# The factor in Mott formula for relativistic Coulomb scattering

I have confronted a following problem while deriving the scattering on the Coulomb potential with charge $- Ze$ for Dirac particle with charge $e$ in Born approximation. The Mott cross-section formula has the following form:

$$\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega}=\frac{Z^2 \alpha^2}{4 p^2 \beta^2 \sin^4\left(\frac{\theta}{2}\right)}\left ( 1 - \beta^2 \sin^2 \frac{\theta}{2} \right )$$

where $\beta = \frac{v}{c}$ (see, for example, Drell, S. D. "Quantum electrodynamics at small distances." Annals of Physics 4.1 (1958): 75-86, $§ 25$).

This formula is very similar to the Rutherford formula for scattring on the Coulomb potential despite the factor $\left ( 1 - \beta^2 \sin^2 \frac{\theta}{2} \right )$. The question is: does the appearance of this factor for the relativistic particle have a simple physical meaning? In "Elements of Quantum Electrodynamics" by A. I. Achiezer it is said that this factor is simply kinematic, but there are no details on the matter nad this explanation does not help me. I tried performing relativistic transformation of the solid angle in the ordinary Rutherford formula and some other things, but nothing seems to work.