How is the speed of nucleons in the nucleus measured?What is an intuitive picture of the motion of nucleons?How do we know that the nucleus isn't a quark-gluon plasma?What do we know about the interactions between the protons and neutrons in a nucleus?binding energy of a nucleus is positive?Oscillation of nucleons mediated by polarizing electron charge on opposite ends of the nucleus?Are nucleons discrete within a nucleus?Properties of the nucleons and the nucleus from the interactions between quarks and gluonsRest mass of the nucleus and its constituent nucleonsIs the arrangement of nucleons within a nucleus identical between heavy elements of the same type?formation of atomic nuclei from nucleons

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How is the speed of nucleons in the nucleus measured?


What is an intuitive picture of the motion of nucleons?How do we know that the nucleus isn't a quark-gluon plasma?What do we know about the interactions between the protons and neutrons in a nucleus?binding energy of a nucleus is positive?Oscillation of nucleons mediated by polarizing electron charge on opposite ends of the nucleus?Are nucleons discrete within a nucleus?Properties of the nucleons and the nucleus from the interactions between quarks and gluonsRest mass of the nucleus and its constituent nucleonsIs the arrangement of nucleons within a nucleus identical between heavy elements of the same type?formation of atomic nuclei from nucleons






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;









3












$begingroup$


How do you measure or infer the speed of nucleons in the nucleus?



(For Aaron Stevens : link.)










share|cite|improve this question











$endgroup$













  • $begingroup$
    Do you have an example/source of this? More detail is needed.
    $endgroup$
    – Aaron Stevens
    7 hours ago










  • $begingroup$
    Closely related: physics.stackexchange.com/questions/35724/… where I posted some of the same figures.
    $endgroup$
    – dmckee
    7 hours ago

















3












$begingroup$


How do you measure or infer the speed of nucleons in the nucleus?



(For Aaron Stevens : link.)










share|cite|improve this question











$endgroup$













  • $begingroup$
    Do you have an example/source of this? More detail is needed.
    $endgroup$
    – Aaron Stevens
    7 hours ago










  • $begingroup$
    Closely related: physics.stackexchange.com/questions/35724/… where I posted some of the same figures.
    $endgroup$
    – dmckee
    7 hours ago













3












3








3





$begingroup$


How do you measure or infer the speed of nucleons in the nucleus?



(For Aaron Stevens : link.)










share|cite|improve this question











$endgroup$




How do you measure or infer the speed of nucleons in the nucleus?



(For Aaron Stevens : link.)







experimental-physics nuclear-physics measurements neutrons protons






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









dmckee

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asked 8 hours ago









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  • $begingroup$
    Do you have an example/source of this? More detail is needed.
    $endgroup$
    – Aaron Stevens
    7 hours ago










  • $begingroup$
    Closely related: physics.stackexchange.com/questions/35724/… where I posted some of the same figures.
    $endgroup$
    – dmckee
    7 hours ago
















  • $begingroup$
    Do you have an example/source of this? More detail is needed.
    $endgroup$
    – Aaron Stevens
    7 hours ago










  • $begingroup$
    Closely related: physics.stackexchange.com/questions/35724/… where I posted some of the same figures.
    $endgroup$
    – dmckee
    7 hours ago















$begingroup$
Do you have an example/source of this? More detail is needed.
$endgroup$
– Aaron Stevens
7 hours ago




$begingroup$
Do you have an example/source of this? More detail is needed.
$endgroup$
– Aaron Stevens
7 hours ago












$begingroup$
Closely related: physics.stackexchange.com/questions/35724/… where I posted some of the same figures.
$endgroup$
– dmckee
7 hours ago




$begingroup$
Closely related: physics.stackexchange.com/questions/35724/… where I posted some of the same figures.
$endgroup$
– dmckee
7 hours ago










2 Answers
2






active

oldest

votes


















9














$begingroup$

If you shoot an electron or a proton at a nucleus at moderate energies (a few hundred $mathrmMeV$ to a few $mathrmGeV$) it will usually either bounce off the whole nucleus or break up the nucleus. But every once in a while (and this gets rarer and rarer the harder you throw it in) it will actually bounce off of a single nucleon. At the right energies this can happen without exciting the target nucleon (or the beam nucleon if you are using a proton beam), and still knock that nucleon right out of the parent nucleus without doing too much mischief on the way out.



These events are termed "quasi-elastic scattering" (not to be confused with the use of the same term in neutrino scattering).



The "quasi" is because there is some interaction between the beam particle and nucleus on the way in and out and some interaction between the scattered particle and the remnant-nucleus on the way out. But this is fairly modest and and can be computed in simulation.



It is the "elastic" part which we concentrate on. Elastic collisions between two objects are fully constrained by energy and momentum conservation. We know the initial energy and momentum of the beam particle, so if we measure the energy and momentum of the scattered particles we can deduce the energy and momentum of the target particle before it was hit.



Then all that remains is to computationaly remove the effects of the final-state interactions. Well, and we have to allow for the fact that a nuclear proton is slightly different from a free proton, and for that we rely on phenomenological models.



This is exactly the kind of data we took in my disseration experiment (our purpose was a little more subtle than just making the momentum measurement, but we got that for free). We took events like $A(e,e'p)$ using a $5$-$6 ,mathrmGeV$ electron beam on protium, deutrium, carbon, and iron targets.



enter image description here



This figure appeared in my dissertation. It shows the processed result for squared momentum transfer of $3.3 ,mathrmGeV^2$ on a carbon target.



  • In the right-hand panel we plot the magnitude of the initial momentum of the struct particle in units of $mathrmGeV/c$. You can see that the protons are mildly relativistic (recall that their mass is nearly $1 ,mathrmGeV/c^2$).


  • In the left-hand panel we plot the binding energy of the struck proton.


  • We see the shell structure of the nucleus. The s-shell is responsible for the smaller, more-tightly bound energy hump and the non-zero density at the center of the momentum graph. The p-shell is responsible for the less tightly bound energy hump and the two-lobed structure of the momentum graph.






share|cite|improve this answer











$endgroup$














  • $begingroup$
    Yeah. And thanks for the blue pencil work. I read what I meant instead of what I wrote for hours or days.
    $endgroup$
    – dmckee
    7 hours ago










  • $begingroup$
    Not a problem. Decades of coding has taught me to never underestimate the elusiveness of a determined typo. ;)
    $endgroup$
    – PM 2Ring
    7 hours ago










  • $begingroup$
    This is a very nice answer, I learnt something today! +1
    $endgroup$
    – Gert
    2 hours ago


















3














$begingroup$

There are a variety of ways of getting this kind of information. The simplest way to get a rough estimate is just to use the size of the nucleus and the de Broglie relation. A moderate-sized nucleus has a diameter of roughly 7 fm, so the lowest-energy standing wave would have a wavelength of 14 fm along any given axis $x$. Plugging in to the de Broglie relation gives $v_x/capprox 0.1$ in this example.






share|cite|improve this answer









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    2 Answers
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    2 Answers
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    active

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    9














    $begingroup$

    If you shoot an electron or a proton at a nucleus at moderate energies (a few hundred $mathrmMeV$ to a few $mathrmGeV$) it will usually either bounce off the whole nucleus or break up the nucleus. But every once in a while (and this gets rarer and rarer the harder you throw it in) it will actually bounce off of a single nucleon. At the right energies this can happen without exciting the target nucleon (or the beam nucleon if you are using a proton beam), and still knock that nucleon right out of the parent nucleus without doing too much mischief on the way out.



    These events are termed "quasi-elastic scattering" (not to be confused with the use of the same term in neutrino scattering).



    The "quasi" is because there is some interaction between the beam particle and nucleus on the way in and out and some interaction between the scattered particle and the remnant-nucleus on the way out. But this is fairly modest and and can be computed in simulation.



    It is the "elastic" part which we concentrate on. Elastic collisions between two objects are fully constrained by energy and momentum conservation. We know the initial energy and momentum of the beam particle, so if we measure the energy and momentum of the scattered particles we can deduce the energy and momentum of the target particle before it was hit.



    Then all that remains is to computationaly remove the effects of the final-state interactions. Well, and we have to allow for the fact that a nuclear proton is slightly different from a free proton, and for that we rely on phenomenological models.



    This is exactly the kind of data we took in my disseration experiment (our purpose was a little more subtle than just making the momentum measurement, but we got that for free). We took events like $A(e,e'p)$ using a $5$-$6 ,mathrmGeV$ electron beam on protium, deutrium, carbon, and iron targets.



    enter image description here



    This figure appeared in my dissertation. It shows the processed result for squared momentum transfer of $3.3 ,mathrmGeV^2$ on a carbon target.



    • In the right-hand panel we plot the magnitude of the initial momentum of the struct particle in units of $mathrmGeV/c$. You can see that the protons are mildly relativistic (recall that their mass is nearly $1 ,mathrmGeV/c^2$).


    • In the left-hand panel we plot the binding energy of the struck proton.


    • We see the shell structure of the nucleus. The s-shell is responsible for the smaller, more-tightly bound energy hump and the non-zero density at the center of the momentum graph. The p-shell is responsible for the less tightly bound energy hump and the two-lobed structure of the momentum graph.






    share|cite|improve this answer











    $endgroup$














    • $begingroup$
      Yeah. And thanks for the blue pencil work. I read what I meant instead of what I wrote for hours or days.
      $endgroup$
      – dmckee
      7 hours ago










    • $begingroup$
      Not a problem. Decades of coding has taught me to never underestimate the elusiveness of a determined typo. ;)
      $endgroup$
      – PM 2Ring
      7 hours ago










    • $begingroup$
      This is a very nice answer, I learnt something today! +1
      $endgroup$
      – Gert
      2 hours ago















    9














    $begingroup$

    If you shoot an electron or a proton at a nucleus at moderate energies (a few hundred $mathrmMeV$ to a few $mathrmGeV$) it will usually either bounce off the whole nucleus or break up the nucleus. But every once in a while (and this gets rarer and rarer the harder you throw it in) it will actually bounce off of a single nucleon. At the right energies this can happen without exciting the target nucleon (or the beam nucleon if you are using a proton beam), and still knock that nucleon right out of the parent nucleus without doing too much mischief on the way out.



    These events are termed "quasi-elastic scattering" (not to be confused with the use of the same term in neutrino scattering).



    The "quasi" is because there is some interaction between the beam particle and nucleus on the way in and out and some interaction between the scattered particle and the remnant-nucleus on the way out. But this is fairly modest and and can be computed in simulation.



    It is the "elastic" part which we concentrate on. Elastic collisions between two objects are fully constrained by energy and momentum conservation. We know the initial energy and momentum of the beam particle, so if we measure the energy and momentum of the scattered particles we can deduce the energy and momentum of the target particle before it was hit.



    Then all that remains is to computationaly remove the effects of the final-state interactions. Well, and we have to allow for the fact that a nuclear proton is slightly different from a free proton, and for that we rely on phenomenological models.



    This is exactly the kind of data we took in my disseration experiment (our purpose was a little more subtle than just making the momentum measurement, but we got that for free). We took events like $A(e,e'p)$ using a $5$-$6 ,mathrmGeV$ electron beam on protium, deutrium, carbon, and iron targets.



    enter image description here



    This figure appeared in my dissertation. It shows the processed result for squared momentum transfer of $3.3 ,mathrmGeV^2$ on a carbon target.



    • In the right-hand panel we plot the magnitude of the initial momentum of the struct particle in units of $mathrmGeV/c$. You can see that the protons are mildly relativistic (recall that their mass is nearly $1 ,mathrmGeV/c^2$).


    • In the left-hand panel we plot the binding energy of the struck proton.


    • We see the shell structure of the nucleus. The s-shell is responsible for the smaller, more-tightly bound energy hump and the non-zero density at the center of the momentum graph. The p-shell is responsible for the less tightly bound energy hump and the two-lobed structure of the momentum graph.






    share|cite|improve this answer











    $endgroup$














    • $begingroup$
      Yeah. And thanks for the blue pencil work. I read what I meant instead of what I wrote for hours or days.
      $endgroup$
      – dmckee
      7 hours ago










    • $begingroup$
      Not a problem. Decades of coding has taught me to never underestimate the elusiveness of a determined typo. ;)
      $endgroup$
      – PM 2Ring
      7 hours ago










    • $begingroup$
      This is a very nice answer, I learnt something today! +1
      $endgroup$
      – Gert
      2 hours ago













    9














    9










    9







    $begingroup$

    If you shoot an electron or a proton at a nucleus at moderate energies (a few hundred $mathrmMeV$ to a few $mathrmGeV$) it will usually either bounce off the whole nucleus or break up the nucleus. But every once in a while (and this gets rarer and rarer the harder you throw it in) it will actually bounce off of a single nucleon. At the right energies this can happen without exciting the target nucleon (or the beam nucleon if you are using a proton beam), and still knock that nucleon right out of the parent nucleus without doing too much mischief on the way out.



    These events are termed "quasi-elastic scattering" (not to be confused with the use of the same term in neutrino scattering).



    The "quasi" is because there is some interaction between the beam particle and nucleus on the way in and out and some interaction between the scattered particle and the remnant-nucleus on the way out. But this is fairly modest and and can be computed in simulation.



    It is the "elastic" part which we concentrate on. Elastic collisions between two objects are fully constrained by energy and momentum conservation. We know the initial energy and momentum of the beam particle, so if we measure the energy and momentum of the scattered particles we can deduce the energy and momentum of the target particle before it was hit.



    Then all that remains is to computationaly remove the effects of the final-state interactions. Well, and we have to allow for the fact that a nuclear proton is slightly different from a free proton, and for that we rely on phenomenological models.



    This is exactly the kind of data we took in my disseration experiment (our purpose was a little more subtle than just making the momentum measurement, but we got that for free). We took events like $A(e,e'p)$ using a $5$-$6 ,mathrmGeV$ electron beam on protium, deutrium, carbon, and iron targets.



    enter image description here



    This figure appeared in my dissertation. It shows the processed result for squared momentum transfer of $3.3 ,mathrmGeV^2$ on a carbon target.



    • In the right-hand panel we plot the magnitude of the initial momentum of the struct particle in units of $mathrmGeV/c$. You can see that the protons are mildly relativistic (recall that their mass is nearly $1 ,mathrmGeV/c^2$).


    • In the left-hand panel we plot the binding energy of the struck proton.


    • We see the shell structure of the nucleus. The s-shell is responsible for the smaller, more-tightly bound energy hump and the non-zero density at the center of the momentum graph. The p-shell is responsible for the less tightly bound energy hump and the two-lobed structure of the momentum graph.






    share|cite|improve this answer











    $endgroup$



    If you shoot an electron or a proton at a nucleus at moderate energies (a few hundred $mathrmMeV$ to a few $mathrmGeV$) it will usually either bounce off the whole nucleus or break up the nucleus. But every once in a while (and this gets rarer and rarer the harder you throw it in) it will actually bounce off of a single nucleon. At the right energies this can happen without exciting the target nucleon (or the beam nucleon if you are using a proton beam), and still knock that nucleon right out of the parent nucleus without doing too much mischief on the way out.



    These events are termed "quasi-elastic scattering" (not to be confused with the use of the same term in neutrino scattering).



    The "quasi" is because there is some interaction between the beam particle and nucleus on the way in and out and some interaction between the scattered particle and the remnant-nucleus on the way out. But this is fairly modest and and can be computed in simulation.



    It is the "elastic" part which we concentrate on. Elastic collisions between two objects are fully constrained by energy and momentum conservation. We know the initial energy and momentum of the beam particle, so if we measure the energy and momentum of the scattered particles we can deduce the energy and momentum of the target particle before it was hit.



    Then all that remains is to computationaly remove the effects of the final-state interactions. Well, and we have to allow for the fact that a nuclear proton is slightly different from a free proton, and for that we rely on phenomenological models.



    This is exactly the kind of data we took in my disseration experiment (our purpose was a little more subtle than just making the momentum measurement, but we got that for free). We took events like $A(e,e'p)$ using a $5$-$6 ,mathrmGeV$ electron beam on protium, deutrium, carbon, and iron targets.



    enter image description here



    This figure appeared in my dissertation. It shows the processed result for squared momentum transfer of $3.3 ,mathrmGeV^2$ on a carbon target.



    • In the right-hand panel we plot the magnitude of the initial momentum of the struct particle in units of $mathrmGeV/c$. You can see that the protons are mildly relativistic (recall that their mass is nearly $1 ,mathrmGeV/c^2$).


    • In the left-hand panel we plot the binding energy of the struck proton.


    • We see the shell structure of the nucleus. The s-shell is responsible for the smaller, more-tightly bound energy hump and the non-zero density at the center of the momentum graph. The p-shell is responsible for the less tightly bound energy hump and the two-lobed structure of the momentum graph.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 5 hours ago

























    answered 7 hours ago









    dmckeedmckee

    77.2k6 gold badges140 silver badges281 bronze badges




    77.2k6 gold badges140 silver badges281 bronze badges














    • $begingroup$
      Yeah. And thanks for the blue pencil work. I read what I meant instead of what I wrote for hours or days.
      $endgroup$
      – dmckee
      7 hours ago










    • $begingroup$
      Not a problem. Decades of coding has taught me to never underestimate the elusiveness of a determined typo. ;)
      $endgroup$
      – PM 2Ring
      7 hours ago










    • $begingroup$
      This is a very nice answer, I learnt something today! +1
      $endgroup$
      – Gert
      2 hours ago
















    • $begingroup$
      Yeah. And thanks for the blue pencil work. I read what I meant instead of what I wrote for hours or days.
      $endgroup$
      – dmckee
      7 hours ago










    • $begingroup$
      Not a problem. Decades of coding has taught me to never underestimate the elusiveness of a determined typo. ;)
      $endgroup$
      – PM 2Ring
      7 hours ago










    • $begingroup$
      This is a very nice answer, I learnt something today! +1
      $endgroup$
      – Gert
      2 hours ago















    $begingroup$
    Yeah. And thanks for the blue pencil work. I read what I meant instead of what I wrote for hours or days.
    $endgroup$
    – dmckee
    7 hours ago




    $begingroup$
    Yeah. And thanks for the blue pencil work. I read what I meant instead of what I wrote for hours or days.
    $endgroup$
    – dmckee
    7 hours ago












    $begingroup$
    Not a problem. Decades of coding has taught me to never underestimate the elusiveness of a determined typo. ;)
    $endgroup$
    – PM 2Ring
    7 hours ago




    $begingroup$
    Not a problem. Decades of coding has taught me to never underestimate the elusiveness of a determined typo. ;)
    $endgroup$
    – PM 2Ring
    7 hours ago












    $begingroup$
    This is a very nice answer, I learnt something today! +1
    $endgroup$
    – Gert
    2 hours ago




    $begingroup$
    This is a very nice answer, I learnt something today! +1
    $endgroup$
    – Gert
    2 hours ago













    3














    $begingroup$

    There are a variety of ways of getting this kind of information. The simplest way to get a rough estimate is just to use the size of the nucleus and the de Broglie relation. A moderate-sized nucleus has a diameter of roughly 7 fm, so the lowest-energy standing wave would have a wavelength of 14 fm along any given axis $x$. Plugging in to the de Broglie relation gives $v_x/capprox 0.1$ in this example.






    share|cite|improve this answer









    $endgroup$



















      3














      $begingroup$

      There are a variety of ways of getting this kind of information. The simplest way to get a rough estimate is just to use the size of the nucleus and the de Broglie relation. A moderate-sized nucleus has a diameter of roughly 7 fm, so the lowest-energy standing wave would have a wavelength of 14 fm along any given axis $x$. Plugging in to the de Broglie relation gives $v_x/capprox 0.1$ in this example.






      share|cite|improve this answer









      $endgroup$

















        3














        3










        3







        $begingroup$

        There are a variety of ways of getting this kind of information. The simplest way to get a rough estimate is just to use the size of the nucleus and the de Broglie relation. A moderate-sized nucleus has a diameter of roughly 7 fm, so the lowest-energy standing wave would have a wavelength of 14 fm along any given axis $x$. Plugging in to the de Broglie relation gives $v_x/capprox 0.1$ in this example.






        share|cite|improve this answer









        $endgroup$



        There are a variety of ways of getting this kind of information. The simplest way to get a rough estimate is just to use the size of the nucleus and the de Broglie relation. A moderate-sized nucleus has a diameter of roughly 7 fm, so the lowest-energy standing wave would have a wavelength of 14 fm along any given axis $x$. Plugging in to the de Broglie relation gives $v_x/capprox 0.1$ in this example.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 4 hours ago









        Ben CrowellBen Crowell

        61.2k6 gold badges179 silver badges343 bronze badges




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            François Viète Contents Biography Work and thought Bibliography See also Notes Further reading External links Navigation menup. 21Google Bookspp. 75–77Google BooksDe thou (from University of Saint Andrews)ArchivedGoogle BooksGoogle BooksGoogle BooksGoogle booksGoogle Bookscc-parthenay.frL'histoire universelle (fr)Universal History (en)ArchivedAdsabs.harvard.eduPagesperso-orange.frArchive.orgChikara Sasaki. Descartes' mathematical thought p.259Google BooksGoogle BooksGoogle Bookspp. 152 and onwardGoogle BooksGoogle BooksScribd.comGoogle Books1257-7979Google BooksGoogle BooksGoogle BooksGoogle BooksGoogle BooksGoogle BooksGallica.bnf.frGoogle BooksGoogle Books"François Viète"Francois Viète: Father of Modern Algebraic NotationThe Lawyer and the GamblerAbout TarporleySite de Jean-Paul GuichardL'algèbre nouvelle"About the Harmonicon"cb120511976(data)1188044800000 0001 0913 5903n82164680ola2013766880073431702w6vt1sb70287374827140948071409480