1
$\begingroup$

I’m studying Kosterlitz transition on this book: https://tinymachines.weebly.com/uploads/5/1/8/8/51885267/kardar._statistical_physics_of_fields__2007_.pdf#page173 . At page 165 it says:” The gradient expansion applies to configuration that can be continuosly deformed to the uniformly ordered state”. My question is why? I know that at low temperature there are vortex antivortex pair (that can be transformed into ordered state) and an higher temperature vortex that cannot transformed into ordered state.

$\endgroup$
  • $\begingroup$ Also in this case mit.edu/~levitov/8.334/notes/XYnotes1.pdf at page 5 he calculated the correlation function ignoring the vortex. So this calculation cannot be done with vortex. Is this linked to the fact that whit a vortex we have a request on vorticity? Probably yes $\endgroup$ – MementoMori Oct 11 '18 at 9:51
  • $\begingroup$ So we can do that calculation only in the case of a low temperature where we have a total charge =0 so there is not a request of the $\theta$ $\endgroup$ – MementoMori Oct 11 '18 at 9:54
  • $\begingroup$ At high temperature then we ignore again the vortex because we expect to have a disordered state, correct? $\endgroup$ – MementoMori Oct 11 '18 at 9:57
  • $\begingroup$ Minor comment to the post (v1): Please consider to mention explicitly author, title, etc. of links, so it is possible to reconstruct link in case of link rot. $\endgroup$ – Qmechanic Nov 9 '18 at 23:55
0
$\begingroup$

When doing gradient expansion, we are implicitly assuming $\theta(x)$ is a single valued function of $x$ and there is no discontinuity. Otherwise the gradient of $\theta$ diverges.

As the textbook says, any single-valued function can be continuously deformed to a uniformly ordered state. Here is how. Define a uniformly ordered state by $\theta_{uni}(x) = 0$ for all $x$. Then we can define a continuous deformation from $\theta_0(x)$ by

$$ \theta(x,\lambda) = (1-\lambda)\theta_0(x) - \lambda\theta_{uni}(x) = (1-\lambda)\theta_0(x) $$

where $\lambda = 0$ correspond to the initial state and $\lambda=1$ corresponds to the final state. As long as $\theta_0(x)$ is smooth and continuous, this transformation is also smooth and continuous.

Vortex configurations cannot be described as a single valued function. For any single valued function $f(x)$ we have

$$ \int_C \nabla f(x) dx = 0 $$

for any closed loop C. However, for a loop C containing a vortex we have

$$ \int_C \nabla \theta(x) dx = 2\pi n $$

where n is defined to be the vorticity of the vortex (+1 for vortex, -1 for anti vortex and so on). Thus, $\theta(x)$ cannot be a single valued function, and we need to put extra care in order to take vortices into account.

That explains why gradient approximation fails. The more interesting questions is why gradient approximation succeeds in some cases. As you have pointed out, at low temperature there are very few vortices present, so we can get away with ignoring their presence and using gradient expansion.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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