Why are protons and neutrons the “right” degrees of freedom of nuclei?
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 Dec 6 at 14:51
tonydo
939619
|
show 3 more comments
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 Dec 6 at 14:51
tonydo
939619
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
Dec 6 at 19:29
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
Dec 6 at 19:34
4
@jamesqf That is a perfectly valid answer.
– dmckee♦
Dec 6 at 23:29
4
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
Dec 7 at 16:24
1
Could someone please explain how the term "degrees of freedom" is applicable to the components of a system, as in this question, please? I'm used to the term in statistics (~unconstrained set of variables) and robotics (~independent movements), but don't quite see how to abstract it to fit its use here. Is it just another way to say "analyse the system? If Weinberg and others use it in this way, I have something useful to learn...
– iSeeker
Dec 11 at 17:39
|
show 3 more comments
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 Dec 6 at 14:51
tonydo
939619
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 Dec 6 at 14:51
tonydo
939619
asked Dec 6 at 14:51
tonydo
939619
asked Dec 6 at 14:51
tonydo
939619
asked Dec 6 at 14:51
tonydo
939619
asked Dec 6 at 14:51
tonydo
939619
939619
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
Dec 6 at 19:29
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
Dec 6 at 19:34
4
@jamesqf That is a perfectly valid answer.
– dmckee♦
Dec 6 at 23:29
4
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
Dec 7 at 16:24
1
Could someone please explain how the term "degrees of freedom" is applicable to the components of a system, as in this question, please? I'm used to the term in statistics (~unconstrained set of variables) and robotics (~independent movements), but don't quite see how to abstract it to fit its use here. Is it just another way to say "analyse the system? If Weinberg and others use it in this way, I have something useful to learn...
– iSeeker
Dec 11 at 17:39
|
show 3 more comments
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
Dec 6 at 19:29
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
Dec 6 at 19:34
4
@jamesqf That is a perfectly valid answer.
– dmckee♦
Dec 6 at 23:29
4
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
Dec 7 at 16:24
1
Could someone please explain how the term "degrees of freedom" is applicable to the components of a system, as in this question, please? I'm used to the term in statistics (~unconstrained set of variables) and robotics (~independent movements), but don't quite see how to abstract it to fit its use here. Is it just another way to say "analyse the system? If Weinberg and others use it in this way, I have something useful to learn...
– iSeeker
Dec 11 at 17:39
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
Dec 6 at 19:29
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
Dec 6 at 19:29
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
Dec 6 at 19:34
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
Dec 6 at 19:34
4
4
@jamesqf That is a perfectly valid answer.
– dmckee♦
Dec 6 at 23:29
@jamesqf That is a perfectly valid answer.
– dmckee♦
Dec 6 at 23:29
4
4
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
Dec 7 at 16:24
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
Dec 7 at 16:24
1
1
Could someone please explain how the term "degrees of freedom" is applicable to the components of a system, as in this question, please? I'm used to the term in statistics (~unconstrained set of variables) and robotics (~independent movements), but don't quite see how to abstract it to fit its use here. Is it just another way to say "analyse the system? If Weinberg and others use it in this way, I have something useful to learn...
– iSeeker
Dec 11 at 17:39
Could someone please explain how the term "degrees of freedom" is applicable to the components of a system, as in this question, please? I'm used to the term in statistics (~unconstrained set of variables) and robotics (~independent movements), but don't quite see how to abstract it to fit its use here. Is it just another way to say "analyse the system? If Weinberg and others use it in this way, I have something useful to learn...
– iSeeker
Dec 11 at 17:39
|
show 3 more comments
2 Answers
2
active
oldest
votes
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 Dec 6 at 15:13
dmckee♦
73.7k6131266
add a comment |
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 Dec 6 at 15:08
Ben Crowell
48.3k4151292
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
Dec 6 at 17:35
@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
Dec 6 at 19:34
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 Dec 6 at 15:13
dmckee♦
73.7k6131266
add a comment |
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 Dec 6 at 15:13
dmckee♦
73.7k6131266
add a comment |
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 Dec 6 at 15:13
dmckee♦
73.7k6131266
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 Dec 6 at 15:13
dmckee♦
73.7k6131266
answered Dec 6 at 15:13
dmckee♦
73.7k6131266
answered Dec 6 at 15:13
dmckee♦
73.7k6131266
answered Dec 6 at 15:13
dmckee♦
73.7k6131266
73.7k6131266
add a comment |
add a comment |
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 Dec 6 at 15:08
Ben Crowell
48.3k4151292
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
Dec 6 at 17:35
@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
Dec 6 at 19:34
add a comment |
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 Dec 6 at 15:08
Ben Crowell
48.3k4151292
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
Dec 6 at 17:35
@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
Dec 6 at 19:34
add a comment |
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 Dec 6 at 15:08
Ben Crowell
48.3k4151292
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 Dec 6 at 15:08
Ben Crowell
48.3k4151292
answered Dec 6 at 15:08
Ben Crowell
48.3k4151292
answered Dec 6 at 15:08
Ben Crowell
48.3k4151292
answered Dec 6 at 15:08
Ben Crowell
48.3k4151292
48.3k4151292
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
Dec 6 at 17:35
@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
Dec 6 at 19:34
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
Dec 6 at 17:35
@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
Dec 6 at 19:34
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
Dec 6 at 17:35
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
Dec 6 at 17:35
@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
Dec 6 at 19:34
@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
Dec 6 at 19:34
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
Dec 6 at 19:29
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
Dec 6 at 19:34
4
@jamesqf That is a perfectly valid answer.
– dmckee♦
Dec 6 at 23:29
4
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
Dec 7 at 16:24
1
Could someone please explain how the term "degrees of freedom" is applicable to the components of a system, as in this question, please? I'm used to the term in statistics (~unconstrained set of variables) and robotics (~independent movements), but don't quite see how to abstract it to fit its use here. Is it just another way to say "analyse the system? If Weinberg and others use it in this way, I have something useful to learn...
– iSeeker
Dec 11 at 17:39