I'm giving a report on the LHCb for a detector physics class. I don't quite understand how b-bbar-pairs are produced at the LHCb. LHCb collides two symmetric proton beams and is supposed to work as a b factory. In many papers concerning the LHCb, for instance http://arxiv.org/abs/0910.1740v1" [Broken] a picure of the b-bbar-production cross section can be found. They all say that most b-bbar-pairs are produced under small or large polar angles, i.e. very close to the beam, thus justifying the forward detector design. It seems that such predictions can be obtained using a physics generator (Pythia) that uses some hadron interaction models. However, I would like to understand, if possible, the angular distribution of the products as a direct result of the p-p interaction, so that I can give a ("handwaving") explanation in my presentation. Unfortounately, I'm not very firm in theoretical hadron physics. I imagined something like this: A valence quark q from the first proton annihilates with a sea quark qbar from the second photon into a virtual Z0, which then creates a b-bbar-pair. The reason for the b-bbar-pair to move close to the beam in either direction would then be, that the first proton (with the valence quark) contributes much more momentum to the quark pair then the second proton (with the sea quark). The only Feynman diagram for the b-bbar-production at LHCb that I could find is from page 4 of http://www.fz-juelich.de/ikp/ghp2009/Talks/talk_ruf_thomas.pdf" [Broken]. However, here it seems that both protons simply radiate a gluon, and these then form the b-bbar-pair. This would, however, not explain the mentioned angular distribution. Is there any remotely easy way to understand this? What is the actual reaction taking place? Is there a simple Feynman diagram for this that I could show? Thanks in advance.