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Maxwell's equations in linear homogeneous matter: $$ \begin{eqnarray} \nabla\cdot\mathbf{E}&=&\frac{\rho}{\epsilon},\tag{1}\\ \nabla\times\mathbf{E}&=&-\frac{\partial\mathbf{B}}{\partial t},\tag{2}\\ \nabla\cdot\mathbf{B}&=&0,\tag{3}\\ \nabla\times\mathbf{B}&=&\mu\,\mathbf{J}+\frac{1}{c^2}\frac{\partial\mathbf{E}}{\partial t}.\tag{4} \end{eqnarray} $$ Equations $(2)$ and $(3)$ have solutions in terms of the scalar potential $\phi$ and vector potential $\mathbf{A}$: $$ \begin{eqnarray} \mathbf{E}&=&-\nabla\phi-\frac{\partial\mathbf{A}}{\partial t},\tag{5}\\ \mathbf{B}&=&\nabla\times\mathbf{A}.\tag{6} \end{eqnarray} $$ By substituting equations $(5)$ and $(6)$ into equations $(1)$ and $(4)$ using the Lorenz gauge $$\nabla\cdot\mathbf{A}+\frac{1}{c^2}\frac{\partial\phi}{\partial t}=0\tag{7}$$ and the vector identity $$\nabla\times(\nabla\times\mathbf{A})=\nabla(\nabla\cdot\mathbf{A})-\nabla^2\mathbf{A}\tag{8}$$ we obtain wave equations for $\phi$ and $\mathbf{A}$: $$ \begin{eqnarray} \nabla^2\phi-\frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2}&=&-\frac{\rho}{\epsilon},\tag{9}\\ \nabla^2\mathbf{A}-\frac{1}{c^2}\frac{\partial^2\mathbf{A}}{\partial t^2}&=&-\mu\mathbf{J}.\tag{10} \end{eqnarray} $$ Imagine two linear homogeneous dielectrics meet at a smooth interface: region $1$ has permittivity $\epsilon_1$ and region $2$ has permittivity $\epsilon_2$.

By using a pill-box across the interface Gauss's law, equation $(1)$, implies the following condition for the component of the electric field normal to the interface: $$\epsilon_2\,\mathbf{E}_{2n}-\epsilon_1\,\mathbf{E}_{1n}=\rho_A\tag{11}$$ where $\rho_A$ is the surface density of free charges.

Substituting equation $(5)$ into equation $(11)$ gives $$\epsilon_2\left(-\nabla\phi-\frac{\partial\mathbf{A}}{\partial t}\right)_{2n}-\epsilon_1\left(-\nabla\phi-\frac{\partial\mathbf{A}}{\partial t}\right)_{1n}=\rho_A.\tag{12}$$

However the wave equation for the scalar potential $\phi$, equation $(9)$, also has a matching condition across the interface which can be derived using a pill-box. Noting that $\nabla^2\phi=\nabla\cdot\nabla\phi$ one can use the divergence theorem in the limit of zero pill-box height to obtain $$\epsilon_2(-\nabla\phi)_{2n}-\epsilon_1(-\nabla\phi)_{1n}=\rho_A.\tag{13}$$ The contradiction between equations $(12)$ and $(13)$ does not seem to be due to writing equations in terms of $\phi$ and $\mathbf{A}$ since $\mathbf{E}$ and $\mathbf{B}$ are gauge invariant.

Does this show that Maxwell's equations are inconsistent?

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    $\begingroup$ Your starting point is incorrect. You can't just replace $\epsilon_0$ and $\mu_0$ with $\epsilon$ and $\mu$ when the latter vary in space. $\endgroup$ Commented Aug 20 at 18:34
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    $\begingroup$ I don't really get the downvote - while "Are Maxwell's equations inconsistent?" might not be the best phrasing (the answer is "obviously" no), the OP seems to be asking about a somewhat subtle point when trying to look at the interface between two media, which seems like quite an appropriate thing to ask about here. $\endgroup$ Commented Aug 20 at 18:43
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    $\begingroup$ @FusRoDah I downvoted because the title seems like clickbait to me. It’s as unreasonable as the OP’s previous question about whether charge conservation could be violated just by moving a charge. When my math doesn’t seem to work, I don’t wonder whether a century and a half of physics is wrong; I wonder where my mistake is. $\endgroup$ Commented Aug 20 at 19:36
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    $\begingroup$ @JohnEastmond: No, since the vector potential $\mathbf{A}$ and its time derivative are continuous across the boundary. So in that case the two terms involving $\partial \mathbf{A}/\partial t$ cancel in your Eq. (12), which is therefore consistent with your eq. (13). $\endgroup$ Commented Aug 20 at 21:02
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    $\begingroup$ (Really, the displacement current term has just the permeability $\epsilon$). In other words, the Maxwell equations in linear medium are $\nabla\cdot (\epsilon \mathbf E) = \rho_f$ (density of free charge), and $\nabla \times (\mu\mathbf H) = \mu_0 \mathbf j_f + \mu_0\frac{\partial (\epsilon \mathbf E)}{\partial t}$ ($\mu_0$ times (free current density plus displacement current density)). $\endgroup$ Commented Aug 21 at 2:09

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As @knzhou pointed out in the comments, Maxwell's equations in linear matter of spatially varying permittivity/permeability are properly written as \begin{align} \nabla \cdot (\epsilon \mathbf{E}) = \rho_f \tag{1a} \\ \nabla \times \left(\frac{1}{\mu} \mathbf{B} \right) - \frac{\partial}{\partial t} (\epsilon \mathbf{E}) = \mathbf{J}_f \tag{4a}. \end{align} Equations (2) and (3) are still valid, so we can define the potentials as before and plug them in, yielding the equation $$ \nabla \cdot \left[ \epsilon \left( -\nabla\phi-\frac{\partial\mathbf{A}}{\partial t} \right) \right] = \rho_f. \tag{*} $$ In vacuum ($\epsilon = 1$) the second term on the left-hand side can be replaced with a term proportional to $\partial^2 \phi/\partial t^2$ via the Lorenz gauge condition. But if $\epsilon$ varies in space another term appears, since $$ \nabla \cdot \left[ \epsilon \frac{\partial\mathbf{A}}{\partial t} \right] = \epsilon \frac{\partial (\nabla \cdot \mathbf{A})}{\partial t} + \frac{\partial \mathbf{A}}{\partial t} \cdot \nabla \epsilon = -\frac{\epsilon}{c^2} \frac{\partial^2 \phi}{\partial t^2} + \frac{\partial \mathbf{A}}{\partial t} \cdot \nabla \epsilon $$ if $\epsilon$ varies in space.

So your equation (9) is incorrect. It should instead read $$ - \nabla \cdot(\epsilon \nabla \phi) + \frac{\epsilon}{c^2} \frac{\partial^2 \phi}{\partial t^2} - \frac{\partial \mathbf{A}}{\partial t} \cdot \nabla \epsilon = \rho_f \tag{9a} $$ Integrating the third term (which you had neglected) over an infinitesimally thin Gaussian pillbox and using the identity $$ \int_\mathcal{V} \nabla \phi \, dV = \int_{\partial \mathcal{V}} \phi \mathbf{\hat{n}} \, dA $$ then yields terms of the form $$ - \epsilon_2 \frac{\partial \mathbf{A}}{\partial t} \cdot \mathbf{\hat{n}} + \epsilon_1 \frac{\partial \mathbf{A}}{\partial t} \cdot \mathbf{\hat{n}} $$ which are exactly what are needed to resolve the apparent discrepancy.

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The discrepancy between your boundary conditions (12) and (13) is due to the neglect of the term $\frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2}$ in your pill-box derivation of (13) from equ. (9). The correct boundary condition is (11) or (12)!

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  • $\begingroup$ The integral of the $\frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2}$ term should go to zero as the height of the pill-box goes to zero. $\endgroup$ Commented Aug 20 at 20:31
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    $\begingroup$ Note that $\partial^2 \phi/\partial t^2 = -c^2\partial_t( \nabla \cdot {\bf A})$, so you cannot assume that its pillbox goes to zero. $\endgroup$ Commented Aug 20 at 20:51

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