Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,851 answers , 20,616 comments
1,470 users with positive rep
501 active unimported users
More ...

Axion strings and spontaneously broken symmetry

+ 2 like - 0 dislike
64 views

I have two question about axion strings:

  1. Why their appearance is connected with spontaneously broken symmetry? How to demonstrate that?
  2. Why they are stable topological configurations (look to the "Addition" text below)?
  3. Why when we choose string located along $z-$axis and set solution for Peccei-Quinn scalar field $\varphi $ in a string-like form $\varphi = ve^{i\theta}$, where $v$ is VEV of $\varphi$, $\theta$ is axion, then we have $$ [\partial_{x}, \partial_{y}]\theta = 2\pi \delta (x) \delta (y)? $$ How to demonstrate that?

Addition

Let's assume axion "bare" lagrangian $$ \tag 1 L = \frac{1}{2}|\partial_{\mu}\varphi |^{2} - \frac{\lambda}{4} (|\varphi |^{2} - v^{2})^{2} $$ One of solution of corresponding e.o.m. is axion string - stable topological configuration. If string is located along z-axis and if it is static, then corresponding solution is simply ($\rho$ is polar radius, $\varphi$ corresponds to polar angle and, in fact, to axion) $$ \varphi (x) = f(\rho ) e^{i n \varphi}, \quad f(0) = 0, \quad f(\infty ) = v, $$ where $n$ is winding number.

Statement that configurations with different winding numbers are stable means that they are separated by infinite potential barriers. But I don't understand how $(1)$ creates barriers for different $n$.

Addition 2 Thank to the Meng Cheng comment. The first and the third questions are closed. Explicit proof of the statement of the third question: $$ [\partial_{x}, \partial_{y}]e^{iarctg\left[\frac{y}{x}\right]} = i\partial_{x}\left[ \frac{x}{x^{2} + y^{2} + a^{2}}\right]_{\lim a \to 0} + i\partial_{y}\left[ \frac{y}{x^{2} + y^{2} + a^{2}}\right]_{\lim a \to 0} = $$ $$ =i\left[\frac{2a^{2}}{(x^{2} + y^{2} + a^{2})^{2}}\right]_{\lim a \to 0} = 2 \pi i \left[\frac{a^{2}}{\pi}\frac{1}{(r^{2} + a^{2})} \right]_{\lim a =0} = 2 \pi i \delta_{a}(\mathbf r) $$

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user Name YYY
asked Jun 29, 2015 in Theoretical Physics by NAME_XXX (1,010 points) [ no revision ]
Most voted comments show all comments
What's an "axion string"? Googling gives me only "axions in string theory" stuff.

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user ACuriousMind
@ACuriousMind : string is topologically stable solution of equation of motion for case of one space dimension in a case of spontaneously broken continuous symmetry. As for axion string, some outlook is here: arxiv.org/pdf/hep-ph/9807374v2.pdf . However, I don't understand explicitly how spontaneously broken symmetry causes existence of string as topologically stable solution.

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user Name YYY
Sounds like what you are talking about is just vortex lines in a superfluid.

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user Meng Cheng
@MengCheng : maybe you're right, and this case is very similar to superconductors. But I still have problem with understanding its topological nature.

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user Name YYY
@ACuriousMind : section II in an article.

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user Name YYY
Most recent comments show all comments
@MengCheng : thank you! And the last question, if you please: we add boundary conditions for $\varphi = f(r)e^{i\theta}$ such that $f(0) = 0, f(\infty ) = v$ to avoid bad behaviour of $\varphi$ at $r = 0$ and to satisfy the minimum condition of potential?

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user Name YYY
@NameYYY That is exactly what happens for vortices: the amplitude of the order parameter is suppressed at the core (so there is an energy associated), and far away from the core the amplitude is just $v$. I think in my previous comments I was abusing $v$ for $f$, so all I mean by $v=0$ is that the amplitude has to vanish. Sorry for the confusion.

This post imported from StackExchange Physics at 2015-06-30 15:36 (UTC), posted by SE-user Meng Cheng

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverf$\varnothing$ow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...