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...
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...