In both cases they had a resonance decaying into two leptons, and the detecting system consisted of two arms, covering a relatively narrow angular acceptance.

When they motivate the choice of the opening angle between the arms of the spectrometer, they do the calculation in the extreme assumption that, in the center-of-mass rest frame, the leptons are produced at 90 degrees with respect to the beam. Then of course you boost the system by the proton momentum, and you get the MAXIMUM angle of the leptons in the lab frame (tg theta < 1/gamma*beta = M_proton / P_proton).

Fine.

But of course the vast majority of the leptons are not produced at 90 degrees, and are so lost.

The authors of those experiments motivate this choice by the fact that at central angles the hadronic background is huge and so they have to restrict the acceptance as much as possible to the extreme corners of the angular distribution. Fine.

Now, my question is very stupid and only concerns basic kinematics: I would like to estimate the fraction of leptons inside the detector acceptance (that I know from the original articles), knowing their angular distribution in the center-of-mass frame.

(It should be ~ 1+cos^2 theta, in the case of J/psi and Y, but I can maybe start with the simplified assumption of isotropy, just to get the order of magnitude.)

My strategy is the following:

I write dGamma/dCosTheta(lab) = [dGamma/dCosTheta(cm)]*[dCosTheta(cm)/dCosTheta(lab)], and since I already know dGamma/dCosTheta(cm), the game becomes finding CosTheta(cm) as a function of CosTheta(lab).

This should be some shallow algebra after applying the Lorentz transformations, but I'm having difficulties.

Applying the Lorentz transformation I found that tgTheta(lab)=(1/gamma)*senTheta(cm)/[cosTheta(cm)+beta].

I rewrote all the sines into sqrt(1-cos^2) and tried to do some manipulation, but I get a very ugly second order equation with cumbersome coefficients.

Is there an easy way of performing this calculation?

Thanks a lot!