I've been following this paper for a treatment of electrodynamics using differential forms. In particular, they demonstrate that Maxwell's equations expressed using differential forms are form-invariant with regards to dimensionality. They then develop 2D electrodynamics with $\mathcal{E}$, a 1-form representing the electric field intensity and $\mathcal{D}$, being a 1-form representing the electric flux density, related by the constitutive relation $\mathcal{D} = \varepsilon {\star}\mathcal{E}$, where $\star$ represents the 2D Hodge star operator, which notably behaves as ${\star}dx = dy, {\star}dy = -dx$. This is in contrast to the 3D case, where $\mathcal{E}$ is a 1-form and $\mathcal{D}$ is a 2-form.

In 3.5, the paper notes that $\mathcal{D}$ is actually a twisted 1-form. This is due to the fact that, while both $\mathcal{E}$ and $\mathcal{D}$ can be represented by their equipotential lines, and integrals count the number of such lines pierced along the path, a path integral of $\mathcal{E}$ represents potential change along the path, while a path integral of $\mathcal{D}$ represents the amount of flux flowing perpendicularly across the path.

My confusion arises from not being able to make sense of a sign convention for a very simple example of this scenario. Consider $\mathcal{E} = dx + dy$, giving $\mathcal{D} = -\varepsilon dx + \varepsilon dy$. Let $\mathcal{C}$ be a curve along the $x$-axis parameterized by $\mathbf{r}(t) = t\widehat{\mathbf{i}}$, $0 \leq t \leq 1$. Then we have $$ \int_{\mathcal{C}} \mathcal{E} = \int_0^1 \mathcal{E}(\mathbf{r}'(t)) dt = \int_0^1 \mathcal{E}(\widehat{\mathbf{i}}) dt = \int_0^1 dt = 1 $$

while $$ \int_{\mathcal{C}} \mathcal{D} = \int_0^1 \mathcal{D}(\mathbf{r}'(t)) dt = \int_0^1 \mathcal{D}(\widehat{\mathbf{i}}) dt = \int_0^1 - \varepsilon dt = - \varepsilon $$ However, it seems from visualizing the situation that the potential change along $\mathcal{C}$ and the flux across $\mathcal{C}$ should both be positive (electric field lines are going up and to the right).

I expect that the confusion arises since, in 3D, we cannot simply re-use the same manifold on which we integrated $\mathcal{E}$ to integrate $\mathcal{D}$, since the former must be a curve, while the latter must be a surface.

It seems that, perhaps, we must choose the opposite orientation for $\mathcal{C}$ when integrating $\mathcal{D}$ instead of $\mathcal{E}$. It seems then that, in 2D, there may be a difference between a "path-like" curve and a "surface-like" curve, with regards to orientation. Is there a way to make sense of this?


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