Monday 26 October 2020

quantum mechanics - Basic question on the Aharonov-Bohm effect


I have a very basic question on the Aharonov-Bohm effect.


Why is the curve integral $\oint_\Gamma {A}\cdot d{r}$ non-zero ? $\Gamma$ is the "difference" of both paths $P_1$ and $P_2$. If the magnetic field is limited to the interior of the solenoid $\operatorname{curl} {A}=0$ along the integral path $\Gamma$, so I can conclude that I can write ${A}=\nabla f$. A closed curve integral of a gradient function is zero.


I guess it is related with a possible singularity of $A$ in the very center of the solenoid.


Nevertheless if I travel around a point-like source of a gravitational field and compute the integral $\oint_\Gamma {F}\cdot d{r}$ where $F=-\nabla V(r)$ the closed curve integral over an conservative force field is certainly zero, whereas $V(r)$ even has a singularity and $F$ consequently too. I would be very grateful for an explanation.



Answer




From $$\operatorname{curl} A = 0 \tag{1} $$ in a region $U$, you can in general not conclude that $$A = \nabla f \tag{2}$$ for some function $f$ defined on all of $U$. Indeed this is related to the singularity, which removes a line through the origin. The degree to which (1) fails to imply (2) depends on the topology of $U$, more specifically it's de Rham cohomology. (1) implies (2) iff the de Rham cohomology is trivial. For $U = \mathbb{R} \times (\mathbb{R^2} \setminus \{ 0 \} )$, it turns out that the cohomology is not trivial, and hence your integral need not vanish.


To show that (1) does not imply (2) for 3-space with a line removed, we can assume that the line is the $z$-axis and, consider $$A = \begin{cases} \nabla \operatorname{arctan2} (y,x) & (x,y) \neq (0,-1) \\ (0, -1, 0 ) & (x,y) = (0,-1) \end{cases}$$ where also $(x,y) \neq (0,0)$. $A$ is smooth, and $\operatorname{curl} A = 0$, but you cannot find a function $f$, continuous away from the $z$-axis, such that $\nabla f = A$ everywhere.


It less trivial (or rather, one first needs to construct some sledgehammers, and then it's a 2-line proof) to show that for 3-space with a point (1) does indeed imply (2).




Update: If you want to understand why a point can be removed but not a line, one can think of it as needing to make a hole that is big enough. Suppose we have two curves $\gamma_1$ and $\gamma_2$ and $\int_{\gamma_i} A$, where $\operatorname{curl} A = 0$. If $\gamma_1$ can be deformed to $\gamma_2$ one can apply Stokes's theorem to argue that the integrals are equal. Suppose that the space is such that any curve can be deformed into any other curve. Then a standard prescription exists to find $f$: pick any point $x_0$ and let $$f(x) = \int_\gamma A $$ where $\gamma$ is any curve connecting $x_0$ and $x$. By the assumption the integral doesn't depend on the particular $\gamma$ chosen so $f$ is unambiguously defined. To make the argument fail you have to make a "hole" big enough that curves can't be deformed into each other. Removing a point isn't enough in three dimensions, but a line is: make a loop around the line singularity, you can't get it to not encircle the line without dragging it across the singularity. (In 2 dimensions a point is enough, because you can consider the projection of this example onto a plane.)


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...