Respuesta :
Answer:
Chapter 1
1. Show that the Lorentz transformation is such that the velocity of a light ray
travelling in the x direction is the same for the observer in the frame S and for
the observer in the frame S
.
Solution: Consider a light ray travelling in the x direction. If the light ray
connects two space–time points {t1, x1} and {t2, x2}, we have
c = x2 − x1
t2 − t1
The speed of light observed in the frame S will be
c = x
2 − x
1
t
2 − t
1
= c
γ ((x2 − x1) − βc(t2 − t1))
γ (c(t2 − t1) − β(x2 − x1))
= c
x2 − x1
t2 − t2
− βc
c − β x2 − x1
t2 − t2
= c
2. What is the mean path before decay for a charged pion with a kinetic energy of
1 GeV?
Solution: The pion has a lifetime 2.6 × 10–8 s and a mass of 139.6 MeV. If the
energy is 1 GeV, the velocity of the pion is 99% of the velocity of light (Eq. 1.4).
The mean path before decay is
= 0.99 c γ τ
= 0.99 c
1000 + 139.6
139.6
2.6 10−8 = 63 m
S. Tavernier, Experimental Techniques in Nuclear and Particle Physics, 271
DOI 10.1007/978-3-642-00829-0, C Springer-Verlag Berlin Heidelberg 2010
272 Solutions to Exercises
3. Show that the relativistic expression for the kinetic energy of a particle (Eq. 1.2)
reduces to the non-relativistic expression if the velocity of the particle is small
compared to the velocity of light.
Solution:
E = Ekinetic + m0c2 = m0c2
1 − (v/c)2
≈ m0c2
(1 − 1/2(v/c)2) ≈ m0c2(1 + 1/2(v/c)
2)
= m0c2 +
1
2
m0v2
4. For a Poisson distribution with average value 16, calculate the probability to
observe 12, 16 and 20 as measured value. Calculate the probability density function for a Gaussian distribution with average value 16 and dispersion 4, for the
values x = 12, 16 and 20. Compare the results.
Solution: For a Poisson distribution P(12) = 0.0829, P(16) = 0.1024, P(20) =
0.0418
For a Gaussian distribution, f(12) = 0.0605, f(16) = 0.0997, f(20) = 0.0605
5. Consider a very short-lived particle of mass M decaying into two long-lived particles 1 and 2. Assume you can measure accurately the energies and momenta of
the two long-lived particles. How will you calculate the mass of the short-lived
particle from the known energies and momenta of the two long-lived objects?
Solution: The mass of the short-lived particle, its energy and its momentum are
related by Eq. (1.1). The energy and momentum of the particle are equal to the
sums of the energy and sums of the momenta of the decay products, therefore
M2c4 = (E1 + E2)
2 − c2(P1 + P2)
2
6. Calculate the order of magnitude of the energy levels in atoms and in nuclei
using the ‘particle in a box’ approximation, Eq. (1.9). Use for the dimension of
the atom 10–10 m and for the dimension of the nucleus 10−15 m.
Solution: Atomic energy levels: ≈40 eV; nuclear energy levels: ≈400 MeV.
7 . Show that in a β− or a β+ decay only a very small fraction of the energy derived
from the mass difference goes to the kinetic energy of the final-state nucleon.
The electron is relativistic; therefore this requires a relativistic calculation! Hint:
the 3-body problem can be reduced to a 2-body problem by considering the
electron–neutrino system as one object with a mass of a few MeV.
Solution. Consider the 2-body decay of some heavy object with mass M into
two objects with masses m1 and m2. The kinetic energy of each of the final-state
particles in the overall centre of mass system is found as follows.
Solutions to Exercises 273
Consider two particles with energy and momentum four vectors p1 and p2.
The symbol pi stands for the four-vector {Ei,cpi}. The energy E appearing in this
expression is the total energy E, i.e. the rest energy mc2 plus the kinetic energy.
The four-vector product (p1.p2) is defined as
(p1.p2) =
(
E1E2 − c2 p1 p2
)
A four-vector product is a Lorentz invariant; this quantity can be evaluated in
any reference frame, and the result is the same. Consider now the quantity
(p1.p2)
m1c2
This is a Lorentz invariant. Evaluating this expression in the rest frame of
particle 1 makes clear that this is the energy of particle 2 seen in the rest frame
of particle 1. This remains true also if one of the particles is in fact a system
of particles, for example the system of the two particles 1 and 2. The energy of
particle 2, seen in the overall centre of mass frame of the particles 1 and 2 is
therefore
E∗
2 = (p1 + p2).p2
(p1 + p2)2
We have the following relations:
(p1 + p2)
2 = M2c4
(p1.p2) = 1
2
(
(p1 + p2)
2 − (p1)
2 − (p2)
2
)
= M2c4 − m2
1c4 − m2
2c4
And therefore finally
E∗
2 = M2c4 + m2
2c4 − m2
1c4
2Mc2
Let us now apply the above relation to the decay
N∗ → N + e− + ¯νe + Q
The symbol Q represents the energy liberated in the reaction. Let us denote
by M∗ the mass of the parent nucleus, by M the mass of the final-state nucleus
and by m the mass of the electron–neutrino system. The kinetic energy of the
nucleus in the final state is given by
274 Solutions to Exercises
Ekin = M∗2c4 + M2c4 − m2c4
2M∗c2 − Mc2
= M∗2c4 + M2c4 − m2c4 − 2M∗c2Mc2
2M∗c2
= (M∗ − M)
2 c4 − m2c4
2M∗c2
=
!
mc2 + Q
