Ionisation from charged particles

 Transition radiation photons are emitted in the in the same direction as the charged particle creating the radiation, hence the energy deposition in the detection gas will be from both the transition radiation photons and the ionisation caused by the same particle. As both pions and electrons will ionise the gas a detailed description of the ionisation is required for the understanding of the performance of a transition radiation detector for particle identification.

The Bethe-Bloch formula for the energy loss of a charged particle in a medium is well known but gives only the mean integrated energy loss. This is shown for a selection of materials in fig. 3.8. The relativistic rise makes the energy loss increase for $\beta$$\gamma$ above the minimum ionising point at $\beta$$\gamma$ = 3.5 . The rise is limited by the polarisation of the media which depends on the electron density and the relativistic rise is thus most suppressed for high density media. The result is that gases, with low electron density, have a large relativistic rise. For xenon at 1 atm the relativistic rise is around 75% [36].

  

Figure 3.8: The energy loss of a charged particle in a range of materials.

The mean value of the total energy loss is given from the Bethe-Bloch formula but the fluctuations are large due to a low number of high energy ionisations. The energy loss distribution is given by the Landau distribution and shown in fig. 3.9. The long Landau tail towards high energies is a result of the large fluctuations in the high energy ionisations.

  

Figure 3.9: The Landau distribution of the total energy loss across a gas volume. Note the long Landau tail on the distribution.

To simulate the true signal in a proportional chamber it is necessary to use a much more detailed model that gives the distribution of the individual ionisations along the track and their energies.

The photo absorption ionisation (PAI) model was first in detail described in [37]. It was developed using a semi-classical approach, that starts with the Maxwell equations for a charged particle traversing a medium with dielectric constant $\varepsilon$ . In this way the energy loss is expressed as

 

$${\frac{{d}E}{{d}x}} {\frac{e \vec{E} \cdot \vec{\beta}}{\beta}}$$ (66)

where $\vec{\beta}$ is the velocity vector of the charged particle and $\vec{E}$ the electric field created by the particle itself evaluated at the point of the particle. Making a Fourier transform of the electric field the energy loss can instead be described as a continuous energy loss in different frequency regions [37]. The result is
 $$\begin{align} \frac{{d}E}{{d}x} &= - \frac{\alpha}{\beta^2 \pi} \int_0^\infty... ... \int_0^\infty {d}\omega' \frac{\sigma(\omega')}{Z} \right] \notag\end{align}$$
with $\varepsilon$($\omega$) = $\varepsilon_{1}^{}$ + i $\varepsilon_{2}^{}$ the dielectric constant, n_e the electron density, and Z the atomic number. The dielectric constant can be expressed by the atomic photo absorption cross section $\sigma$($\omega$) through the relations
 $$\begin{align} \varepsilon_2(\omega) &= \frac{n_{e}}{Z \omega} \sigma(\omega) \\... ...^\infty {d}\omega' \frac{\sigma(\omega')} {{\omega'}^2 - \omega^2}.\end{align}$$
The only remaining term is $\Theta$ = arg(1 - $\varepsilon^{\ast}_{}$$\beta^{2}_{}$) which is responsible for the Cherenkov radiation. The photo absorption cross section is related directly to the absorption length in the gas through the relation

 

$$\lambda {\frac{A}{N_A \sigma}}$$ (67)

where A is the atomic weight and N_A Avogadros number. For composite gases a mean cross section weighted with the atomic weight of the different components will give an accuracy of a few percent in the absorption length [38]. Both the cross section and the absorption length for xenon is shown in fig. 3.10.

  

Figure 3.10: The photo absorption cross section (a) and the absorption length at 1 atm (b) for xenon.

The energy loss in (3.33) describes a continuous energy loss. It can, however, be reinterpreted as a number of discrete collisions with energy transfer $\omega$ . This is the semi-classical approach following the same line as Plancks derivation of the black body spectrum. The result following directly from (3.33) is
  $$\begin{align} \left\langle \frac{{d}E}{{d}x} \right\rangle &= - \int_0^\infty... ...\int_0^\infty {d}\omega' \frac{\sigma(\omega')}{Z} \right]. \notag\end{align}$$