Some LHC Calculations

The LHC is currently running at 7 TeV, giving a relativistic gamma factor of 3730:

E=γmc2→γ=Emc2→γ(3.5TeV)=3730E = \gamma mc^2 \rightarrow \gamma = {E \over mc^2} \rightarrow \gamma(3.5\text{TeV}) = 3730

This means the protons are moving at 99.999996% the speed of light:

γ=11−β2→β=1−1γ2≈0.99999996\gamma = {1 \over \sqrt{1 - \beta^2}} \rightarrow \beta = \sqrt{1 - {1 \over \gamma^2}} \approx 0.99999996

Alright, now let’s look at the collision rate, assuming a recent luminosity of 1e33:

Rtotal=Lσ=(1033cm−2s)(110mb)(10−27cm2mb)R_\text{total} = \mathcal{L} \sigma = (10^{33} {\text{cm}^{-2} \over \text{s}})(110\text{mb})({10^{-27}\text{cm}^2 \over \text{mb}})

which gives approximately 100 million interactions per second.

The cross section for the events I’m looking at is $\sigma_{WZ\rightarrow 3\ell\nu} \approx 0.5 \text{pb}$, so

R(WZ→3ℓν)=(1033cm−2s)(0.5pb)(10−36cm2mb)R(WZ\to 3\ell\nu) = (10^{33} {\text{cm}^{-2} \over \text{s}})(0.5\text{pb})({10^{-36}\text{cm}^2 \over \text{mb}})

yielding 0.0005 Hz or 40 WZ events per day.

So how many of these events will we have collected by the end of the 2011 pp run?

NWZ→3ℓν≈(5fb−1)(0.5pb)=2500 eventsN_{WZ\rightarrow 3\ell\nu} \approx (5 {\text{fb}^{-1}}) (0.5 \text{pb}) = \text{2500 events}

By the time we make all the required cuts to account for the acceptance of the detector and rejection of backgrounds, however, we end up with something on the order of only 100 events in our final sample.