The torque contribution due to the uniaxial anisotropy is given by the equation below(adsbygoogle = window.adsbygoogle || []).push({});

[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

**Physics Forums - The Fusion of Science and Community**

# Derivation of LLG equation in polar coordinates

Have something to add?

- Similar discussions for: Derivation of LLG equation in polar coordinates

Loading...

**Physics Forums - The Fusion of Science and Community**