Transit time factor,
T
t
r
{\displaystyle T_{tr}}
, characterizes the energy gain of a particle passing through an acceleration gap. The energy change
Δ
E
{\displaystyle \Delta E}
of the particle is given by
Δ
E
=
q
V
0
T
t
r
cos
ϕ
{\displaystyle \Delta E=qV_{0}T_{tr}\cos \phi }
where
q
{\displaystyle q}
is the charge of the particle,
V
0
{\displaystyle V_{0}}
the maximum voltage difference between the gap,
ϕ
{\displaystyle \phi }
the initial phase, defined as the phase of the oscillating field at
t
=
0
{\displaystyle t=0}
, compared to the crest. The expression of
T
t
r
{\displaystyle T_{tr}}
is given by
T
t
r
=
sin
(
π
L
/
β
λ
)
π
L
/
β
λ
=
sin
(
ω
L
/
2
v
)
ω
L
/
2
v
{\displaystyle T_{tr}={\frac {\sin(\pi L/\beta \lambda )}{\pi L/\beta \lambda }}={\frac {\sin(\omega L/2v)}{\omega L/2v}}}
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
=
v
t
{\displaystyle z=vt}
. The energy change
Δ
E
{\displaystyle \Delta E}
is given by the integration:
Δ
E
=
q
V
0
L
∫
−
L
/
2
L
/
2
cos
(
ω
z
v
+
ϕ
)
d
z
=
q
V
0
L
∫
−
L
/
2
L
/
2
cos
(
ω
z
v
)
cos
ϕ
−
q
V
0
L
∫
−
L
/
2
L
/
2
sin
(
ω
z
v
)
sin
ϕ
d
z
=
q
V
0
L
∫
−
L
/
2
L
/
2
cos
(
ω
z
v
)
cos
ϕ
−
0
=
q
V
0
cos
ϕ
sin
(
ω
L
/
2
v
)
ω
L
/
2
v
{\displaystyle {\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
{\displaystyle z}
.
T
t
r
{\displaystyle T_{tr}}
as a function of
L
/
β
λ
{\displaystyle 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
t
r
{\displaystyle T_{tr}}
be as close to unity as possible. This can be done by choosing
L
{\displaystyle L}
to be
β
λ
/
2
{\displaystyle \beta \lambda /2}
, for example.
However, if
L
{\displaystyle 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
β
λ
/
4
{\displaystyle \beta \lambda /4}
.
The equations above assumed a uniform and constant electric field
E
z
{\displaystyle E_{z}}
in the derivation. Realistically, the field is a function of
r
,
z
,
t
{\displaystyle r,z,t}
, where
r
{\displaystyle r}
is the radius of the particle trajectory w.r.t. the center of acceleration gap,
z
{\displaystyle z}
the longitudinal position, and
t
{\displaystyle t}
the time.
The modified transit time factor is then
T
t
r
=
T
(
k
)
I
0
(
K
r
)
=
I
0
(
K
r
)
J
0
(
2
π
a
/
λ
)
I
0
(
K
a
)
sin
(
π
g
/
β
λ
)
π
g
/
β
λ
{\displaystyle 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
{\displaystyle I_{0}}
is the zeroth order modified Bessel function,
J
0
{\displaystyle J_{0}}
the zeroth order Bessel function,
K
=
2
π
γ
β
λ
{\displaystyle K={\frac {2\pi }{\gamma \beta \lambda }}}
, and
a
{\displaystyle a}
the drift-tube bore radius.
Let's consider the general expression for
E
z
(
r
=
0
)
=
E
(
0
,
z
)
cos
(
ω
t
(
z
)
+
ϕ
)
{\displaystyle E_{z}(r=0)=E(0,z)\cos(\omega t(z)+\phi )}
The integration in the simple model becomes
Δ
E
=
q
∫
−
L
/
2
L
/
2
E
(
0
,
z
)
cos
(
ω
t
(
z
)
+
ϕ
)
d
z
=
q
∫
−
L
/
2
L
/
2
E
(
0
,
z
)
d
z
∫
−
L
/
2
L
/
2
E
(
0
,
z
)
cos
ω
t
(
z
)
d
z
∫
−
L
/
2
L
/
2
E
(
0
,
z
)
d
z
{\displaystyle {\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
{\displaystyle V_{0}}
as
V
0
=
∫
−
L
/
2
L
/
2
E
(
0
,
z
)
d
z
{\displaystyle V_{0}=\int _{-L/2}^{L/2}E(0,z)dz}
, and the transit time factor as
T
t
r
=
∫
−
L
/
2
L
/
2
E
(
0
,
z
)
cos
ω
t
(
z
)
d
z
∫
−
L
/
2
L
/
2
E
(
0
,
z
)
d
z
{\displaystyle 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
Δ
E
{\displaystyle \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
−
L
2
{\displaystyle -{\frac {L}{2}}}
to
L
2
{\displaystyle {\frac {L}{2}}}
, the limits can actually be expanded to infinity, such that
V
0
T
t
r
=
∫
−
∞
∞
E
(
0
,
z
)
cos
(
ω
t
(
z
)
)
d
z
=
∫
−
∞
∞
E
(
0
,
z
)
cos
(
k
z
)
d
z
{\displaystyle {\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
π
/
β
λ
{\displaystyle 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
E
(
0
,
z
)
=
V
0
2
π
∫
−
∞
∞
T
(
k
)
cos
(
k
z
)
d
k
{\displaystyle {\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
E
(
r
,
z
)
=
V
0
2
π
∫
−
∞
∞
T
(
r
,
k
)
cos
(
k
z
)
d
k
{\displaystyle {\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
∂
2
E
z
∂
r
2
+
1
r
∂
E
z
∂
r
+
1
r
2
∂
2
E
z
∂
φ
2
+
∂
2
E
z
∂
z
2
−
1
c
2
∂
2
E
z
∂
t
2
=
0
{\displaystyle {\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
E
(
r
,
z
)
{\displaystyle {\begin{aligned}E(r,z)\end{aligned}}}
, the wave equation becomes
∂
2
T
t
r
∂
ρ
2
+
1
ρ
∂
T
t
r
∂
ρ
+
T
t
r
=
0
{\displaystyle {\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
ρ
2
=
K
2
r
2
{\displaystyle \rho ^{2}=K^{2}r^{2}}