$$b=r_0\cos\frac{\vartheta_0}{2}.$$

At the vertex $p_r$ becomes zero for a moment, so that during the interaction process the kinetic energy is given only by the component orthogonal to it, namely $\overline{\rm
BA}=p\cos\frac{\vartheta_0}{2}$. The energy, contributed from both ingoing particles to the virtual exchange particle is therefore equal to $${\cal
E}=2\left(\sqrt{p^2+m^2}-\sqrt{p^2\cos^2\frac{\vartheta_0}{2}+m^2}\,\right).$$

Under the assumption $p\gg m$ the exchanged energy $\cal E$ can be expanded in two different ways, according to the scattering angle $\vartheta_0$. If $\vartheta_0$ is small, ${\cal
E}\approx2p(1-\cos\frac{\vartheta_0}{2})$ is small, too. The particles do not slow down much and remain relativistic and the considerations of the foregoing subsection, involving a Newtonian potential, become inappropriate.

In the other case, when the collision is almost central and $\vartheta_0$ is close enough to $180^\circ$, so that $\cos\frac{\vartheta_0}{2}\ll\frac{m}{p}$, the particles slow down to nonrelativistic speed and the calculations with Newtonian gravity are more reliable. Now the energy transfer $$\label{et}
{\cal
E}\approx2p\left[1-\frac{m}{p}\left(1+\frac{1}{2}\frac{p^2}{m^2}\cos^2\frac{\vartheta}{2}
\right)\right]$$ is large and for the wavelength associated with the exchange particle we can again use the relativistic relation (\[l\]), $\lambda_0=1/\cal E,$ giving an estimate for the minimal distance $2r_0$ of the particles.

An expansion of the potential energy $\Delta E_1$ of the rest masses of the particles in the gravitational field of 