I'm reading Griffiths E&M book and I just need some clarification about the electric displacement $\mathbf{D}$ given by $$\mathbf{D}=\epsilon_0\mathbf{E}+\mathbf{P}$$ where $\mathbf{E}$ is the electric field and $\mathbf{P}$ is my dipole moment per unit volume. Gauss's law in integral form then states $$\iint_{\partial s}\mathbf{D}\cdot d\mathbf{A}=Q_{free}$$ He later claims that we can not always simply draw up a Gaussian surface to solve for $\mathbf{D}$ given that $\nabla\times\mathbf{D}=\nabla\times\mathbf{P}$, which is not necessarily zero. My question is why does the non-zero curl of $\mathbf{D}$ mean that we can not rely on symmetry to draw up Gaussian surfaces and find $\mathbf{D}$ easily? Does a non-zero curl imply non-symmetric fields?
3 Answers
A nonzero Curl of $\bf{P}$ implies that the curl of $\bf{D}$ is also nonzero, since $$ \nabla \times \bf{D} = \nabla \times \bf{P} $$ which means that the components parallel to the Gaussian surface are not guaranteed to be zero, and therefore you cannot calculate the field $\bf{D}$ that way.
If you can find a Gaussian surface that allows finding $\vec{D}$ from Gauss's theorem, then knowledge of the free charges is indeed enough to find $\vec{D}$. But these cases of very strong symmetry are very rare.
In the general case, when Gauss's theorem alone is not sufficient, the vector $\vec{D}$ does not only depend on the free charges. This is the point on which Griffith's text wants to insist. In particular, as Griffith points out, there is no "Coulomb's law" which would make it possible to find $\vec{D}$ by an integration when we know the distribution of free charges.
Hope it can help and sorry for my poor english.
In general, the displacement (or any smooth vector field) can be separated into a part that has a divergence but no curl and a part that has curl and no divergence (known as Helmholtz decomposition). i.e. $$ \vec{D} = \vec{D}_1 + \vec{D}_2 $$ where $$\nabla \times \vec{D}_1 = 0, \ \ \ \nabla \cdot \vec{D}_2 = 0 $$ and thus $$\nabla \cdot \vec{D}=\nabla \cdot \vec{D}_1, \ \ \ \nabla \times \vec{D} = \nabla \times \vec{D}_2\ .$$
Gauss's law, where the closed surface integral of the displacement equals the volume integral of its divergence, can only tell you about $\vec{D_1}$, since $\vec{D}_2$ contributes nothing to the divergence of $\vec{D}$. But for $\vec{D}_1$ to represent all of $\vec{D}$ implies that $\vec{D}_2=0$ and hence $\nabla \times \vec{D} = 0$.
To put it another way, if the curl of $\vec{D}$ is non-zero, which would be the case if $\nabla \times \vec{P}\neq 0$, then it implies that $\vec{D}_2 \neq 0$ and so Gauss's law, which only tells you about $\vec{D}_1$ cannot give you the total D-field.
