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In the lecture note Track Reconstruction and Pattern Recongnition in High-Energy Physics written by Prof. Ivan Kisel, there is a figure (in page 7), which describes the different pieces of modern detectors. enter image description here

He explained, Electromagnetic Calorimeter measures the total energy of $e^{+}$, $e^{-}$ and photons, and only muons and neutrinos can get to Muon Chambers.

Theoretically, muons and electrons are very similar except for their mass, both of them can interact with electromagnetic field. So my question is, why only muons can get to Muon Chambers, or why muons can get to Muon Chambers but not be detected in EM Calorimeter.

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Muons can more easily penetrate more material. Typically in most detectors there is a distinguishable pattern between muons and electrons. A clear example would be data from the Super Kamiokande detector where one detects Cerenkov light coming from electrons/muons.

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The "fuzzines" of the right circle means that the light came from an electron which got scattered and emitted a few Bremsstrahlung photons.

The point of all this is to show you that just the difference in mass is enough to make the Muon penetrate a lot more material than the electron. This is due to the fact that the (simplified) formulas for Bremsstrahlung are:

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or

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and both are proportional to the acceleration SQUARED! The force applied on both particles from the material in the calorimeter for example is the same (just an E-field), but due to the different masses, the electron has a lot bigger acceleration, thus loses more energy. This of course is simplified, since if you are calculating it precisely you need to take relativistic effects into account, but the intuition is the same.

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  • $\begingroup$ Interesting. Btw, the site supports formulas in latex e.g. $E = mc^2$ gives $E=mc^2$, and it's preferred over pictures of formulas. $\endgroup$ – innisfree Sep 6 '17 at 2:00
  • $\begingroup$ Luthelin: "the (simplified) formulas for Bremsstrahlung are [...]" -- These are apparently the formulas of total radiated power $P$ for a charged particle in a vacuum. As far as I understand, if the particle underwent acceleration $a$ (perhaps due to interaction with virtual photon emitted by "at least some material, somewhere") then it looses power strongly depending on "its $\gamma = E / (m c^2)$". Thus, at equal energies $E$ and equal $a$, far more power loss for the particle of less mass $m$. [contd.] $\endgroup$ – user12262 Sep 8 '17 at 3:14
  • $\begingroup$ Luthelin: "The force applied on both particles from the material in the calorimeter for example is the same (just an E-field), but due to the different masses, the electron has a lot bigger acceleration, thus loses more energy." -- At least, this intuition amplifies the conclusion: how "specific energy loss (in material) $1/E~1/\rho~d/dx[ E ]$" or stopping power depends on mass, at relevant HEP energies. Also, note the usual spelling of P. A. Cherenkov's surname. +1. (+2?!) $\endgroup$ – user12262 Sep 8 '17 at 3:17
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Nit picking:

why only muons can get to Muon Chambers, or why muons can get to Muon Chambers but not be detected in EM Calorimeter.

Muons are detected in the electromagnetic calorimeter as charged tracks. The are not identified as muons and go through as possible hadrons: protons, charged kaons, pions . The hadronic calorimeter detects the hadrons by their strong interactions with the material, and the muon detector makes sure that the track going through has only electromagnetic and weak interactions as it has gone through so much hadronic mass without interaction. Thus it is identified as a muon, by exclusion of other possibilities and use of the standard model which has no other charged weakly interacting particles.

This article covers in detail the subject of particle energy loss as they pass through various matter. This is mass dependent, for the same momentum the smaller the mass the higher the energy loss as shown in other answers.

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Muons are about 200 times heavier than electrons; muons are about 100 MeV, whereas electrons are about 0.5 MeV. It follows that whereas an electron is stopped in the ECAL, a muon just ploughs through it and into the muon chamber, as illustrated by this cartoon from this blog post about the muon.

enter image description here

This figure, though, shouldn't be interpreted as saying that muons create a huge disturbance and deposit lots of their energy and momentum in the ECAL; quite the opposite.

In fact, we may model the elastic collision of a muon with an electron at rest and of two electrons using conservation of energy and momentum, e.g., $$ \text{Initial muon momentum} = \text{Final muon momentum} + \text{Electron momentum} $$ and $$ \text{Initial muon energy} = \text{Final muon energy} + \text{Electron energy} $$ With the usual expressions (e.g. $p = mv$), we find that a muon retains about $99\%$ of its initial velocity, $v_i$, $$ \frac{v_f}{v_i} = \frac{m_\mu - m_e}{m_\mu + m_e} \approx 99\% $$ where $v_f$ is its final velocity. An electron, on the other hand, retains none of its velocity, $v_f = 0$, as in an elastic collision between balls of equal mass, the balls simply 'swap' velocities.

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  • $\begingroup$ I'm afraid the figure is actually really bad. The muon dumps less energy and momentum into the medium as it passes through, resulting is less disturbance. At this level particles don't really act like little billiard balls at all. $\endgroup$ – dmckee Sep 6 '17 at 1:36
  • $\begingroup$ Ah I see your point, yes the figure could be misinterpreted. You don't think the heuristic calculation is relevant? $\endgroup$ – innisfree Sep 6 '17 at 1:37
  • $\begingroup$ The calculation represents the results of a hard scattering event, but most of the energy loss for muons comes from multiple soft scatterings. It's not wrong, it's just rarish. Such event appear in tracking detectors as delta rays, and a long track will have several. $\endgroup$ – dmckee Sep 6 '17 at 1:41
  • $\begingroup$ Hmm. The only difference between muons and electrons is that $m_\mu \gg m_e$, so I think a simple explanation of their different behaviours in detectors based on this fact should exist. I thought this was it. Now not so sure. $\endgroup$ – innisfree Sep 6 '17 at 1:43
  • $\begingroup$ I think you're on a useful road, but the big difference is in the response to acceleration: electrons generate a lot more bremstrallung. $\endgroup$ – dmckee Sep 6 '17 at 1:58

protected by Qmechanic Sep 6 '17 at 5:53

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