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Is the Higgs boson an elementary particle? If so, why does it decay? [duplicate]
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This question already has an answer here:
Why are muons considered to be “elementary particles” in the Standard Model?
5 answers
The Higgs boson is an excitation of the Higgs field and is very massive and short lived. It also interacts with the Higgs field and thus is able to experience mass.
Why does it decay if it is supposed to be an elementary particle according to the standard model?
standard-model higgs elementary-particles
edited Jan 20 at 3:24
Peter Mortensen
1,93611323
asked Jan 15 at 3:42
user6760user6760
2,80611940
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marked as duplicate by Chair, ACuriousMind♦ Jan 20 at 14:29
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Why are muons considered to be “elementary particles” in the Standard Model?
5 answers
The Higgs boson is an excitation of the Higgs field and is very massive and short lived. It also interacts with the Higgs field and thus is able to experience mass.
Why does it decay if it is supposed to be an elementary particle according to the standard model?
standard-model higgs elementary-particles
edited Jan 20 at 3:24
Peter Mortensen
1,93611323
asked Jan 15 at 3:42
user6760user6760
2,80611940
$endgroup$
marked as duplicate by Chair, ACuriousMind♦ Jan 20 at 14:29
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
5
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I've removed some comments that were answering the question. Please keep in mind that comments are meant for requesting clarification or suggesting improvements to the question, not for answering.
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– David Z♦
Jan 16 at 7:37
1
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Essentially a duplicate of Why are muons considered to be “elementary particles” in the Standard Model? with "muon" replaced by "Higgs" (which was also a HNQ). Are we going to do a HNQ for every particle of the Standard Model now?
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– ACuriousMind♦
Jan 20 at 13:14
add a comment |
$begingroup$
This question already has an answer here:
Why are muons considered to be “elementary particles” in the Standard Model?
5 answers
The Higgs boson is an excitation of the Higgs field and is very massive and short lived. It also interacts with the Higgs field and thus is able to experience mass.
Why does it decay if it is supposed to be an elementary particle according to the standard model?
standard-model higgs elementary-particles
edited Jan 20 at 3:24
Peter Mortensen
1,93611323
asked Jan 15 at 3:42
user6760user6760
2,80611940
$endgroup$
This question already has an answer here:
Why are muons considered to be “elementary particles” in the Standard Model?
5 answers
The Higgs boson is an excitation of the Higgs field and is very massive and short lived. It also interacts with the Higgs field and thus is able to experience mass.
Why does it decay if it is supposed to be an elementary particle according to the standard model?
This question already has an answer here:
Why are muons considered to be “elementary particles” in the Standard Model?
5 answers
standard-model higgs elementary-particles
standard-model higgs elementary-particles
edited Jan 20 at 3:24
Peter Mortensen
1,93611323
asked Jan 15 at 3:42
user6760user6760
2,80611940
edited Jan 20 at 3:24
Peter Mortensen
1,93611323
asked Jan 15 at 3:42
user6760user6760
2,80611940
edited Jan 20 at 3:24
Peter Mortensen
1,93611323
edited Jan 20 at 3:24
Peter Mortensen
1,93611323
edited Jan 20 at 3:24
Peter Mortensen
1,93611323
1,93611323
asked Jan 15 at 3:42
user6760user6760
2,80611940
asked Jan 15 at 3:42
user6760user6760
2,80611940
asked Jan 15 at 3:42
user6760user6760
2,80611940
2,80611940
marked as duplicate by Chair, ACuriousMind♦ Jan 20 at 14:29
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Chair, ACuriousMind♦ Jan 20 at 14:29
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
5
$begingroup$
I've removed some comments that were answering the question. Please keep in mind that comments are meant for requesting clarification or suggesting improvements to the question, not for answering.
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– David Z♦
Jan 16 at 7:37
1
$begingroup$
Essentially a duplicate of Why are muons considered to be “elementary particles” in the Standard Model? with "muon" replaced by "Higgs" (which was also a HNQ). Are we going to do a HNQ for every particle of the Standard Model now?
$endgroup$
– ACuriousMind♦
Jan 20 at 13:14
add a comment |
5
$begingroup$
I've removed some comments that were answering the question. Please keep in mind that comments are meant for requesting clarification or suggesting improvements to the question, not for answering.
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– David Z♦
Jan 16 at 7:37
1
$begingroup$
Essentially a duplicate of Why are muons considered to be “elementary particles” in the Standard Model? with "muon" replaced by "Higgs" (which was also a HNQ). Are we going to do a HNQ for every particle of the Standard Model now?
$endgroup$
– ACuriousMind♦
Jan 20 at 13:14
5
5
$begingroup$
I've removed some comments that were answering the question. Please keep in mind that comments are meant for requesting clarification or suggesting improvements to the question, not for answering.
$endgroup$
– David Z♦
Jan 16 at 7:37
$begingroup$
I've removed some comments that were answering the question. Please keep in mind that comments are meant for requesting clarification or suggesting improvements to the question, not for answering.
$endgroup$
– David Z♦
Jan 16 at 7:37
1
1
$begingroup$
Essentially a duplicate of Why are muons considered to be “elementary particles” in the Standard Model? with "muon" replaced by "Higgs" (which was also a HNQ). Are we going to do a HNQ for every particle of the Standard Model now?
$endgroup$
– ACuriousMind♦
Jan 20 at 13:14
$begingroup$
Essentially a duplicate of Why are muons considered to be “elementary particles” in the Standard Model? with "muon" replaced by "Higgs" (which was also a HNQ). Are we going to do a HNQ for every particle of the Standard Model now?
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– ACuriousMind♦
Jan 20 at 13:14
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6 Answers
6
active
oldest
votes
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Most fundamental particles in the standard model decay: muons, tau leptons, the heavy quarks, W and Z bosons. There’s nothing problematic about that, nor about Higgs decays.
Your question may come from a misconception about particle decay: that it’s somehow the particle ‘coming apart’ into preexisting constituents. It’s not like that. Decays are transformations into things that weren’t there before.
answered Jan 15 at 3:55
Bob JacobsenBob Jacobsen
5,2421017
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1
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Hi I'm still not clear about this transformation, I just read it is probabilistic so higgs boson can in fact decay into many things including 2 photons so can the same 2 photons cannot transform back into higgs boson? I highly doubt so but dunno why?
$endgroup$
– user6760
Jan 15 at 4:46
21
$begingroup$
Generally, particle physics reactions can go either way. Yes, if you had sufficiently energetic photons appropriately arranged, the SM says they could combine to form a Higgs particle.
$endgroup$
– Bob Jacobsen
Jan 15 at 4:49
2
$begingroup$
@safesphere the diagram for 2gamma to e+e- certainly exists and has a non-zero amplitude. I agree that phase factors make it small (that’s the “appropriately arranged” bit). But it was a major part of Big Bang thermalization before freeze-out, and its the mechanism for photon to e+e- pair production (via a photon from a nucleus)
$endgroup$
– Bob Jacobsen
Jan 15 at 5:11
1
$begingroup$
@safesphere as G.Smith points out, the amplitude of the diagram is exactly the same. The cross section (probability) and event rate can differ because of the actual ongoing & outgoing particles: it’s easier to make electrons than high energy photons in a small space.
$endgroup$
– Bob Jacobsen
Jan 15 at 6:11
2
$begingroup$
Even more on point: Telnov proposed building a collider to study photon-photon Higgs production. arxiv.org/abs/1409.5563.
$endgroup$
– Bob Jacobsen
Jan 15 at 6:19
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show 5 more comments
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Another way to answer this question is that particles are not "elementary," not even in a given quantum field theory. Quantum field theories (like the Standard Model) are expressed in terms of fields, not particles. Particles are phenomena that the model predicts; some of them are stable, some are transient (they decay). The Standard Model is constructed using an elementary Higgs field, and it predicts a Higgs particle, which is unstable.
Although the language "elementary particle" is very common and probably can't be revised at this point, it might be less confusing and more accurate to talk about the elementary fields used to express a model. Even that language isn't perfect, though, because some models can be expressed in more than one way, using seemingly-unrelated sets of fields. Quantum field theory is a rich subject with many surprises!
answered Jan 15 at 4:15
Dan YandDan Yand
11k21540
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1
$begingroup$
are u saying the excitation of the field can disturb other fields too? So the reality is just fields interacting with one another.
$endgroup$
– user6760
Jan 15 at 4:53
5
$begingroup$
@user6760 I'll shy away from using the word "reality" here (because different-looking descriptions can make equivalent predictions), but yes: The way quantum field theory describes things is as quantum fields interacting with each other. A particle is one manifestation of all those fields interacting with each other. The Higgs particle involves more than just the Higgs field.
$endgroup$
– Dan Yand
Jan 15 at 4:56
$begingroup$
@user6760 A common approximation method in QFT involves starting with a different model that has only non-interacting fields, then adding a series of "corrections" to gradually scootch the results closer to what the real model with interacting fields would predict. That's what Feynman diagrams are about, and that's what the "virtual particle" langauge is about. In a model with non-interacting fields, there is a relatively direct correspondence between fields and particles; but that correspondence becomes less direct (to say the least) in models where the fields interact.
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– Dan Yand
Jan 15 at 5:08
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Yes it makes sense to me now the virtual particle that you have mentioned.
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– user6760
Jan 15 at 5:23
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To add to what Dan mentioned - which is impeccable according to QFT - here a couple of examples of Higgs decay expressed in Feynman diagrams: inspirehep.net/record/929477/plots that involve more than photons
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– virtual82
Jan 15 at 7:20
add a comment |
$begingroup$
"Decay" is just the name given to an interaction where one particle goes in and two or more particles go out.
The rule in quantum mechanics is "anything not forbidden is compulsory" – that is, any process (decay or otherwise) can happen unless it violates a conservation law. As a result, most particles, whether fundamental or composite, do decay.
The exceptions are particles for which there's literally no set of outputs you can choose that doesn't violate some conservation law. For example, electrons can't decay because the decay would have to conserve electric charge – so at least one output would have to be charged – and would also have to conserve mass/energy – so the total mass of all the outputs would have to be no larger than the electron's mass – and this is impossible because there are no electrically charged particles with lower mass. So it's stable not because it's elementary but because everything that could make it unstable is forbidden.
The muon, which is also fundamental and is almost identical to the electron except for its mass, can and does decay, because the higher input mass means that you can find outputs that conserve mass while also satisfying all other constraints.
The proton, which is not a fundamental particle, can't decay because it's the lightest particle with another conserved property called baryon number.
But, again, the particles that can't decay are the exceptions. As a rule almost everything can and does, and this applies to fundamental and composite particles alike.
answered Jan 15 at 8:51
benrgbenrg
2,138616
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2
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And even those that "can't" decay, could potentially decay extremely rarely, by some extremely unlikely quantum phenomenon. Quantum tunnelling (moving between states that on the surface apparently can't be reached from each other) is a term used here. So even if a proton can't decay, there may be processes that for all practical purposes never happen in an incomprehensible timescale of 10^35 years, which happen often on an even more incomprehensible timescale of 10^100 or 10^1000 years.
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– Stilez
Jan 15 at 9:38
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All fundamental or elementary particles decay after being born. Take, for example, electron. While being created in some process, it "decays" into "another electron" and many soft photons. As it is unlikely that "another electron" may stay without further interactions with its environment, it continues to interact, i.e., generally speaking, absorb and emit soft photons.
answered Jan 15 at 8:35
Vladimir KalitvianskiVladimir Kalitvianski
11k11334
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Are you referring to the emission of photon when an electron goes from excited state to ground state as a form of decay(transformation which I just learnt today)? So everything is just interaction and decay is one of them right?
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– user6760
Jan 15 at 8:51
1
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Right. One may safely say that there are changes of "occupation numbers" of "elementary" particles (excitations) due to interactions (that never end).
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– Vladimir Kalitvianski
Jan 15 at 8:55
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A particle is elementary when there aren't subcomponents that we can identify.
This has nothing to do with the concept of decay, and you can easily convince yourself of this fact by observing that whereas a particle (elementary or not) may decay in many different ways, the number and type of its constituents is univocally determined.
answered Jan 15 at 4:42
Francesco BernardiniFrancesco Bernardini
49915
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add a comment |
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The decay of quanta is not like solid matter breaking apart. When we think of the word "decay" it is often associated with rotting flesh or physical separation. Decay could also be simply a depreciation.
It might help one to consider of Bosons and Fermions as states of Energy which follow stochastic rules. As the energy is conserved from the initial state to the post-"decay" state, you can think of decay as a change from an unstable energetic state to an alternative energetic state within its intrinsic nature by way of an increase in entropy (and probably other speculated or mathematically-derived phenomena).
Furthermore, the nature of all particles as fundamental is still up for debate.
answered Jan 17 at 15:00
Daniel ThorneDaniel Thorne
1
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add a comment |
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Most fundamental particles in the standard model decay: muons, tau leptons, the heavy quarks, W and Z bosons. There’s nothing problematic about that, nor about Higgs decays.
Your question may come from a misconception about particle decay: that it’s somehow the particle ‘coming apart’ into preexisting constituents. It’s not like that. Decays are transformations into things that weren’t there before.
answered Jan 15 at 3:55
Bob JacobsenBob Jacobsen
5,2421017
$endgroup$
1
$begingroup$
Hi I'm still not clear about this transformation, I just read it is probabilistic so higgs boson can in fact decay into many things including 2 photons so can the same 2 photons cannot transform back into higgs boson? I highly doubt so but dunno why?
$endgroup$
– user6760
Jan 15 at 4:46
21
$begingroup$
Generally, particle physics reactions can go either way. Yes, if you had sufficiently energetic photons appropriately arranged, the SM says they could combine to form a Higgs particle.
$endgroup$
– Bob Jacobsen
Jan 15 at 4:49
2
$begingroup$
@safesphere the diagram for 2gamma to e+e- certainly exists and has a non-zero amplitude. I agree that phase factors make it small (that’s the “appropriately arranged” bit). But it was a major part of Big Bang thermalization before freeze-out, and its the mechanism for photon to e+e- pair production (via a photon from a nucleus)
$endgroup$
– Bob Jacobsen
Jan 15 at 5:11
1
$begingroup$
@safesphere as G.Smith points out, the amplitude of the diagram is exactly the same. The cross section (probability) and event rate can differ because of the actual ongoing & outgoing particles: it’s easier to make electrons than high energy photons in a small space.
$endgroup$
– Bob Jacobsen
Jan 15 at 6:11
2
$begingroup$
Even more on point: Telnov proposed building a collider to study photon-photon Higgs production. arxiv.org/abs/1409.5563.
$endgroup$
– Bob Jacobsen
Jan 15 at 6:19
|
show 5 more comments
$begingroup$
Most fundamental particles in the standard model decay: muons, tau leptons, the heavy quarks, W and Z bosons. There’s nothing problematic about that, nor about Higgs decays.
Your question may come from a misconception about particle decay: that it’s somehow the particle ‘coming apart’ into preexisting constituents. It’s not like that. Decays are transformations into things that weren’t there before.
answered Jan 15 at 3:55
Bob JacobsenBob Jacobsen
5,2421017
$endgroup$
1
$begingroup$
Hi I'm still not clear about this transformation, I just read it is probabilistic so higgs boson can in fact decay into many things including 2 photons so can the same 2 photons cannot transform back into higgs boson? I highly doubt so but dunno why?
$endgroup$
– user6760
Jan 15 at 4:46
21
$begingroup$
Generally, particle physics reactions can go either way. Yes, if you had sufficiently energetic photons appropriately arranged, the SM says they could combine to form a Higgs particle.
$endgroup$
– Bob Jacobsen
Jan 15 at 4:49
2
$begingroup$
@safesphere the diagram for 2gamma to e+e- certainly exists and has a non-zero amplitude. I agree that phase factors make it small (that’s the “appropriately arranged” bit). But it was a major part of Big Bang thermalization before freeze-out, and its the mechanism for photon to e+e- pair production (via a photon from a nucleus)
$endgroup$
– Bob Jacobsen
Jan 15 at 5:11
1
$begingroup$
@safesphere as G.Smith points out, the amplitude of the diagram is exactly the same. The cross section (probability) and event rate can differ because of the actual ongoing & outgoing particles: it’s easier to make electrons than high energy photons in a small space.
$endgroup$
– Bob Jacobsen
Jan 15 at 6:11
2
$begingroup$
Even more on point: Telnov proposed building a collider to study photon-photon Higgs production. arxiv.org/abs/1409.5563.
$endgroup$
– Bob Jacobsen
Jan 15 at 6:19
|
show 5 more comments
$begingroup$
Most fundamental particles in the standard model decay: muons, tau leptons, the heavy quarks, W and Z bosons. There’s nothing problematic about that, nor about Higgs decays.
Your question may come from a misconception about particle decay: that it’s somehow the particle ‘coming apart’ into preexisting constituents. It’s not like that. Decays are transformations into things that weren’t there before.
answered Jan 15 at 3:55
Bob JacobsenBob Jacobsen
5,2421017
$endgroup$
Most fundamental particles in the standard model decay: muons, tau leptons, the heavy quarks, W and Z bosons. There’s nothing problematic about that, nor about Higgs decays.
Your question may come from a misconception about particle decay: that it’s somehow the particle ‘coming apart’ into preexisting constituents. It’s not like that. Decays are transformations into things that weren’t there before.
answered Jan 15 at 3:55
Bob JacobsenBob Jacobsen
5,2421017
answered Jan 15 at 3:55
Bob JacobsenBob Jacobsen
5,2421017
answered Jan 15 at 3:55
Bob JacobsenBob Jacobsen
5,2421017
answered Jan 15 at 3:55
Bob JacobsenBob Jacobsen
5,2421017
5,2421017
1
$begingroup$
Hi I'm still not clear about this transformation, I just read it is probabilistic so higgs boson can in fact decay into many things including 2 photons so can the same 2 photons cannot transform back into higgs boson? I highly doubt so but dunno why?
$endgroup$
– user6760
Jan 15 at 4:46
21
$begingroup$
Generally, particle physics reactions can go either way. Yes, if you had sufficiently energetic photons appropriately arranged, the SM says they could combine to form a Higgs particle.
$endgroup$
– Bob Jacobsen
Jan 15 at 4:49
2
$begingroup$
@safesphere the diagram for 2gamma to e+e- certainly exists and has a non-zero amplitude. I agree that phase factors make it small (that’s the “appropriately arranged” bit). But it was a major part of Big Bang thermalization before freeze-out, and its the mechanism for photon to e+e- pair production (via a photon from a nucleus)
$endgroup$
– Bob Jacobsen
Jan 15 at 5:11
1
$begingroup$
@safesphere as G.Smith points out, the amplitude of the diagram is exactly the same. The cross section (probability) and event rate can differ because of the actual ongoing & outgoing particles: it’s easier to make electrons than high energy photons in a small space.
$endgroup$
– Bob Jacobsen
Jan 15 at 6:11
2
$begingroup$
Even more on point: Telnov proposed building a collider to study photon-photon Higgs production. arxiv.org/abs/1409.5563.
$endgroup$
– Bob Jacobsen
Jan 15 at 6:19
|
show 5 more comments
1
$begingroup$
Hi I'm still not clear about this transformation, I just read it is probabilistic so higgs boson can in fact decay into many things including 2 photons so can the same 2 photons cannot transform back into higgs boson? I highly doubt so but dunno why?
$endgroup$
– user6760
Jan 15 at 4:46
21
$begingroup$
Generally, particle physics reactions can go either way. Yes, if you had sufficiently energetic photons appropriately arranged, the SM says they could combine to form a Higgs particle.
$endgroup$
– Bob Jacobsen
Jan 15 at 4:49
2
$begingroup$
@safesphere the diagram for 2gamma to e+e- certainly exists and has a non-zero amplitude. I agree that phase factors make it small (that’s the “appropriately arranged” bit). But it was a major part of Big Bang thermalization before freeze-out, and its the mechanism for photon to e+e- pair production (via a photon from a nucleus)
$endgroup$
– Bob Jacobsen
Jan 15 at 5:11
1
$begingroup$
@safesphere as G.Smith points out, the amplitude of the diagram is exactly the same. The cross section (probability) and event rate can differ because of the actual ongoing & outgoing particles: it’s easier to make electrons than high energy photons in a small space.
$endgroup$
– Bob Jacobsen
Jan 15 at 6:11
2
$begingroup$
Even more on point: Telnov proposed building a collider to study photon-photon Higgs production. arxiv.org/abs/1409.5563.
$endgroup$
– Bob Jacobsen
Jan 15 at 6:19
1
1
$begingroup$
Hi I'm still not clear about this transformation, I just read it is probabilistic so higgs boson can in fact decay into many things including 2 photons so can the same 2 photons cannot transform back into higgs boson? I highly doubt so but dunno why?
$endgroup$
– user6760
Jan 15 at 4:46
$begingroup$
Hi I'm still not clear about this transformation, I just read it is probabilistic so higgs boson can in fact decay into many things including 2 photons so can the same 2 photons cannot transform back into higgs boson? I highly doubt so but dunno why?
$endgroup$
– user6760
Jan 15 at 4:46
21
21
$begingroup$
Generally, particle physics reactions can go either way. Yes, if you had sufficiently energetic photons appropriately arranged, the SM says they could combine to form a Higgs particle.
$endgroup$
– Bob Jacobsen
Jan 15 at 4:49
$begingroup$
Generally, particle physics reactions can go either way. Yes, if you had sufficiently energetic photons appropriately arranged, the SM says they could combine to form a Higgs particle.
$endgroup$
– Bob Jacobsen
Jan 15 at 4:49
2
2
$begingroup$
@safesphere the diagram for 2gamma to e+e- certainly exists and has a non-zero amplitude. I agree that phase factors make it small (that’s the “appropriately arranged” bit). But it was a major part of Big Bang thermalization before freeze-out, and its the mechanism for photon to e+e- pair production (via a photon from a nucleus)
$endgroup$
– Bob Jacobsen
Jan 15 at 5:11
$begingroup$
@safesphere the diagram for 2gamma to e+e- certainly exists and has a non-zero amplitude. I agree that phase factors make it small (that’s the “appropriately arranged” bit). But it was a major part of Big Bang thermalization before freeze-out, and its the mechanism for photon to e+e- pair production (via a photon from a nucleus)
$endgroup$
– Bob Jacobsen
Jan 15 at 5:11
1
1
$begingroup$
@safesphere as G.Smith points out, the amplitude of the diagram is exactly the same. The cross section (probability) and event rate can differ because of the actual ongoing & outgoing particles: it’s easier to make electrons than high energy photons in a small space.
$endgroup$
– Bob Jacobsen
Jan 15 at 6:11
$begingroup$
@safesphere as G.Smith points out, the amplitude of the diagram is exactly the same. The cross section (probability) and event rate can differ because of the actual ongoing & outgoing particles: it’s easier to make electrons than high energy photons in a small space.
$endgroup$
– Bob Jacobsen
Jan 15 at 6:11
2
2
$begingroup$
Even more on point: Telnov proposed building a collider to study photon-photon Higgs production. arxiv.org/abs/1409.5563.
$endgroup$
– Bob Jacobsen
Jan 15 at 6:19
$begingroup$
Even more on point: Telnov proposed building a collider to study photon-photon Higgs production. arxiv.org/abs/1409.5563.
$endgroup$
– Bob Jacobsen
Jan 15 at 6:19
|
show 5 more comments
$begingroup$
Another way to answer this question is that particles are not "elementary," not even in a given quantum field theory. Quantum field theories (like the Standard Model) are expressed in terms of fields, not particles. Particles are phenomena that the model predicts; some of them are stable, some are transient (they decay). The Standard Model is constructed using an elementary Higgs field, and it predicts a Higgs particle, which is unstable.
Although the language "elementary particle" is very common and probably can't be revised at this point, it might be less confusing and more accurate to talk about the elementary fields used to express a model. Even that language isn't perfect, though, because some models can be expressed in more than one way, using seemingly-unrelated sets of fields. Quantum field theory is a rich subject with many surprises!
answered Jan 15 at 4:15
Dan YandDan Yand
11k21540
$endgroup$
1
$begingroup$
are u saying the excitation of the field can disturb other fields too? So the reality is just fields interacting with one another.
$endgroup$
– user6760
Jan 15 at 4:53
5
$begingroup$
@user6760 I'll shy away from using the word "reality" here (because different-looking descriptions can make equivalent predictions), but yes: The way quantum field theory describes things is as quantum fields interacting with each other. A particle is one manifestation of all those fields interacting with each other. The Higgs particle involves more than just the Higgs field.
$endgroup$
– Dan Yand
Jan 15 at 4:56
$begingroup$
@user6760 A common approximation method in QFT involves starting with a different model that has only non-interacting fields, then adding a series of "corrections" to gradually scootch the results closer to what the real model with interacting fields would predict. That's what Feynman diagrams are about, and that's what the "virtual particle" langauge is about. In a model with non-interacting fields, there is a relatively direct correspondence between fields and particles; but that correspondence becomes less direct (to say the least) in models where the fields interact.
$endgroup$
– Dan Yand
Jan 15 at 5:08
$begingroup$
Yes it makes sense to me now the virtual particle that you have mentioned.
$endgroup$
– user6760
Jan 15 at 5:23
$begingroup$
To add to what Dan mentioned - which is impeccable according to QFT - here a couple of examples of Higgs decay expressed in Feynman diagrams: inspirehep.net/record/929477/plots that involve more than photons
$endgroup$
– virtual82
Jan 15 at 7:20
add a comment |
$begingroup$
Another way to answer this question is that particles are not "elementary," not even in a given quantum field theory. Quantum field theories (like the Standard Model) are expressed in terms of fields, not particles. Particles are phenomena that the model predicts; some of them are stable, some are transient (they decay). The Standard Model is constructed using an elementary Higgs field, and it predicts a Higgs particle, which is unstable.
Although the language "elementary particle" is very common and probably can't be revised at this point, it might be less confusing and more accurate to talk about the elementary fields used to express a model. Even that language isn't perfect, though, because some models can be expressed in more than one way, using seemingly-unrelated sets of fields. Quantum field theory is a rich subject with many surprises!
answered Jan 15 at 4:15
Dan YandDan Yand
11k21540
$endgroup$
1
$begingroup$
are u saying the excitation of the field can disturb other fields too? So the reality is just fields interacting with one another.
$endgroup$
– user6760
Jan 15 at 4:53
5
$begingroup$
@user6760 I'll shy away from using the word "reality" here (because different-looking descriptions can make equivalent predictions), but yes: The way quantum field theory describes things is as quantum fields interacting with each other. A particle is one manifestation of all those fields interacting with each other. The Higgs particle involves more than just the Higgs field.
$endgroup$
– Dan Yand
Jan 15 at 4:56
$begingroup$
@user6760 A common approximation method in QFT involves starting with a different model that has only non-interacting fields, then adding a series of "corrections" to gradually scootch the results closer to what the real model with interacting fields would predict. That's what Feynman diagrams are about, and that's what the "virtual particle" langauge is about. In a model with non-interacting fields, there is a relatively direct correspondence between fields and particles; but that correspondence becomes less direct (to say the least) in models where the fields interact.
$endgroup$
– Dan Yand
Jan 15 at 5:08
$begingroup$
Yes it makes sense to me now the virtual particle that you have mentioned.
$endgroup$
– user6760
Jan 15 at 5:23
$begingroup$
To add to what Dan mentioned - which is impeccable according to QFT - here a couple of examples of Higgs decay expressed in Feynman diagrams: inspirehep.net/record/929477/plots that involve more than photons
$endgroup$
– virtual82
Jan 15 at 7:20
add a comment |
$begingroup$
Another way to answer this question is that particles are not "elementary," not even in a given quantum field theory. Quantum field theories (like the Standard Model) are expressed in terms of fields, not particles. Particles are phenomena that the model predicts; some of them are stable, some are transient (they decay). The Standard Model is constructed using an elementary Higgs field, and it predicts a Higgs particle, which is unstable.
Although the language "elementary particle" is very common and probably can't be revised at this point, it might be less confusing and more accurate to talk about the elementary fields used to express a model. Even that language isn't perfect, though, because some models can be expressed in more than one way, using seemingly-unrelated sets of fields. Quantum field theory is a rich subject with many surprises!
answered Jan 15 at 4:15
Dan YandDan Yand
11k21540
$endgroup$
Another way to answer this question is that particles are not "elementary," not even in a given quantum field theory. Quantum field theories (like the Standard Model) are expressed in terms of fields, not particles. Particles are phenomena that the model predicts; some of them are stable, some are transient (they decay). The Standard Model is constructed using an elementary Higgs field, and it predicts a Higgs particle, which is unstable.
Although the language "elementary particle" is very common and probably can't be revised at this point, it might be less confusing and more accurate to talk about the elementary fields used to express a model. Even that language isn't perfect, though, because some models can be expressed in more than one way, using seemingly-unrelated sets of fields. Quantum field theory is a rich subject with many surprises!
answered Jan 15 at 4:15
Dan YandDan Yand
11k21540
answered Jan 15 at 4:15
Dan YandDan Yand
11k21540
answered Jan 15 at 4:15
Dan YandDan Yand
11k21540
answered Jan 15 at 4:15
Dan YandDan Yand
11k21540
11k21540
1
$begingroup$
are u saying the excitation of the field can disturb other fields too? So the reality is just fields interacting with one another.
$endgroup$
– user6760
Jan 15 at 4:53
5
$begingroup$
@user6760 I'll shy away from using the word "reality" here (because different-looking descriptions can make equivalent predictions), but yes: The way quantum field theory describes things is as quantum fields interacting with each other. A particle is one manifestation of all those fields interacting with each other. The Higgs particle involves more than just the Higgs field.
$endgroup$
– Dan Yand
Jan 15 at 4:56
$begingroup$
@user6760 A common approximation method in QFT involves starting with a different model that has only non-interacting fields, then adding a series of "corrections" to gradually scootch the results closer to what the real model with interacting fields would predict. That's what Feynman diagrams are about, and that's what the "virtual particle" langauge is about. In a model with non-interacting fields, there is a relatively direct correspondence between fields and particles; but that correspondence becomes less direct (to say the least) in models where the fields interact.
$endgroup$
– Dan Yand
Jan 15 at 5:08
$begingroup$
Yes it makes sense to me now the virtual particle that you have mentioned.
$endgroup$
– user6760
Jan 15 at 5:23
$begingroup$
To add to what Dan mentioned - which is impeccable according to QFT - here a couple of examples of Higgs decay expressed in Feynman diagrams: inspirehep.net/record/929477/plots that involve more than photons
$endgroup$
– virtual82
Jan 15 at 7:20
add a comment |
1
$begingroup$
are u saying the excitation of the field can disturb other fields too? So the reality is just fields interacting with one another.
$endgroup$
– user6760
Jan 15 at 4:53
5
$begingroup$
@user6760 I'll shy away from using the word "reality" here (because different-looking descriptions can make equivalent predictions), but yes: The way quantum field theory describes things is as quantum fields interacting with each other. A particle is one manifestation of all those fields interacting with each other. The Higgs particle involves more than just the Higgs field.
$endgroup$
– Dan Yand
Jan 15 at 4:56
$begingroup$
@user6760 A common approximation method in QFT involves starting with a different model that has only non-interacting fields, then adding a series of "corrections" to gradually scootch the results closer to what the real model with interacting fields would predict. That's what Feynman diagrams are about, and that's what the "virtual particle" langauge is about. In a model with non-interacting fields, there is a relatively direct correspondence between fields and particles; but that correspondence becomes less direct (to say the least) in models where the fields interact.
$endgroup$
– Dan Yand
Jan 15 at 5:08
$begingroup$
Yes it makes sense to me now the virtual particle that you have mentioned.
$endgroup$
– user6760
Jan 15 at 5:23
$begingroup$
To add to what Dan mentioned - which is impeccable according to QFT - here a couple of examples of Higgs decay expressed in Feynman diagrams: inspirehep.net/record/929477/plots that involve more than photons
$endgroup$
– virtual82
Jan 15 at 7:20
1
1
$begingroup$
are u saying the excitation of the field can disturb other fields too? So the reality is just fields interacting with one another.
$endgroup$
– user6760
Jan 15 at 4:53
$begingroup$
are u saying the excitation of the field can disturb other fields too? So the reality is just fields interacting with one another.
$endgroup$
– user6760
Jan 15 at 4:53
5
5
$begingroup$
@user6760 I'll shy away from using the word "reality" here (because different-looking descriptions can make equivalent predictions), but yes: The way quantum field theory describes things is as quantum fields interacting with each other. A particle is one manifestation of all those fields interacting with each other. The Higgs particle involves more than just the Higgs field.
$endgroup$
– Dan Yand
Jan 15 at 4:56
$begingroup$
@user6760 I'll shy away from using the word "reality" here (because different-looking descriptions can make equivalent predictions), but yes: The way quantum field theory describes things is as quantum fields interacting with each other. A particle is one manifestation of all those fields interacting with each other. The Higgs particle involves more than just the Higgs field.
$endgroup$
– Dan Yand
Jan 15 at 4:56
$begingroup$
@user6760 A common approximation method in QFT involves starting with a different model that has only non-interacting fields, then adding a series of "corrections" to gradually scootch the results closer to what the real model with interacting fields would predict. That's what Feynman diagrams are about, and that's what the "virtual particle" langauge is about. In a model with non-interacting fields, there is a relatively direct correspondence between fields and particles; but that correspondence becomes less direct (to say the least) in models where the fields interact.
$endgroup$
– Dan Yand
Jan 15 at 5:08
$begingroup$
@user6760 A common approximation method in QFT involves starting with a different model that has only non-interacting fields, then adding a series of "corrections" to gradually scootch the results closer to what the real model with interacting fields would predict. That's what Feynman diagrams are about, and that's what the "virtual particle" langauge is about. In a model with non-interacting fields, there is a relatively direct correspondence between fields and particles; but that correspondence becomes less direct (to say the least) in models where the fields interact.
$endgroup$
– Dan Yand
Jan 15 at 5:08
$begingroup$
Yes it makes sense to me now the virtual particle that you have mentioned.
$endgroup$
– user6760
Jan 15 at 5:23
$begingroup$
Yes it makes sense to me now the virtual particle that you have mentioned.
$endgroup$
– user6760
Jan 15 at 5:23
$begingroup$
To add to what Dan mentioned - which is impeccable according to QFT - here a couple of examples of Higgs decay expressed in Feynman diagrams: inspirehep.net/record/929477/plots that involve more than photons
$endgroup$
– virtual82
Jan 15 at 7:20
$begingroup$
To add to what Dan mentioned - which is impeccable according to QFT - here a couple of examples of Higgs decay expressed in Feynman diagrams: inspirehep.net/record/929477/plots that involve more than photons
$endgroup$
– virtual82
Jan 15 at 7:20
add a comment |
$begingroup$
"Decay" is just the name given to an interaction where one particle goes in and two or more particles go out.
The rule in quantum mechanics is "anything not forbidden is compulsory" – that is, any process (decay or otherwise) can happen unless it violates a conservation law. As a result, most particles, whether fundamental or composite, do decay.
The exceptions are particles for which there's literally no set of outputs you can choose that doesn't violate some conservation law. For example, electrons can't decay because the decay would have to conserve electric charge – so at least one output would have to be charged – and would also have to conserve mass/energy – so the total mass of all the outputs would have to be no larger than the electron's mass – and this is impossible because there are no electrically charged particles with lower mass. So it's stable not because it's elementary but because everything that could make it unstable is forbidden.
The muon, which is also fundamental and is almost identical to the electron except for its mass, can and does decay, because the higher input mass means that you can find outputs that conserve mass while also satisfying all other constraints.
The proton, which is not a fundamental particle, can't decay because it's the lightest particle with another conserved property called baryon number.
But, again, the particles that can't decay are the exceptions. As a rule almost everything can and does, and this applies to fundamental and composite particles alike.
answered Jan 15 at 8:51
benrgbenrg
2,138616
$endgroup$
2
$begingroup$
And even those that "can't" decay, could potentially decay extremely rarely, by some extremely unlikely quantum phenomenon. Quantum tunnelling (moving between states that on the surface apparently can't be reached from each other) is a term used here. So even if a proton can't decay, there may be processes that for all practical purposes never happen in an incomprehensible timescale of 10^35 years, which happen often on an even more incomprehensible timescale of 10^100 or 10^1000 years.
$endgroup$
– Stilez
Jan 15 at 9:38
add a comment |
$begingroup$
"Decay" is just the name given to an interaction where one particle goes in and two or more particles go out.
The rule in quantum mechanics is "anything not forbidden is compulsory" – that is, any process (decay or otherwise) can happen unless it violates a conservation law. As a result, most particles, whether fundamental or composite, do decay.
The exceptions are particles for which there's literally no set of outputs you can choose that doesn't violate some conservation law. For example, electrons can't decay because the decay would have to conserve electric charge – so at least one output would have to be charged – and would also have to conserve mass/energy – so the total mass of all the outputs would have to be no larger than the electron's mass – and this is impossible because there are no electrically charged particles with lower mass. So it's stable not because it's elementary but because everything that could make it unstable is forbidden.
The muon, which is also fundamental and is almost identical to the electron except for its mass, can and does decay, because the higher input mass means that you can find outputs that conserve mass while also satisfying all other constraints.
The proton, which is not a fundamental particle, can't decay because it's the lightest particle with another conserved property called baryon number.
But, again, the particles that can't decay are the exceptions. As a rule almost everything can and does, and this applies to fundamental and composite particles alike.
answered Jan 15 at 8:51
benrgbenrg
2,138616
$endgroup$
2
$begingroup$
And even those that "can't" decay, could potentially decay extremely rarely, by some extremely unlikely quantum phenomenon. Quantum tunnelling (moving between states that on the surface apparently can't be reached from each other) is a term used here. So even if a proton can't decay, there may be processes that for all practical purposes never happen in an incomprehensible timescale of 10^35 years, which happen often on an even more incomprehensible timescale of 10^100 or 10^1000 years.
$endgroup$
– Stilez
Jan 15 at 9:38
add a comment |
$begingroup$
"Decay" is just the name given to an interaction where one particle goes in and two or more particles go out.
The rule in quantum mechanics is "anything not forbidden is compulsory" – that is, any process (decay or otherwise) can happen unless it violates a conservation law. As a result, most particles, whether fundamental or composite, do decay.
The exceptions are particles for which there's literally no set of outputs you can choose that doesn't violate some conservation law. For example, electrons can't decay because the decay would have to conserve electric charge – so at least one output would have to be charged – and would also have to conserve mass/energy – so the total mass of all the outputs would have to be no larger than the electron's mass – and this is impossible because there are no electrically charged particles with lower mass. So it's stable not because it's elementary but because everything that could make it unstable is forbidden.
The muon, which is also fundamental and is almost identical to the electron except for its mass, can and does decay, because the higher input mass means that you can find outputs that conserve mass while also satisfying all other constraints.
The proton, which is not a fundamental particle, can't decay because it's the lightest particle with another conserved property called baryon number.
But, again, the particles that can't decay are the exceptions. As a rule almost everything can and does, and this applies to fundamental and composite particles alike.
answered Jan 15 at 8:51
benrgbenrg
2,138616
$endgroup$
"Decay" is just the name given to an interaction where one particle goes in and two or more particles go out.
The rule in quantum mechanics is "anything not forbidden is compulsory" – that is, any process (decay or otherwise) can happen unless it violates a conservation law. As a result, most particles, whether fundamental or composite, do decay.
The exceptions are particles for which there's literally no set of outputs you can choose that doesn't violate some conservation law. For example, electrons can't decay because the decay would have to conserve electric charge – so at least one output would have to be charged – and would also have to conserve mass/energy – so the total mass of all the outputs would have to be no larger than the electron's mass – and this is impossible because there are no electrically charged particles with lower mass. So it's stable not because it's elementary but because everything that could make it unstable is forbidden.
The muon, which is also fundamental and is almost identical to the electron except for its mass, can and does decay, because the higher input mass means that you can find outputs that conserve mass while also satisfying all other constraints.
The proton, which is not a fundamental particle, can't decay because it's the lightest particle with another conserved property called baryon number.
But, again, the particles that can't decay are the exceptions. As a rule almost everything can and does, and this applies to fundamental and composite particles alike.
answered Jan 15 at 8:51
benrgbenrg
2,138616
answered Jan 15 at 8:51
benrgbenrg
2,138616
answered Jan 15 at 8:51
benrgbenrg
2,138616
answered Jan 15 at 8:51
benrgbenrg
2,138616
2,138616
2
$begingroup$
And even those that "can't" decay, could potentially decay extremely rarely, by some extremely unlikely quantum phenomenon. Quantum tunnelling (moving between states that on the surface apparently can't be reached from each other) is a term used here. So even if a proton can't decay, there may be processes that for all practical purposes never happen in an incomprehensible timescale of 10^35 years, which happen often on an even more incomprehensible timescale of 10^100 or 10^1000 years.
$endgroup$
– Stilez
Jan 15 at 9:38
add a comment |
2
$begingroup$
And even those that "can't" decay, could potentially decay extremely rarely, by some extremely unlikely quantum phenomenon. Quantum tunnelling (moving between states that on the surface apparently can't be reached from each other) is a term used here. So even if a proton can't decay, there may be processes that for all practical purposes never happen in an incomprehensible timescale of 10^35 years, which happen often on an even more incomprehensible timescale of 10^100 or 10^1000 years.
$endgroup$
– Stilez
Jan 15 at 9:38
2
2
$begingroup$
And even those that "can't" decay, could potentially decay extremely rarely, by some extremely unlikely quantum phenomenon. Quantum tunnelling (moving between states that on the surface apparently can't be reached from each other) is a term used here. So even if a proton can't decay, there may be processes that for all practical purposes never happen in an incomprehensible timescale of 10^35 years, which happen often on an even more incomprehensible timescale of 10^100 or 10^1000 years.
$endgroup$
– Stilez
Jan 15 at 9:38
$begingroup$
And even those that "can't" decay, could potentially decay extremely rarely, by some extremely unlikely quantum phenomenon. Quantum tunnelling (moving between states that on the surface apparently can't be reached from each other) is a term used here. So even if a proton can't decay, there may be processes that for all practical purposes never happen in an incomprehensible timescale of 10^35 years, which happen often on an even more incomprehensible timescale of 10^100 or 10^1000 years.
$endgroup$
– Stilez
Jan 15 at 9:38
add a comment |
$begingroup$
All fundamental or elementary particles decay after being born. Take, for example, electron. While being created in some process, it "decays" into "another electron" and many soft photons. As it is unlikely that "another electron" may stay without further interactions with its environment, it continues to interact, i.e., generally speaking, absorb and emit soft photons.
answered Jan 15 at 8:35
Vladimir KalitvianskiVladimir Kalitvianski
11k11334
$endgroup$
$begingroup$
Are you referring to the emission of photon when an electron goes from excited state to ground state as a form of decay(transformation which I just learnt today)? So everything is just interaction and decay is one of them right?
$endgroup$
– user6760
Jan 15 at 8:51
1
$begingroup$
Right. One may safely say that there are changes of "occupation numbers" of "elementary" particles (excitations) due to interactions (that never end).
$endgroup$
– Vladimir Kalitvianski
Jan 15 at 8:55
add a comment |
$begingroup$
All fundamental or elementary particles decay after being born. Take, for example, electron. While being created in some process, it "decays" into "another electron" and many soft photons. As it is unlikely that "another electron" may stay without further interactions with its environment, it continues to interact, i.e., generally speaking, absorb and emit soft photons.
answered Jan 15 at 8:35
Vladimir KalitvianskiVladimir Kalitvianski
11k11334
$endgroup$
$begingroup$
Are you referring to the emission of photon when an electron goes from excited state to ground state as a form of decay(transformation which I just learnt today)? So everything is just interaction and decay is one of them right?
$endgroup$
– user6760
Jan 15 at 8:51
1
$begingroup$
Right. One may safely say that there are changes of "occupation numbers" of "elementary" particles (excitations) due to interactions (that never end).
$endgroup$
– Vladimir Kalitvianski
Jan 15 at 8:55
add a comment |
$begingroup$
All fundamental or elementary particles decay after being born. Take, for example, electron. While being created in some process, it "decays" into "another electron" and many soft photons. As it is unlikely that "another electron" may stay without further interactions with its environment, it continues to interact, i.e., generally speaking, absorb and emit soft photons.
answered Jan 15 at 8:35
Vladimir KalitvianskiVladimir Kalitvianski
11k11334
$endgroup$
All fundamental or elementary particles decay after being born. Take, for example, electron. While being created in some process, it "decays" into "another electron" and many soft photons. As it is unlikely that "another electron" may stay without further interactions with its environment, it continues to interact, i.e., generally speaking, absorb and emit soft photons.
answered Jan 15 at 8:35
Vladimir KalitvianskiVladimir Kalitvianski
11k11334
answered Jan 15 at 8:35
Vladimir KalitvianskiVladimir Kalitvianski
11k11334
answered Jan 15 at 8:35
Vladimir KalitvianskiVladimir Kalitvianski
11k11334
answered Jan 15 at 8:35
Vladimir KalitvianskiVladimir Kalitvianski
11k11334
11k11334
$begingroup$
Are you referring to the emission of photon when an electron goes from excited state to ground state as a form of decay(transformation which I just learnt today)? So everything is just interaction and decay is one of them right?
$endgroup$
– user6760
Jan 15 at 8:51
1
$begingroup$
Right. One may safely say that there are changes of "occupation numbers" of "elementary" particles (excitations) due to interactions (that never end).
$endgroup$
– Vladimir Kalitvianski
Jan 15 at 8:55
add a comment |
$begingroup$
Are you referring to the emission of photon when an electron goes from excited state to ground state as a form of decay(transformation which I just learnt today)? So everything is just interaction and decay is one of them right?
$endgroup$
– user6760
Jan 15 at 8:51
1
$begingroup$
Right. One may safely say that there are changes of "occupation numbers" of "elementary" particles (excitations) due to interactions (that never end).
$endgroup$
– Vladimir Kalitvianski
Jan 15 at 8:55
$begingroup$
Are you referring to the emission of photon when an electron goes from excited state to ground state as a form of decay(transformation which I just learnt today)? So everything is just interaction and decay is one of them right?
$endgroup$
– user6760
Jan 15 at 8:51
$begingroup$
Are you referring to the emission of photon when an electron goes from excited state to ground state as a form of decay(transformation which I just learnt today)? So everything is just interaction and decay is one of them right?
$endgroup$
– user6760
Jan 15 at 8:51
1
1
$begingroup$
Right. One may safely say that there are changes of "occupation numbers" of "elementary" particles (excitations) due to interactions (that never end).
$endgroup$
– Vladimir Kalitvianski
Jan 15 at 8:55
$begingroup$
Right. One may safely say that there are changes of "occupation numbers" of "elementary" particles (excitations) due to interactions (that never end).
$endgroup$
– Vladimir Kalitvianski
Jan 15 at 8:55
add a comment |
$begingroup$
A particle is elementary when there aren't subcomponents that we can identify.
This has nothing to do with the concept of decay, and you can easily convince yourself of this fact by observing that whereas a particle (elementary or not) may decay in many different ways, the number and type of its constituents is univocally determined.
answered Jan 15 at 4:42
Francesco BernardiniFrancesco Bernardini
49915
$endgroup$
add a comment |
$begingroup$
A particle is elementary when there aren't subcomponents that we can identify.
This has nothing to do with the concept of decay, and you can easily convince yourself of this fact by observing that whereas a particle (elementary or not) may decay in many different ways, the number and type of its constituents is univocally determined.
answered Jan 15 at 4:42
Francesco BernardiniFrancesco Bernardini
49915
$endgroup$
add a comment |
$begingroup$
A particle is elementary when there aren't subcomponents that we can identify.
This has nothing to do with the concept of decay, and you can easily convince yourself of this fact by observing that whereas a particle (elementary or not) may decay in many different ways, the number and type of its constituents is univocally determined.
answered Jan 15 at 4:42
Francesco BernardiniFrancesco Bernardini
49915
$endgroup$
A particle is elementary when there aren't subcomponents that we can identify.
This has nothing to do with the concept of decay, and you can easily convince yourself of this fact by observing that whereas a particle (elementary or not) may decay in many different ways, the number and type of its constituents is univocally determined.
answered Jan 15 at 4:42
Francesco BernardiniFrancesco Bernardini
49915
answered Jan 15 at 4:42
Francesco BernardiniFrancesco Bernardini
49915
answered Jan 15 at 4:42
Francesco BernardiniFrancesco Bernardini
49915
answered Jan 15 at 4:42
Francesco BernardiniFrancesco Bernardini
49915
49915
add a comment |
add a comment |
$begingroup$
The decay of quanta is not like solid matter breaking apart. When we think of the word "decay" it is often associated with rotting flesh or physical separation. Decay could also be simply a depreciation.
It might help one to consider of Bosons and Fermions as states of Energy which follow stochastic rules. As the energy is conserved from the initial state to the post-"decay" state, you can think of decay as a change from an unstable energetic state to an alternative energetic state within its intrinsic nature by way of an increase in entropy (and probably other speculated or mathematically-derived phenomena).
Furthermore, the nature of all particles as fundamental is still up for debate.
answered Jan 17 at 15:00
Daniel ThorneDaniel Thorne
1
$endgroup$
add a comment |
$begingroup$
The decay of quanta is not like solid matter breaking apart. When we think of the word "decay" it is often associated with rotting flesh or physical separation. Decay could also be simply a depreciation.
It might help one to consider of Bosons and Fermions as states of Energy which follow stochastic rules. As the energy is conserved from the initial state to the post-"decay" state, you can think of decay as a change from an unstable energetic state to an alternative energetic state within its intrinsic nature by way of an increase in entropy (and probably other speculated or mathematically-derived phenomena).
Furthermore, the nature of all particles as fundamental is still up for debate.
answered Jan 17 at 15:00
Daniel ThorneDaniel Thorne
1
$endgroup$
add a comment |
$begingroup$
The decay of quanta is not like solid matter breaking apart. When we think of the word "decay" it is often associated with rotting flesh or physical separation. Decay could also be simply a depreciation.
It might help one to consider of Bosons and Fermions as states of Energy which follow stochastic rules. As the energy is conserved from the initial state to the post-"decay" state, you can think of decay as a change from an unstable energetic state to an alternative energetic state within its intrinsic nature by way of an increase in entropy (and probably other speculated or mathematically-derived phenomena).
Furthermore, the nature of all particles as fundamental is still up for debate.
answered Jan 17 at 15:00
Daniel ThorneDaniel Thorne
1
$endgroup$
The decay of quanta is not like solid matter breaking apart. When we think of the word "decay" it is often associated with rotting flesh or physical separation. Decay could also be simply a depreciation.
It might help one to consider of Bosons and Fermions as states of Energy which follow stochastic rules. As the energy is conserved from the initial state to the post-"decay" state, you can think of decay as a change from an unstable energetic state to an alternative energetic state within its intrinsic nature by way of an increase in entropy (and probably other speculated or mathematically-derived phenomena).
Furthermore, the nature of all particles as fundamental is still up for debate.
answered Jan 17 at 15:00
Daniel ThorneDaniel Thorne
1
answered Jan 17 at 15:00
Daniel ThorneDaniel Thorne
1
answered Jan 17 at 15:00
Daniel ThorneDaniel Thorne
1
answered Jan 17 at 15:00
Daniel ThorneDaniel Thorne
1
1
add a comment |
add a comment |
5
$begingroup$
I've removed some comments that were answering the question. Please keep in mind that comments are meant for requesting clarification or suggesting improvements to the question, not for answering.
$endgroup$
– David Z♦
Jan 16 at 7:37
1
$begingroup$
Essentially a duplicate of Why are muons considered to be “elementary particles” in the Standard Model? with "muon" replaced by "Higgs" (which was also a HNQ). Are we going to do a HNQ for every particle of the Standard Model now?
$endgroup$
– ACuriousMind♦
Jan 20 at 13:14