Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I am reading on magnetic monopoles from a variety of sources, eg. the Jeff Harvey lectures.. It talks about something called the winding $N$, which is used to calculate the magnetic flux. I searched the internet but am not being able to understand the calculation done in this particular case.

$g=-\frac{1}{8}\int_{S^2_\infty} Tr([d\hat{\Phi},d\hat{\Phi}],\hat{\Phi})$

Then the author says that

Now $\Phi$ restricts to a map $\Phi : S_\infty^2 → S^2$ , where the target is the unit sphere in $su(2)$. This map has some degree $N$ , and it is easy to verify that the right-hand side of the above equation is $−2\pi$ times this. Therefore $g = −2\pi N$ .

What is $N$, the winding number also called as the degree on the map? By what i have learnt, it is the number of times you wind an object unto the another, then shouldn't the integral be $N*4\pi$, as $4\pi$ is the surface area of $S^2$.

share|improve this question
    
I suppose that you are working with the SU(2)-Yang-Mills-Higgs theory, right? what is $\hat{\Phi}$? Is this scalar field the time component of a static gauge potential in Minkowski, after making dimensional reduction? –  c.p. Jun 30 '12 at 1:17
    
@JorgeCampos, I am working with SU(2) yang-mills. $\hat{\Phi}$, is the angular part of the higgs field. The total higgs field is given by $\Phi=h\hat{\Phi}$, where $h$, is a function of the radial distance $r$, where norm $\hat{\Phi}=1$. –  ramanujan_dirac Jun 30 '12 at 5:47
add comment

2 Answers

up vote 3 down vote accepted

I didn't find the equation and the argument you quoted in that paper. But, yes, it is the Brouwer degree, deg$(\hat{\Phi})$, which equals the monopole number $$ N\equiv\frac{1}{4\pi}\int_{\mathbb{R}^3} \mathrm{Tr}(F_A\wedge D_A(\Phi))=\frac{1}{4\pi}\int_{\mathbb{R}^3} d(\mathrm{Tr}(\Phi)F_A) =\frac{1}{4\pi}\int_{S^2_\infty} \mathrm{Tr}(\Phi F_A)$$ $$ =\frac{1}{4\pi}\int_{S^2_\infty} \mathrm{Tr}(\hat{\Phi} F_A) $$

where the one has used Bianchi identity, Stokes' theorem, to obtain the first two equalities, and Jaffe & Taubes show in their book that one can replace $\Phi$ by $\hat{\Phi}$. Now this coincides with the brower degree, for which there is an explicit formula:

$$N=\mathrm{deg}(\hat{\Phi})=-\frac{1}{4\pi}\int_{S^2_\infty} \mathrm{Tr(\hat{\Phi} d\hat{\Phi}\wedge d\hat{\Phi}})\in \mathbb{Z}=\pi_2(S^2)=[S^2,S^2]$$ (what you wrote.) This is physically understood as an infinite wall potential, separating the monopole sectors corresponding to different integers. Now, to actually answer your question, you can compute this integral for the t'Hooft-Polyakov monopole solution, for which $$ \hat{\phi}=(\sin(\theta)\cos\phi,\sin\theta \sin\phi,\cos\theta)_i\cdot \sigma^i, $$ and you will find $$N=-\frac{1}{4\pi}\int_{S^2_\infty} \mathrm{Tr(\hat{\Phi} d\hat{\Phi}\wedge d\hat{\Phi}})=+\frac{1}{4\pi}\int_{[0,2\pi]} \int_{[0,\pi]}\sin\theta d\theta\wedge d\phi=1.$$

share|improve this answer
    
I asked this question exactly because I don't know what is brower degree, I coudn't understand it from the internet due to reference to algebraic topology which i have not studies. Please could you explain what this is, and how did you arrive at the explicit formula of the brower degree. –  ramanujan_dirac Jul 2 '12 at 8:09
    
I'll answer that soon; I'll reedit... –  c.p. Jul 2 '12 at 19:28
add comment

N is equal to the number of points in the $S_2$ sphere at infinity mapped to the same point of the $S_2$ Higgs vacuum manifold.The integral is a topological invariant depending only on this number and not on the details of the map. In the following, I'll describe to you a family of these maps:

One way to perform the integral is to use the stereographic projection coordinate:

$ z = tan(\frac{\theta}{2}) e^{i\phi}$

In this coordinate system,a map of winding number $N$ will look as:

$ z \rightarrow Z= z^N$

The surface element of the sphere in these coordinates is:

$ d \mu = \frac{dzd\bar{z}}{1+z \bar{z}}$

Remark: In these coordinates, the Higgs components are given by:

$\Phi_x = 2\frac{Re(Z)}{1+Z\bar{Z}}$

$\Phi_y = 2\frac{Im(Z)}{1+Z\bar{Z}}$

$\Phi_z = \frac{1-Z\bar{Z}}{1+Z\bar{Z}}$

Using these coordinates,it is not difficult to see that the integral (1.43) value is N.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.