I am quite confused about this.

Imagine we put a perfectly diamagnetic sphere in homogenuous magnetic field $$ \vec{B} = B_0 \, \vec{e}_z $$

You can verify yourself that the solution outside the ball is given by $$ \vec{H} = \frac{B_0}{\mu_0} \left( - \frac{3}{2} \frac{x z}{r^5}, - \frac{3}{2} \frac{y z}{r^5}, 1 + \frac{R^3}{2} \frac{r^2 - 3 z^2}{r^5} \right) $$ (yes, the ball distorts magnetic field in its vicinity)

Now inside the ball we have the equation $$ \vec{\nabla} \times \vec{H} = \vec{0} $$

since there are no free currents. We can now introduce scalar magnetic potential $\phi$ so that $$ \vec{H} = - \vec{\nabla} \phi $$

And solve for the $\phi$ using the boundary condition (calculated from $H$) $$ \left. \phi \right|_{r = R} = - \frac{3 B_0}{2 \mu_0} \, r \cos \theta $$

The solution is homogenous field in $z$ direction $$ \vec{H} = \frac{3 B_0}{2 \mu_0} \vec{e}_z $$

However, if we take Maxwell's equation $$ \vec{\nabla} \cdot \vec{H} = 0 $$

Make a gradient of it and utilize $\Delta \vec{H} = \vec{\nabla} (\vec{\nabla} \cdot \vec{H}) - \vec{\nabla} \times \vec{\nabla} \times \vec{H}$ we obtain $$ \Delta \vec{H} = 0 $$

Now using the boundary condition by evaluating $\left. \vec{H} \right|_{r = R}$, we obtain a whole different solution which is not homogenous inside the ball.

Following is the image of the magnetic $H$ field strength obtained from the scalar potential (constant function)


And now the second solution, obtained from the vector Laplace equation


(blue for regions where $|H| = 0$, orange for value equal to maximum |H| = $3 B_0 / 2 \mu_0$)

Which one (if ever) is the correct one and why? Thank you.

P.S.: please note that the second solution is also obtained from London equation in the limit of penetration depth going to infinity ($\lambda \to \infty$) which occurs when $T \to T_c$. London equation is basically a vector Helmholtz equation $$ \Delta \vec{H} = \frac{1}{\lambda} \vec{H} $$

However, I might not have interpreted boundary conditions correctly.


1 Answer 1


I believe that you are making an incorrect assumption in your second solution. It is true that Maxwell's equations say that $\nabla \cdot \vec{B} = 0$, but Maxwell's equations do not say that $\nabla \cdot \vec{H} = 0$. Inside the ball (really on the surface of it) $\nabla \cdot \vec{H} \neq 0$ because the divergence of the magnetization $\vec{M}$ is not zero: $\nabla \cdot \vec{H} = - \nabla \cdot \vec{M}$.

So, I say that your first solution is correct and your second solution is not.

  • $\begingroup$ Thank you, this is probably the culprit. However, today I realized there must be more to this whole topic of superconducting piece of geometrical object. I'm not sure how to obtain a magnetic field in case there are two different theories governing the field 1) inside the object (London) and 2) outside the object (magnetostatics). I think it is a self-consistent problem (SC currents, magnetostatics and London equation with boundary condition must align and be consistent) really and I'm not sure how to tackle it, even numerically. $\endgroup$
    – user16320
    Apr 13, 2018 at 20:38
  • $\begingroup$ Imagine $T = 0$, $H_{ext.} \ll H_c$. In this situation the penetration depth is very very small so you get very fast attenuation of any field trying to penetrate the ball. However if $H_{ext}$ is, say, $0.9 H_c$, field the ball experiences at the "equator" is larger than critical, i.e. penetration depth is infinity in these places, while on poles it is still finite. How to find such mixed state, what equations govern this situation? The field is clearly partially penetrating the ball (at sides), so you can't use that solution obtained from scalar potential... $\endgroup$
    – user16320
    Apr 13, 2018 at 20:41
  • $\begingroup$ Unfortunately, I can't help with the superconducting part of the problem because I don't have experience with the London equations. All I know is that assuming $\nabla \cdot \vec{H} = 0$ is a common mistake (there's a section in Griffiths about it). It wouldn't surprise me if you have to numerically solve something self-consistently, since that happens in a lot of non-textbook problems. $\endgroup$
    – lnmaurer
    Apr 15, 2018 at 14:56
  • $\begingroup$ I've been thinking about it and yes, self-consistent solution (perhaps implying iterative approach so that the consistency is ensured in limiting case of numer of steps going to infinity) is probably the only way to tackle it. :( $\endgroup$
    – user16320
    Apr 15, 2018 at 17:59

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