# Derivation of LLG equation in polar coordinates

1. Aug 28, 2014

### apervaiz

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
$$\begin{bmatrix} \theta \ ' \\ \phi \ ' \\ \end{bmatrix} = \begin{bmatrix} \theta \\ \phi \\ \end{bmatrix}$$

Now the result of this derivation is already given as

$$\begin{bmatrix} \theta \ ' \\ \phi \ ' \\ \end{bmatrix} = \begin{bmatrix} \alpha \sin \theta \cos \theta \\ \cos \theta \\ \end{bmatrix}$$

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. Aug 31, 2014

### Hypersphere

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.

3. Aug 31, 2014

### apervaiz

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.

4. Sep 8, 2014