\Phi(\mbox{\boldmath$\varrho$})=
\frac{1}{(2\pi)^2}\int d^2q \exp(-i{\bf q}
\mbox{\boldmath$\varrho$})\exp\left[-\frac{q_z^2}{4\Sigma_z^2}
-\frac{q_y^2}{4\Sigma_y^2}\right]=
\frac{\Sigma_z\Sigma_y}{\pi}\exp[-\varrho_z^2\Sigma_z^2-\varrho_y^2\Sigma_y^2];
\nonumber \\
\hspace{-5mm}&& \Sigma_z^2=\frac{1}{2(\sigma_z^2+\Delta_z^2)},~
\Sigma_y^2=\frac{1}{2(\sigma_y^2+\Delta_y^2)},
\label{4.2}\end{aligned}$$

Below we consider the general situation when the axes of colliding beams are displaced with respect each other in the transverse plane by the vector ${\bf x}_0$ with the coordinates $z_0,y_0$. This displacement has essential influence on the luminosity. For the processes for which the short distances are essential only (e.g. double bremsstrahlung [@BK1]) the probability of process is the product of the cross section and luminosity. The geometrical luminosity per bunch, not taking into account the disruption effects, is given by $${\cal L}=N_cN_r \Phi({\bf x}_0),
\label{4.3}$$ where as above $N_r$ and $N_c$ are the number of particles in the radiating and target beams correspondingly. We will use the same definition for our case. Then we have $$dw_{\gamma}= \Phi({\bf x}_0)d\sigma_{\gamma} ,\quad d\sigma_1=
\Phi^{-1}({\bf x}_0) dw_1, \label{4.4}$$ where $dw_1$ is defined in Eq.(\[3.11\]).

We calculate first the function $F^{(1)}(\omega, \zeta)$ in Eq.(\[3.12\]) for the case of coaxial beams when ${\bf x}_0=0$. Passing on to the momentum representation with the help of formula (\[4.1\]) we find $$F^{(1)}(\omega, \zeta)= - \frac{1}{2\pi\zeta^2}\int w_r({\bf q})
w_c({\bf q}) F_2\left(\frac{q}{2\eta}\right) qdq d\varphi,
\label{4.5}$$ where $\eta=q_{min}\zeta$ is introduced in (\[3.12\]), $$\begin{aligned}
&& F_2\left(\frac{q}{2\eta}\right) = 