# Accelerator Physics/Physics of linear accelerators (Focusing on longitudinal dynamics)/Important concepts and definitions/Transit Time Factor

Transit time factor, $T_{tr}$, characterizes the energy gain of a particle passing through an acceleration gap. The energy change $\Delta E$ of the particle is given by
$\Delta E=qV_{0}T_{tr}\cos \phi$
where $q$ is the charge of the particle, $V_{0}$ the maximum voltage difference between the gap, $\phi$ the initial phase, defined as the phase of the oscillating field at $t=0$, compared to the crest. The expression of $T_{tr}$ is given by
$T_{tr}={\frac {\sin(\pi L/\beta \lambda )}{\pi L/\beta \lambda }}={\frac {\sin(\omega L/2v)}{\omega L/2v}}$

## Derivation

Consider a relativistic particle passing through an acceleration gap, and ignore the velocity change of the particle during this acceleration, such that the time is related with its longitudinal position by $z=vt$. The energy change $\Delta E$ is given by the integration:
${\begin{aligned}\Delta E&={\frac {qV_{0}}{L}}\int _{-L/2}^{L/2}\cos({\frac {\omega z}{v}}+\phi )dz\\&={\frac {qV_{0}}{L}}\int _{-L/2}^{L/2}\cos({\frac {\omega z}{v}})\cos \phi -{\frac {qV_{0}}{L}}\int _{-L/2}^{L/2}\sin({\frac {\omega z}{v}})\sin \phi dz\\&={\frac {qV_{0}}{L}}\int _{-L/2}^{L/2}\cos({\frac {\omega z}{v}})\cos \phi -0\\&=qV_{0}\cos \phi {\frac {\sin(\omega L/2v)}{\omega L/2v}}\end{aligned}}$
where the second integration vanishes because the odd-function of $z$.

## Property

$T_{tr}$ as a function of $L/\beta \lambda$ is shown in the figure to the right. To maximize the acceleration efficiency the length of the gap needs to be chose wisely to let $T_{tr}$be as close to unity as possible. This can be done by choosing $L$ to be $\beta \lambda /2$, for example.
However, if $L$ is too small, sparks will appear in the acceleration device as the gradient increases. There is little to be gained by reducing it to less than, say $\beta \lambda /4$.

## More complicated but more realistic model

The equations above assumed a uniform and constant electric field $E_{z}$in the derivation. Realistically, the field is a function of $r,z,t$, where $r$ is the radius of the particle trajectory w.r.t. the center of acceleration gap, $z$ the longitudinal position, and $t$ the time.
The modified transit time factor is then
$T_{tr}=T(k)I_{0}(Kr)=I_{0}(Kr){\frac {J_{0}(2\pi a/\lambda )}{I_{0}(Ka)}}{\frac {\sin(\pi g/\beta \lambda )}{\pi g/\beta \lambda }}$
where $I_{0}$ is the zeroth order modified Bessel function, $J_{0}$ the zeroth order Bessel function, $K={\frac {2\pi }{\gamma \beta \lambda }}$, and $a$ the drift-tube bore radius.

## Derivation of the above model

Let's consider the general expression for $E_{z}(r=0)=E(0,z)\cos(\omega t(z)+\phi )$
The integration in the simple model becomes
${\begin{aligned}\Delta E&=q\int _{-L/2}^{L/2}E(0,z)\cos(\omega t(z)+\phi )dz\\&=q\int _{-L/2}^{L/2}E(0,z)dz{\frac {\int _{-L/2}^{L/2}E(0,z)\cos \omega t(z)dz}{\int _{-L/2}^{L/2}E(0,z)dz}}\end{aligned}}$
Define the axial RF voltage $V_{0}$ as $V_{0}=\int _{-L/2}^{L/2}E(0,z)dz$, and the transit time factor as
$T_{tr}={\frac {\int _{-L/2}^{L/2}E(0,z)\cos \omega t(z)dz}{\int _{-L/2}^{L/2}E(0,z)dz}}$,
the expression of $\Delta E$ is then the same as the first equation of this first section's first equation.
Now since the integration is only effective in the field region between $-{\frac {L}{2}}$ to ${\frac {L}{2}}$, the limits can actually be expanded to infinity, such that
${\begin{aligned}V_{0}T_{tr}&=\int _{-\infty }^{\infty }E(0,z)\cos(\omega t(z))dz\\&=\int _{-\infty }^{\infty }E(0,z)\cos(kz)dz\end{aligned}}$
where $k=2\pi /\beta \lambda$ is the wave number. Noticing that the integral has a form of the Fourier cosine integral, the Fourier transform can be calculated by
${\begin{aligned}E(0,z)&={\frac {V_{0}}{2\pi }}\int _{-\infty }^{\infty }T(k)\cos(kz)dk\end{aligned}}$
expanding this expression to off-axis regions, then
${\begin{aligned}E(r,z)&={\frac {V_{0}}{2\pi }}\int _{-\infty }^{\infty }T(r,k)\cos(kz)dk\end{aligned}}$
This expression has to satisfy the wave equation, which is given by
${\begin{aligned}{\frac {\partial ^{2}E_{z}}{\partial r^{2}}}+{\frac {1}{r}}{\frac {\partial E_{z}}{\partial r}}+{\frac {1}{r^{2}}}{\frac {\partial ^{2}E_{z}}{\partial \varphi ^{2}}}+{\frac {\partial ^{2}E_{z}}{\partial z^{2}}}-{\frac {1}{c^{2}}}{\frac {\partial ^{2}E_{z}}{\partial t^{2}}}=0\end{aligned}}$
in cylindrical coordinates. Noticing the azimuthal symmetry of the system, and applying expression of ${\begin{aligned}E(r,z)\end{aligned}}$, the wave equation becomes
${\begin{aligned}{\frac {\partial ^{2}T_{tr}}{\partial \rho ^{2}}}+{\frac {1}{\rho }}{\frac {\partial T_{tr}}{\partial \rho }}+T_{tr}=0\end{aligned}}$
where $\rho ^{2}=K^{2}r^{2}$