Derivation of LLG equation in polar coordinates

  1. The torque contribution due to the uniaxial anisotropy is given by the equation below

    [tex] \frac{\Gamma}{l_m K} = (2 \sin\theta \cos\theta)[\sin\phi e_x - \cos\phi e_y] (3)[/tex]

    This contribution can be taken in the LLG equation to derive the LLG equation in polar coordinates

    [tex] \frac{\partial n_m}{\partial t} + ( n_m \times \frac{\partial n_m}{\partial t})=\frac{1}{2}\Omega_K \frac{\Gamma}{l_m K} (9) [/tex]
    where [itex]n_m= [r,\theta,\phi][/itex]. Since r is unity for magnetization a differential equation in the two angles should be possible which should have the form
    [tex] \begin{bmatrix}
    \theta \ ' \\
    \phi \ ' \\
    \end{bmatrix}
    = \begin{bmatrix}
    \theta \\
    \phi \\
    \end{bmatrix}
    [/tex]

    Now the result of this derivation is already given as

    [tex] \begin{bmatrix}
    \theta \ ' \\
    \phi \ ' \\
    \end{bmatrix}
    = \begin{bmatrix}
    \alpha \sin \theta \cos \theta \\
    \cos \theta \\
    \end{bmatrix}
    [/tex]

    I'm having a hard time deriving this result from (3) using (9). Could anyone help me with this?
    here is the link of this paper
    http://journals.aps.org/prb/abstract/10.1103/PhysRevB.62.570
     
  2. jcsd
  3. Can you be more specific about where you are having a hard time? Is it the change of coordinate system?

    Also, when asking a question like this, it's quite reasonable to mention that you're setting [itex]h_p=h=h_s=0[/itex]. Makes it easier to relate your equations to those in the paper.
     
  4. Thanks for the reply. You are right this equation only caters for uniaxial anisotropy. the equation has to be derived in polar angles while (3) is in cartesian. I think going from (3) to polar is not hard, but how to get a clean expression like the final one.
    I'm not sure how to proceed for the derivation at all.
     
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