# Connection between Maxwell's equations and Cauchy-Riemann equation

How to get Cauchy-Riemann equation from Maxwell's equations ?

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Could you precise your question a little bit? Cauchy-Riemann equations deal with functions defined in $\mathbb R^2$, whereas Maxwell's equations concern functions $\vec E$, $\vec B$, $\rho$, $\vec j$ on $\mathbb R^4$ with values in $\mathbb R^3$, $\mathbb R^3$, $\mathbb R$ and $\mathbb R^3$. So they are not so directly related. So it would help to know what type of relation between those two sets of equations you are searching. – Sebastien B Feb 11 '13 at 22:23
Maxwell's equations implies Cauchy-Riemann equation. – ziang chen Feb 11 '13 at 23:27

## 1 Answer

If you take $z = x + iy$, so that $f(z) = u(x,y) + i v(x,y)$ is your holomorphic function, then we can interpret $f$ as a vector field $F: \mathbb{R}^2 \rightarrow \mathbb{R}^2$, by $F(x,y) = (u(x,y), -v(x,y)$. Then we can notice that $\nabla \cdot F = 0$ and $\nabla \times F = 0$ are equivalent to the Cauchy-Riemann equations.

If $E$ is an electrostatic field on the plane with no electric charges, then Gauss's law tells us that $E$ is divergence free, and in the absence of a magnetic field Faraday's law says that $E$ has zero curl; thus $E$ can be interpreted as representing a holomorphic function.

Similarly, if $B$ is a magnetic field on the plane with no current, then Ampere's law and whichever is the fourth Maxwell equation (I don't remember the name) gives us zero curl and divergence, in the same way, so then such a magnetic field could also represent a holomorphic function.

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Gauss's law for magnetism? Or no-magnetic-monopoles equation... – user53153 Feb 12 '13 at 5:34