How do you translate from separate vector potential gradients to magnetic field components?

Question

In an attempt at calculating anisotropic London penetration depths, I have obtained equations of the form

$$ \begin{split} -\mu_0^{-1}\nabla^2B_x &= Z\partial_yA_z - T\partial_zA_x-Y\partial_zA_y,\\ -\mu_0^{-1}\nabla^2B_y &= X\partial_xA_x + T\partial_zA_y - Z\partial_xA_z,\\ -\mu_0^{-1}\nabla^2B_z &= T(\partial_xA_x-\partial_yA_x) + Y\partial_xA_y - X\partial_yA_x, \end{split} $$

where $X$, $Y$, $Z$ and $T$ are constant coefficients and we have assumed time-independence. These equations are a result of applying the electrostatic Maxwell's equation $\nabla\times\mathbf{B} = \mu_0\mathbf{j}$ to the current density

$$ \begin{split} j_x &= XA_x+TA_y,\\ j_y &= TA_x+YA_y,\\ j_z &= ZA_z, \end{split} $$ which was obtained by extremizing the free energy of an unconventional superconductor.

My question is how to translate this into a set of equations for the magnetic field component $B_i$ only, i.e., how to eliminate the vector potential components from these partial differential equations? Is it through some gauge choice in general possible to create a one to one relationship going between vector potential gradients and magnetic field components?

I'm following the notes on p. 43 of "Introduction to Unconventional Superconductivity", and according to Sigrist I should be able to put these equations on the form $$ \begin{split} \nabla^2B_x &= \lambda_1^{-2}B_x + \lambda_3^{-2}B_y\\ \nabla^2B_y &= \lambda_3^{-2}B_x + \lambda_2^{-2}B_y\\ \nabla^2B_z &= \lambda_4^{-2}B_z. \end{split} $$

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What I've tried so far

Coulomb Gauge

Applying the Coulomb gauge we get $\mu_0\mathbf{j} = -\nabla^2\mathbf{A}$., which yields the set of equations $$ \begin{split} \nabla^2A_x &= -\mu_0(XA_x+TA_y)\\ \nabla^2A_y &= -\mu_0(TA_x+YA_y)\\ \nabla^2A_z &= -\mu_0ZA_z, \end{split} $$ when the current density above is inserted. This has exactly the same form as the equations of Sigrist, but for the vector potential components, and not the magnetic field ones. Could it be that Sigrist just mis-typed and meant to write this instead?

Axial gauge

Using the axial gauge, we can assume that $A_z=0$. This implies that $B_x = -\partial_zA_y$ and $B_y = \partial_zA_x$, by insertion in the equation $\mathbf{B} = \nabla\times\mathbf{A}$. Inserting this above then yields the set of equations $$ \begin{split} -\mu_0^{-1}\nabla^2B_x &= -TB_y + YB_x\\ -\mu_0^{-1}\nabla^2B_y &= XB_y - TB_x\\ -\mu_0^{-1}\nabla^2B_z &= T(\partial_xA_x-\partial_yA_y) + YB_z + (Y-X)\partial_yA_x. \end{split} $$ I thus get equations for $B_x$ and $B_y$ as desired, but with remaining vector potential components in the equations for $B_z$. Would it be possible to eliminate these with a residual gauge condition? Maybe I've misunderstood the axial gauge?

Answer 1

I think I've found a gauge-independent solution (which makes sense considering $\mathbf{B}$ should be independent of the gauge).

Starting by writing the current density on matrix form such that $\mathbf{j}=P\mathbf{A}$, we simply invert this equation such that $\mathbf{A} = P^{-1}\mathbf{j}$. Taking the curl of both sides we get $\mathbf{B} = \nabla\times\mathbf{A} = \nabla\times(P^{-1}\mathbf{j})$. Now we insert Ampere's law, which says that $\mu_0\mathbf{j} = \nabla\times\mathbf{B}$ to finally get

$$ \mu_0\mathbf{B} = \nabla\times\big[P^{-1}(\nabla\times\mathbf{B})\big], $$ which in index notation (with summation over repeated indices) takes the complicated form $$ \mu_0B_i = P^{-1}_{kl}\epsilon_{ijk}\epsilon_{lmn}\partial_j\partial_mB_n, $$ where $\epsilon_{ijk}$ is the Levi-Civita symbol and we have (as in the question) used the notation $\partial_i = \frac{\partial}{\partial x_i}$ for the partial derivative of coordinate $x_i$.

By rotating the coordinates by defining new coordinates $\xi_i = R_{ij}x_j$, we can slightly simplify this expression by rotating to where $P$ is diagonal, i.e., to the principal axis. In terms of the rotated components $B_i' = R_{ij}B_j$ and partial derivatives $\partial'_i = R^{-1}_{ji}\partial_j$, we then get the equation $$ \mu_0B_i' = (RP^{-1}R^{-1})_{jl}\epsilon_{ijk}\epsilon_{lmn}\partial_j'\partial_m'B_n'. $$ Writing the diagonal components $RP^{-1}R^{-1}$ as $\mathrm{diag}(\lambda_1^2,\lambda_2^2,\lambda_3^2)/\mu_0$, we get the coupled set of PDEs $$ \mathbf{B}' = \begin{pmatrix} -(\lambda_1^{2}\partial_y^{'2}+\lambda_2^{2}\partial_z^{' 2}) & \lambda_3^{2}\partial_x'\partial_y' & \lambda_2^{2}\partial_x'\partial_z'\\ \lambda_3^{2}\partial_x'\partial_y' & -(\lambda_3^{2}\partial_x'{}^2 + \lambda_1^{2}\partial_z'{}^2) & \lambda_1^{2}\partial_y'\partial_z'\\ \lambda_2^{2}\partial_x'\partial_z' & \lambda_1^{2}\partial_y'\partial_z' & -(\lambda_2^{2}\partial_x'{}^2+\lambda_1^{2}\partial_y'{}^2) \end{pmatrix}\mathbf{B}'. $$ We see that we can get $3$ different length scales by defining them as the eigenvalues of $P$, however, the form of these equations are certainly not the same as the one that Sigrist has...