Why are protons and neutrons the “right” degrees of freedom of nuclei?
up vote
23
down vote
favorite
This question may sound stupid but why do we visualize nuclei as composed of a bunch of neutrons and protons?
Wouldn't the nucleons be too close together to be viewed as different particles? Isn't the whole nucleus just a complicated low energy state of QCD?
nuclear-physics
asked yesterday
tonydo
949619
add a comment |
up vote
23
down vote
favorite
This question may sound stupid but why do we visualize nuclei as composed of a bunch of neutrons and protons?
Wouldn't the nucleons be too close together to be viewed as different particles? Isn't the whole nucleus just a complicated low energy state of QCD?
nuclear-physics
asked yesterday
tonydo
949619
5
Some "stupid sounding" questions can be rather fundamental. This one is fundamental enough that'd I'd have to ask which direction do you want this to be handled? Are you looking at it the historical way, which went from chemical properties towards atomic weights and numbers? Or are you looking backwards, from the perspective of QM, to see that QM predicts these to be meaningful visualizations? For the latter, it's useful to remember that QM was invented to fit the data we had already observed, such as the meaningfulness of protons and neutrons in a nucleus.
– Cort Ammon
yesterday
3
Perhaps because we like to simplify things to make them easier to understand & talk about? In a lot of contexts, we don't have to understand what a nucleus actually "is" inside, we just need to know that it has so many units of charge (protons), such and such an atomic weight that's not the same as units of charge (neutrons), and that if we break it apart by e.g. hitting it with high-energy particles, protons & neutrons come out.
– jamesqf
yesterday
4
@jamesqf That is a perfectly valid answer.
– dmckee♦
yesterday
Related, related.
– rob♦
14 hours ago
2
This is a straightforward application of Weinberg’s Third Law of Progress in Theoretical Physics: "You may use any degrees of freedom you like to describe a physical system, but if you use the wrong ones, you’ll be sorry!"
– Xerxes
12 hours ago
add a comment |
up vote
23
down vote
favorite
up vote
23
down vote
favorite
This question may sound stupid but why do we visualize nuclei as composed of a bunch of neutrons and protons?
Wouldn't the nucleons be too close together to be viewed as different particles? Isn't the whole nucleus just a complicated low energy state of QCD?
nuclear-physics
asked yesterday
tonydo
949619
This question may sound stupid but why do we visualize nuclei as composed of a bunch of neutrons and protons?
Wouldn't the nucleons be too close together to be viewed as different particles? Isn't the whole nucleus just a complicated low energy state of QCD?
nuclear-physics
nuclear-physics
asked yesterday
tonydo
949619
asked yesterday
tonydo
949619
asked yesterday
tonydo
949619
asked yesterday
tonydo
949619
asked yesterday
tonydo
949619
949619
5
Some "stupid sounding" questions can be rather fundamental. This one is fundamental enough that'd I'd have to ask which direction do you want this to be handled? Are you looking at it the historical way, which went from chemical properties towards atomic weights and numbers? Or are you looking backwards, from the perspective of QM, to see that QM predicts these to be meaningful visualizations? For the latter, it's useful to remember that QM was invented to fit the data we had already observed, such as the meaningfulness of protons and neutrons in a nucleus.
– Cort Ammon
yesterday
3
Perhaps because we like to simplify things to make them easier to understand & talk about? In a lot of contexts, we don't have to understand what a nucleus actually "is" inside, we just need to know that it has so many units of charge (protons), such and such an atomic weight that's not the same as units of charge (neutrons), and that if we break it apart by e.g. hitting it with high-energy particles, protons & neutrons come out.
– jamesqf
yesterday
4
@jamesqf That is a perfectly valid answer.
– dmckee♦
yesterday
Related, related.
– rob♦
14 hours ago
2
This is a straightforward application of Weinberg’s Third Law of Progress in Theoretical Physics: "You may use any degrees of freedom you like to describe a physical system, but if you use the wrong ones, you’ll be sorry!"
– Xerxes
12 hours ago
add a comment |
5
Some "stupid sounding" questions can be rather fundamental. This one is fundamental enough that'd I'd have to ask which direction do you want this to be handled? Are you looking at it the historical way, which went from chemical properties towards atomic weights and numbers? Or are you looking backwards, from the perspective of QM, to see that QM predicts these to be meaningful visualizations? For the latter, it's useful to remember that QM was invented to fit the data we had already observed, such as the meaningfulness of protons and neutrons in a nucleus.
– Cort Ammon
yesterday
3
Perhaps because we like to simplify things to make them easier to understand & talk about? In a lot of contexts, we don't have to understand what a nucleus actually "is" inside, we just need to know that it has so many units of charge (protons), such and such an atomic weight that's not the same as units of charge (neutrons), and that if we break it apart by e.g. hitting it with high-energy particles, protons & neutrons come out.
– jamesqf
yesterday
4
@jamesqf That is a perfectly valid answer.
– dmckee♦
yesterday
Related, related.
– rob♦
14 hours ago
2
This is a straightforward application of Weinberg’s Third Law of Progress in Theoretical Physics: "You may use any degrees of freedom you like to describe a physical system, but if you use the wrong ones, you’ll be sorry!"
– Xerxes
12 hours ago
5
5
Some "stupid sounding" questions can be rather fundamental. This one is fundamental enough that'd I'd have to ask which direction do you want this to be handled? Are you looking at it the historical way, which went from chemical properties towards atomic weights and numbers? Or are you looking backwards, from the perspective of QM, to see that QM predicts these to be meaningful visualizations? For the latter, it's useful to remember that QM was invented to fit the data we had already observed, such as the meaningfulness of protons and neutrons in a nucleus.
– Cort Ammon
yesterday
Some "stupid sounding" questions can be rather fundamental. This one is fundamental enough that'd I'd have to ask which direction do you want this to be handled? Are you looking at it the historical way, which went from chemical properties towards atomic weights and numbers? Or are you looking backwards, from the perspective of QM, to see that QM predicts these to be meaningful visualizations? For the latter, it's useful to remember that QM was invented to fit the data we had already observed, such as the meaningfulness of protons and neutrons in a nucleus.
– Cort Ammon
yesterday
3
3
Perhaps because we like to simplify things to make them easier to understand & talk about? In a lot of contexts, we don't have to understand what a nucleus actually "is" inside, we just need to know that it has so many units of charge (protons), such and such an atomic weight that's not the same as units of charge (neutrons), and that if we break it apart by e.g. hitting it with high-energy particles, protons & neutrons come out.
– jamesqf
yesterday
Perhaps because we like to simplify things to make them easier to understand & talk about? In a lot of contexts, we don't have to understand what a nucleus actually "is" inside, we just need to know that it has so many units of charge (protons), such and such an atomic weight that's not the same as units of charge (neutrons), and that if we break it apart by e.g. hitting it with high-energy particles, protons & neutrons come out.
– jamesqf
yesterday
4
4
@jamesqf That is a perfectly valid answer.
– dmckee♦
yesterday
@jamesqf That is a perfectly valid answer.
– dmckee♦
yesterday
Related, related.
– rob♦
14 hours ago
Related, related.
– rob♦
14 hours ago
2
2
This is a straightforward application of Weinberg’s Third Law of Progress in Theoretical Physics: "You may use any degrees of freedom you like to describe a physical system, but if you use the wrong ones, you’ll be sorry!"
– Xerxes
12 hours ago
This is a straightforward application of Weinberg’s Third Law of Progress in Theoretical Physics: "You may use any degrees of freedom you like to describe a physical system, but if you use the wrong ones, you’ll be sorry!"
– Xerxes
12 hours ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
24
down vote
This is basically a matter of energy scales. By analogy, you could ask why we don't take into account nuclear structure when we talk about chemistry. The answer is that the eV energy scale of chemistry is mismatched with the MeV energy scale of nuclear structure.
Nuclear matter has two phases. One is the phase we normally see, and the other is a quark-gluon plasma. The phase transition happens at a temperature on the order of 100 GeV per nucleon (at standard nuclear densities). Below the temperature of the phase transition, the quarks are strongly correlated, and those correlated groups behave in a way that's very similar to free neutrons and protons. To the extent that they don't quite have those properties, often we can subsume the discrepancies within adjustments to the parameters of the model. It's helpful in terms of practical computation that the fictitious neutrons and protons are nonrelativistic, which makes the theory much more tractable than QCD. If there are small relativistic effects, because the nucleons are moving at a few percent of $c$, these can also be subsumed within adjustments to the parameters.
By the way, it is actually possible to consider larger clusters to be the relevant degrees of freedom for nuclear structure. There is a model called the interacting boson approximation (IBA, also known as the interacting boson model, IBM), in which pairs of nucleons coupled to spin 0 or 1 are considered the degrees of freedom. It does pretty well in phenomonologically fitting the properties of many nuclei that are intractable in other models.
In a similar vein, there are alpha cluster models and ideas like explaining alpha decay in terms of preformation of an alpha particle, which then tunnels out through the Coulomb barrier. Pictures like these go back to the 1940's, and have considerable utility and explanatory power, although they can't really be microscopically correct, because they violate the Pauli exclusion principle.
answered yesterday
Ben Crowell
47.6k3151289
1
Besides the quark-gluon plasma at high temperature, a quark soup picture should be appropriate at extreme density. It is unclear whether stable neutron stars actually achieve extreme density, or whether it is only achieved ephemerally, in the course of collapse to a black hole.
– Bert Barrois
yesterday
@BertBarrois: Yes, my description of two phases is definitely a little oversimplified. The WP article on quark-gluon plasma has a complicated phase diagram, with a bunch of different phases marked in, but I don't know whether some of these are purely theoretical.
– Ben Crowell
yesterday
add a comment |
up vote
17
down vote
We can measure the form-factors of bound nucleons. For instance by doing quasi-elastic scattering of a proton out of the nucleus $A(e,e'p)$ at low energy loss (my dissertation work involved this reaction for deuterium, helium, carbon and iron).
The result are quite similar to (but measurably not identical to) the equivalent results on free protons. That similarity make the choice of nucleons as the degrees of freedom a good starting point.
answered yesterday
dmckee♦
73.5k6131266
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
24
down vote
This is basically a matter of energy scales. By analogy, you could ask why we don't take into account nuclear structure when we talk about chemistry. The answer is that the eV energy scale of chemistry is mismatched with the MeV energy scale of nuclear structure.
Nuclear matter has two phases. One is the phase we normally see, and the other is a quark-gluon plasma. The phase transition happens at a temperature on the order of 100 GeV per nucleon (at standard nuclear densities). Below the temperature of the phase transition, the quarks are strongly correlated, and those correlated groups behave in a way that's very similar to free neutrons and protons. To the extent that they don't quite have those properties, often we can subsume the discrepancies within adjustments to the parameters of the model. It's helpful in terms of practical computation that the fictitious neutrons and protons are nonrelativistic, which makes the theory much more tractable than QCD. If there are small relativistic effects, because the nucleons are moving at a few percent of $c$, these can also be subsumed within adjustments to the parameters.
By the way, it is actually possible to consider larger clusters to be the relevant degrees of freedom for nuclear structure. There is a model called the interacting boson approximation (IBA, also known as the interacting boson model, IBM), in which pairs of nucleons coupled to spin 0 or 1 are considered the degrees of freedom. It does pretty well in phenomonologically fitting the properties of many nuclei that are intractable in other models.
In a similar vein, there are alpha cluster models and ideas like explaining alpha decay in terms of preformation of an alpha particle, which then tunnels out through the Coulomb barrier. Pictures like these go back to the 1940's, and have considerable utility and explanatory power, although they can't really be microscopically correct, because they violate the Pauli exclusion principle.
answered yesterday
Ben Crowell
47.6k3151289
1
Besides the quark-gluon plasma at high temperature, a quark soup picture should be appropriate at extreme density. It is unclear whether stable neutron stars actually achieve extreme density, or whether it is only achieved ephemerally, in the course of collapse to a black hole.
– Bert Barrois
yesterday
@BertBarrois: Yes, my description of two phases is definitely a little oversimplified. The WP article on quark-gluon plasma has a complicated phase diagram, with a bunch of different phases marked in, but I don't know whether some of these are purely theoretical.
– Ben Crowell
yesterday
add a comment |
up vote
24
down vote
This is basically a matter of energy scales. By analogy, you could ask why we don't take into account nuclear structure when we talk about chemistry. The answer is that the eV energy scale of chemistry is mismatched with the MeV energy scale of nuclear structure.
Nuclear matter has two phases. One is the phase we normally see, and the other is a quark-gluon plasma. The phase transition happens at a temperature on the order of 100 GeV per nucleon (at standard nuclear densities). Below the temperature of the phase transition, the quarks are strongly correlated, and those correlated groups behave in a way that's very similar to free neutrons and protons. To the extent that they don't quite have those properties, often we can subsume the discrepancies within adjustments to the parameters of the model. It's helpful in terms of practical computation that the fictitious neutrons and protons are nonrelativistic, which makes the theory much more tractable than QCD. If there are small relativistic effects, because the nucleons are moving at a few percent of $c$, these can also be subsumed within adjustments to the parameters.
By the way, it is actually possible to consider larger clusters to be the relevant degrees of freedom for nuclear structure. There is a model called the interacting boson approximation (IBA, also known as the interacting boson model, IBM), in which pairs of nucleons coupled to spin 0 or 1 are considered the degrees of freedom. It does pretty well in phenomonologically fitting the properties of many nuclei that are intractable in other models.
In a similar vein, there are alpha cluster models and ideas like explaining alpha decay in terms of preformation of an alpha particle, which then tunnels out through the Coulomb barrier. Pictures like these go back to the 1940's, and have considerable utility and explanatory power, although they can't really be microscopically correct, because they violate the Pauli exclusion principle.
answered yesterday
Ben Crowell
47.6k3151289
1
Besides the quark-gluon plasma at high temperature, a quark soup picture should be appropriate at extreme density. It is unclear whether stable neutron stars actually achieve extreme density, or whether it is only achieved ephemerally, in the course of collapse to a black hole.
– Bert Barrois
yesterday
@BertBarrois: Yes, my description of two phases is definitely a little oversimplified. The WP article on quark-gluon plasma has a complicated phase diagram, with a bunch of different phases marked in, but I don't know whether some of these are purely theoretical.
– Ben Crowell
yesterday
add a comment |
up vote
24
down vote
up vote
24
down vote
This is basically a matter of energy scales. By analogy, you could ask why we don't take into account nuclear structure when we talk about chemistry. The answer is that the eV energy scale of chemistry is mismatched with the MeV energy scale of nuclear structure.
Nuclear matter has two phases. One is the phase we normally see, and the other is a quark-gluon plasma. The phase transition happens at a temperature on the order of 100 GeV per nucleon (at standard nuclear densities). Below the temperature of the phase transition, the quarks are strongly correlated, and those correlated groups behave in a way that's very similar to free neutrons and protons. To the extent that they don't quite have those properties, often we can subsume the discrepancies within adjustments to the parameters of the model. It's helpful in terms of practical computation that the fictitious neutrons and protons are nonrelativistic, which makes the theory much more tractable than QCD. If there are small relativistic effects, because the nucleons are moving at a few percent of $c$, these can also be subsumed within adjustments to the parameters.
By the way, it is actually possible to consider larger clusters to be the relevant degrees of freedom for nuclear structure. There is a model called the interacting boson approximation (IBA, also known as the interacting boson model, IBM), in which pairs of nucleons coupled to spin 0 or 1 are considered the degrees of freedom. It does pretty well in phenomonologically fitting the properties of many nuclei that are intractable in other models.
In a similar vein, there are alpha cluster models and ideas like explaining alpha decay in terms of preformation of an alpha particle, which then tunnels out through the Coulomb barrier. Pictures like these go back to the 1940's, and have considerable utility and explanatory power, although they can't really be microscopically correct, because they violate the Pauli exclusion principle.
answered yesterday
Ben Crowell
47.6k3151289
This is basically a matter of energy scales. By analogy, you could ask why we don't take into account nuclear structure when we talk about chemistry. The answer is that the eV energy scale of chemistry is mismatched with the MeV energy scale of nuclear structure.
Nuclear matter has two phases. One is the phase we normally see, and the other is a quark-gluon plasma. The phase transition happens at a temperature on the order of 100 GeV per nucleon (at standard nuclear densities). Below the temperature of the phase transition, the quarks are strongly correlated, and those correlated groups behave in a way that's very similar to free neutrons and protons. To the extent that they don't quite have those properties, often we can subsume the discrepancies within adjustments to the parameters of the model. It's helpful in terms of practical computation that the fictitious neutrons and protons are nonrelativistic, which makes the theory much more tractable than QCD. If there are small relativistic effects, because the nucleons are moving at a few percent of $c$, these can also be subsumed within adjustments to the parameters.
By the way, it is actually possible to consider larger clusters to be the relevant degrees of freedom for nuclear structure. There is a model called the interacting boson approximation (IBA, also known as the interacting boson model, IBM), in which pairs of nucleons coupled to spin 0 or 1 are considered the degrees of freedom. It does pretty well in phenomonologically fitting the properties of many nuclei that are intractable in other models.
In a similar vein, there are alpha cluster models and ideas like explaining alpha decay in terms of preformation of an alpha particle, which then tunnels out through the Coulomb barrier. Pictures like these go back to the 1940's, and have considerable utility and explanatory power, although they can't really be microscopically correct, because they violate the Pauli exclusion principle.
answered yesterday
Ben Crowell
47.6k3151289
answered yesterday
Ben Crowell
47.6k3151289
answered yesterday
Ben Crowell
47.6k3151289
answered yesterday
Ben Crowell
47.6k3151289
47.6k3151289
1
Besides the quark-gluon plasma at high temperature, a quark soup picture should be appropriate at extreme density. It is unclear whether stable neutron stars actually achieve extreme density, or whether it is only achieved ephemerally, in the course of collapse to a black hole.
– Bert Barrois
yesterday
@BertBarrois: Yes, my description of two phases is definitely a little oversimplified. The WP article on quark-gluon plasma has a complicated phase diagram, with a bunch of different phases marked in, but I don't know whether some of these are purely theoretical.
– Ben Crowell
yesterday
add a comment |
1
Besides the quark-gluon plasma at high temperature, a quark soup picture should be appropriate at extreme density. It is unclear whether stable neutron stars actually achieve extreme density, or whether it is only achieved ephemerally, in the course of collapse to a black hole.
– Bert Barrois
yesterday
@BertBarrois: Yes, my description of two phases is definitely a little oversimplified. The WP article on quark-gluon plasma has a complicated phase diagram, with a bunch of different phases marked in, but I don't know whether some of these are purely theoretical.
– Ben Crowell
yesterday
1
1
Besides the quark-gluon plasma at high temperature, a quark soup picture should be appropriate at extreme density. It is unclear whether stable neutron stars actually achieve extreme density, or whether it is only achieved ephemerally, in the course of collapse to a black hole.
– Bert Barrois
yesterday
Besides the quark-gluon plasma at high temperature, a quark soup picture should be appropriate at extreme density. It is unclear whether stable neutron stars actually achieve extreme density, or whether it is only achieved ephemerally, in the course of collapse to a black hole.
– Bert Barrois
yesterday
@BertBarrois: Yes, my description of two phases is definitely a little oversimplified. The WP article on quark-gluon plasma has a complicated phase diagram, with a bunch of different phases marked in, but I don't know whether some of these are purely theoretical.
– Ben Crowell
yesterday
@BertBarrois: Yes, my description of two phases is definitely a little oversimplified. The WP article on quark-gluon plasma has a complicated phase diagram, with a bunch of different phases marked in, but I don't know whether some of these are purely theoretical.
– Ben Crowell
yesterday
add a comment |
up vote
17
down vote
We can measure the form-factors of bound nucleons. For instance by doing quasi-elastic scattering of a proton out of the nucleus $A(e,e'p)$ at low energy loss (my dissertation work involved this reaction for deuterium, helium, carbon and iron).
The result are quite similar to (but measurably not identical to) the equivalent results on free protons. That similarity make the choice of nucleons as the degrees of freedom a good starting point.
answered yesterday
dmckee♦
73.5k6131266
add a comment |
up vote
17
down vote
We can measure the form-factors of bound nucleons. For instance by doing quasi-elastic scattering of a proton out of the nucleus $A(e,e'p)$ at low energy loss (my dissertation work involved this reaction for deuterium, helium, carbon and iron).
The result are quite similar to (but measurably not identical to) the equivalent results on free protons. That similarity make the choice of nucleons as the degrees of freedom a good starting point.
answered yesterday
dmckee♦
73.5k6131266
add a comment |
up vote
17
down vote
up vote
17
down vote
We can measure the form-factors of bound nucleons. For instance by doing quasi-elastic scattering of a proton out of the nucleus $A(e,e'p)$ at low energy loss (my dissertation work involved this reaction for deuterium, helium, carbon and iron).
The result are quite similar to (but measurably not identical to) the equivalent results on free protons. That similarity make the choice of nucleons as the degrees of freedom a good starting point.
answered yesterday
dmckee♦
73.5k6131266
We can measure the form-factors of bound nucleons. For instance by doing quasi-elastic scattering of a proton out of the nucleus $A(e,e'p)$ at low energy loss (my dissertation work involved this reaction for deuterium, helium, carbon and iron).
The result are quite similar to (but measurably not identical to) the equivalent results on free protons. That similarity make the choice of nucleons as the degrees of freedom a good starting point.
answered yesterday
dmckee♦
73.5k6131266
answered yesterday
dmckee♦
73.5k6131266
answered yesterday
dmckee♦
73.5k6131266
answered yesterday
dmckee♦
73.5k6131266
73.5k6131266
add a comment |
add a comment |
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5
Some "stupid sounding" questions can be rather fundamental. This one is fundamental enough that'd I'd have to ask which direction do you want this to be handled? Are you looking at it the historical way, which went from chemical properties towards atomic weights and numbers? Or are you looking backwards, from the perspective of QM, to see that QM predicts these to be meaningful visualizations? For the latter, it's useful to remember that QM was invented to fit the data we had already observed, such as the meaningfulness of protons and neutrons in a nucleus.
– Cort Ammon
yesterday
3
Perhaps because we like to simplify things to make them easier to understand & talk about? In a lot of contexts, we don't have to understand what a nucleus actually "is" inside, we just need to know that it has so many units of charge (protons), such and such an atomic weight that's not the same as units of charge (neutrons), and that if we break it apart by e.g. hitting it with high-energy particles, protons & neutrons come out.
– jamesqf
yesterday
4
@jamesqf That is a perfectly valid answer.
– dmckee♦
yesterday
Related, related.
– rob♦
14 hours ago
2
This is a straightforward application of Weinberg’s Third Law of Progress in Theoretical Physics: "You may use any degrees of freedom you like to describe a physical system, but if you use the wrong ones, you’ll be sorry!"
– Xerxes
12 hours ago