Physics of Collective Beam Instabilities in High Energy Accelerators: Wake Fields and Impedances (Draft)
Following [1], Chapter 2, Wake Fields and Impedances. (i) Calculate the electromagnetic wake fields generated by a relativistic beam in a vacuum chamber. (ii) Define the wake function and its Fourier transform, the impedance.
Mode expansion of the charge and current distributions
We will use cylindrical coordiantes \(r\), \(\theta\), \(s\), where \(s\) is the axial coordinate. Consider a charge charge distribution \(\rho(r, \theta, s)\) and current distribution \(\mathbf{J}(r, \theta, s)\) of the form:
\[ \begin{aligned} \rho(r, \theta, s) &= \sum_{m = 0}^{\infty} \rho_m(r, \theta, s) , \\ \mathbf{J}(r, \theta, s) &= \sum_{m = 0}^{\infty} \mathbf{J}_m(r, \theta, s) , \end{aligned} \tag{1}\]
with the modes \(\rho_m\) and \(\mathbf{J}_m\) defined as follows:
\[ \begin{aligned} \rho_m(r, \theta, s) &= \frac{I_m}{\pi a^{m + 1} (1 + \delta_{m0})} \delta(s - ct) \delta(r - a) \cos(m\theta) , \\ \mathbf{J}_m(r, \theta, s) &= c \rho_m(r, \theta, s) \hat{s} , \end{aligned} \tag{2}\]
where \(a\) is a radius, \(\delta_{m0} = 1\) if \(m = 0\) and \(0\) otherwise, \(I_m\) is the \(m\)th moment of the distribution, \(c\) is the speed of light, and \(t\) is time. Equation 2 describes a ring of radius \(a\) with density proportional to \(\cos(m\theta)\). The longitudinal density is a pancake, enforced by the \(\delta(s - ct)\) term.
Wake functions
What is the force experienced by a particle trailing the beam? Assume the particle moves only along the \(s\) axis. In the cylindrical coordinates, the Lorentz force \(\mathbf{F} = q(\mathbf{E} + \hat{s} \times \mathbf{B})\) becomes
\[ \begin{aligned} F_s &= q E_s , \\ F_\theta &= q \left( E_\theta + B_r \right) , \\ F_r &= q \left( E_r + B_\theta \right) . \end{aligned} \tag{3}\]
At high energies, we may assume the beam remains unchanged by the wake field and focus on the integrated force \(\bar{\mathbf{F}}\),
\[ \bar{\mathbf{F}} = \int_{-L/2}^{+L/2} \mathbf{F} ds, \]
where \(L\) is some distance longer than the region of interest. The result is (2.49), which relates the averaged force to the averaged field. For an axially symmetric environment, (2.49) can be solved to yield:
\[ \begin{aligned} \bar{F}_r &= +q I_m W_m(z)_ m r^{m-1} \cos(m\theta) , \\ \bar{F}_\theta &= -q I_m W_m(z)_ m r^{m-1} \sin(m\theta) , \\ \bar{F}_s &= -q I_m W'_m(z) r^m \cos(m\theta) , \\ \bar{B}_s &= +q I_m W'_m(z) r^m \sin(m\theta) . \end{aligned} \tag{4}\]
\(W_m(z)\) is called the wake function and \(W'_m(z)\) is the derivative of the wake function with respect to \(s\). Equation 4 is valid for any axially symmetric boundary, but the wake function can only be determined once boundary conditions are applied.
Panofsky-Wenzel Theorem. Define the transverse wake potential as \(\bar{\mathbf{F}}_{\perp} = (\bar{F}_r, \bar{F}_\theta)\) and the longitudinal wake potential as \(\bar{F}_{\parallel} = \bar{F}_s\). Equation 4 implies:
\[ \nabla_{\perp} \bar{F}_{\parallel} = \nabla_z \bar{\mathbf{F}}_{\perp} = \frac{\partial}{\partial z} \bar{\mathbf{F}}_{\perp} . \tag{5}\]
Properties of wake functions:
The wake function is determined by the environment, not the beam.
\(W'_m(z) = 0\) if \(z > 0\). (Wake fields always trail the beam.)
\(W'_m(0^-) \ge 0\). (We expect a deccelerating field just behind the beam. It follows that a single particle experiences a longitudinal force but no transverse force from its own wake field.)
\(W'_m(0^-) \ge |W_m'(z)|\) for all \(z\). (Follows from the two-particle example; the system cannot gain energy.)
…
Table 2.2 shows the wake potentials for \(m = 0, 1, 2, 3\) in a Cartesian coordinate system.
The section ends with a discussion of wake field acceleration.
Impedances
Define the longitudinal impedance \(Z_{m}^{\parallel}(\omega)\) and transverse impedance \(Z_{m}^{\perp}(\omega)\) as the Fourier transforms of the longitudinal and transverse wake functions:
\[ Z_{m}^{\parallel}(\omega) = \frac{1}{c} \int_{-\infty}^{+\infty} e^{-i \omega z / c} W_m'(z) dz. \tag{6}\]
\[ Z_{m}^{\perp}(\omega) = \frac{i}{c} \int_{-\infty}^{+\infty} e^{-i \omega z / c} W_m (z) dz. \tag{7}\]
It’s often the case that one is most interested in the \(m=0\) longitudinal mode and \(m = 1\) transverse modes; thus longitudinal impedance and transverse impedance are sometimes used to refer only to \(Z_{0}^{\parallel}\) and \(Z_{1}^{\perp}\).
Equation 5 states the relationship between the transverse and longitudinal impedance:
\[ Z_{m}^{\parallel}(\omega) = \frac{\omega}{c} Z_{m}^{\perp}(\omega). \]
Space charge impedance
For the ring beam described above (Equation 1 and eq-ring-beam-moments), the electric field is given by (1.32) and (1.51) […]. The fields in the region \(r < a\) can be expressed using a wake function (2.55):
\[ W_m(z) = \frac{2 L}{\gamma^2} \delta(z) \begin{cases} \ln\left(\frac{b}{a}\right) & \text{if } m = 0, \\ \frac{1}{m} \left( \frac{1}{a^{2m}} - \frac{1}{b^{2m}} \right) & \text{if } m > 0. \end{cases} \tag{8}\]
To cast Equation 8 in the impedance framework, we assume a circular accelerator of radius \(R\) and set the integration window to \(L = 2 \pi R\). This gives the following expression for the longitudinal space charge impedance:
\[ Z_{m}^{\parallel}(\omega) = i Z_0 \frac{R \omega}{c \gamma^2} \begin{cases} \ln\left(\frac{b}{a}\right) & \text{if } m = 0, \\ \frac{1}{m} \left( \frac{1}{a^{2m}} - \frac{1}{b^{2m}} \right) & \text{if } m > 0., \end{cases} \tag{9}\]
where \(Z_0 = 4 \pi / c\). For a uniform disk beam, the space charge forces cannot be described using a wake function. It is only possible to do it for the \(m = 0\) longitudinal case. The wake function is
\[ W_0(z) = \frac{4 \pi R}{\gamma^2} \delta(z) \left( \ln\left(\frac{b}{a}\right) + \frac{1}{2} \right), \]
and the corresponding impedance is
\[ Z_{0}^{\parallel}(\omega) = i Z_0 \frac{R\omega}{c \gamma^2} \left( \ln\left(\frac{b}{a}\right) + \frac{1}{2} \right) . \]
LRC circuit
We can model the longitudinal impedance using an LRC circuit. Recall the definition of impedance from electrical engineering. […]
Properties of impedance
Symmetries
In most cases, the integrals of imaginary impedance are zero.
There cannot be any singularities in the upper half of the complex plane.
The energy loss for a longitudinal distribution \(\rho(z)\) is given by
\[ \begin{align} \Delta \mathcal{E} &= -\int_{-\infty}^{+\infty} dz' \rho(z') \int_{z'}^{\infty} dz \rho(z) W_m'(z' - z) \\ \Delta \mathcal{E} &= -\frac{1}{2\pi} \int_{-\infty}^{+\infty} d\omega \left| \tilde{\rho}(\omega) \right|^2 \text{Re} Z_{m}^{\parallel}(\omega) . \end{align} \]
The term \(\tilde{\rho}(\omega)\) is the Fourier transform of the charge distribution:
\[ \tilde{\rho}(\omega) = \int_{\infty}^{+\infty} dz e^{-i \omega z / c} \rho(z). \]
There can be no net energy gain from the pipe structure; therefore, \(\text{Re} Z_{m}^{\parallel}(\omega) \ge 0\) for all \(\omega\).
Broad-band resonator model of cavity impedance
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Diffraction model of cavity impedance
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Calculation of wake fields and impedances
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Parasitic loss
Parasitic loss refers to the transfer of energy from the beam to the wake fields. The reverse process is ultimately responsible for beam instabilities. […]