Given a vector potential describing an infinitely thin line of flux,

$$\vec{A} = \frac{\Phi}{2\pi r} \vec{e}_\varphi,$$

How can I calculate the curl so that the magnetic field is given by

$$\vec{B} = \delta(x) \delta(y) \Phi \vec{e}_z,$$

or similar expression? The expression for the curl (in cylindrical coordinates) comes down to this:

$$\nabla \times \vec{A} = \vec{e}_z \frac{1}{r} \frac{\partial}{\partial r} (r A_\varphi).$$

I am aware there is a singularity at $r=0$ which will probably give me the answer. I just don't see how. I'm asking here, because well, a mathematician might get a fit caused by my liberal use of Dirac delta's above. Any help is appreciated.

  • 1
    $\begingroup$ Use the 2-dimensional Green's function for the Laplacian to find an alternative representation of the 2-dimensional delta function. $\endgroup$ – Ryan Unger Feb 9 '15 at 16:49

The explanation of my comment: Recall $$\Delta\log r=2\pi\delta^2(x,y)$$ On the right we have $$\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial\log r}{\partial r}\right)$$ Your potential can be written as $$\vec A=\frac{\Phi}{2\pi}\frac{\partial\log r}{\partial r}\vec e_\varphi$$ The curl is then $$\vec\nabla\times\vec A=\frac{\Phi}{2\pi}\frac{1}{r}\left(r\frac{\partial\log r}{\partial r}\right)\vec e_z=\frac{\Phi}{2\pi}\Delta\log r\,\vec e_z=\delta^2(x,y)\Phi\vec e_z$$ As was to be shown.

  • $\begingroup$ Right, and where does the naive approach (i.e. filling in the curl directly) actually go wrong? I'm still unable to pinpoint exactly which step is illegal/wrong and how to perform that step generically in a mathematically correct way. Thanks. $\endgroup$ – rubenvb Feb 10 '15 at 15:21
  • $\begingroup$ @rubenvb: Filling in the curl directly? $\endgroup$ – Ryan Unger Feb 10 '15 at 16:37
  • $\begingroup$ $\nabla \times \boldsymbol A = \frac{1}{r} \partial_r r \frac{\Phi}{2\pi r} = \frac{\Phi}{2\pi r} \partial_r 1 = 0$. Or something like that. $\endgroup$ – rubenvb Feb 10 '15 at 19:40
  • $\begingroup$ Of course that is only the $\boldsymbol e_z$ part, the rest is identically zero. $\endgroup$ – rubenvb Feb 10 '15 at 19:47
  • $\begingroup$ @rubenvb: The fraction $\frac{1}{r}\partial_r 1$ is not well defined at $r=0$. It is the "peak" of the delta function, which is clear from the identity $\Delta\log r=2\pi\delta^2(x,y)$. $\endgroup$ – Ryan Unger Feb 10 '15 at 20:36

You have to take into account that your vector potential is defined only for $r >0$. For $r=0$, the vector potential actually vanishes, $\vec{A}(0) = \vec{0}$. You can show this by considering a thick solenoid and letting its width go to zero. Therefore, the correct expression for the vector potential of a thin solenoid is $$\vec{A} = \theta(r) \frac{\Phi}{2\pi r} \vec{e}_\varphi.$$ Here, $\theta(\cdot)$ is the Heaviside step function. This will give you a correct magnetic field which is proportional to a delta function.


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