 Solution

For two identical, head-on beams with Gaussian distributions, integrating the product of their normalized transverse densities gives an overlap $1/(4\pi {\sigma }_x{\sigma }_y)$. Each of the $n_b$ colliding bunch pairs crosses once per revolution, with $f_{\mathrm{rev}}\simeq c/C$. ThusThereforeThe factor $4\pi$ accounts for the convolution of two beams, rather than the $2\pi$ normalization of just one beam. The result assumes all stated bunch pairs collide at the chosen interaction point, speeds approximately $c$, zero crossing angle and negligible hourglass reduction. Additional interaction points do not multiply the local collider luminosity at a single detector.