LHCb’s Ring Imaging CHerenkov detector (RICH) is aimed at telling different charged particles apart by measuring their velocity, which, together with an independent measurement of their momentum, is enough to derive their masses. It has been built in order to achieve a minimum angular acceptance of $10$ mrad and its medium has refractive index $n=1.0014$. As far as I know there is also RICH2, with a lower refractive index $n=1.0005$, to measure high energy particles, since it postpones the saturation effect.

The problem is that I don't really understand what "postpones the saturation effect" really means. The lower refraction index is supposed to allow me to distinguish between particles at higher energies, but I don't really understand why the lower index should help here.

Can somebody maybe explain this, or make an example that illustrates how the refractive index helps to distinguish particles with higher energy?


1 Answer 1


I am not entirely sure what ``saturation effect'' means in this case. But let us go back to the expression for the Cherenkov angle: $$ \cos\theta = \frac{1}{n \beta} $$ we can obtain two limits:

  • first, on the minimum $\beta$ to have Cherenkov effect: $$ \beta_{\rm th} = \frac{1}{n}, $$ and - as you noticed - the smaller $n$, the lower the velocity of the particle we can detect;

  • but also on the maximum aperture angle of the Cherenkov cone, in case the particle is ultrarelativistic $\beta \rightarrow 1$: $$ \cos\theta_{\rm max} = \frac{1}{n}. $$

I thought the ``saturation'' might be related to this second point, the fastest particles will generate cones with apertures:

\begin{equation} \begin{split} \theta_{\rm max}^{\rm RICH} &= \arccos \left( \frac{1}{1.0014} \right) \approx 3.03^{\circ}, \\ \theta_{\rm max}^{\rm RICH2} &= \arccos \left( \frac{1}{1.0005} \right) \approx 1.81^{\circ}, \end{split} \end{equation}

In the two different detectors. For ultra-relativistic particles you will have smaller rings in the detector with the smaller $n$. If you receive many particles it would be easier to ``saturate'' (to create more overlapping rings, maybe?) the detector with the larger $n$.
Just my speculation though!


After the useful suggestions of @dukwon I understood the saturation was not due to the number of particles hitting the detector but by the impossibility to distinguish their momentum when their $\beta \rightarrow 1$. Therefore I created this plot. Here I show the values of $\theta_{\rm C}$, i.e. the Cherenkov angle for different values of the particle energy. I consider three particles (kaon, pion and muon) and two mediums with the indexes, $n_1=1.0014$ and $n_2=1.0005$ as in @Sito's original question. enter image description here I placed the vertical dashed line by eye at the point where I am not capable of distinguishing anymore kaons and pions angles. As you can see in the material with the worst (higher) refraction index this happens already at $80\,{\rm GeV}$. In the material with the better (lower) refraction index you can push the indistinguishability of kaons and pions at $E > 100\,{\rm GeV}$. and that's exactly why $n_2=1.0005$

postpones the saturation effect .

Thanks @dukwon!

@Sito, here the snippet if you want to reproduce the figure

import numpy as np
import matplotlib.pyplot as plt

def beta(E, mec2):
    gamma = E / mec2
    return np.sqrt(1 - 1 / np.power(gamma, 2))

def theta(beta, n):
    return np.arccos(1 / (beta * n))

mec2_kaon = 493.68  # MeV / c^2
mec2_pion = 139.57  # MeV / c^2
mec2_muon = 105.65  # MeV / c^2

E = np.logspace(np.log10(5e3), 6, 100) # MeV

beta_pion = beta(E, mec2_pion)
beta_kaon = beta(E, mec2_kaon)
beta_muon = beta(E, mec2_muon)

n_1 = 1.0014
n_2 = 1.0005

fig, ax = plt.subplots()

plt.semilogx(E, theta(beta_kaon, n_1), color="crimson", ls="-", label=r"$K, n_1=1.0014$")
plt.semilogx(E, theta(beta_pion, n_1), color="crimson", ls="--", label=r"$\pi, n_1=1.0014$")
plt.semilogx(E, theta(beta_muon, n_1), color="crimson", ls="-.", label=r"$\mu, n_1=1.0014$")

plt.semilogx(E, theta(beta_kaon, n_2), color="k", ls="-", label=r"$K, n_2=1.0005$")
plt.semilogx(E, theta(beta_pion, n_2), color="k", ls="--", label=r"$\pi, n_2=1.0005$")
plt.semilogx(E, theta(beta_muon, n_2), color="k", ls="-.", label=r"$\mu, n_2=1.0005$")

plt.axvline(8e4, color="crimson", ls=":")
plt.axvline(1.1e5, color="k", ls=":")

plt.xlabel("E / MeV")
plt.ylabel(r"$\theta_{\rm C}\,/\,rad$")
  • $\begingroup$ Thank you for the answer! I think point two is related to saturation, at least I remember my professor mentioning $\beta \to 1$... The problem with you answer is just that I still don't really see how the smaller refractive index helps keeping high energy particles apart... Could you maybe elaborate on that a bit? $\endgroup$
    – Sito
    Commented Apr 10, 2019 at 18:13
  • 1
    $\begingroup$ Almost there with regards to saturation. It is indeed when the angle is at the maximum, but it's not due to overlapping rings. It's that there's no distinction between particles of different mass. $\endgroup$
    – dukwon
    Commented Apr 10, 2019 at 19:10
  • $\begingroup$ @dukwon could you maybe expand on that a bit? I'm still having some trouble with the concept here and it doesn't seem like this answer will get any updates anymore... $\endgroup$
    – Sito
    Commented Apr 16, 2019 at 15:02
  • 2
    $\begingroup$ Saturation is where the lines meet: twiki.cern.ch/twiki/pub/LHCb/RICHPicturesAndFigures/… $\endgroup$
    – dukwon
    Commented Apr 16, 2019 at 15:57
  • $\begingroup$ Hi, improved the answer thanks to @dukwon clarifications. $\endgroup$ Commented Apr 17, 2019 at 12:07

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