Luminosity for storage ring Why is the Luminosity $\mathcal{L}$ for a storage ring $\mathcal{L}=fn\frac{N_1N_2}{A}$, where $f$ is the rotation frequency, $N_i$ is the amount of particles in the bunches, $A$ is the cross sectional area and $n$ is the number of bunches per beam. Isn‘t the Luminosity $\mathcal{L}=\Phi N_t$ defined as the product of the flux with the number of targets? Because then I would say that the flux is $\Phi= fn\frac{N_1}{A}$ and the number of targets is $N_t=nN_2$ but that would give $\mathcal{L}=fn^2\frac{N_1N_2}{A}$.
 A: If the beams overlapped all around a collider ring, then
$$\mathcal{L}=fn^2\frac{N_1N_2}{A}$$
would be correct, but the interactions would occur at $n$ locations around the ring, not just inside the detectors, so most of that luminosity would be wasted. (Not to mention that the beam lifetimes would likely be horribly degraded by all the beam-beam interactions.) In real colliders, the beams are separated except at the collision points inside the detectors.
There is only ever one "target" bunch at a time at a single collision point, so the number of targets inside a single detector is always just $N_2$, not $n N_2$, and the luminosity is just
$$\mathcal{L}=fn\frac{N_1 N_2}{A}$$
Another way to think about it is to imagine you are filming the collision point with a frame rate of $fn$. What you would see is an unchanging image of one bunch colliding with another bunch. (The actual bunches would change every frame, but since they are all identical, it doesn't matter.) Each frame would correspond to an integrated luminosity of
$$\mathcal{L}_{int}=\frac{N_1 N_2}{A}$$
Since this is happening with a frequency $fn$, the luminosity is
$$\mathcal{L}=fn\,\mathcal{L}_{int}=fn\frac{N_1 N_2}{A}$$.
