Here is exercise 2.3 in the original list:
A beam of protons with energy E1=20 GeV is brought to collide head-on with another beam of protons of energy E2=5 GeV. Determine:
a) The CM energy;
b) the CM boost in the lab frame;
c) the angle which a ultra-relativistic particle makes in the lab frame if produced at 90 degrees in the CM;
d) the energy which a proton beam must possess in order to produce the same CM energy in a collision with a fixed target.
So the problem is one of simple relativistic kinematics. Let us work our way through the four questions.
a) In the laboratory frame E(lab)=20+5=25 GeV, while P(lab) can be obtained by taking the difference of the momenta of the protons in each beam. We can use the relationship between energies and masses, P^2=E^2-M^2, and remembering that the proton mass is M=0.938 GeV: P(lab)=(400-0.938^2)^0.5-(25-0.938^2)^0.5=15.066 GeV.
In the CM the energy can be obtained recalling that P(CM)=0 by definition, so that E(CM)^2-P(CM)^2=E(CM)^2=E(lab)^2-P(lab)^2 , since the effective invariant mass E^2-P^2 is a relativistically invariant quantity. This yields E(CM)=19.95 GeV.
b) To find the boost of the CM in the laboratory frame, we may take the 20 GeV proton: it has a four-momentum given by (E,px,py,pz)=(20,19.978,0,0), where we have arbitrarily chosen x as the axis coincident with the beam direction. In the CM instead the 20 GeV four-momentum can be written as (E(CM)/2,(E(CM)^2/4-M^2,0,0) = (9.975,9.93,0,0). Using Lorentz transformations we may write
E1* = gamma (E1-beta P1),
P1* = gamma (p1-beta E1)
where starred variables refer to the CM frame, and unstarred ones are measured in the laboratory frame. Gamma and beta are the usual relativistic factors.
By simple algebra, eliminating gamma one finds p1*E1 - beta p1 p1*= p1 E1*- beta E1 E1*, from which one immediately gets beta = 0.608.
c) For an ultra-relativistic particle, which has beta=1, and p*=E*, emitted at 90 degrees in the CM frame, in the laboratory frame one measures the angle theta=atan(Py/Px) between the particle and the beam axis. If the particle in the CM has four-momentum (E*,0,E*,0) (having taken the y direction as the one of 90 degree emission), in the laboratory it will be
px = gamma (px* + beta E*)=beta gamma E*
py = py*=E*.
One thus finds that the CM angle is theta=atan[1/(beta gamma)]=0.92 radians.
d) The energy released in a fixed-targed collision is given by (2 M E(beam))^0.5, and since this must equate to 19.95 GeV, E(beam) must equate to 19.95^2/1.976=201.4 GeV. One thus realizes the importance of head-on collisions and their advantage with respect to fixed target collisions, which provide much smaller energy for given beam energy.
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