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The skyrmion number is defined as $$n=\frac{1}{4\pi}\int\mathbf{M}\cdot\left(\frac{\partial\mathbf{M}}{\partial x}\times\frac{\partial\mathbf{M}}{\partial y}\right)dxdy$$ where $n$ is the topological index, $\mathbf {M}$ is the unit vector in the direction of the local magnetization within the magnetic thin, ultra-thin or bulk film, and the integral is taken over a two dimensional space.

It is known that $\mathbf{r}=\left(r\cos\alpha,r\sin\alpha\right)$ and $\mathbf{m}=\left(\cos\phi \sin\theta,\sin\phi \sin\theta,\cos\theta\right)$. In skyrmion configurations the spatial dependence of the magnetisation can be simplified by setting the perpendicular magnetic variable independent of the in-plane angle ($ \theta \left(r\right)$) and the in-plane magnetic variable independent of the radius ($ \phi \left(\alpha\right)$). Then the skyrmion number reads: $$n=\frac{1}{4\pi}\int_0^\infty dr\int_0^{2\pi}d\alpha\ \frac{d\theta\left(r\right)}{dr}\frac{d\phi\left(\alpha\right)}{d\alpha}\sin\theta\left(r\right)=\frac{1}{4\pi}\ [\cos\theta\left(r\right)]_{\theta\left(r=0\right)}^{\theta\left(r=\infty\right)}[\phi\left(\alpha\right)]_{\theta\left(\alpha=0\right)}^{\theta\left(\alpha=2\pi\right)}$$

My question is: is $\frac{\partial\mathbf{M}}{\partial x}\times \frac{\partial\mathbf{M}}{\partial y}$ a curl product and what is the output of this term? How to reach to the final equation then?

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It is not a curl. This can be seen by expressing the curl in vector components. $$\nabla \times \mathbf M=\begin{pmatrix} \partial_yM_z-\partial_z M_y\\ \partial_zM_x-\partial_x M_z\\ \partial_xM_y-\partial_y M_x \end{pmatrix}$$ Here $\partial_x$ denotes the partial derivative with respect to $x$. The quantity $\partial_x\mathbf M$ is a vector just like $\mathbf M$. It has components $$\partial_x \mathbf M=\begin{pmatrix} \partial_xM_x\\ \partial_xM_y\\ \partial_xM_z \end{pmatrix}$$ Calculating the quantity $\partial_x\mathbf M\times\partial_y\mathbf M$ is then just a matter of applying the cross product. $$\partial_x\mathbf M\times\partial_y\mathbf M=\begin{pmatrix} \partial_xM_y\partial_yM_z-\partial_xM_z\partial_yM_y\\ \partial_xM_z\partial_yM_x-\partial_xM_x\partial_yM_z\\ \partial_xM_x\partial_yM_y-\partial_xM_y\partial_yM_x \end{pmatrix}$$ This is a daunting expression and you probably won't get a lot of intuition from looking at the components. What you can say about it is that $\mathbf A\cdot(\mathbf B\times \mathbf C)$ forms the vector triple product. This gives the volume spanned by (the parallelepiped of) $\mathbf A,\mathbf B$ and $\mathbf C$. So the quantity you're integrating is the volume spanned by $\mathbf M,\partial_x \mathbf M$ and $\partial_y \mathbf M$.

To calculate the integral in your last equation is just a matter of plugging everything in in my last expression for $\partial_x\mathbf M\times\partial_y\mathbf M$. This is tedious but should be doable.


EDIT I'll add some more info to make the calculation less tedious. The partial derivatives can be expanded using the chain rule $\partial_x=\frac{\partial r}{\partial x}\partial_r+\frac{\partial \alpha}{\partial x}\partial_\alpha$. These can be calculated to be $$\partial_x=\cos\alpha\partial_r-\frac{\sin\alpha}r\partial_\alpha\\ \partial_y=\sin\alpha\partial_r+\frac{\cos\alpha}r\partial_\alpha$$ Next note that $\partial_r\mathbf M=\frac{d\theta}{dr}\partial_\theta\mathbf M$ and $\partial_\alpha\mathbf M=\frac{d\phi}{d\alpha}\partial_\phi\mathbf M$. If we name these partial derivative vectors $\mathbf e_\theta=\partial_\theta\mathbf M$ and $\mathbf e_\phi=\partial_\phi\mathbf M$ then the cross product becomes $$\partial_x\mathbf M\times\partial_y\mathbf M=\left(\cos\alpha\frac{d\theta}{dr}\mathbf e_\theta-\frac{\sin\alpha}r\frac{d\phi}{d\alpha}\mathbf e_\phi\right)\times\left(\sin\alpha\frac{d\theta}{dr}\mathbf e_\theta + \frac{\cos\alpha}r\frac{d\phi}{d\alpha}\mathbf e_\phi\right)$$ Finally you can calculate that $\mathbf e_\theta\times \mathbf e_\phi=\sin\theta \,\mathbf M$ and you should be able to do this calculation without explicitly calculating all the components.

And yes you should add the factor $r$ when you switch to polar coordinates like you mentioned in your comment.

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  • $\begingroup$ Thanks. The most difficult part is to understand for example, $\partial_{x}\mathbf{M}_y$, namely, $\frac{\partial [sin\phi\left(\alpha\right) sin\theta\left(r\right)]}{\partial x}$. I am really confused by this term. Is this term can be calculated as $sin\phi\left(\alpha\right)\ \frac{\partial\ sin[\theta\left(r\right)]}{\partial\ [\theta\left(r\right)]}\ \frac{\partial\ [\theta\left(r\right)]}{\partial\ x}$? Whether yes or no, I dont know what to do next with this term. $\endgroup$
    – Eric Z
    Oct 15, 2020 at 10:40
  • $\begingroup$ Also note that in the last equation in my question, the integration transformed from Cartesian coordinate to polar coordinate, seems they used this transform: $\int\int_D\ f\left(x,y\right)\ dxdy=\int\int_{D^\prime}\ f\left(r\ cos\alpha,r sin\alpha\right)\ rdrd\alpha$ $\endgroup$
    – Eric Z
    Oct 15, 2020 at 11:07
  • $\begingroup$ Brilliant! Thanks a lot, now I see how it works. $\endgroup$
    – Eric Z
    Oct 16, 2020 at 2:30

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