Not back-to-back emission of decay products

In summary, the data shows that muons are not emitted back to back in the LHC. There is a small deviation from this behavior.
  • #1
malawi_glenn
Science Advisor
Homework Helper
Gold Member
6,735
2,448
Hi!

I am reading about the [itex] Z^0 [/itex] production and decay processes at the moment.

I have understand that the [itex] Z^0 [/itex] is produced at rest, and when it decays to two leptons (or quarks) they must be emitted back to back to conserve momentum.

But when I look at real and simulated data, the muons are not emitted back to back, there is a small deviation.

I just wonder "WHY"? Have I missed something fundamental?

I mean, I don't think it has to do with the bending in the central detector magnetic field.

http://www.particle.kth.se/zlab/project/images/event11/event11_side.gif [Broken]

http://www4.tsl.uu.se/~Atlas/Webuppgift/simevent2.html [Broken]
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Neither Z is produced at rest. Why do you think they are?
 
  • #3
Vanadium 50 said:
Neither Z is produced at rest. Why do you think they are?

Ok so the explanation is that the Z is not produced in rest?

I can't see it =(

using four-vectors:

[tex] p(e^+) + p(e^-) = E_{cm} + \vec{0} = 2E_{beam} = p_Z [/tex]

So the Z boson's 3-momenta is equal to 0. I assue that the beams are totally head-on (same energy)
 
  • #4
I think your sentence should read:
"I have understand thatif/in the frame in which the Z0 is produced at rest, then/there when it decays to two leptons (or quarks) they must be emitted back to back to conserve momentum."

I don't know what kind of event your first link is. But the 2nd one is LHC in the lab frame. The cms frame of the relevant partons that make up your process (i.e. ignoring the remants of the initial protons) does not have to coincide with the lab frame.

EDIT: Since I am late with my response: Ebeam is the energy of the proton beam (again@LHC) but you effectively collide quarks and gluons; not the whole protons.
 
  • #5
Ok, then I see why they are not back to back @ LHC :)

But the first image is electron-postiron @ LEP.

In lab-frame if one has equal energy beams, then Z must be produced at rest also in lab-frame? That is what my equation above refers to, sorry if I wasn't clear enough.
 
  • #6
Note that in the 1st link in addition to what supposedly is the two lepton tracks there is a green quad on the right side that indicates a detector signal and that the header sais "Hcal(N=2, SumE=4.5)" which might mean hadron calorimeter. Spontaneously from that I'd suspect that the process of the event is not "e+ e- -> s-channel Z -> mu+ mu-" but something different - but that's only a guess of mine.

EDIT: I've taken a look at the standard LEP91 setup of the Sherpa event generator. That setup includes processes with up to two QCD vertices. Assuming it wasn't put there just for fun that strengthens my idea that the missing p comes from hadronic stuff.
 
Last edited:
  • #7
malawi_glenn said:
In lab-frame if one has equal energy beams, then Z must be produced at rest also in lab-frame?

No. That would be true only if the beam energy were half the Z mass. As you can see from the e+e- event display, the beam energy is 65.129 GeV, significantly above half the Z mass.
 
  • #8
Vanadium 50 said:
No. That would be true only if the beam energy were half the Z mass. As you can see from the e+e- event display, the beam energy is 65.129 GeV, significantly above half the Z mass.


Vanadium 50 said:
No. That would be true only if the beam energy were half the Z mass. As you can see from the e+e- event display, the beam energy is 65.129 GeV, significantly above half the Z mass.




Timo: Ok, but if the beam have 65.129 GeV, the Z boson will be emited with some kinetic energy. But what will be emitted to compensate for the momentum violation? (since we must have overall momentum conservation here?)

V-50: So what reaction can give away two muons and a hadronic signal in an ee-bar collider @ 65GeV beam energy? :S
 
  • #9
malawi_glenn said:
What reaction can give away two muons and a hadronic signal in an ee-bar collider @ 65GeV beam energy? :S

e+e- goes to Z + something else. The most likely "something else" is a photon going down the beampipe. In fact, you can see it in the picture - it's the green thing on the right. It says 32.8 GeV of energy was detected in the forward calorimeter: 65*2 - 33 = 97 GeV, pretty close to 91 GeV.
 
  • #10
Vanadium 50 said:
e+e- goes to Z + something else. The most likely "something else" is a photon going down the beampipe. In fact, you can see it in the picture - it's the green thing on the right. It says 32.8 GeV of energy was detected in the forward calorimeter: 65*2 - 33 = 97 GeV, pretty close to 91 GeV.


Well yes that is of course possble, so the remaining 6GeV could be kinetic energy of the Z.

So the feynman - diag would be: https://www.physicsforums.com/attachment.php?attachmentid=13824&stc=1&d=1209838584

?
 

Attachments

  • zdiag.jpg
    zdiag.jpg
    5.3 KB · Views: 595
  • #11
Yes, a diagram like that.
 
  • #12
Vanadium 50 said:
Yes, a diagram like that.

cool! :biggrin:

Thanx a lot Van-50 and Timo! See you around, and enjoy PF and life
 

1. What is "not back-to-back emission of decay products"?

Not back-to-back emission of decay products refers to the process in which two particles are not emitted in opposite directions during radioactive decay. This is in contrast to back-to-back emission, where the two particles are emitted in opposite directions.

2. Why is "not back-to-back emission of decay products" important?

This phenomenon is important because it gives insight into the internal structure and dynamics of the nucleus. The direction of particle emission can reveal information about the spin and energy of the particles involved in the decay process.

3. How is "not back-to-back emission of decay products" observed?

This phenomenon is typically observed through the use of detectors that can detect the direction and energy of emitted particles. By analyzing the data from these detectors, scientists can determine whether the emission of particles is back-to-back or not.

4. What are some examples of "not back-to-back emission of decay products"?

One example is the decay of a neutron into a proton, electron, and antineutrino. The electron and antineutrino are not emitted in opposite directions, but rather at an angle. Another example is the alpha decay of a nucleus, where the alpha particle is not emitted in the exact opposite direction of the daughter nucleus.

5. How does "not back-to-back emission of decay products" relate to other areas of physics?

This phenomenon is not limited to nuclear physics but is also observed in other areas such as particle physics and astrophysics. It is a fundamental aspect of the decay process and is important in understanding the behavior of particles and the structure of matter.

Similar threads

  • High Energy, Nuclear, Particle Physics
Replies
11
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
17
Views
5K
  • High Energy, Nuclear, Particle Physics
Replies
4
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
4
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
1K
  • High Energy, Nuclear, Particle Physics
2
Replies
49
Views
9K
  • High Energy, Nuclear, Particle Physics
Replies
3
Views
3K
  • High Energy, Nuclear, Particle Physics
Replies
2
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
2
Views
4K
  • High Energy, Nuclear, Particle Physics
2
Replies
41
Views
8K
Back
Top