Four-Momentum

The four-momentum is a four-vector frequently in particle physics to describe collisions of electrons or protons. It is defined as $$p := \begin{pmatrix}E\\p_x\\p_y\\p_z\end{pmatrix}$$ according to the laws of the special theory of relativity. In this definition, $E$ stands for the total energy of the object (e.g. a single particle) and the three columns below represent the components of the relativistic momentum of the particle. Because it is a four-vector, the arrow above the letter is usually omitted. It can also be written as $$\boxed{p := \begin{pmatrix}E\\ \vec{p}\end{pmatrix}}$$ The absolute value of this vector is given as $$|p| = \sqrt{E^2 - \vec{p}^2} = m_0$$ and therefore equal to the invariant mass $m_0$ of the particle. This equation is very important for all fields of particle physics. If two or more particles collide, the center-of-mass energy is then given as the sum of the four-momentum vectors of all particles: $$E_\mathrm{CMS} = \sqrt{s} = \sqrt{\begin{pmatrix}E_1\\ \vec{p_1}\end{pmatrix} + \begin{pmatrix}E_2\\ \vec{p_2}\end{pmatrix}}$$ For a particle collider with two identical particles colliding e.g. in a detector, this results in $$E_\mathrm{CMS} = 2E$$ For a fixed-target experiment, the result changes to $$E_\mathrm{CMS} = \sqrt{2Em_0 + 2m_0^2}$$ However, if the energy of the beam is significantly larger than the rest mass of the particles, the square root can be simplified to $$E_\mathrm{CMS} = \sqrt{2Em_0}$$ by omitting the square of $m_0$. This shows very clearly the advantages of colliding two beams compared to a fixed-target collision.

• exercises:electron_positron_pair (++)