\gamma^2\vartheta^2> (\chi/u)^{2/3} \gg 1$. Under these conditions the formation length $l_{f}=l_{f0}/\zeta$ decreases as $(\chi/u)^{2/3}$. The same inhibition factor acquires the bremsstrahlung probability [@BKS4].

We consider now the spectral distribution of radiation probability in the case $\chi \ll 1$ (this condition is fulfilled in all existing installations), so $$\chi \sim \alpha N_c \gamma
\frac{\lambda_c^2}{(\sigma_z+\sigma_y)l} \ll 1. \label{2.4}$$ Only the soft photons $(\omega \leq \chi \varepsilon \ll
\varepsilon)$ contribute to the [*coherent*]{} radiation (“beamstrahlung”) while the hard photon region $\omega \gg \chi
\varepsilon$ is suppressed exponentially as it is known from the classical radiation theory. As it was mentioned in the soft photon region $(\omega \leq \chi \varepsilon \ll \varepsilon)$, the spectral probability of [*bremsstrahlung*]{} is suppressed by the factor $(\omega/\varepsilon\chi)^{2/3}$ only. On the contrary, the spectral probability of the bremsstrahlung is negligible comparing with the beamstrahlung taking into consideration that the mean square of multiple scattering angle during all time of beam collisions is small comparing with the value $1/\gamma^2$: $$\gamma^2 <\vartheta_s^2>=\frac{<q_s^2>}{m^2} \simeq
\frac{8\alpha^2N_c \lambda_c^2}{\sigma_z\sigma_y}L \ll 1,
\label{2.5}$$ where $L$ is the characteristic logarithm of scattering problem (in the typical experimental condition $L \sim 10$).

It was supposed in the above estimations of beamstrahlung probability that the radiation formation length is 