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Kinetic Energy in Classical treatment of Larmor Precession

  1. May 21, 2012 #1
    This question pertains to the classical treatment of Larmor precession.I don't know whether to put it in the Classical Physics forum or this so I am putting it here.

    In the treatment we assume that the potential energy of the dipole-magnetic field system remains constant because there is no way for the electron to lose that potential energy.But I was wondering that there is some kinetic energy associated with the precession. Where does that come from?

    Is it because initially before the application of the field the potential energy of the electron was zero and after the application it reduces to -μ.B.Is that the source of the kinetic energy considering that the magnetic field can not possibly do any work on the system.

    Considering this I calculated the kinetic energy of the precession.It came out to be +0.5μ.B.Now isn't this a problem.The atom can not possibly lose energy and hence there is no way for it to dissipate that energy term.

    Am I missing something basic?I am not sure whether I have calculated the thing right.I would appreciate if someone can look into this.

    (Since I am just beginning this topic I am familiar with only the classical treatment so will appreciate if the answer is in those terms)
  2. jcsd
  3. May 21, 2012 #2


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    I am somewhat confused regarding the treatment of Larmor Precession in QM, but in the classical terms you are using, I have a couple of comments (not sure that they are right and would be pleased to be corrected).

    Regarding the source of the kinetic energy, just like a gyroscope in a gravitational field, the electron will have tried to move a bit (tilts - ie. falls slightly in the magnetic field), with the application of the field.

    Regarding where the kinetic energy goes, I too cannot see what happens to stop the precession. I do suspect that only the free and/or unpaired electrons would be subject to precession. My conclussion is that free and/or unpaired electrons must be merrily precessing along virtually all the time.
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