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Why spinning protons have precession under magnetic field?

  1. Apr 15, 2005 #1
    And this phenomenon also occurs in neutrons?
  2. jcsd
  3. Apr 15, 2005 #2


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    What is spinning ?The particle?Why and how do you know that?What angular velocity does it have??

  4. Apr 15, 2005 #3
    from classical electrodynamics, we know that a charged spinning object generates a magnetic field denoted by the magnetic moment. The reason for this effect is the 'symmetry' between electric and magnetic phenomena. This means that a magnetic dipole is equivalent to a charged particle that makes a circular movement (in anorbit for the orbital momentum and around it's axis for the spin momentum)

    An angular momentum and a magnetic moment could indeed arise from a spinning sphere of charge, but this classical picture cannot fit the size or quantized nature of the electron spin.

    The property called electron spin must be considered to be a quantum concept without detailed classical analogy. So, in QM, the object does not actually spin around its axis. the reason why spin exists really is the Zeemann-effect and the theoretical work of Goudsmit and Uhlenbeck.

    A neutron indeed has the same property : a non-zero magnetic moment. The reason for this is the presence of the three constituent quarks and many other virtual quark anti-quark-pairs, called dynamical quarks. Each quark has a charge. As a matter of fact, the non-zero magnetic moment of a neutron was one of the concepts that suggested that there had to be somthing inside the neutron.

    Last edited: Apr 15, 2005
  5. Apr 15, 2005 #4
    I would like to know how the magnetic field generated by spinning protons interacts with the external B-field.

    Do we need to use the term, torque, to explain the precession?
  6. Apr 15, 2005 #5
    the philosophy is to look how the magnetic dipole (denoted by the magnetic moment mu) interacts with the extern B-field
    Yes : http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/larmor.html

  7. Apr 15, 2005 #6
    What makes the magnetic moment with a tilting angle?
    (shown in green)

    Attached Files:

  8. Apr 15, 2005 #7
    Well the circular orbit of the particle is always perpendicular to the orbital angular momentum vector L because of it's definition. Now suppose that the plane of this circular orbit is NOT perpendicular to the B-field, then you acquire the tilt. Thus, the reason is that we treat this phenomenon as generally as possible. the particle does NOT need (but it can ofcourse) to 'rotate' in a plane perpendicular to B
    Last edited: Feb 9, 2006
  9. Apr 15, 2005 #8
    So, the angle is caused by the orientation of plane of this circular orbit but not due to any forces, right?
  10. Apr 15, 2005 #9
    The angle is not caused by forces, however, when given a magnetic moment that makes an angle with the B-field there will be an associated orientation potential energy [tex]- \vec{\mu}. \vec{B}[/tex]

    This energy arises from the fact that a force (ie the Lorentzforce) will act on the moving charged particles. Also, a torque will be exerted on this L-vector because of this very reason. It is this torque that is directly responsable for the precession of L because a torque is equal to a change in L with respect to time : http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magmom.html#c2

  11. Apr 16, 2005 #10
    How about if the orientation of plane of this circular orbit is exactly perpendicular to the B-field, will there be no precession?
  12. Apr 16, 2005 #11
    No, because in that case, B and mu are parallell and therefore their vector-product will be zero, thus the torque exerted on the loop will be zero. If the torque is zero then L does not change wtr to time and it remains constant.

  13. Apr 16, 2005 #12
    So, no torque, no precession...but why protons do not align their magnetic moments to the Bo?
    I mean what factors make them not to align, just like the case in NMR.
    Last edited: Apr 16, 2005
  14. Apr 16, 2005 #13
    What exactly do you mean ?

    Look, suppose we have a ferromagnetic material which is caracterized by the fact thal all atom spins will align due to magnetic interactions between neigboring atoms. This results in a permanent magnetization beneath some critical temperature. You need to know that the specimen is actually devided into many local areas called the domains and the atoms in each domain are all aligned in the same direction. However, two different domains each can have a different direction in which all atoms are aligned.

    Now, we apply a magnetic field B along the z-axis

    The potential energy U (see above formula) will be lowest when the magnetization (or the magnetic moments) is parallel to the B-field. So indeed, there will be an attempt of the atoms to align their spin in the B-direction.
    But the specimen is actually divided into many domains in which the atoms are all aligned in the same direction. However each domain has a different magnetic moment (ie : magnetization). So some magnetizations will already be aligned along the z-axis and others will have an angle with the B-field. However, and i think that is what you mean, the B-field will try to lower the potential energy by directing the magnetic moments of each domain (denoted by the magnetization vector) along the B-field.

    The alignment of the magnetic moments towards the B-field can be made difficult by a concept called anisotropy.There is indeed an influence of the crystal structure and the shape of grains on the direction of magnetization. The dependence of magnetic properties on a preferred direction is called magnetic anisotropy. There are several different types of anisotropy:

    Type depends on

    1. magnetocrystalline- crystal structure

    2. shape- grain shape

    3. stress- applied or residual stresses

    Last edited: Apr 16, 2005
  15. Apr 18, 2005 #14
    Does the tilted angle relate to the nuclear spin quantum number, 1/2?
  16. Apr 18, 2005 #15


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    I see several levels of misunderstanding here.

    First, when we look at the NMR case, we need to keep in mind that the magnetization here is the result of the BULK magnetization from the whole material. This is the sum of ALL the individual magnetic moment of nucleus in that material, and at a finite temperature. What this means is that there is a population statistics of magnetic moments that are alligned and anti-alligned with the external field. So even in principle, at a non-zero temperature, the net magnetization is NEVER the total sum of ALL the individual magnetic moment.

    Secondly, there is a problem with the question: "I would like to know how the magnetic field generated by spinning protons interacts with the external B-field."

    If you look at the history of this thing, you would have seen that what is discovered FIRST is the magnetic moment of particles such as electrons and protons, etc. In other words, we first detected that they interact with an external magnetic field. Thus, we know they have magnetic moments. Since our knowledge back then only state that anything with a magnetic moment should have a charge "spinning", we gave a property of spin to these particles. Only later did we know better, that what we thought we knew isn't really the case, but the name stuck anyway since anyone who have studied QM would have realized that it is just a LABEL with only a historical context.

    So what you are asking for is to explain why something that doesn't exist explain something that does exist. Do you see why this is almost impossible to answer? If you ask "what is the origin of the magnetic moment in protons, electrons, neutrons, etc.", then that would have some meaning, and spin physics is a very active topic of reserach. However, asking how the spinning proton generates a magnetic moment cannot elicit any reasonable answer since you are trying to connect something that doesn't exist (spinning proton) with something that is (magnetic moment).

  17. Apr 18, 2005 #16
    Why would you think that if you know that the spin 1/2 value is the socalled z-component of the spin operator. Look, it's not that i don't want to answer your question, but if you start asking questions like that it is fundamental that you know concepts like spin (and how this arises in QM) very well...
    Otherwise, this thread will degenerate in a sequence of posts that only continue based upon misconceptions and poor interpretations due to lack of knowledge. I think you know very well what you have to do now...

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