Derivation of LLG equation in polar coordinates

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SUMMARY

The discussion centers on deriving the Landau-Lifshitz-Gilbert (LLG) equation in polar coordinates, specifically addressing the torque contribution from uniaxial anisotropy. The equation provided, \(\frac{\Gamma}{l_m K} = (2 \sin\theta \cos\theta)[\sin\phi e_x - \cos\phi e_y]\), is crucial for this derivation. Participants note the need to transition from Cartesian coordinates to polar angles, emphasizing the challenge of obtaining a clean expression for the angles \(\theta\) and \(\phi\) as shown in the final result. The conversation highlights the importance of setting parameters such as \(h_p=h=h_s=0\) for clarity in the derivation process.

PREREQUISITES
  • Understanding of the Landau-Lifshitz-Gilbert (LLG) equation
  • Familiarity with uniaxial anisotropy in magnetic systems
  • Knowledge of polar and Cartesian coordinate transformations
  • Basic differential equations involving angular variables
NEXT STEPS
  • Study the derivation of the LLG equation in polar coordinates
  • Review the mathematical treatment of uniaxial anisotropy in magnetic materials
  • Examine the implications of setting \(h_p=h=h_s=0\) in LLG equations
  • Explore related literature on angular dynamics in magnetic systems, particularly the referenced papers
USEFUL FOR

Physicists, researchers in magnetism, and graduate students studying magnetic dynamics and the LLG equation will benefit from this discussion.

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

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

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

\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)
where n_m= [r,\theta,\phi]. Since r is unity for magnetization a differential equation in the two angles should be possible which should have the form
\begin{bmatrix}<br /> \theta \ &#039; \\<br /> \phi \ &#039; \\<br /> \end{bmatrix}<br /> = \begin{bmatrix}<br /> \theta \\<br /> \phi \\<br /> \end{bmatrix}<br />

Now the result of this derivation is already given as

\begin{bmatrix}<br /> \theta \ &#039; \\<br /> \phi \ &#039; \\<br /> \end{bmatrix}<br /> = \begin{bmatrix}<br /> \alpha \sin \theta \cos \theta \\<br /> \cos \theta \\<br /> \end{bmatrix}<br />

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
 
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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 h_p=h=h_s=0. Makes it easier to relate your equations to those in the paper.
 
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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