Is the Higgs boson an elementary particle? If so, why does it decay?
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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. My question is why if according to the standard model it is supposed to be an elementary particle does it decay?
standard-model higgs elementary-particles
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add a comment |
$begingroup$
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. My question is why if according to the standard model it is supposed to be an elementary particle does it decay?
standard-model higgs elementary-particles
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This is an excellent question
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– Gaius
2 hours ago
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Take a complicated mechanical system: some bodies connected with springs. One can describe it with so called normal (independent, fundamental, elementary) modes in a linear approximation. Beyond this approximation the elementary modes "interact" with each other, and this is what we are dealing in QFT too, in my humble opinion.
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– Vladimir Kalitvianski
2 hours ago
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Decay is just how we (with our linear understanding of time) see the interaction between some number of particles. Some of the particles come to the interaction from the past, some come from the future. They meet and interact, and that's it. Our brains interpret it as some particles coming into the interaction, and some other particles leaving.
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– Arthur
2 hours ago
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@Arthur: are you referring to antimatter that the positron is electron going backward in time which I have heard before.
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– user6760
2 hours ago
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@user6760 That is one common instance of trying to break free of the linear understanding of time, yes. Electron-positron annihilation is an electron and a photon meeting up and them both changing direction as a result. In many instances, however, particles' paths will end at the interaction instead of just going through it. I'm no physicist, though, and this is purely based on my understanding of quantum electrodynamics from Feynman's popular lectures, extrapolated to the rest of the standard model. So take it with a grain of salt.
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– Arthur
1 hour ago
add a comment |
$begingroup$
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. My question is why if according to the standard model it is supposed to be an elementary particle does it decay?
standard-model higgs elementary-particles
$endgroup$
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. My question is why if according to the standard model it is supposed to be an elementary particle does it decay?
standard-model higgs elementary-particles
standard-model higgs elementary-particles
edited 14 mins ago
user545424
30516
30516
asked 7 hours ago
user6760user6760
2,56411739
2,56411739
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This is an excellent question
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– Gaius
2 hours ago
$begingroup$
Take a complicated mechanical system: some bodies connected with springs. One can describe it with so called normal (independent, fundamental, elementary) modes in a linear approximation. Beyond this approximation the elementary modes "interact" with each other, and this is what we are dealing in QFT too, in my humble opinion.
$endgroup$
– Vladimir Kalitvianski
2 hours ago
$begingroup$
Decay is just how we (with our linear understanding of time) see the interaction between some number of particles. Some of the particles come to the interaction from the past, some come from the future. They meet and interact, and that's it. Our brains interpret it as some particles coming into the interaction, and some other particles leaving.
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– Arthur
2 hours ago
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@Arthur: are you referring to antimatter that the positron is electron going backward in time which I have heard before.
$endgroup$
– user6760
2 hours ago
$begingroup$
@user6760 That is one common instance of trying to break free of the linear understanding of time, yes. Electron-positron annihilation is an electron and a photon meeting up and them both changing direction as a result. In many instances, however, particles' paths will end at the interaction instead of just going through it. I'm no physicist, though, and this is purely based on my understanding of quantum electrodynamics from Feynman's popular lectures, extrapolated to the rest of the standard model. So take it with a grain of salt.
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– Arthur
1 hour ago
add a comment |
$begingroup$
This is an excellent question
$endgroup$
– Gaius
2 hours ago
$begingroup$
Take a complicated mechanical system: some bodies connected with springs. One can describe it with so called normal (independent, fundamental, elementary) modes in a linear approximation. Beyond this approximation the elementary modes "interact" with each other, and this is what we are dealing in QFT too, in my humble opinion.
$endgroup$
– Vladimir Kalitvianski
2 hours ago
$begingroup$
Decay is just how we (with our linear understanding of time) see the interaction between some number of particles. Some of the particles come to the interaction from the past, some come from the future. They meet and interact, and that's it. Our brains interpret it as some particles coming into the interaction, and some other particles leaving.
$endgroup$
– Arthur
2 hours ago
$begingroup$
@Arthur: are you referring to antimatter that the positron is electron going backward in time which I have heard before.
$endgroup$
– user6760
2 hours ago
$begingroup$
@user6760 That is one common instance of trying to break free of the linear understanding of time, yes. Electron-positron annihilation is an electron and a photon meeting up and them both changing direction as a result. In many instances, however, particles' paths will end at the interaction instead of just going through it. I'm no physicist, though, and this is purely based on my understanding of quantum electrodynamics from Feynman's popular lectures, extrapolated to the rest of the standard model. So take it with a grain of salt.
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– Arthur
1 hour ago
$begingroup$
This is an excellent question
$endgroup$
– Gaius
2 hours ago
$begingroup$
This is an excellent question
$endgroup$
– Gaius
2 hours ago
$begingroup$
Take a complicated mechanical system: some bodies connected with springs. One can describe it with so called normal (independent, fundamental, elementary) modes in a linear approximation. Beyond this approximation the elementary modes "interact" with each other, and this is what we are dealing in QFT too, in my humble opinion.
$endgroup$
– Vladimir Kalitvianski
2 hours ago
$begingroup$
Take a complicated mechanical system: some bodies connected with springs. One can describe it with so called normal (independent, fundamental, elementary) modes in a linear approximation. Beyond this approximation the elementary modes "interact" with each other, and this is what we are dealing in QFT too, in my humble opinion.
$endgroup$
– Vladimir Kalitvianski
2 hours ago
$begingroup$
Decay is just how we (with our linear understanding of time) see the interaction between some number of particles. Some of the particles come to the interaction from the past, some come from the future. They meet and interact, and that's it. Our brains interpret it as some particles coming into the interaction, and some other particles leaving.
$endgroup$
– Arthur
2 hours ago
$begingroup$
Decay is just how we (with our linear understanding of time) see the interaction between some number of particles. Some of the particles come to the interaction from the past, some come from the future. They meet and interact, and that's it. Our brains interpret it as some particles coming into the interaction, and some other particles leaving.
$endgroup$
– Arthur
2 hours ago
$begingroup$
@Arthur: are you referring to antimatter that the positron is electron going backward in time which I have heard before.
$endgroup$
– user6760
2 hours ago
$begingroup$
@Arthur: are you referring to antimatter that the positron is electron going backward in time which I have heard before.
$endgroup$
– user6760
2 hours ago
$begingroup$
@user6760 That is one common instance of trying to break free of the linear understanding of time, yes. Electron-positron annihilation is an electron and a photon meeting up and them both changing direction as a result. In many instances, however, particles' paths will end at the interaction instead of just going through it. I'm no physicist, though, and this is purely based on my understanding of quantum electrodynamics from Feynman's popular lectures, extrapolated to the rest of the standard model. So take it with a grain of salt.
$endgroup$
– Arthur
1 hour ago
$begingroup$
@user6760 That is one common instance of trying to break free of the linear understanding of time, yes. Electron-positron annihilation is an electron and a photon meeting up and them both changing direction as a result. In many instances, however, particles' paths will end at the interaction instead of just going through it. I'm no physicist, though, and this is purely based on my understanding of quantum electrodynamics from Feynman's popular lectures, extrapolated to the rest of the standard model. So take it with a grain of salt.
$endgroup$
– Arthur
1 hour ago
add a comment |
5 Answers
5
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.
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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?
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– user6760
6 hours ago
4
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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.
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– Bob Jacobsen
6 hours ago
2
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@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)
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– Bob Jacobsen
5 hours ago
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@safesphere I don't understand why ATLAS should be left out of scope? Photon fusion does occur - (dde.web.cern.ch/dde/presentations/fp420_dec09/…) but if you want to say that photon fusion too rare; there's always the more common gluon fusion? nikhef.nl/pub/services/biblio/preprints/05-007.pdf Or is your problem more with the on-shell-ness of things?
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– Joshua Lin
5 hours ago
1
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@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.
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– Bob Jacobsen
4 hours ago
|
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!
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are u saying the excitation of the field can disturb other fields too? So the reality is just fields interacting with one another.
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– user6760
6 hours ago
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@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.
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– Dan Yand
6 hours ago
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@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
6 hours ago
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Yes it makes sense to me now the virtual particle that you have mentioned.
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– user6760
5 hours ago
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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
3 hours ago
add a comment |
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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.
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add a comment |
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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.
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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
2 hours ago
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
2 hours ago
add a comment |
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"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.
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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
1 hour ago
add a comment |
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5 Answers
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5 Answers
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$begingroup$
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.
$endgroup$
$begingroup$
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
6 hours ago
4
$begingroup$
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
6 hours ago
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
5 hours ago
$begingroup$
@safesphere I don't understand why ATLAS should be left out of scope? Photon fusion does occur - (dde.web.cern.ch/dde/presentations/fp420_dec09/…) but if you want to say that photon fusion too rare; there's always the more common gluon fusion? nikhef.nl/pub/services/biblio/preprints/05-007.pdf Or is your problem more with the on-shell-ness of things?
$endgroup$
– Joshua Lin
5 hours ago
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
4 hours ago
|
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.
$endgroup$
$begingroup$
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
6 hours ago
4
$begingroup$
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
6 hours ago
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
5 hours ago
$begingroup$
@safesphere I don't understand why ATLAS should be left out of scope? Photon fusion does occur - (dde.web.cern.ch/dde/presentations/fp420_dec09/…) but if you want to say that photon fusion too rare; there's always the more common gluon fusion? nikhef.nl/pub/services/biblio/preprints/05-007.pdf Or is your problem more with the on-shell-ness of things?
$endgroup$
– Joshua Lin
5 hours ago
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
4 hours ago
|
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.
$endgroup$
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 7 hours ago
Bob JacobsenBob Jacobsen
4,576617
4,576617
$begingroup$
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
6 hours ago
4
$begingroup$
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
6 hours ago
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
5 hours ago
$begingroup$
@safesphere I don't understand why ATLAS should be left out of scope? Photon fusion does occur - (dde.web.cern.ch/dde/presentations/fp420_dec09/…) but if you want to say that photon fusion too rare; there's always the more common gluon fusion? nikhef.nl/pub/services/biblio/preprints/05-007.pdf Or is your problem more with the on-shell-ness of things?
$endgroup$
– Joshua Lin
5 hours ago
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
4 hours ago
|
show 5 more comments
$begingroup$
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
6 hours ago
4
$begingroup$
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
6 hours ago
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
5 hours ago
$begingroup$
@safesphere I don't understand why ATLAS should be left out of scope? Photon fusion does occur - (dde.web.cern.ch/dde/presentations/fp420_dec09/…) but if you want to say that photon fusion too rare; there's always the more common gluon fusion? nikhef.nl/pub/services/biblio/preprints/05-007.pdf Or is your problem more with the on-shell-ness of things?
$endgroup$
– Joshua Lin
5 hours ago
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
4 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
4
4
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
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
5 hours ago
$begingroup$
@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
5 hours ago
$begingroup$
@safesphere I don't understand why ATLAS should be left out of scope? Photon fusion does occur - (dde.web.cern.ch/dde/presentations/fp420_dec09/…) but if you want to say that photon fusion too rare; there's always the more common gluon fusion? nikhef.nl/pub/services/biblio/preprints/05-007.pdf Or is your problem more with the on-shell-ness of things?
$endgroup$
– Joshua Lin
5 hours ago
$begingroup$
@safesphere I don't understand why ATLAS should be left out of scope? Photon fusion does occur - (dde.web.cern.ch/dde/presentations/fp420_dec09/…) but if you want to say that photon fusion too rare; there's always the more common gluon fusion? nikhef.nl/pub/services/biblio/preprints/05-007.pdf Or is your problem more with the on-shell-ness of things?
$endgroup$
– Joshua Lin
5 hours ago
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
4 hours ago
$begingroup$
@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
4 hours ago
|
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!
$endgroup$
$begingroup$
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
6 hours ago
$begingroup$
@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
6 hours ago
$begingroup$
@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
6 hours ago
$begingroup$
Yes it makes sense to me now the virtual particle that you have mentioned.
$endgroup$
– user6760
5 hours ago
$begingroup$
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
3 hours ago
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!
$endgroup$
$begingroup$
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
6 hours ago
$begingroup$
@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
6 hours ago
$begingroup$
@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
6 hours ago
$begingroup$
Yes it makes sense to me now the virtual particle that you have mentioned.
$endgroup$
– user6760
5 hours ago
$begingroup$
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
3 hours ago
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!
$endgroup$
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 6 hours ago
Dan YandDan Yand
8,41011235
8,41011235
$begingroup$
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
6 hours ago
$begingroup$
@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
6 hours ago
$begingroup$
@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
6 hours ago
$begingroup$
Yes it makes sense to me now the virtual particle that you have mentioned.
$endgroup$
– user6760
5 hours ago
$begingroup$
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
3 hours ago
add a comment |
$begingroup$
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
6 hours ago
$begingroup$
@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
6 hours ago
$begingroup$
@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
6 hours ago
$begingroup$
Yes it makes sense to me now the virtual particle that you have mentioned.
$endgroup$
– user6760
5 hours ago
$begingroup$
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
3 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
@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
6 hours ago
$begingroup$
@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
6 hours ago
$begingroup$
@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
6 hours ago
$begingroup$
@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
6 hours ago
$begingroup$
Yes it makes sense to me now the virtual particle that you have mentioned.
$endgroup$
– user6760
5 hours ago
$begingroup$
Yes it makes sense to me now the virtual particle that you have mentioned.
$endgroup$
– user6760
5 hours ago
$begingroup$
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
3 hours ago
$begingroup$
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
3 hours ago
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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 6 hours ago
Francesco BernardiniFrancesco Bernardini
4545
4545
add a comment |
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.
$endgroup$
$begingroup$
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
2 hours ago
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
2 hours ago
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.
$endgroup$
$begingroup$
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
2 hours ago
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
2 hours ago
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.
$endgroup$
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 2 hours ago
Vladimir KalitvianskiVladimir Kalitvianski
10.6k11234
10.6k11234
$begingroup$
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
2 hours ago
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
2 hours ago
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
2 hours ago
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
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
2 hours ago
$begingroup$
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
2 hours ago
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.
$endgroup$
$begingroup$
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
1 hour ago
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.
$endgroup$
$begingroup$
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
1 hour ago
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.
$endgroup$
"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 2 hours ago
benrgbenrg
1,900613
1,900613
$begingroup$
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
1 hour ago
add a comment |
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
add a comment |
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$begingroup$
This is an excellent question
$endgroup$
– Gaius
2 hours ago
$begingroup$
Take a complicated mechanical system: some bodies connected with springs. One can describe it with so called normal (independent, fundamental, elementary) modes in a linear approximation. Beyond this approximation the elementary modes "interact" with each other, and this is what we are dealing in QFT too, in my humble opinion.
$endgroup$
– Vladimir Kalitvianski
2 hours ago
$begingroup$
Decay is just how we (with our linear understanding of time) see the interaction between some number of particles. Some of the particles come to the interaction from the past, some come from the future. They meet and interact, and that's it. Our brains interpret it as some particles coming into the interaction, and some other particles leaving.
$endgroup$
– Arthur
2 hours ago
$begingroup$
@Arthur: are you referring to antimatter that the positron is electron going backward in time which I have heard before.
$endgroup$
– user6760
2 hours ago
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@user6760 That is one common instance of trying to break free of the linear understanding of time, yes. Electron-positron annihilation is an electron and a photon meeting up and them both changing direction as a result. In many instances, however, particles' paths will end at the interaction instead of just going through it. I'm no physicist, though, and this is purely based on my understanding of quantum electrodynamics from Feynman's popular lectures, extrapolated to the rest of the standard model. So take it with a grain of salt.
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– Arthur
1 hour ago