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

$$\Delta E = q V_0 T_{tr} \cos \phi$$

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:

$$\begin{align} \Delta E &= \frac{qV_0}{L}\int^{L/2}_{-L/2}\cos(\frac{\omega z}{v} + \phi)dz \\ &=\frac{qV_0}{L}\int^{L/2}_{-L/2}\cos(\frac{\omega z}{v})\cos\phi - \frac{qV_0}{L}\int^{L/2}_{-L/2}\sin(\frac{\omega z}{v})\sin\phi dz \\ &=\frac{qV_0}{L}\int^{L/2}_{-L/2}\cos(\frac{\omega z}{v})\cos\phi - 0 \\ &=qV_0\cos\phi\frac{\sin(\omega L/2v)}{\omega L/2v} \end{align}$$

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

$$\begin{align} \Delta E &= q\int^{L/2}_{-L/2}E(0,z)\cos(\omega t(z) + \phi)dz \\ &=q\int^{L/2}_{-L/2}E(0,z)dz\frac{\int^{L/2}_{-L/2}E(0,z)\cos\omega t(z)dz}{\int^{L/2}_{-L/2}E(0,z)dz} \end{align}$$

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

$$\begin{align} V_0 T_{tr} &= \int^{\infty}_{-\infty}E(0,z)\cos(\omega t(z)) dz\\ &=\int^{\infty}_{-\infty}E(0,z)\cos(kz) dz \end{align}$$

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

$$\begin{align} E(0,z) &= \frac{V_0}{2\pi}\int^{\infty}_{-\infty}T(k)\cos(kz)dk \end{align}$$

expanding this expression to off-axis regions, then

$$\begin{align} E(r,z) &= \frac{V_0}{2\pi}\int^{\infty}_{-\infty}T(r, k)\cos(kz)dk \end{align}$$

This expression has to satisfy the wave equation, which is given by

$$\begin{align} \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{align}$$

in cylindrical coordinates. Noticing the azimuthal symmetry of the system, and applying expression of $$\begin{align} E(r,z) \end{align}$$, the wave equation becomes

$$\begin{align} \frac{\partial^2 T_{tr}}{\partial \rho^2} + \frac{1}{\rho}\frac{\partial T_{tr}}{\partial \rho} + T_{tr} = 0

\end{align}$$

where $$\rho^2 = K^2 r^2$$