Time of flight experiment to distinguish muons from pions?

1. May 1, 2015

1a2

If I wanted to design an experiment which would allow to distinguish muons from pions in the energy range from 200 MeV to 2 GeV, how would I do it in a simple way? Could I just use the Time of Flight experiment? What quantitative arguments can one make for the design of the detectors?

2. May 1, 2015

ChrisVer

I'd guess you could shoot them on iron for example... The muons would not leave a large signal, at that energy range their energy deposit in the matterial will be around dE/dx~ 1.5 MeV cm^2 /gr ....

On the other hand the pions will interact strongly with the matterial and lead to a hadronic shower. By installing scintillator layers within your iron walls, you'll be able to get signals from them.
(check slides 3 ,for what happens with the hadronic shower, and 9 ,for the muon energy deposit as a function of its energy)

3. May 1, 2015

Staff: Mentor

A RICH detector is very good in that range. Time of flight can be possible, but at 2 GeV the required timing resolution can get tricky.

There is no way to say what would be best without more details of the goal of the experiment.

4. May 1, 2015

Staff Emeritus
ChrtisVer's suggestion will work fine at 2 GeV, but will be utterly useless at 200 MeV. Time of flight will work beautifully at 200 MeV, but will be difficult or impossible at 2 GeV. mfb's RICH suggestion will struggle at 200 MeV as well. To get a signal you need to go to a very high index, and that means glass, and that means your particles will be absorbed in a very short distance.

5. May 1, 2015

Staff: Mentor

You could use a very thin layer of glass, but combining time of flight with RICH is probably easier.
Measuring dE/dx in material can work as well in the very low energy range.

6. May 1, 2015

Staff Emeritus
Doesn't help. A layer of glass half as thick produces only half the photons. However much glass you need is the amount of glass you need. This is why the LHCb RICHes are so big: they need them to be that big to collect enough light.

7. May 1, 2015

Staff: Mentor

You should know I don't make those claims without numbers backing them up.
Muons at 200 MeV have a velocity of 0.835, in glass with a refractive index of 2 this gives $\theta=0.93$, in the wavelength range of 400 to 800 nm this gives about 35 photons per millimeter. Even with losses in the material and the detectors 1mm should be sufficient to make a nice ring. Particles at higher energy give fewer photons but they can get a second medium if required (exactly what LHCb is doing with the aerogel).

LHCb is focusing on particles at higher energy, where you need a smaller refractive index (and therefore theta) to separate particles, which means the detectors have to be larger. Also, the overall LHCb RICH size is given by the detector geometry and eta range covered there. A RICH for lower energies works as well, it just needs a different design.

8. May 1, 2015

Staff Emeritus
Glass with a refractive index of 2 is hard to find. I think the OPAL calorimeter had an index around 1.8, and that is probably the record for things actually used.

If you have PMTs with QE's of 20% and cover half the area - which is doing pretty good - you need to produce 10x as many photons as you detect. I think you need at least a couple of millimeters, and at 200 MeV you can't afford to lose much energy before you both a) hit the Bragg peak and b) drop under threshold. Your dE/dx is already 50% or so above the MIP peak and that's only going to get bigger.

If I wanted to go with Cerenkov counters, I'd be thinking threshold and not imaging.

LHCb's RICH is much longer than BTeV's. This is because LHCb needs to go to higher energy, which means lower index, which means less light per unit distance, which means a longer distance. That's why they are the behemoths that they are. They are driven to it by running at 14 TeV and not 2.

9. May 1, 2015

Staff Emeritus
One other thing - this particular problem is hard in part because 200 MeV and 2 GeV are on opposite sides of the Bethe-Block curve, and so any solution needs to encompass a wide range of behavior.

10. May 1, 2015

ChrisVer

How would a time-of-flight experiment work? Put them into a magnetic field and see the kink of the pions?

11. May 1, 2015

ChrisVer

Also here in Fig1 (pg 6) if I interpret it right, the 0.2-2 GeV are not so different (for Cu)

Last edited by a moderator: May 7, 2017
12. May 1, 2015

Envelope

I agree that a more fleshed out context would be needed to give a reasonable muon/pion separation strategy.

In its simplest form, which could be appropriate for (say) a fixed-target test beam, you just place two thin but fast detectors some distance apart and let the particles pass through them. At 200 MeV kinetic energy and a one meter separation, you are already only looking at a 200 picosecond time-of-flight difference. At higher energies, this approach becomes impractical fast due to limited timing resolution.

13. May 1, 2015

Staff: Mentor

The numbers are not that different with 1.8, and glass with 2 exists (you just need a thin layer, not a large calorimeter).

Performance of the LHCb RICH detector at the LHC
The LHCb velo gets ~5 photoelectrons per particle (table 1) with several mrad resolution for each photon (figure 10).
The active area fraction is 82% and the quantum efficiency is 30% at 270nm (page 5), combined ~1/4.
Our 35 photons are better than the LHC aerogel.

200 MeV are sufficient to go through a few millimeters of glass. Even protons could do this, and our muons and pions are faster.

Threshold mode for 200 MeV would mean n between 1.2 and 1.4. Using 1.2 (muon threshold 200) would set the threshold for pions to 255 MeV. But then you need a really long radiator to get photons.

@ChrisVer: the time of flight concept gives you velocity via the flight time and distance. Combine it with a momentum measurement (usually: curvature in a magnetic field) to get the particle type.
Nice for 200 MeV where the difference between 0.71c and 0.84c is massive, harder for 2 GeV between 0.9976c and 0.9985c (100ps over 30m).

14. May 1, 2015

Staff Emeritus
The problem is that energy loss behaves differently on the two sides of the curve - on the high side, energy loss means the next ionization loss will be smaller, but on the low side, it means it will be bigger. That means the particle is going to stop in just a little bit of material. So, if I were trying to separate a 2 GeV muon from a 2 GeV pion, I can put up a hunk of material and the pion will stop and the muon will go through. If I try that with a 200 MeV pion, they both stop, With the energy range given, you need to handle both behaviors. The point is not so much that this particular method won't work, it's that the interaction with material is changing a lot over this range and that makes it hard to find a single process that will work well.

Onto time of flight. Let's say you know exactly when the particle is produced - perhaps when the beam hit the target is known from a signal. You then measure the time a detector some distance away responds, and that tells you the particle's velocity. For example, if you had particles with E=200 MeV, a pion would take 24.4 ns to go 5 meters, but a muon would only take 19.7, so you have a difference of 4.7 ns. This is doable. If you go to 2 GeV, the difference drops to 21 ps. This is much harder. Most TOF systems do pi-K and pi-p separation and not pi-mu.

If you don't know when it was produced, you can measure the particle at two points. This means that your resolution is sqrt(2) worse.

15. May 1, 2015

1a2

I know that to distinguish electrons from muons, we can use the time of flight method. We can determine the time if the momentum is known. We can measure the momentum can be done by mounting a magnet on the way of particles, and the muons and electrons will be bent with the same curvature since they have the same charge. The particle will go through two paddles

But I'm not sure how to distringuish muons from pions?

16. May 1, 2015

Staff Emeritus
At 2 GeV, the velocity difference between electrons and muons is 0.1%. That means you get one picosecond of difference per foot of flight. It's hard to get time of flight to work with this small a difference.

Is this a real experiment?

17. May 2, 2015

ChrisVer

Why the same curvature? Their masses are different.

18. May 2, 2015

Envelope

1a2 assumed equal momentum, so equal curvature.

19. May 2, 2015

Ah OK.