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RFTM110

Tracks through a TM110-mode (deflecting) rf cavity with all magnetic and electric field components. NOT RECOMMENDED--See below.
Parallel capable? : yes
Parameter Name Units Type Default Description
PHASE $DEG$ double 0.0 phase
TILT $RAD$ double 0.0 rotation about longitudinal axis
FREQUENCY $HZ$ double 2856000000 frequency
VOLTAGE $V$ double 0.0 peak deflecting voltage
PHASE_REFERENCE   long 0 phase reference number (to link with other time-dependent elements)
VOLTAGE_WAVEFORM   STRING NULL $<$filename$>$=$<$x$>$+$<$y$>$ form specification of input file giving voltage waveform factor vs time
VOLTAGE_PERIODIC   long 0 If non-zero, voltage waveform is periodic with period given by time span.
ALIGN_WAVEFORMS   long 0 If non-zero, waveforms' t=0 is aligned with first bunch arrival time.
VOLTAGE_NOISE   double 0.0 Rms fractional noise level for voltage.
PHASE_NOISE $DEG$ double 0.0 Rms noise level for phase.
GROUP_VOLTAGE_NOISE   double 0.0 Rms fractional noise level for voltage linked to group.
GROUP_PHASE_NOISE $DEG$ double 0.0 Rms noise level for phase linked to group.
VOLTAGE_NOISE_GROUP   long 0 Group number for voltage noise.
PHASE_NOISE_GROUP   long 0 Group number for phase noise.





NB: Although this element is correct insofar as it uses the fields for a pure TM110 mode, it is recommended that the RFDF element be used instead. In a real deflecting cavity with entrance and exit tubes, the deflecting mode is a hybrid TE/TM mode, in which the deflection has no dependence on the radial coordinate.

To derive the field expansion, we start with some results from Jackson[16], section 8.7. The longitudinal electric field for a TM mode is just

\begin{displaymath}
E_z = - 2 i E_0 \Psi(\rho, \phi) \cos \left(\frac{p \pi z}{d}\right) e^{-i\omega t},
\end{displaymath} (14)

where $p$ is an integer, $d$ is the length of the cavity, and we use cylindrical coordinates $(\rho, \phi, z)$. The factor of $-2i$ represents a choice of sign and phase convention. We are interested in the TM110 mode, so we set $p=0$. In this case, we have
\begin{displaymath}
E_x = E_y = 0
\end{displaymath} (15)

and (using CGS units)
\begin{displaymath}
\vec{H} = - 2 i E_0 \frac{i \epsilon \omega}{c k^2} \hat{z} \times \nabla \Psi e^{-i \omega t}.
\end{displaymath} (16)

For a cylindrical cavity, the function $\Psi$ for the $m=1$ aximuthal mode is
\begin{displaymath}
\Psi(\rho, \phi) = J_1 (k \rho) \cos \phi,
\end{displaymath} (17)

where $k = x_{11}/R$, $x_{11}$ is the first zero of $J_1(x)$, and $R$ is the cavity radius. We don't need to know the cavity radius, since $k = \omega/c$, where $\omega$ is the resonant frequency. By choosing $\cos\phi$ for the aximuthal dependence, we'll get a magnetic field primarily in the vertical direction.

In MKS units, the magnetic field is

\begin{displaymath}
\vec{B} = \frac{2 E_0}{k c} e^{-i \omega t} \left( \hat{\rho...
...hi} \cos\phi \frac{\partial J_1(k\rho)}{\partial \rho}\right).
\end{displaymath} (18)

Using mathematica, we expanded these expressions to sixth order in $k*\rho$. Here, we present only the expressions to second order. Taking the real parts only, we now have

$\displaystyle E_z$ $\textstyle \approx$ $\displaystyle E_0 k \rho \cos \phi \sin \omega t$ (19)
$\displaystyle c B_\rho$ $\textstyle \approx$ $\displaystyle E_0 \left(1 - \frac{k^2 \rho^2}{8}\right)\sin\phi \cos\omega t$ (20)
$\displaystyle c B_\phi$ $\textstyle \approx$ $\displaystyle E_0 \left(1 - \frac{3 k^2 \rho^2}{8}\right)\cos\phi \cos\omega t$ (21)

The Cartesian components of $\vec{B}$ can be computed easily
$\displaystyle c B_x$ $\textstyle =$ $\displaystyle c B_\rho\cos\phi - c B_\phi\sin\phi$ (22)
  $\textstyle =$ $\displaystyle \frac{E_0}{4} \rho^2 k^2 \cos\phi \sin\phi \cos\omega t$ (23)
$\displaystyle c B_y$ $\textstyle =$ $\displaystyle c B_\rho\sin\phi + c B_\phi\cos\phi$ (24)
  $\textstyle =$ $\displaystyle E_0 \left(1 - \frac{k^2\rho^2 (2 \cos^2\phi + 1)}{8}\right) \cos\omega t$ (25)

The Lorentz force on an electron is $F = -e E_z \hat{z} - e c \vec{\beta} \times \vec{B}$, giving

$\displaystyle F_x/e$ $\textstyle =$ $\displaystyle \beta_z c B_y$ (26)
$\displaystyle F_y/e$ $\textstyle =$ $\displaystyle -\beta_z c B_x$ (27)
$\displaystyle F_z/e$ $\textstyle =$ $\displaystyle -E_z - \beta_x c B_y + \beta_y c B_x$ (28)

We see that for $\rho \rightarrow 0$, we have $E_z = 0$, $B_x = 0$, and
\begin{displaymath}
c B_y = E_0 \cos \omega t.
\end{displaymath} (29)

Hence, for $\omega t=0$ and $E_0>0$ we have $F_x>0$. This explains our choice of sign and phase convention above. Indeed, owing to the factor of $2$, we have a peak deflection of $e E_0 L/E$, where $L$ is the cavity length and $E$ the beam energy. Thus, if $V = E_0 L$ is specified in volts, and the beam energy expressed in electron volts, the deflection is simply the ratio of the two. As a result, we've chosen to parametrize the deflection strength simply by referring to the ``deflecting voltage,'' $V$.
next up previous
Next: RFTMEZ0 Up: Element Dictionary Previous: RFMODE
Robert Soliday 2007-04-02