Is higgs boson an elementary particle? If so why does it decays?
$begingroup$
Higgs boson is excitation of higgs field and is very massive and short lived, it also interact with the higgs field and thus is able to experience mass. My question is if according to standard model it is supposedly to be an elementary particle then why can it decays?
standard-model higgs elementary-particles
asked 1 hour ago
user6760user6760
2,53611738
$endgroup$
add a comment |
$begingroup$
Higgs boson is excitation of higgs field and is very massive and short lived, it also interact with the higgs field and thus is able to experience mass. My question is if according to standard model it is supposedly to be an elementary particle then why can it decays?
standard-model higgs elementary-particles
asked 1 hour ago
user6760user6760
2,53611738
$endgroup$
add a comment |
$begingroup$
Higgs boson is excitation of higgs field and is very massive and short lived, it also interact with the higgs field and thus is able to experience mass. My question is if according to standard model it is supposedly to be an elementary particle then why can it decays?
standard-model higgs elementary-particles
asked 1 hour ago
user6760user6760
2,53611738
$endgroup$
Higgs boson is excitation of higgs field and is very massive and short lived, it also interact with the higgs field and thus is able to experience mass. My question is if according to standard model it is supposedly to be an elementary particle then why can it decays?
standard-model higgs elementary-particles
standard-model higgs elementary-particles
asked 1 hour ago
user6760user6760
2,53611738
asked 1 hour ago
user6760user6760
2,53611738
asked 1 hour ago
user6760user6760
2,53611738
asked 1 hour ago
user6760user6760
2,53611738
asked 1 hour ago
user6760user6760
2,53611738
2,53611738
add a comment |
add a comment |
3 Answers
3
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 1 hour ago
Bob JacobsenBob Jacobsen
4,406616
$endgroup$
$begingroup$
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
26 mins 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
24 mins ago
$begingroup$
"they could combine to form a Higgs particle" - This is incorrect. Let's take a simpler example, an annihilation of electron and positron into two photons. This happens nearly every time an electron meats a positron. Now, let's try turning this around. Please show a single evidence where two free photons in an otherwise empty space hit each other to produce an electron-positron pair. This never happens. Photons don't interact with each other, because photons interact only with electrically charged particles. (I will leave the two-photon physics and the Atlas experiment out of scope here.)
$endgroup$
– safesphere
11 mins ago
$begingroup$
+ I think I got it
$endgroup$
– user6760
11 mins 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
2 mins 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!
answered 57 mins ago
Dan YandDan Yand
8,31211234
$endgroup$
$begingroup$
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
19 mins 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
16 mins 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
4 mins 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.
answered 30 mins ago
Francesco BernardiniFrancesco Bernardini
4345
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
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 1 hour ago
Bob JacobsenBob Jacobsen
4,406616
$endgroup$
$begingroup$
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
26 mins 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
24 mins ago
$begingroup$
"they could combine to form a Higgs particle" - This is incorrect. Let's take a simpler example, an annihilation of electron and positron into two photons. This happens nearly every time an electron meats a positron. Now, let's try turning this around. Please show a single evidence where two free photons in an otherwise empty space hit each other to produce an electron-positron pair. This never happens. Photons don't interact with each other, because photons interact only with electrically charged particles. (I will leave the two-photon physics and the Atlas experiment out of scope here.)
$endgroup$
– safesphere
11 mins ago
$begingroup$
+ I think I got it
$endgroup$
– user6760
11 mins 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
2 mins ago
add a comment |
$begingroup$
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 1 hour ago
Bob JacobsenBob Jacobsen
4,406616
$endgroup$
$begingroup$
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
26 mins 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
24 mins ago
$begingroup$
"they could combine to form a Higgs particle" - This is incorrect. Let's take a simpler example, an annihilation of electron and positron into two photons. This happens nearly every time an electron meats a positron. Now, let's try turning this around. Please show a single evidence where two free photons in an otherwise empty space hit each other to produce an electron-positron pair. This never happens. Photons don't interact with each other, because photons interact only with electrically charged particles. (I will leave the two-photon physics and the Atlas experiment out of scope here.)
$endgroup$
– safesphere
11 mins ago
$begingroup$
+ I think I got it
$endgroup$
– user6760
11 mins 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
2 mins ago
add a comment |
$begingroup$
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 1 hour ago
Bob JacobsenBob Jacobsen
4,406616
$endgroup$
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 1 hour ago
Bob JacobsenBob Jacobsen
4,406616
answered 1 hour ago
Bob JacobsenBob Jacobsen
4,406616
answered 1 hour ago
Bob JacobsenBob Jacobsen
4,406616
answered 1 hour ago
Bob JacobsenBob Jacobsen
4,406616
4,406616
$begingroup$
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
26 mins 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
24 mins ago
$begingroup$
"they could combine to form a Higgs particle" - This is incorrect. Let's take a simpler example, an annihilation of electron and positron into two photons. This happens nearly every time an electron meats a positron. Now, let's try turning this around. Please show a single evidence where two free photons in an otherwise empty space hit each other to produce an electron-positron pair. This never happens. Photons don't interact with each other, because photons interact only with electrically charged particles. (I will leave the two-photon physics and the Atlas experiment out of scope here.)
$endgroup$
– safesphere
11 mins ago
$begingroup$
+ I think I got it
$endgroup$
– user6760
11 mins 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
2 mins ago
add a comment |
$begingroup$
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
26 mins 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
24 mins ago
$begingroup$
"they could combine to form a Higgs particle" - This is incorrect. Let's take a simpler example, an annihilation of electron and positron into two photons. This happens nearly every time an electron meats a positron. Now, let's try turning this around. Please show a single evidence where two free photons in an otherwise empty space hit each other to produce an electron-positron pair. This never happens. Photons don't interact with each other, because photons interact only with electrically charged particles. (I will leave the two-photon physics and the Atlas experiment out of scope here.)
$endgroup$
– safesphere
11 mins ago
$begingroup$
+ I think I got it
$endgroup$
– user6760
11 mins 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
2 mins 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
26 mins 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
26 mins 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
24 mins 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
24 mins ago
$begingroup$
"they could combine to form a Higgs particle" - This is incorrect. Let's take a simpler example, an annihilation of electron and positron into two photons. This happens nearly every time an electron meats a positron. Now, let's try turning this around. Please show a single evidence where two free photons in an otherwise empty space hit each other to produce an electron-positron pair. This never happens. Photons don't interact with each other, because photons interact only with electrically charged particles. (I will leave the two-photon physics and the Atlas experiment out of scope here.)
$endgroup$
– safesphere
11 mins ago
$begingroup$
"they could combine to form a Higgs particle" - This is incorrect. Let's take a simpler example, an annihilation of electron and positron into two photons. This happens nearly every time an electron meats a positron. Now, let's try turning this around. Please show a single evidence where two free photons in an otherwise empty space hit each other to produce an electron-positron pair. This never happens. Photons don't interact with each other, because photons interact only with electrically charged particles. (I will leave the two-photon physics and the Atlas experiment out of scope here.)
$endgroup$
– safesphere
11 mins ago
$begingroup$
+ I think I got it
$endgroup$
– user6760
11 mins ago
$begingroup$
+ I think I got it
$endgroup$
– user6760
11 mins 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
2 mins 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
2 mins 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!
answered 57 mins ago
Dan YandDan Yand
8,31211234
$endgroup$
$begingroup$
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
19 mins 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
16 mins 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
4 mins 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!
answered 57 mins ago
Dan YandDan Yand
8,31211234
$endgroup$
$begingroup$
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
19 mins 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
16 mins 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
4 mins 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!
answered 57 mins ago
Dan YandDan Yand
8,31211234
$endgroup$
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 57 mins ago
Dan YandDan Yand
8,31211234
answered 57 mins ago
Dan YandDan Yand
8,31211234
answered 57 mins ago
Dan YandDan Yand
8,31211234
answered 57 mins ago
Dan YandDan Yand
8,31211234
8,31211234
$begingroup$
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
19 mins 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
16 mins 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
4 mins 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
19 mins 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
16 mins 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
4 mins 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
19 mins 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
19 mins 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
16 mins 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
16 mins 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
4 mins 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
4 mins 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.
answered 30 mins ago
Francesco BernardiniFrancesco Bernardini
4345
$endgroup$
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 30 mins ago
Francesco BernardiniFrancesco Bernardini
4345
$endgroup$
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 30 mins ago
Francesco BernardiniFrancesco Bernardini
4345
$endgroup$
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 30 mins ago
Francesco BernardiniFrancesco Bernardini
4345
answered 30 mins ago
Francesco BernardiniFrancesco Bernardini
4345
answered 30 mins ago
Francesco BernardiniFrancesco Bernardini
4345
answered 30 mins ago
Francesco BernardiniFrancesco Bernardini
4345
4345
add a comment |
add a comment |
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