326
6 Special Theory of Relativity
6.68 A proton of momentum p large compared with its rest mass M, collides with
a proton inside a target nucleus with Fermi momentum p f . Find the available
kinetic energy in the collision, as compared with that for a free-nucleon target,
when p and p f are (a) parallel (b) anti parallel (c) orthogonal.
6.69 An antiproton of momentum 5 GeV/c suffers a scattering. The angles of the
recoil proton and scattered antiproton are found to be 82
◦ and 2
◦ 30
with
respect to the incident direction. Show that the event is consistent with an
elastic scattering of an antiproton with a free proton.
6.70 Show that if E is the ultra-relativistic laboratory energy of electrons incident
on a nucleus of mass M, the nucleus will acquire kinetic energy
E N = (E
2
/Mc
2 )(1 − cos θ)/(1 + E(1 − cos θ )/Mc
2 )
where θ is the scattering angle.
6.71 A particle of mass M m e scatters elastically from an electron. If the incident particle’s momentum is p and the scattered electron’s relativistic energy
is E and φ is the angle the electron makes with the incident particle, show that
M = P[{[E + m e ]/[E − m e ]} cos
2
φ − 1]]
1/2
6.72 A neutrino of energy 2 GeV collides with an electron. Calculate the maximum
momentum transfer to the electron.
6.73 A particle of mass m 1 collides elastically target particle of mass m 2 at relativistic energy. Show that the maximum angle at which m 1 is scattered in the
lab system is dependent only on the masses of particles provided m 1 > m 2
6.74 Show that if energy ν(> m e c
2 ) and momentum q are transferred to a free stationary electron the four-momentum transfer squared is given by q
2
= −2m e ν
6.75 A photon of energy E travelling in the +x direction collides elastically with
an electron of mass m moving in the opposite direction. After the collision,
the photon travels back along the –x direction with the same energy E.
(a) Use the conservation of energy and momentum to demonstrate that the
initial and final electron momenta are equal and opposite and of magnitude E/c.
(b) Hence show that the electron speed is given by
v/c = (1 + (m c
2
/E)
2 )
−1/2
[adapted from the University of Manchester 2008]
6.2.4 Invariance Principle
6.76 Use the invariance of scalar product of two four-vectors under Lorentz transformation to obtain the expression for Compton scattering wavelength shift.
6.77 Show that for a high energy electron scattering at an angle θ , the value of
the squared four-momentum transfer is given approximately by Q
2
= 2E
2
6 Special Theory of Relativity
6.68 A proton of momentum p large compared with its rest mass M, collides with
a proton inside a target nucleus with Fermi momentum p f . Find the available
kinetic energy in the collision, as compared with that for a free-nucleon target,
when p and p f are (a) parallel (b) anti parallel (c) orthogonal.
6.69 An antiproton of momentum 5 GeV/c suffers a scattering. The angles of the
recoil proton and scattered antiproton are found to be 82
◦ and 2
◦ 30
with
respect to the incident direction. Show that the event is consistent with an
elastic scattering of an antiproton with a free proton.
6.70 Show that if E is the ultra-relativistic laboratory energy of electrons incident
on a nucleus of mass M, the nucleus will acquire kinetic energy
E N = (E
2
/Mc
2 )(1 − cos θ)/(1 + E(1 − cos θ )/Mc
2 )
where θ is the scattering angle.
6.71 A particle of mass M m e scatters elastically from an electron. If the incident particle’s momentum is p and the scattered electron’s relativistic energy
is E and φ is the angle the electron makes with the incident particle, show that
M = P[{[E + m e ]/[E − m e ]} cos
2
φ − 1]]
1/2
6.72 A neutrino of energy 2 GeV collides with an electron. Calculate the maximum
momentum transfer to the electron.
6.73 A particle of mass m 1 collides elastically target particle of mass m 2 at relativistic energy. Show that the maximum angle at which m 1 is scattered in the
lab system is dependent only on the masses of particles provided m 1 > m 2
6.74 Show that if energy ν(> m e c
2 ) and momentum q are transferred to a free stationary electron the four-momentum transfer squared is given by q
2
= −2m e ν
6.75 A photon of energy E travelling in the +x direction collides elastically with
an electron of mass m moving in the opposite direction. After the collision,
the photon travels back along the –x direction with the same energy E.
(a) Use the conservation of energy and momentum to demonstrate that the
initial and final electron momenta are equal and opposite and of magnitude E/c.
(b) Hence show that the electron speed is given by
v/c = (1 + (m c
2
/E)
2 )
−1/2
[adapted from the University of Manchester 2008]
6.2.4 Invariance Principle
6.76 Use the invariance of scalar product of two four-vectors under Lorentz transformation to obtain the expression for Compton scattering wavelength shift.
6.77 Show that for a high energy electron scattering at an angle θ , the value of
the squared four-momentum transfer is given approximately by Q
2
= 2E
2