# Vortices - differentiation of the field in abelian Higgs

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1. Apr 10, 2013

### Lapidus

In Zee's "Nutshell QFT" (chapter V.7) or Shifman's book "Avanced Topics in QFT" (section 10) when they talk about vortices, they claim:

if $\phi(r,\theta)$ goes to $\nu$exp( i$\theta$) as r goes to infinity

then $\partial_{i}$$\phi$ becomes $\nu$(1/r)

I do not see how? Is the phase equal to 1/r if r goes to r???

thank you

Last edited: Apr 10, 2013
2. Apr 10, 2013

### fzero

Just express the gradient in polar coordinates:

$$\nabla f =( \partial_r f )\hat{e}_r + \frac{1}{r} (\partial_\theta f )\hat{e}_\theta.$$

3. Apr 10, 2013

### Lapidus

thanks, fzero!

4. Apr 11, 2013

### Lapidus

Wait, still not clear...

Especially, what Shifman does or Rubakov in "Classical Theory of Gauge Theories" on page 159, of which I attached a pdf copy down below. Where does the minus sign come from? Are the partial derivatives in polar or x-y coordinates? I know that the 1/r comes from the gradient somehow, as fzero pointed out, but why the Levi-Cita symbol?

I'm sorry, scine I know that it must be super simple and "obvious", but I can't see it.

Very grateful in advance for any extra hints!

#### Attached Files:

• ###### Rubakov.pdf
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5. Apr 11, 2013

### fzero

You should work out $\partial_i \theta$ using the usual relation from polar coordinates that $\theta = \tan^{-1}(y/x)$. The relative minus sign between the $x$ and $y$ components is related to the appearance of $x/y$ in this formula.

6. Apr 12, 2013

### Lapidus

Ok, after all not so super obvious:

Given z = arctan(y/x) = arctan(yx⁻¹):

∂z/∂x = [1/(1 + (y/x)²)] * (∂/∂x)(yx⁻¹)
= [1/(1 + (y/x)²)] * (-yx⁻²)
= [1/(1 + y²/x²)] * (-y/x²)
= -y / (x² + y²).

∂z/∂y = [1/(1 + (y/x)²)] * (∂/∂y)(yx⁻¹)
= [1/(1 + y²/x²)] * (x⁻¹)
= [1/(1 + y²/x²)] * (1/x)
= 1/(x + y²/x)
= x / (x² + y²).

Arghhh, those textbooks that throw just equations at you without explaining anything.