# Is curl of a vector a scalar quantity in 2 spatial dimensions? If it is so, then somebody help me understanding Maxwell's equations in 2+1 D

I have seen on wikipedia that in 2 spatial dimensions, Green's theorem, Gauss's divergence and Stokes theorems are equivalent and it makes sense. When I tried to write Maxwell's equations in 2+1 spacetime dimensions, I got many questions and confusions. Here I give a brief summary.

In 3+1 D we have $$\oint_{\bf{A}} {\bf E}\cdot d{\bf A}=\int_V \nabla\cdot\bf {E} dV= \int_V \rho dV.$$ In 2+1D it might be written as $$\oint_\ell {\bf E}\cdot d\bf{\ell}=\int_A \nabla\cdot\bf {E} dA=\int_A \rho dA.$$ $A$ is scalar area in 2+1 D. In 3+1 D $E\cdot dA$ gives the flux. What does ${\bf E}\cdot d\bf{\ell}$ give us?

Secondly in 3+1 D we have $$\oint {\bf E}\cdot d\vec{\ell}=\int_A (\nabla\times\bf {E})\cdot dA=-\frac{\partial}{\partial t}\int_A \bf B\cdot dA.$$ I think it can be written in 2+1 D as $$\oint {\bf E}\cdot d\vec{\ell}=\int_A (\nabla\times\bf {E}) dA=-\frac{\partial}{\partial t}\int_A B dA,$$ where $B$ is a scalar in 2+1 D and I think the cross product term is also a scalar.

Either I am wrong at writing these expressions or some thing goes messy in 2+1 D. From above expressions we can see that $$\nabla \cdot \bf E = \nabla \times \bf E$$ which is the main prob for me to understand.

• Related: physics.stackexchange.com/q/21678/2451 , physics.stackexchange.com/q/32685/2451 and links therein. – Qmechanic Aug 19 '17 at 7:27
• @Qmechanic in 2+1 dimensions we can expand $\bigl(\vec\nabla\times\vec{V}\bigr)^\alpha = \epsilon^{\alpha\beta\gamma}\frac{\partial V_\beta}{\partial x^\gamma}$ for $\alpha =0$, will it be a scalar for sure? What about $\nabla \cdot \bf E = \nabla \times \bf E$? Please describe the probs with above relations. – Sami Khan Aug 19 '17 at 10:11

In 3D, the vector area element $d\bf{A}$ points perpendicularly to the plane of the surface element. So, in the 2D version of Gauss' Law, the vector associated with the line element $d\vec{\ell}$ needs to be perpendicular to that line element. Thus, $\vec{E}\cdot{}d\ell$ in the 2D version of Gauss' Law gives the component of the electric field that crosses the boundary of the surface. The expression $\vec{E}\cdot{}d\vec\ell$ in Faraday's Law gives the component of the field parallel to the boundary. You should probably use a different symbol for each of these.
$$\oint_\ell {\bf E}\cdot \mathrm d{\boldsymbol\ell}=\int_A \nabla\cdot{\bf E}\:\mathrm d\mathbf A. \tag{wrong!}$$ This is not how Green's theorem works in two dimensions, and the circulation and flux integrals, while notated similarly, are completely different beasts. To see why this needs to be the case, simply note that $d{\boldsymbol\ell}$ is tangent to the line while $\mathrm d\mathbf A$ is normal to the surface, so they are radically different geometric concepts. (In two dimensions, the integral $\oint_\ell {\bf E}\cdot \mathrm d{\boldsymbol\ell}$ is the circulation of $\mathbf E$ over $\ell$, and it is equal to the area integral of the curl of $\mathbf E$.) For more details, see your vector calculus textbook (or, if you don't have a good one, H. Schey's Div, Grad, Curl, and All That).
So, is the curl a scalar quantity in two dimensions? Well, yes and no. If you see your two-dimensional field as an object that's embedded in a plane inside 3D space (or, even better, a field that's translationally invariant along the $z$ dimension), then you have a field that only has $x$ and $y$ components and only depends on $x$ and $y$, so its only meaningful curl term is $$\nabla\times\mathbf E = \left(\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}\right)\hat{\mathbf e}_z,$$ i.e. a single vectorial component sticking out of the plane. This can, often enough, be treated as a signed scalar, once you decide on an orientation for the plane, but it always pays to remember that it's actually a component of a vector that's been frozen as orthogonal to the plane. Similarly, when this interplays with Faraday's law, it tells you that the only relevant magnetic field component must be $B_z$, - and that's just as well, since that's the field component that produces circular motion in the $x,y$ plane.