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Wikipedia

 

The muon (/ˈmjuːɒn/; from the Greek letter mu (μ) used to represent it) is an elementary particle similar to the electron, with electric charge of −1 e and a spin of 1⁄2, but with a much greater mass. It is classified as a lepton. As is the case with other leptons, the muon is not believed to have any sub-structure—that is, it is not thought to be composed of any simpler particles.

 

The muon is an unstable subatomic particle with a mean lifetime of 2.2 µs. Among all known unstable subatomic particles, only the neutron (lasting around 15 minutes) and some atomic nuclei have a longer decay lifetime; others decay significantly faster. The decay of the muon (as well as of the neutron, the longest-lived unstable baryon), is mediated by the weak interaction exclusively. Muon decay always produces at least three particles, which must include an electron of the same charge as the muon and two neutrinos of different types.

Two statements seem contradictory, "the muon is not believed to have any sub-structure—that is, it is not thought to be composed of any simpler particles" and "Muon decay always produces at least three particles, which must include an electron of the same charge as the muon and two neutrinos of different types." If a particle decays, it must be composed of something that can decay, instead of having no substructure. What is really happening?

 

 

Posted (edited)

No particle decay doesn't require a substructure. Particles can decay into less massive particles but must obey several conservation rules such as conservation of Lepton, charge, etc.

 

It's incorrect to think that the particles that are involved in the decay are part of the particle itself. The muon isn't composed of the particles it can decay into

Edited by Mordred
Posted
  On 3/9/2016 at 7:02 PM, Mordred said:

No particle decay doesn't require a substructure. Particles can decay into less massive particles but must obey several conservation rules such as conservation of Lepton, charge, etc.

 

It's incorrect to think that the particles that are involved in the decay are part of the particle itself. The muon isn't composed of the particles it can decay into

I expect the decay involves E=mc2. Maybe insight into why the specific particle masses would be helpful.

Posted

It's total energy that's involved

 

E^2=(pc)^2+(m_oc^2)^2

 

A particle can only decay into particles that have less total energy.

 

For example the LHC smashes protons and yet can cause decays of particles such as the Higgs boson which has a higher rest mass than the protons themselves.

 

It does this via the total energy gain via accelerating the particle to near light speed.

here is the various conservation laws involved

 

http://en.wikipedia....ge_conservation

http://en.wikipedia....rticle_physics)

http://en.wikipedia....i/Lepton_number

http://en.wikipedia....ki/Weak_isospin

These two free articles will help. Though they are lengthy the first article gives a good coverage of the decay rules

 

 

http://arxiv.org/abs/0810.3328

A Simple Introduction to Particle Physics

 

Part 2 is here

http://arxiv.org/abs/0908.1395

Posted
  On 3/9/2016 at 10:16 PM, Mordred said:

It's total energy that's involved

 

E^2=(pc)^2+(m_oc^2)^2

 

A particle can only decay into particles that have less total energy.

 

 

 

It's simpler for decay, though. It has to be able to happen at rest, so we can set p=0. If it can't happen in one frame, it can't happen in any frame.

Posted

Yeah your right I had forgotten that the sum of total energy of the decayed products must be less than or equal to the rest mass of the decaying particle. For others reading.

 

Mc^2=\sum_{products}(mc^2+E_k)

Posted
  On 3/10/2016 at 5:22 AM, Mordred said:

Yeah your right I had forgotten that the sum of total energy of the decayed products must be less than or equal to the rest mass of the decaying particle. For others reading.

 

Mc^2=\sum_{products}(mc^2+E_k)

What is the situation happening in the particle accelerators then? http://www.scienceforums.net/topic/93886-muon-no-substructure-yet-decays/#entry910222 Is that not decay then? Is it more like "making" these particles?

Posted

When particles collide, other particles are created. Some of the created particles, e.g., Higgs Boson (mean lifetime 1.56×10−22 s) will decay and may be imaged by the detectors.

Posted

Is there a general principle that if a less massive particle with the same properties exists, then the more massive particle "must" decay into the lighter equivalent (plus something to take away the excess energy)?

 

So a muon decays into an electron, which conserves all properties except mass-energy, and so a pair of neutrino/antineutrino take away the excess kinetic energy. Does this happen simply because the electron exists as a less massive equivalent of the muon?

Posted
  On 3/10/2016 at 6:21 AM, Robittybob1 said:

What is the situation happening in the particle accelerators then? http://www.scienceforums.net/topic/93886-muon-no-substructure-yet-decays/#entry910222 Is that not decay then? Is it more like "making" these particles?

 

 

Colliders have induced reactions involving more than one particle. Decays are spontaneous and involve a single particle.

  On 3/10/2016 at 11:46 AM, Strange said:

Is there a general principle that if a less massive particle with the same properties exists, then the more massive particle "must" decay into the lighter equivalent (plus something to take away the excess energy)?

 

So a muon decays into an electron, which conserves all properties except mass-energy, and so a pair of neutrino/antineutrino take away the excess kinetic energy. Does this happen simply because the electron exists as a less massive equivalent of the muon?

 

Basically, yes. There's a particle physics saying that goes something like "All that is not forbidden is mandatory" A particular result may not happen because it violates some other conservation law — it's not just energy you have to worry about — but if there is some other pathway, then the decay will occur. It just may be less likely (so it would take longer, on average, for a spontaneous decay, or have a smaller cross-section, for an induced reaction).

Posted

Thanks.

 

I realise now that I have hugely over-simplified because, for example, a tau does not decay into a muon that decays into an electron. There are many other (more likely) possibilities. But the general principle still applies.

  • 4 weeks later...
Posted
  On 3/9/2016 at 6:14 PM, EdEarl said:

Two statements seem contradictory, "the muon is not believed to have any sub-structure—that is, it is not thought to be composed of any simpler particles" and "Muon decay always produces at least three particles, which must include an electron of the same charge as the muon and two neutrinos of different types." If a particle decays, it must be composed of something that can decay, instead of having no substructure. What is really happening?

 

 

One "simple" example is 40K which can decay via beta-plus or beta-minus (or electron capture too), suggesting that decay is not a composite particle splitting into smaller compounds. That is, if the neutron were an electron and a proton close to an other, as wrong theories supposed long ago, then the emission of a positron would be impossible.

 

So, new particles are created from nothing, provided the energy is available and some quantities are conserved. Just like a photon is created or destroyed.

  On 3/10/2016 at 11:46 AM, Strange said:

Is there a general principle that if a less massive particle with the same properties exists, then the more massive particle "must" decay into the lighter equivalent (plus something to take away the excess energy)?

 

So a muon decays into an electron, which conserves all properties except mass-energy, and so a pair of neutrino/antineutrino take away the excess kinetic energy. Does this happen simply because the electron exists as a less massive equivalent of the muon?

I imagine this is because our universe is cold now.

 

If the decay with emission of a neutrino pair takes place, the inverse reaction must be possible too, but it doesn't happen because the random presence of a suitable pair of energetic neutrinos is rare these days.

 

With the proper reformulation, thermodynamics should model that. Put the proper temperature, and you get an equilibrium between the lighter and heavier particle. But at 3K, the equilibrium is so unfavourable to the heavy particle that only one reaction direction is observed.

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