Solutions to the exercises

Solutions to Exercise 1:

a) Outer radius of the vertex detector: 13 cm

b) The magnetic field comes out of the picture

c) Five particles are produced right after the interaction. One of them, the Ks, decays into two pions. 

d) The Ks is neutral because it's trajectory is strait.

e) If the Ks escapes from the interaction point with the velocity of light it gets  s = c*t = 2.68 cm, where t is the lifetime of the particle.

f) it actually gets about 17 cm. The lifetime of the particle is refered to the system in which the particle is at rest. In the detector system, the lifetime of the particle is longer due to time dilatation! Or from the point of Lorentz contraction: looking from a inertial system which moves with the Ks the distance of 17 cm looks  like 

g)  l' = l / g    g: gamma factor. For l = 2.68 cm and l' = 17 cm         g = 6,34   ==>   b =   0.918 c

h) The track in the picture is only a projection in the x-y-plane. The Ks has also a component in the z-direction. So the track of the Ks and therefore it' velocity is even bigger!

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Solutions to Exercise 2a:

Sagitta s = 1.0306 mm and L = 13.687 cm

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Solutions to Exercise 2b:

                   

but for small q you find also the relation        

Combining the two relations gives the requested equation   

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Solutions to exercise 2c:     r = 227 cm

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Solutions to exercise 3a:

centripetal force:   

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Solution to exercise 3b:

Lorentz force:   

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Solution to Exercise 3c:

q v B = p v/r    =>    q B = p / r        == >    p = r q B

pt = 1,047 GeV/c   

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Solution to Exercise 3d:

pt / p = sin q    =>    p = pt / sin q    = 1,129 GeV/c

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Solution to exercise 3e:

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Solution to exercise 4:

The mass of the top quark is about 175 GeV, and is therefore much too big for a top-antitop pair to have been produced in a 200 GeV collision. The masses of all other quarks are all small enough. In particular, the mass of the bottom quark is about 4.4 GeV So the total of the energy of a bottom antibottom pair would be around 9 GeV. There is a fairly high probability, that a bottom-antibottom-state has been observed.

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Solutions to exercise 5:

a) A single muon cannot be produced on its own as charge and muon-lepton number must be conserved. 

b)      

The mass of the muon is very small compared to the total energy of the particle. p = 99.99994 GeV/c

c) Combining the formulas for the centripetal and the Lorentz force, you get an equation for the momentum  p = r q B. You can calculate the radius r = p /qB = 222 m 

d) In the tracking problem you derived the relation between the radius of  the track, the sagitta and L:   . The radial diameter of the time projection chamber has to be more than 3 meters!

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