# 2D Gauss law vs residue theorem

I used to have a vague feeling that the residue theorem is a close analogy to 2D electrostatics in which the residues themselves play a role of point charges. However, the equations don't seem to add up. If we start from 2D electrostatics given by $$\frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} = \frac{\rho}{\epsilon_0},$$ where the charge density $\rho = \sum_i q_i \delta(\vec{r}-\vec{r}_i)$ consists of point charges $q_i$ located at positions $\vec{r}_i$, and integrate over the area bounded by some curve $\mathcal{C}$, we find (using Green's theorem) $$\int_\mathcal{C} (E_x\, dy - E_y\, dx) = \frac{1}{\epsilon_0} \sum_i q_i.$$ Now, I would like to interpret the RHS as a sum of residua $2\pi i\sum_i \text{Res}\, f(z_i)$ of some analytic function $f(z_i)$ so that I would have the correspondence $$q_i = 2\pi i\epsilon_0 \text{Res}\, f(z_i).$$ For this to hold, the LHS would have to satisfy $$\int_\mathcal{C} (E_x\, dy - E_y\, dx) = \int_\mathcal{C} f(z)\, dz,$$ however, it is painfully obvious that the differential form $$E_x\, dy - E_y\, dx = -\frac{1}{2}(E_y+iE_x)dz + \frac{1}{2}(-E_y + iE_x)dz^*$$ can never be brought to the form $f(z)dz$ for an analytic $f(z)$.

So, it would appear that there really isn't any direct analogy between 2D Gauss law and the residue theorem? Or am I missing something?

## 3 Answers

There is indeed a connection. The holomorphy is easily seen in the electrostatic potential.

In a charge free (two-dimensional) region, the electrostatic potential solves Laplace's equation and hence is a harmonic function. The real and imaginary parts of a holomorphic function are harmonic functions and thus the electrostatic potential can be identified with, say, the real part of a holomorphic function. In more detail, let us write (with $z=x+iy$) $$f(z) = \phi(x,y) + i\ \psi(x,y)\ ,$$ where we choose to identify the real part with the electrostatic potential. You can check that Cauchy-Riemann conditions imply that $\mathbf{E}\cdot \nabla \psi=0$. This implies that the $\psi=$ constant lines are the electrostatic field lines.

Adding a point charge, implies that it is harmonic everywhere except at the location of the charge. The relevant function is $f(z) = \lambda \log (z-z_0)$, where $z_0$ is the location of the charge and $\lambda$ is proportional to the charge. The connection with the residue theorem follows since $f'(z)$ has a simple pole at $z_0$ with residue $\lambda$.

• Not a point charge but a homogeneous linear distribution, which is perpendicular to the complex plane. Apr 22, 2014 at 11:20
• The question is on 2D electrostatics. Yes, one can view a two-dimensional point charge as a line charge distribution in three dimensions (or a planar charge distribution in four dimensions and so on) Apr 22, 2014 at 12:07
• How do you find the Coulomb potential in dimension 2? Apr 22, 2014 at 12:57
• @auxsvr Good point. It is taken to be the solution to Laplace's equation. For "point" charges in 2d, it is $\log(x)$ and $d>2$, it is $1/r^{d-2}$. Apr 22, 2014 at 13:50

The analogy follows with the right definitions. The "flux" of the "vector" $E(z)$ through a contour $\Gamma$ is $\mathrm{Re}\left(\int_\Gamma E(z)^*\,\mathrm{d} z\right)$.

I think you may have forgotten the conjugate in the relationship between the "Electric field" and the complex potential $\Omega$: $E(z) = -(\mathrm{d}_z \Omega(z))^*$.

So it is the conjugate of $E(z)$, not $E(z)$ itself, that is holomorphic, being given by $E(z)^* = -\mathrm{d}_z\Omega(z)$. Work out the direction of the vectors from the Cauchy-Riemann relations and you'll see that you need a conjugate to make $-\nabla \phi$ equal to the real part of the derivative $-(\mathrm{d}_z\Omega(z))^*$.

Hence $\mathrm{Re}\left(\int_\Gamma E(z)^*\,\mathrm{d} z\right)$ is the real part of an everyday contour integral and the flux is thus the real part of $2\,\pi\,i$ times the sum of residues at the poles of $E(z)^*$, whence the 2D Gauss law follows.

There is a very good discussion of this in Tristan Needham's Visual Complex Analysis final chapter, but I'd like to share one other view point of it. When we have some charge configuration or so, we define it after removing some set of points in $$\mathbb{R^d}$$. Eg: if we define the electric field of a charge, we don't talk about field at charge's location, for a line of charge we exclude the potential/ field on line itself.

## Deeper idea of Gauss law

When we have such a situation, one will find that if a differential $$p$$ form defined on space, then generally exact forms won't be closed. If we are working in $$\mathbb{R^2}$$, then this leads us to define the first de Rham cohomology group $$\mathbb{H^1}(\mathbb{R^2} -\text{origin})$$. The elements of this group are given by equivalence class of one forms where two one form belong to the same equivalence class if their difference is equal to a ckised form i.e:

$$\tilde{\phi} - \phi = df$$

Now, it will be that each form in this equivalence class have the same circulation over any loop. If we add two forms $$\phi_1$$ and $$\phi_2$$ of different classes evaluated on any loop, we see we have the following structure:

$$C (\phi_1 + \phi_2) =C(\phi_1) + C(\phi_2)$$

Hence, we will find that the first deRham cohomology group is isomorphic to $$\mathbb{R}$$ as a group. This is the idea underlying the fact that we can take different surface/ loops and still get the same answer when applying gauss law.

### Analogous idea of Cauchy's theorem

The above I talked about $$\mathbb{R^2}$$ but you can similarly talk about complex valued differential forms and define everything analogously. Infact Cauchy Riemann equation are just a statement of complex valued form being closed. The deformation theorem of closed 1-form says that the integral of a form over a loop is invariant under a continous deformation as long as we don't cross any removed point (eg: the removed singularity at origin).

Hence, the exact same idea as the previous case comes. For more detials, you can check chapter-37 of Tristan Needham's Visual Differential Geometry.