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In a collider experiment, the Luminosity is defined as the proportionality factor between the interaction rate and the interaction cross section $dN/dt = L\times\sigma$, with units of $cm^{-2}s^{-1}$. I was curious to see how this quantity is related to the flux of incoming particles since it has the same units as the flux. Doing some research I came across this paper that states:

A priori the two beams have different distribution functions and a different number of particles in the beams. The overlap integral which is proportional to the luminosity ($L$) we can then write as: $$ L ∝ K ·\int\int\int\int_{-\infty}^{\infty}\ \rho_{1}(x, y, s, −s_0) \rho_2(x, y, s, s_0) dxdydsds_0 $$ where $\rho_{1,2}$ are the ) are the time-dependent beam density distribution functions. We assume, that the two bunches meet at $s_0 = 0$. Because the beams are moving against each other, we have to multiply this expression with a kinematic factor $K$

here they used $s_0 = ct$ as a "time" coordinate. I believe in this equation is established the relation between the fluxes of the two beams but it seems to have probability theory too. However, I do not get the basic reasoning that leads to this equation. Could someone explain me?

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It is related to the density distribution and the kinematic factor $K$. For head-on collision we have $K = 2v$ with $v$ the beam velocity. Now the particle flux is (as seen by the other beam):

$$ \Phi_{1, 2}(x, y, s, s_0) = 2v\cdot\rho_{1,2}(x, y, s, \pm s_0) $$

It depends on the observer's position $(x, y, s)$ as well as how far the beam/bunch is away ($s_0$).

Following the paper a bit further, they calculate the luminosity for two Gaussian beam profiles (section 4). It turns out to be:

$$ \mathcal{L} = \frac{N_1 N_2 f N_b}{4\pi\sigma_x\sigma_y} $$

Now the average particle flux is (averaged over one revolution period as well as the transverse dimensions), as seen by the other beam:

$$ \phi_{1,2} = \frac{N_{1,2}fN_b}{4\pi\sigma_x\sigma_y} $$

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The idea here is that the probability of a particle to collide with an incoming distributions will increase if it goes deeper or stays longer within the distribution. The overlap integral only quantifies this increment, you can see it as purely geometrical if you want.

The actual probabilistic part of a collision to take place, is factored out into the cross section, which you need if you want to compute the average number of a particular events per bunch crossing (or unit time).

For example if you want to get the total number of collision produced by the LHC over some period of time T you will do:

$$N_\text{evt}=\int_\text{T} \mathcal L(t) \, \sigma_\text{pp} \, d\text{t} = \bar{\mathcal L}\, \sigma_\text{pp}\,T$$ where $\sigma_\text{pp}$ is the total proton-proton cross section at the given energy.

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