I am currently studying 'Introduction to Electromagnetism' by David Griffiths, and I was reading about the electric displacement $\boldsymbol{D}$. I decided to try to extract eq. 4.27, which states:
$\mathbf{D}_{\text {above }}^{\|}-\mathbf{D}_{\text {below }}^{\|}=\mathbf{P}_{\text {above }}^{\|}-\mathbf{P}_{\text {below }}^{\|}$
Where $\textbf{P}$ is the polarization.
To derive this I looked at eq. 4.25, which states:
$\nabla \times \mathbf{D}=\nabla \times \mathbf{P}$
And I know that:
$\nabla \times \mathbf{D}=\nabla \times \mathbf{P} \leftrightarrow \nabla \times \left(\mathbf{D}-\mathbf{P}\right)=\epsilon_0\nabla \times \mathbf{E}=0$
Now if $\nabla \times \mathbf{E}=0$ then $\boldsymbol{E_2}^{\parallel}=\boldsymbol{E_1}^{\parallel}$. If we consider the gaussian pillbox as a cylinder with height epsilon, then in the limit of epsilon$\rightarrow$0, we get that $\boldsymbol{E_2}$ is the electric field above and $\boldsymbol{E_1}$ is the electric field below. From this we get eq. 4.27; $\mathbf{D}_{\text{above}}^{\|}-\mathbf{D}_{\text{below}}^{\|}=\mathbf{P}_{\text{above}}^{\|}-\mathbf{P}_{\text{below}}^{\|}$
But my question is; why does $\nabla \times \mathbf{E}=0$ imply $\boldsymbol{E_2}^{\parallel}=\boldsymbol{E_1}^{\parallel}$?
Update: My understanding is that if the curl of the electric field is zero-vector, then electric field lines will propagate parallel next to each other through space. We consider the gaussian pillbox as a gaussian cylinder, where we denote the height of the cylinder as epsilon. In the limit of epsilon$\rightarrow$0, only the electric field directly above and under the cylinder remains. Since these field lines are parallel, we have that the parallel component of the electric field directly above is equal to the parallel component of the electric field directly belov, i.e. $\boldsymbol{E_2}^{\parallel}=\boldsymbol{E_1}^{\parallel}$. Perhaps someone can elaborate if this is a correct understanding?
As far as I know you are right, I also always thought of it that way.
You can also get something very similar with the divergence, for example knowing that $\nabla \cdot \mathbf{B}=0$ you can say that $\mathbf{B}^{\perp}_1 = \mathbf{B}^{\perp}_2$ with a very similar argument.