# Poisson equation in non-homogenous media

If the dielectric constant be different in x, y, and z directions, how the Poisson equation would change. Is it right to say: $$\epsilon_{xx} d^2V/dx^2 + \epsilon_{yy} d^2V/dy^2 + \epsilon_{zz} d^2V/dz^2 = -\rho$$

• You should distinguish between homogeneity and isotropy. If your medium is anisotropic but homogeneous, then the dielectric constant is different in different directions, but there is no variation from point to point. If your medium is not homogeneous, then the dielectric constant will vary from point to point, and you may need to consider that variation somehow. Mar 11, 2015 at 18:41

Linear anisotropic materials can be modeled as having a field $P^i = \epsilon_0 ~ \chi^i_j ~ E^j$. This follows two conventions: the first is "Einstein summation rules"; you sum over any index which is repeated both upper and lower. The second convention is that upper indices correspond to a vector space while lower indices correspond to its dual space. This difference is huge when you're talking about crystals, where the atoms have a certain set of real-space vectors $\hat b_i$ which tell you how the crystal is laid out, so any vector is $\vec v = \hat b_i v^i$, while the dual-space vectors perpendicular to them $\hat b^i$ are defined so that $\hat b^i \cdot \hat b_j = \delta^i_j$ is the Kronecker delta (1 if $i = j$, 0 otherwise). If you work through it you will find that there is also a matrix called the metric $g^{ij}$ which relates the two components via $v^i = g^{ij} v_j.$
So we see that we replace the normal electric susceptibility $\chi$ with a susceptibility tensor $\chi^i_j$. This causes $D^i = \epsilon^i_j ~E^j$ for $\epsilon^i_j = \epsilon_0 (\delta^i_j + \chi^i_j)$. This means that if $E_k = -\partial_k \phi$ for some potential $\phi$, the relevant equation for a bulk with no free charges is $$\partial_i D^i = \partial_i (\epsilon^i_j ~ g^{jk} ~ \partial_k\phi) = (\partial_i \epsilon^{ik}) \partial_k \phi + \epsilon^{ik} \partial_i \partial_k \phi = 0.$$ For a homogeneous ($\chi$ not depending on position) linear anisotropic substance, we know $\partial_n \epsilon^{ik} = 0$ and so we can confidently discard that first term.
The second term is trickier, but there will be certainly some basis which diagonalizes $\epsilon^i_j = e^{i} ~ \delta^i_j$, and it will usually be the crystal basis, so you will still have the expression $\epsilon^j ~ g^{jk} ~ \partial_j ~ \partial_k \phi ~=~ \epsilon^j ~ \partial_j ~ \partial^j \phi = 0.$ In general it looks like this expression will "tweak" the $\epsilon^j$ coefficients with the metric before they enter into an equation substantially similar to yours.
If the material is non-homogeneous then of course the first term in the above expression is some vector field $t^k = \partial_i (\epsilon^i_j g^{jk})$ which also contributes a $(\vec t \cdot \nabla) \phi$ term to the mix.