Timeline for A question on the nature of the vortex number
Current License: CC BY-SA 4.0
10 events
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| Nov 19, 2019 at 22:57 | comment | added | Michael Engelhardt | That's incompatible with the boundary conditions - as long as $f$ still varies, you have $F\neq 0$. If $f$ becomes a constant, but not equal to an integer, then $A$ at the boundary isn't generated by a single-valued $U$. | |
| Nov 19, 2019 at 22:53 | comment | added | Paul | I have one last doubt. What happens if I consider $A=(f(r)/r)e_\phi$, but with $f$ not going to 1 at infinity? | |
| Nov 19, 2019 at 21:40 | vote | accept | Paul | ||
| Nov 19, 2019 at 21:40 | comment | added | Paul | Got it! Thank you again! | |
| Nov 19, 2019 at 21:24 | comment | added | Michael Engelhardt | You can of course decompose any 𝐴 into 𝑛 vortices and a homotopically trivial rest, and in that sense, you're counting precisely the vortices. But you don't have to think that way - really, it's about the winding of $U$, whatever may be happening at finite $r$. | |
| Nov 19, 2019 at 21:04 | comment | added | Michael Engelhardt | The vortices are just specific representatives of forms $A$ with nontrivial winding number - really, $A$ can do whatever it likes at finite distances, you can deform it as you please, and all that really matters is the $U$ that describes it at the boundary via $A=iU\partial U^{\dagger} $. These $U$s can be classified according to their winding number $n$ - what phase $U$ accumulates in its exponent as one walks around the boundary, which must be an integer multiple of $2\pi $. Which is just another way of saying $\pi_{1} (S^1 ) =Z$. | |
| Nov 19, 2019 at 19:55 | comment | added | Paul | Thank you so much for the explanation! I think I understood the point. I don't know if we could use thin vortices here since $A$ is a 1-form, so it's supposed to have smooth coefficients (or else they could just be considered as long as the integral is finite), anyway your point on the thick vortices convinced me. My only question is: how do we know that the forms of the form $A=d\phi f(r)/r$, with $f\to1$ at $\infty$ (and maybe such that $f$ cancels the singularity at zero), are the only objects that $n$ is counting? | |
| Nov 19, 2019 at 16:46 | history | edited | Michael Engelhardt | CC BY-SA 4.0 |
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| Nov 19, 2019 at 16:03 | history | edited | Michael Engelhardt | CC BY-SA 4.0 |
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| Nov 19, 2019 at 15:42 | history | answered | Michael Engelhardt | CC BY-SA 4.0 |