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PC1672 Advanced dynamics


1.6 Relativistic kinematics

In studying the collisions of particles at speeds which are significant fractions of the speed of light, we can use two principles:

These 4-scalars can be calculated in any inertial frame since their values are invariants. For processes involving a collision between two particles, it is often convenient to work in the centre-of-momentum (c.m.) frame in which the total 3-momentum is zero.

A common process of interest is a collision between two particles leading to creation of new particles. An example of this is annihilation of an electron and a positron to form two electron-positron pairs:

\begin{displaymath}e^++e^-\rightarrow e^++e^++e^-+e^-\end{displaymath}

(The positron is the antiparticle for the electron, predicted by Dirac. It has exactly the same mass as the electron but the opposite charge.)

The minimum energy is required to create a set of particles when they are all at rest with respect to each other. In the c.m. frame this means that they are all at rest and so their total energy is just the sum of their rest energies. For the example of two $e^+e^-$ pairs, the minimum total energy is $4m_ec^2$. By conservation of energy this must be equal to the total energy of the incoming electron and positron. Since the momenta of the two incoming particles are equal and opposite in the c.m. frame (and they have equal masses) their energies are also equal and are given by

\begin{displaymath}E_{e^+}=E_{e^-}=2m_ec^2\end{displaymath}

In a general frame we need to use conservation of the total 4-momentum, combined with the fact that the square of this 4-momentum is invariant. For production of two $e^+e^-$ pairs, the minimum energy corresponds to a squared 4-momentum of

\begin{displaymath}{\sf p}_{\rm tot}^2=-16m_e^2c^2\end{displaymath}

In the lab frame where the initial electron is at rest and the incoming positron has energy $E_{e^+}$, the square of the 4-momentum is given by

\begin{displaymath}{\sf p}_{\rm tot}^2=-2m_ec^2-2m_eE_{e^+}\end{displaymath}

Equating these two gives a minimum positron energy of

\begin{displaymath}E_{e^+}=7m_ec^2\end{displaymath}

A rather different example is Compton scattering of a photon from an electron intitally at rest. In this case we are interested in the relation between the energy and angle of the outgoing photon. The initial 4-momenta of the photon and electron are

\begin{displaymath}{\sf p}_\gamma=\left({E_\gamma\over c},0,0,{\rm i}{E_\gamma\over c}\right)\end{displaymath}


\begin{displaymath}{\sf p}_e=(0,0,0,{\rm i}m_ec)\end{displaymath}

and the final photon 4-momentum is

\begin{displaymath}{\sf p}_\gamma'=\left({E_\gamma'\over c}\cos\theta,{E_\gamma'\over c}\sin
\theta,0,{\rm i}{E_\gamma'\over c}\right)\end{displaymath}

We are not interested in the recoiling electron. All we need is the fact that its 4-momentum squared is the invariant

\begin{displaymath}{\sf p}_e^{\prime 2}=-m_e^2c^2\end{displaymath}

and conservation of total 4-momentum which implies

\begin{displaymath}{\sf p}_e'={\sf p}_\gamma+{\sf p}_e-{\sf p}_\gamma'\end{displaymath}

This leads to the relation

\begin{displaymath}E_\gamma'\left[mc^2+E_\gamma(1-\cos\theta)\right]=E_\gamma mc^2\end{displaymath}

which is equivalent to the usual expression for the shift of the wavelength in Compton scattering.

The same method can also be applied to the decay of a particle into two particles, where we are interested in the energy and direction of only one of the final particles. Again we first use conservation of 4-momentum to express the 4-momentum of the one of the final particles in terms of the 4-momenta of the other final particle (the one we are interested in) and the original particle. We then construct the square of the 4-momentum of the final particle whose energy we do not want. This gives us an equation which we can solve to find the energy of the particle of interest.

Textbook references


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Mike Birse
17th May 2000