- Email: [email protected]
A simple model for electron pair production by muons
NUCLEAR INSTRUMENTS &METHODS IN PHYSICS RESEARCH
Nuclear Instruments and Methods in Physics Research A315 (1992) 65-66 North-Holland
Section A
A simple model for electron pair production by muons Presented by L.S. Osborne L.S . Osborne
Massachusetts Institute of Technology, Laboratory for Nuclear Science, Cambridge, MA, USA
This paper offers a relatively simple explanation for the mechanics of electron pair production by muons which enables a quantitative description .
1. The mechanism
Energy co,iservation gives,
A semi-sophisticated approach to electron pair production by muons would be to use the Weizs5ckerWilliams (WW) formalism to obtain the equivalent photon spectrum carried by the muon and have these photons impinge on the target material with the electron pail production gross section . Let's do it. WW gives for the photon spectrum [1], drt =
2a 7r
log(183Z-113) dk . k
Under this assumption the muon energy loss by this mechanism would be, dE _ - E(number of radiation lengths) 2a x -log(183Z 7r
).
Going through 2 m of iron (112 radiation lengths) d E/E = -1 .69. Two orders of magnitude too high (see fig . 1)! What is wrong? The answer lies in recognizing the sensitivity of cross sections which involve the Coulomb field to the momentum transfer (call it a propagator if you want, the energy component is negligible). In these da a d;q/q' =
w
I =w,+k,
where k (equivalent photon energy) = E + + E _ . Then, in mw inc m~ + + qr. = 2co, 2co t 2E + 2E_ m~
1 in ; 2(otco, k + 2k a(1 -a)
where. E += ak and E _ = (1 - a)k . It is clear that the the momentum transfer remains "electron pair like" provided, k<
in,
in
wj,
but as the photon energy increases beyond this limit, the muon term in eq. (6) takes over and q, increases 20 N
E
15
d qT dqr
2 qT+qf) . (2
After integrating with respect to dqT we get
w
5
dgel9é . We now look at the momentum delivered to the Coulomb field using the approximation, P = E` - m' = E - in -/2E for the reaction, Ws -_~
R2
(6)
+ e + + e-.
ionization
0
100
1000 EN, (GeV)
1500
Fig. 1 . Muon mean energy loss per unit length in iron. Taken from ref. [2].
0168-9002/92/$05 .00 © 1992 - Elsevier Science Publishers B.V. All righi, reserved
1. HIGH LUMINOSITY TRACKING
66
L.S. Osborne / Electron pair production by muons
the muon is at rest and the nucleus is travelling with the muon laboratory velocity and carrying the photons expressed by eq. (1). If you transform the electrons to this frame you find that condition (7) gives the electrons going along the photon direction . The opposite of condition (7) is pair production in the back direction which results in a large momentum transfer to the muon Coulomb field and so a diminishing cross section.
2. Implicat ons EIL (GeV)
Fig. 2. The number of events with energy loss greater than E in I m of iron for 10 TeV and 100 GeV muons from various interactions (ref. [2]) and' A.V. Lanyov, private communication. into the "muon like" region thereby decrasing the cross section . This is in fact what is observed in an accurate calculation [2] (see fig. 2). The equivalent photon energy, k, goes like 1/k as predicted from eqs. (1) and (2) up to an energy of about m ew i /m, after which : falls roughly as k-3. The point of fall-off is predicted by this simple model though not the power of the fall-off. It would be handy to use this model as a simple measure of pair production . We do this by taking eq. (1) with a cutoff at, me .8-0) 1, wl, w where ß is a parameter to be fixed to get the right energy loss from an exact calculation . We have done this by fitting the curve in fig . 1 and obtained, ß = 0 .80. There is a. other way to look at this phenomenon, leading to the same result . Consider the system where
It is clear from eq. (2) that the energy loss by a muon is proportional to its energy from this pair production mechanism . We can expect that a muon exiting from a material of many radiation lengths will bring with it the tails of the high energy showers produced by these pairs and that this will be more aggravated the higher the energy of the muon. However, there is another problem; the muon in traversing detectors of a fraction of a radiation length will also produce pairs. In this case it will be the number of pairs that provide the aggravation since the multiplication (the begining of the shower) will be minimal . In this case, we get the number of pairs by integrating eq. (1) over the equivalent photon energies that can cause trouble in the detector . This number varies only logarithmically with muon energy.
References [1] S. Gasiorowicz, Elementary Particle Physics (Wiley, New York) 180. [2] A.V. Lanyov, Inst. of High Energy Physics Publication IHEP 89-145 and EMPACT Note 321 ; and references therein. [3] EMPACT/TEXAS LOI submitted to the SSC Dec. 1, 1990.
Nuclear Instruments and Methods in Physics Research A315 (1992) 65-66 North-Holland
Section A
A simple model for electron pair production by muons Presented by L.S. Osborne L.S . Osborne
Massachusetts Institute of Technology, Laboratory for Nuclear Science, Cambridge, MA, USA
This paper offers a relatively simple explanation for the mechanics of electron pair production by muons which enables a quantitative description .
1. The mechanism
Energy co,iservation gives,
A semi-sophisticated approach to electron pair production by muons would be to use the Weizs5ckerWilliams (WW) formalism to obtain the equivalent photon spectrum carried by the muon and have these photons impinge on the target material with the electron pail production gross section . Let's do it. WW gives for the photon spectrum [1], drt =
2a 7r
log(183Z-113) dk . k
Under this assumption the muon energy loss by this mechanism would be, dE _ - E(number of radiation lengths) 2a x -log(183Z 7r
).
Going through 2 m of iron (112 radiation lengths) d E/E = -1 .69. Two orders of magnitude too high (see fig . 1)! What is wrong? The answer lies in recognizing the sensitivity of cross sections which involve the Coulomb field to the momentum transfer (call it a propagator if you want, the energy component is negligible). In these da a d;q/q' =
w
I =w,+k,
where k (equivalent photon energy) = E + + E _ . Then, in mw inc m~ + + qr. = 2co, 2co t 2E + 2E_ m~
1 in ; 2(otco, k + 2k a(1 -a)
where. E += ak and E _ = (1 - a)k . It is clear that the the momentum transfer remains "electron pair like" provided, k<
in,
in
wj,
but as the photon energy increases beyond this limit, the muon term in eq. (6) takes over and q, increases 20 N
E
15
d qT dqr
2 qT+qf) . (2
After integrating with respect to dqT we get
w
5
dgel9é . We now look at the momentum delivered to the Coulomb field using the approximation, P = E` - m' = E - in -/2E for the reaction, Ws -_~
R2
(6)
+ e + + e-.
ionization
0
100
1000 EN, (GeV)
1500
Fig. 1 . Muon mean energy loss per unit length in iron. Taken from ref. [2].
0168-9002/92/$05 .00 © 1992 - Elsevier Science Publishers B.V. All righi, reserved
1. HIGH LUMINOSITY TRACKING
66
L.S. Osborne / Electron pair production by muons
the muon is at rest and the nucleus is travelling with the muon laboratory velocity and carrying the photons expressed by eq. (1). If you transform the electrons to this frame you find that condition (7) gives the electrons going along the photon direction . The opposite of condition (7) is pair production in the back direction which results in a large momentum transfer to the muon Coulomb field and so a diminishing cross section.
2. Implicat ons EIL (GeV)
Fig. 2. The number of events with energy loss greater than E in I m of iron for 10 TeV and 100 GeV muons from various interactions (ref. [2]) and' A.V. Lanyov, private communication. into the "muon like" region thereby decrasing the cross section . This is in fact what is observed in an accurate calculation [2] (see fig. 2). The equivalent photon energy, k, goes like 1/k as predicted from eqs. (1) and (2) up to an energy of about m ew i /m, after which : falls roughly as k-3. The point of fall-off is predicted by this simple model though not the power of the fall-off. It would be handy to use this model as a simple measure of pair production . We do this by taking eq. (1) with a cutoff at, me .8-0) 1, wl, w where ß is a parameter to be fixed to get the right energy loss from an exact calculation . We have done this by fitting the curve in fig . 1 and obtained, ß = 0 .80. There is a. other way to look at this phenomenon, leading to the same result . Consider the system where
It is clear from eq. (2) that the energy loss by a muon is proportional to its energy from this pair production mechanism . We can expect that a muon exiting from a material of many radiation lengths will bring with it the tails of the high energy showers produced by these pairs and that this will be more aggravated the higher the energy of the muon. However, there is another problem; the muon in traversing detectors of a fraction of a radiation length will also produce pairs. In this case it will be the number of pairs that provide the aggravation since the multiplication (the begining of the shower) will be minimal . In this case, we get the number of pairs by integrating eq. (1) over the equivalent photon energies that can cause trouble in the detector . This number varies only logarithmically with muon energy.
References [1] S. Gasiorowicz, Elementary Particle Physics (Wiley, New York) 180. [2] A.V. Lanyov, Inst. of High Energy Physics Publication IHEP 89-145 and EMPACT Note 321 ; and references therein. [3] EMPACT/TEXAS LOI submitted to the SSC Dec. 1, 1990.