## Show Solution

Inside the superconducter, the **Proca equation** \[\left(\Delta-\mu_{\mathrm{L}}^{2}\right)\mathbf{A}\left(\mathbf{r}\right) = 0\] holds for the vector potential \(\mathbf{A}\left(\mathbf{r}\right)\) with the **London penetration depth** \(\lambda_{\mathrm{L}}=1/\mu_{\mathrm{L}}\). Now, we might take the **curl** of it and find that it also holds for the magnetic induction, \[\begin{eqnarray*}0&=&\nabla\times\left(\Delta-\mu_{\mathrm{L}}^{2}\right)\mathbf{A}\left(\mathbf{r}\right)\\&=&\left(\Delta-\mu_{\mathrm{L}}^{2}\right)\nabla\times\mathbf{A}\left( \mathbf{r}\right) \\ & = &\left(\Delta-\mu_{\mathrm{L}}^{2}\right)\mathbf{B}\left(\mathbf{r}\right)\ .\end{eqnarray*}\]Here, we have made use of **Schwarz's theorem** which allows us to interchange partial derivatives. It also becomes evident that without external currents, \(\mathbf{j}_{\mathrm{ext}}\left(\mathbf{r}\right)=0\), the effect of the superconducting material is included in \(\mu_{\mathrm{L}}\) - the tangential part of the magnetic induction has to be **continuous**.

## The Superconducting Half-Space

This is the main information we have to use for the first part of the problem. So, if an external magnetic field \(\mathbf{B}_{0}\left(\mathbf{r}\right)=B_{0}\mathbf{e}_{y}\) is present, we have to match it to the physical solution of the equation we just derived. Because of the **translation symmetry** of the problem, the variation can only be in \(z\) direction, hence we know that the latter equation reduces to \[\left(\partial_{zz}-\mu_{\mathrm{L}}^{2}\right)B_{y}\left(z\right) = 0\ \mathrm{for}\ z\leq0\ .\]Using the ansatz \(B_{y}\left(z\right)=B_{y}^{0}\exp\left(\kappa z\right)\), we find\[B_{y}\left(z\right) = B_{1}e^{\mu_{\mathrm{L}}z}+B_{2}e^{-\mu_{\mathrm{L}}z}\ \mathrm{for}\ z\leq0\ .\]Requiring **finiteness** of the magnetic field for \(z\rightarrow-\infty\), we can drop the second term. At \(z=0\), the magnitude is given by \(B_{0}\) and hence we find the required \[\mathbf{B}\left(\mathbf{r}\right) = B_{0}\mathbf{e}_{y}\begin{cases}1 & z>0\\e^{-\mu_{\mathrm{L}}\left|z\right|} & z\leq0\end{cases}\ .\]So the field inside the superconductor falls off **exponentially**. We remind ourselves that \(\mu_{\mathrm{L}}=1/\lambda_{\mathrm{L}}\) and can see directly that the **London penetration depth** is really some kind of penetration distance of the magnetic field into the superconductor.

## Slab and Wire Geometries

Now that we know the easiest possible example we can advance to the case of a thin superconducting **slab**. Then again, inside the superconductor the equation\[\left(\partial_{zz}-\mu_{\mathrm{L}}^{2}\right)B_{y}\left(z\right) = 0\ \mathrm{for}\ \left|z\right|\leq d/2\]holds and we find the solutions to be again of the form \[B_{y}\left(z\right) = B_{1}e^{\mu_{\mathrm{L}}z}+B_{2}e^{-\mu_{\mathrm{L}}z}\ \mathrm{for}\ \left|z\right|\leq d/2\ .\]Since at \(z=\pm d/2\) and \(B_{y}=B_{0}\), \(B_{1}=B_{2}=B_{0}\) we conclude \[\begin{eqnarray*}B_{y}\left(z\right)&=&B_{0}\frac{e^{\mu_{\mathrm{L}}z}+e^{-\mu_{\mathrm{L}}z}}{e^{\mu_{\mathrm{L}}d/2}+e^{-\mu_{\mathrm{L}}d/2}}\\&=&B_{0}\frac{\cosh\mu_{\mathrm{L}}z}{\cosh\mu_{L}d/2}\ \mathrm{for}\ \left|z\right|\leq d/2\end{eqnarray*}\]using an appropriate normalization. Finally, \[\mathbf{B}\left(\mathbf{r}\right) = B_{0}\mathbf{e}_{y}\begin{cases}1 & \left|z\right|>d/2\\\frac{\cosh\mu_{\mathrm{L}}z}{\cosh\mu_{L}d/2} & \left|z\right|\leq d/2\end{cases}\ .\]In the model of a superconductor we have employed so far, we have no explicit expression for the current. However, since we found the magnetic induction we might assign this field a conductive current using Ampère's law\[\begin{eqnarray*}\nabla\times\mathbf{B}\left(\mathbf{r}\right)&=&\mu_{0}\mathbf{j}_{\mathrm{cond}}\left(\mathbf{r}\right)\ ,\ \mathrm{so}\\\mathbf{j}_{\mathrm{cond}}\left(\mathbf{r}\right) & = &\frac{B_{0}}{\mu_{0}}\nabla\times\left(\frac{\cosh\mu_{\mathrm{L}}z}{\cosh\mu_{L}d/2}\mathbf{e}_{y}\right)\\&=&-\frac{B_{0}}{\mu_{0}}\frac{\mu_{\mathrm{L}}\sinh\mu_{\mathrm{L}}z}{\cosh\mu_{L}d/2}\mathbf{e}_{x}\ \mathrm{for}\ \left|z\right|\leq d/2\ .\end{eqnarray*}\]So we find that the current is strongest at the termination of the superconducter and vanishing in the middle. Furthermore it has only a component in \(x\)-direction, **parallel to the surface** of the superconductor effectively **pushing charges perpendicular to the magnetic field**. Of course there is no such thing as an infinitely extended superconductor but we may still get a feeling what happens if the geometry changes. Assuming a **superconducting wire**, the current would flow along the radial termination with a negative sense of rotation with respect to the magnetic field effectively pushing it out of the material, see figure. This is **different to a perfect diamagnet** where the magnetic field in- and outside of the material just remain unchanged if one could somehow turn off the material's resistance.