Topology of Aharonov Bohm Effect - Lewis Ryder's QFT book.

[maverick280857]

Hi,

I am reading through Section 3.4 of Lewis Ryder's QFT book, where he makes the statement,

The configuration space of the Bohm-Aharonov experiment is the plane $\mathbb{R}^2$ with a hole in, and this is topologically, the direct product of the line $\mathbb{R}^1$ and the circle $S^1$: $\mathbb{R}^1 \times \mathbb{S}^1$.

This makes some sense intuitively, but can someone please explain this direct product equivalence to me as I do not have a firm background in topology (unfortunately, I need some of it for a project so I am reading it as I go along -- a bad idea, but I have no time :().

Specifically, how does this direct product come about?

Thanks in advance!

 

[Physics Monkey]

Think about the plane in polar coordinates: $ds^2 = dr^2 + r^2 d\theta^2$. If you remove the point r=0 then you can cover the the remaining space with the coordinates $-\infty < u < \infty$ ($r = e^u$) and $\theta$. The metric is then $ds^2 = e^{2u} (du^2 + d\theta^2 )$. The appearance of this overall prefactor in the metric means that the plane with a point removed is not only topologically equivalent but also conformally equivalent to a cylinder $\mathbb{R}_u \times S^1_\theta$. In plain terms, by locally deforming the punctured plane you can endow it with the metric of a flat cylinder which should make the topological equivalence clear.

 

[maverick280857]

Thanks Physics Monkey, that small substitution step helped quite a bit. Can you suggest some good references where I could learn more about these things without getting too caught up in formalism (which I have appreciation for, but just no time ). Ryder makes all these statements which I am sure make sense if one thinks about them the way you suggest, but that does require some background in topology which I lack (I am concurrently reading topology but I am way behind all this.)

 

[tom.stoer]

Try Nakahara: Geometry, Topology and Physics