Charged tracks in a uniform magnetic field in the barrel geometry follow the track equations:

\[\phi(r) = \phi_{0} - \sin^{-1}\left(\frac{r}{2R}\right)\] \[z(r) = z_{0} + \cot{\theta} \left[2R \sin^{-1}\left(\frac{r}{2R}\right)\right]\]

where $R = \frac{p_{\mathrm{T}}}{0.003 B}$, $R$ and $r$ are in cm and $B$ is the magnetic field in Tesla. The constant 0.003 is the speed of light multiplied by the necessary conversion factors. A muon with $p_{\mathrm{T}} = 2~\text{GeV}$ in the CMS experiment ($B = 3.8~\text{T}$) has $R = 1.75~\text{m}$.

In the endcap geometry, the track equations are:

\[\phi(z) = \phi_{0} + \frac{z - z_{0}}{2R \cot{\theta}}\] \[r(z) = \left|2R \sin\left( \frac{z - z_{0}}{2R \cot{\theta}} \right) \right|\]

where $R \cot{\theta} = \frac{p_{\mathrm{z}}}{0.003 B}$, $z$ is in cm. By the way, $\cot{\theta}$ can be easily calculated from $\eta$ via the equation $\cot{\theta} = \sinh(\eta)$.

When considering a vertex-unconstrained fit, with transverse impact parameter $d_{0} \neq 0$, the $\phi_{r}$ equation becomes:

\[\begin{align*} \sin(\phi - \phi_{0}) &= -\frac{r}{2R} \frac{\left(1-{d_{0}}^2/r^2\right)}{\left(1 + d_{0}/R\right)} - \frac{d_{0}}{r} \\ &\approx -\frac{r}{2R} - \frac{d_{0}}{r} \end{align*}\]

For the endcap muon trigger in CMS, one has to take into account several considerations:

  • Detector planes perpendicular (instead of parallel) to the solenoid magnetic field;
  • Significant multiple scattering;
  • Significant energy loss;
  • Neutron background;
  • Non-uniform magnetic field, including a non-zero radial component.