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Precession in B field

  1. Jan 27, 2013 #1
    There is a problem about precession in general that I am trying to understand, and it can be exampled by electron orbiting around nucleus in external [itex]\vec{B}[/itex] field.

    We could consider electron motion as loop current which is characterized by the magnetic moment [itex]\vec{μ}[/itex], [itex]\vec{μ}[/itex]=-[itex]\frac{e}{2m}[/itex][itex]\vec{L}[/itex]. Suppose that external [itex]\vec{B}[/itex] field is orthogonal to [itex]\vec{μ}[/itex].

    The torque due to [itex]\vec{B}[/itex] is: [itex]\vec{τ}[/itex]=[itex]\vec{μ}[/itex]×[itex]\vec{B}[/itex].

    Assume that [itex]\vec{B}[/itex] was suddenly turned on. What happens after that? I am not sure.

    If you write [itex]\vec{τ}[/itex]=[itex]\frac{d\vec{L}}{dt}[/itex], you get [itex]d\vec{L}[/itex]=[itex]\vec{τ}dt[/itex]. Therefore, since [itex]\vec{τ}[/itex] is orthogonal to both [itex]\vec{μ}[/itex] and [itex]\vec{B}[/itex] increment of [itex]\vec{L}[/itex] is orthogonal to itself. From this reasoning I conclude that [itex]\vec{L}[/itex] would rotate in a plane that is orthogonal on vector [itex]\vec{B}[/itex].

    But, when I sketch all the forces in this example, I think this "current loop's" vector [itex]\vec{L}[/itex] would change, and become antiparalel to vector [itex]\vec{B}[/itex] in the end.
    Last edited: Jan 27, 2013
  2. jcsd
  3. Jan 27, 2013 #2


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    Substitute some macroscopic body instead of electron in your question, because the issue with the electron is more complicated since you need relativistic quantum theory (Dirac equation) for a full understanding of it. The non-relativistic treatment in general does not lead to the correct gyrofactor of around 2 for the electron-spin magnetic-momentum relationship.

    With this caveat consider a body with magnetic moment [itex]\vec{\mu}[/itex] in an homogeneous magnetic field. Then you have already written the equation of motion correctly. The only thing you need to consider now is that the magnetic moment is proportional to the angular momentum, as you have written for the example of a classical ring current (which in fact is Ampere's ingeneous model for a permanent magnet). Let's only write it in terms of a general proportionality constant
    [tex]\vec{\mu}=\gamma \vec{L}.[/tex]
    Here [itex]\gamma=q g/(2m)[/itex], with the gyromagnetic factor, which is approximately 2 for the case, where the angular momentum is given by the spin of a spin-1/2 particle.

    Now you equation of motion is
    [tex]\dot{\vec{L}}=-\gamma \vec{B} \times \vec{L}.[/tex]
    To meet your initial conditions let's put [itex]\vec{B}=B \vec{e}_z[/itex] and [tex]\vec{L}(t=0)=\vec{L}_0=L_0 \vec{e}_x.[/tex]
    Then your equations of motion read
    [tex]\begin{pmatrix} \dot{L}_x \\ \dot{L}_y \\ \dot{L}_z\end{pmatrix}=-\gamma B
    -L_y \\ L_x \\ 0
    This immediately gives
    So we need to bother only with the [itex]xy[/itex] components.

    The trick with such equations is to use complex notation. Let's define
    [tex]L_c=L_x+\mathrm{i} L_y[/tex]
    Then you have from the equations of motion
    [tex]\dot{L}_c=-\gamma B (-L_y+\mathrm{i} L_x=-\mathrm{i} \gamma B (L_x+\mathrm{i} L_y)=-\mathrm{i} \gamma B L_c.[/tex]
    This differential equation is readily solved by separation of coordinates or by an exponential ansatz:
    [tex]L_c(t)=L_{0} \exp(-\mathrm{i} \omega_L t) \quad \text{with} \quad \omega_L=\gamma B.[/tex]
    Split again in real and imaginary part you get
    [tex]L_x(t)=L_{0} \cos(\omega_L t), \quad L_y(t)=-L_{0} \sin(\omega_L t).[/tex]
    The angular momentum (and thus also the magnetic moment) thus precesses around the direction of the magnetic field with a precession frequency
    [tex]\omega_L=\gamma B,[/tex]
    the Larmor frequency.
  4. Jan 27, 2013 #3
    Thank you very much for such a detailed answer, it really helped me :). There is something I would like to ask you. Could you please try to explain precession in terms of only forces that are acting on parts of the body that is rotating in an external torque in the easiest way for you, and hopefully for me too.

    Every explaination that I saw on precession included angular momentum, but I must say after many explainations I still do not understand deep enough the basic physics underlying precession.

    Thank you in advance.
  5. Jan 27, 2013 #4


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    That's a hard question for me to answer since for me forces are not very intuitive. What we have used here is
    If you think in terms of a rigid body, the torque on the body is excerted by a pair of forces trying to rotate the body, and that's precisely the meaning of the cross product. Writing our equation in terms of differential it says
    [tex]\mathrm{d} \vec{L}=\mathrm{d} t \vec{\tau}=-\mathrm{d} t \gamma \vec{B} \times \vec{L}.[/tex]
    This equation describes exactly what we got out by solving the differential equation in a local form: The infinitesimal change of the angular momentum is given as the change of [itex]\vec{L}[/itex] by rotating it around the direction given by [itex]-\vec{B}[/itex] by the infintesimal angle [itex]\mathrm{d} \varphi=\gamma B \mathrm{d} t[/itex]. The rotation direction is given by the right-hand rule: Put the thumb of your right hand in the direction of [itex]-\vec{B}[/itex]. Then your fingers give the sense of rotation around this direction.

    This is also exactly what our solution tells us: The end of [itex]\vec{L}[/itex] describes a circle in this direction, which we have chosen as [itex]-\vec{e}_z[/itex] (because we took [itex]\vec{B}=+|\vec{B}| \vec{e}_z[/itex].

    I hope this geometrical picture helps you a bit in understanding what's going on.
  6. Jan 27, 2013 #5
    Well everything you have said is ok with me and understandable. What makes me difficulties I think I could give you in an example.

    Let's we consider a rotating wheel that is rotating with constant angular velocity [itex]\vec{ω}[/itex], and no external torque. After some time you could spin it up with your own hands, which could be described with torque [itex]\vec{τ}[/itex]. From [itex]\vec{τ}[/itex]=I[itex]\frac{d\vec{ω}}{dt}[/itex] you get how the angular velocity would change.

    The same conclusion could be made if you consider the force acting on the part of the wheel. The most further part of the wheel has the velocity [itex]\vec{v}[/itex], and it has the mass [itex]dm[/itex]. From the second Newton's law and known distance from the axis you could conclude how would wheel change in spinning.

    What I think is of importance is the centripetal acceleration of the rotational motion.

    In the end, whole rotational part of mechanics is derived from basics Newton laws for tinny parts of the whole rigid body.

    This is analogy to what I am trying to find about precession phenomenon.
    Last edited: Jan 27, 2013
  7. Jan 27, 2013 #6
    Some silly idea crossed my mind. If you are looking on a current loop whose magnetic moment (angular momentum) is orthogonal to the magnetic field, the maximal force acting on loop are on two places that are intersected by magnetic field (only one magnetic field line is drawn, the one going through radius of the loop).

    If the force acted in that moment would be manifested after the current carriers have moved for 90 degrees, I think there wouldn't be any problem if we looked at it from this (different) point of view. But this suggestion opens another problem.
  8. Jan 27, 2013 #7
    On a current loop in uniform magnetic field there is no force on it but a torque.however I am not getting what is the main problem.
  9. Jan 27, 2013 #8
    Well, I agree that there is no resulting force acting on the current loop in the magnetic field, but only torque. But that torque emerges from the forces acting on the different parts of the loop, even though they cancel each other when added at the center of the mass.

    Think of a coin that you spin on a table with your fingers. When you spin it, you actually act force to its edge with yours two fingers, and those two forces create torque why coin spins.

    Whole motion of a coin could be said without using any formula or things such as angular momentum. Those explainations I understand, but I truly don't understand the basic physics of the precession.

    If you still don't get the problem, try to think how to explain precession to some child that haven't went to school yet.
  10. Jan 28, 2013 #9
    if a body has an angular momentum about z axis and you want to rotate the body around another axis(y),then you will need a torque along the x axis.Conversely,if a body in rotation is acted upon by a torque along second axis then the body will precess around the third axis,cross product of precession angular velocity and body's angular velocity will determine the torque axis.
  11. Jan 28, 2013 #10
    probably I am not clear about what is problem for me to understand or you don't understand what is the problem.

    All the quantities of rotational motion could be described in terms of mass, velocity, momentum and force instead of moment of inertia, angular velocity, angular momentum and torque, but we use the latter ones since it is much simpler for describing.

    I would be really grateful if you could describe me a precession in terms of mass, velocity, momentum or force; without using terms of moment of inertia, angular velocity, angular momentum or torque.

    There is I think another parallel, if you consider laws of conservation. Conservation laws tell what will happen in some system, but whole machinery (all physics) is hidden within them.

    You could very easily conclude what would happen if you are holding a spinning wheel horizontally, and suddenly you lift it vertically while standing on a bigger wheel, by using only the law of conservation of angular momentum. But, I think that real interesting physics is under those laws.
    Last edited: Jan 28, 2013
  12. Jan 28, 2013 #11


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    Maybe this is what you want?

    Precession without vectors


    Can we explain precession without vectors? Qualitatively, the answer is yes. At right, below the movie, are two stills taken from it: look at the top one. The cord on the right hand side (looking at the photo) is pulling the shaft up, the weight of the shaft is pulling it down. If it were not spinning, we know that the whole apparatus would tip anticlockwise: the top of the wheel would move left and the bottom right. So how does the spinning make it precess instead of falling?



    Let's consider a small part of the rim of the wheel at the top – call it the top piece, and let's colour it red. At the instant of the top photograph, the top piece is travelling (for a very short time) horizontally, outwards from the photo, at high speed. But the combined effect of weight and the tension in the cord, as we mentioned above, makes it tend to move to the left. In fact, it does go to the left a little bit, but it also comes very rapidly out towards us. Just for this explanation, let's say that, after it has moved a quarter turn around the shaft, it will have come out towards us and be now the closest part of the wheel to us (and going downwards), but it will be displaced a tiny bit left with respect to left side of the rim in the (top) photo.

    Similarly, let's consider a small part of the rim of the wheel at the bottom, and let's colour it green. At the instant of the photo, the bottom piece is travelling horizontally, into the photo. This time, the combined effect of weight and the tension in the cord makes it tend to move to the right. So again, let's say that, after it has moved a quarter turn around the shaft, it will have gone inwards, away from us and be now the furthest part of the wheel from us (and going upwards), but it will be displaced a tiny bit right with respect to the rim in the upper still photo.

    So, after a quarter of a turn, the closest bit of the wheel will have moved slightly to the left, and the far side of the wheel will have moved a bit to the right. So we look at the bottom still – and then run the movie – and we see that that is exactly what happens. The two coloured bits of the wheel are moving as expected from their weight and the tension in the cord (i.e. the external torque), but they don't have time to move very far because of their rapid rotation. The combination of these motions, and those of all the other parts, gives us the precession we see. Note, too, that the faster the wheel turns, the less time there is for the pieces to move sideways, so the slower the precession, as given by the equation above.

  13. Jan 28, 2013 #12
    you can not use momentum for angular momentum,it is moment of momentum.While momentum can be zero,angular momentum can not be.Simply a disc spinning about it's axis.
  14. Jan 28, 2013 #13


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    I'm supposing he means if you tried describing the motion of individual particles you could get away without ever using angular quantities?

    The net momentum of disk CG might be zero yet every particle has a linear momentum.
  15. Jan 28, 2013 #14

    Well, it is absurd to say what you have. Those are not my words, at least not the meaning of them. If you have a spinning disk that is not moving, you certainly can't use its momentum to describe its angular momentum, that is really absurd to think of.

    But could you please define me its angular momentum?

    If not, tinny parts of the disk that is spinning are moving. And therefore have their momentum. With that momentum and known radius of that (and any other) tinny part of disk you can describe total momentum of the spinning disk.


    I think that is what I was looking for. I will certainly read it as soon as I can. Thank you.
  16. Jan 28, 2013 #15
    rollingsteinm, it is something that I was looking for, a different point of view on precession. But that is not all I was looking for.

    I will clearly write down what I am interested in. The simplest example is I think current loop in a magnetic field. Lets consider the moving of only tiny part of the loop current (just think of the electron in H atom, and Borh model of the atom). Could someone help me to write down all the forces that are acting on the electron while being in magnetic field, and how it leads to precession. Without using things such as angular momentum, torque, moment of inertia or angular velocity.
  17. Jan 28, 2013 #16
    Well, I found a solution to my problem. I considered H atom, and electron orbiting nucleus. After writing centripetal force (electrostatic due to the electric field of nucleus) and "magnetic" force, I have succeeded in visualizing it's precessional motion without using rotational quantities.

    One similarity came up in my mind. This procession problem reminded me of time (when I was much younger) when I tried to figure out why Earth don't fall on the Sun due to the gravitational force. For me these two problems are similar.

    Thank you all for help, I really appreciate it :)
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