\approx 1.6 - 1.2 = 0.4$ and $r_i = 1.2$ in Fig. 1 one gets $$\delta \approx 0.1.        \label{delta}$$ The ring of Cherenkov gluons is really quite narrow. Following Proposal 1, its width can be directly measured by plotting the pseudorapidity distribution of particles with the away-side jet direction chosen as $z$-axis.

Actually, Eq. (\[cos\]) implies that the ring is squeezed to a circle. There are three physical reasons which can lead to the finite width of the ring. First, it is the dispersion, i.e. the energy dependence of the refractive index. Its contribution to the width is well known $$\delta _d=\int _0^{\delta _d}d\theta =\cot \theta _c\int _0^{\infty }
\frac {1}{n} \frac{dn}{dE} dE.      \label{lan}$$ If the Breit-Wigner expression (\[ren\]) for $n(E)$ is used, the result is $$\delta _d=0.   \label{deld}$$ It is amazing that there is no widening of the Cherenkov cone due to the dispersion of $n(E)$ described by the formula (\[ren\]) with Breit-Wigner resonances.

Second, the width of the Cherenkov ring can be due to the finite free path length for partons. Qualitatively, it can be estimated as the ratio of the parton wavelength $\lambda $ to the free path length $R_f$ $$\delta _f \sim \frac {\lambda }{R_f}.    \label{rf}$$ 