# vortex anti-vortex configuration to ground state [closed]

i'm studying kosterlitz transition. I'm reading this:

Now at page 5 it says :

"The right panel shows a vortex anti-vortex configuration which can be smoothly transformed to the ground state " where the ground state is the state where all the spins are aligned.

Now i'm trying to figure out this thing.

Pay attention in my picture the vectors represent the velocities, and not the phase $$\theta$$ as in figure 3 at page 5, like in the picture below

The link between this image and the figure 3 at page 6 is that $$v \propto \nabla \theta$$

## closed as unclear what you're asking by ZeroTheHero, Kyle Kanos, glS, Aaron Stevens, AccidentalFourierTransformOct 3 '18 at 2:56

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I think it may be worth describing this in a mathematical way, rather than trying to give sketches, although the animation referred to in the previous answer, and also in my answer to your previous question, helps give a good picture. You can find plenty of discussion of the Kosterlitz-Thouless transition online, for instance here.

The angular field around a $$+1$$ defect located at position $$(-a,0)$$ has the form $$\theta^+(x,y) = \tan^{-1}\left(\frac{y}{x+a}\right)$$ and around a $$-1$$ defect at position $$(+a,0)$$ $$\theta^-(x,y) = -\tan^{-1}\left(\frac{y}{x-a}\right)$$ where $$\tan^{-1}$$ is the arctangent function. If you combine these two (simply add them) you get $$\theta^{\pm}(x,y) = \tan^{-1}\left(\frac{2ay}{a^2-x^2-y^2}\right)$$ for the field around a pair of opposite defects separated by $$2a$$ on the $$x$$ axis. This form is given here and you can derive it yourself using an identity for the difference between two arctangent functions.

Now you can plot the relevant fields in any form you like, either as the gradient $$\nabla\theta^{\pm}(x,y)$$ or as the direction field $$(\cos\theta^{\pm}(x,y),\sin\theta^{\pm}(x,y))$$. If you vary the separation $$2a$$, allowing $$a\rightarrow 0$$ you should be able to convince yourself that the separated $$+1$$ and $$-1$$ defects can be smoothly converted back to the ground state, when all the angles are the same.

There is a practical issue here (the ambiguity of the $$\tan^{-1}$$ function with respect to which quadrant the result is in) which you may need to tackle. One possible approach is to do the calculation numerically on a grid, using the above formula, but employing a version of "arctan" which allows you to supply the numerator and denominator separately. Many programming languages provide this function, for exactly this reason. Also, as usual, any constant may be added uniformly to the angle field without changing its validity.

Anyway, I hope this will help you sort out what is going on.

• Thank you very much. So is it not possible to figure out this thing with simple sketches like mine ? – MementoMori Sep 28 '18 at 14:06
• Oh no, I didn't mean to imply that you can't do it like that! I just wanted to make clear that there is a systematic way of doing it based on known formulae, and you can generate the necessary plots accurately, using a plotting package or program, if you want to. – user197851 Sep 28 '18 at 14:17
• Ok thanks. Your explanation is very clear. I was trying to figure out with sketches because with a single vortex is very easy to understand that there isn't a continuous rotations of the spins that transform them in a ground state. Instead, in the case of vortex anti-vortex is more difficult. Probably i should consider more vectors in my sketches. – MementoMori Sep 28 '18 at 14:25
• Yes, I think this would help ..... but this is where drawing the configurations by hand gets more tiresome! And it is now clear what your source means by "continuous rotation" (footnote on p5). They do not mean rotating all the spins by the same angle. They mean rotating the spins by an amount that depends in a continuous way on the position in $(x,y)$ coordinates. That's a little harder to express, if you are sketching by hand, but it's clear from the mathematical expressions (assuming you avoid the exact points where the defects are located). – user197851 Sep 28 '18 at 14:34

Say $$d$$ is the distance vortex-antivortex. When $$d\rightarrow 0$$ the vortex and antivortex annihilate each other. Here's a gif I found on google which makes it clear.

When the vortex and antivortex are at the same position, you get back the ground state of the system (a ferromagnet here if the arrows represent spins).