Martin & Shaw, appendix A.

The marriage of quantum mechanics with special relativity leads to quantum field theory.

The sources for particles are cosmic rays and accelerators.

We use high energies to achieve the best spatial resolution and to create massive particles.

It is convenient to work in units appropriate to the problem at hand.
In particle physics it is convenient to work in units in which ħ = c = 1.
Additionally, setting ε_{0} = μ_{0} = 1, we have Heaviside-Lorentz units.
In these units, the fine structure constant (from quantum mechanics applied to the hydrogen atom) is given by

α = e^{2} / ħ c = 1/137.

We still have one more unit to choose, the unit of mass. Masses in particle physics are normally expressed in MeV (millions of electron volts) or GeV (billions of electron volts). The proton mass is 938MeV, and the electron mass is 0.511MeV.

We will normally deal with processes where at least some particles are moving at speeds close to the speed of light. In such situations, the rules of special relativity must be followed in order that the process be independent of the reference frame.

In relativity, space and time are united in a 4-dimensional space-time. A vector, must now be a 4-component space-time vector, a 4-vector, rather than a 3-component spatial vector. The first vector I will introduce is the momentum vector:

p = (E, p_{vec})

where p
p_{0} = E, p_{1} = p_{x}, p_{2} = p_{y}, p_{3} = p_{z}.

As for 3-vectors, the dot product of a 4-vector with itself is a scalar quantity, invariant under rotations and translations of the coordinate system. In special relativity, the dot product (or scalar product) of any two 4-vectors measured in any inertial reference frame is invariant.

The dot product of p with itself is defined as:

p·p = E^{2} - p_{vec}^{2} = m^{2}

where m is the mass of the particle with 4-vector p.
Since the dot product is an invariant quantity, we get the comforting result that the mass of a particle is the same in all inertial reference frames.
To transform p to another reference frame (the primed frame, with quantities indicated with a prime) moving with velocity v_{x} = βc with respect to the original, we use the Lorentz transformation:

p'_{0} = γp_{0} - γβp_{1}

p'_{1} = γp_{1} - γβp_{0}

p'_{2} = p_{2}

p'_{3} = p_{3}

where γ = 1/sqrt(1-βp'

p'

p'

We use the momentum vector to describe the direction and momentum of particles in scattering and decay events. In these cases, it is sometimes helpful to transform between frames, for instance from the laboratory frame to the center of mass frame, and back. Another 4-vector that is used is the space-time vector:

x = (t, x_{vec})

(x_{0}, x_{1}, x_{2}, x_{3}) = ( t, x, y, z)

Under a change of inertial reference frames, this vector Lorentz transforms in the same manner as the energy-momentum vector.
(x

As a first example of the use of the Lorentz transformation and dot product for 4-vectors, let's consider the energy available for particle creation in two common experimental arrangements.

The fixed target arrangement is the easiest to conceive and was the first to be used. A high energy beam of particles is made to impact on a target at rest in the laboratory, a thin foil, wire, or other solid, liquid, or gas. In a fixed target experiment, a wide variety of nuclei can be used as targets. The experimental apparatus is placed behind the target to catch the debris that emerges.

The colliding beam arrangement is harder to conceive and came later. It consists of two beams of particles that are made to collide, usually head on. The experimental apparatus is built around the point of collision. It is common for one beam to consist of the anti-particle of the other, and that they are counter-rotating with the same energy (symmetric collisions).

Let us say that the incoming beam consists of particles of mass m_{a}, total energy E_{a}, and momentum p_{a}.
Of course these 3 quantities satisfy the relation m_{a}^{2} = E_{a}^{2} - p_{a}^{2}.
Let the target consist of particles of mass m_{b}, total energy E_{b}, and momentum p_{b}, that satisfy a relation as above.

Our system consists of a particle of the incoming beam colliding with a particle of the target.
The total 4-momentum of the system is p = p_{a} + p_{b}.
The square of the 4-momentum is an invariant quantity given by:

p^{2} = (E_{a} + E_{b})^{2} - (p_{a vec} + p_{b vec})^{2}

= E_{a}^{2} - p_{a}^{2} + E_{b}^{2} - p_{b}^{2} + 2E_{a}E_{b} - 2p_{a vec}·p_{b vec}

= m_{a}^{2} + m_{b}^{2} + 2E_{a}E_{b} - 2p_{a vec}·p_{b vec}

The center-of-momentum frame (indicated by a *) is the frame in which the total 3-momentum is zero: p= E

= m

p^{2} = (E^{*}_{a} + E^{*}_{b})^{2} = E^{*2},

where EIn a fixed target configuration the target particle is at rest in the lab frame such that p_{b} = 0 and E_{b} = m_{b}.
Then the (square of) energy available in the collision is:

E^{*2} = p^{2} = m_{a}^{2} + m_{b}^{2} + 2m_{b}E_{a}.

The energy available increases as the square root of the incident particle energy.
Suppose we have a colliding beam configuration with particle and anti-particle colliding with equal energy. In this case

E^{*2} = p^{2} = 2(E_{a}E_{b} + p_{a}p_{b}) + (m_{a}^{2} + m_{b}^{2})

=~ 4E_{a}E_{b}.

If the particles are identical and collide in the CM, then the energy available increases linearly with the energy of the beam particles.
=~ 4E

Copyright © Robert Harr 2005