \hat{t}^{4}\;f_{\mathrm{ R}}^{2}\;\frac{\vert Q\vert \;\hat{V }\;h\;2\pi W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} {\vert \eta _{\mathrm{R}}\vert }.}$$

We expand the fraction with 2π h in order to get ω_RF:

$$\displaystyle{A_{\mathrm{b,stat}}^{2} =\pi ^{2}\Delta \hat{t}^{4}\;\omega _{ \mathrm{RF}}^{2}\;\frac{\vert Q\vert \;\hat{V }\;W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} {2\pi h\vert \eta _{\mathrm{R}}\vert }.}$$

Finally, we take

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

into account and obtain

$$\displaystyle\begin{array}{rcl} & A_{\mathrm{b,stat}}^{2} =\pi ^{2}\Delta \hat{\varphi }_{\mathrm{RF}}^{4}\;\frac{\vert Q\vert \;\hat{V }\;W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} {2\pi h\vert \eta _{\mathrm{R}}\vert \omega _{\mathrm{RF}}^{2}} & \\ & \Rightarrow \hat{ V } = \frac{2h\vert \eta _{\mathrm{R}}\vert \;\omega _{\mathrm{RF}}^{2}\;A_{\mathrm{ b,stat}}^{2}} {\pi \;\vert Q\vert \;W_{\mathrm{R}}\;\beta _{\mathrm{R}}^{2}\;\Delta \hat{\varphi }_{\mathrm{RF}}^{4}}. &{}\end{array}$$

(3.70)

Therefore, the required voltage decreases as the fourth power of the bunch length; for reducing the bunch length to one-half its value, the voltage has to be increased by a factor of 16. Equation (3.70) can also be found in Edwards/Syphers [4] as formula (2.76).

3.21 Coasting Beam

Often, the (relative) momentum spread

$$\displaystyle{\frac{\Delta p} {p_{\mathrm{R}}} }$$

of the coasting beam, i.e., of the nonbunched beam, is given. Due to

$$\displaystyle{\frac{\mathrm{d}\gamma } {\gamma } =\beta ^{2}\frac{\mathrm{d}(\gamma \beta )} {\gamma \beta },}$$

one obtains

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

so that the energy deviation can be calculated with the help of

$$\displaystyle{W_{\mathrm{R}} = \mathit{mc}_{0}^{2} =\gamma _{\mathrm{ R}}\;m_{0}\;c_{0}^{2}.}$$

It is important to note that W_R is the total energy. If, for example,²³⁸U²⁸⁺ with a kinetic energy of 11. 4 MeV∕u is given, one obtains

$$\displaystyle\begin{array}{rcl} W_{\mathrm{R}}& =& A_{\mathrm{r}}\left (W_{\mathrm{kin,u}} + W_{\mathrm{rest,u}}\right ) = A_{\mathrm{r}}\left (W_{\mathrm{kin,u}} + m_{\mathrm{u}}c_{0}^{2}\right ) = {}\\ & =& 238.05 \cdot (11.4\,\mathrm{MeV} + 931.49\,\mathrm{MeV}) = 224.455\;\mathrm{GeV} {}\\ \end{array}$$

for the total energy. By means of

$$\displaystyle{A_{\mathrm{b,DC}} = 2\;\Delta W\;\Delta t,}$$

one may now calculate the longitudinal emittance of the coasting beam. The factor 2 is due to the fact that the energy is in the range from $W_{\mathrm{R}} - \Delta W$ to $W_{\mathrm{R}} + \Delta W$. The time deviation corresponds to the revolution time of the particles:

$$\displaystyle{A_{\mathrm{b,DC}} = 2\;\Delta W\;T_{\mathrm{R}} = \frac{2\;\Delta W} {f_{\mathrm{R}}} = \frac{2\;W_{\mathrm{R}}\beta _{\mathrm{R}}^{2}} {f_{\mathrm{R}}} \frac{\Delta p} {p_{\mathrm{R}}}.}$$

We know that the phase space area is preserved (Liouville’s theorem). Therefore, it should be possible to convert the coasting beam with emittance A_b,DC into a bunched beam produced by a cavity operating with harmonic number h.

In this case, the phase space area will be split among the h bunches according to

$$\displaystyle{A_{\mathrm{b}} = \frac{A_{\mathrm{b,DC}}} {h}.}$$

In Sect. 5.3, we will see that this conversion from a coasting beam to a bunched beam may be accomplished by an adiabatic capture process.

3.22 Ramps

We have seen that according to Eq. (3.67), the revolution frequency f_R is determined in a unique way as soon as the magnetic field B is given. With respect to Eq. (1.4), it is clear that the RF frequency f_RF is then also determined if a certain harmonic number h is chosen. Finally, Eq. (3.22) shows that the voltage $\hat{V }$ is also influenced by the choice of the magnetic field.

In other words, all these quantities have to be in synchrony, whence the name “synchrotron.” A synchrotron is usually operated by transmitting time functions such as B(t), f_RF(t), $\hat{V }(t)$ to the synchrotron devices. These time functions are often called ramps.

We will now discuss how ramps may be generated for a simple accelerating cycle. Such a machine cycle has to satisfy certain requirements:

• The momentum of the particles, and hence also the energy, is directly related to the magnetic rigidity. Therefore, the magnetic field of the bending magnets is low when the particles are injected into the synchrotron, and it must be high after the particles are accelerated to the final energy until they are extracted. Therefore, the magnetic field B has to be increased.

• Some time before the acceleration starts at injection energy and some time after the acceleration is completed at extraction energy, the beam must be bunched. On the one hand, we have $\dot{B} = 0$ during these phases, and on the other hand, we need some voltage amplitude $\hat{V } > 0$. According to Eq. (3.49), we have V_R = 0, and due to Eq. (3.50), this leads to φ_R = 0 or $\varphi _{\mathrm{R}} = \pm \pi$, depending on whether particles with positive or negative charge are considered and on whether operation takes place below or above the transition energy.

• During the accelerating phase, i.e., when $\dot{B} > 0$ holds, we need a larger voltage according to Eq. (3.49). Furthermore, we must have $\varphi _{\mathrm{R}}\neq 0,\pm \pi$, because the reference particle has to experience an accelerating voltage V_R ≠ 0 (Eq. (3.50)).

• In other words, we have a stationary bucket at injection energy, an acceleration bucket during acceleration, and a stationary bucket at extraction energy.

• During the whole process, we would, of course, like to have matched bunches,⁹ because in that case, the phase space area that is occupied by the bunches will remain constant. A larger phase space area caused by the filamentation of unmatched bunches is undesired, because more voltage would then be needed. The strategy for having matched bunches permanently is to control the bucket area by means of the voltage $\hat{V }$.

One typically begins with the magnetic field B(t) of the dipole magnets. As a first approximation, one would start with a piecewise linear curve (especially for this function, the name “ramp” is justified because the magnetic dipole field is ramped up):

$$\displaystyle{ B(t) = \left \{\begin{array}{ll} B_{\mathrm{inj}} & 0 < t < t_{1}, \\ \frac{B_{\mathrm{extr}}(t-t_{1})+B_{\mathrm{inj}}(t_{2}-t)} {t_{2}-t_{1}} & t_{1} < t < t_{2}, \\ B_{\mathrm{extr}} & t_{2} < t < t_{3}, \\ \frac{B_{\mathrm{extr}}(t_{4}-t)+B_{\mathrm{inj}}(t-t_{3})} {t_{4}-t_{3}} & t_{3} < t < t_{4}.\\ \end{array} \right. }$$

(3.71)

Such a choice, however, leads to a ramp $\dot{B}(t)$ that is not continuous. One result is that the time function for the reference phase $\varphi _{\mathrm{R}}(t)$ would contain jumps. The bunches would be unable to follow those jumps (unmatched bunch), and undesired longitudinal beam oscillations and filamentation would be the consequence.

Therefore, one may insert transitions at the times t₁, t₂, and t₃ in the ramp B(t) specified above. These transitions may be defined in such a way that the ramp rate $\dot{B}$ increases or decreases to the desired value of the next segment. If this is done in a linear way, the ramp $\dot{B}(t)$ will be piecewise linear. The ramp B(t) will then look similar to the one specified by Eq. (3.71), but its edges will be rounded off.

In conclusion, we have to keep in mind that transitions are needed in Eq. (3.71). Nevertheless, we will now discuss the meaning of the different phases.

• During the phase 0 < t < t₁, the beam is injected into the synchrotron while the magnetic field is constant (phase A in Fig. 3.13). The magnetic field, of course, has to fit to the energy of the particles (injection energy/plateau). The RF voltage may already be switched on (so-called injection into stationary buckets), or it is slowly¹⁰ ramped up¹¹ (adiabatic capture; see Sect. 5.3.2). The latter case is shown as phase B in Fig. 3.13. In phase C, the beam is kept bunched at constant energy for a while.

• The phase t₁ < t < t₂ is the accelerating phase (phase D in Fig. 3.13).

  Fig. 3.13

  

  An example of ramps in a machine cycle

• The phase t₂ < t < t₃ is called the “flat top” (phase E in Fig. 3.13). Here, at maximum energy, the beam is extracted from the synchrotron (extraction energy/plateau). Both fast extraction of bunches and slow “spill” extraction in order to get a steady beam current are possible (cf. [11, Sect. 7.4.1]).

• In the last phase, t₃ < t < t₄, the magnetic field is ramped down again in order to have the same initial conditions for the next machine cycle (phase F in Fig. 3.13). The actual shape of the ramps in this phase is unimportant, because no beam is present.

• Afterward, the next machine cycle begins. Therefore, t₄ is the cycle time (if we neglect the above-mentioned transitions), which should be kept as low as possible if a large intensity, i.e., a large number of bunches per time, is to be offered.

The magnetic field in the bending magnets is restricted for technical reasons. This is valid for both the minimum of the magnetic field B_inj at injection and the maximum of the magnetic field B_extr at extraction. In general, the dipole fields have to be controlled to very high accuracy (e.g., 10⁻⁵ or 10⁻⁶).

We now derive all relevant ramps based on the definition of the magnetic field ramp B(t). First of all, we may directly calculate the revolution frequency with the help of Eq. (3.67):

$$\displaystyle{f_{\mathrm{R}} = \frac{c_{0}} {l_{\mathrm{R}}}\; \frac{\frac{\mathit{Qr}_{\mathrm{R}}B} {m_{0}c_{0}} } {\sqrt{1 + \left (\frac{\mathit{Qr } _{\mathrm{R} } B} {m_{0}c_{0}} \right )^{2}}}.}$$

The RF frequency is then given by

$$\displaystyle{f_{\mathrm{RF}} = h\;f_{\mathrm{R}}.}$$

It is easy to determine other quantities such as the velocity, the relativistic Lorentz factors β_R and γ_R, and kinetic and total energy.

Let us now assume that an adiabatic capture is performed, in which case, one would calculate the longitudinal emittance A_b,DC of the coasting beam based on the known momentum spread $\Delta p/p$ of the coasting beam (see Sect. 3.21). At the end of the capture process, the sum h ⋅ A_B of the phase space areas of the h buckets must, of course, be larger than this emittance in order to allow all particles to move into the buckets.¹² In reality, however, some losses must be accepted. One may increase the voltage even further if bunching is to be enhanced, i.e., if shorter bunches are to be generated that fill a smaller fraction of the final bucket area A_B.

In the simplest case, one may increase the voltage slowly (more slowly than the period of the synchrotron oscillation, i.e., adiabatically) in a linear fashion. However, one may also use the isoadiabatic ramps derived in Sect. 5.3.2 in order to save time.

For the sake of simplicity, we assume a linear increase from 0 to $\hat{V }_{\mathrm{inj}}$ in the time interval t_A < t < t_B, where 0 < t_A < t_B < t₁ holds. The time t_A is needed to inject the beam into the synchrotron and to give it enough time to form a real coasting beam with respect to the original momentum spread. For t_B < t < t₁, the voltage is kept constant at $\hat{V } =\hat{ V }_{\mathrm{inj}}$.

Now we have to calculate how the voltage $\hat{V }$ varies in the time interval t₁ < t < t₂. For this purpose, we assume that a constant area in phase space is available for the beam:¹³

$$\displaystyle{A_{\mathrm{B}} =\mathrm{ const}\qquad \mbox{ and}\quad A_{\mathrm{B}} = A_{\mathrm{B,stat}}.}$$

We use Eq. (3.41),

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

(3.72)

which allows us to calculate A_B for the stationary case with $\varphi _{\mathrm{R}} = 0$ (or $\varphi _{\mathrm{R}} = \pm \pi$) and $\alpha (\varphi _{\mathrm{R}}) = 1$ and $\hat{V } =\hat{ V }_{\mathrm{inj}}$. For the time interval t₁ < t < t₂, this value A_B remains constant, according to the requirements, whereas $\varphi _{\mathrm{R}}$ and $\hat{V }$ may change. The devolution of the other quantities is already determined, as mentioned above.

We now return to Eqs. (3.22) and (3.23), which lead to

$$\displaystyle{ \hat{V }\;\sin \varphi _{\mathrm{R}} = l_{\mathrm{R}}r_{\mathrm{R}}\dot{B} }$$

(3.73)

for harmonic gap voltages. With the exception of $\varphi _{\mathrm{R}}$ and $\hat{V }$, all other quantities are already determined. Therefore, the last two equations allow us to determine both $\varphi _{\mathrm{R}}$ and $\hat{V }$ numerically.

At the flat top, the voltage $\hat{V }$ will be reduced, due to $\varphi _{\mathrm{R}} = 0$. One may even decrease it to zero in order to create a coasting beam again—this time, however, with a significantly higher (extraction) energy.

We now show how a numerical calculation of $\varphi _{\mathrm{R}}$ may be performed. For this purpose, we insert Eq. (3.73) into Eq. (3.72) in order to eliminate $\hat{V }$:

$$\displaystyle{A_{\mathrm{B}}\sqrt{\sin \varphi _{\mathrm{R}}} = \frac{4\sqrt{2}} {\pi h} T_{\mathrm{R}}\;\sqrt{\frac{W_{\mathrm{R} } \beta _{\mathrm{R} }^{2 }\vert Q\vert \;l_{\mathrm{R} } r_{\mathrm{R} } \dot{B}} {\pi h\vert \eta _{\mathrm{R}}\vert }} \;\alpha (\varphi _{\mathrm{R}}).}$$

$$\displaystyle{ \Leftrightarrow \sqrt{\sin \varphi _{\mathrm{R}}} - q\;\alpha (\varphi _{\mathrm{R}}) = 0 }$$

(3.74)

Here we defined

$$\displaystyle{q = \frac{4\sqrt{2}\;T_{\mathrm{R}}} {\pi h\;A_{\mathrm{B}}} \;\sqrt{\frac{W_{\mathrm{R} } \beta _{\mathrm{R} }^{2 }\vert Q\vert \;l_{\mathrm{R} } r_{\mathrm{R} } \dot{B}} {\pi h\vert \eta _{\mathrm{R}}\vert }}.}$$

Now we see that the left-hand side of Eq. (3.74) equals − q for $\varphi _{\mathrm{R}} = 0$, i.e., it is negative. For $\varphi _{\mathrm{R}} =\pi /2$, it equals 1, which is positive. Therefore, the root may easily be determined by means of a bisection method (i.e., by repeatedly bisecting the interval $\varphi _{\mathrm{R}} \in [0,\pi /2]$).

3.23 Multicavity Operation

Let us assume that several cavities are distributed along the synchrotron. A cavity is said to work at harmonic number h if its operating frequency equals h times the revolution frequency of the reference particle. The circumference of the synchrotron is denoted by l_R, and s will be a coordinate that describes the path length. For a specific harmonic number h, we have a number n_h of cavities that work at this harmonic number h and that are installed at the positions s_h, k ($k = 1,\ldots,n_{h}$). The number of bunches circulating in the synchrotron is denoted by h_b.

Each of the n_h cavities produces an RF signal

$$\displaystyle{V _{h,k}(t) =\hat{ V }_{h,k}\;\sin (h\omega _{\mathrm{R}}t -\varphi _{h,k}).}$$

One usually wants to ensure that a particle bunch passes all these cavities in such a way that it reaches them at the same RF phase. The particle bunch will reach the kth cavity at the time

$$\displaystyle{t_{h,k} = \frac{s_{h,k}} {u_{\mathrm{R}}},}$$

where u_R is determined by

$$\displaystyle{\omega _{\mathrm{R}} = \frac{2\pi } {T_{\mathrm{R}}} = \frac{2\pi u_{\mathrm{R}}} {l_{\mathrm{R}}}.}$$

Therefore, we have

$$\displaystyle{t_{h,k} = 2\pi \;\frac{s_{h,k}} {\omega _{\mathrm{R}}l_{\mathrm{R}}}.}$$

The requirement that the bunch see the same phase at the different cavities means that

$$\displaystyle{2\pi h\;\frac{s_{h,k}} {l_{\mathrm{R}}} -\varphi _{h,k} = 2\pi p,}$$

where the left-hand side is obtained by inserting t_h, k into the argument of the sine function. The right-hand side with an integer p results from the fact that all periods of the sine function are equivalent.

We now define

$$\displaystyle{\theta _{h,k} = 2\pi \;\frac{s_{h,k}} {l_{\mathrm{R}}} }$$

as the angle that describes the position of the cavity in the synchrotron. This leads to

$$\displaystyle{\varphi _{h,k} = h\theta _{h,k} - 2\pi p.}$$

As an example, we now consider the synchrotron SIS18 at GSI. It has two cavities that are exactly on opposite positions in the ring, so that we have θ_h, 1 = 0 and θ_h, 2 = π. For the standard operation at h = 4 or any other even harmonic number h, one gets

$$\displaystyle{\varphi _{h,1} = 0\qquad \mbox{ and }\varphi _{h,2} = 0,}$$

where p is chosen in such a way that $\varphi _{h,k}$ is located in the interval $\left ]-\pi,\pi \right ]$. For odd values of h, one immediately sees that one must choose

$$\displaystyle{\varphi _{h,1} = 0\qquad \mbox{ and }\varphi _{h,2} =\pi.}$$

In addition, we have to ensure that all bunches are treated equally. The distance between the h_b buckets is l_R∕h_b. Assuming that two adjacent buckets are filled, their time difference equals

$$\displaystyle{\Delta t = \frac{l_{\mathrm{R}}} {h_{\mathrm{b}}\;u_{\mathrm{R}}} = \frac{2\pi } {h_{\mathrm{b}}\;\omega _{\mathrm{R}}}.}$$

If we insert this into the argument of the sine function, we see that

$$\displaystyle{2\pi \frac{h} {h_{\mathrm{b}}}}$$

must be an integer multiple of 2π if the same RF phase is to be obtained for all bunches. Hence, only integer multiples of h_b are allowed for h.

Furthermore, at least one group of cavities with h = h_b must exist in order to create the h_b bunches.

3.24 Bunch Shape

Several topics that we have presented thus far were based on the simple model that the bunch occupies a well-defined area in phase space. This allowed us, for example, to easily compare the longitudinal emittance (i.e., the bunch area in phase space) with the bucket area. Arguments of this type are implicitly based on the assumption that the particles have a distribution in phase space with clear margins. In reality, this is, of course, not true. For realistic bunches, one has to use rms values to describe the dimensions of the bunches in phase space. For example, one may define the longitudinal emittance¹⁴ by

$$\displaystyle{A_{\mathrm{b}} \sim \pi \Delta t_{\mathrm{rms}}\Delta W_{\mathrm{rms}}.}$$

It is obvious that this formula is obtained if the area of an ellipse is calculated by means of the two semiaxes. In practice, different constants are used in the literature to make the definition complete (cf. [6, Sect. 5.4.4]).

Now we return to our beam current example, which was shown in Fig. 1.3 (Measurement No. 43 dated August, 21, 2008,⁴⁰Ar¹⁸⁺, 11. 4 MeV∕u, 6 kV, h = 4, N_beam = 1. 5 ⋅ 10⁹ particles). According to

$$\displaystyle{ \fbox{$\bar{I}_{\mathrm{beam}} = \frac{N_{\mathrm{beam}}z_{q}e} {T_{\mathrm{R}}} = N_{\mathrm{beam}}z_{q}ef_{\mathrm{R}},$} }$$

(3.75)

this corresponds to an average beam current of about 0. 9 mA. Since the quality of the beam was very good during this experiment, the measured pulse shape may be used for an analysis of the distribution.

Figure 3.14 shows different curves in comparison with the measurement.

Fig. 3.14

Comparison of beam current measurement with different curves

If one discards the problem that the zero line of the beam current is hard to identify, the Gaussian distribution (second curve) is an excellent approximation of the measured pulse shape (with σ = 78 ns). Please note that the Gaussian distribution cannot, in spite of the good results, represent the “absolute truth,” since a bunch that is captured inside a bucket must always have a finite size. The Gaussian distribution, however, represents the typical shape of the bunches of a low-intensity ion beam when space-charge effects are low (cf. [6, Sect. 5.4.7]).

Also, the cos² distribution (third curve) provides good results—except that at the margins of the bunch, larger deviations are visible (the cos² function must, of course, be clipped beyond the minima on both sides—this was not done in the figure in order to show the periodicity of this function).

The fourth curve shows the results of a simulation in which a coasting beam with a constant distribution of 30, 000 macroparticles in phase space with $\left.\Delta p/p\right \vert _{\mathrm{max}} = \pm 3.5 \cdot 10^{-4}$ was assumed. In the simulation, the complete capture process was simulated. The projection that is needed to calculate the bunch shape was based on 200 equidistant bins. One sees that the constant distribution does not describe the reality very well, but for qualitative analyses it is often sufficient.

The fifth curve shows the simulation results in which a coasting beam with a Gaussian distribution ($\left.\Delta p/p\right \vert _{\sigma } = 2 \cdot 10^{-4}$) was assumed at the beginning. Again, 30, 000 macroparticles with 200 bins were used, and the complete capture process was simulated. The good agreement with the measurement is obvious.

Finally, the theoretical curve for a constant particle density inside an ellipse in phase space is shown. In this case, the projection is an ellipse as well.

Of course, there exist other models for the distribution of the particles (e.g., the parabolic bunch, which is an important distribution for space-charge dominated beams [6, 12]).