1. Units and special relativity
Learning Objective: be able to calculate the kinematics of 2-body interactions and decays.
1.1 Elementary particles
From a historical point of view, particle physics begins in the 3rd century BC, with the
Democritus particle theory. Democritus, a Greek philosopher, believed that the world was
made of indivisible “atoms”. This was, in fact, correct as we now know that the basic
components of matter are “elementary particles”. Such elementary particles do not have a
known substructure and are considered to be point-like. In the past century, we saw great
advancements in our understanding of the structure of matter. Broadly speaking, these
advancements were achieved by bombarding matter with “probes”.
In the case of a microscope, such a probe is a photon, which will be able to study structures
larger than the wavelength 𝜆𝜆 of the photons. If the probe is another type of particles, we
should use the associated De Broglie wavelength 𝜆𝜆 = ℎ/𝑝𝑝, where 𝑝𝑝 is the particle
momentum.
As an example, if we use visible light as a probe, such as in an optical microscope we can
explore structure of the size of ~500 nm, much larger than the size of an atom (compare with
the Bohr radius of 0.5 nm), but sufficient to study most biological objects. Using X-rays, we
can observe structures of the size of a few nm, which is why we use X-rays in applications
such as crystallography. If we want to probe the subatomic structure of the atom, we will
need shorter wavelengths and use probes of higher momentum. In the case of the Rutherford
experiment, alpha-particles were used with a kinetic energy of 5.30 MeV, corresponding to a
wavelength of about 6.3 fm, much smaller than the size of the atom, and sufficient to discover
the presence of a hard nucleus inside an atom.
Alpha particles in the Rutherford experiment, however, did not have sufficient momentum to
probe the nucleus structure. The story of particle physics and the quest for more energetic
probes and “higher resolution” experiments began over a century ago and is still ongoing with
the modern Large Hadron Collider.
1.2 Units of energy, momentum and mass
As in atomic and solid state physics, a useful unit of energy in particle and nuclear physics is
the electron volt (eV). This is the amount of kinetic energy gained by an electron when it is
accelerated through a potential difference of one volt. Normally the energies involved in
nuclear reactions are millions of electron volts (MeV) and in high energy particle interactions
they may be billions of electron volts or Giga electron volts (GeV = 109 eV). A convenient unit
for particle masses makes use of the Einstein mass-energy relationship
𝐸𝐸 = 𝑚𝑚𝑐𝑐 2
This yields a unit for mass expressed as energy divided by the square of the velocity of light,
MeV/c2 or GeV/c2. For example
proton mass = 938.3 MeV/c2
electron mass = 0.511 MeV/c2
This system of units is extended to momentum through the relativistic relationship for the
energy of a particle of rest mass m moving with momentum p,
𝐸𝐸 2 = 𝑝𝑝2 𝑐𝑐 2 + 𝑚𝑚2 𝑐𝑐 4 .
From this, it follows that if we express momentum in the units of energy divided by the
5
, velocity of light (GeV/c), we have a self-consistent system in which the velocity of light is
implicitly used, but its value does not have to be explicitly put in to the calculations.
In several textbooks, you will find that 𝑐𝑐 = 1, e.g. the speed of light is not explicitly indicated
in the formulae. This is a common notation used in particle physics, where particle masses
are expressed in GeV by implicitly multiplying their mass by 𝑐𝑐 2 . For clarity, I will try to avoid
this notation in my lectures.
A useful value to remember is that of ℏ𝑐𝑐 = 197 MeV fm
Example 1.1. Evaluate the de Broglie wavelength of an alpha particle of kinetic energy
𝐸𝐸𝐾𝐾 = 5.3 MeV.
[𝑚𝑚𝛼𝛼 = 6.64 × 10−27 kg = 3.73 GeV⁄c 2 ]
Solution:
Important: The alpha particle is non-relativistic. So we use the Galilean expression for its
kinetic energy 𝐸𝐸𝐾𝐾 = 𝑝𝑝2 ⁄2𝑚𝑚 . We will soon see that this is not the case when particles are
relativistic!
1. First we solve using SI units for all the quantities
We calculate the momentum of the alpha particle as
𝑝𝑝 = �2 𝑚𝑚 𝐸𝐸𝐾𝐾
= �2 × ( 6.64 × 10−27 ) × ( 5.3 × 106 ) × (1.6 × 10−19 ) kg m s −1
= 1.06 × 10−19 kg m s −1
We use the De Broglie formula to evaluate 𝜆𝜆
ℎ 6.63 × 10−34 kg m2 s −1
𝜆𝜆 = = = 6.25 × 10−15 m = 6.25 fm
𝑝𝑝 1.06 × 10−19 kg m s −1
2. We can also calculate the De Broglie wavelength using particle physics units
Instead of 𝑝𝑝 we work with 𝑝𝑝𝑝𝑝 which is
𝑝𝑝𝑝𝑝 = �2 𝑚𝑚𝑐𝑐 2 𝐸𝐸𝐾𝐾 = �2 × (3730 MeV) × 5.3 MeV = 198 MeV
ℎ 2𝜋𝜋 ℏ𝑐𝑐 197 MeV fm
𝜆𝜆 = = = 6.28 = 6.25 fm
𝑝𝑝 𝑝𝑝𝑝𝑝 198 MeV
6
Learning Objective: be able to calculate the kinematics of 2-body interactions and decays.
1.1 Elementary particles
From a historical point of view, particle physics begins in the 3rd century BC, with the
Democritus particle theory. Democritus, a Greek philosopher, believed that the world was
made of indivisible “atoms”. This was, in fact, correct as we now know that the basic
components of matter are “elementary particles”. Such elementary particles do not have a
known substructure and are considered to be point-like. In the past century, we saw great
advancements in our understanding of the structure of matter. Broadly speaking, these
advancements were achieved by bombarding matter with “probes”.
In the case of a microscope, such a probe is a photon, which will be able to study structures
larger than the wavelength 𝜆𝜆 of the photons. If the probe is another type of particles, we
should use the associated De Broglie wavelength 𝜆𝜆 = ℎ/𝑝𝑝, where 𝑝𝑝 is the particle
momentum.
As an example, if we use visible light as a probe, such as in an optical microscope we can
explore structure of the size of ~500 nm, much larger than the size of an atom (compare with
the Bohr radius of 0.5 nm), but sufficient to study most biological objects. Using X-rays, we
can observe structures of the size of a few nm, which is why we use X-rays in applications
such as crystallography. If we want to probe the subatomic structure of the atom, we will
need shorter wavelengths and use probes of higher momentum. In the case of the Rutherford
experiment, alpha-particles were used with a kinetic energy of 5.30 MeV, corresponding to a
wavelength of about 6.3 fm, much smaller than the size of the atom, and sufficient to discover
the presence of a hard nucleus inside an atom.
Alpha particles in the Rutherford experiment, however, did not have sufficient momentum to
probe the nucleus structure. The story of particle physics and the quest for more energetic
probes and “higher resolution” experiments began over a century ago and is still ongoing with
the modern Large Hadron Collider.
1.2 Units of energy, momentum and mass
As in atomic and solid state physics, a useful unit of energy in particle and nuclear physics is
the electron volt (eV). This is the amount of kinetic energy gained by an electron when it is
accelerated through a potential difference of one volt. Normally the energies involved in
nuclear reactions are millions of electron volts (MeV) and in high energy particle interactions
they may be billions of electron volts or Giga electron volts (GeV = 109 eV). A convenient unit
for particle masses makes use of the Einstein mass-energy relationship
𝐸𝐸 = 𝑚𝑚𝑐𝑐 2
This yields a unit for mass expressed as energy divided by the square of the velocity of light,
MeV/c2 or GeV/c2. For example
proton mass = 938.3 MeV/c2
electron mass = 0.511 MeV/c2
This system of units is extended to momentum through the relativistic relationship for the
energy of a particle of rest mass m moving with momentum p,
𝐸𝐸 2 = 𝑝𝑝2 𝑐𝑐 2 + 𝑚𝑚2 𝑐𝑐 4 .
From this, it follows that if we express momentum in the units of energy divided by the
5
, velocity of light (GeV/c), we have a self-consistent system in which the velocity of light is
implicitly used, but its value does not have to be explicitly put in to the calculations.
In several textbooks, you will find that 𝑐𝑐 = 1, e.g. the speed of light is not explicitly indicated
in the formulae. This is a common notation used in particle physics, where particle masses
are expressed in GeV by implicitly multiplying their mass by 𝑐𝑐 2 . For clarity, I will try to avoid
this notation in my lectures.
A useful value to remember is that of ℏ𝑐𝑐 = 197 MeV fm
Example 1.1. Evaluate the de Broglie wavelength of an alpha particle of kinetic energy
𝐸𝐸𝐾𝐾 = 5.3 MeV.
[𝑚𝑚𝛼𝛼 = 6.64 × 10−27 kg = 3.73 GeV⁄c 2 ]
Solution:
Important: The alpha particle is non-relativistic. So we use the Galilean expression for its
kinetic energy 𝐸𝐸𝐾𝐾 = 𝑝𝑝2 ⁄2𝑚𝑚 . We will soon see that this is not the case when particles are
relativistic!
1. First we solve using SI units for all the quantities
We calculate the momentum of the alpha particle as
𝑝𝑝 = �2 𝑚𝑚 𝐸𝐸𝐾𝐾
= �2 × ( 6.64 × 10−27 ) × ( 5.3 × 106 ) × (1.6 × 10−19 ) kg m s −1
= 1.06 × 10−19 kg m s −1
We use the De Broglie formula to evaluate 𝜆𝜆
ℎ 6.63 × 10−34 kg m2 s −1
𝜆𝜆 = = = 6.25 × 10−15 m = 6.25 fm
𝑝𝑝 1.06 × 10−19 kg m s −1
2. We can also calculate the De Broglie wavelength using particle physics units
Instead of 𝑝𝑝 we work with 𝑝𝑝𝑝𝑝 which is
𝑝𝑝𝑝𝑝 = �2 𝑚𝑚𝑐𝑐 2 𝐸𝐸𝐾𝐾 = �2 × (3730 MeV) × 5.3 MeV = 198 MeV
ℎ 2𝜋𝜋 ℏ𝑐𝑐 197 MeV fm
𝜆𝜆 = = = 6.28 = 6.25 fm
𝑝𝑝 𝑝𝑝𝑝𝑝 198 MeV
6
